Linear Algebra_ Theory_Jim Hefferon

Linear Algebra
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Jim Hefferon

Notation
R real numbers
N natural numbers: {0, 1, 2, . . . }
C complex numbers
{. . .
¯
¯ . . . } set of . . . such that . . .
h. . . i sequence; like a set but order matters
V, W, U vector spaces
~
v, ~
w vectors
~
0, ~
0V zero vector, zero vector of V
B, D bases
En = h~
e1, . . . , ~
eni standard basis for Rn
~
β,~
δ basis vectors
RepB(~
v) matrix representing the vector
Pn set of n-th degree polynomials
Mn×m set of n×m matrices
[S] span of the set S
M ⊕ N direct sum of subspaces
V ∼
= W isomorphic spaces
h, g homomorphisms
H, G matrices
t, s transformations; maps from a space to itself
T, S square matrices
RepB,D(h) matrix representing the map h
hi,j matrix entry from row i, column j
|T| determinant of the matrix T
R(h), N (h) rangespace and nullspace of the map h
R∞(h), N∞(h) generalized rangespace and nullspace
Lower case Greek alphabet
name symbol name symbol name symbol
alpha α iota ι rho ρ
beta β kappa κ sigma σ
gamma γ lambda λ tau τ
delta δ mu µ upsilon υ
epsilon ² nu ν phi φ
zeta ζ xi ξ chi χ
eta η omicron o psi ψ
theta θ pi π omega ω
Cover. This is Cramer’s Rule applied to the system x + 2y = 6, 3x + y = 8. The area
of the first box is the determinant shown. The area of the second box is x times that,
and equals the area of the final box. Hence, x is the final determinant divided by the
first determinant.

Preface
In most mathematics programs linear algebra is taken in the first or second
year, following or along with at least one course in calculus. While the location
of this course is stable, lately the content has been under discussion. Some in-
structors have experimented with varying the traditional topics, trying courses
focused on applications, or on the computer. Despite this (entirely healthy)
debate, most instructors are still convinced, I think, that the right core material
is vector spaces, linear maps, determinants, and eigenvalues and eigenvectors.
Applications and computations certainly can have a part to play but most math-
ematicians agree that the themes of the course should remain unchanged.
Not that all is fine with the traditional course. Most of us do think that
the standard text type for this course needs to be reexamined. Elementary
texts have traditionally started with extensive computations of linear reduction,
matrix multiplication, and determinants. These take up half of the course.
Finally, when vector spaces and linear maps appear, and definitions and proofs
start, the nature of the course takes a sudden turn. In the past, the computation
drill was there because, as future practitioners, students needed to be fast and
accurate with these. But that has changed. Being a whiz at 5×5 determinants
just isn’t important anymore. Instead, the availability of computers gives us an
opportunity to move toward a focus on concepts.
This is an opportunity that we should seize. The courses at the start of
most mathematics programs work at having students correctly apply formulas
and algorithms, and imitate examples. Later courses require some mathematical
maturity: reasoning skills that are developed enough to follow different types
of proofs, a familiarity with the themes that underly many mathematical in-
vestigations like elementary set and function facts, and an ability to do some
independent reading and thinking, Where do we work on the transition?
Linear algebra is an ideal spot. It comes early in a program so that progress
made here pays off later. The material is straightforward, elegant, and acces-
sible. The students are serious about mathematics, often majors and minors.
There are a variety of argument styles—proofs by contradiction, if and only if
statements, and proofs by induction, for instance—and examples are plentiful.
The goal of this text is, along with the development of undergraduate linear
algebra, to help an instructor raise the students’ level of mathematical sophis-
tication. Most of the differences between this book and others follow straight
from that goal.
One consequence of this goal of development is that, unlike in many compu-
tational texts, all of the results here are proved. On the other hand, in contrast
with more abstract texts, many examples are given, and they are often quite
detailed.
Another consequence of the goal is that while we start with a computational
topic, linear reduction, from the first we do more than just compute. The
solution of linear systems is done quickly but it is also done completely, proving
i

everything (really these proofs are just verifications), all the way through the
uniqueness of reduced echelon form. In particular, in this first chapter, the
opportunity is taken to present a few induction proofs, where the arguments
just go over bookkeeping details, so that when induction is needed later (e.g., to
prove that all bases of a finite dimensional vector space have the same number
of members), it will be familiar.
Still another consequence is that the second chapter immediately uses this
background as motivation for the definition of a real vector space. This typically
occurs by the end of the third week. We do not stop to introduce matrix
multiplication and determinants as rote computations. Instead, those topics
appear naturally in the development, after the definition of linear maps.
To help students make the transition from earlier courses, the presentation
here stresses motivation and naturalness. An example is the third chapter,
on linear maps. It does not start with the definition of homomorphism, as
is the case in other books, but with the definition of isomorphism. That’s
because this definition is easily motivated by the observation that some spaces
are just like each other. After that, the next section takes the reasonable step of
defining homomorphisms by isolating the operation-preservation idea. A little
mathematical slickness is lost, but it is in return for a large gain in sensibility
to students.
Having extensive motivation in the text helps with time pressures. I ask
students to, before each class, look ahead in the book, and they follow the
classwork better because they have some prior exposure to the material. For
example, I can start the linear independence class with the definition because I
know students have some idea of what it is about. No book can take the place
of an instructor, but a helpful book gives the instructor more class time for
examples and questions.
Much of a student’s progress takes place while doing the exercises; the exer-
cises here work with the rest of the text. Besides computations, there are many
proofs. These are spread over an approachability range, from simple checks
to some much more involved arguments. There are even a few exercises that
are reasonably challenging puzzles taken, with citation, from various journals,
competitions, or problems collections (as part of the fun of these, the original
wording has been retained as much as possible). In total, the questions are
aimed to both build an ability at, and help students experience the pleasure of,
doing mathematics.
Applications, and Computers. The point of view taken here, that linear
algebra is about vector spaces and linear maps, is not taken to the exclusion
of all other ideas. Applications, and the emerging role of the computer, are
interesting, important, and vital aspects of the subject. Consequently, every
chapter closes with a few application or computer-related topics. Some of the
topics are: network flows, the speed and accuracy of computer linear reductions,
Leontief Input/Output analysis, dimensional analysis, Markov chains, voting
paradoxes, analytic projective geometry, and solving difference equations.
These are brief enough to be done in a day’s class or to be given as indepen-
ii

dent projects for individuals or small groups. Most simply give a reader a feel
for the subject, discuss how linear algebra comes in, point to some accessible
further reading, and give a few exercises. I have kept the exposition lively and
given an overall sense of breadth of application. In short, these topics invite
readers to see for themselves that linear algebra is a tool that a professional
must have.
For people reading this book on their own. The emphasis on motivation
and development make this book a good choice for self-study. While a pro-
fessional mathematician knows what pace and topics suit a class, perhaps an
independent student would find some advice helpful. Here are two timetables
for a semester. The first focuses on core material.
week Mon. Wed. Fri.
1 1.I.1 1.I.1, 2 1.I.2, 3
2 1.I.3 1.II.1 1.II.2
3 1.III.1, 2 1.III.2 2.I.1
4 2.I.2 2.II 2.III.1
5 2.III.1, 2 2.III.2 exam
6 2.III.2, 3 2.III.3 3.I.1
7 3.I.2 3.II.1 3.II.2
8 3.II.2 3.II.2 3.III.1
9 3.III.1 3.III.2 3.IV.1, 2
10 3.IV.2, 3, 4 3.IV.4 exam
11 3.IV.4, 3.V.1 3.V.1, 2 4.I.1, 2
12 4.I.3 4.II 4.II
13 4.III.1 5.I 5.II.1
14 5.II.2 5.II.3 review
The second timetable is more ambitious (it presupposes 1.II, the elements of
vectors, usually covered in third semester calculus).
week Mon. Wed. Fri.
1 1.I.1 1.I.2 1.I.3
2 1.I.3 1.III.1, 2 1.III.2
3 2.I.1 2.I.2 2.II
4 2.III.1 2.III.2 2.III.3
5 2.III.4 3.I.1 exam
6 3.I.2 3.II.1 3.II.2
7 3.III.1 3.III.2 3.IV.1, 2
8 3.IV.2 3.IV.3 3.IV.4
9 3.V.1 3.V.2 3.VI.1
10 3.VI.2 4.I.1 exam
11 4.I.2 4.I.3 4.I.4
12 4.II 4.II, 4.III.1 4.III.2, 3
13 5.II.1, 2 5.II.3 5.III.1
14 5.III.2 5.IV.1, 2 5.IV.2
See the table of contents for the titles of these subsections.
iii

For guidance, in the table of contents I have marked some subsections as
optional if, in my opinion, some instructors will pass over them in favor of
spending more time elsewhere. These subsections can be dropped or added, as
desired. You might also adjust the length of your study by picking one or two
Topics that appeal to you from the end of each chapter. You’ll probably get
more out of these if you have access to computer software that can do the big
calculations.
Do many exercises. (The answers are available.) I have marked a good sam-
ple with X’s. Be warned about the exercises, however, that few inexperienced
people can write correct proofs. Try to find a knowledgeable person to work
with you on this aspect of the material.
Finally, if I may, a caution: I cannot overemphasize how much the statement
(which I sometimes hear), “I understand the material, but it’s only that I can’t
do any of the problems.” reveals a lack of understanding of what we are up
to. Being able to do particular things with the ideas is the entire point. The
quote below expresses this sentiment admirably, and captures the essence of
this book’s approach. It states what I believe is the key to both the beauty and
the power of mathematics and the sciences in general, and of linear algebra in
particular.
I know of no better tactic
than the illustration of exciting principles
by well-chosen particulars.
–Stephen Jay Gould
Jim Hefferon
Saint Michael’s College
Colchester, Vermont USA
jim@joshua.smcvt.edu
April 20, 2000
Author’s Note. Inventing a good exercise, one that enlightens as well as tests,
is a creative act, and hard work (at least half of the the effort on this text
has gone into exercises and solutions). The inventor deserves recognition. But,
somehow, the tradition in texts has been to not give attributions for questions.
I have changed that here where I was sure of the source. I would greatly appre-
ciate hearing from anyone who can help me to correctly attribute others of the
questions. They will be incorporated into later versions of this book.
iv

Contents
1 Linear Systems 1
1.I Solving Linear Systems . . . . . . . . . . . . . . . . . . . . . . . . 1
1.I.1 Gauss’ Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.I.2 Describing the Solution Set . . . . . . . . . . . . . . . . . . . . 11
1.I.3 General = Particular + Homogeneous . . . . . . . . . . . . . . 20
1.II Linear Geometry of n-Space . . . . . . . . . . . . . . . . . . . . . . 32
1.II.1 Vectors in Space . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.II.2 Length and Angle Measures∗
. . . . . . . . . . . . . . . . . . . 38
1.III Reduced Echelon Form . . . . . . . . . . . . . . . . . . . . . . . . 45
1.III.1 Gauss-Jordan Reduction . . . . . . . . . . . . . . . . . . . . . . 45
1.III.2 Row Equivalence . . . . . . . . . . . . . . . . . . . . . . . . . . 51
Topic: Computer Algebra Systems . . . . . . . . . . . . . . . . . . . . . 61
Topic: Input-Output Analysis . . . . . . . . . . . . . . . . . . . . . . . 63
Topic: Accuracy of Computations . . . . . . . . . . . . . . . . . . . . . 67
Topic: Analyzing Networks . . . . . . . . . . . . . . . . . . . . . . . . . 72
2 Vector Spaces 79
2.I Definition of Vector Space . . . . . . . . . . . . . . . . . . . . . . . 80
2.I.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 80
2.I.2 Subspaces and Spanning Sets . . . . . . . . . . . . . . . . . . . 91
2.II Linear Independence . . . . . . . . . . . . . . . . . . . . . . . . . . 102
2.II.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 102
2.III Basis and Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . 113
2.III.1 Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
2.III.2 Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
2.III.3 Vector Spaces and Linear Systems . . . . . . . . . . . . . . . . 124
2.III.4 Combining Subspaces∗
. . . . . . . . . . . . . . . . . . . . . . . 131
Topic: Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
Topic: Crystals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143
Topic: Voting Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . 147
Topic: Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . 152
v

3 Maps Between Spaces 159
3.I Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
3.I.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 159
3.I.2 Dimension Characterizes Isomorphism . . . . . . . . . . . . . . 169
3.II Homomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
3.II.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176
3.II.2 Rangespace and Nullspace . . . . . . . . . . . . . . . . . . . . . 184
3.III Computing Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . 194
3.III.1 Representing Linear Maps with Matrices . . . . . . . . . . . . 194
3.III.2 Any Matrix Represents a Linear Map∗
. . . . . . . . . . . . . . 204
3.IV Matrix Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . 211
3.IV.1 Sums and Scalar Products . . . . . . . . . . . . . . . . . . . . . 211
3.IV.2 Matrix Multiplication . . . . . . . . . . . . . . . . . . . . . . . 214
3.IV.3 Mechanics of Matrix Multiplication . . . . . . . . . . . . . . . . 221
3.IV.4 Inverses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
3.V Change of Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
3.V.1 Changing Representations of Vectors . . . . . . . . . . . . . . . 238
3.V.2 Changing Map Representations . . . . . . . . . . . . . . . . . . 242
3.VI Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250
3.VI.1 Orthogonal Projection Into a Line∗
. . . . . . . . . . . . . . . . 250
3.VI.2 Gram-Schmidt Orthogonalization∗
. . . . . . . . . . . . . . . . 255
3.VI.3 Projection Into a Subspace∗
. . . . . . . . . . . . . . . . . . . . 260
Topic: Line of Best Fit . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
Topic: Geometry of Linear Maps . . . . . . . . . . . . . . . . . . . . . . 274
Topic: Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . 280
Topic: Orthonormal Matrices . . . . . . . . . . . . . . . . . . . . . . . . 286
4 Determinants 293
4.I Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
4.I.1 Exploration∗
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
4.I.2 Properties of Determinants . . . . . . . . . . . . . . . . . . . . 299
4.I.3 The Permutation Expansion . . . . . . . . . . . . . . . . . . . . 303
4.I.4 Determinants Exist∗
. . . . . . . . . . . . . . . . . . . . . . . . 312
4.II Geometry of Determinants . . . . . . . . . . . . . . . . . . . . . . 319
4.II.1 Determinants as Size Functions . . . . . . . . . . . . . . . . . . 319
4.III Other Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326
4.III.1 Laplace’s Expansion∗
. . . . . . . . . . . . . . . . . . . . . . . 326
Topic: Cramer’s Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331
Topic: Speed of Calculating Determinants . . . . . . . . . . . . . . . . . 334
Topic: Projective Geometry . . . . . . . . . . . . . . . . . . . . . . . . . 337
5 Similarity 347
5.I Complex Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . 347
5.I.1 Factoring and Complex Numbers; A Review∗
. . . . . . . . . . 348
5.I.2 Complex Representations . . . . . . . . . . . . . . . . . . . . . 350
5.II Similarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351
vi

5.II.1 Definition and Examples . . . . . . . . . . . . . . . . . . . . . . 351
5.II.2 Diagonalizability . . . . . . . . . . . . . . . . . . . . . . . . . . 353
5.II.3 Eigenvalues and Eigenvectors . . . . . . . . . . . . . . . . . . . 357
5.III Nilpotence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
5.III.1 Self-Composition∗
. . . . . . . . . . . . . . . . . . . . . . . . . 365
5.III.2 Strings∗
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
5.IV Jordan Form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379
5.IV.1 Polynomials of Maps and Matrices∗
. . . . . . . . . . . . . . . 379
5.IV.2 Jordan Canonical Form∗
. . . . . . . . . . . . . . . . . . . . . . 386
Topic: Computing Eigenvalues—the Method of Powers . . . . . . . . . 399
Topic: Stable Populations . . . . . . . . . . . . . . . . . . . . . . . . . . 403
Topic: Linear Recurrences . . . . . . . . . . . . . . . . . . . . . . . . . 405
Appendix A-1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1
Propositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-1
Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-3
Techniques of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . A-5
Sets, Functions, and Relations . . . . . . . . . . . . . . . . . . . . . . . A-6
∗
Note: starred subsections are optional.
vii

Chapter 1
Linear Systems
1.I Solving Linear Systems
Systems of linear equations are common in science and mathematics. These two
examples from high school science [Onan] give a sense of how they arise.
The first example is from Physics. Suppose that we are given three objects,
one with a mass of 2 kg, and are asked to find the unknown masses. Suppose
further that experimentation with a meter stick produces these two balances.
c
h 2
15
40 50
c h
2
25 50
25
Now, since the sum of moments on the left of each balance equals the sum of
moments on the right (the moment of an object is its mass times its distance
from the balance point), the two balances give this system of two equations.
40h + 15c = 100
25c = 50 + 50h
The second example of a linear system is from Chemistry. We can mix,
under controlled conditions, toluene C7H8 and nitric acid HNO3 to produce
trinitrotoluene C7H5O6N3 along with the byproduct water (conditions have to
be controlled very well, indeed — trinitrotoluene is better known as TNT). In
what proportion should those components be mixed? The number of atoms of
each element present before the reaction
x C7H8 + y HNO3 −→ z C7H5O6N3 + w H2O
must equal the number present afterward. Applying that principle to the ele-
ments C, H, N, and O in turn gives this system.
7x = 7z
8x + 1y = 5z + 2w
1y = 3z
3y = 6z + 1w
1

2 Chapter 1. Linear Systems
To finish each of these examples requires solving a system of equations. In
each, the equations involve only the first power of the variables. This chapter
shows how to solve any such system.
1.I.1 Gauss’ Method
1.1 Definition A linear equation in variables x1, x2, . . . , xn has the form
a1x1 + a2x2 + a3x3 + · · · + anxn = d
where the numbers a1, . . . , an ∈ R are the equation’s coefficients and d ∈ R is
the constant. An n-tuple (s1, s2, . . . , sn) ∈ Rn
is a solution of, or satisfies, that
equation if substituting the numbers s1, . . . , sn for the variables gives a true
statement: a1s1 + a2s2 + . . . + ansn = d.
A system of linear equations
a1,1x1 + a1,2x2 + · · · + a1,nxn = d1
a2,1x1 + a2,2x2 + · · · + a2,nxn = d2
.
.
.
am,1x1 + am,2x2 + · · · + am,nxn = dm
has the solution (s1, s2, . . . , sn) if that n-tuple is a solution of all of the equations
in the system.
1.2 Example The ordered pair (−1, 5) is a solution of this system.
3x1 + 2x2 = 7
−x1 + x2 = 6
In contrast, (5, −1) is not a solution.
Finding the set of all solutions is solving the system. No guesswork or good
fortune is needed to solve a linear system. There is an algorithm that always
works. The next example introduces that algorithm, called Gauss’ method. It
transforms the system, step by step, into one with a form that is easily solved.
1.3 Example To solve this system
3x3 = 9
x1 + 5x2 − 2x3 = 2
1
3 x1 + 2x2 = 3

Section I. Solving Linear Systems 3
we repeatedly transform it until it is in a form that is easy to solve.
swap row 1 with row 3
−→
1
3 x1 + 2x2 = 3
x1 + 5x2 − 2x3 = 2
3x3 = 9
multiply row 1 by 3
−→
x1 + 6x2 = 9
x1 + 5x2 − 2x3 = 2
3x3 = 9
add −1 times row 1 to row 2
−→
x1 + 6x2 = 9
−x2 − 2x3 = −7
3x3 = 9
The third step is the only nontrivial one. We’ve mentally multiplied both sides
of the first row by −1, mentally added that to the old second row, and written
the result in as the new second row.
Now we can find the value of each variable. The bottom equation shows
that x3 = 3. Substituting 3 for x3 in the middle equation shows that x2 = 1.
Substituting those two into the top equation gives that x1 = 3 and so the system
has a unique solution: the solution set is { (3, 1, 3) }.
Most of this subsection and the next one consists of examples of solving
linear systems by Gauss’ method. We will use it throughout this book. It is
fast and easy. But, before we get to those examples, we will first show that
this method is also safe in that it never loses solutions or picks up extraneous
solutions.
1.4 Theorem (Gauss’ method) If a linear system is changed to another by
one of these operations
(1) an equation is swapped with another
(2) an equation has both sides multiplied by a nonzero constant
(3) an equation is replaced by the sum of itself and a multiple of another
then the two systems have the same set of solutions.
Each of those three operations has a restriction. Multiplying a row by 0 is
not allowed because obviously that can change the solution set of the system.
Similarly, adding a multiple of a row to itself is not allowed because adding −1
times the row to itself has the effect of multiplying the row by 0. Finally, swap-
ping a row with itself is disallowed to make some results in the fourth chapter
easier to state and remember (and besides, self-swapping doesn’t accomplish
anything).
Proof. We will cover the equation swap operation here and save the other two
cases for Exercise 29.

Consider this swap of row i with row j.
a1,1x1 + a1,2x2 + · · · a1,nxn = d1
.
.
.
ai,1x1 + ai,2x2 + · · · ai,nxn = di
.
.
.
aj,1x1 + aj,2x2 + · · · aj,nxn = dj
.
.
.
am,1x1 + am,2x2 + · · · am,nxn = dm
−→
a1,1x1 + a1,2x2 + · · · a1,nxn = d1
.
.
.
aj,1x1 + aj,2x2 + · · · aj,nxn = dj
.
.
.
ai,1x1 + ai,2x2 + · · · ai,nxn = di
.
.
.
am,1x1 + am,2x2 + · · · am,nxn = dm
The n-tuple (s1, . . . , sn) satisfies the system before the swap if and only if
substituting the values, the s’s, for the variables, the x’s, gives true statements:
a1,1s1+a1,2s2+· · ·+a1,nsn = d1 and . . . ai,1s1+ai,2s2+· · ·+ai,nsn = di and . . .
aj,1s1 + aj,2s2 + · · · + aj,nsn = dj and . . . am,1s1 + am,2s2 + · · · + am,nsn = dm.
In a requirement consisting of statements and-ed together we can rearrange
the order of the statements, so that this requirement is met if and only if a1,1s1+
a1,2s2 + · · · + a1,nsn = d1 and . . . aj,1s1 + aj,2s2 + · · · + aj,nsn = dj and . . .
ai,1s1 + ai,2s2 + · · · + ai,nsn = di and . . . am,1s1 + am,2s2 + · · · + am,nsn = dm.
This is exactly the requirement that (s1, . . . , sn) solves the system after the row
swap. QED
1.5 Definition The three operations from Theorem 1.4 are the elementary re-
duction operations, or row operations, or Gaussian operations. They are swap-
ping, multiplying by a scalar or rescaling, and pivoting.
When writing out the calculations, we will abbreviate ‘row i’ by ‘ρi’. For
instance, we will denote a pivot operation by kρi + ρj, with the row that is
changed written second. We will also, to save writing, often list pivot steps
together when they use the same ρi.
1.6 Example A typical use of Gauss’ method is to solve this system.
x + y = 0
2x − y + 3z = 3
x − 2y − z = 3
The first transformation of the system involves using the first row to eliminate
the x in the second row and the x in the third. To get rid of the second row’s
2x, we multiply the entire first row by −2, add that to the second row, and
write the result in as the new second row. To get rid of the third row’s x, we
multiply the first row by −1, add that to the third row, and write the result in
as the new third row.
−ρ1+ρ3
−→
−2ρ1+ρ2
x + y = 0
−3y + 3z = 3
−3y − z = 3
(Note that the two ρ1 steps −2ρ1 + ρ2 and −ρ1 + ρ3 are written as one opera-
tion.) In this second system, the last two equations involve only two unknowns.

To finish we transform the second system into a third system, where the last
equation involves only one unknown. This transformation uses the second row
to eliminate y from the third row.
−ρ2+ρ3
−→
x + y = 0
−3y + 3z = 3
−4z = 0
Now we are set up for the solution. The third row shows that z = 0. Substitute
that back into the second row to get y = −1, and then substitute back into the
first row to get x = 1.
1.7 Example For the Physics problem from the start of this chapter, Gauss’
method gives this.
40h + 15c = 100
−50h + 25c = 50
5/4ρ1+ρ2
−→
40h + 15c = 100
(175/4)c = 175
So c = 4, and back-substitution gives that h = 1. (The Chemistry problem is
solved later.)
1.8 Example The reduction
x + y + z = 9
2x + 4y − 3z = 1
3x + 6y − 5z = 0
−2ρ1+ρ2
−→
−3ρ1+ρ3
x + y + z = 9
2y − 5z = −17
3y − 8z = −27
−(3/2)ρ2+ρ3
−→
x + y + z = 9
2y − 5z = −17
−1
2 z = −3
2
shows that z = 3, y = −1, and x = 7.
As these examples illustrate, Gauss’ method uses the elementary reduction
operations to set up back-substitution.
1.9 Definition In each row, the first variable with a nonzero coefficient is the
row’s leading variable. A system is in echelon form if each leading variable is
to the right of the leading variable in the row above it (except for the leading
variable in the first row).
1.10 Example The only operation needed in the examples above is pivoting.
Here is a linear system that requires the operation of swapping equations. After
the first pivot
x − y = 0
2x − 2y + z + 2w = 4
y + w = 0
2z + w = 5
−2ρ1+ρ2
−→
x − y = 0
z + 2w = 4
y + w = 0
2z + w = 5

the second equation has no leading y. To get one, we look lower down in the
system for a row that has a leading y and swap it in.
ρ2↔ρ3
−→
x − y = 0
y + w = 0
z + 2w = 4
2z + w = 5
(Had there been more than one row below the second with a leading y then we
could have swapped in any one.) The rest of Gauss’ method goes as before.
−2ρ3+ρ4
−→
x − y = 0
y + w = 0
z + 2w = 4
−3w = −3
Back-substitution gives w = 1, z = 2 , y = −1, and x = −1.
Strictly speaking, the operation of rescaling rows is not needed to solve linear
systems. We have included it because we will use it later in this chapter as part
of a variation on Gauss’ method, the Gauss-Jordan method.
All of the systems seen so far have the same number of equations as un-
knowns. All of them have a solution, and for all of them there is only one
solution. We finish this subsection by seeing for contrast some other things that
can happen.
1.11 Example Linear systems need not have the same number of equations
as unknowns. This system
x + 3y = 1
2x + y = −3
2x + 2y = −2
has more equations than variables. Gauss’ method helps us understand this
system also, since this
−2ρ1+ρ2
−→
−2ρ1+ρ3
x + 3y = 1
−5y = −5
−4y = −4
shows that one of the equations is redundant. Echelon form
−(4/5)ρ2+ρ3
−→
x + 3y = 1
−5y = −5
0 = 0
gives y = 1 and x = −2. The ‘0 = 0’ is derived from the redundancy.

That example’s system has more equations than variables. Gauss’ method
is also useful on systems with more variables than equations. Many examples
are in the next subsection.
Another way that linear systems can differ from the examples shown earlier
is that some linear systems do not have a unique solution. This can happen in
two ways.
The first is that it can fail to have any solution at all.
1.12 Example Contrast the system in the last example with this one.
x + 3y = 1
2x + y = −3
2x + 2y = 0
−2ρ1+ρ2
−→
−2ρ1+ρ3
x + 3y = 1
−5y = −5
−4y = −2
Here the system is inconsistent: no pair of numbers satisfies all of the equations
simultaneously. Echelon form makes this inconsistency obvious.
−(4/5)ρ2+ρ3
−→
x + 3y = 1
−5y = −5
0 = 2
The solution set is empty.
1.13 Example The prior system has more equations than unknowns, but that
is not what causes the inconsistency — Example 1.11 has more equations than
unknowns and yet is consistent. Nor is having more equations than unknowns
necessary for inconsistency, as is illustrated by this inconsistent system with the
same number of equations as unknowns.
x + 2y = 8
2x + 4y = 8
−2ρ1+ρ2
−→
x + 2y = 8
0 = −8
The other way that a linear system can fail to have a unique solution is to
have many solutions.
1.14 Example In this system
x + y = 4
2x + 2y = 8
any pair of numbers satisfying the first equation automatically satisfies the sec-
ond. The solution set {(x, y)
¯
¯ x + y = 4} is infinite — some of its members
are (0, 4), (−1, 5), and (2.5, 1.5). The result of applying Gauss’ method here
contrasts with the prior example because we do not get a contradictory equa-
tion.
−2ρ1+ρ2
−→
x + y = 4
0 = 0

Don’t be fooled by the ‘0 = 0’ equation in that example. It is not the signal
that a system has many solutions.
1.15 Example The absence of a ‘0 = 0’ does not keep a system from having
many different solutions. This system is in echelon form
x + y + z = 0
y + z = 0
has no ‘0 = 0’, and yet has infinitely many solutions. (For instance, each of
these is a solution: (0, 1, −1), (0, 1/2, −1/2), (0, 0, 0), and (0, −π, π). There are
infinitely many solutions because any triple whose first component is 0 and
whose second component is the negative of the third is a solution.)
Nor does the presence of a ‘0 = 0’ mean that the system must have many
solutions. Example 1.11 shows that. So does this system, which does not have
many solutions — in fact it has none — despite that when it is brought to
echelon form it has a ‘0 = 0’ row.
2x − 2z = 6
y + z = 1
2x + y − z = 7
3y + 3z = 0
−ρ1+ρ3
−→
2x − 2z = 6
y + z = 1
y + z = 1
3y + 3z = 0
−ρ2+ρ3
−→
−3ρ2+ρ4
2x − 2z = 6
y + z = 1
0 = 0
0 = −3
We will finish this subsection with a summary of what we’ve seen so far
about Gauss’ method.
Gauss’ method uses the three row operations to set a system up for back
substitution. If any step shows a contradictory equation then we can stop
with the conclusion that the system has no solutions. If we reach echelon form
without a contradictory equation, and each variable is a leading variable in its
row, then the system has a unique solution and we find it by back substitution.
Finally, if we reach echelon form without a contradictory equation, and there is
not a unique solution (at least one variable is not a leading variable) then the
system has many solutions.
The next subsection deals with the third case — we will see how to describe
the solution set of a system with many solutions.
Exercises
X 1.16 Use Gauss’ method to find the unique solution for each system.
(a)
2x + 3y = 13
x − y = −1
(b)
x − z = 0
3x + y = 1
−x + y + z = 4
X 1.17 Use Gauss’ method to solve each system or conclude ‘many solutions’ or ‘no
solutions’.

(a) 2x + 2y = 5
x − 4y = 0
(b) −x + y = 1
x + y = 2
(c) x − 3y + z = 1
x + y + 2z = 14
(d) −x − y = 1
−3x − 3y = 2
(e) 4y + z = 20
2x − 2y + z = 0
x + z = 5
x + y − z = 10
(f) 2x + z + w = 5
y − w = −1
3x − z − w = 0
4x + y + 2z + w = 9
X 1.18 There are methods for solving linear systems other than Gauss’ method. One
often taught in high school is to solve one of the equations for a variable, then
substitute the resulting expression into other equations. That step is repeated
until there is an equation with only one variable. From that, the first number in
the solution is derived, and then back-substitution can be done. This method both
takes longer than Gauss’ method, since it involves more arithmetic operations and
is more likely to lead to errors. To illustrate how it can lead to wrong conclusions,
we will use the system
x + 3y = 1
2x + y = −3
2x + 2y = 0
from Example 1.12.
(a) Solve the first equation for x and substitute that expression into the second
equation. Find the resulting y.
(b) Again solve the first equation for x, but this time substitute that expression
into the third equation. Find this y.
What extra step must a user of this method take to avoid erroneously concluding
a system has a solution?
X 1.19 For which values of k are there no solutions, many solutions, or a unique
solution to this system?
x − y = 1
3x − 3y = k
X 1.20 This system is not linear:
2 sin α − cos β + 3 tan γ = 3
4 sin α + 2 cos β − 2 tan γ = 10
6 sin α − 3 cos β + tan γ = 9
but we can nonetheless apply Gauss’ method. Do so. Does the system have a
solution?
X 1.21 What conditions must the constants, the b’s, satisfy so that each of these
systems has a solution? Hint. Apply Gauss’ method and see what happens to the
right side.
(a) x − 3y = b1
3x + y = b2
x + 7y = b3
2x + 4y = b4
(b) x1 + 2x2 + 3x3 = b1
2x1 + 5x2 + 3x3 = b2
x1 + 8x3 = b3
1.22 True or false: a system with more unknowns than equations has at least one
solution. (As always, to say ‘true’ you must prove it, while to say ‘false’ you must
produce a counterexample.)
1.23 Must any Chemistry problem like the one that starts this subsection — a
balance the reaction problem — have infinitely many solutions?
X 1.24 Find the coefficients a, b, and c so that the graph of f(x) = ax2
+bx+c passes
through the points (1, 2), (−1, 6), and (2, 3).

1.25 Gauss’ method works by combining the equations in a system to make new
equations.
(a) Can the equation 3x−2y = 5 be derived, by a sequence of Gaussian reduction
steps, from the equations in this system?
x + y = 1
4x − y = 6
(b) Can the equation 5x−3y = 2 be derived, by a sequence of Gaussian reduction
steps, from the equations in this system?
2x + 2y = 5
3x + y = 4
(c) Can the equation 6x − 9y + 5z = −2 be derived, by a sequence of Gaussian
reduction steps, from the equations in the system?
2x + y − z = 4
6x − 3y + z = 5
1.26 Prove that, where a, b, . . . , e are real numbers and a 6= 0, if
ax + by = c
has the same solution set as
ax + dy = e
then they are the same equation. What if a = 0?
X 1.27 Show that if ad − bc 6= 0 then
ax + by = j
cx + dy = k
has a unique solution.
X 1.28 In the system
ax + by = c
dx + ey = f
each of the equations describes a line in the xy-plane. By geometrical reasoning,
show that there are three possibilities: there is a unique solution, there is no
solution, and there are infinitely many solutions.
1.29 Finish the proof of Theorem 1.4.
1.30 Is there a two-unknowns linear system whose solution set is all of R2
?
X 1.31 Are any of the operations used in Gauss’ method redundant? That is, can
any of the operations be synthesized from the others?
1.32 Prove that each operation of Gauss’ method is reversible. That is, show that if
two systems are related by a row operation S1 ↔ S2 then there is a row operation
to go back S2 ↔ S1.
1.33 A box holding pennies, nickels and dimes contains thirteen coins with a total
value of 83 cents. How many coins of each type are in the box?
1.34 [Con. Prob. 1955] Four positive integers are given. Select any three of the
integers, find their arithmetic average, and add this result to the fourth integer.
Thus the numbers 29, 23, 21, and 17 are obtained. One of the original integers
is:

(a) 19 (b) 21 (c) 23 (d) 29 (e) 17
X 1.35 [Am. Math. Mon., Jan. 1935] Laugh at this: AHAHA + TEHE = TEHAW.
It resulted from substituting a code letter for each digit of a simple example in
addition, and it is required to identify the letters and prove the solution unique.
1.36 [Wohascum no. 2] The Wohascum County Board of Commissioners, which has
20 members, recently had to elect a President. There were three candidates (A, B,
and C); on each ballot the three candidates were to be listed in order of preference,
with no abstentions. It was found that 11 members, a majority, preferred A over
B (thus the other 9 preferred B over A). Similarly, it was found that 12 members
preferred C over A. Given these results, it was suggested that B should withdraw,
to enable a runoff election between A and C. However, B protested, and it was
then found that 14 members preferred B over C! The Board has not yet recovered
from the resulting confusion. Given that every possible order of A, B, C appeared
on at least one ballot, how many members voted for B as their first choice?
1.37 [Am. Math. Mon., Jan. 1963] “This system of n linear equations with n un-
knowns,” said the Great Mathematician, “has a curious property.”
“Good heavens!” said the Poor Nut, “What is it?”
“Note,” said the Great Mathematician, “that the constants are in arithmetic
progression.”
“It’s all so clear when you explain it!” said the Poor Nut. “Do you mean like
6x + 9y = 12 and 15x + 18y = 21?”
“Quite so,” said the Great Mathematician, pulling out his bassoon. “Indeed,
the system has a unique solution. Can you find it?”
“Good heavens!” cried the Poor Nut, “I am baffled.”
Are you?
1.I.2 Describing the Solution Set
A linear system with a unique solution has a solution set with one element.
A linear system with no solution has a solution set that is empty. In these cases
the solution set is easy to describe. Solution sets are a challenge to describe
only when they contain many elements.
2.1 Example This system has many solutions because in echelon form
2x + z = 3
x − y − z = 1
3x − y = 4
−(1/2)ρ1+ρ2
−→
−(3/2)ρ1+ρ3
2x + z = 3
−y − (3/2)z = −1/2
−y − (3/2)z = −1/2
−ρ2+ρ3
−→
2x + z = 3
−y − (3/2)z = −1/2
0 = 0
not all of the variables are leading variables. The Gauss’ method theorem
showed that a triple satisfies the first system if and only if it satisfies the third.
Thus, the solution set {(x, y, z)
¯
¯ 2x + z = 3 and x − y − z = 1 and 3x − y = 4}

can also be described as {(x, y, z)
¯
¯ 2x + z = 3 and −y − 3z/2 = −1/2}. How-
ever, this second description is not much of an improvement. It has two equa-
tions instead of three, but it still involves some hard-to-understand interaction
among the variables.
To get a description that is free of any such interaction, we take the vari-
able that does not lead any equation, z, and use it to describe the variables
that do lead, x and y. The second equation gives y = (1/2) − (3/2)z and
the first equation gives x = (3/2) − (1/2)z. Thus, the solution set can be de-
scribed as {(x, y, z) = ((3/2) − (1/2)z, (1/2) − (3/2)z, z)
¯
¯ z ∈ R}. For instance,
(1/2, −5/2, 2) is a solution because taking z = 2 gives a first component of 1/2
and a second component of −5/2.
The advantage of this description over the ones above is that the only variable
appearing, z, is unrestricted — it can be any real number.
2.2 Definition The non-leading variables in an echelon-form linear system are
free variables.
In the echelon form system derived in the above example, x and y are leading
variables and z is free.
2.3 Example A linear system can end with more than one variable free. This
row reduction
x + y + z − w = 1
y − z + w = −1
3x + 6z − 6w = 6
−y + z − w = 1
−3ρ1+ρ3
−→
x + y + z − w = 1
y − z + w = −1
−3y + 3z − 3w = 3
−y + z − w = 1
3ρ2+ρ3
−→
ρ2+ρ4
x + y + z − w = 1
y − z + w = −1
0 = 0
0 = 0
ends with x and y leading, and with both z and w free. To get the description
that we prefer we will start at the bottom. We first express y in terms of
the free variables z and w with y = −1 + z − w. Next, moving up to the
top equation, substituting for y in the first equation x + (−1 + z − w) + z −
w = 1 and solving for x yields x = 2 − 2z + 2w. Thus, the solution set is
{2 − 2z + 2w, −1 + z − w, z, w)
¯
¯ z, w ∈ R}.
We prefer this description because the only variables that appear, z and w,
are unrestricted. This makes the job of deciding which four-tuples are system
solutions into an easy one. For instance, taking z = 1 and w = 2 gives the
solution (4, −2, 1, 2). In contrast, (3, −2, 1, 2) is not a solution, since the first
component of any solution must be 2 minus twice the third component plus
twice the fourth.

2.4 Example After this reduction
2x − 2y = 0
z + 3w = 2
3x − 3y = 0
x − y + 2z + 6w = 4
−(3/2)ρ1+ρ3
−→
−(1/2)ρ1+ρ4
2x − 2y = 0
z + 3w = 2
0 = 0
2z + 6w = 4
−2ρ2+ρ4
−→
2x − 2y = 0
z + 3w = 2
0 = 0
0 = 0
x and z lead, y and w are free. The solution set is {(y, y, 2 − 3w, w)
¯
¯ y, w ∈ R}.
For instance, (1, 1, 2, 0) satisfies the system — take y = 1 and w = 0. The
four-tuple (1, 0, 5, 4) is not a solution since its first coordinate does not equal its
second.
We refer to a variable used to describe a family of solutions as a parameter
and we say that the set above is paramatrized with y and w. (The terms
‘parameter’ and ‘free variable’ do not mean the same thing. Above, y and w
are free because in the echelon form system they do not lead any row. They
are parameters because they are used in the solution set description. We could
have instead paramatrized with y and z by rewriting the second equation as
w = 2/3 − (1/3)z. In that case, the free variables are still y and w, but the
parameters are y and z. Notice that we could not have paramatrized with x and
y, so there is sometimes a restriction on the choice of parameters. The terms
‘parameter’ and ‘free’ are related because, as we shall show later in this chapter,
the solution set of a system can always be paramatrized with the free variables.
Consequenlty, we shall paramatrize all of our descriptions in this way.)
2.5 Example This is another system with infinitely many solutions.
x + 2y = 1
2x + z = 2
3x + 2y + z − w = 4
−2ρ1+ρ2
−→
−3ρ1+ρ3
x + 2y = 1
−4y + z = 0
−4y + z − w = 1
−ρ2+ρ3
−→
x + 2y = 1
−4y + z = 0
−w = 1
The leading variables are x, y, and w. The variable z is free. (Notice here that,
although there are infinitely many solutions, the value of one of the variables is
fixed — w = −1.) Write w in terms of z with w = −1 + 0z. Then y = (1/4)z.
To express x in terms of z, substitute for y into the first equation to get x =
1 − (1/2)z. The solution set is {(1 − (1/2)z, (1/4)z, z, −1)
¯
¯ z ∈ R}.
We finish this subsection by developing the notation for linear systems and
their solution sets that we shall use in the rest of this book.
2.6 Definition An m×n matrix is a rectangular array of numbers with m rows
and n columns. Each number in the matrix is an entry,

Matrices are usually named by upper case roman letters, e.g. A. Each entry is
denoted by the corresponding lower-case letter, e.g. ai,j is the number in row i
and column j of the array. For instance,
A =
µ
1 2.2 5
3 4 −7
¶
has two rows and three columns, and so is a 2×3 matrix. (Read that “two-
by-three”; the number of rows is always stated first.) The entry in the second
row and first column is a2,1 = 3. Note that the order of the subscripts matters:
a1,2 6= a2,1 since a1,2 = 2.2. (The parentheses around the array are a typo-
graphic device so that when two matrices are side by side we can tell where one
ends and the other starts.)
2.7 Example We can abbreviate this linear system
x1 + 2x2 = 4
x2 − x3 = 0
x1 + 2x3 = 4
with this matrix.


1 2 0 4
0 1 −1 0
1 0 2 4


The vertical bar just reminds a reader of the difference between the coefficients
on the systems’s left hand side and the constants on the right. When a bar
is used to divide a matrix into parts, we call it an augmented matrix. In this
notation, Gauss’ method goes this way.


1 2 0 4
0 1 −1 0
1 0 2 4

 −ρ1+ρ3
−→


1 2 0 4
0 1 −1 0
0 −2 2 0

 2ρ2+ρ3
−→


1 2 0 4
0 1 −1 0
0 0 0 0


The second row stands for y − z = 0 and the first row stands for x + 2y = 4 so
the solution set is {(4 − 2z, z, z)
¯
¯ z ∈ R}. One advantage of the new notation is
that the clerical load of Gauss’ method — the copying of variables, the writing
of +’s and =’s, etc. — is lighter.
We will also use the array notation to clarify the descriptions of solution
sets. A description like {(2 − 2z + 2w, −1 + z − w, z, w)
¯
¯ z, w ∈ R} from Ex-
ample 2.3 is hard to read. We will rewrite it to group all the constants together,
all the coefficients of z together, and all the coefficients of w together. We will
write them vertically, in one-column wide matrices.
{




2
−1
0
0



 +




−2
1
1
0



 · z +




2
−1
0
1



 · w
¯
¯ z, w ∈ R}

For instance, the top line says that x = 2 − 2z + 2w. The next section gives a
geometric interpretation that will help us picture the solution sets when they
are written in this way.
2.8 Definition A vector (or column vector) is a matrix with a single column.
A matrix with a single row is a row vector. The entries of a vector are its
components.
Vectors are an exception to the convention of representing matrices with
capital roman letters. We use lower-case roman or greek letters overlined with
an arrow: ~
a, ~
b, . . . or ~
α, ~
β, . . . (boldface is also common: a or α). For
instance, this is a column vector with a third component of 7.
~
v =


1
3
7


2.9 Definition The linear equation a1x1 + a2x2 + · · · + anxn = d with un-
knowns x1, . . . , xn is satisfied by
~
s =



s1
.
.
.
sn



if a1s1 + a2s2 + · · · + ansn = d. A vector satisfies a linear system if it satisfies
each equation in the system.
The style of description of solution sets that we use involves adding the
vectors, and also multiplying them by real numbers, such as the z and w. We
need to define these operations.
2.10 Definition The vector sum of ~
u and ~
v is this.
~
u + ~
v =



u1
.
.
.
un


 +



v1
.
.
.
vn


 =



u1 + v1
.
.
.
un + vn



In general, two matrices with the same number of rows and the same number
of columns add in this way, entry-by-entry.
2.11 Definition The scalar multiplication of the real number r and the vector
~
v is this.
r · ~
v = r ·



v1
.
.
.
vn


 =



rv1
.
.
.
rvn



In general, any matrix is multiplied by a real number in this entry-by-entry way.

Scalar multiplication can be written in either order: r ·~
v or ~
v · r, or without
the ‘·’ symbol: r~
v. (Do not refer to scalar multiplication as ‘scalar product’
because that name is used for a different operation.)
2.12 Example


2
3
1

 +


3
−1
4

 =


2 + 3
3 − 1
1 + 4

 =


5
2
5

 7 ·




1
4
−1
−3



 =




7
28
−7
−21




Notice that the definitions of vector addition and scalar multiplication agree
where they overlap, for instance, ~
v + ~
v = 2~
v.
With the notation defined, we can now solve systems in the way that we will
use throughout this book.
2.13 Example This system
2x + y − w = 4
y + w + u = 4
x − z + 2w = 0
reduces in this way.


2 1 0 −1 0 4
0 1 0 1 1 4
1 0 −1 2 0 0

 −(1/2)ρ1+ρ3
−→


2 1 0 −1 0 4
0 1 0 1 1 4
0 −1/2 −1 5/2 0 −2


(1/2)ρ2+ρ3
−→


2 1 0 −1 0 4
0 1 0 1 1 4
0 0 −1 3 1/2 0


The solution set is {(w + (1/2)u, 4 − w − u, 3w + (1/2)u, w, u)
¯
¯ w, u ∈ R}. We
write that in vector form.
{






x
y
z
w
u






=






0
4
0
0
0






+






1
−1
3
1
0






w +






1/2
−1
1/2
0
1






u
¯
¯ w, u ∈ R}
Note again how well vector notation sets off the coefficients of each parameter.
For instance, the third row of the vector form shows plainly that if u is held
fixed then z increases three times as fast as w.
That format also shows plainly that there are infinitely many solutions. For
example, we can fix u as 0, let w range over the real numbers, and consider the
first component x. We get infinitely many first components and hence infinitely
many solutions.

Another thing shown plainly is that setting both w and u to zero gives that
this






x
y
z
w
u






=






0
4
0
0
0






is a particular solution of the linear system.
2.14 Example In the same way, this system
x − y + z = 1
3x + z = 3
5x − 2y + 3z = 5
reduces


1 −1 1 1
3 0 1 3
5 −2 3 5

 −3ρ1+ρ2
−→
−5ρ1+ρ3


1 −1 1 1
0 3 −2 0
0 3 −2 0

 −ρ2+ρ3
−→


1 −1 1 1
0 3 −2 0
0 0 0 0


to a one-parameter solution set.
{


1
0
0

 +


−1/3
2/3
1

 z
¯
¯ z ∈ R}
Before the exercises, we pause to point out some things that we have yet to
do.
The first two subsections have been on the mechanics of Gauss’ method.
Except for one result, Theorem 1.4 — without which developing the method
doesn’t make sense since it says that the method gives the right answers — we
have not stopped to consider any of the interesting questions that arise.
For example, can we always describe solution sets as above, with a particular
solution vector added to an unrestricted linear combination of some other vec-
tors? The solution sets we described with unrestricted parameters were easily
seen to have infinitely many solutions so an answer to this question could tell
us something about the size of solution sets. An answer to that question could
also help us picture the solution sets — what do they look like in R2
, or in R3
,
etc?
Many questions arise from the observation that Gauss’ method can be done
in more than one way (for instance, when swapping rows, we may have a choice
of which row to swap with). Theorem 1.4 says that we must get the same
solution set no matter how we proceed, but if we do Gauss’ method in two
different ways must we get the same number of free variables both times, so
that any two solution set descriptions have the same number of parameters?

Must those be the same variables (e.g., is it impossible to solve a problem one
way and get y and w free or solve it another way and get y and z free)?
In the rest of this chapter we answer these questions. The answer to each
is ‘yes’. The first question is answered in the last subsection of this section. In
the second section we give a geometric description of solution sets. In the final
section of this chapter we tackle the last set of questions.
Consequently, by the end of the first chapter we will not only have a solid
grounding in the practice of Gauss’ method, we will also have a solid grounding
in the theory. We will be sure of what can and cannot happen in a reduction.
Exercises
X 2.15 Find the indicated entry of the matrix, if it is defined.
A =
µ
1 3 1
2 −1 4
¶
(a) a2,1 (b) a1,2 (c) a2,2 (d) a3,1
X 2.16 Give the size of each matrix.
(a)
µ
1 0 4
2 1 5
¶
(b)
Ã
1 1
−1 1
3 −1
!
(c)
µ
5 10
10 5
¶
X 2.17 Do the indicated vector operation, if it is defined.
(a)
Ã
2
1
1
!
+
Ã
3
0
4
!
(b) 5
µ
4
−1
¶
(c)
Ã
1
5
1
!
−
Ã
3
1
1
!
(d) 7
µ
2
1
¶
+ 9
µ
3
5
¶
(e)
µ
1
2
¶
+
Ã
1
2
3
!
(f) 6
Ã
3
1
1
!
− 4
Ã
2
0
3
!
+ 2
Ã
1
1
5
!
X 2.18 Solve each system using matrix notation. Express the solution using vec-
tors.
(a) 3x + 6y = 18
x + 2y = 6
(b) x + y = 1
x − y = −1
(c) x1 + x3 = 4
x1 − x2 + 2x3 = 5
4x1 − x2 + 5x3 = 17
(d) 2a + b − c = 2
2a + c = 3
a − b = 0
(e) x + 2y − z = 3
2x + y + w = 4
x − y + z + w = 1
(f) x + z + w = 4
2x + y − w = 2
3x + y + z = 7
X 2.19 Solve each system using matrix notation. Give each solution set in vector
notation.
(a) 2x + y − z = 1
4x − y = 3
(b) x − z = 1
y + 2z − w = 3
x + 2y + 3z − w = 7
(c) x − y + z = 0
y + w = 0
3x − 2y + 3z + w = 0
−y − w = 0
(d) a + 2b + 3c + d − e = 1
3a − b + c + d + e = 3
X 2.20 The vector is in the set. What value of the parameters produces that vec-
tor?
(a)
µ
5
−5
¶
, {
µ
1
−1
¶
k
¯
¯ k ∈ R}

(b)
Ã
−1
2
1
!
, {
Ã
−2
1
0
!
i +
Ã
3
0
1
!
j
¯
¯ i, j ∈ R}
(c)
Ã
0
−4
2
!
, {
Ã
1
1
0
!
m +
Ã
2
0
1
!
n
¯
¯ m, n ∈ R}
2.21 Decide if the vector is in the set.
(a)
µ
3
−1
¶
, {
µ
−6
2
¶
k
¯
¯ k ∈ R}
(b)
µ
5
4
¶
, {
µ
5
−4
¶
j
¯
¯ j ∈ R}
(c)
Ã
2
1
−1
!
, {
Ã
0
3
−7
!
+
Ã
1
−1
3
!
r
¯
¯ r ∈ R}
(d)
Ã
1
0
1
!
, {
Ã
2
0
1
!
j +
Ã
−3
−1
1
!
k
¯
¯ j, k ∈ R}
2.22 Paramatrize the solution set of this one-equation system.
x1 + x2 + · · · + xn = 0
X 2.23 (a) Apply Gauss’ method to the left-hand side to solve
x + 2y − w = a
2x + z = b
x + y + 2w = c
for x, y, z, and w, in terms of the constants a, b, and c.
(b) Use your answer from the prior part to solve this.
x + 2y − w = 3
2x + z = 1
x + y + 2w = −2
X 2.24 Why is the comma needed in the notation ‘ai,j’ for matrix entries?
X 2.25 Give the 4×4 matrix whose i, j-th entry is
(a) i + j; (b) −1 to the i + j power.
2.26 For any matrix A, the transpose of A, written Atrans
, is the matrix whose
columns are the rows of A. Find the transpose of each of these.
(a)
µ
1 2 3
4 5 6
¶
(b)
µ
2 −3
1 1
¶
(c)
µ
5 10
10 5
¶
(d)
Ã
1
1
0
!
X 2.27 (a) Describe all functions f(x) = ax2
+ bx + c such that f(1) = 2 and
f(−1) = 6.
(b) Describe all functions f(x) = ax2
+ bx + c such that f(1) = 2.
2.28 Show that any set of five points from the plane R2
lie on a common conic
section, that is, they all satisfy some equation of the form ax2
+ by2
+ cxy + dx +
ey + f = 0 where some of a, . . . , f are nonzero.
2.29 Make up a four equations/four unknowns system having
(a) a one-parameter solution set;
(b) a two-parameter solution set;
(c) a three-parameter solution set.

2.30 [USSR Olympiad no. 174]
(a) Solve the system of equations.
ax + y = a2
x + ay = 1
For what values of a does the system fail to have solutions, and for what values
of a are there infinitely many solutions?
(b) Answer the above question for the system.
ax + y = a3
x + ay = 1
2.31 [Math. Mag., Sept. 1952] In air a gold-surfaced sphere weighs 7588 grams. It
is known that it may contain one or more of the metals aluminum, copper, silver,
or lead. When weighed successively under standard conditions in water, benzene,
alcohol, and glycerine its respective weights are 6588, 6688, 6778, and 6328 grams.
How much, if any, of the forenamed metals does it contain if the specific gravities
of the designated substances are taken to be as follows?
Aluminum 2.7 Alcohol 0.81
Copper 8.9 Benzene 0.90
Gold 19.3 Glycerine 1.26
Lead 11.3 Water 1.00
Silver 10.8
1.I.3 General = Particular + Homogeneous
The prior subsection has many descriptions of solution sets. They all fit a
pattern. They have a vector that is a particular solution of the system added
to an unrestricted combination of some other vectors. The solution set from
Example 2.13 illustrates.
{






0
4
0
0
0






| {z }
particular
solution
+ w






1
−1
3
1
0






+ u






1/2
−1
1/2
0
1






| {z }
unrestricted
combination
¯
¯ w, u ∈ R}
The combination is unrestricted in that w and u can be any real numbers —
there is no condition like “such that 2w−u = 0” that would restrict which pairs
w, u can be used to form combinations.
That example shows an infinite solution set conforming to the pattern. We
can think of the other two kinds of solution sets as also fitting the same pat-
tern. A one-element solution set fits in that it has a particular solution, and
the unrestricted combination part is a trivial sum (that is, instead of being a
combination of two vectors, as above, or a combination of one vector, it is a

combination of no vectors). A zero-element solution set fits the pattern since
there is no particular solution, and so the set of sums of that form is empty.
We will show that the examples from the prior subsection are representative,
in that the description pattern discussed above holds for every solution set.
3.1 Theorem For any linear system there are vectors ~
β1, . . . , ~
βk such that
the solution set can be described as
{~
p + c1
~
β1 + · · · + ck
~
βk
¯
¯ c1, . . . , ck ∈ R}
where ~
p is any particular solution, and where the system has k free variables.
This description has two parts, the particular solution ~
p and also the un-
restricted linear combination of the ~
β’s. We shall prove the theorem in two
corresponding parts, with two lemmas.
We will focus first on the unrestricted combination part. To do that, we
consider systems that have the vector of zeroes as one of the particular solutions,
so that ~
p + c1
~
β1 + · · · + ck
~
βk can be shortened to c1
~
β1 + · · · + ck
~
βk.
3.2 Definition A linear equation is homogeneous if it has a constant of zero,
that is, if it can be put in the form a1x1 + a2x2 + · · · + anxn = 0.
(These are ‘homogeneous’ because all of the terms involve the same power of
their variable — the first power — including a ‘0x0’ that we can imagine is on
the right side.)
3.3 Example With any linear system like
3x + 4y = 3
2x − y = 1
we associate a system of homogeneous equations by setting the right side to
zeros.
3x + 4y = 0
2x − y = 0
Our interest in the homogeneous system associated with a linear system can be
understood by comparing the reduction of the system
3x + 4y = 3
2x − y = 1
−(2/3)ρ1+ρ2
−→
3x + 4y = 3
−(11/3)y = −1
with the reduction of the associated homogeneous system.
3x + 4y = 0
2x − y = 0
−(2/3)ρ1+ρ2
−→
3x + 4y = 0
−(11/3)y = 0
Obviously the two reductions go in the same way. We can study how linear sys-
tems are reduced by instead studying how the associated homogeneous systems
are reduced.

Studying the associated homogeneous system has a great advantage over
studying the original system. Nonhomogeneous systems can be inconsistent.
But a homogeneous system must be consistent since there is always at least one
solution, the vector of zeros.
3.4 Definition A column or row vector of all zeros is a zero vector, denoted ~
0.
There are many different zero vectors, e.g., the one-tall zero vector, the two-tall
zero vector, etc. Nonetheless, people often refer to “the” zero vector, expecting
that the size of the one being discussed will be clear from the context.
3.5 Example Some homogeneous systems have the zero vector as their only
solution.
3x + 2y + z = 0
6x + 4y = 0
y + z = 0
−2ρ1+ρ2
−→
3x + 2y + z = 0
−2z = 0
y + z = 0
ρ2↔ρ3
−→
3x + 2y + z = 0
y + z = 0
−2z = 0
3.6 Example Some homogeneous systems have many solutions. One example
is the Chemistry problem from the first page of this book.
7x − 7j = 0
8x + y − 5j − 2k = 0
y − 3j = 0
3y − 6j − k = 0
−(8/7)ρ1+ρ2
−→
7x − 7z = 0
y + 3z − 2w = 0
y − 3z = 0
3y − 6z − w = 0
−ρ2+ρ3
−→
−3ρ2+ρ4
7x − 7z = 0
y + 3z − 2w = 0
−6z + 2w = 0
−15z + 5w = 0
−(5/2)ρ3+ρ4
−→
7x − 7z = 0
y + 3z − 2w = 0
−6z + 2w = 0
0 = 0
The solution set:
{




1/3
1
1/3
1



 w
¯
¯ k ∈ R}
has many vectors besides the zero vector (if we interpret w as a number of
molecules then solutions make sense only when w is a nonnegative multiple of
3).
We now have the terminology to prove the two parts of Theorem 3.1. The
first lemma deals with unrestricted combinations.

3.7 Lemma For any homogeneous linear system there exist vectors ~
β1, . . . ,
~
βk such that the solution set of the system is
{c1
~
β1 + · · · + ck
~
βk
¯
¯ c1, . . . , ck ∈ R}
where k is the number of free variables in an echelon form version of the system.
Before the proof, we will recall the back substitution calculations that were
done in the prior subsection. Imagine that we have brought a system to this
echelon form.
x + 2y − z + 2w = 0
−3y + z = 0
−w = 0
We next perform back-substitution to express each variable in terms of the
free variable z. Working from the bottom up, we get first that w is 0 · z,
next that y is (1/3) · z, and then substituting those two into the top equation
x + 2((1/3)z) − z + 2(0) = 0 gives x = (1/3) · z. So, back substitution gives
a paramatrization of the solution set by starting at the bottom equation and
using the free variables as the parameters to work row-by-row to the top. The
proof below follows this pattern.
Comment: That is, this proof just does a verification of the bookkeeping in
back substitution to show that we haven’t overlooked any obscure cases where
this procedure fails, say, by leading to a division by zero. So this argument,
while quite detailed, doesn’t give us any new insights. Nevertheless, we have
written it out for two reasons. The first reason is that we need the result — the
computational procedure that we employ must be verified to work as promised.
The second reason is that the row-by-row nature of back substitution leads to a
proof that uses the technique of mathematical induction.∗
This is an important,
and non-obvious, proof technique that we shall use a number of times in this
book. Doing an induction argument here gives us a chance to see one in a setting
where the proof material is easy to follow, and so the technique can be studied.
Readers who are unfamiliar with induction arguments should be sure to master
this one and the ones later in this chapter before going on to the second chapter.
Proof. First use Gauss’ method to reduce the homogeneous system to echelon
form. We will show that each leading variable can be expressed in terms of free
variables. That will finish the argument because then we can use those free
variables as the parameters. That is, the ~
β’s are the vectors of coefficients of
the free variables (as in Example 3.6, where the solution is x = (1/3)w, y = w,
z = (1/3)w, and w = w).
We will proceed by mathematical induction, which has two steps. The base
step of the argument will be to focus on the bottom-most non-‘0 = 0’ equation
and write its leading variable in terms of the free variables. The inductive step
of the argument will be to argue that if we can express the leading variables from
∗ More information on mathematical induction is in the appendix.

the bottom t rows in terms of free variables, then we can express the leading
variable of the next row up — the t + 1-th row up from the bottom — in terms
of free variables. With those two steps, the theorem will be proved because by
the base step it is true for the bottom equation, and by the inductive step the
fact that it is true for the bottom equation shows that it is true for the next
one up, and then another application of the inductive step implies it is true for
third equation up, etc.
For the base step, consider the bottom-most non-‘0 = 0’ equation (the case
where all the equations are ‘0 = 0’ is trivial). We call that the m-th row:
am,`m
x`m
+ am,`m+1x`m+1 + · · · + am,nxn = 0
where am,`m
6= 0. (The notation here has ‘`’ stand for ‘leading’, so am,`m
means
“the coefficient, from the row m of the variable leading row m”.) Either there
are variables in this equation other than the leading one x`m
or else there are
not. If there are other variables x`m+1, etc., then they must be free variables
because this is the bottom non-‘0 = 0’ row. Move them to the right and divide
by am,`m
x`m
= (−am,`m+1/am,`m
)x`m+1 + · · · + (−am,n/am,`m
)xn
to expresses this leading variable in terms of free variables. If there are no free
variables in this equation then x`m
= 0 (see the “tricky point” noted following
this proof).
For the inductive step, we assume that for the m-th equation, and for the
(m − 1)-th equation, . . . , and for the (m − t)-th equation, we can express the
leading variable in terms of free variables (where 0 ≤ t < m). To prove that the
same is true for the next equation up, the (m − (t + 1))-th equation, we take
each variable that leads in a lower-down equation x`m
, . . . , x`m−t
and substitute
its expression in terms of free variables. The result has the form
am−(t+1),`m−(t+1)
x`m−(t+1)
+ sums of multiples of free variables = 0
where am−(t+1),`m−(t+1)
6= 0. We move the free variables to the right-hand side
and divide by am−(t+1),`m−(t+1)
, to end with x`m−(t+1)
expressed in terms of free
variables.
Because we have shown both the base step and the inductive step, by the
principle of mathematical induction the proposition is true. QED
We say that the set {c1
~
β1 + · · · + ck
~
βk
¯
¯ c1, . . . , ck ∈ R} is generated by or
spanned by the set of vectors {~
β1, . . . , ~
βk}. There is a tricky point to this
definition. If a homogeneous system has a unique solution, the zero vector,
then we say the solution set is generated by the empty set of vectors. This fits
with the pattern of the other solution sets: in the proof above the solution set is
derived by taking the c’s to be the free variables and if there is a unique solution
then there are no free variables.
This proof incidentally shows, as discussed after Example 2.4, that solution
sets can always be paramatrized using the free variables.

The next lemma finishes the proof of Theorem 3.1 by considering the par-
ticular solution part of the solution set’s description.
3.8 Lemma For a linear system, where ~
p is any particular solution, the solu-
tion set equals this set.
{~
p + ~
h
¯
¯ ~
h satisfies the associated homogeneous system}
Proof. We will show mutual set inclusion, that any solution to the system is
in the above set and that anything in the set is a solution to the system.∗
For set inclusion the first way, that if a vector solves the system then it is in
the set described above, assume that ~
s solves the system. Then ~
s − ~
p solves the
associated homogeneous system since for each equation index i between 1 and
n,
ai,1(s1 − p1) + · · · + ai,n(sn − pn) = (ai,1s1 + · · · + ai,nsn)
− (ai,1p1 + · · · + ai,npn)
= di − di
= 0
where pj and sj are the j-th components of ~
p and ~
s. We can write ~
s − ~
p as ~
h,
where ~
h solves the associated homogeneous system, to express ~
s in the required
~
p + ~
h form.
For set inclusion the other way, take a vector of the form ~
p + ~
h, where ~
p
solves the system and ~
h solves the associated homogeneous system, and note
that it solves the given system: for any equation index i,
ai,1(p1 + h1) + · · · + ai,n(pn + hn) = (ai,1p1 + · · · + ai,npn)
+ (ai,1h1 + · · · + ai,nhn)
= di + 0
= di
where hj is the j-th component of ~
h. QED
The two lemmas above together establish Theorem 3.1. We remember that
theorem with the slogan “General = Particular + Homogeneous”.
3.9 Example This system illustrates Theorem 3.1.
x + 2y − z = 1
2x + 4y = 2
y − 3z = 0
Gauss’ method
−2ρ1+ρ2
−→
x + 2y − z = 1
2z = 0
y − 3z = 0
ρ2↔ρ3
−→
x + 2y − z = 1
y − 3z = 0
2z = 0
∗ More information on equality of sets is in the appendix.

shows that the general solution is a singleton set.
{


1
0
0

}
That single vector is, of course, a particular solution. The associated homoge-
neous system reduces via the same row operations
x + 2y − z = 0
2x + 4y = 0
y − 3z = 0
−2ρ1+ρ2
−→
ρ2↔ρ3
−→
x + 2y − z = 0
y − 3z = 0
2z = 0
to also give a singleton set.
{


0
0
0

}
As the theorem states, and as discussed at the start of this subsection, in this
single-solution case the general solution results from taking the particular solu-
tion and adding to it the unique solution of the associated homogeneous system.
3.10 Example Also discussed there is that the case where the general solution
set is empty fits the ‘General = Particular+Homogeneous’ pattern. This system
illustrates. Gauss’ method
x + z + w = −1
2x − y + w = 3
x + y + 3z + 2w = 1
−2ρ1+ρ2
−→
−ρ1+ρ3
x + z + w = −1
−y − 2z − w = 5
y + 2z + w = 2
shows that it has no solutions. The associated homogeneous system, of course,
has a solution.
x + z + w = 0
2x − y + w = 0
x + y + 3z + 2w = 0
−2ρ1+ρ2
−→
−ρ1+ρ3
ρ2+ρ3
−→
x + z + w = 0
−y − 2z − w = 0
0 = 0
In fact, the solution set of the homogeneous system is infinite.
{




−1
−2
1
0



 z +




−1
−1
0
1



 w
¯
¯ z, w ∈ R}
However, because no particular solution of the original system exists, the general
solution set is empty — there are no vectors of the form ~
p+~
h because there are
no ~
p ’s.
3.11 Corollary Solution sets of linear systems are either empty, have one
element, or have infinitely many elements.

Proof. We’ve seen examples of all three happening so we need only prove that
those are the only possibilities.
First, notice a homogeneous system with at least one non-~
0 solution ~
v has
infinitely many solutions because the set of multiples s~
v is infinite — if s 6= 1
then s~
v − ~
v = (s − 1)~
v is easily seen to be non-~
0, and so s~
v 6= ~
v.
Now, apply Lemma 3.8 to conclude that a solution set
{~
p + ~
h
¯
¯ ~
h solves the associated homogeneous system}
is either empty (if there is no particular solution ~
p), or has one element (if there
is a ~
p and the homogeneous system has the unique solution ~
0), or is infinite (if
there is a ~
p and the homogeneous system has a non-~
0 solution, and thus by the
prior paragraph has infinitely many solutions). QED
This table summarizes the factors affecting the size of a general solution.
number of solutions of the
associated homogeneous system
particular
solution
exists?
one infinitely many
yes
unique
solution
infinitely many
solutions
no
no
solutions
no
solutions
The factor on the top of the table is the simpler one. When we perform
Gauss’ method on a linear system, ignoring the constants on the right side and
so paying attention only to the coefficients on the left-hand side, we either end
with every variable leading some row or else we find that some variable does not
lead a row, that is, that some variable is free. (Of course, “ignoring the constants
on the right” is formalized by considering the associated homogeneous system.
We are simply putting aside for the moment the possibility of a contradictory
equation.)
A nice insight into the factor on the top of this table at work comes from con-
sidering the case of a system having the same number of equations as variables.
This system will have a solution, and the solution will be unique, if and only if it
reduces to an echelon form system where every variable leads its row, which will
happen if and only if the associated homogeneous system has a unique solution.
Thus, the question of uniqueness of solution is especially interesting when the
system has the same number of equations as variables.
3.12 Definition A square matrix is nonsingular if it is the matrix of coeffi-
cients of a homogeneous system with a unique solution. It is singular otherwise,
that is, if it is the matrix of coefficients of a homogeneous system with infinitely
many solutions.

3.13 Example The systems from Example 3.3, Example 3.5, and Example 3.9
each have an associated homogeneous system with a unique solution. Thus these
matrices are nonsingular.
µ
3 4
2 −1
¶


3 2 1
6 −4 0
0 1 1




1 2 −1
2 4 0
0 1 −3


The Chemistry problem from Example 3.6 is a homogeneous system with more
than one solution so its matrix is singular.




7 0 −7 0
8 1 −5 −2
0 1 −3 0
0 3 −6 −1




3.14 Example The first of these matrices is nonsingular while the second is
singular
µ
1 2
3 4
¶ µ
1 2
3 6
¶
because the first of these homogeneous systems has a unique solution while the
second has infinitely many solutions.
x + 2y = 0
3x + 4y = 0
x + 2y = 0
3x + 6y = 0
We have made the distinction in the definition because a system (with the same
number of equations as variables) behaves in one of two ways, depending on
whether its matrix of coefficients is nonsingular or singular. A system where
the matrix of coefficients is nonsingular has a unique solution for any constants
on the right side: for instance, Gauss’ method shows that this system
x + 2y = a
3x + 4y = b
has the unique solution x = b − 2a and y = (3a − b)/2. On the other hand, a
system where the matrix of coefficients is singular never has a unique solutions —
it has either no solutions or else has infinitely many, as with these.
x + 2y = 1
3x + 6y = 2
x + 2y = 1
3x + 6y = 3
Thus, ‘singular’ can be thought of as connoting “troublesome”, or at least “not
ideal”.
The above table has two factors. We have already considered the factor
along the top: we can tell which column a given linear system goes in solely by

considering the system’s left-hand side — the the constants on the right-hand
side play no role in this factor. The table’s other factor, determining whether a
particular solution exists, is tougher. Consider these two
3x + 2y = 5
3x + 2y = 5
3x + 2y = 5
3x + 2y = 4
with the same left sides but different right sides. Obviously, the first has a
solution while the second does not, so here the constants on the right side
decide if the system has a solution. We could conjecture that the left side of a
linear system determines the number of solutions while the right side determines
if solutions exist, but that guess is not correct. Compare these two systems
3x + 2y = 5
4x + 2y = 4 and
3x + 2y = 5
3x + 2y = 4
with the same right sides but different left sides. The first has a solution but
the second does not. Thus the constants on the right side of a system don’t
decide alone whether a solution exists; rather, it depends on some interaction
between the left and right sides.
For some intuition about that interaction, consider this system with one of
the coefficients left as the parameter c.
x + 2y + 3z = 1
x + y + z = 1
cx + 3y + 4z = 0
If c = 2 this system has no solution because the left-hand side has the third row
as a sum of the first two, while the right-hand does not. If c 6= 2 this system has
a unique solution (try it with c = 1). For a system to have a solution, if one row
of the matrix of coefficients on the left is a linear combination of other rows,
then on the right the constant from that row must be the same combination of
constants from the same rows.
More intuition about the interaction comes from studying linear combina-
tions. That will be our focus in the second chapter, after we finish the study of
Gauss’ method itself in the rest of this chapter.
Exercises
X 3.15 Solve each system. Express the solution set using vectors. Identify the par-
ticular solution and the solution set of the homogeneous system.
(a) 3x + 6y = 18
x + 2y = 6
(b) x + y = 1
x − y = −1
(c) x1 + x3 = 4
x1 − x2 + 2x3 = 5
4x1 − x2 + 5x3 = 17
(d) 2a + b − c = 2
2a + c = 3
a − b = 0
(e) x + 2y − z = 3
2x + y + w = 4
x − y + z + w = 1
(f) x + z + w = 4
2x + y − w = 2
3x + y + z = 7
3.16 Solve each system, giving the solution set in vector notation. Identify the
particular solution and the solution of the homogeneous system.

(a) 2x + y − z = 1
4x − y = 3
(b) x − z = 1
y + 2z − w = 3
x + 2y + 3z − w = 7
(c) x − y + z = 0
y + w = 0
3x − 2y + 3z + w = 0
−y − w = 0
(d) a + 2b + 3c + d − e = 1
3a − b + c + d + e = 3
X 3.17 For the system
2x − y − w = 3
y + z + 2w = 2
x − 2y − z = −1
which of these can be used as the particular solution part of some general solu-
tion?
(a)



0
−3
5
0


 (b)



2
1
1
0


 (c)



−1
−4
8
−1



X 3.18 Lemma 3.8 says that any particular solution may be used for ~
p. Find, if
possible, a general solution to this system
x − y + w = 4
2x + 3y − z = 0
y + z + w = 4
that uses the given vector as its particular solution.
(a)



0
0
0
4


 (b)



−5
1
−7
10


 (c)



2
−1
1
1



3.19 One of these is nonsingular while the other is singular. Which is which?
(a)
µ
1 3
4 −12
¶
(b)
µ
1 3
4 12
¶
X 3.20 Singular or nonsingular?
(a)
µ
1 2
1 3
¶
(b)
µ
1 2
−3 −6
¶
(c)
µ
1 2 1
1 3 1
¶
(Careful!)
(d)
Ã
1 2 1
1 1 3
3 4 7
!
(e)
Ã
2 2 1
1 0 5
−1 1 4
!
X 3.21 Is the given vector in the set generated by the given set?
(a)
µ
2
3
¶
, {
µ
1
4
¶
,
µ
1
5
¶
}
(b)
Ã
−1
0
1
!
, {
Ã
2
1
0
!
,
Ã
1
0
1
!
}
(c)
Ã
1
3
0
!
, {
Ã
1
0
4
!
,
Ã
2
1
5
!
,
Ã
3
3
0
!
,
Ã
4
2
1
!
}
(d)



1
0
1
1


 , {



2
1
0
1


 ,



3
0
0
2


}

3.22 Prove that any linear system with a nonsingular matrix of coefficients has a
solution, and that the solution is unique.
3.23 To tell the whole truth, there is another tricky point to the proof of Lemma 3.7.
What happens if there are no non-‘0 = 0’ equations? (There aren’t any more tricky
points after this one.)
X 3.24 Prove that if ~
s and ~
t satisfy a homogeneous system then so do these vec-
tors.
(a) ~
s + ~
t (b) 3~
s (c) k~
s + m~
t for k, m ∈ R
What’s wrong with: “These three show that if a homogeneous system has one
solution then it has many solutions — any multiple of a solution is another solution,
and any sum of solutions is a solution also — so there are no homogeneous systems
with exactly one solution.”?
3.25 Prove that if a system with only rational coefficients and constants has a
solution then it has at least one all-rational solution. Must it have infinitely many?

1.II Linear Geometry of n-Space
For readers who have seen the elements of vectors before, in calculus or physics,
this section is an optional review. However, later work in this book will refer to
this material often, so this section is not optional if it is not a review.
In the first section, we had to do a bit of work to show that there are only
three types of solution sets — singleton, empty, and infinite. But for systems
with two equations and two unknowns, we can just see this. We picture each
two-unknowns equation as a line in R2
and then the two lines could have a
unique intersection, be parallel, or be the same.
One solution
3x + 2y = 7
x − y = −1
No solutions
3x + 2y = 7
3x + 2y = 4
Infinitely many
solutions
3x + 2y = 7
6x + 4y = 14
As this shows, sometimes our results are expressed clearly in a picture. In this
section we develop the terminology and ideas we need to express our results
from the prior section, and from some future sections, geometrically. The two-
dimensional case is familiar enough, but to extend to systems with more than
two unknowns we shall also need some higher-dimensional geometry.
1.II.1 Vectors in Space
“Higher-dimensionsional geometry” sounds exotic. It is exotic — interesting
and eye-opening. But it isn’t distant or unreachable.
As a start, we define one-dimensional space to be the set R1
. To see that
definition is reasonable, draw a one-dimensional space
and make the usual correspondence with R: pick a point to label 0 and another
to label 1.
0 1
Now, armed with a scale and a direction, finding the point corresponding to,
say +2.17, is easy — start at 0, head in the direction of 1 (i.e., the positive
direction), but don’t stop there, go 2.17 times as far.
The basic idea here, combining magnitude with direction, is the key to ex-
tending to higher dimensions.

Section II. Linear Geometry of n-Space 33
An object comprised of a magnitude and a direction is a vector (we will use
the same word as in the previous section because we shall show below how to
describe such an object with a column vector). We can draw a vector as having
some length, and pointing somewhere.
There is a subtlety here — these
are equal, even though they start in different places, because they have equal
lengths and equal directions. Again: those vectors are not just alike, they are
equal.
How can things that are in different places be equal? Think of a vector as
representing a displacement (‘vector’ is Latin for “carrier” or “traveler”). These
squares undergo the same displacement, despite that those displacements start
in different places.
Sometimes, to emphasize this property vectors have of not being anchored, they
are referred to as free vectors.
These two, as free vectors, are equal;
we can think of each as a displacement of one over and two up. More generally,
two vectors in the plane are the same if and only if they have the same change
in first components and the same change in second components: the vector
extending from (a1, a2) to (b1, b2) equals the vector from (c1, c2) to (d1, d2) if
and only if b1 − a1 = d1 − c1 and b2 − a2 = d2 − c2.
An expression like ‘the vector that, were it to start at (a1, a2), would stretch
to (b1, b2)’ is awkward. Instead of that terminology, from among all of these
we single out the one starting at the origin as being in canonical (or natural)
position and we describe a vector by stating its endpoint when it is in canonical

position, as a column. For instance, the ‘one over and two up’ vectors above are
denoted in this way.
µ
1
2
¶
More generally, the plane vector starting at (a1, a2) and stretching to (b1, b2) is
denoted
µ
b1 − a1
b2 − a2
¶
since the prior paragraph shows that when the vector starts at the origin, it
ends at this location.
We often just say “the point
µ
1
2
¶
”
rather than “the endpoint of the canonical position of” that vector. That is, we
shall find it convienent to blur the distinction between a point in space and the
vector that, if it starts at the origin, ends at that point. Thus, we will refer to
both of these as Rn
.
{(x1, x2)
¯
¯ x1, x2 ∈ R} {
µ
x1
x2
¶
¯
¯ x1, x2 ∈ R}
In the prior section we defined vectors and vector operations with an alge-
braic motivation;
r ·
µ
v1
v2
¶
=
µ
rv1
rv2
¶ µ
v1
v2
¶
+
µ
w1
w2
¶
=
µ
v1 + w1
v2 + w2
¶
we can now interpret those operations geometrically. For instance, if ~
v repre-
sents a displacement then 3~
v represents a displacement in the same direction
but three times as far, and −1~
v represents a displacement of the same distance
as ~
v but in the opposite direction.
~
v
3~
v
−~
v
And, where ~
v and ~
w represent displacements, ~
v + ~
w represents those displace-
ments combined.
~
v
~
w
~
v + ~
w

The long arrow is the combined displacement in this sense: if, in one minute, a
ship’s motion gives it the displacement relative to the earth of ~
v and a passen-
ger’s motion gives a displacement relative to the ship’s deck of ~
w, then ~
v + ~
w is
the displacement of the passenger relative to the earth.
Another way to understand the vector sum is with the parallelogram rule.
Draw the parallelogram formed by the vectors ~
v1,~
v2 and then the sum ~
v1 + ~
v2
extends along the diagonal to the far corner.
³
x1
y1
´
³
x2
y2
´ ³
x1 + x2
y1 + y2
´
The above drawings show how vectors and vector operations behave in R2
.
We can extend to R3
, or to even higher-dimensional spaces where we have no
pictures, with the obvious generalization: the free vector that, if it starts at
(a1, . . . , an), ends at (b1, . . . , bn), is represented by this column



b1 − a1
.
.
.
bn − an



(vectors are equal if they have the same representation), we aren’t too careful
to distinguish between a point and the vector whose canonical representation
ends at that point,
Rn
= {



v1
.
.
.
vn



¯
¯ v1, . . . , vn ∈ R}
and addition and scalar multiplication are component-wise.
Having considered points, we now turn to the lines. In R2
, the line through
(1, 2) and (3, 1) is comprised of (the endpoints of) the vectors in this set
{
µ
1
2
¶
+ t ·
µ
2
−1
¶
¯
¯ t ∈ R}
That description expresses this picture.
³
2
−1
´
=
³
3
1
´
−
³
1
2
´
The vector associated with the parameter t has its whole body in the line — it
is a direction vector for the line. Note that points on the line to the left of x = 1
are described using negative values of t.

In R3
, the line through (1, 2, 3) and (5, 5, 5) is the set of (endpoints of)
vectors of this form
{


1
2
3

 + t ·


4
3
2


¯
¯ t ∈ R}
and lines in even higher-dimensional spaces work in the same way.
If a line uses one parameter, so that there is freedom to move back and
forth in one dimension, then a plane must involve two. For example, the plane
through the points (1, 0, 5), (2, 1, −3), and (−2, 4, 0.5) consists of (endpoints of)
the vectors in
{


1
0
5

 + t ·


1
1
−8

 + s ·


−3
4
−4.5


¯
¯ t, s ∈ R}
(the column vectors associated with the parameters


1
1
−8

 =


2
1
−3

 −


1
0
5




−3
4
−4.5

 =


−2
4
0.5

 −


1
0
5


are two vectors whose whole bodies lie in the plane). As with the line, note that
some points in this plane are described with negative t’s or negative s’s or both.
A description of planes that is often encountered in algebra and calculus uses
a single equation
P = {


x
y
z


¯
¯ 2x + 3y − z = 4}
as the condition that describes the relationship among the first, second, and
third coordinates of points in a plane. The translation from such a description
to the vector description that we favor in this book is to think of the condition
as a one-equation linear system and paramatrize x = (1/2)(4 − 3y + z).
P = {


2
0
0

 +


−3/2
1
0

 y +


1/2
0
1

 z
¯
¯ y, z ∈ R}
Generalizing from lines and planes, we define a k-dimensional linear sur-
face (or k-flat) in Rn
to be {~
p + t1~
v1 + t2~
v2 + · · · + tk~
vk
¯
¯ t1, . . . , tk ∈ R} where
~
v1, . . . ,~
vk ∈ Rn
. For example, in R4
,
{




2
π
3
−0.5



 + t




1
0
0
0




¯
¯ t ∈ R}

is a line,
{




0
0
0
0



 + t




1
1
0
−1



 + s




2
0
1
0




¯
¯ t, s ∈ R}
is a plane, and
{




3
1
−2
0.5



 + r




0
0
0
−1



 + s




1
0
1
0



 + t




2
0
1
0




¯
¯ r, s, t ∈ R}
is a three-dimensional linear surface. Again, the intuition is that a line per-
mits motion in one direction, a plane permits motion in combinations of two
directions, etc.
A linear surface description can be misleading about the dimension — this
L = {




1
0
−1
−2



 + t




1
1
0
−1



 + s




2
2
0
−2




¯
¯ t, s ∈ R}
is a degenerate plane because it is actually a line.
L = {




1
0
−1
−2



 + r




1
1
0
−1




¯
¯ r ∈ R}
We shall see in the Linear Independence section of Chapter Two what relation-
ships among vectors causes the linear surface they generate to be degenerate.
We finish this subsection by restating our conclusions from the first section
in geometric terms. First, the solution set of a linear system with n unknowns
is a linear surface in Rn
. Specifically, it is an k-dimensional linear surface,
where k is the number of free variables in an echelon form version of the system.
Second, the solution set of a homogeneous linear system is a linear surface
passing through the origin. Finally, we can view the general solution set of any
linear system as being the solution set of its associated homogeneous system
offset from the origin by a vector, namely by any particular solution.
Exercises
X 1.1 Find the canonical name for each vector.
(a) the vector from (2, 1) to (4, 2) in R2
(b) the vector from (3, 3) to (2, 5) in R2
(c) the vector from (1, 0, 6) to (5, 0, 3) in R3
(d) the vector from (6, 8, 8) to (6, 8, 8) in R3

X 1.2 Decide if the two vectors are equal.
(a) the vector from (5, 3) to (6, 2) and the vector from (1, −2) to (1, 1)
(b) the vector from (2, 1, 1) to (3, 0, 4) and the vector from (5, 1, 4) to (6, 0, 7)
X 1.3 Does (1, 0, 2, 1) lie on the line through (−2, 1, 1, 0) and (5, 10, −1, 4)?
X 1.4 (a) Describe the plane through (1, 1, 5, −1), (2, 2, 2, 0), and (3, 1, 0, 4).
(b) Is the origin in that plane?
1.5 Describe the plane that contains this point and line.
Ã
2
0
3
!
{
Ã
−1
0
−4
!
+
Ã
1
1
2
!
t
¯
¯ t ∈ R}
X 1.6 Intersect these planes.
{
Ã
1
1
1
!
t +
Ã
0
1
3
!
s
¯
¯ t, s ∈ R} {
Ã
1
1
0
!
+
Ã
0
3
0
!
k +
Ã
2
0
4
!
m
¯
¯ k, m ∈ R}
X 1.7 Intersect each pair, if possible.
(a) {
Ã
1
1
2
!
+ t
Ã
0
1
1
!
¯
¯ t ∈ R}, {
Ã
1
3
−2
!
+ s
Ã
0
1
2
!
¯
¯ s ∈ R}
(b) {
Ã
2
0
1
!
+ t
Ã
1
1
−1
!
¯
¯ t ∈ R}, {s
Ã
0
1
2
!
+ w
Ã
0
4
1
!
¯
¯ s, w ∈ R}
1.8 Show that the line segments (a1, a2)(b1, b2) and (c1, c2)(d1, d2) have the same
lengths and slopes if b1 − a1 = d1 − c1 and b2 − a2 = d2 − c2. Is that only if?
1.9 How should R0
be defined?
X 1.10 [Math. Mag., Jan. 1957] A person traveling eastward at a rate of 3 miles per
hour finds that the wind appears to blow directly from the north. On doubling his
speed it appears to come from the north east. What was the wind’s velocity?
1.11 Euclid describes a plane as “a surface which lies evenly with the straight lines
on itself”. Commentators (e.g., Heron) have interpreted this to mean “(A plane
surface is) such that, if a straight line pass through two points on it, the line
coincides wholly with it at every spot, all ways”. (Translations from [Heath], pp.
171-172.) Do planes, as described in this section, have that property? Does this
description adequately define planes?
1.II.2 Length and Angle Measures
We’ve translated the first section’s results about solution sets into geometric
terms for insight into how those sets look. But we must watch out not to be
mislead by our own terms; labeling subsets of Rk
of the forms {~
p + t~
v
¯
¯ t ∈ R}
and {~
p + t~
v + s~
w
¯
¯ t, s ∈ R} as “lines” and “planes” doesn’t make them act like
the lines and planes of our prior experience. Rather, we must ensure that the
names suit the sets. While we can’t prove that the sets satisfy our intuition —
we can’t prove anything about intuition — in this subsection we’ll observe that

a result familiar from R2
and R3
, when generalized to arbitrary Rk
, supports
the idea that a line is straight and a plane is flat. Specifically, we’ll see how to
do Euclidean geometry in a “plane” by giving a definition of the angle between
two Rn
vectors in the plane that they generate.
2.1 Definition The length of a vector ~
v ∈ Rn
is this.
k~
v k =
q
v2
1 + · · · + v2
n
2.2 Remark This is a natural generalization of the Pythagorean Theorem. A
classic discussion is in [Polya].
We can use that definition to derive a formula for the angle between two
vectors. For a model of what to do, consider two vectors in R3
.
~
u
~
v
Put them in canonical position and, in the plane that they determine, consider
the triangle formed by ~
u, ~
v, and ~
u − ~
v.
To that triangle, apply the Law of Cosines,
k~
u − ~
v k2
= k~
u k2
+ k~
v k2
− 2 k~
u k k~
v k cos θ
where θ is the angle between ~
u and ~
v. Expand both sides
(u1 − v1)2
+ (u2 − v2)2
+ (u3 − v3)2
= (u2
1 + u2
2 + u2
3) + (v2
1 + v2
2 + v2
3) − 2 k~
u k k~
v k cos θ
and simplify.
θ = arccos(
u1v1 + u2v2 + u3v3
k~
u k k~
v k
)
In higher dimensions no picture suffices but we can make the same argument
analytically. First, the form of the numerator is clear — it comes from the middle
terms of the squares (u1 − v1)2
, (u2 − v2)2
, etc.
2.3 Definition The dot product (or inner product, or scalar product) of two
n-component real vectors is the linear combination of their components.
~
u ~
v = u1v1 + u2v2 + · · · + unvn

Notice that the dot product of two vectors is a real number, not a vector, and
that the dot product of a vector from Rn
with a vector from Rm
is defined
only when n equals m. Notice also this relationship between dot product and
length: dotting a vector with itself gives its length squared ~
u ~
u = u1u1 + · · · +
unun = k~
u k2
.
2.4 Remark The wording in that definition allows one or both of the two to
be a row vector instead of a column vector. Some books require that the first
vector be a row vector and that the second vector be a column vector. We shall
not be that strict.
Still reasoning with letters, but guided by the pictures, we use the next
theorem to argue that the triangle formed by ~
u, ~
v, and ~
u − ~
v in Rn
lies in the
planar subset of Rn
generated by ~
u and ~
v.
2.5 Theorem (Triangle Inequality) For any ~
u,~
v ∈ Rn
,
k~
u + ~
v k ≤ k~
u k + k~
v k
with equality if and only if one of the vectors is a nonnegative scalar multiple
of the other one.
This inequality is the source of the familiar saying, “The shortest distance
between two points is in a straight line.”
~
u
~
v
~
u + ~
v
start .
. finish
Proof. We’ll use some algebraic properties of dot product that we have not
shown, for instance that ~
u·(~
a+~
b) = ~
u·~
a+~
u·~
b and that ~
u·~
v = ~
v ·~
u. Verification
of those properties is Exercise 17. The desired inequality holds if and only if its
square holds.
k~
u + ~
v k2
≤ (k~
uk + k~
vk)2
(~
u + ~
v) (~
u + ~
v) ≤ k~
u k2
+ 2 k~
u k k~
v k + k~
v k2
~
u ~
u + ~
u ~
v + ~
v ~
u + ~
v ~
v ≤ ~
u ~
u + 2 k~
u k k~
v k + ~
v ~
v
2 ~
u ~
v ≤ 2 k~
u k k~
v k
That, in turn, holds if and only if the relationship obtained by multiplying both
sides by the nonnegative numbers k~
u k and k~
v k
2 (k~
v k~
u) (k~
u k~
v) ≤ 2 k~
u k2
k~
v k2
and rewriting
0 ≤ k~
u k2
k~
v k2
− 2 (k~
v k~
u) (k~
u k~
v) + k~
u k2
k~
v k2

is true. But factoring
0 ≤ (k~
u k~
v − k~
v k~
u) (k~
u k~
v − k~
v k~
u)
shows that this certainly is true since it only says that the square of the length
of the vector k~
u k~
v − k~
v k~
u is not negative.
As for equality, it holds when, and only when, k~
u k~
v − k~
v k~
u is ~
0. The check
that k~
u k~
v = k~
v k~
u if and only if one vector is a nonnegative real scalar multiple
of the other is easy. QED
This result supports the intuition that even in higher-dimensional spaces,
lines are straight and planes are flat. For any two points in a linear surface, the
line segment connecting them is contained in that surface (this is easily checked
from the definition). But if the surface has a bend then that would allow for a
shortcut (shown here dotted, while the line segment from P to Q, contained in
the linear surface, is solid).
. P
. Q
Because the Triangle Inequality says that in any Rn
, the shortest cut between
two endpoints is simply the line segment connecting them, linear surfaces have
no such bends.
Back to the definition of angle measure. The heart of the Triangle Inequal-
ity’s proof is the ‘~
u · ~
v ≤ k~
u k k~
v k’ line. At first glance, a reader might wonder
if some pairs of vectors satisfy the inequality in this way: while ~
u · ~
v is a large
number, with absolute value bigger than the right-hand side, it is a negative
large number. The next result says that no such pair of vectors exists.
2.6 Corollary (Cauchy-Schwartz Inequality) For any ~
u,~
v ∈ Rn
,
|~
u · ~
v | ≤ k~
u k k~
v k
with equality if and only if one vector is a scalar multiple of the other.
Proof. The Triangle Inequality’s proof shows that ~
u ~
v ≤ k~
u k k~
v k so if ~
u ~
v is
positive or zero then we are done. If ~
u ~
v is negative then this holds.
|~
u ~
v| = −(~
u ~
v) = (−~
u) ~
v ≤ k − ~
u k k~
v k = k~
u k k~
v k
The equality condition is Exercise 18. QED
The Cauchy-Schwartz inequality assures us that the next definition makes
sense because the fraction has absolute value less than or equal to one.

2.7 Definition The angle between two nonzero vectors ~
u,~
v ∈ Rn
is
θ = arccos(
~
u ~
v
k~
u k k~
v k
)
(the angle between the zero vector and any other vector is defined to be a right
angle).
Thus vectors from Rn
are orthogonal if and only if their dot product is zero.
2.8 Example These vectors are orthogonal.
µ
1
−1
¶ µ
1
1
¶
= 0
Although they are shown away from canonical position so that they don’t appear
to touch, nonetheless they are orthogonal.
2.9 Example The R3
angle formula given at the start of this subsection is a
special case of the definition. Between these two
µ
0
3
2
¶
µ
1
1
0
¶
the angle is
arccos(
(1)(0) + (1)(3) + (0)(2)
√
12 + 12 + 02
√
02 + 32 + 22
) = arccos(
3
√
2
√
13
)
approximately 0.94 radians. Notice that these vectors are not orthogonal. Al-
though the yz-plane may appear to be perpendicular to the xy-plane, in fact
the two planes are that way only in the weak sense that there are vectors in each
orthogonal to all vectors in the other. Not every vector in each is orthogonal to
all vectors in the other.
Exercises
X 2.10 Find the length of each vector.
(a)
µ
3
1
¶
(b)
µ
−1
2
¶
(c)
Ã
4
1
1
!
(d)
Ã
0
0
0
!
(e)



1
−1
1
0



X 2.11 Find the angle between each two, if it is defined.

(a)
µ
1
2
¶
,
µ
1
4
¶
(b)
Ã
1
2
0
!
,
Ã
0
4
1
!
(c)
µ
1
2
¶
,
Ã
1
4
−1
!
X 2.12 During maneuvers preceding the Battle of Jutland, the British battle cruiser
Lion moved as follows (in nautical miles): 1.2 miles north, 6.1 miles 38 degrees
east of south, 4.0 miles at 89 degrees east of north, and 6.5 miles at 31 degrees
east of north. Find the distance between starting and ending positions.
2.13 Find k so that these two vectors are perpendicular.
µ
k
1
¶ µ
4
3
¶
2.14 Describe the set of vectors in R3
orthogonal to this one.
Ã
1
3
−1
!
X 2.15 (a) Find the angle between the diagonal of the unit square in R2
and one of
the axes.
(b) Find the angle between the diagonal of the unit cube in R3
and one of the
axes.
(c) Find the angle between the diagonal of the unit cube in Rn
and one of the
axes.
(d) What is the limit, as n goes to ∞, of the angle between the diagonal of the
unit cube in Rn
and one of the axes?
2.16 Is there any vector that is perpendicular to itself?
X 2.17 Describe the algebraic properties of dot product.
(a) Is it right-distributive over addition: (~
u + ~
v) ~
w = ~
u ~
w + ~
v ~
w?
(b) Is is left-distributive (over addition)?
(c) Does it commute?
(d) Associate?
(e) How does it interact with scalar multiplication?
As always, any assertion must be backed by either a proof or an example.
2.18 Verify the equality condition in Corollary 2.6, the Cauchy-Schwartz Inequal-
ity.
(a) Show that if ~
u is a negative scalar multiple of ~
v then ~
u ~
v and ~
v ~
u are less
than or equal to zero.
(b) Show that |~
u ~
v| = k~
u k k~
v k if and only if one vector is a scalar multiple of
the other.
2.19 Suppose that ~
u ~
v = ~
u ~
w and ~
u 6= ~
0. Must ~
v = ~
w?
X 2.20 Does any vector have length zero except a zero vector? (If “yes”, produce an
example. If “no”, prove it.)
X 2.21 Find the midpoint of the line segment connecting (x1, y1) with (x2, y2) in R2
.
Generalize to Rn
.
2.22 Show that if ~
v 6= ~
0 then ~
v/k~
v k has length one. What if ~
v = ~
0?
2.23 Show that if r ≥ 0 then r~
v is r times as long as ~
v. What if r < 0?
X 2.24 A vector ~
v ∈ Rn
of length one is a unit vector. Show that the dot product
of two unit vectors has absolute value less than or equal to one. Can ‘less than’
happen? Can ‘equal to’?

2.25 Prove that k~
u + ~
v k2
+ k~
u − ~
v k2
= 2k~
u k2
+ 2k~
v k2
.
2.26 Show that if ~
x ~
y = 0 for every ~
y then ~
x = ~
0.
2.27 Is k~
u1 + · · · + ~
unk ≤ k~
u1k + · · · + k~
unk? If it is true then it would generalize
the Triangle Inequality.
2.28 What is the ratio between the sides in the Cauchy-Schwartz inequality?
2.29 Why is the zero vector defined to be perpendicular to every vector?
2.30 Describe the angle between two vectors in R1
.
2.31 Give a simple necessary and sufficient condition to determine whether the
angle between two vectors is acute, right, or obtuse.
X 2.32 Generalize to Rn
the converse of the Pythagorean Theorem, that if ~
u and ~
v
are perpendicular then k~
u + ~
v k2
= k~
u k2
+ k~
v k2
.
2.33 Show that k~
u k = k~
v k if and only if ~
u + ~
v and ~
u − ~
v are perpendicular. Give
an example in R2
.
2.34 Show that if a vector is perpendicular to each of two others then it is perpen-
dicular to each vector in the plane they generate. (Remark. They could generate
a degenerate plane — a line or a point — but the statement remains true.)
2.35 Prove that, where ~
u,~
v ∈ Rn
are nonzero vectors, the vector
~
u
k~
u k
+
~
v
k~
v k
bisects the angle between them. Illustrate in R2
.
2.36 Verify that the definition of angle is dimensionally correct: (1) if k > 0 then
the cosine of the angle between k~
u and ~
v equals the cosine of the angle between
~
u and ~
v, and (2) if k < 0 then the cosine of the angle between k~
u and ~
v is the
negative of the cosine of the angle between ~
u and ~
v.
X 2.37 Show that the inner product operation is linear: for ~
u,~
v, ~
w ∈ Rn
and k, m ∈ R,
~
u (k~
v + m~
w) = k(~
u ~
v) + m(~
u ~
w).
X 2.38 The geometric mean of two positive reals x, y is
√
xy. It is analogous to the
arithmetic mean (x + y)/2. Use the Cauchy-Schwartz inequality to show that the
geometric mean of any x, y ∈ R is less than or equal to the arithmetic mean.
2.39 [Am. Math. Mon., Feb. 1933] A ship is sailing with speed and direction ~
v1;
the wind blows apparently (judging by the vane on the mast) in the direction of
a vector ~
a; on changing the direction and speed of the ship from ~
v1 to ~
v2 the
apparent wind is in the direction of a vector ~
b.
Find the vector velocity of the wind.
2.40 Verify the Cauchy-Schwartz inequality by first proving Lagrange’s identity:
Ã
X
1≤j≤n
ajbj
!2
=
Ã
X
1≤j≤n
a2
j
! Ã
X
1≤j≤n
b2
j
!
−
X
1≤k<j≤n
(akbj − ajbk)2
and then noting that the final term is positive. (Recall the meaning
X
1≤j≤n
ajbj = a1b1 + a2b2 + · · · + anbn
and X
1≤j≤n
aj
2
= a1
2
+ a2
2
+ · · · + an
2
of the Σ notation.) This result is an improvement over Cauchy-Schwartz because
it gives a formula for the difference between the two sides. Interpret that difference
in R2
.

Section III. Reduced Echelon Form 45
1.III Reduced Echelon Form
After developing the mechanics of Gauss’ method, we observed that it can be
done in more than one way. One example is that we sometimes have to swap
rows and there can be more than one row to choose from. Another example is
that from this matrix
µ
2 2
4 3
¶
Gauss’ method could derive any of these echelon form matrices.
µ
2 2
0 −1
¶ µ
1 1
0 −1
¶ µ
2 0
0 −1
¶
The first results from −2ρ1 +ρ2. The second comes from following (1/2)ρ1 with
−4ρ1 + ρ2. The third comes from −2ρ1 + ρ2 followed by 2ρ2 + ρ1 (after the first
pivot the matrix is already in echelon form so the second one is extra work but
it is nonetheless a legal row operation).
The fact that the echelon form outcome of Gauss’ method is not unique
leaves us with some questions. Will any two echelon form versions of a system
have the same number of free variables? Will they in fact have exactly the same
variables free? In this section we will answer both questions “yes”. We will
do more than answer the questions. We will give a way to decide if one linear
system can be derived from another by row operations. The answers to the two
questions will follow from this larger result.
1.III.1 Gauss-Jordan Reduction
Gaussian elimination coupled with back-substitution solves linear systems,
but it’s not the only method possible. Here is an extension of Gauss’ method
that has some advantages.
1.1 Example To solve
x + y − 2z = −2
y + 3z = 7
x − z = −1
we can start by going to echelon form as usual.
−ρ1+ρ3
−→


1 1 −2 −2
0 1 3 7
0 −1 1 1

 ρ2+ρ3
−→


1 1 −2 −2
0 1 3 7
0 0 4 8



We can keep going to a second stage by making the leading entries into ones
(1/4)ρ3
−→


1 1 −2 −2
0 1 3 7
0 0 1 2


and then to a third stage that uses the leading entries to eliminate all of the
other entries in each column by pivoting upwards.
−3ρ3+ρ2
−→
2ρ3+ρ1


1 1 0 2
0 1 0 1
0 0 1 2

 −ρ2+ρ1
−→


1 0 0 1
0 1 0 1
0 0 1 2


The answer is x = 1, y = 1, and z = 2.
Note that the pivot operations in the first stage proceed from column one to
column three while the pivot operations in the third stage proceed from column
three to column one.
1.2 Example We often combine the operations of the middle stage into a
single step, even though they are operations on different rows.
µ
2 1 7
4 −2 6
¶
−2ρ1+ρ2
−→
µ
2 1 7
0 −4 −8
¶
(1/2)ρ1
−→
(−1/4)ρ2
µ
1 1/2 7/2
0 1 2
¶
−(1/2)ρ2+ρ1
−→
µ
1 0 5/2
0 1 2
¶
The answer is x = 5/2 and y = 2.
This extension of Gauss’ method is Gauss-Jordan reduction. It goes past
echelon form to a more refined, more specialized, matrix form.
1.3 Definition A matrix is in reduced echelon form if, in addition to being in
echelon form, each leading entry is a one and is the only nonzero entry in its
column.
The disadvantage of using Gauss-Jordan reduction to solve a system is that the
additional row operations mean additional arithmetic. The advantage is that
the solution set can just be read off.
In any echelon form, plain or reduced, we can read off when a system has
an empty solution set because there is a contradictory equation, we can read off
when a system has a one-element solution set because there is no contradiction
and every variable is the leading variable in some row, and we can read off when
a system has an infinite solution set because there is no contradiction and at
least one variable is free.
In reduced echelon form we can read off not just what kind of solution set
the system has, but also its description. Whether or not the echelon form

is reduced, we have no trouble describing the solution set when it is empty,
of course. The two examples above show that when the system has a single
solution then the solution can be read off from the right-hand column. In the
case when the solution set is infinite, its parametrization can also be read off
of the reduced echelon form. Consider, for example, this system that is shown
brought to echelon form and then to reduced echelon form.


2 6 1 2 5
0 3 1 4 1
0 3 1 2 5

 −ρ2+ρ3
−→


2 6 1 2 5
0 3 1 4 1
0 0 0 −2 4


(1/2)ρ1
−→
(1/3)ρ2
−(1/2)ρ3
(4/3)ρ3+ρ2
−→
−ρ3+ρ1
−3ρ2+ρ1
−→


1 0 −1/2 0 −9/2
0 1 1/3 0 3
0 0 0 1 −2


Starting with the middle matrix, the echelon form version, back substitution
produces −2x4 = 4 so that x4 = −2, then another back substitution gives
3x2 + x3 + 4(−2) = 1 implying that x2 = 3 − (1/3)x3, and then the final
back substitution gives 2x1 + 6(3 − (1/3)x3) + x3 + 2(−2) = 5 implying that
x1 = −(9/2) + (1/2)x3. Thus the solution set is this.
S = {




x1
x2
x3
x4



 =




−9/2
3
0
−2



 +




1/2
−1/3
1
0



 x3
¯
¯ x3 ∈ R}
Now, considering the final matrix, the reduced echelon form version, note that
adjusting the parametrization by moving the x3 terms to the other side does
indeed give the description of this infinite solution set.
Part of the reason that this works is straightforward. While a set can have
many parametrizations that describe it, e.g., both of these also describe the
above set S (take t to be x3/6 and s to be x3 − 1)
{




−9/2
3
0
−2



 +




3
−2
6
0



 t
¯
¯ t ∈ R} {




−4
8/3
1
−2



 +




1/2
−1/3
1
0



 s
¯
¯ s ∈ R}
nonetheless we have in this book stuck to a convention of parametrizing using
the unmodified free variables (that is, x3 = x3 instead of x3 = 6t). We can
easily see that a reduced echelon form version of a system is equivalent to a
parametrization in terms of unmodified free variables. For instance,
x1 = 4 − 2x3
x2 = 3 − x3
⇐⇒


1 0 2 4
0 1 1 3
0 0 0 0


(to move from left to right we also need to know how many equations are in the
system). So, the convention of parametrizing with the free variables by solving

each equation for its leading variable and then eliminating that leading variable
from every other equation is exactly equivalent to the reduced echelon form
conditions that each leading entry must be a one and must be the only nonzero
entry in its column.
Not as straightforward is the other part of the reason that the reduced
echelon form version allows us to read off the parametrization that we would
have gotten had we stopped at echelon form and then done back substitution.
The prior paragraph shows that reduced echelon form corresponds to some
parametrization, but why the same parametrization? A solution set can be
parametrized in many ways, and Gauss’ method or the Gauss-Jordan method
can be done in many ways, so a first guess might be that we could derive many
different reduced echelon form versions of the same starting system and many
different parametrizations. But we never do. Experience shows that starting
with the same system and proceeding with row operations in many different
ways always yields the same reduced echelon form and the same parametrization
(using the unmodified free variables).
In the rest of this section we will show that the reduced echelon form version
of a matrix is unique. It follows that the parametrization of a linear system in
terms of its unmodified free variables is unique because two different ones would
give two different reduced echelon forms.
We shall use this result, and the ones that lead up to it, in the rest of the
book but perhaps a restatement in a way that makes it seem more immediately
useful may be encouraging. Imagine that we solve a linear system, parametrize,
and check in the back of the book for the answer. But the parametrization there
appears different. Have we made a mistake, or could these be different-looking
descriptions of the same set, as with the three descriptions above of S? The prior
paragraph notes that we will show here that different-looking parametrizations
(using the unmodified free variables) describe genuinely different sets.
Here is an informal argument that the reduced echelon form version of a
matrix is unique. Consider again the example that started this section of a
matrix that reduces to three different echelon form matrices. The first matrix
of the three is the natural echelon form version. The second matrix is the same
as the first except that a row has been halved. The third matrix, too, is just a
cosmetic variant of the first. The definition of reduced echelon form outlaws this
kind of fooling around. In reduced echelon form, halving a row is not possible
because that would change the row’s leading entry away from one, and neither
is combining rows possible, because then a leading entry would no longer be
alone in its column.
This informal justification is not a proof; we have argued that no two different
reduced echelon form matrices are related by a single row operation step, but
we have not ruled out the possibility that multiple steps might do. Before we go
to that proof, we finish this subsection by rephrasing our work in a terminology
that will be enlightening.
Many different matrices yield the same reduced echelon form matrix. The
three echelon form matrices from the start of this section, and the matrix they

were derived from, all give this reduced echelon form matrix.
µ
1 0
0 1
¶
We think of these matrices as related to each other. The next result speaks to
this relationship.
1.4 Lemma Elementary row operations are reversible.
Proof. For any matrix A, the effect of swapping rows is reversed by swapping
them back, multiplying a row by a nonzero k is undone by multiplying by 1/k,
and adding a multiple of row i to row j (with i 6= j) is undone by subtracting
the same multiple of row i from row j.
A
ρi↔ρj
−→
ρj ↔ρi
−→ A A
kρi
−→
(1/k)ρi
−→ A A
kρi+ρj
−→
−kρi+ρj
−→ A
(The i 6= j conditions is needed. See Exercise 13.) QED
This lemma suggests that ‘reduces to’ is misleading — where A −→ B, we
shouldn’t think of B as “after” A or “simpler than” A. Instead we should think
of them as interreducible or interrelated. Below is a picture of the idea. The
matrices from the start of this section and their reduced echelon form version
are shown in a cluster. They are all related; some of the interrelationships are
shown also.
µ
1 0
0 1
¶
↔
↔
↔
µ
2 2
4 3
¶
↔
↔
↔
↔
µ
2 0
0 −1
¶
↔ ↔
↔ ↔ ↔
µ
2 2
0 −1
¶
↔
↔
↔
↔
µ
1 1
0 −1
¶
The technical phrase in this situation is that matrices that reduce to each other
are ‘equivalent with respect to the relationship of row reducibility’. The next
result verifies this statement using the definition of an equivalence.∗
1.5 Lemma Between matrices, ‘reduces to’ is an equivalence relation.
Proof. We must check the conditions (i) reflexivity, that any matrix reduces to
itself, (ii) symmetry, that if A reduces to B then B reduces to A, and (iii) tran-
sitivity, that if A reduces to B and B reduces to C then A reduces to C.
Reflexivity is easy; any matrix reduces to itself in zero row operations.
That the relationship is symmetric is Lemma 1.4 — if A reduces to B by
some row operations then also B reduces to A by reversing those operations.
For transitivity, suppose that A reduces to B and that B reduces to C.
Linking the reduction steps from A → · · · → B with those from B → · · · → C
gives a reduction from A to C. QED
∗ More information on equivalence relations is in the appendix.

1.6 Definition Two matrices that are interreducible by the elementary row
operations are row equivalent.
The diagram below has the collection of all matrices as a box. Inside that
box, each matrix lies in some class. Matrices are in the same class if and only if
they are interreducible. The classes are disjoint — no matrix is in two distinct
classes. The collection of matrices has been partitioned into row equivalence
classes. One of the reasons that showing the row equivalence relation is an
equivalence is useful is that any equivalence relation gives rise to a partition.∗
All matrices: %
$
Ã
!
¿
À
. . .
.A
B.
A row equivalent
to B.
One of the classes in this partition is the cluster of matrices shown above,
expanded to include all of the nonsingular 2×2 matrices.
The next subsection proves that the reduced echelon form of a matrix is
unique; that every matrix reduces to one and only one reduced echelon form
matrix. Rephrased in the relation language, we shall prove that every matrix is
row equivalent to one and only one reduced echelon form matrix. In terms of the
partition in the picture what we shall prove is: every equivalence class contains
one and only one reduced echelon form matrix. So each reduced echelon form
matrix serves as a representative of its class.
After that proof we shall, as mentioned in the introduction to this section,
have a way to decide if one matrix can be derived from another by row reduction.
We can just apply the Gauss-Jordan procedure to both and see whether or not
they come to the same reduced echelon form.
Exercises
X 1.7 Use Gauss-Jordan reduction to solve each system.
(a) x + y = 2
x − y = 0
(b) x − z = 4
2x + 2y = 1
(c) 3x − 2y = 1
6x + y = 1/2
(d) 2x − y = −1
x + 3y − z = 5
y + 2z = 5
X 1.8 Find the reduced echelon form of each matrix.
(a)
µ
2 1
1 3
¶
(b)
Ã
1 3 1
2 0 4
−1 −3 −3
!
(c)
Ã
1 0 3 1 2
1 4 2 1 5
3 4 8 1 2
!
(d)
Ã
0 1 3 2
0 0 5 6
1 5 1 5
!
X 1.9 Find each solution set by using Gauss-Jordan reduction, then reading off the
parametrization.
∗ More information on partitions and class representatives is in the appendix.

(a) 2x + y − z = 1
4x − y = 3
(b) x − z = 1
y + 2z − w = 3
x + 2y + 3z − w = 7
(c) x − y + z = 0
y + w = 0
3x − 2y + 3z + w = 0
−y − w = 0
(d) a + 2b + 3c + d − e = 1
3a − b + c + d + e = 3
1.10 Give two distinct echelon form versions of this matrix.
Ã
2 1 1 3
6 4 1 2
1 5 1 5
!
X 1.11 List the reduced echelon forms possible for each size.
(a) 2×2 (b) 2×3 (c) 3×2 (d) 3×3
X 1.12 What results from applying Gauss-Jordan reduction to a nonsingular matrix?
1.13 The proof of Lemma 1.4 contains a reference to the i 6= j condition on the
row pivoting operation.
(a) The definition of row operations has an i 6= j condition on the swap operation
ρi ↔ ρj. Show that in A
ρi↔ρj
−→
ρi↔ρj
−→ A this condition is not needed.
(b) Write down a 2×2 matrix with nonzero entries, and show that the −1·ρ1 +ρ1
operation is not reversed by 1 · ρ1 + ρ1.
(c) Expand the proof of that lemma to make explicit exactly where the i 6= j
condition on pivoting is used.
1.III.2 Row Equivalence
We will close this section and this chapter by proving that every matrix is
row equivalent to one and only one reduced echelon form matrix. The ideas
that appear here will reappear, and be further developed, in the next chapter.
The underlying theme here is that one way to understand a mathematical
situation is by being able to classify the cases that can happen. We have met this
theme several times already. We have classified solution sets of linear systems
into the no-elements, one-element, and infinitely-many elements cases. We have
also classified linear systems with the same number of equations as unknowns
into the nonsingular and singular cases. We adopted these classifications because
they give us a way to understand the situations that we were investigating. Here,
where we are investigating row equivalence, we know that the set of all matrices
breaks into the row equivalence classes. When we finish the proof here, we will
have a way to understand each of those classes — its matrices can be thought
of as derived by row operations from the unique reduced echelon form matrix
in that class.
To understand how row operations act to transform one matrix into another,
we consider the effect that they have on the parts of a matrix. The crucial
observation is that row operations combine the rows linearly.

2.1 Definition A linear combination of x1, . . . , xm is an expression of the form
c1x1 + c2x2 + · · · + cmxm where the c’s are scalars.
(We have already used the phrase ‘linear combination’ in this book. The mean-
ing is unchanged, but the next result’s statement makes a more formal definition
in order.)
2.2 Lemma (Linear Combination Lemma) A linear combination of linear
combinations is a linear combination.
Proof. Given the linear combinations c1,1x1 + · · · + c1,nxn through cm,1x1 +
· · · + cm,nxn, consider a combination of those
d1(c1,1x1 + · · · + c1,nxn) + · · · + dm(cm,1x1 + · · · + cm,nxn)
where the d’s are scalars along with the c’s. Distributing those d’s and regroup-
ing gives
= d1c1,1x1 + · · · + d1c1,nxn + d2c2,1x1 + · · · + dmc1,1x1 + · · · + dmc1,nxn
= (d1c1,1 + · · · + dmcm,1)x1 + · · · + (d1c1,n + · · · + dmcm,n)xn
which is indeed a linear combination of the x’s. QED
In this subsection we will use the convention that, where a matrix is named
with an upper case roman letter, the matching lower-case greek letter names
the rows.
A =






α1
α2
.
.
.
αm






B =







β1
β2
.
.
.
βm







2.3 Corollary Where one matrix row reduces to another, each row of the
second is a linear combination of the rows of the first.
The proof below uses induction on the number of row operations used to
reduce one matrix to the other. Before we proceed, here is an outline of the ar-
gument (readers unfamiliar with induction may want to compare this argument
with the one used in the ‘General = Particular+Homogeneous’ proof).∗
First,
for the base step of the argument, we will verify that the proposition is true
when reduction can be done in zero row operations. Second, for the inductive
step, we will argue that if being able to reduce the first matrix to the second
in some number t ≥ 0 of operations implies that each row of the second is a
linear combination of the rows of the first, then being able to reduce the first to
the second in t + 1 operations implies the same thing. Together, this base step
and induction step prove this result because by the base step the proposition
∗ More information on mathematical induction is in the appendix.

is true in the zero operations case, and by the inductive step the fact that it is
true in the zero operations case implies that it is true in the one operation case,
and the inductive step applied again gives that it is therefore true in the two
operations case, etc.
Proof. We proceed by induction on the minimum number of row operations
that take a first matrix A to a second one B.
In the base step, that zero reduction operations suffice, the two matrices
are equal and each row of B is obviously a combination of A’s rows: ~
βi =
0 · ~
α1 + · · · + 1 · ~
αi + · · · + 0 · ~
αm.
For the inductive step, assume the inductive hypothesis: with t ≥ 0, if a
matrix can be derived from A in t or fewer operations then its rows are linear
combinations of the A’s rows. Consider a B that takes t+1 operations. Because
there are more than zero operations, there must be a next-to-last matrix G so
that A −→ · · · −→ G −→ B. This G is only t operations away from A and so the
inductive hypothesis applies to it, that is, each row of G is a linear combination
of the rows of A.
If the last operation, the one from G to B, is a row swap then the rows
of B are just the rows of G reordered and thus each row of B is also a linear
combination of the rows of A. The other two possibilities for this last operation,
that it multiplies a row by a scalar and that it adds a multiple of one row to
another, both result in the rows of B being linear combinations of the rows of
G. But therefore, by the Linear Combination Lemma, each row of B is a linear
combination of the rows of A.
With that, we have both the base step and the inductive step, and so the
proposition follows. QED
2.4 Example In the reduction
µ
0 2
1 1
¶
ρ1↔ρ2
−→
µ
1 1
0 2
¶
(1/2)ρ2
−→
µ
1 1
0 1
¶
−ρ2+ρ1
−→
µ
1 0
0 1
¶
,
call the matrices A, D, G, and B. The methods of the proof show that there
are three sets of linear relationships.
δ1 = 0 · α1 + 1 · α2
δ2 = 1 · α1 + 0 · α2
γ1 = 0 · α1 + 1 · α2
γ2 = (1/2)α1 + 0 · α2
β1 = (−1/2)α1 + 1 · α2
β2 = (1/2)α1 + 0 · α2
The prior result gives us the insight that Gauss’ method works by taking
linear combinations of the rows. But to what end; why do we go to echelon
form as a particularly simple, or basic, version of a linear system? The answer,
of course, is that echelon form is suitable for back substitution, because we have
isolated the variables. For instance, in this matrix
R =




2 3 7 8 0 0
0 0 1 5 1 1
0 0 0 3 3 0
0 0 0 0 2 1





x1 has been removed from x5’s equation. That is, Gauss’ method has made x5’s
row independent of x1’s row.
Independence of a collection of row vectors, or of any kind of vectors, will
be precisely defined and explored in the next chapter. But a first take on it is
that we can show that, say, the third row above is not comprised of the other
rows, that ρ3 6= c1ρ1 + c2ρ2 + c4ρ4. For, suppose that there are scalars c1, c2,
and c4 such that this relationship holds.
¡
0 0 0 3 3 0
¢
= c1
¡
2 3 7 8 0 0
¢
+ c2
¡
0 0 1 5 1 1
¢
+ c4
¡
0 0 0 0 2 1
¢
The first row’s leading entry is in the first column and narrowing our considera-
tion of the above relationship to consideration only of the entries from the first
column 0 = 2c1+0c2+0c4 gives that c1 = 0. The second row’s leading entry is in
the third column and the equation of entries in that column 0 = 7c1 +1c2 +0c4,
along with the knowledge that c1 = 0, gives that c2 = 0. Now, to finish, the
third row’s leading entry is in the fourth column and the equation of entries
in that column 3 = 8c1 + 5c2 + 0c4, along with c1 = 0 and c2 = 0, gives an
impossibility.
The following result shows that this effect always holds. It shows that what
Gauss’ linear elimination method eliminates is linear relationships among the
rows.
2.5 Lemma In an echelon form matrix, no nonzero row is a linear combination
of the other rows.
Proof. Let R be in echelon form. Suppose, to obtain a contradiction, that
some nonzero row is a linear combination of the others.
ρi = c1ρ1 + . . . + ci−1ρi−1 + ci+1ρi+1 + . . . + cmρm
We will first use induction to show that the coefficients c1, . . . , ci−1 associated
with rows above ρi are all zero. The contradiction will come from consideration
of ρi and the rows below it.
The base step of the induction argument is to show that the first coefficient
c1 is zero. Let the first row’s leading entry be in column number `1 be the
column number of the leading entry of the first row and consider the equation
of entries in that column.
ρi,`1
= c1ρ1,`1
+ . . . + ci−1ρi−1,`1
+ ci+1ρi+1,`1
+ . . . + cmρm,`1
The matrix is in echelon form so the entries ρ2,`1
, . . . , ρm,`1
, including ρi,`1
, are
all zero.
0 = c1ρ1,`1
+ · · · + ci−1 · 0 + ci+1 · 0 + · · · + cm · 0
Because the entry ρ1,`1
is nonzero as it leads its row, the coefficient c1 must be
zero.

The inductive step is to show that for each row index k between 1 and i − 2,
if the coefficient c1 and the coefficients c2, . . . , ck are all zero then ck+1 is also
zero. That argument, and the contradiction that finishes this proof, is saved for
Exercise 21. QED
We can now prove that each matrix is row equivalent to one and only one
reduced echelon form matrix. We will find it convenient to break the first half
of the argument off as a preliminary lemma. For one thing, it holds for any
echelon form whatever, not just reduced echelon form.
2.6 Lemma If two echelon form matrices are row equivalent then the leading
entries in their first rows lie in the same column. The same is true of all the
nonzero rows — the leading entries in their second rows lie in the same column,
etc.
For the proof we rephrase the result in more technical terms. Define the form
of an m×n matrix to be the sequence h`1, `2, . . . , `mi where ì is the column
number of the leading entry in row i and ì = ∞ if there is no leading entry
in that column. The lemma says that if two echelon form matrices are row
equivalent then their forms are equal sequences.
Proof. Let B and D be echelon form matrices that are row equivalent. Because
they are row equivalent they must be the same size, say n×m. Let the column
number of the leading entry in row i of B be ì and let the column number of
the leading entry in row j of D be kj. We will show that `1 = k1, that `2 = k2,
etc., by induction.
This induction argument relies on the fact that the matrices are row equiv-
alent, because the Linear Combination Lemma and its corollary therefore give
that each row of B is a linear combination of the rows of D and vice versa:
βi = si,1δ1 + si,2δ2 + · · · + si,mδm and δj = tj,1β1 + tj,2β2 + · · · + tj,mβm
where the s’s and t’s are scalars.
The base step of the induction is to verify the lemma for the first rows of
the matrices, that is, to verify that `1 = k1. If either row is a zero row then
the entire matrix is a zero matrix since it is in echelon form, and hterefore both
matrices are zero matrices (by Corollary 2.3), and so both `1 and k1 are ∞. For
the case where neither β1 nor δ1 is a zero row, consider the i = 1 instance of
the linear relationship above.
β1 = s1,1δ1 + s1,2δ2 + · · · + s1,mδm
¡
0 · · · b1,`1
· · ·
¢
= s1,1
¡
0 · · · d1,k1
· · ·
¢
+ s1,2
¡
0 · · · 0 · · ·
¢
.
.
.
+ s1,m
¡
0 · · · 0 · · ·
¢

First, note that `1 < k1 is impossible: in the columns of D to the left of column
k1 the entries are are all zeroes (as d1,k1
leads the first row) and so if `1 < k1
then the equation of entries from column `1 would be b1,`1
= s1,1·0+· · ·+s1,m·0,
but b1,`1
isn’t zero since it leads its row and so this is an impossibility. Next,
a symmetric argument shows that k1 < `1 also is impossible. Thus the `1 = k1
base case holds.
The inductive step is to show that if `1 = k1, and `2 = k2, . . . , and `r = kr,
then also `r+1 = kr+1 (for r in the interval 1 .. m − 1). This argument is saved
for Exercise 22. QED
That lemma answers two of the questions that we have posed (i) any two
echelon form versions of a matrix have the same free variables, and consequently
(ii) any two echelon form versions have the same number of free variables. There
is no linear system and no combination of row operations such that, say, we could
solve the system one way and get y and z free but solve it another way and get
y and w free, or solve it one way and get two free variables while solving it
another way yields three.
We finish now by specializing to the case of reduced echelon form matrices.
2.7 Theorem Each matrix is row equivalent to a unique reduced echelon form
matrix.
Proof. Clearly any matrix is row equivalent to at least one reduced echelon
form matrix, via Gauss-Jordan reduction. For the other half, that any matrix
is equivalent to at most one reduced echelon form matrix, we will show that if
a matrix Gauss-Jordan reduces to each of two others then those two are equal.
Suppose that a matrix is row equivalent the two reduced echelon form ma-
trices B and D, which are therefore row equivalent to each other. The Linear
Combination Lemma and its corollary allow us to write the rows of one, say
B, as a linear combination of the rows of the other βi = ci,1δ1 + · · · + ci,mδm.
The preliminary result, Lemma 2.6, says that in the two matrices, the same
collection of rows are nonzero. Thus, if β1 through βr are the nonzero rows of
B then the nonzero rows of D are δ1 through δr. Zero rows don’t contribute to
the sum so we can rewrite the relationship to include just the nonzero rows.
βi = ci,1δ1 + · · · + ci,rδr (∗)
The preliminary result also says that for each row j between 1 and r, the
leading entries of the j-th row of B and D appear in the same column, denoted
`j. Rewriting the above relationship to focus on the entries in the `j-th column
¡
· · · bi,`j
· · ·
¢
= ci,1
¡
· · · d1,`j
· · ·
¢
+ ci,2
¡
· · · d2,`j
· · ·
¢
.
.
.
+ ci,r
¡
· · · dr,`j
· · ·
¢

gives this set of equations for i = 1 up to i = r.
b1,`j
= c1,1d1,`j
+ · · · + c1,jdj,`j
+ · · · + c1,rdr,`j
.
.
.
bj,`j
= cj,1d1,`j
+ · · · + cj,jdj,`j
+ · · · + cj,rdr,`j
.
.
.
br,`j
= cr,1d1,`j
+ · · · + cr,jdj,`j
+ · · · + cr,rdr,`j
Since D is in reduced echelon form, all of the d’s in column `j are zero except for
dj,`j
, which is 1. Thus each equation above simplifies to bi,`j
= ci,jdj,`j
= ci,j ·1.
But B is also in reduced echelon form and so all of the b’s in column `j are zero
except for bj,`j
, which is 1. Therefore, each ci,j is zero, except that c1,1 = 1,
and c2,2 = 1, . . . , and cr,r = 1.
We have shown that the only nonzero coefficient in the linear combination
labelled (∗) is cj,j, which is 1. Therefore βj = δj. Because this holds for all
nonzero rows, B = D. QED
We end with a recap. In Gauss’ method we start with a matrix and then
derive a sequence of other matrices. We defined two matrices to be related if one
can be derived from the other. That relation is an equivalence relation, called
row equivalence, and so partitions the set of all matrices into row equivalence
classes.
All matrices: %
$
Ã
!
¿
À
. . .
. ( 1 3
2 7 )
. ( 1 3
0 1 )
each class
consists of
row equivalent
matrices
(There are infinitely many matrices in the pictured class, but we’ve only got
room to show two.) We have proved there is one and only one reduced echelon
form matrix in each row equivalence class. So the reduced echelon form is a
canonical form∗
for row equivalence: the reduced echelon form matrices are
representatives of the classes.
All matrices: %
$
Ã
!
¿
À
. . .
?
?
?
?
?
( 1 0
0 1 )
one reduced
echelon form matrix
from each class
We can answer questions about the classes by translating them into questions
about the representatives.
∗ More information on canonical representatives is in the appendix.

2.8 Example We can decide if matrices are interreducible by seeing if Gauss-
Jordan reduction produces the same reduced echelon form result. Thus, these
are not row equivalent
µ
1 −3
−2 6
¶ µ
1 −3
−2 5
¶
because their reduced echelon forms are not equal.
µ
1 −3
0 0
¶ µ
1 0
0 1
¶
2.9 Example Any nonsingular 3×3 matrix Gauss-Jordan reduces to this.


1 0 0
0 1 0
0 0 1


2.10 Example We can describe the classes by listing all possible reduced ech-
elon form matrices. Any 2×2 matrix lies in one of these: the class of matrices
row equivalent to this,
µ
0 0
0 0
¶
the infinitely many classes of matrices row equivalent to one of this type
µ
1 a
0 0
¶
where a ∈ R (including a = 0), the class of matrices row equivalent to this,
µ
0 1
0 0
¶
and the class of matrices row equivalent to this
µ
1 0
0 1
¶
(this the class of nonsingular 2×2 matrices).
Exercises
X 2.11 Decide if the matrices are row equivalent.
(a)
µ
1 2
4 8
¶
,
µ
0 1
1 2
¶
(b)
Ã
1 0 2
3 −1 1
5 −1 5
!
,
Ã
1 0 2
0 2 10
2 0 4
!
(c)
Ã
2 1 −1
1 1 0
4 3 −1
!
,
µ
1 0 2
0 2 10
¶
(d)
µ
1 1 1
−1 2 2
¶
,
µ
0 3 −1
2 2 5
¶
(e)
µ
1 1 1
0 0 3
¶
,
µ
0 1 2
1 −1 1
¶
2.12 Describe the matrices in each of the classes represented in Example 2.10.
2.13 Describe all matrices in the row equivalence class of these.

(a)
µ
1 0
0 0
¶
(b)
µ
1 2
2 4
¶
(c)
µ
1 1
1 3
¶
2.14 How many row equivalence classes are there?
2.15 Can row equivalence classes contain different-sized matrices?
2.16 How big are the row equivalence classes?
(a) Show that the class of any zero matrix is finite.
(b) Do any other classes contain only finitely many members?
X 2.17 Give two reduced echelon form matrices that have their leading entries in the
same columns, but that are not row equivalent.
X 2.18 Show that any two n×n nonsingular matrices are row equivalent. Are any
two singular matrices row equivalent?
X 2.19 Describe all of the row equivalence classes containing these.
(a) 2 × 2 matrices (b) 2 × 3 matrices (c) 3 × 2 matrices
(d) 3×3 matrices
2.20 (a) Show that a vector ~
β0 is a linear combination of members of the set
{~
β1, . . . , ~
βn} if and only there is a linear relationship ~
0 = c0
~
β0 + · · · + cn
~
βn
where c0 is not zero. (Watch out for the ~
β0 = ~
0 case.)
(b) Derive Lemma 2.5.
X 2.21 Finish the proof of Lemma 2.5.
(a) First illustrate the inductive step by showing that `2 = k2.
(b) Do the full inductive step: assume that ck is zero for 1 ≤ k < i − 1, and
deduce that ck+1 is also zero.
(c) Find the contradiction.
2.22 Finish the induction argument in Lemma 2.6.
(a) State the inductive hypothesis, Also state what must be shown to follow from
that hypothesis.
(b) Check that the inductive hypothesis implies that in the relationship βr+1 =
sr+1,1δ1 + sr+2,2δ2 + · · · + sr+1,mδm the coefficients sr+1,1, . . . , sr+1,r are each
zero.
(c) Finish the inductive step by arguing, as in the base case, that `r+1 < kr+1
and kr+1 < `r+1 are impossible.
2.23 Why, in the proof of Theorem 2.7, do we bother to restrict to the nonzero rows?
Why not just stick to the relationship that we began with, βi = ci,1δ1+· · ·+ci,mδm,
with m instead of r, and argue using it that the only nonzero coefficient is ci,i,
which is 1?
X 2.24 [Trono] Three truck drivers went into a roadside cafe. One truck driver pur-
chased four sandwiches, a cup of coffee, and seven doughnuts for $8.45. Another
driver purchased three sandwiches, a cup of coffee, and seven doughnuts for $6.30.
What did the third truck driver pay for a sandwich, a cup of coffee, and a dough-
nut?
2.25 The fact that Gaussian reduction disallows multiplication of a row by zero is
needed for the proof of uniqueness of reduced echelon form, or else every matrix
would be row equivalent to a matrix of all zeros. Where is it used?
X 2.26 The Linear Combination Lemma says which equations can be gotten from
Gaussian reduction from a given linear system.
(1) Produce an equation not implied by this system.
3x + 4y = 8
2x + y = 3

(2) Can any equation be derived from an inconsistent system?
2.27 Extend the definition of row equivalence to linear systems. Under your defi-
nition, do equivalent systems have the same solution set?
X 2.28 In this matrix
Ã
1 2 3
3 0 3
1 4 5
!
the first and second columns add to the third.
(a) Show that remains true under any row operation.
(b) Make a conjecture.
(c) Prove that it holds.

Topic: Computer Algebra Systems 61
Topic: Computer Algebra Systems
The linear systems in this chapter are small enough that their solution by hand
is easy. But large systems are easiest, and safest, to do on a computer. There
are special purpose programs such as LINPACK for this job. Another popular
tool is a general purpose computer algebra system, including both commercial
packages such as Maple, Mathematica, or MATLAB, or free packages such as
SciLab, or Octave.
For example, in the Topic on Networks, we need to solve this.
i0 − i1 − i2 = 0
i1 − i3 − i5 = 0
i2 − i4 + i5 = 0
i3 + i4 − i6 = 0
5i1 + 10i3 = 10
2i2 + 4i4 = 10
5i1 − 2i2 + 50i5 = 0
It can be done by hand, but it would take a while and be error-prone. Using a
computer is better.
We illustrate by solving that system under Maple (for another system, a
user’s manual would obviously detail the exact syntax needed). The array of
coefficients can be entered in this way
> A:=array( [[1,-1,-1,0,0,0,0],
[0,1,0,-1,0,-1,0],
[0,0,1,0,-1,1,0],
[0,0,0,1,1,0,-1],
[0,5,0,10,0,0,0],
[0,0,2,0,4,0,0],
[0,5,-1,0,0,10,0]] );
(putting the rows on separate lines is not necessary, but is done for clarity).
The vector of constants is entered similarly.
> u:=array( [0,0,0,0,10,10,0] );
Then the system is solved, like magic.
> linsolve(A,u);
7 2 5 2 5 7
[ -, -, -, -, -, 0, - ]
3 3 3 3 3 3
Systems with infinitely many solutions are solved in the same way — the com-
puter simply returns a parametrization.
Exercises
1 Use the computer to solve the two problems that opened this chapter.
(a) This is the Statics problem.
40h + 15c = 100
25c = 50 + 50h

(b) This is the Chemistry problem.
7h = 7j
8h + 1i = 5j + 2k
1i = 3j
3i = 6j + 1k
2 Use the computer to solve these systems from the first subsection, or conclude
‘many solutions’ or ‘no solutions’.
(a) 2x + 2y = 5
x − 4y = 0
(b) −x + y = 1
x + y = 2
(c) x − 3y + z = 1
x + y + 2z = 14
(d) −x − y = 1
−3x − 3y = 2
(e) 4y + z = 20
2x − 2y + z = 0
x + z = 5
x + y − z = 10
(f) 2x + z + w = 5
y − w = −1
3x − z − w = 0
4x + y + 2z + w = 9
3 Use the computer to solve these systems from the second subsection.
(a) 3x + 6y = 18
x + 2y = 6
(b) x + y = 1
x − y = −1
(c) x1 + x3 = 4
x1 − x2 + 2x3 = 5
4x1 − x2 + 5x3 = 17
(d) 2a + b − c = 2
2a + c = 3
a − b = 0
(e) x + 2y − z = 3
2x + y + w = 4
x − y + z + w = 1
(f) x + z + w = 4
2x + y − w = 2
3x + y + z = 7
4 What does the computer give for the solution of the general 2×2 system?
ax + cy = p
bx + dy = q

Topic: Input-Output Analysis 63
Topic: Input-Output Analysis
An economy is an immensely complicated network of interdependences. Changes
in one part can ripple out to affect other parts. Economists have struggled to
be able to describe, and to make predictions about, such a complicated object.
Mathematical models using systems of linear equations have emerged as a key
tool. One is Input-Output Analysis, pioneered by W. Leontief, who won the
1973 Nobel Prize in Economics.
Consider an economy with many parts, two of which are the steel industry
and the auto industry. As they work to meet the demand for their product from
other parts of the economy, that is, from users external to the steel and auto
sectors, these two interact tightly. For instance, should the external demand
for autos go up, that would lead to an increase in the auto industry’s usage of
steel. Or, should the external demand for steel fall, then it would lead to a fall
in steel’s purchase of autos. The type of Input-Output model we will consider
takes in the external demands and then predicts how the two interact to meet
those demands.
We start with a listing of production and consumption statistics. (These
numbers, giving dollar values in millions, are excerpted from [Leontief 1965],
describing the 1958 U.S. economy. Today’s statistics would be quite different,
both because of inflation and because of technical changes in the industries.)
used by
steel
used by
auto
used by
others total
value of
steel 5 395 2 664 25 448
value of
auto 48 9 030 30 346
For instance, the dollar value of steel used by the auto industry in this year is
2, 664 million. Note that industries may consume some of their own output.
We can fill in the blanks for the external demand. This year’s value of the
steel used by others this year is 17, 389 and this year’s value of the auto used
by others is 21, 268. With that, we have a complete description of the external
demands and of how auto and steel interact, this year, to meet them.
Now, imagine that the external demand for steel has recently been going up
by 200 per year and so we estimate that next year it will be 17, 589. Imagine
also that for similar reasons we estimate that next year’s external demand for
autos will be down 25 to 21, 243. We wish to predict next year’s total outputs.
That prediction isn’t as simple as adding 200 to this year’s steel total and
subtracting 25 from this year’s auto total. For one thing, a rise in steel will
cause that industry to have an increased demand for autos, which will mitigate,
to some extent, the loss in external demand for autos. On the other hand, the
drop in external demand for autos will cause the auto industry to use less steel,
and so lessen somewhat the upswing in steel’s business. In short, these two
industries form a system, and we need to predict the totals at which the system
as a whole will settle.

For that prediction, let s be next years total production of steel and let a be
next year’s total output of autos. We form these equations.
next year’s production of steel = next year’s use of steel by steel
+ next year’s use of steel by auto
+ next year’s use of steel by others
next year’s production of autos = next year’s use of autos by steel
+ next year’s use of autos by auto
+ next year’s use of autos by others
On the left side of those equations go the unknowns s and a. At the ends of the
right sides go our external demand estimates for next year 17, 589 and 21, 243.
For the remaining four terms, we look to the table of this year’s information
about how the industries interact.
For instance, for next year’s use of steel by steel, we note that this year the
steel industry used 5395 units of steel input to produce 25, 448 units of steel
output. So next year, when the steel industry will produce s units out, we
expect that doing so will take s · (5395)/(25 448) units of steel input — this is
simply the assumption that input is proportional to output. (We are assuming
that the ratio of input to output remains constant over time; in practice, models
may try to take account of trends of change in the ratios.)
Next year’s use of steel by the auto industry is similar. This year, the auto
industry uses 2664 units of steel input to produce 30346 units of auto output. So
next year, when the auto industry’s total output is a, we expect it to consume
a · (2664)/(30346) units of steel.
Filling in the other equation in the same way, we get this system of linear
equation.
5 395
25 448
· s +
2 664
30 346
· a + 17 589 = s
48
25 448
· s +
9 030
30 346
· a + 21 243 = a
Rounding to four decimal places and putting it into the form for Gauss’ method
gives this.
0.7880s − 0.0879a = 17 589
−0.0019s + 0.7024a = 21 268
The solution is s = 25 708 and a = 30 350.
Looking back, recall that above we described why the prediction of next
year’s totals isn’t as simple as adding 200 to last year’s steel total and subtract-
ing 25 from last year’s auto total. In fact, comparing these totals for next year
to the ones given at the start for the current year shows that, despite the drop
in external demand, the total production of the auto industry is predicted to
rise. The increase in internal demand for autos caused by steel’s sharp rise in
business more than makes up for the loss in external demand for autos.

Topic: Input-Output Analysis 65
One of the advantages of having a mathematical model is that we can ask
“What if . . . ?” questions. For instance, we can ask “What if the estimates for
next year’s external demands are somewhat off?” To try to understand how
much the model’s predictions change in reaction to changes in our estimates, we
can try revising our estimate of next year’s external steel demand from 17, 589
down to 17, 489, while keeping the assumption of next year’s external demand
for autos fixed at 21, 243. The resulting system
0.7880s − 0.0879a = 17 489
−0.0019s + 0.7024a = 21 243
when solved gives s = 25 577 and a = 30 314. This kind of exploration of the
model is sensitivity analysis. We are seeing how sensitive the predictions of our
model are to the accuracy of the assumptions.
Obviously, we can consider larger models that detail the interactions among
more sectors of an economy. These models are typically solved on a computer,
using the techniques of matrix algebra that we will develop in Chapter Three.
Some examples are given in the exercises. Obviously also, a single model does
not suit every case; expert judgment is needed to see if the assumptions under-
lying the model can are reasonable ones to apply to a particular case. With
those caveats, however, this model has proven in practice to be a useful and ac-
curate tool for economic analysis. For further reading, try [Leontief 1951] and
[Leontief 1965].
Exercises
Hint: these systems are easiest to solve on a computer.
1 With the steel-auto system given above, estimate next year’s total productions
in these cases.
(a) Next year’s external demands are: up 200 from this year for steel, and un-
changed for autos.
(b) Next year’s external demands are: up 100 for steel, and up 200 for autos.
(c) Next year’s external demands are: up 200 for steel, and up 200 for autos.
2 Imagine a new process for making autos is pioneered. The ratio for use of steel
by the auto industry falls to .0500 (that is, the new process is more efficient in its
use of steel).
(a) How will the predictions for next year’s total productions change compared
to the first example discussed above (i.e., taking next year’s external demands
to be 17, 589 for steel and 21, 243 for autos)?
(b) Predict next year’s totals if, in addition, the external demand for autos rises
to be 21, 500 because the new cars are cheaper.
3 This table gives the numbers for the auto-steel system from a different year, 1947
(see [Leontief 1951]). The units here are billions of 1947 dollars.
used by
steel
used by
auto
used by
others total
value of
steel 6.90 1.28 18.69
value of
autos 0 4.40 14.27

(a) Fill in the missing external demands, and compute the ratios.
(b) Solve for total output if next year’s external demands are: steel’s demand
up 10% and auto’s demand up 15%.
(c) How do the ratios compare to those given above in the discussion for the
1958 economy?
(d) Solve these equations with the 1958 external demands (note the difference
in units; a 1947 dollar buys about what $1.30 in 1958 dollars buys). How far off
are the predictions for total output?
4 Predict next year’s total productions of each of the three sectors of the hypothet-
ical economy shown below
used by
farm
used by
rail
used by
shipping
used by
others total
value of
farm 25 50 100 800
value of
rail 25 50 50 300
value of
shipping 15 10 0 500
if next year’s external demands are as stated.
(a) 625 for farm, 200 for rail, 475 for shipping
(b) 650 for farm, 150 for rail, 450 for shipping
5 This table gives the interrelationships among three segments of an economy (see
[Clark & Coupe]).
used by
food
used by
wholesale
used by
retail
used by
others total
value of
food 0 2 318 4 679 11 869
value of
wholesale 393 1 089 22 459 122 242
value of
retail 3 53 75 116 041
We will do an Input-Output analysis on this system.
(a) Fill in the numbers for this year’s external demands.
(b) Set up the linear system, leaving next year’s external demands blank.
(c) Solve the system where next year’s external demands are calculated by tak-
ing this year’s external demands and inflating them 10%. Do all three sectors
increase their total business by 10%? Do they all even increase at the same
rate?
(d) Solve the system where next year’s external demands are calculated by taking
this year’s external demands and reducing them 7%. (The study from which
these numbers are taken concluded that because of the closing of a local military
facility, overall personal income in the area would fall 7%, so this might be a
first guess at what would actually happen.)

Topic: Accuracy of Computations 67
Topic: Accuracy of Computations
Gauss’ method lends itself nicely to computerization. The code below illus-
trates. It operates on an n×n matrix a, pivoting with the first row, then with
the second row, etc. (This code is in the C language. For readers unfamil-
iar with this concise language, here is a brief translation. The loop construct
for(pivot row=1;pivot row<=n-1;pivot row++){· · · } sets pivot row to be
1 and then iterates while pivot row is less than or equal to n − 1, each time
through incrementing pivot row by one with the ‘++’ operation. The other
non-obvious construct is that the ‘-=’ in the innermost loop amounts to the
a[row below,col] = −multiplier ∗ a[pivot row,col] + a[row below,col]
operation.)
for(pivot_row=1;pivot_row<=n-1;pivot_row++){
for(row_below=pivot_row+1;row_below<=n;row_below++){
multiplier=a[row_below,pivot_row]/a[pivot_row,pivot_row];
for(col=pivot_row;col<=n;col++){
a[row_below,col]-=multiplier*a[pivot_row,col];
}
}
}
While this code provides a first take on how Gauss’ method can be mechanized,
it is not ready to use. It is naive in many ways. The most glaring way is that
it assumes that a nonzero number is always found in the pivot row, pivot row
position for use as the pivot entry. To make it practical, one way in which this
code needs to be reworked is to cover the case where finding a zero in that
location leads to a row swap, or to the conclusion that the matrix is singular.
Adding some if · · · statements to cover those cases is not hard, but we
won’t pursue that here. Instead, we will consider some more subtle ways in
which the code is naive. There are pitfalls arising from the computer’s reliance
on finite-precision floating point arithmetic.
For example, we have seen above that we must handle as a separate case a
system that is singular. But systems that are nearly singular also require great
care. Consider this one.
x + 2y = 3
1.000 000 01x + 2y = 3.000 000 01
By eye we get the solution x = 1 and y = 1. But a computer has more trouble. A
computer that represents real numbers to eight significant places (as is common,
usually called single precision) will represent the second equation internally as
1.000 000 0x + 2y = 3.000 000 0, losing the digits in the ninth place. Instead of
reporting the correct solution, this computer will report something that is not
even close — this computer thinks that the system is singular because the two
equations are represented internally as equal.
For some intuition about how the computer could think something that is
so far off, we can graph the system.

-1
0
1
2
3
4
-1 0 1 2 3 4
(1,1)
At the scale of this graph, the two lines are hard to resolve apart. This system
is nearly singular in the sense that the two lines are nearly the same line. Near-
singularity gives this system the property that a small change in the system
can cause a large change in its solution; for instance, changing the 3.000 000 01
to 3.000 000 03 changes the intersection point from (1, 1) to (3, 0). This system
changes radically depending on a ninth digit, which explains why the eight-
place computer is stumped. A problem that is very sensitive to inaccuracy or
uncertainties in the input values is ill-conditioned.
The above example gives one way in which a system can be difficult to solve
on a computer. It has the advantage that the picture of nearly-equal lines
gives a memorable insight into one way that numerical difficulties can arise.
Unfortunately, though, this insight isn’t very useful when we wish to solve some
large system. We cannot, typically, hope to understand the geometry of an
arbitrary large system. And, in addition, the reasons that the computer’s results
may be unreliable are more complicated than only that the angle between some
of the linear surfaces is quite small.
For an example, consider the system below, from [Hamming].
0.001x + y = 1
x − y = 0
The second equation gives x = y, so x = y = 1/1.001 and thus both variables
have values that are just less than 1. A computer using two digits represents
the system internally in this way (we will do this example in two-digit floating
point arithmetic, but a similar one with eight digits is easy to invent).
(1.0 × 10−2
)x + (1.0 × 100
)y = 1.0 × 100
(1.0 × 100
)x − (1.0 × 100
)y = 0.0 × 100
The computer’s row reduction step −1000ρ1 + ρ2 produces a second equation
−1001y = −999, which the computer rounds to two places as (−1.0 × 103
)y =
−1.0 × 103
. Then the computer decides from the second equation that y = 1
and from the first equation that x = 0. This y value is fairly good, but the x
is way off. Thus, another cause of unreliable output is the mixture of floating
point arithmetic and a reliance on pivots that are small.

An experienced programmer may respond that we should go to double pre-
cision where, usually, sixteen significant digits are retained. It is true, this will
solve many problems. However, there are some difficulties with it as a general
approach. For one thing, double precision takes longer than single precision (on
a ’486 chip, multiplication takes eleven ticks in single precision but fourteen in
double precision [Programmer’s Ref.]) and has twice the memory requirements.
So attempting to do all calculations in double precision is just not practical. And
besides, the above systems can obviously be tweaked to give the same trouble in
the seventeenth digit, so double precision won’t fix all problems. What we need
is a strategy to minimize the numerical trouble arising from solving systems
on a computer, and some guidance as to how far the reported solutions can be
trusted.
Mathematicians have made a careful study of how to get the most reliable
results. A basic improvement on the naive code above is to not simply take
the entry in the pivot row, pivot row position for the pivot, but rather to look
at all of the entries in the pivot row column below the pivot row row, and take
the one that is most likely to give reliable results (e.g., take one that is not too
small). This strategy is partial pivoting. For example, to solve the troublesome
system (∗) above, we start by looking at both equations for a best first pivot,
and taking the 1 in the second equation as more likely to give good results.
Then, the pivot step of −.001ρ2 + ρ1 gives a first equation of 1.001y = 1, which
the computer will represent as (1.0×100
)y = 1.0×100
, leading to the conclusion
that y = 1 and, after back-substitution, x = 1, both of which are close to right.
The code from above can be adapted to this purpose.
for(pivot_row=1;pivot_row<=n-1;pivot_row++){
/* find the largest pivot in this column (in row max) */
max=pivot_row;
for(row_below=pivot_row+1;pivot_row<=n;row_below++){
if (abs(a[row_below,pivot_row]) > abs(a[max,row_below]))
max=row_below;
}
/* swap rows to move that pivot entry up */
temp=a[pivot_row,col];
a[pivot_row,col]=a[max,col];
a[max,col]=temp;
}
/* proceed as before */
for(row_below=pivot_row+1;row_below<=n;row_below++){
multiplier=a[row_below,pivot_row]/a[pivot_row,pivot_row];
a[row_below,col]-=multiplier*a[pivot_row,col];
}
}
}
A full analysis of the best way to implement Gauss’ method is outside the
scope of the book (see [Wilkinson 1965]), but the method recommended by most

experts is a variation on the code above that first finds the best pivot among
the candidates, and then scales it to a number that is less likely to give trouble.
This is scaled partial pivoting.
In addition to returning a result that is likely to be reliable, most well-
done code will return a number, called the conditioning number of the matrix,
that describes the factor by which uncertainties in the input numbers could be
magnified to become possible inaccuracies in the results returned (see [Rice]).
The lesson of this discussion is that just because Gauss’ method always works
in theory, and just because computer code correctly implements that method,
and just because the answer appears on green-bar paper, doesn’t mean that the
answer is reliable. In practice, always use a package where experts have worked
hard to counter what can go wrong.
Exercises
1 Using two decimal places, add 253 and 2/3.
2 This intersect-the-lines problem contrasts with the example discussed above.
-1
0
1
2
3
4
-1 0 1 2 3 4
(1,1) x + 2y = 3
3x − 2y = 1
Illustrate that, in the resulting system, some small change in the numbers will
produce only a small change in the solution by changing the constant in the bot-
tom equation to 1.008 and solving. Compare it to the solution of the unchanged
system.
3 Solve this system by hand ([Rice]).
0.000 3x + 1.556y = 1.569
0.345 4x − 2.346y = 1.018
(a) Solve it accurately, by hand. (b) Solve it by rounding at each step to
four significant digits.
4 Rounding inside the computer often has an effect on the result. Assume that
your machine has eight significant digits.
(a) Show that the machine will compute (2/3) + ((2/3) − (1/3)) as unequal to
((2/3) + (2/3)) − (1/3). Thus, computer arithmetic is not associative.
(b) Compare the computer’s version of (1/3)x + y = 0 and (2/3)x + 2y = 0. Is
twice the first equation the same as the second?
5 Ill-conditioning is not only dependent on the matrix of coefficients. This example
[Hamming] shows that it can arise from an interaction between the left and right
sides of the system. Let ε be a small real.
3x + 2y + z = 6
2x + 2εy + 2εz = 2 + 4ε
x + 2εy − εz = 1 + ε

(a) Solve the system by hand. Notice that the ε’s divide out only because there
is an exact cancelation of the integer parts on the right side as well as on the
left.
(b) Solve the system by hand, rounding to two decimal places, and with ε =
0.001.

Topic: Analyzing Networks
This is the diagram of an electrical circuit. It happens to describe some of the
connections between a car’s battery and lights, but it is typical of such diagrams.
To read it, we can think of the electricity as coming out of one end of the battery
(labeled 6V OR 12V), flowing through the wires (drawn as straight lines to make
the diagram more readable), and back into the other end of the battery. If, in
making its way from one end of the battery to the other through the network of
wires, some electricity flows through a light bulb (drawn as a circle enclosing a
loop of wire), then that light lights. For instance, when the driver steps on the
brake at point A then the switch makes contact and electricity flows through
the brake lights at point B.
This network of connections and components is complicated enough that to
analyze it — for instance, to find out how much electricity is used when both
the headlights and the brake lights are on — then we need systematic tools.
One such tool is linear systems. To illustrate this application, we first need a
few facts about electricity and networks.
The two facts that we need about electricity concern how the electrical com-
ponents act. First, the battery is like a pump for electricity; it provides a force
or push so that the electricity will flow, if there is at least one available path for
it. The second fact about the components is the observation that (in the mate-
rials commonly used in components) the amount of current flow is proportional
to the force pushing it. For each electrical component there is a constant of
proportionality, called its resistance, satisfying that potential = flow·resistance.
(The units are: potential to flow is described in volts, the rate of flow itself is
given in amperes, and resistance to the flow is in ohms. These units are set up
so that volts = amperes · ohms.)
For example, suppose a bulb has a resistance of 25 ohms. Wiring its ends
to a battery with 12 volts results in a flow of electrical current of 12/25 =
0.48 amperes. Conversely, with that same bulb, if we have flow of electrical
current of 2 amperes through it, then the potential difference between one end

Topic: Analyzing Networks 73
of the bulb and the other end will be 2 · 25 = 50 volts. This is the voltage drop
across this bulb. One way to think of the above circuit is that the battery is a
voltage source, or rise, and the other components are voltage sinks, or drops,
that use up the force provided by the battery.
The two facts that we need about networks are Kirchhoff’s Laws.
First Law. The flow into any spot equals the flow out.
Second Law. Around a circuit the total drop equals the total rise.
(In the above circuit the only voltage rise is at the one battery, but some circuits
have more than one rise.)
We can use these facts for a simple analysis of the circuit shown below.
There are three components; they might be bulbs, or they might be some other
component that resists the flow of electricity (resistors are drawn as zig-zags ).
When components are wired one after another, as these are, they are said to be
in series.
20 volt
potential
3 ohm
resistance
2 ohm
resistance
5 ohm
resistance
By Kirchhoff’s Second Law, because the voltage rise in this circuit is 20 volts, so
too, the total voltage drop around this circuit is 20 volts. Since the resistance in
total, from start to finish, in this circuit is 10 ohms (we can take the resistance
of a wire to be negligible), we get that the current is (20/10) = 2 amperes. Now,
Kirchhoff’s First Law says that there are 2 amperes through each resistor, and
so the voltage drops are 4 volts, 10 volts, and 6 volts.
Linear systems appear in the analysis of the next network. In this one, the
resistors are not in series. They are instead in parallel. This network is more
like the car’s lighting diagram.
20 volts 12 ohm 8 ohm

We begin by labeling the branches of the network. Call the flow of current
coming out of the top of the battery and through the top wire i0, call the
current through the left branch of the parallel portion i1, that through the right
branch i2, and call the current flowing through the bottom wire and into the
bottom of the battery i3. (Remark: in labeling, we don’t have to know the
actual direction of flow. We arbitrarily choose a direction to establish a sign
convention for the equations.)
i0
i3
i1 i2
The fact that i0 splits into i1 and i2, on application of Kirchhoff’s First Law,
gives that i1 + i2 = i0. Similarly, we have that i1 + i2 = i3. In the circuit that
loops out of the top of the battery, down the left branch of the parallel portion,
and back into the bottom of the battery, the voltage rise is 20 and the voltage
drop is i1 ·12, so Kirchoff’s Second Law gives that 12i1 = 20. In the circuit from
the battery to the right branch and back to the battery there is a voltage rise of
20 and a voltage drop of i2 ·8, so Kirchoff’s Second law gives that 8i2 = 20. And
finally, in the circuit that just loops around in the left and right branches of the
parallel portion (taken clockwise), there is a voltage rise of 0 and a voltage drop
of 8i2 − 12i1 so Kirchoff’s Second Law gives 8i2 − 12i1 = 0.
All of these equations taken together make this system.
i0 − i1 − i2 = 0
− i1 − i2 + i3 = 0
12i1 = 20
8i2 = 20
−12i1 + 8i2 = 0
The solution is i0 = 25/6, i1 = 5/3, i2 = 5/2, and i3 = 25/6 (all in amperes).
(Incidentally, this illustrates that redundant equations do arise in practice, since
the fifth equation here is redundant.)
Kirchhoff’s laws can be used to establish the electrical properties of networks
of great complexity. The next circuit has five resistors, wired in a combination
of series and parallel. It is said to be a series-parallel circuit.

10 volts
5 ohm
10 ohm
2 ohm
4 ohm
50 ohm
This circuit is a Wheatstone bridge. It is used to measure the resistance of an
component placed at, say, the location labeled 5 ohms, against known resistances
placed in the other positions (see Exercise 7). To analyze it, we can establish
the arrows in this way.
i0
i6
i1
i3
i2
i4
i5
Kirchoff’s First Law, applied to the top node, the left node, the right node, and
the bottom node gives these equations.
i0 = i1 + i2
i1 = i3 + i5
i2 + i5 = i4
i3 + i4 = i6
Kirchhoff’s Second Law, applied to the inside loop (i0-i1-i3-i6), the outside loop,
and the upper loop not involving the battery, gives these equations.
5i1 + 10i3 = 10
2i2 + 4i4 = 10
5i1 + 50i5 − 2i2 = 0
We could get more equations, but these are enough to produce a solution: i0 =
7/3, i1 = 2/3, i2 = 5/3, i3 = 2/3, i4 = 5/3, i5 = 0, and i6 = 7/3.
Networks of other kinds, not just electrical ones, can also be analyzed in this
way. For instance, a network of streets in given in the exercises.
Exercises
Hint: Most of the linear systems are large enough that they are best solved on a
computer.

1 Calculate the amperages in each part of each network.
(a) This is a relatively simple network.
9 volt
2 ohm
3 ohm
2 ohm
(b) Compare this one with the parallel case discussed above.
9 volt
2 ohm
3 ohm
2 ohm 2 ohm
(c) This is a reasonably complicated network.
9 volt
2 ohm
3 ohm
3 ohm 2 ohm
2 ohm
3 ohm
4 ohm
2 Kirchhoff’s laws can apply to a network of streets, as here. On Cape Cod, in
Massachusetts, there are many intersections that involve traffic circles like this
one.
North Ave
Pier Bvd
Main St
Assume the traffic is as below.
North Pier Main
into 100 150 25
out of 75 150 50

We can use Kirchhoff’s Law, that the flow into any intersection equals the flow
out, to establish some equations modeling how traffic flows work here.
(a) Label each of the three arcs of road in the circle with a variable. For each of
the three in-out intersections, get an equation describing the traffic flow at that
node.
(b) Solve that system.
3 This is a map of a network of streets. Below we will describe the flow of cars
into, and out of, this network.
west Winooski Ave
east
Shelburne St
Willow
Jay
Ln
The hourly flow of cars into this network’s entrances, and out of its exits can be
observed.
east Winooski west Winooski Willow Jay Shelburne
into 100 150 25 – 200
out of 125 150 50 25 125
(The total in must approximately equal the total out over a long period of time.)
Once inside the network, the traffic may proceed in different ways, perhaps
filling Willow and leaving Jay mostly empty, or perhaps flowing in some other
way. We can use Kirchhoff’s Law that the flow into any intersection equals the
flow out.
(a) Determine the restrictions on the flow inside this network of streets by setting
up a variable for each block, establishing the equations, and solving them. Notice
that some streets are one-way only. (Hint: this will not yield a unique solution,
since traffic can flow through this network in various ways. You should get at
least one free variable.)
(b) Suppose some construction is proposed for Winooski Avenue East between
Willow and Jay, so traffic on that block will be reduced. What is the least
amount of traffic flow that can be allowed on that block without disrupting the
hourly flow into and out of the network?
4 Calculate the amperages in this network with more than one voltage rise.
1.5 volt
10 ohm
5 ohm
2 ohm 3 volt
3 ohm
6 ohm
5 In the circuit with the 8 ohm and 12 ohm resistors in parallel, the electric current
away from and back to the battery was found to be 25/6 amperes. Thus, the

parallel pair can be said to be equivalent to a single resistor having a value of
20/(25/6) = 24/5 = 4.8 ohms.
(a) What is the equivalent resistance if the two resistors in parallel are 8 ohms
and 5 ohms? Has the equivalent resistance risen or fallen?
(b) What is the equivalent resistance if the two are both 8 ohms?
(c) Find the formula for the equivalent resistance R if the two resistors in parallel
are R1 ohms and R2 ohms.
(d) What is the formula for more than two resistors in parallel?
6 In the car dashboard example that begins the discussion, solve for these amper-
ages. Assume all resistances are 15 ohms.
(a) If the driver is stepping on the brakes, so the brake lights are on, and no
other circuit is closed.
(b) If all the switches are closed (suppose both the high beams and the low beams
rate 15 ohms).
7 Show that, in the Wheatstone Bridge, if r2r6 = r3r5 then i4 = 0. (The way this
device is used in practice is that an unknown resistance, say at r1, is compared
to three known resistances. At r3 is placed a meter that shows the current. The
known resistances are varied until the current is read as 0, and then from the above
equation the value of the resistor at r1 can be calculated.)

Chapter 2
Vector Spaces
The first chapter began by introducing Gauss’ method and finished with a fair
understanding, keyed on the Linear Combination Lemma, of how it finds the
solution set of a linear system. Gauss’ method systematically takes linear com-
binations of the rows. With that insight, we now move to a general study of
linear combinations.
We need a setting for this study. At times in the first chapter, we’ve com-
bined vectors from R2
, at other times vectors from R3
, and at other times vectors
from even higher-dimensional spaces. Thus, our first impulse might be to work
in Rn
, leaving n unspecified. This would have the advantage that any of the
results would hold for R2
and for R3
and for many other spaces, simultaneously.
But, if having the results apply to many spaces at once is advantageous then
sticking only to Rn
’s is overly restrictive. We’d like the results to also apply to
combinations of row vectors, as in the final section of the first chapter. We’ve
even seen some spaces that are not just a collection of all of the same-sized
column vectors or row vectors. For instance, we’ve seen a solution set of a
homogeneous system that is a plane, inside of R3
. This solution set is a closed
system in the sense that a linear combination of these solutions is also a solution.
But it is not just a collection of all of the three-tall column vectors; only some
of them are in this solution set.
We want the results about linear combinations to apply anywhere that linear
combinations are sensible. We shall call any such set a vector space. Our results,
instead of being phrased as “Whenever we have a collection in which we can
sensibly take linear combinations . . . ”, will be stated as “In any vector space
. . . ”.
Such a statement describes at once what happens in many spaces. The step
up in abstraction from studying a single space at a time to studying a class
of spaces can be hard to make. To understand its advantages, consider this
analogy. Imagine that the government made laws one person at a time: “Leslie
Jones can’t jay walk.” That would be a bad idea; statements have the virtue of
economy when they apply to many cases at once. Or, suppose that they ruled,
“Kim Ke must stop when passing the scene of an accident.” Contrast that with,
“Any doctor must stop when passing the scene of an accident.” More general
statements, in some ways, are clearer.
79

80 Chapter 2. Vector Spaces
2.I Definition of Vector Space
We shall study structures with two operations, an addition and a scalar multi-
plication, that are subject to some simple conditions. We will reflect more on
the conditions later, but on first reading notice how reasonable they are. For
instance, surely any operation that can be called an addition (e.g., column vec-
tor addition, row vector addition, or real number addition) will satisfy all the
conditions in (1) below.
2.I.1 Definition and Examples
1.1 Definition A vector space (over R) consists of a set V along with two
operations ‘+’ and ‘·’ such that
(1) if ~
v, ~
w ∈ V then their vector sum ~
v + ~
w is in V and
• ~
v + ~
w = ~
w + ~
v
• (~
v + ~
w) + ~
u = ~
v + (~
w + ~
u) (where ~
u ∈ V )
• there is a zero vector ~
0 ∈ V such that ~
v +~
0 = ~
v for all ~
v ∈ V
• each ~
v ∈ V has an additive inverse ~
w ∈ V such that ~
w + ~
v = ~
0
(2) if r, s are scalars (members of R) and ~
v, ~
w ∈ V then each scalar multiple
r · ~
v is in V and
• (r + s) · ~
v = r · ~
v + s · ~
v
• r · (~
v + ~
w) = r · ~
v + r · ~
w
• (rs) · ~
v = r · (s · ~
v)
• 1 · ~
v = ~
v.
1.2 Remark Because it involves two kinds of addition and two kinds of mul-
tiplication, that definition may seem confused. For instance, in ‘(r + s) · ~
v =
r · ~
v + s · ~
v ’, the first ‘+’ is the real number addition operator while the ‘+’ to
the right of the equals sign represents vector addition in the structure V . These
expressions aren’t ambiguous because, e.g., r and s are real numbers so ‘r + s’
can only mean real number addition.
The best way to go through the examples below is to check all of the con-
ditions in the definition. That check is written out in the first example. Use
it as a model for the others. Especially important are the two: ‘~
v + ~
w is in
V ’ and ‘r · ~
v is in V ’. These are the closure conditions. They specify that the
addition and scalar multiplication operations are always sensible — they must
be defined for every pair of vectors, and every scalar and vector, and the result
of the operation must be a member of the set (see Example 1.4).

Section I. Definition of Vector Space 81
1.3 Example The set R2
is a vector space if the operations ‘+’ and ‘·’ have
their usual meaning.
µ
x1
x2
¶
+
µ
y1
y2
¶
=
µ
x1 + y1
x2 + y2
¶
r ·
µ
x1
x2
¶
=
µ
rx1
rx2
¶
We shall check all of the conditions in the definition.
There are five conditions in item (1). First, for closure of addition, note that
for any v1, v2, w1, w2 ∈ R the result of the sum
µ
v1
v2
¶
+
µ
w1
w2
¶
=
µ
v1 + w1
v2 + w2
¶
is a column array with two real entries, and so is in R2
. Second, to show that
addition of vectors commutes, take all entries to be real numbers and compute
µ
v1
v2
¶
+
µ
w1
w2
¶
=
µ
v1 + w1
v2 + w2
¶
=
µ
w1 + v1
w2 + v2
¶
=
µ
w1
w2
¶
+
µ
v1
v2
¶
(the second equality follows from the fact that the components of the vectors
are real numbers, and the addition of real numbers is commutative). The third
condition, associativity of vector addition, is similar.
(
µ
v1
v2
¶
+
µ
w1
w2
¶
) +
µ
u1
u2
¶
=
µ
(v1 + w1) + u1
(v2 + w2) + u2
¶
=
µ
v1 + (w1 + u1)
v2 + (w2 + u2)
¶
=
µ
v1
v2
¶
+ (
µ
w1
w2
¶
+
µ
u1
u2
¶
)
For the fourth we must produce a zero element — the vector of zeroes is it.
µ
v1
v2
¶
+
µ
0
0
¶
=
µ
v1
v2
¶
Fifth, to produce an additive inverse, note that for any v1, v2 ∈ R we have
µ
−v1
−v2
¶
+
µ
v1
v2
¶
=
µ
0
0
¶
so the first vector is the desired additive inverse of the second.
The checks for the five conditions in item (2) are just as routine. First, for
closure under scalar multiplication, where r, v1, v2 ∈ R,
r ·
µ
v1
v2
¶
=
µ
rv1
rv2
¶
is a column array with two real entries, and so is in R2
. This checks the second
condition.
(r + s) ·
µ
v1
v2
¶
=
µ
(r + s)v1
(r + s)v2
¶
=
µ
rv1 + sv1
rv2 + sv2
¶
= r ·
µ
v1
v2
¶
+ s ·
µ
v1
v2
¶

For the third condition, that scalar multiplication distributes from the left over
vector addition, the check is also straightforward.
r · (
µ
v1
v2
¶
+
µ
w1
w2
¶
) =
µ
r(v1 + w1)
r(v2 + w2)
¶
=
µ
rv1 + rw1
rv2 + rw2
¶
= r ·
µ
v1
v2
¶
+ r ·
µ
w1
w2
¶
The fourth
(rs) ·
µ
v1
v2
¶
=
µ
(rs)v1
(rs)v2
¶
=
µ
r(sv1)
r(sv2)
¶
= r · (s ·
µ
v1
v2
¶
)
and fifth conditions are also easy.
1 ·
µ
v1
v2
¶
=
µ
1v1
1v2
¶
=
µ
v1
v2
¶
In a similar way, each Rn
is a vector space with the usual operations of
vector addition and scalar multiplication. (In R1
, we usually do not write the
members as column vectors, i.e., we usually do not write ‘(π)’. Instead we just
write ‘π’.)
1.4 Example This subset of R3
that is a plane through the origin
P = {


x
y
z


¯
¯ x + y + z = 0}
is a vector space if ‘+’ and ‘·’ are interpreted in this way.


x1
y1
z1

 +


x2
y2
z2

 =


x1 + x2
y1 + y2
z1 + z2

 r ·


x
y
z

 =


rx
ry
rz


The addition and scalar multiplication operations here are just the ones of R3
,
reused on its subset P. We say P inherits these operations from R3
. Here is a
typical addition in P.


1
1
−2

 +


−1
0
1

 =


0
1
−1


This illustrates that P is closed under addition. We’ve added two vectors from
P — that is, with the property that the sum of their three entries is zero —
and we’ve gotten a vector also in P. Of course, this example of closure is not
a proof of closure. To prove that P is closed under addition, take two elements
of P


x1
y1
z1

 ,


x2
y2
z2



(membership in P means that x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0), and
observe that their sum


x1 + x2
y1 + y2
z1 + z2


is also in P since (x1+x2)+(y1+y2)+(z1+z2) = (x1+y1+z1)+(x2+y2+z2) = 0.
To show that P is closed under scalar multiplication, start with a vector from
P


x
y
z


(so that x + y + z = 0), and then for r ∈ R observe that the scalar multiple
r ·


x
y
z

 =


rx
ry
rz


satisfies that rx + ry + rz = r(x + y + z) = 0. Thus the two closure conditions
are satisfied. The checks for the other conditions in the definition of a vector
space are just as easy.
1.5 Example Example 1.3 shows that the set of all two-tall vectors with real
entries is a vector space. Example 1.4 gives a subset of an Rn
that is also a
vector space. In contrast with those two, consider the set of two-tall columns
with entries that are integers (under the obvious operations). This is a subset
of a vector space, but it is not itself a vector space. The reason is that this set is
not closed under scalar multiplication, that is, it does not satisfy requirement (2)
in the definition. Here is a column with integer entries, and a scalar, such that
the outcome of the operation
0.5 ·
µ
4
3
¶
=
µ
2
1.5
¶
is not a member of the set, since its entries are not all integers.
1.6 Example The singleton set
{




0
0
0
0



}
is a vector space under the operations




0
0
0
0



 +




0
0
0
0



 =




0
0
0
0



 r ·




0
0
0
0



 =




0
0
0
0




that it inherits from R4
.

A vector space must have at least one element, its zero vector. Thus a
one-element vector space is the smallest one possible.
1.7 Definition A one-element vector space is a trivial space.
Warning! The examples so far involve sets of column vectors with the usual
operations. But vector spaces need not be collections of column vectors, or even
of row vectors. Below are some other types of vector spaces. The term ‘vector
space’ does not mean ‘collection of columns of reals’. It means something more
like ‘collection in which any linear combination is sensible’.
1.8 Example Consider P3 = {a0 + a1x + a2x2
+ a3x3
¯
¯ a0, . . . , a3 ∈ R}, the
set of polynomials of degree three or less (in this book, we’ll take constant
polynomials, including the zero polynomial, to be of degree zero). It is a vector
space under the operations
(a0 + a1x + a2x2
+ a3x3
) + (b0 + b1x + b2x2
+ b3x3
)
= (a0 + b0) + (a1 + b1)x + (a2 + b2)x2
+ (a3 + b3)x3
and
r · (a0 + a1x + a2x2
+ a3x3
) = (ra0) + (ra1)x + (ra2)x2
+ (ra3)x3
(the verification is easy). This vector space is worthy of attention because these
are the polynomial operations familiar from high school algebra. For instance,
3 · (1 − 2x + 3x2
− 4x3
) − 2 · (2 − 3x + x2
− (1/2)x3
) = −1 + 7x2
− 11x3
.
Although this space is not a subset of any Rn
, there is a sense in which we
can think of P3 as “the same” as R4
. If we identify these two spaces’s elements
in this way
a0 + a1x + a2x2
+ a3x3
corresponds to




a0
a1
a2
a3




then the operations also correspond. Here is an example of corresponding ad-
ditions.
1 − 2x + 0x2
+ 1x3
+ 2 + 3x + 7x2
− 4x3
3 + 1x + 7x2
− 3x3
corresponds to




1
−2
0
1



 +




2
3
7
−4



 =




3
1
7
−3




Things we are thinking of as “the same” add to “the same” sum. Chapter Three
makes precise this idea of vector space correspondence. For now we shall just
leave it as an intuition.

1.9 Example The set {f
¯
¯ f : N → R} of all real-valued functions of one nat-
ural number variable is a vector space under the operations
(f1 + f2) (n) = f1(n) + f2(n) (r · f) (n) = r f(n)
so that if, for example, f1(n) = n2
+ 2 sin(n) and f2(n) = − sin(n) + 0.5 then
(f1 + 2f2) (n) = n2
+ 1.
We can view this space as a generalization of Example 1.3 by thinking of
these functions as “the same” as infinitely-tall vectors:
n f(n) = n2
+ 1
0 1
1 2
2 5
3 10
.
.
.
.
.
.
corresponds to







1
2
5
10
.
.
.







with addition and scalar multiplication are component-wise, as before. (The
“infinitely-tall” vector can be formalized as an infinite sequence, or just as a
function from N to R, in which case the above correspondence is an equality.)
1.10 Example The set of polynomials with real coefficients
{a0 + a1x + · · · + anxn
¯
¯ n ∈ N and a0, . . . , an ∈ R}
makes a vector space when given the natural ‘+’
(a0 + a1x + · · · + anxn
) + (b0 + b1x + · · · + bnxn
)
= (a0 + b0) + (a1 + b1)x + · · · + (an + bn)xn
and ‘·’.
r · (a0 + a1x + . . . anxn
) = (ra0) + (ra1)x + . . . (ran)xn
This space differs from the space P3 of Example 1.8. This space contains not just
degree three polynomials, but degree thirty polynomials and degree three hun-
dred polynomials, too. Each individual polynomial of course is of a finite degree,
but the set has no single bound on the degree of all of its members.
This example, like the prior one, can be thought of in terms of infinite-tuples.
For instance, we can think of 1 + 3x + 5x2
as corresponding to (1, 3, 5, 0, 0, . . . ).
However, don’t confuse this space with the one from Example 1.9. Each member
of this set has a bounded degree, so under our correspondence there are no
elements from this space matching (1, 2, 5, 10, . . . ). The vectors in this space
correspond to infinite-tuples that end in zeroes.
1.11 Example The set {f
¯
¯ f : R → R} of all real-valued functions of one real
variable is a vector space under these.
(f1 + f2) (x) = f1(x) + f2(x) (r · f) (x) = r f(x)
The difference between this and Example 1.9 is the domain of the functions.

1.12 Example The set F = {a cos θ+b sin θ
¯
¯ a, b ∈ R} of real-valued functions
of the real variable θ is a vector space under the operations
(a1 cos θ + b1 sin θ) + (a2 cos θ + b2 sin θ) = (a1 + a2) cos θ + (b1 + b2) sin θ
and
r · (a cos θ + b sin θ) = (ra) cos θ + (rb) sin θ
inherited from the space in the prior example. (We can think of F as “the same”
as R2
in that a cos θ + b sin θ corresponds to the vector with components a and
b.)
1.13 Example The set
{f : R → R
¯
¯ d2
f
dx2
+ f = 0}
is a vector space under the, by now natural, interpretation.
(f + g) (x) = f(x) + g(x) (r · f) (x) = r f(x)
In particular, notice that closure is a consequence:
d2
(f + g)
dx2
+ (f + g) = (
d2
f
dx2
+ f) + (
d2
g
dx2
+ g)
and
d2
(rf)
dx2
+ (rf) = r(
d2
f
dx2
+ f)
of basic Calculus. This turns out to equal the space from the prior example —
functions satisfying this differential equation have the form a cos θ + b sin θ —
but this description suggests an extension to solutions sets of other differential
equations.
1.14 Example The set of solutions of a homogeneous linear system in n vari-
ables is a vector space under the operations inherited from Rn
. For closure
under addition, if
~
v =



v1
.
.
.
vn


 ~
w =



w1
.
.
.
wn



both satisfy the condition that their entries add to zero then ~
v + ~
w also satisfies
that condition: c1(v1 + w1) + · · · + cn(vn + wn) = (c1v1 + · · · + cnvn) + (c1w1 +
· · · + cnwn) = 0. The checks of the other conditions are just as routine.

As we’ve done in those equations, we often omit the multiplication symbol ‘·’.
We can distinguish the multiplication in ‘c1v1’ from that in ‘r~
v ’ since if both
multiplicands are real numbers then real-real multiplication must be meant,
while if one is a vector then scalar-vector multiplication must be meant.
The prior example has brought us full circle since it is one of our motivating
examples.
1.15 Remark Now, with some feel for the kinds of structures that satisfy the
definition of a vector space, we can reflect on that definition. For example, why
specify in the definition the condition that 1 · ~
v = ~
v but not a condition that
0 · ~
v = ~
0?
One answer is that this is just a definition — it gives the rules of the game
from here on, and if you don’t like it, put the book down and walk away.
Another answer is perhaps more satisfying. People in this area have worked
hard to develop the right balance of power and generality. This definition has
been shaped so that it contains the conditions needed to prove all of the inter-
esting and important properties of spaces of linear combinations, and so that it
does not contain extra conditions that only bar as examples spaces where those
properties occur. As we proceed, we shall derive all of the properties natural to
collections of linear combinations from the conditions given in the definition.
The next result is an example. We do not need to include these properties
in the definition of vector space because they follow from the properties already
listed there.
1.16 Lemma In any vector space V ,
(1) 0 · ~
v = ~
0
(2) (−1 · ~
v) + ~
v = ~
0
(3) r ·~
0 = ~
0
for any ~
v ∈ V and r ∈ R.
Proof. For the first item, note that ~
v = (1 + 0) · ~
v = ~
v + (0 · ~
v). Add to both
sides the additive inverse of ~
v, the vector ~
w such that ~
w + ~
v = ~
0.
~
w + ~
v = ~
w + ~
v + 0 · ~
v
~
0 = ~
0 + 0 · ~
v
~
0 = 0 · ~
v
The second item is easy: (−1 · ~
v) + ~
v = (−1 + 1) · ~
v = 0 · ~
v = ~
0 shows that
we can write ‘−~
v ’ for the additive inverse of ~
v without worrying about possible
confusion with (−1) · ~
v.
For the third one, this r ·~
0 = r · (0 ·~
0) = (r · 0) ·~
0 = ~
0 will do. QED
We finish this subsection with an recap, and a comment.

Chapter One studied Gaussian reduction. That lead us here to the study of
collections of linear combinations. We have named any such structure a ‘vector
space’. In a phrase, the point of this material is that vector spaces are the right
context in which to study linearity.
Finally, a comment. From the fact that it forms a whole chapter, and espe-
cially because that chapter is the first one, a reader could come to think that
the study of linear systems is our purpose. The truth is, we will not so much
use vector spaces in the study of linear systems as we will instead have linear
systems lead us into the study of vector spaces. The wide variety of examples
from this subsection shows that the study of vector spaces is interesting and im-
portant in its own right, aside from how it helps us understand linear systems.
Linear systems won’t go away. But from now on our primary objects of study
will be vector spaces.
Exercises
1.17 Give the zero vector from each of these vector spaces.
(a) The space of degree three polynomials under the natural operations
(b) The space of 2×4 matrices
(c) The space {f : [0..1] → R
¯
¯ f is continuous}
(d) The space of real-valued functions of one natural number variable
X 1.18 Find the additive inverse, in the vector space, of the vector.
(a) In P3, the vector −3 − 2x + x2
(b) In the space of 2×2 matrices with real number entries under the usual matrix
addition and scalar multiplication,
µ
1 −1
0 3
¶
(c) In {aex
+ be−x
¯
¯ a, b ∈ R}, a space of functions of the real variable x under
the natural operations, the vector 3ex
− 2e−x
.
X 1.19 Show that each of these is a vector space.
(a) The set of linear polynomials P1 = {a0 + a1x
¯
¯ a0, a1 ∈ R} under the usual
polynomial addition and scalar multiplication operations
(b) The set of 2×2 matrices with real entries under the usual matrix operations
(c) The set of three-component row vectors with their usual operations
(d) The set
L = {



x
y
z
w


 ∈ R4
¯
¯ x + y − z + w = 0}
under the operations inherited from R4
X 1.20 Show that each set is not a vector space. (Hint. Start by listing two members
of each set.)
(a) Under the operations inherited from R3
, this set
{
Ã
x
y
z
!
∈ R3
¯
¯ x + y + z = 1}

(b) Under the operations inherited from R3
, this set
{
Ã
x
y
z
!
∈ R3
¯
¯ x2
+ y2
+ z2
= 1}
(c) Under the usual matrix operations,
{
µ
a 1
b c
¶
¯
¯ a, b, c ∈ R}
(d) Under the usual polynomial operations,
{a0 + a1x + a2x2
¯
¯ a0, a1, a2 ∈ R+
}
where R+
is the set of reals greater than zero
(e) Under the inherited operations,
{
µ
x
y
¶
∈ R2
¯
¯ x + 3y = 4 and 2x − y = 3 and 6x + 4y = 10}
1.21 Define addition and scalar multiplication operations to make the complex
numbers a vector space over R.
X 1.22 Is the set of rational numbers a vector space over R under the usual addition
and scalar multiplication operations?
1.23 Show that the set of linear combinations of the variables x, y, z is a vector
space under the natural addition and scalar multiplication operations.
1.24 Prove that this is not a vector space: the set of two-tall column vectors with
real entries subject to these operations.
µ
x1
y1
¶
+
µ
x2
y2
¶
=
µ
x1 − x2
y1 − y2
¶
r ·
µ
x
y
¶
=
µ
rx
ry
¶
1.25 Prove or disprove that R3
is a vector space under these operations.
(a)
Ã
x1
y1
z1
!
+
Ã
x2
y2
z2
!
=
Ã
0
0
0
!
and r
Ã
x
y
z
!
=
Ã
rx
ry
rz
!
(b)
Ã
x1
y1
z1
!
+
Ã
x2
y2
z2
!
=
Ã
0
0
0
!
and r
Ã
x
y
z
!
=
Ã
0
0
0
!
X 1.26 For each, decide if it is a vector space; the intended operations are the natural
ones.
(a) The diagonal 2×2 matrices
{
µ
a 0
0 b
¶
¯
¯ a, b ∈ R}
(b) This set of 2×2 matrices
{
µ
x x + y
x + y y
¶
¯
¯ x, y ∈ R}
(c) This set
{



x
y
z
w


 ∈ R4
¯
¯ x + y + w = 1}

(d) The set of functions {f : R → R
¯
¯ df/dx + 2f = 0}
(e) The set of functions {f : R → R
¯
¯ df/dx + 2f = 1}
X 1.27 Prove or disprove that this is a vector space: the real-valued functions f of
one real variable such that f(7) = 0.
X 1.28 Show that the set R+
of positive reals is a vector space when ‘x + y’ is inter-
preted to mean the product of x and y (so that 2 + 3 is 6), and ‘r · x’ is interpreted
as the r-th power of x.
1.29 Is {(x, y)
¯
¯ x, y ∈ R} a vector space under these operations?
(a) (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) and r(x, y) = (rx, y)
(b) (x1, y1) + (x2, y2) = (x1 + x2, y1 + y2) and r · (x, y) = (rx, 0)
1.30 Prove or disprove that this is a vector space: the set of polynomials of degree
greater than or equal to two, along with the zero polynomial.
1.31 At this point “the same” is only an intuitive notion, but nonetheless for each
vector space identify the k for which the space is “the same” as Rk
.
(a) The 2×3 matrices under the usual operations
(b) The n×m matrices (under their usual operations)
(c) This set of 2×2 matrices
{
µ
a 0
b c
¶
¯
¯ a, b, c ∈ R}
(d) This set of 2×2 matrices
{
µ
a 0
b c
¶
¯
¯ a + b + c = 0}
X 1.32 Using ~
+ to represent vector addition and ~
· for scalar multiplication, restate
the definition of vector space.
X 1.33 Prove these.
(a) Any vector is the additive inverse of the additive inverse of itself.
(b) Vector addition left-cancels: if ~
v,~
s,~
t ∈ V then ~
v + ~
s = ~
v + ~
t implies that
~
s = ~
t.
1.34 The definition of vector spaces does not explicitly say that ~
0 + ~
v = ~
v (check
the order in which the summands appear). Show that it must nonetheless hold in
any vector space.
X 1.35 Prove or disprove that this is a vector space: the set of all matrices, under
the usual operations.
1.36 In a vector space every element has an additive inverse. Can some elements
have two or more?
1.37 (a) Prove that every point, line, or plane thru the origin in R3
is a vector
space under the inherited operations.
(b) What if it doesn’t contain the origin?
X 1.38 Using the idea of a vector space we can easily reprove that the solution set of
a homogeneous linear system has either one element or infinitely many elements.
Assume that ~
v ∈ V is not ~
0.
(a) Prove that r · ~
v = ~
0 if and only if r = 0.
(b) Prove that r1 · ~
v = r2 · ~
v if and only if r1 = r2.
(c) Prove that any nontrivial vector space is infinite.
(d) Use the fact that a nonempty solution set of a homogeneous linear system is
a vector space to draw the conclusion.

1.39 Is this a vector space under the natural operations: the real-valued functions
of one real variable that are differentiable?
1.40 A vector space over the complex numbers C has the same definition as a vector
space over the reals except that scalars are drawn from C instead of from R. Show
that each of these is a vector space over the complex numbers. (Recall how complex
numbers add and multiply: (a0 + a1i) + (b0 + b1i) = (a0 + b0) + (a1 + b1)i and
(a0 + a1i)(b0 + b1i) = (a0b0 − a1b1) + (a0b1 + a1b0)i.)
(a) The set of degree two polynomials with complex coefficients
(b) This set
{
µ
0 a
b 0
¶
¯
¯ a, b ∈ C and a + b = 0 + 0i}
1.41 Find a property shared by all of the Rn
’s not listed as a requirement for a
vector space.
X 1.42 (a) Prove that a sum of four vectors ~
v1, . . . ,~
v4 ∈ V can be associated in
any way without changing the result.
((~
v1 + ~
v2) + ~
v3) + ~
v4 = (~
v1 + (~
v2 + ~
v3)) + ~
v4
= (~
v1 + ~
v2) + (~
v3 + ~
v4)
= ~
v1 + ((~
v2 + ~
v3) + ~
v4)
= ~
v1 + (~
v2 + (~
v3 + ~
v4))
This allows us to simply write ‘~
v1 + ~
v2 + ~
v3 + ~
v4’ without ambiguity.
(b) Prove that any two ways of associating a sum of any number of vectors give
the same sum. (Hint. Use induction on the number of vectors.)
1.43 For any vector space, a subset that is itself a vector space under the inherited
operations (e.g., a plane through the origin inside of R3
) is a subspace.
(a) Show that {a0 + a1x + a2x2
¯
¯ a0 + a1 + a2 = 0} is a subspace of the vector
space of degree two polynomials.
(b) Show that this is a subspace of the 2×2 matrices.
{
µ
a b
c 0
¶
¯
¯ a + b = 0}
(c) Show that a nonempty subset S of a real vector space is a subspace if and only
if it is closed under linear combinations of pairs of vectors: whenever c1, c2 ∈ R
and ~
s1,~
s2 ∈ S then the combination c1~
v1 + c2~
v2 is in S.
2.I.2 Subspaces and Spanning Sets
One of the examples that led us to introduce the idea of a vector space
was the solution set of a homogeneous system. For instance, we’ve seen in
Example 1.4 such a space that is a planar subset of R3
. There, the vector space
R3
contains inside it another vector space, the plane.
2.1 Definition For any vector space, a subspace is a subset that is itself a
vector space, under the inherited operations.

2.2 Example The plane from the prior subsection,
P = {


x
y
z


¯
¯ x + y + z = 0}
is a subspace of R3
. As specified in the definition, the operations are the ones
that are inherited from the larger space, that is, vectors add in P3 as they add
in R3


x1
y1
z1

 +


x2
y2
z2

 =


x1 + x2
y1 + y2
z1 + z2


and scalar multiplication is also the same as it is in R3
. To show that P is a
subspace, we need only note that it is a subset and then verify that it is a space.
Checking that P satisfies the conditions in the definition of a vector space is
routine. For instance, for closure under addition, just note that if the summands
satisfy that x1 + y1 + z1 = 0 and x2 + y2 + z2 = 0 then the sum satisfies that
(x1 + x2) + (y1 + y2) + (z1 + z2) = (x1 + y1 + z1) + (x2 + y2 + z2) = 0.
2.3 Example The x-axis in R2
is a subspace where the addition and scalar
multiplication operations are the inherited ones.
µ
x1
0
¶
+
µ
x2
0
¶
=
µ
x1 + x2
0
¶
r ·
µ
x
0
¶
=
µ
rx
0
¶
As above, to verify that this is a subspace, we simply note that it is a subset
and then check that it satisfies the conditions in definition of a vector space.
For instance, the two closure conditions are satisfied: (1) adding two vectors
with a second component of zero results in a vector with a second component
of zero, and (2) multiplying a scalar times a vector with a second component of
zero results in a vector with a second component of zero.
2.4 Example Another subspace of R2
is
{
µ
0
0
¶
}
its trivial subspace.
Any vector space has a trivial subspace {~
0 }. At the opposite extreme, any
vector space has itself for a subspace. These two are the improper subspaces.
Other subspaces are proper.
2.5 Example The condition in the definition requiring that the addition and
scalar multiplication operations must be the ones inherited from the larger space
is important. Consider the subset {1} of the vector space R1
. Under the opera-
tions 1+1 = 1 and r·1 = 1 that set is a vector space, specifically, a trivial space.
But it is not a subspace of R1
because those aren’t the inherited operations, since
of course R1
has 1 + 1 = 2.

2.6 Example All kinds of vector spaces, not just Rn
’s, have subspaces. The
vector space of cubic polynomials {a + bx + cx2
+ dx3
¯
¯ a, b, c, d ∈ R} has a sub-
space comprised of all linear polynomials {m + nx
¯
¯ m, n ∈ R}.
2.7 Example Another example of a subspace not taken from an Rn
is one
from the examples following the definition of a vector space. The space of all
real-valued functions of one real variable f : R → R has a subspace of functions
satisfying the restriction (d2
f/dx2
) + f = 0.
2.8 Example Being vector spaces themselves, subspaces must satisfy the clo-
sure conditions. The set R+
is not a subspace of the vector space R1
because
with the inherited operations it is not closed under scalar multiplication: if
~
v = 1 then −1 · ~
v 6∈ R+
.
The next result says that Example 2.8 is prototypical. The only way that a
subset can fail to be a subspace (if it is nonempty and the inherited operations
are used) is if it isn’t closed.
2.9 Lemma For a nonempty subset S of a vector space, under the inherited
operations, the following are equivalent statements.∗
(1) S is a subspace of that vector space
(2) S is closed under linear combinations of pairs of vectors: for any vectors
~
s1,~
s2 ∈ S and scalars r1, r2 the vector r1~
s1 + r2~
s2 is in S
(3) S is closed under linear combinations of any number of vectors: for any
vectors ~
s1, . . . ,~
sn ∈ S and scalars r1, . . . , rn the vector r1~
s1 + · · · + rn~
sn is
in S.
Briefly, the way that a subset gets to be a subspace is by being closed under
linear combinations.
Proof. ‘The following are equivalent’ means that each pair of statements are
equivalent.
(1) ⇐⇒ (2) (2) ⇐⇒ (3) (3) ⇐⇒ (1)
We will show this equivalence by establishing that (1) =⇒ (3) =⇒ (2) =⇒
(1). This strategy is suggested by noticing that (1) =⇒ (3) and (3) =⇒ (2)
are easy and so we need only argue the single implication (2) =⇒ (1).
For that argument, assume that S is a nonempty subset of a vector space V
and that S is closed under combinations of pairs of vectors. We will show that
S is a vector space by checking the conditions.
The first item in the vector space definition has five conditions. First, for
closure under addition, if ~
s1,~
s2 ∈ S then ~
s1 +~
s2 ∈ S, as ~
s1 +~
s2 = 1 ·~
s1 + 1 ·~
s2.
Second, for any ~
s1,~
s2 ∈ S, because addition is inherited from V , the sum ~
s1 +~
s2
in S equals the sum ~
s1 +~
s2 in V , and that equals the sum ~
s2 +~
s1 in V (because
V is a vector space, its addition is commutative), and that in turn equals the
sum ~
s2 +~
s1 in S. The argument for the third condition is similar to that for the
∗ More information on equivalence of statements is in the appendix.

second. For the fourth, consider the zero vector of V and note that closure of S
under linear combinations of pairs of vectors gives that (where ~
s is any member
of the nonempty set S) 0 · ~
s + 0 · ~
s = ~
0 is in S; showing that ~
0 acts under the
inherited operations as the additive identity of S is easy. The fifth condition is
satisfied because for any ~
s ∈ S, closure under linear combinations shows that
the vector 0 · ~
0 + (−1) · ~
s is in S; showing that it is the additive inverse of ~
s
under the inherited operations is routine.
The checks for item (2) are similar and are saved for Exercise 32. QED
We usually show that a subset is a subspace with (2) =⇒ (1).
2.10 Remark At the start of this chapter we introduced vector spaces as col-
lections in which linear combinations are “sensible”. The above result speaks
to this.
The vector space definition has ten conditions but eight of them, the ones
stated there with the ‘•’ bullets, simply ensure that referring to the operations
as an ‘addition’ and a ‘scalar multiplication’ is sensible. The proof above checks
that if the nonempty set S satisfies statement (2) then inheritance of the oper-
ations from the surrounding vector space brings with it the inheritance of these
eight properties also (i.e., commutativity of addition in S follows right from
commutativity of addition in V ). So, in this context, this meaning of “sensible”
is automatically satisfied.
In assuring us that this first meaning of the word is met, the result draws
our attention to the second meaning. It has to do with the two remaining
conditions, the closure conditions. Above, the two separate closure conditions
inherent in statement (1) are combined in statement (2) into the single condition
of closure under all linear combinations of two vectors, which is then extended
in statement (3) to closure under combinations of any number of vectors. The
latter two statements say that we can always make sense of an expression like
r1~
s1 + r2~
s2, without restrictions on the r’s — such expressions are “sensible” in
that the vector described is defined and is in the set S.
This second meaning suggests that a good way to think of a vector space
is as a collection of unrestricted linear combinations. The next two examples
take some spaces and describe them in this way. That is, in these examples we
paramatrize, just as we did in Chapter One to describe the solution set of a
homogeneous linear system.
2.11 Example This subset of R3
S = {


x
y
z


¯
¯ x − 2y + z = 0}
is a subspace under the usual addition and scalar multiplication operations of
column vectors (the check that it is nonempty and closed under linear combi-
nations of two vectors is just like the one in Example 2.2). To paramatrize, we

can take x − 2y + z = 0 to be a one-equation linear system and expressing the
leading variable in terms of the free variables x = 2y − z.
S = {


2y − z
y
z


¯
¯ y, z ∈ R} = {y


2
1
0

 + z


−1
0
1


¯
¯ y, z ∈ R}
Now the subspace is described as the collection of unrestricted linear combi-
nations of those two vectors. Of course, in either description, this is a plane
through the origin.
2.12 Example This is a subspace of the 2×2 matrices
L = {
µ
a 0
b c
¶
¯
¯ a + b + c = 0}
(checking that it is nonempty and closed under linear combinations is easy). To
paramatrize, express the condition as a = −b − c.
L = {
µ
−b − c 0
b c
¶
¯
¯ b, c ∈ R} = {b
µ
−1 0
1 0
¶
+ c
µ
−1 0
0 1
¶
¯
¯ b, c ∈ R}
As above, we’ve described the subspace as a collection of unrestricted linear
combinations (by coincidence, also of two elements).
Paramatrization is an easy technique, but it is important. We shall use it
often.
2.13 Definition The span (or linear closure) of a nonempty subset S of a
vector space is the set of all linear combinations of vectors from S.
[S] = {c1~
s1 + · · · + cn~
sn
¯
¯ c1, . . . , cn ∈ R and ~
s1, . . . ,~
sn ∈ S}
The span of the empty subset of a vector space is the trivial subspace.
No notation for the span is completely standard. The square brackets used here
are common, but so are ‘span(S)’ and ‘sp(S)’.
2.14 Remark In Chapter One, after we showed that the solution set of a
homogeneous linear system can written as {c1
~
β1 + · · · + ck
~
βk
¯
¯ c1, . . . , ck ∈ R},
we described that as the set ‘generated’ by the ~
β’s. We now have the technical
term; we call that the ‘span’ of the set {~
β1, . . . , ~
βk}.
Recall also the discussion of the “tricky point” in that proof. The span of
the empty set is defined to be the set {~
0} because we follow the convention that
a linear combination of no vectors sums to ~
0. Besides, defining the empty set’s
span to be the trivial subspace is a convienence in that it keeps results like the
next one from having annoying exceptional cases.
2.15 Lemma In a vector space, the span of any subset is a subspace.

Proof. Call the subset S. If S is empty then by definition its span is the trivial
subspace. If S is not empty then by Lemma 2.9 we need only check that the
span [S] is closed under linear combinations. For a pair of vectors from that
span, ~
v = c1~
s1 +· · ·+cn~
sn and ~
w = cn+1~
sn+1 +· · ·+cm~
sm, a linear combination
p · (c1~
s1 + · · · + cn~
sn) + r · (cn+1~
sn+1 + · · · + cm~
sm)
= pc1~
s1 + · · · + pcn~
sn + rcn+1~
sn+1 + · · · + rcm~
sm
(p, r scalars) is a linear combination of elements of S and so is in [S] (possibly
some of the ~
si’s forming ~
v equal some of the ~
sj’s from ~
w, but it does not
matter). QED
The converse of the lemma holds: any subspace is the span of some set,
because a subspace is obviously the span of the set of its members. Thus a
subset of a vector space is a subspace if and only if it is a span. This fits the
intuition that a good way to think of a vector space is as a collection in which
linear combinations are sensible.
Taken together, Lemma 2.9 and Lemma 2.15 show that the span of a subset
S of a vector space is the smallest subspace containing all the members of S.
2.16 Example In any vector space V , for any vector ~
v, the set {r · ~
v
¯
¯ r ∈ R}
is a subspace of V . For instance, for any vector ~
v ∈ R3
, the line through the
origin containing that vector, {k~
v
¯
¯ k ∈ R} is a subspace of R3
. This is true even
when ~
v is the zero vector, in which case the subspace is the degenerate line, the
trivial subspace.
2.17 Example The span of this set is all of R2
.
{
µ
1
1
¶
,
µ
1
−1
¶
}
Tocheck this we must show that any member of R2
is a linear combination of
these two vectors. So we ask: for which vectors (with real components x and y)
are there scalars c1 and c2 such that this holds?
c1
µ
1
1
¶
+ c2
µ
1
−1
¶
=
µ
x
y
¶
Gauss’ method
c1 + c2 = x
c1 − c2 = y
−ρ1+ρ2
−→
c1 + c2 = x
−2c2 = −x + y
with back substitution gives c2 = (x − y)/2 and c1 = (x + y)/2. These two
equations show that for any x and y that we start with, there are appropriate
coefficients c1 and c2 making the above vector equation true. For instance, for
x = 1 and y = 2 the coefficients c2 = −1/2 and c1 = 3/2 will do. That is, any
vector in R2
can be written as a linear combination of the two given vectors.

Since spans are subspaces, and we know that a good way to understand a
subspace is to paramatrize its description, we can try to understand a set’s span
in that way.
2.18 Example Consider, in P2, the span of the set {3x − x2
, 2x}. By the
definition of span, it is the subspace of unrestricted linear combinations of the
two {c1(3x − x2
) + c2(2x)
¯
¯ c1, c2 ∈ R}. Clearly polynomials in this span must
have a constant term of zero. Is that necessary condition also sufficient?
We are asking: for which members a2x2
+ a1x + a0 of P2 are there c1 and c2
such that a2x2
+ a1x + a0 = c1(3x − x2
) + c2(2x)? Since polynomials are equal
if and only if their coefficients are equal, we are looking for conditions on a2,
a1, and a0 satisfying these.
−c1 = a2
3c1 + 2c2 = a1
0 = a0
Gauss’ method gives that c1 = −a2, c2 = (3/2)a2 + (1/2)a1, and 0 = a0. Thus
the only condition on polynomials in the span is the condition that we knew
of — as long as a0 = 0, we can give appropriate coefficients c1 and c2 to describe
the polynomial a0 + a1x + a2x2
as in the span. For instance, for the polynomial
0 − 4x + 3x2
, the coefficients c1 = −3 and c2 = 5/2 will do. So the span of the
given set is {a1x + a2x2
¯
¯ a1, a2 ∈ R}.
This shows, incidentally, that the set {x, x2
} also spans this subspace. A
space can have more than one spanning set. Two other sets spanning this sub-
space are {x, x2
, −x + 2x2
} and {x, x + x2
, x + 2x2
, . . . }. (Naturally, we usually
prefer to work with spanning sets that have only a few members.)
2.19 Example These are the subspaces of R3
that we now know of, the trivial
subspace, the lines through the origin, the planes through the origin, and the
whole space (of course, the picture shows only a few of the infinitely many
subspaces). In the next section we will prove that R3
has no other type of
subspaces, so in fact this picture shows them all.
{x
³1
0
0
´
+ y
³0
1
0
´
+ z
³0
0
1
´
}
»
»
»
»
»
»
»
»
»
»
{x
³1
0
0
´
+ y
³0
1
0
´
}
³
³
³
³
³
³
{x
³1
0
0
´
+ z
³0
0
1
´
}
¡
¡
{x
³1
1
0
´
+ z
³0
0
1
´
} . . .
¤
¤
³
³
³
³
³
{x
³1
0
0
´
}
A
A
{y
³0
1
0
´
}
H
HHH
{y
³2
1
0
´
}
¡
¡
{y
³1
1
1
´
} . . .
XXXXXXXXXXXX
PPPPPPPP
H
HH
H
H
@
@
{
³0
0
0
´
}

The subsets are described as spans of sets, using a minimal number of members,
and are shown connected to their supersets. Note that these subspaces fall
naturally into levels — planes on one level, lines on another, etc. — according
to how many vectors are in a minimal-sized spanning set.
So far in this chapter we have seen that to study the properties of linear
combinations, the right setting is a collection that is closed under these com-
binations. In the first subsection we introduced such collections, vector spaces,
and we saw a great variety of examples. In this subsection we saw still more
spaces, ones that happen to be subspaces of others. In all of the variety we’ve
seen a commonality. Example 2.19 above brings it out: vector spaces and sub-
spaces are best understood as a span, and especially as a span of a small number
of vectors. The next section studies spanning sets that are minimal.
Exercises
X 2.20 Which of these subsets of the vector space of 2×2 matrices are subspaces
under the inherited operations? For each one that is a subspace, paramatrize its
description. For each that is not, give a condition that fails.
(a) {
µ
a 0
0 b
¶
¯
¯ a, b ∈ R}
(b) {
µ
a 0
0 b
¶
¯
¯ a + b = 0}
(c) {
µ
a 0
0 b
¶
¯
¯ a + b = 5}
(d) {
µ
a c
0 b
¶
¯
¯ a + b = 0, c ∈ R}
X 2.21 Is this a subspace of P2: {a0 + a1x + a2x2
¯
¯ a0 + 2a1 + a2 = 4}? If so, para-
matrize its description.
X 2.22 Decide if the vector lies in the span of the set, inside of the space.
(a)
Ã
2
0
1
!
, {
Ã
1
0
0
!
,
Ã
0
0
1
!
}, in R3
(b) x − x3
, {x2
, 2x + x2
, x + x3
}, in P3
(c)
µ
0 1
4 2
¶
, {
µ
1 0
1 1
¶
,
µ
2 0
2 3
¶
}, in M2
×
2
2.23 Which of these are members of the span [{cos2
x, sin2
x}] in the vector space
of real-valued functions of one real variable?
(a) f(x) = 1 (b) f(x) = 3 + x2
(c) f(x) = sin x (d) f(x) = cos(2x)
X 2.24 Which of these sets spans R3
? That is, which of these sets has the property
that any three-tall vector can be expressed as a suitable linear combination of the
set’s elements?
(a) {
Ã
1
0
0
!
,
Ã
0
2
0
!
,
Ã
0
0
3
!
} (b) {
Ã
2
0
1
!
,
Ã
1
1
0
!
,
Ã
0
0
1
!
} (c) {
Ã
1
1
0
!
,
Ã
3
0
0
!
}
(d) {
Ã
1
0
1
!
,
Ã
3
1
0
!
,
Ã
−1
0
0
!
,
Ã
2
1
5
!
} (e) {
Ã
2
1
1
!
,
Ã
3
0
1
!
,
Ã
5
1
2
!
,
Ã
6
0
2
!
}

X 2.25 Paramatrize each subspace’s description. Then express each subspace as a
span.
(a) The subset {a + bx + cx3
¯
¯ a − 2b + c = 0} of P3
(b) The subset {
¡
a b c
¢ ¯
¯ a − c = 0} of the three-wide row vectors
(c) This subset of M2
×
2
{
µ
a b
c d
¶
¯
¯ a + d = 0}
(d) This subset of M2
×
2
{
µ
a b
c d
¶
¯
¯ 2a − c − d = 0 and a + 3b = 0}
(e) The subset of P2 of quadratic polynomials p such that p(7) = 0
X 2.26 Find a set to span the given subspace of the given space. (Hint. Paramatrize
each.)
(a) the xz-plane in R3
(b) {
Ã
x
y
z
!
¯
¯ 3x + 2y + z = 0} in R3
(c) {



x
y
z
w



¯
¯ 2x + y + w = 0 and y + 2z = 0} in R4
(d) {a0 + a1x + a2x2
+ a3x3
¯
¯ a0 + a1 = 0 and a2 − a3 = 0} in P3
(e) The set P4 in the space P4
(f) M2
×
2 in M2
×
2
2.27 Is R2
a subspace of R3
?
X 2.28 Decide if each is a subspace of the vector space of real-valued functions of one
real variable.
(a) The even functions {f : R → R
¯
¯ f(−x) = f(x) for all x}. For example, two
members of this set are f1(x) = x2
and f2(x) = cos(x).
(b) The odd functions {f : R → R
¯
¯ f(−x) = −f(x) for all x}. Two members are
f3(x) = x3
and f4(x) = sin(x).
2.29 Example 2.16 says that for any vector ~
v in any vector space V , the set
{r · ~
v
¯
¯ r ∈ R} is a subspace of V . (This is of course, simply the span of the
singleton set {~
v}.) Must any such subspace be a proper subspace, or can it be
improper?
2.30 An example following the definition of a vector space shows that the solution
set of a homogeneous linear system is a vector space. In the terminology of this
subsection, it is a subspace of Rn
where the system has n variables. What about
a non-homogeneous linear system; do its solutions form a subspace (under the
inherited operations)?
2.31 Example 2.19 shows that R3
has infinitely many subspaces. Does every non-
trivial space have infinitely many subspaces?
2.32 Finish the proof of Lemma 2.9.
2.33 Show that each vector space has only one trivial subspace.
X 2.34 Show that for any subset S of a vector space, the span of the span equals the
span [[S]] = [S]. (Hint. Members of [S] are linear combinations of members of

S. Members of [[S]] are linear combinations of linear combinations of members of
S.)
2.35 All of the subspaces that we’ve seen use zero in their description in some
way. For example, the subspace in Example 2.3 consists of all the vectors from R2
with a second component of zero. In contrast, the collection of vectors from R2
with a second component of one does not form a subspace (it is not closed under
scalar multiplication). Another example is Example 2.2, where the condition on
the vectors is that the three components add to zero. If the condition were that the
three components add to ong then it would not be a subspace (again, it would fail
to be closed). This exercise shows that a reliance on zero is not strictly necessary.
Consider the set
{
Ã
x
y
z
!
¯
¯ x + y + z = 1}
under these operations.
Ã
x1
y1
z1
!
+
Ã
x2
y2
z2
!
=
Ã
x1 + x2 − 1
y1 + y2
z1 + z2
!
r
Ã
x
y
z
!
=
Ã
rx − r + 1
ry
rz
!
(a) Show that it is not a subspace of R3
. (Hint. See Example 2.5).
(b) Show that it is a vector space. Note that by the prior item, Lemma 2.9 can
not apply.
(c) Show that any subspace of R3
must pass thru the origin, and so any subspace
of R3
must involve zero in its description. Does the converse hold? Does any
subset of R3
that contains the origin become a subspace when given the inherited
operations?
2.36 We can give a justification for the convention that the sum of no vectors equals
the zero vector. Consider this sum of three vectors ~
v1 + ~
v2 + ~
v3.
(a) What is the difference between this sum of three vectors and the sum of the
first two of this three?
(b) What is the difference between the prior sum and the sum of just the first
one vector?
(c) What should be the difference between the prior sum of one vector and the
sum of no vectors?
(d) So what should be the definition of the sum of no vectors?
2.37 Is a space determined by its subspaces? That is, if two vector spaces have the
same subspaces, must the two be equal?
2.38 (a) Give a set that is closed under scalar multiplication but not addition.
(b) Give a set closed under addition but not scalar multiplication.
(c) Give a set closed under neither.
2.39 Show that the span of a set of vectors does not depend on the order in which
the vectors are listed in that set.
2.40 Which trivial subspace is the span of the empty set? Is it
{
Ã
0
0
0
!
} ⊆ R3
, or {0 + 0x} ⊆ P1,
or some other subspace?
2.41 Show that if a vector is in the span of a set then adding that vector to the set
won’t make the span any bigger. Is that also ‘only if’?

X 2.42 Subspaces are subsets and so we naturally consider how ‘is a subspace of’
interacts with the usual set operations.
(a) If A, B are subspaces of a vector space, must A ∩ B be a subspace? Always?
Sometimes? Never?
(b) Must A ∪ B be a subspace?
(c) If A is a subspace, must its complement be a subspace?
(Hint. Try some test subspaces from Example 2.19.)
X 2.43 Does the span of a set depend on the enclosing space? That is, if W is a
subspace of V and S is a subset of W (and so also a subset of V ), might the span
of S in W differ from the span of S in V ?
2.44 Is the relation ‘is a subspace of’ transitive? That is, if V is a subspace of W
and W is a subspace of X, must V be a subspace of X?
X 2.45 Because ‘span of’ is an operation on sets we naturally consider how it interacts
with the usual set operations.
(a) If S ⊆ T are subsets of a vector space, is [S] ⊆ [T]? Always? Sometimes?
Never?
(b) If S, T are subsets of a vector space, is [S ∪ T] = [S] ∪ [T]?
(c) If S, T are subsets of a vector space, is [S ∩ T] = [S] ∩ [T]?
(d) Is the span of the complement equal to the complement of the span?
2.46 Reprove Lemma 2.15 without doing the empty set separately.
2.47 Find a structure that is closed under linear combinations, and yet is not a
vector space. (Remark. This is a bit of a trick question.)

2.II Linear Independence
The prior section shows that a vector space can be understood as an unrestricted
linear combination of some of its elements — that is, as a span. For example,
the space of linear polynomials {a + bx
¯
¯ a, b ∈ R} is spanned by the set {1, x}.
The prior section also showed that a space can have many sets that span it.
The space of linear polynomials is also spanned by {1, 2x} and {1, x, 2x}.
At the end of that section we described some spanning sets as ‘minimal’,
but we never precisely defined that word. We could take ‘minimal’ to mean one
of two things. We could mean that a spanning set is minimal if it contains the
smallest number of members of any set with the same span. With this meaning
{1, x, 2x} is not minimal because it has one member more than the other two.
Or we could mean that a spanning set is minimal when it has no elements that
can be removed without changing the span. Under this meaning {1, x, 2x} is not
minimal because removing the 2x and getting {1, x} leaves the span unchanged.
The first sense of minimality appears to be a global requirement, in that to
check if a spanning set is minimal we seemingly must look at all the spanning sets
of a subspace and find one with the least number of elements. The second sense
of minimality is local in that we need to look only at the set under discussion
and consider the span with and without various elements. For instance, using
the second sense, we could compare the span of {1, x, 2x} with the span of {1, x}
and note that the 2x is a “repeat” in that its removal doesn’t shrink the span.
In this section we will use the second sense of ‘minimal spanning set’ because
of this technical convenience. However, the most important result of this book
is that the two senses coincide; we will prove that in the section after this one.
2.II.1 Definition and Examples
We first characterize when a vector can be removed from a set without
changing its span.
1.1 Lemma Where S is a subset of a vector space,
[S] = [S ∪ {~
v}] if and only if ~
v ∈ [S]
for any ~
v in that space.
Proof. The left to right implication is easy. If [S] = [S ∪ {~
v}] then, since
obviously ~
v ∈ [S ∪ {~
v}], the equality of the two sets gives that ~
v ∈ [S].
For the right to left implication assume that ~
v ∈ [S] to show that [S] = [S ∪
{~
v}] by mutual inclusion. The inclusion [S] ⊆ [S ∪{~
v}] is obvious. For the other
inclusion [S] ⊇ [S∪{~
v}], write an element of [S∪{~
v}] as d0~
v+d1~
s1 +· · ·+dm~
sm,
and substitute ~
v’s expansion as a linear combination of members of the same set
d0(c0~
t0 + · · · + ck~
tk) + d1~
s1 + · · · + dm~
sm. This is a linear combination of linear
combinations, and so after distributing d0 we end with a linear combination of
vectors from S. Hence each member of [S ∪ {~
v}] is also a member of [S]. QED

Section II. Linear Independence 103
1.2 Example In R3
, where
~
v1 =


1
0
0

 , ~
v2 =


0
1
0

 , ~
v3 =


2
1
0


the spans [{~
v1,~
v2}] and [{~
v1,~
v2,~
v3}] are equal since ~
v3 is in the span [{~
v1,~
v2}].
The lemma says that if we have a spanning set then we can remove a ~
v to
get a new set S with the same span if and only if ~
v is a linear combination of
vectors from S. Thus, under the second sense described above, a spanning set
is minimal if and only if it contains no vectors that are linear combinations of
the others in that set. We have a term for this important property.
1.3 Definition A subset of a vector space is linearly independent if none of its
elements is a linear combination of the others. Otherwise it is linearly dependent.
Here is a small but useful observation: although this way of writing one
vector as a combination of the others
~
s0 = c1~
s1 + c2~
s2 + · · · + cn~
sn
visually sets ~
s0 off from the other vectors, algebraically there is nothing special
in that equation about ~
s0. For any ~
si with a coefficient ci that is nonzero, we
can rewrite the relationship to set off ~
si.
~
si = (1/ci)~
s0 + (−c1/ci)~
s1 + · · · + (−cn/ci)~
sn
When we don’t want to single out any vector by writing it alone on one side of
the equation we will instead say that ~
s0,~
s1, . . . ,~
sn are in a linear relationship and
write the relationship with all of the vectors on the same side. The next result
rephrases the linear independence definition in this style. It gives what is usually
the easiest way to compute whether a finite set is dependent or independent.
1.4 Lemma A subset S of a vector space is linearly independent if and only if
for any distinct ~
s1, . . . ,~
sn ∈ S the only linear relationship among those vectors
c1~
s1 + · · · + cn~
sn = ~
0 c1, . . . , cn ∈ R
is the trivial one: c1 = 0, . . . , cn = 0.
Proof. This is a direct consequence of the observation above.
If the set S is linearly independent then no vector ~
si can be written as a linear
combination of the other vectors from S so there is no linear relationship where
some of the ~
s ’s have nonzero coefficients. If S is not linearly independent then
some ~
si is a linear combination ~
si = c1~
s1 +· · ·+ci−1~
si−1 +ci+1~
si+1 +· · ·+cn~
sn of
other vectors from S, and subtracting ~
si from both sides of that equation gives
a linear relationship involving a nonzero coefficient, namely the −1 in front of
~
si. QED

1.5 Example In the vector space of two-wide row vectors, the two-element set
{
¡
40 15
¢
,
¡
−50 25
¢
} is linearly independent. To check this, set
c1 ·
¡
40 15
¢
+ c2 ·
¡
−50 25
¢
=
¡
0 0
¢
and solve the resulting system.
40c1 − 50c2 = 0
15c1 + 25c2 = 0
−(15/40)ρ1+ρ2
−→
40c1 − 50c2 = 0
(175/4)c2 = 0
Therefore c1 and c2 both equal zero and so the only linear relationship between
the two given row vectors is the trivial relationship.
In the same vector space, {
¡
40 15
¢
,
¡
20 7.5
¢
} is linearly dependent since
we can satisfy
c1
¡
40 15
¢
+ c2 ·
¡
20 7.5
¢
=
¡
0 0
¢
with c1 = 1 and c2 = −2.
1.6 Remark Recall the Statics example that began this book. We first set the
unknown-mass objects at 40 cm and 15 cm and got a balance, and then we set
the objects at −50 cm and 25 cm and got a balance. With those two pieces of
information we could compute values of the unknown masses. Had we instead
first set the unknown-mass objects at 40 cm and 15 cm, and then at 20 cm and
7.5 cm, we would not have been able to compute the values of the unknown
masses (try it). Intuitively, the problem is that the
¡
20 7.5
¢
information is a
“repeat” of the
¡
40 15
¢
information — that is,
¡
20 7.5
¢
is in the span of the
set {
¡
40 15
¢
} — and so we would be trying to solve a two-unknowns problem
with what is essentially one piece of information.
1.7 Example The set {1 + x, 1 − x} is linearly independent in P2, the space
of quadratic polynomials with real coefficients, because
0 + 0x + 0x2
= c1(1 + x) + c2(1 − x) = (c1 + c2) + (c1 − c2)x + 0x2
gives
c1 + c2 = 0
c1 − c2 = 0
−ρ1+ρ2
−→
c1 + c2 = 0
2c2 = 0
(since polynomials are equal only if their coefficients are equal). Thus, the only
linear relationship between these two members of P2 is the trivial one.
1.8 Example In R3
, where
~
v1 =


3
4
5

 , ~
v2 =


2
9
2

 , ~
v3 =


4
18
4



the set S = {~
v1,~
v2,~
v3} is linearly dependent because this is a relationship
0 · ~
v1 + 2 · ~
v2 − 1 · ~
v3 = ~
0
where not all of the scalars are zero (the fact that some scalars are zero is
irrelevant).
1.9 Remark That example shows why, although Definition 1.3 is a clearer
statement of what independence is, Lemma 1.4 is more useful for computations.
Working straight from the definition, someone trying to compute whether S is
linearly independent would start by setting ~
v1 = c2~
v2 +c3~
v3 and concluding that
there are no such c2 and c3. But knowing that the first vector is not dependent
on the other two is not enough. Working straight from the definition, this
person would have to go on to try ~
v2 = c3~
v3 in order to find the dependence
c3 = 1/2. Similarly, working straight from the definition, a set with four vectors
would require checking three vector equations. Lemma 1.4 makes the job easier
because it allows us to get the conclusion with only one computation.
1.10 Example The empty subset of a vector space is linearly independent.
There is no nontrivial linear relationship among its members as it has no mem-
bers.
1.11 Example In any vector space, any subset containing the zero vector is
linearly dependent. For example, in the space P2 of quadratic polynomials,
consider the subset {1 + x, x + x2
, 0}.
One way to see that this subset is linearly dependent is to use Lemma 1.4: we
have 0·~
v1 +0·~
v2 +1·~
0 = ~
0, and this is a nontrivial relationship as not all of the
coefficients are zero. Another way to see that this subset is linearly dependent
is to go straight to Definition 1.3: we can express the third member of the subset
as a linear combination of the first two, namely, c1~
v1 + c2~
v2 = ~
0 is satisfied by
taking c1 = 0 and c2 = 0 (in contrast to the lemma, the definition allows all of
the coefficients to be zero).
(There is still another way to see this that is somewhat trickier. The zero
vector is equal to the trivial sum, that is, it is the sum of no vectors. So in
a set containing the zero vector, there is an element that can be written as a
combination of a collection of other vectors from the set, specifically, the zero
vector can be written as a combination of the empty collection.)
Lemma 1.1 suggests how to turn a spanning set into a spanning set that is
minimal. Given a finite spanning set, we can repeatedly pull out vectors that
are a linear combination of the others, until there aren’t any more such vectors
left.
1.12 Example This set spans R3
.
S0 = {


1
0
0

 ,


0
2
0

 ,


1
2
0

 ,


0
−1
1

 ,


3
3
0

}

Looking for a linear relationship
c1


1
0
0

 + c2


0
2
0

 + c3


1
2
0

 + c4


0
−1
1

 + c5


3
3
0

 =


0
0
0


gives a three equations/five unknowns linear system whose solution set can be
paramatrized in this way.
{






c1
c2
c3
c4
c5






= c3






−1
−1
1
0
0






+ c5






−3
−3/2
0
0
1






¯
¯ c3, c5 ∈ R}
Setting, say, c3 = 0 and c5 = 1 shows that the fifth vector is a linear combination
of the first two. Thus, Lemma 1.1 gives that this set
S1 = {


1
0
0

 ,


0
2
0

 ,


1
2
0

 ,


0
−1
1

}
has the same span as S0. Similarly, the third vector of the new set S1 is a linear
combination of the first two and we get
S2 = {


1
0
0

 ,


0
2
0

 ,


0
−1
1

}
with the same span as S1 and S0, but with one difference. This last set is
linearly independent (this is easily checked), and so removal of any of its vectors
will shrink the span.
We finish this subsection by recasting that example as a theorem that any
finite spanning set has a subset with the same span that is linearly independent.
To prove that result we will first need some facts about how linear independence
and dependence, which are properties of sets, interact with the subset relation
between sets.
1.13 Lemma Any subset of a linearly independent set is also linearly inde-
pendent. Any superset of a linearly dependent set is also linearly dependent.
Proof. This is clear. QED
Restated, independence is preserved by subset and dependence is preserved
by superset. Those are two of the four possible cases of interaction that we
can consider. The third case, whether linear dependence is preserved by the
subset operation, is covered by Example 1.12, which gives a linearly dependent
set S0 with a subset S1 that is linearly dependent and another subset S2 that
is linearly independent.
That leaves one case, whether linear independence is preserved by superset.
The next example shows what can happen.

1.14 Example Here are some linearly independent sets from R3
and their
supersets.
(1) If S1 = {


1
0
0

} then the span [S1] is the x-axis.
A linearly dependent superset: {


1
0
0

 ,


−3
0
0

}
A linearly independent superset: {


1
0
0

 ,


0
1
0

}
(2) If S2 = {


1
0
0

 ,


0
1
0

} then [S2] is the xy-plane.


1
0
0

 ,


0
1
0

 ,


3
−2
0

}
A linearly independent superset: {


1
0
0

 ,


0
1
0

 ,


0
0
1

}
(3) If S3 = {


1
0
0

 ,


0
1
0

 ,


0
0
1

} then [S3] is all of R3
.


1
0
0

 ,


0
1
0

 ,


0
0
1

 ,


2
−1
3

}
There are no linearly independent supersets.
(Checking the dependence or independence of these sets is easy.)
So in general a linearly independent set can have some supersets that are de-
pendent and some supersets that are independent. We can characterize when a
superset of a independent set is dependent and when it is independent.
1.15 Lemma Where S is a linearly independent subset of a vector space V ,
S ∪ {~
v} is linearly dependent if and only if ~
v ∈ [S]
for any ~
v ∈ V with ~
v 6∈ S.

Proof. One implication is clear: if ~
v ∈ [S] then ~
v = c1~
s1 + c2~
s2 + · · · + cn~
sn
where each ~
si ∈ S and ci ∈ R, and so ~
0 = c1~
s1 + c2~
s2 + · · · + cn~
sn + (−1)~
v is a
nontrivial linear relationship among elements of S ∪ {~
v}.
The other implication requires the assumption that S is linearly independent.
With S ∪ {~
v} linearly dependent, there is a nontrivial linear relationship c0~
v +
c1~
s1 + c2~
s2 + · · · + cn~
sn = ~
0 and independence of S then implies that c0 6= 0, or
else that would be a nontrivial relationship among members of S. Now rewriting
this equation as ~
v = −(c1/c0)~
s1 − · · · − (cn/c0)~
sn shows that ~
v ∈ [S]. QED
(Compare this result with Lemma 1.1. Note the additional hypothesis here of
linear independence.)
1.16 Corollary A subset S = {~
s1, . . . ,~
sn} of a vector space is linearly depen-
dent if and only if some ~
si is a linear combination of the vectors ~
s1, . . . , ~
si−1
listed before it.
Proof. Consider S0 = {}, S1 = {~
s1}, S2 = {~
s1,~
s2}, etc. Some index i ≥ 1 is
the first one with Si−1 ∪ {~
si} linearly dependent, and there ~
si ∈ [Si−1]. QED
Lemma 1.15 can be restated in terms of independence instead of dependence:
if S is linearly independent (and ~
v 6∈ S) then the set S ∪ {~
v} is also linearly
independent if and only if ~
v 6∈ [S]. Applying Lemma 1.1, we conclude that if
S is linearly independent and ~
v 6∈ S then S ∪ {~
v} is also linearly independent
if and only if [S ∪ {~
v}] 6= [S]. Briefly, to preserve linear independence through
superset we must expand the span.
Example 1.14 shows that some linearly independent sets are maximal —
have as many elements as possible — in that they have no linearly independent
supersets. By the prior paragraph, linearly independent sets are maximal if and
only if their span is the entire space, because then no vector exists that is not
already in the span.
This table summarizes the interaction between the properties of indepen-
dence and dependence and the relations of subset and superset.
K ⊂ S K ⊃ S
S independent K must be independent K may be either
S dependent K may be either K must be dependent
In developing this table we’ve uncovered an intimate relationship between linear
independence and span. Complementing the fact that a spanning set is minimal
if and only if it is linearly independent, a linearly independent set is maximal if
and only if it spans the space.
We close with the result promised earlier that recasts Example 1.12 as a
theorem.
1.17 Theorem In a vector space, any finite subset has a linearly independent
subset with the same span.

Proof. If the finite set S is linearly independent then there is nothing to prove
so assume that S = {~
s1, . . . ,~
sn} is linearly dependent. By Corollary 1.16, there
is a vector ~
si that is a linear combination of ~
s1, . . . , ~
si−1. Define S1 to be the
set S − {~
si}. Lemma 1.1 then says that the span does not shrink: [S1] = [S].
If S1 is linearly independent then we are finished. Otherwise repeat the
prior paragraph to derive S2 ⊂ S1 such that [S2] = [S1]. Repeat this process
until a linearly independent set appears; one must eventually appear because
S is finite. (Formally, this part of the argument uses mathematical induction.
Exercise 37 asks for the details.) QED
In summary, we have introduced the definition of linear independence to
formalize the idea of the minimality of a spanning set. We have developed some
elementary properties of this idea. The most important is Lemma 1.15, which,
complementing that a spanning set is minimal when it linearly independent,
tells us that a linearly independent set is maximal when it spans the space.
Exercises
X 1.18 Decide whether each subset of R3
is linearly dependent or linearly indepen-
dent.
(a) {
Ã
1
−3
5
!
,
Ã
2
2
4
!
,
Ã
4
−4
14
!
}
(b) {
Ã
1
7
7
!
,
Ã
2
7
7
!
,
Ã
3
7
7
!
}
(c) {
Ã
0
0
−1
!
,
Ã
1
0
4
!
}
(d) {
Ã
9
9
0
!
,
Ã
2
0
1
!
,
Ã
3
5
−4
!
,
Ã
12
12
−1
!
}
X 1.19 Which of these subsets of P3 are linearly dependent and which are indepen-
dent?
(a) {3 − x + 9x2
, 5 − 6x + 3x2
, 1 + 1x − 5x2
}
(b) {−x2
, 1 + 4x2
}
(c) {2 + x + 7x2
, 3 − x + 2x2
, 4 − 3x2
}
(d) {8 + 3x + 3x2
, x + 2x2
, 2 + 2x + 2x2
, 8 − 2x + 5x2
}
X 1.20 Prove that each set {f, g} is linearly independent in the vector space of all
functions from R+
to R.
(a) f(x) = x and g(x) = 1/x
(b) f(x) = cos(x) and g(x) = sin(x)
(c) f(x) = ex
and g(x) = ln(x)
X 1.21 Which of these subsets of the space of real-valued functions of one real vari-
able is linearly dependent and which is linearly independent? (Note that we have
abbreviated some constant functions; e.g., in the first item, the ‘2’ stands for the
constant function f(x) = 2.)

(a) {2, 4 sin2
(x), cos2
(x)} (b) {1, sin(x), sin(2x)} (c) {x, cos(x)}
(d) {(1 + x)2
, x2
+ 2x, 3} (e) {cos(2x), sin2
(x), cos2
(x)} (f) {0, x, x2
}
1.22 Does the equation sin2
(x)/ cos2
(x) = tan2
(x) show that this set of functions
{sin2
(x), cos2
(x), tan2
(x)} is a linearly dependent subset of the set of all real-valued
functions with domain (−π/2..π/2)?
1.23 Why does Lemma 1.4 say “distinct”?
X 1.24 Show that the nonzero rows of an echelon form matrix form a linearly inde-
pendent set.
X 1.25 (a) Show that if the set {~
u,~
v, ~
w} linearly independent set then so is the set
{~
u, ~
u + ~
v, ~
u + ~
v + ~
w}.
(b) What is the relationship between the linear independence or dependence of
the set {~
u,~
v, ~
w} and the independence or dependence of {~
u − ~
v,~
v − ~
w, ~
w − ~
u}?
1.26 Example 1.10 shows that the empty set is linearly independent.
(a) When is a one-element set linearly independent?
(b) How about a set with two elements?
1.27 In any vector space V , the empty set is linearly independent. What about all
of V ?
1.28 Show that if {~
x, ~
y, ~
z} is linearly independent then so are all of its proper
subsets: {~
x, ~
y}, {~
x, ~
z}, {~
y, ~
z}, {~
x},{~
y}, {~
z}, and {}. Is that ‘only if’ also?
1.29 (a) Show that this
S = {
Ã
1
1
0
!
,
Ã
−1
2
0
!
}
is a linearly independent subset of R3
.
(b) Show that
Ã
3
2
0
!
is in the span of S by finding c1 and c2 giving a linear relationship.
c1
Ã
1
1
0
!
+ c2
Ã
−1
2
0
!
=
Ã
3
2
0
!
Show that the pair c1, c2 is unique.
(c) Assume that S is a subset of a vector space and that ~
v is in [S], so that ~
v is
a linear combination of vectors from S. Prove that if S is linearly independent
then a linear combination of vectors from S adding to ~
v is unique (that is, unique
up to reordering and adding or taking away terms of the form 0 · ~
s). Thus S
as a spanning set is minimal in this strong sense: each vector in [S] is “hit” a
minimum number of times — only once.
(d) Prove that it can happen when S is not linearly independent that distinct
linear combinations sum to the same vector.
1.30 Prove that a polynomial gives rise to the zero function if and only if it is
the zero polynomial. (Comment. This question is not a Linear Algebra matter,
but we often use the result. A polynomial gives rise to a function in the obvious
way: x 7→ cnxn
+ · · · + c1x + c0.)
1.31 Return to Section 1.2 and redefine point, line, plane, and other linear surfaces
to avoid degenerate cases.

1.32 (a) Show that any set of four vectors in R2
is linearly dependent.
(b) Is this true for any set of five? Any set of three?
(c) What is the most number of elements that a linearly independent subset of
R2
can have?
X 1.33 Is there a set of four vectors in R3
, any three of which form a linearly inde-
pendent set?
1.34 Must every linearly dependent set have a subset that is dependent and a
subset that is independent?
1.35 In R4
, what is the biggest linearly independent set you can find? The smallest?
The biggest linearly dependent set? The smallest? (‘Biggest’ and ‘smallest’ mean
that there are no supersets or subsets with the same property.)
X 1.36 Linear independence and linear dependence are properties of sets. We can
thus naturally ask how those properties act with respect to the familiar elementary
set relations and operations. In this body of this subsection we have covered the
subset and superset relations. We can also consider the operations of intersection,
complementation, and union.
(a) How does linear independence relate to intersection: can an intersection of
linearly independent sets be independent? Must it be?
(b) How does linear independence relate to complementation?
(c) Show that the union of two linearly independent sets need not be linearly
independent.
(d) Characterize when the union of two linearly independent sets is linearly in-
dependent, in terms of the intersection of the span of each.
X 1.37 For Theorem 1.17,
(a) fill in the induction for the proof;
(b) give an alternate proof that starts with the empty set and builds a sequence
of linearly independent subsets of the given finite set until one appears with the
same span as the given set.
1.38 With a little calculation we can get formulas to determine whether or not a
set of vectors is linearly independent.
(a) Show that this subset of R2
{
µ
a
c
¶
,
µ
b
d
¶
}
is linearly independent if and only if ad − bc 6= 0.
(b) Show that this subset of R3
{
Ã
a
d
g
!
,
Ã
b
e
h
!
,
Ã
c
f
i
!
}
is linearly independent iff aei + bfg + cdh − hfa − idb − gec 6= 0.
(c) When is this subset of R3
{
Ã
a
d
g
!
,
Ã
b
e
h
!
}
linearly independent?
(d) This is an opinion question: for a set of four vectors from R4
, must there be
a formula involving the sixteen entries that determines independence of the set?
(You needn’t produce such a formula, just decide if one exists.)

X 1.39 (a) Prove that a set of two perpendicular nonzero vectors from Rn
is linearly
independent when n > 1.
(b) What if n = 1? n = 0?
(c) Generalize to more than two vectors.
1.40 Consider the set of functions from the open interval (−1..1) to R.
(a) Show that this set is a vector space under the usual operations.
(b) Recall the formula for the sum of an infinite geometric series: 1+x+x2
+· · · =
1/(1−x) for all x ∈ (−1..1). Why does this not express a dependence inside of the
set {g(x) = 1/(1 − x), f0(x) = 1, f1(x) = x, f2(x) = x2
, . . . } (in the vector space
that we are considering)? (Hint. Review the definition of linear combination.)
(c) Show that the set in the prior item is linearly independent.
This shows that some vector spaces exist with linearly independent subsets that
are infinite.
1.41 Show that, where S is a subspace of V , if a subset T of S is linearly indepen-
dent in S then T is also linearly independent in V . Is that ‘only if’?

Section III. Basis and Dimension 113
2.III Basis and Dimension
The prior section ends with the statement that a spanning set is minimal when
it is linearly independent and that a linearly independent set is maximal when
it spans the space. So the notions of minimal spanning set and maximal inde-
pendent set coincide. In this section we will name this notion and study some
of its properties.
2.III.1 Basis
1.1 Definition A basis for a vector space is a sequence of vectors that form a
set that is linearly independent and that spans the space.
We denote a basis with angle brackets h~
β1, ~
β2, . . . i to signify that this collec-
tion is a sequence∗
— the order of the elements is significant. (The requirement
that a basis be ordered will be needed, for instance, in Definition 1.13.)
1.2 Example This is a basis for R2
.
h
µ
2
4
¶
,
µ
1
1
¶
i
It is linearly independent
c1
µ
2
4
¶
+ c2
µ
1
1
¶
=
µ
0
0
¶
=⇒
2c1 + 1c2 = 0
4c1 + 1c2 = 0
=⇒ c1 = c2 = 0
and it spans R2
.
2c1 + 1c2 = x
4c1 + 1c2 = y
=⇒ c2 = 2x − y and c1 = (y − x)/2
1.3 Example This basis for R2
h
µ
1
1
¶
,
µ
2
4
¶
i
differs from the prior one because of its different order. The verification that it
is a basis is just as in the prior example.
1.4 Example The space R2
has many bases. Another one is this.
h
µ
1
0
¶
,
µ
0
1
¶
i
The verification is easy.
∗ More information on sequences is in the appendix.

1.5 Definition For any Rn
,
En = h





1
0
.
.
.
0





,





0
1
.
.
.
0





, . . . ,





0
0
.
.
.
1





i
is the standard (or natural) basis. We denote these vectors by ~
e1, . . . ,~
en.
Note that the symbol ‘~
e1’ means something different in a discussion of R3
than
it means in a discussion of R2
. (Calculus books call R2
’s standard basis vectors
~
ı and ~
 instead of ~
e1 and ~
e2, and they call R3
’s standard basis vectors ~
ı, ~
, and
~
k instead of ~
e1, ~
e2, and ~
e3.)
1.6 Example We can give bases for spaces other than just those comprised of
column vectors. For instance, consider the space {a · cos θ + b · sin θ
¯
¯ a, b ∈ R}
of function of the real variable θ. This is a natural basis
h1 · cos θ + 0 · sin θ, 0 · cos θ + 1 · sin θi = hcos θ, sin θi
while, another, more generic, basis is hcos θ − sin θ, 2 cos θ + 3 sin θi. Verfication
that these two are bases is Exercise 22.
1.7 Example A natural basis for the vector space of cubic polynomials P3 is
h1, x, x2
, x3
i. Two other bases for this space are hx3
, 3x2
, 6x, 6i and h1, 1+x, 1+
x + x2
, 1 + x + x2
+ x3
i. Checking that these are linearly independent and span
the space is easy.
1.8 Example The trivial space {~
0} has only one basis, the empty one hi.
1.9 Example The space of finite degree polynomials has a basis with infinitely
many elements h1, x, x2
, . . . i.
1.10 Example We have seen bases before. For instance, we have described
the solution set of homogeneous systems such as this one
x + y − w = 0
z + w = 0
by paramatrizing.
{




−1
1
0
0



 y +




1
0
−1
1



 w
¯
¯ y, w ∈ R}
That is, we have described the vector space of solutions as the span of a two-
element set. We can easily check that this two-vector set is also linearly inde-
pendent. Thus the solution set is a subspace of R4
with a two-element basis.

1.11 Example Parameterization helps find bases for other vector spaces, not
just for solution sets of homogeneous systems. To find a basis for this subspace
of M2×2
{
µ
a b
c 0
¶
¯
¯ a + b − 2c = 0}
we rewrite the condition as a = −b + 2c to get this.
{
µ
−b + 2c b
c 0
¶
¯
¯ b, c ∈ R} = {b
µ
−1 1
0 0
¶
+ c
µ
2 0
1 0
¶
¯
¯ b, c ∈ R}
Thus, this is a natural candidate for a basis.
h
µ
−1 1
0 0
¶
,
µ
2 0
1 0
¶
i
The above work shows that it spans the space. To show that it is linearly
independent is routine.
Consider Example 1.2 again. To show that the basis spans the space we
looked at a general vector
¡x
y
¢
from R2
. We found a formula for coefficients c1
and c2 in terms of x and y. Although we did not mention it in the example,
the formula shows that for each vector there is only one suitable coefficient pair.
This always happens.
1.12 Theorem In any vector space, a subset is a basis if and only if each
vector in the space can be expressed as a linear combination of elements of the
subset in a unique way. (We consider combinations to be the same if they differ
only in the order of summands or in the addition or deletion of terms of the
form ‘0 · ~
β’.)
Proof. By definition, a sequence is a basis if and only if its vectors form both
a spanning set and a linearly independent set. A subset is a spanning set if
and only if each vector in the space is a linear combination of elements of that
subset in at least one way.
Thus, to finish we need only show that a subset is linearly independent if
and only if every vector in the space is a linear combination of elements from
the subset in at most one way. Consider two expressions of a vector as a linear
combination of the members of the basis. We can rearrange the two sums, and
if necessary add some 0~
βi’s, so that the two combine the same ~
β’s in the same
order: ~
v = c1
~
β1 + c2
~
β2 + · · · + cn
~
βn and ~
v = d1
~
β1 + d2
~
β2 + · · · + dn
~
βn. Now,
equality
c1
~
β1 + c2
~
β2 + · · · + cn
~
βn = d1
~
β1 + d2
~
β2 + · · · + dn
~
βn
holds if and only if
(c1 − d1)~
β1 + · · · + (cn − dn)~
βn = ~
0
holds, and so asserting that each coefficient in the lower equation is zero is the
same thing as asserting that ci = di for each i. QED

1.13 Definition In a vector space with basis B the representation of ~
v with
respect to B is the column vector of the coefficients used to express ~
v as a linear
combination of the basis vectors. That is,
RepB(~
v) =





c1
c2
.
.
.
cn





B
where B = h~
β1, . . . , ~
βni and ~
v = c1
~
β1 + c2
~
β2 + · · · + cn
~
βn. The c’s are the
coordinates of ~
v with respect to B.
1.14 Example In P3, with respect to the basis B = h1, 2x, 2x2
, 2x3
i, the rep-
resentation of x + x2
is
RepB(x + x2
) =




0
1/2
1/2
0




B
(note that the coordinates are scalars, not vectors). With respect to a different
basis D = h1 + x, 1 − x, x + x2
, x + x3
i, the representation
RepD(x + x2
) =




0
0
1
0




D
is different.
1.15 Remark This use of column notation and the term ‘coordinates’ has
both a down side and an up side.
The down side is that representations look like vectors from Rn
, and that
can be confusing when the vector space we are working with is Rn
, especially
since we sometimes omit the subscript base. We must then infer the intent from
the context. For example, the phrase ‘in R2
, where
~
v =
µ
3
2
¶
, . . . ’
refers to the plane vector that, when in canonical position, ends at (3, 2). To
find the coordinates of that vector with respect to the basis
B = h
µ
1
1
¶
,
µ
0
2
¶
i
we solve
c1
µ
1
1
¶
+ c2
µ
0
2
¶
=
µ
3
2
¶

to get that c1 = 3 and c2 = 1/2. Then we have this.
RepB(~
v) =
µ
3
−1/2
¶
Here, although we’ve ommited the subscript B from the column, the fact that
the right side it is a representation is clear from the context.
The up side of the notation and the term ‘coordinates’ is that they generalize
the use that we are familiar with: in Rn
and with respect to the standard
basis En, the vector starting at the origin and ending at (v1, . . . , vn) has this
representation.
RepEn
(



v1
.
.
.
vn


) =



v1
.
.
.
vn



En
Our main use of representations will come in the third chapter. The defini-
tion appears here because the fact that every vector is a linear combination of
basis vectors in a unique way is a crucial property of bases, and also to help make
two points. First, we put the elements of a basis in a fixed order so that coor-
dinates can stated in that order. Second, for calculation of coordinates, among
other things, we shall want our bases to have only finitely many elements. We
will see that in the next subsection.
Exercises
X 1.16 Decide if each is a basis for R3
.
(a) h
Ã
1
2
3
!
,
Ã
3
2
1
!
,
Ã
0
0
1
!
i (b) h
Ã
1
2
3
!
,
Ã
3
2
1
!
i (c) h
Ã
0
2
−1
!
,
Ã
1
1
1
!
,
Ã
2
5
0
!
i
(d) h
Ã
0
2
−1
!
,
Ã
1
1
1
!
,
Ã
1
3
0
!
i
X 1.17 Represent the vector with respect to the basis.
(a)
µ
1
2
¶
, B = h
µ
1
1
¶
,
µ
−1
1
¶
i ⊆ R2
(b) x2
+ x3
, D = h1, 1 + x, 1 + x + x2
, 1 + x + x2
+ x3
i ⊆ P3
(c)



0
−1
0
1


, E4 ⊆ R4
1.18 Find a basis for P2, the space of all quadratic polynomials. Must any such
basis contain a polynomial of each degree: degree zero, degree one, and degree
two?
1.19 Find a basis for the solution set of this system.
x1 − 4x2 + 3x3 − x4 = 0
2x1 − 8x2 + 6x3 − 2x4 = 0
X 1.20 Find a basis for M2
×
2, the space of 2×2 matrices.

X 1.21 Find a basis for each.
(a) The subspace {a2x2
+ a1x + a0
¯
¯ a2 − 2a1 = a0} of P2
(b) The space of three-wide row vectors whose first and second components add
to zero
(c) This subspace of the 2×2 matrices
{
µ
a b
0 c
¶
¯
¯ c − 2b = 0}
1.22 Check Example 1.6.
X 1.23 Find the span of each set and then find a basis for that span.
(a) {1 + x, 1 + 2x} in P2 (b) {2 − 2x, 3 + 4x2
} in P2
X 1.24 Find a basis for each of these subspaces of the space P3 of cubic polynomi-
als.
(a) The subspace of cubic polynomials p(x) such that p(7) = 0
(b) The subspace of polynomials p(x) such that p(7) = 0 and p(5) = 0
(c) The subspace of polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) = 0
(d) The space of polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0,
and p(1) = 0
1.25 We’ve seen that it is possible for a basis to remain a basis when it is reordered.
Must it remain a basis?
1.26 Can a basis contain a zero vector?
X 1.27 Let h~
β1, ~
β2, ~
β3i be a basis for a vector space.
(a) Show that hc1
~
β1, c2
~
β2, c3
~
β3i is a basis when c1, c2, c3 6= 0. What happens
when at least one ci is 0?
(b) Prove that h~
α1, ~
α2, ~
α3i is a basis where ~
αi = ~
β1 + ~
βi.
1.28 Give one more vector ~
v that will make each into a basis for the indicated
space.
(a) h
µ
1
1
¶
,~
vi in R2
(b) h
Ã
1
1
0
!
,
Ã
0
1
0
!
,~
vi in R3
(c) hx, 1 + x2
,~
vi in P2
X 1.29 Where h~
β1, . . . , ~
βni is a basis, show that in this equation
c1
~
β1 + · · · + ck
~
βk = ck+1
~
βk+1 + · · · + cn
~
βn
each of the ci’s is zero. Generalize.
1.30 A basis contains some of the vectors from a vector space; can it contain them
all?
1.31 Theorem 1.12 shows that, with respect to a basis, every linear combination is
unique. If a subset is not a basis, can linear combinations be not unique? If so,
must they be?
X 1.32 A square matrix is symmetric if for all indices i and j, entry i, j equals entry
j, i.
(a) Find a basis for the vector space of symmetric 2×2 matrices.
(b) Find a basis for the space of symmetric 3×3 matrices.
(c) Find a basis for the space of symmetric n×n matrices.
X 1.33 We can show that every basis for R3
contains the same number of vectors,
specifically, three of them.
(a) Show that no linearly independent subset of R3
contains more than three
vectors.

(b) Show that no spanning subset of R3
contains fewer than three vectors. (Hint.
Recall how to calculate the span of a set and show that this method, when applied
to two vectors, cannot yield all of R3
.)
1.34 One of the exercises in the Subspaces subsection shows that the set
{
Ã
x
y
z
!
¯
¯ x + y + z = 1}
is a vector space under these operations.
Ã
x1
y1
z1
!
+
Ã
x2
y2
z2
!
=
Ã
x1 + x2 − 1
y1 + y2
z1 + z2
!
r
Ã
x
y
z
!
=
Ã
rx − r + 1
ry
rz
!
Find a basis.
2.III.2 Dimension
In the prior subsection we saw that a vector space can have many different
bases. For example, following the definition of a basis, we saw three differ-
ent bases for R2
. So we cannot talk about “the” basis for a vector space.
True, some vector spaces have bases that strike us as more natural than oth-
ers, for instance, R2
’s basis E2 or R3
’s basis E3 or P2’s basis h1, x, x2
i. But
the idea of “natural” is hard to make formal. For example, with the space
{a2x2
+ a1x + a0
¯
¯ 2a2 − a0 = a1}, no particular basis leaps out at us as “the”
natural one. We cannot, in general, associate with a space any single basis that
best describes that space.
We can, however, find something about the bases that is uniquely associated
with the space. This subsection shows that any two bases for a space have the
same number of elements. So, with each space we can associate a number, the
number of vectors in any of its bases.
This brings us back to when we considered the two things that could be
meant by the term ‘minimal spanning set’. At that point we defined ‘minimal’
as linearly independent, but we noted that another reasonable interpretation of
the term is that a spanning set is ‘minimal’ when it has the fewest number of
elements of any set with the same span. At the end of this subsection, after we
have shown that all bases have the same number of elements, then we will have
shown that the two senses of ‘minimal’ are equivalent.
Before we start, we first limit our attention to spaces where at least one basis
has only finitely many members.
2.1 Definition A vector space is finite-dimensional if it has a basis with only
finitely many vectors.
(One reason for sticking to finite-dimensional spaces is so that the representation
of a vector with respect to a basis is a finitely-tall vector, and so can be easily
written out. A further remark is at the end of this subsection.) From now on

we study only finite-dimensional vector spaces. We shall take the term ‘vector
space’ to mean ‘finite-dimensional vector space’. Infinite-dimensional spaces are
interesting and important, but they lie outside of our scope.
To prove the main theorem we shall use a technical result.
2.2 Lemma (Exchange Lemma) Assume that B = h~
β1, . . . , ~
βni is a basis
for a vector space, and that for the vector ~
v the relationship ~
v = c1
~
β1 + c2
~
β2 +
· · · + cn
~
βn has ci 6= 0. Then exchanging ~
βi for ~
v yields another basis for the
space.
Proof. Call the outcome of the exchange B̂ = h~
β1, . . . , ~
βi−1,~
v, ~
βi+1, . . . , ~
βni.
We first show that B̂ is linearly independent. Any relationship d1
~
β1 + · · · +
di~
v + · · · + dn
~
βn = ~
0 among the members of B̂, after substitution for ~
v,
d1
~
β1 + · · · + di · (c1
~
β1 + · · · + ci
~
βi + · · · + cn
~
βn) + · · · + dn
~
βn = ~
0 (∗)
gives a linear relationship among the members of B. The basis B is linearly
independent, so the coefficient dici of ~
βi is zero. Because ci is assumed to be
nonzero, di = 0. Using this in equation (∗) above gives that all of the other d’s
are also zero. Therefore B̂ is linearly independent.
We finish by showing that B̂ has the same span as B. Half of this argument,
that [B̂] ⊆ [B], is easy; any member d1
~
β1 + · · · + di~
v + · · · + dn
~
βn of [B̂] can
be written d1
~
β1 + · · · + di · (c1
~
β1 + · · · + cn
~
βn) + · · · + dn
~
βn, which is a linear
combination of linear combinations of members of B, and hence is in [B]. For
the [B] ⊆ [B̂] half of the argument, recall that when ~
v = c1
~
β1 + · · · + cn
~
βn with
ci 6= 0, then the equation can be rearranged to ~
βi = (−c1/ci)~
β1+· · ·+(−1/ci)~
v+
· · · + (−cn/ci)~
βn. Now, consider any member d1
~
β1 + · · · + di
~
βi + · · · + dn
~
βn of
[B], substitute for ~
βi its expression as a linear combination of the members
of B̂, and recognize (as in the first half of this argument) that the result is a
linear combination of linear combinations, of members of B̂, and hence is in
[B̂]. QED
2.3 Theorem In any finite-dimensional vector space, all of the bases have the
same number of elements.
Proof. Fix a vector space with at least one finite basis. Choose, from among all
of this space’s bases, B = h~
β1, . . . , ~
βni of minimal size. We will show that any
other basis D = h~
δ1,~
δ2, . . . i also has the same number of members, n. Because
B has minimal size, D has no fewer than n vectors. We will argue that it cannot
have more.
The basis B spans the space and ~
δ1 is in the space, so ~
δ1 is a nontrivial linear
combination of elements of B. By the Exchange Lemma, ~
δ1 can be swapped for
a vector from B, resulting in a basis B1, where one element is ~
δ and all of the
n − 1 other elements are ~
β’s.
The prior paragraph forms the basis step for an induction argument. The
inductive step starts with a basis Bk (for 1 ≤ k < n) containing k members of D
and n − k members of B. We know that D has at least n members so there is a

~
δk+1. Represent it as a linear combination of elements of Bk. The key point: in
that representation, at least one of the nonzero scalars must be associated with
a ~
βi or else that representation would be a nontrivial linear relationship among
elements of the linearly independent set D. Exchange ~
δk+1 for ~
βi to get a new
basis Bk+1 with one ~
δ more and one ~
β fewer than the previous basis Bk.
Repeat the inductive step until no ~
β’s remain, so that Bn contains ~
δ1, . . . ,~
δn.
Now, D cannot have more than these n vectors because any ~
δn+1 that remains
would be in the span of Bn (since it is a basis) and hence would be a linear com-
bination of the other ~
δ’s, contradicting that D is linearly independent. QED
2.4 Definition The dimension of a vector space is the number of vectors in
any of its bases.
2.5 Example Any basis for Rn
has n vectors since the standard basis En has
n vectors. Thus, this definition generalizes the most familiar use of term, that
Rn
is n-dimensional.
2.6 Example The space Pn of polynomials of degree at most n has dimension
n + 1. We can show this by exhibiting any basis — h1, x, . . . , xn
i comes to
mind — and counting its members.
2.7 Example A trivial space is zero-dimensional since its basis is empty.
Again, although we sometimes say ‘finite-dimensional’ as a reminder, in the
rest of this book all vector spaces are assumed to be finite-dimensional. An
instance of this is that in the next result the word ‘space’ should be taken to
mean ‘finite-dimensional vector space’.
2.8 Corollary No linearly independent set can have a size greater than the
dimension of the enclosing space.
Proof. Inspection of the above proof shows that it never uses that D spans the
space, only that D is linearly independent. QED
2.9 Example Recall the subspace diagram from the prior section showing the
subspaces of R3
. Each subspaces shown is described with a minimal spanning
set, for which we now have the term ‘basis’. The whole space has a basis with
three members, the plane subspaces have bases with two members, the line
subspaces have bases with one member, and the trivial subspace has a basis
with zero members. When we saw that diagram we could not show that these
are the only subspaces that this space has. We can show it now. The prior
corollary proves the only subspaces of R3
are either three-, two-, one-, or zero-
dimensional. Therefore, the diagram indicates all of the subspaces. There are
no subspaces somehow, say, between lines and planes.
2.10 Corollary Any linearly independent set can be expanded to make a basis.
Proof. If a linearly independent set is not already a basis then it must not
span the space. Adding to it a vector that is not in the span preserves linear
independence. Keep adding, until the resulting set does span the space, which
the prior corollary shows will happen after only a finite number of steps. QED

2.11 Corollary Any spanning set can be shrunk to a basis.
Proof. Call the spanning set S. If S is empty then it is already a basis. If
S = {~
0} then it can be shrunk to the empty basis without changing the span.
Otherwise, S contains a vector ~
s1 with ~
s1 6= ~
0 and we can form a basis
B1 = h~
s1i. If [B1] = [S] then we are done.
If not then there is a ~
s2 ∈ [S] such that ~
s2 6∈ [B1]. Let B2 = h~
s1, ~
s2i; if
[B2] = [S] then we are done.
We can repeat this process until the spans are equal, which must happen in
at most finitely many steps. QED
2.12 Corollary In an n-dimensional space, a set of n vectors is linearly inde-
pendent if and only if it spans the space.
Proof. First we will show that a subset with n vectors is linearly independent
if and only if it is a basis. ‘If’ is trivially true — bases are linearly independent.
‘Only if’ holds because a linearly independent set can be expanded to a basis,
but a basis has n elements, so that this expansion is actually the set we began
with.
To finish, we will show that any subset with n vectors spans the space if and
only if it is a basis. Again, ‘if’ is trivial. ‘Only if’ holds because any spanning
set can be shrunk to a basis, but a basis has n elements and so this shrunken
set is just the one we started with. QED
The main result of this subsection, that all of the bases in a finite-dimensional
vector space have the same number of elements, is the single most important
result in this book because, as Example 2.9 shows, it describes what vector
spaces and subspaces there can be. We will see more in the next chapter.
2.13 Remark The case of infinite-dimensional vector spaces is somewhat con-
troversial. The statement ‘any infinite-dimensional vector space has a basis’
is known to be equivalent to a statement called the Axiom of Choice (see
[Blass 1984]). Mathematicians differ philosophically on whether to accept or
reject this statement as an axiom on which to base mathematics. Consequently
the question about infinite-dimensional vector spaces is still somewhat up in the
air. (A discussion of the Axiom of Choice can be found in the Frequently Asked
Questions list for the Usenet group sci.math. Another accessible reference is
[Rucker].)
Exercises
Assume that all spaces are finite-dimensional unless otherwise stated.
X 2.14 Find a basis for, and the dimension of, P2.
2.15 Find a basis for, and the dimension of, the solution set of this system.
x1 − 4x2 + 3x3 − x4 = 0
2x1 − 8x2 + 6x3 − 2x4 = 0
X 2.16 Find a basis for, and the dimension of, M2
×
2, the vector space of 2×2 matrices.

2.17 Find the dimension of the vector space of matrices
µ
a b
c d
¶
subject to each condition.
(a) a, b, c, d ∈ R
(b) a − b + 2c = 0 and d ∈ R
(c) a + b + c = 0, a + b − c = 0, and d ∈ R
X 2.18 Find the dimension of each.
(a) The space of cubic polynomials p(x) such that p(7) = 0
(b) The space of cubic polynomials p(x) such that p(7) = 0 and p(5) = 0
(c) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, and p(3) =
0
(d) The space of cubic polynomials p(x) such that p(7) = 0, p(5) = 0, p(3) = 0,
and p(1) = 0
2.19 What is the dimension of the span of the set {cos2
θ, sin2
θ, cos 2θ, sin 2θ}? This
span is a subspace of the space of all real-valued functions of one real variable.
2.20 Find the dimension of C47
, the vector space of 47-tuples of complex numbers.
2.21 What is the dimension of the vector space M3
×
5 of 3×5 matrices?
X 2.22 Show that this is a basis for R4
.
h



1
0
0
0


 ,



1
1
0
0


 ,



1
1
1
0


 ,



1
1
1
1


i
(The results of this subsection can be used to simplify this job.)
2.23 Refer to Example 2.9.
(a) Sketch a similar subspace diagram for P2.
(b) Sketch one for M2
×
2.
X 2.24 Observe that, where S is a set, the functions f : S → R form a vector space
under the natural operations: f + g (s) = f(s) + g(s) and r · f (s) = r · f(s). What
is the dimension of the space resulting for each domain?
(a) S = {1} (b) S = {1, 2} (c) S = {1, . . . , n}
2.25 (See Exercise 24.) Prove that this is an infinite-dimensional space: the set of
all functions f : R → R under the natural operations.
2.26 (See Exercise 24.) What is the dimension of the vector space of functions
f : S → R, under the natural operations, where the domain S is the empty set?
2.27 Show that any set of four vectors in R2
is linearly dependent.
2.28 Show that the set h~
α1, ~
α2, ~
α3i ⊂ R3
is a basis if and only if there is no plane
through the origin containing all three vectors.
2.29 (a) Prove that any subspace of a finite dimensional space has a basis.
(b) Prove that any subspace of a finite dimensional space is finite dimensional.
2.30 Where is the finiteness of B used in Theorem 2.3?
X 2.31 Prove that if U and W are both three-dimensional subspaces of R5
then U ∩W
is non-trivial. Generalize.
2.32 Because a basis for a space is a subset of that space, we are naturally led to
how the property ‘is a basis’ interacts with set operations.
(a) Consider first how bases might be related by ‘subset’. Assume that U, W are

subspaces of some vector space and that U ⊆ W. Can there exist bases BU for
U and BW for W such that BU ⊆ BW ? Must such bases exist?
For any basis BU for U, must there be a basis BW for W such that BU ⊆ BW ?
For any basis BW for W, must there be a basis BU for U such that BU ⊆ BW ?
For any bases BU , BW for U and W, must BU be a subset of BW ?
(b) Is the intersection of bases a basis? For what space?
(c) Is the union of bases a basis? For what space?
(d) What about complement?
(Hint. Test any conjectures against some subspaces of R3
.)
X 2.33 Consider how ‘dimension’ interacts with ‘subset’. Assume U and W are both
subspaces of some vector space, and that U ⊆ W.
(a) Prove that dim(U) ≤ dim(W).
(b) Prove that equality of dimension holds if and only if U = W.
(c) Show that the prior item does not hold if they are infinite-dimensional.
2.34 [Wohascum no. 47] For any vector ~
v in Rn
and any permutation σ of the
numbers 1, 2, . . . , n (that is, σ is a rearrangement of those numbers into a new
order), define σ(~
v) to be the vector whose components are vσ(1), vσ(2), . . . , and
vσ(n) (where σ(1) is the first number in the rearrangement, etc.). Now fix ~
v and
let V be the span of {σ(~
v)
¯
¯ σ permutes 1, . . . , n}. What are the possibilities for
the dimension of V ?
2.III.3 Vector Spaces and Linear Systems
We will now reconsider linear systems and Gauss’ method, aided by the tools
and terms of this chapter. We will make three points.
For the first point, recall the Linear Combination Lemma and its corollary: if
two matrices are related by row operations A −→ · · · −→ B then each row of B
is a linear combination of the rows of A. That is, Gauss’ method works by taking
linear combinations of rows. Therefore, the right setting in which to study row
operations in general, and Gauss’ method in particular, is the following vector
space.
3.1 Definition The row space of a matrix is the span of the set of its rows. The
row rank is the dimension of the row space, the number of linearly independent
rows.
3.2 Example If
A =
µ
2 3
4 6
¶
then Rowspace(A) is this subspace of the space of two-component row vectors.
{c1 ·
¡
2 3
¢
+ c2 ·
¡
4 6
¢ ¯
¯ c1, c2 ∈ R}
The linear dependence of the second on the first is obvious and so we can simplify
this description to {c ·
¡
2 3
¢ ¯
¯ c ∈ R}.

3.3 Lemma If the matrices A and B are related by a row operation
A
ρi↔ρj
−→ B or A
kρi
−→ B or A
kρi+ρj
−→ B
(for i 6= j and k 6= 0) then their row spaces are equal. Hence, row-equivalent
matrices have the same row space, and hence also, the same row rank.
Proof. The row space of A is the set of all linear combinations of the rows
of A. By the Linear Combination Lemma then, each row of B is in the row
space of A. Further, Rowspace(B) ⊆ Rowspace(A) because a member of the
set Rowspace(B) is a linear combination of the rows of B, which means it is a
combination of a combination of the rows of A, and hence is also a member of
Rowspace(A).
For the other containment, recall that row operations are reversible: A −→ B
if and only if B −→ A. With that, Rowspace(A) ⊆ Rowspace(B) also follows
from the prior paragraph, and hence the two sets are equal. QED
So, row operations leave the row space unchanged. But of course, Gauss’
method performs the row operations systematically, with a specific goal in mind,
echelon form.
3.4 Lemma The nonzero rows of an echelon form matrix make up a linearly
independent set.
Proof. A result in the first chapter, Lemma III.2.5, states that in an echelon
form matrix, no nonzero row is a linear combination of the other rows. This is
a restatement of that result into new terminology. QED
Thus, in the language of this chapter, Gaussian reduction works by elim-
inating linear dependences among rows, leaving the span unchanged, until no
nontrivial linear relationships remain (among the nonzero rows). That is, Gauss’
method produces a basis for the row space.
3.5 Example From any matrix, we can produce a basis for the row space by
performing Gauss’ method and taking the nonzero rows of the resulting echelon
form matrix. For instance,


1 3 1
1 4 1
2 0 5

 −ρ1+ρ2
−→
−2ρ1+ρ3
6ρ2+ρ3
−→


1 3 1
0 1 0
0 0 3


produces the basis h
¡
1 3 1
¢
,
¡
0 1 0
¢
,
¡
0 0 3
¢
i for the row space. This
is a basis for the row space of both the starting and ending matrices, since the
two row spaces are equal.
Using this technique, we can also find bases for spans not directly involving
row vectors.

3.6 Definition The column space of a matrix is the span of the set of its
columns. The column rank is the dimension of the column space, the number
of linearly independent columns.
Our interest in column spaces stems from our study of linear systems. An
example is that this system
c1 + 3c2 + 7c3 = d1
2c1 + 3c2 + 8c3 = d2
c2 + 2c3 = d3
4c1 + 4c3 = d4
has a solution if and only if the vector of d’s is a linear combination of the other
column vectors,
c1




1
2
0
4



 + c2




3
3
1
0



 + c3




7
8
2
4



 =




d1
d2
d3
d4




meaning that the vector of d’s is in the column space of the matrix of coefficients.
3.7 Example Given this matrix,




1 3 7
2 3 8
0 1 2
4 0 4




to get a basis for the column space, temporarily turn the columns into rows and
reduce.


1 2 0 4
3 3 1 0
7 8 2 4

 −3ρ1+ρ2
−→
−7ρ1+ρ3
−2ρ2+ρ3
−→


1 2 0 4
0 −3 1 −12
0 0 0 0


Now turn the rows back to columns.
h




1
2
0
4



 ,




0
−3
1
−12



i
The result is a basis for the column space of the given matrix.
3.8 Definition The transpose of a matrix is the result of interchanging the
rows and columns of that matrix. That is, column j of the matrix A is row j of
Atrans
, and vice versa.

So the instructions for the prior example are “transpose, reduce, and transpose
back”.
We can even, at the price of tolerating the as-yet-vague idea of vector spaces
being “the same”, use Gauss’ method to find bases for spans in other types of
vector spaces.
3.9 Example To get a basis for the span of {x2
+ x4
, 2x2
+ 3x4
, −x2
− 3x4
}
in the space P4, think of these three polynomials as “the same” as the row
vectors
¡
0 0 1 0 1
¢
,
¡
0 0 2 0 3
¢
, and
¡
0 0 −1 0 −3
¢
, apply
Gauss’ method


0 0 1 0 1
0 0 2 0 3
0 0 −1 0 −3

 −2ρ1+ρ2
−→
ρ1+ρ3
2ρ2+ρ3
−→


0 0 1 0 1
0 0 0 0 1
0 0 0 0 0


and translate back to get the basis hx2
+ x4
, x4
i. (As mentioned earlier, we will
make the phrase “the same” precise at the start of the next chapter.)
Thus, our first point in this subsection is that the tools of this chapter give
us a more conceptual understanding of Gaussian reduction.
For the second point of this subsection, consider the effect on the column
space of this row reduction.
µ
1 2
2 4
¶
−2ρ1+ρ2
−→
µ
1 2
0 0
¶
The column space of the left-hand matrix contains vectors with a second compo-
nent that is nonzero. But the column space of the right-hand matrix is different
because it contains only vectors whose second component is zero. It is this
knowledge that row operations can change the column space that makes next
result surprising.
3.10 Lemma Row operations do not change the column rank.
Proof. Restated, if A reduces to B then the column rank of B equals the
column rank of A.
We will be done if we can show that row operations do not affect linear re-
lationships among columns (e.g., if the fifth column is twice the second plus the
fourth before a row operation then that relationship still holds afterwards), be-
cause the column rank is just the size of the largest set of unrelated columns. But
this is exactly the first theorem of this book: in a relationship among columns,
c1 ·





a1,1
a2,1
.
.
.
am,1





+ · · · + cn ·





a1,n
a2,n
.
.
.
am,n





=





0
0
.
.
.
0





row operations leave unchanged the set of solutions (c1, . . . , cn). QED

Another way, besides the prior result, to state that Gauss’ method has some-
thing to say about the column space as well as about the row space is to consider
again Gauss-Jordan reduction. Recall that it ends with the reduced echelon form
of a matrix, as here.


1 3 1 6
2 6 3 16
1 3 1 6

 −→ · · · −→


1 3 0 2
0 0 1 4
0 0 0 0


Consider the row space and the column space of this result. Our first point
made above says that a basis for the row space is easy to get: simply collect
together all of the rows with leading entries. However, because this is a reduced
echelon form matrix, a basis for the column space is just as easy: take the
columns containing the leading entries, that is, h~
e1,~
e2i. (Linear independence
is obvious. The other columns are in the span of this set, since they all have
a third component of zero.) Thus, for a reduced echelon form matrix, bases
for the row and column spaces can be found in essentially the same way —
by taking the parts of the matrix, the rows or columns, containing the leading
entries.
3.11 Theorem The row rank and column rank of a matrix are equal.
Proof. First bring the matrix to reduced echelon form. At that point, the
row rank equals the number of leading entries since each equals the number
of nonzero rows. Also at that point, the number of leading entries equals the
column rank because the set of columns containing leading entries consists of
some of the ~
ei’s from a standard basis, and that set is linearly independent and
spans the set of columns. Hence, in the reduced echelon form matrix, the row
rank equals the column rank, because each equals the number of leading entries.
But Lemma 3.3 and Lemma 3.10 show that the row rank and column rank
are not changed by using row operations to get to reduced echelon form. Thus
the row rank and the column rank of the original matrix are also equal. QED
3.12 Definition The rank of a matrix is its row rank or column rank.
So our second point in this subsection is that the column space and row
space of a matrix have the same dimension. Our third and final point is that
the concepts that we’ve seen arising naturally in the study of vector spaces are
exactly the ones that we have studied with linear systems.
3.13 Theorem For linear systems with n unknowns and with matrix of coef-
ficients A, the statements
(1) the rank of A is r
(2) the space of solutions of the associated homogeneous system has dimen-
sion n − r
are equivalent.

So if the system has at least one particular solution then for the set of solutions,
the number of parameters equals n − r, the number of variables minus the rank
of the matrix of coefficients.
Proof. The rank of A is r if and only if Gaussian reduction on A ends with r
nonzero rows. That’s true if and only if echelon form matrices row equivalent
to A have r-many leading variables. That in turn holds if and only if there are
n − r free variables. QED
3.14 Remark [Munkres] Sometimes that result is mistakenly remembered to
say that the general solution of an n unknown system of m equations uses n−m
parameters. The number of equations is not the relevant figure, rather, what
matters is the number of independent equations (the number of equations in
a maximal independent set). Where there are r independent equations, the
general solution involves n − r parameters.
3.15 Corollary Where the matrix A is n×n, the statements
(1) the rank of A is n
(2) A is nonsingular
(3) the rows of A form a linearly independent set
(4) the columns of A form a linearly independent set
(5) any linear system whose matrix of coefficients is A has one and only one
solution
are equivalent.
Proof. Clearly (1) ⇐⇒ (2) ⇐⇒ (3) ⇐⇒ (4). The last, (4) ⇐⇒ (5), holds
because a set of n column vectors is linearly independent if and only if it is a
basis for Rn
, but the system
c1





a1,1
a2,1
.
.
.
am,1





+ · · · + cn





a1,n
a2,n
.
.
.
am,n





=





d1
d2
.
.
.
dn





has a unique solution for all choices of d1, . . . , dn ∈ R if and only if the vectors
of a’s form a basis. QED
Exercises
3.16 Transpose each.
(a)
µ
2 1
3 1
¶
(b)
µ
2 1
1 3
¶
(c)
µ
1 4 3
6 7 8
¶
(d)
Ã
0
0
0
!
(e)
¡
−1 −2
¢
X 3.17 Decide if the vector is in the row space of the matrix.
(a)
µ
2 1
3 1
¶
,
¡
1 0
¢
(b)
Ã
0 1 3
−1 0 1
−1 2 7
!
,
¡
1 1 1
¢
X 3.18 Decide if the vector is in the column space.

(a)
µ
1 1
1 1
¶
,
µ
1
3
¶
(b)
Ã
1 3 1
2 0 4
1 −3 −3
!
,
Ã
1
0
0
!
X 3.19 Find a basis for the row space of this matrix.



2 0 3 4
0 1 1 −1
3 1 0 2
1 0 −4 −1



X 3.20 Find the rank of each matrix.
(a)
Ã
2 1 3
1 −1 2
1 0 3
!
(b)
Ã
1 −1 2
3 −3 6
−2 2 −4
!
(c)
Ã
1 3 2
5 1 1
6 4 3
!
(d)
Ã
0 0 0
0 0 0
0 0 0
!
X 3.21 Find a basis for the span of each set.
(a) {
¡
1 3
¢
,
¡
−1 3
¢
,
¡
1 4
¢
,
¡
2 1
¢
} ⊆ M1
×
2
(b) {
Ã
1
2
1
!
,
Ã
3
1
−1
!
,
Ã
1
−3
−3
!
} ⊆ R3
(c) {1 + x, 1 − x2
, 3 + 2x − x2
} ⊆ P3
(d) {
µ
1 0 1
3 1 −1
¶
,
µ
1 0 3
2 1 4
¶
,
µ
−1 0 −5
−1 −1 −9
¶
} ⊆ M2
×
3
3.22 Which matrices have rank zero? Rank one?
X 3.23 Given a, b, c ∈ R, what choice of d will cause this matrix to have the rank of
one?
µ
a b
c d
¶
3.24 Find the column rank of this matrix.
µ
1 3 −1 5 0 4
2 0 1 0 4 1
¶
3.25 Show that a linear system with at least one solution has at most one solution if
and only if the matrix of coefficients has rank equal to the number of its columns.
X 3.26 If a matrix is 5×9, which set must be dependent, its set of rows or its set of
columns?
3.27 Give an example to show that, despite that they have the same dimension,
the row space and column space of a matrix need not be equal. Are they ever
equal?
3.28 Show that the set {(1, −1, 2, −3), (1, 1, 2, 0), (3, −1, 6, −6)} does not have the
same span as {(1, 0, 1, 0), (0, 2, 0, 3)}. What, by the way, is the vector space?
X 3.29 Show that this set of column vectors
(Ã
d1
d2
d3
!
¯
¯ there are x, y, and z such that
3x + 2y + 4z = d1
x − z = d2
2x + 2y + 5z = d3
)
is a subspace of R3
. Find a basis.

3.30 Show that the transpose operation is linear:
(rA + sB)trans
= rAtrans
+ sBtrans
for r, s ∈ R and A, B ∈ Mm
×
n,
X 3.31 In this subsection we have shown that Gaussian reduction finds a basis for
the row space.
(a) Show that this basis is not unique — different reductions may yield different
bases.
(b) Produce matrices with equal row spaces but unequal numbers of rows.
(c) Prove that two matrices have equal row spaces if and only if after Gauss-
Jordan reduction they have the same nonzero rows.
3.32 Why is there not a problem with Remark 3.14 in the case that r is bigger
than n?
3.33 Show that the row rank of an m×n matrix is at most m. Is there a better
bound?
X 3.34 Show that the rank of a matrix equals the rank of its transpose.
3.35 True or false: the column space of a matrix equals the row space of its trans-
pose.
X 3.36 We have seen that a row operation may change the column space. Must it?
3.37 Prove that a linear system has a solution if and only if that system’s matrix
of coefficients has the same rank as its augmented matrix.
3.38 An m×n matrix has full row rank if its row rank is m, and it has full column
rank if its column rank is n.
(a) Show that a matrix can have both full row rank and full column rank only
if it is square.
(b) Prove that the linear system with matrix of coefficients A has a solution for
any d1, . . . , dn’s on the right side if and only if A has full row rank.
(c) Prove that a homogeneous system has a unique solution if and only if its
matrix of coefficients A has full column rank.
(d) Prove that the statement “if a system with matrix of coefficients A has any
solution then it has a unique solution” holds if and only if A has full column
rank.
3.39 How would the conclusion of Lemma 3.3 change if Gauss’ method is changed
to allow multiplying a row by zero?
X 3.40 What is the relationship between rank(A) and rank(−A)? Between rank(A)
and rank(kA)? What, if any, is the relationship between rank(A), rank(B), and
rank(A + B)?
2.III.4 Combining Subspaces
This subsection is optional. It is required only for the last sections of Chapter
Three and Chapter Five and for occasional exercises, and can be passed over
without loss of continuity.
This chapter opened with the definition of a vector space, and the mid-
dle consisted of a first analysis of the idea. This subsection closes the chapter

by finishing the analysis, in the sense that ‘analysis’ means “method of de-
termining the . . . essential features of something by separating it into parts”
[Macmillan Dictionary].
A common way to understand things is to see how they can be built from
component parts. For instance, we think of R3
as put together, in some way,
from the x-axis, the y-axis, and z-axis. In this subsection we will make this
precise; we will describe how to decompose a vector space into a combination of
some of its subspaces. In developing this idea of subspace combination, we will
keep the R3
example in mind as a benchmark model.
Subspaces are subsets and sets combine via union. But taking the combi-
nation operation for subspaces to be the simple union operation isn’t what we
want. For one thing, the union of the x-axis, the y-axis, and z-axis is not all of
R3
, so the benchmark model would be left out. Besides, union is all wrong for
this reason: a union of subspaces need not be a subspace (it need not be closed;
for instance, this R3
vector


1
0
0

 +


0
1
0

 +


0
0
1

 =


1
1
1


is in none of the three axes and hence is not in the union). In addition to the
members of the subspaces, we must at a minimum also include all possible linear
combinations.
4.1 Definition Where W1, . . . , Wk are subspaces of a vector space, their sum
is the span of their union W1 + W2 + · · · + Wk = [W1 ∪ W2 ∪ . . . Wk].
(The notation, writing the ‘+’ between sets in addition to using it between
vectors, fits with the practice of using this symbol for any natural accumulation
operation.)
4.2 Example The R3
model fits with this operation. Any vector ~
w ∈ R3
can
be written as a linear combination c1~
v1 + c2~
v2 + c3~
v3 where ~
v1 is a member of
the x-axis, etc., in this way


w1
w2
w3

 = 1 ·


w1
0
0

 + 1 ·


0
w2
0

 + 1 ·


0
0
w3


and so R3
= x-axis + y-axis + z-axis.
4.3 Example A sum of subspaces can be less than the entire space. Inside of
P4, let L be the subspace of linear polynomials {a + bx
¯
¯ a, b ∈ R} and let C be
the subspace of purely-cubic polynomials {cx3
¯
¯ c ∈ R}. Then L + C is not all
of P4. Instead, it is the subspace L + C = {a + bx + cx3
¯
¯ a, b, c ∈ R}.
4.4 Example A space can be described as a combination of subspaces in more
than one way. Besides the decomposition R3
= x-axis + y-axis + z-axis, we can

also write R3
= xy-plane + yz-plane. To check this, we simply note that any
~
w ∈ R3
can be written


w1
w2
w3

 = 1 ·


w1
w2
0

 + 1 ·


0
0
w3


as a linear combination of a member of the xy-plane and a member of the
yz-plane.
The above definition gives one way in which a space can be thought of as a
combination of some of its parts. However, the prior example shows that there is
at least one interesting property of our benchmark model that is not captured by
the definition of the sum of subspaces. In the familiar decomposition of R3
, we
often speak of a vector’s ‘x part’ or ‘y part’ or ‘z part’. That is, in this model,
each vector has a unique decomposition into parts that come from the parts
making up the whole space. But in the decomposition used in Example 4.4, we
cannot refer to the “xy part” of a vector — these three sums


1
2
3

 =


1
2
0

 +


0
0
3

 =


1
0
0

 +


0
2
3

 =


1
1
0

 +


0
1
3


all describe the vector as comprised of something from the first plane plus some-
thing from the second plane, but the “xy part” is different in each.
That is, when we consider how R3
is put together from the three axes “in
some way”, we might mean “in such a way that every vector has at least one
decomposition”, and that leads to the definition above. But if we take it to
mean “in such a way that every vector has one and only one decomposition”
then we need another condition on combinations. To see what this condition
is, recall that vectors are uniquely represented in terms of a basis. We can use
this to break a space into a sum of subspaces such that any vector in the space
breaks uniquely into a sum of members of those subspaces.
4.5 Example The benchmark is R3
with its standard basis E3 = h~
e1,~
e2,~
e3i.
The subspace with the basis B1 = h~
e1i is the x-axis. The subspace with the
basis B2 = h~
e2i is the y-axis. The subspace with the basis B3 = h~
e3i is the
z-axis. The fact that any member of R3
is expressible as a sum of vectors from
these subspaces


x
y
z

 =


x
0
0

 +


0
y
0

 +


0
0
z


is a reflection of the fact that E3 spans the space — this equation


x
y
z

 = c1


1
0
0

 + c2


0
1
0

 + c3


0
0
1



has a solution for any x, y, z ∈ R. And, the fact that each such expression is
unique reflects that fact that E3 is linearly independent — any equation like the
one above has a unique solution.
4.6 Example We don’t have to take the basis vectors one at a time, the same
idea works if we conglomerate them into larger sequences. Consider again the
space R3
and the vectors from the standard basis E3. The subspace with the
basis B1 = h~
e1,~
e3i is the xz-plane. The subspace with the basis B2 = h~
e2i is
the y-axis. As in the prior example, the fact that any member of the space is a
sum of members of the two subspaces in one and only one way


x
y
z

 =


x
0
z

 +


0
y
0


is a reflection of the fact that these vectors form a basis — this system


x
y
z

 = (c1


1
0
0

 + c3


0
0
1

) + c2


0
1
0


has one and only one solution for any x, y, z ∈ R.
These examples illustrate a natural way to decompose a space into a sum
of subspaces in such a way that each vector decomposes uniquely into a sum of
vectors from the parts. The next result says that this way is the only way.
4.7 Definition The concatenation of the sequences B1 = h~
β1,1, . . . , ~
β1,n1
i, . . . ,
Bk = h~
βk,1, . . . , ~
βk,nk
i is their adjoinment.
B1
_
B2
_
· · ·
_
Bk = h~
β1,1, . . . , ~
β1,n1
, ~
β2,1, . . . , ~
βk,nk
i
4.8 Lemma Let V be a vector space that is the sum of some of its subspaces
V = W1 + · · · + Wk. Let B1, . . . , Bk be any bases for these subspaces. Then
the following are equivalent.
(1) For every ~
v ∈ V , the expression ~
v = ~
w1 + · · · + ~
wk (with ~
wi ∈ Wi) is
unique.
(2) The concatenation B1
_
· · ·
_
Bk is a basis for V .
(3) The nonzero members of {~
w1, . . . , ~
wk} (with ~
wi ∈ Wi) form a linearly
independent set — among nonzero vectors from different Wi’s, every linear
relationship is trivial.
Proof. We will show that (1) =⇒ (2), that (2) =⇒ (3), and finally that
(3) =⇒ (1). For these arguments, observe that we can pass from a combination
of ~
w’s to a combination of ~
β’s
d1 ~
w1 + · · · + dk ~
wk
= d1(c1,1
~
β1,1 + · · · + c1,n1
~
β1,n1
) + · · · + dk(ck,1
~
βk,1 + · · · + ck,nk
~
βk,nk
)
= d1c1,1 · ~
β1,1 + · · · + dkck,nk
· ~
βk,nk
(∗)

and vice versa.
For (1) =⇒ (2), assume that all decompositions are unique. We will show
that B1
_
· · ·
_
Bk spans the space and is linearly independent. It spans the
space because the assumption that V = W1 + · · · + Wk means that every ~
v
can be expressed as ~
v = ~
w1 + · · · + ~
wk, which translates by equation (∗) to an
expression of ~
v as a linear combination of the ~
β’s from the concatenation. For
linear independence, consider this linear relationship.
~
0 = c1,1
~
β1,1 + · · · + ck,nk
~
βk,nk
Regroup as in (∗) (that is, take d1, . . . , dk to be 1 and move from bottom to
top) to get the decomposition ~
0 = ~
w1 + · · · + ~
wk. Because of the assumption
that decompositions are unique, and because the zero vector obviously has the
decomposition ~
0 = ~
0+· · ·+~
0, we now have that each ~
wi is the zero vector. This
means that ci,1
~
βi,1 + · · · + ci,ni
~
βi,ni
= ~
0. Thus, since each Bi is a basis, we have
the desired conclusion that all of the c’s are zero.
For (2) =⇒ (3), assume that B1
_
· · ·
_
Bk is a basis for the space. Consider
a linear relationship among nonzero vectors from different Wi’s,
~
0 = · · · + di ~
wi + · · ·
in order to show that it is trivial. (The relationship is written in this way
because we are considering a combination of nonzero vectors from only some of
the Wi’s; for instance, there might not be a ~
w1 in this combination.) As in (∗),
~
0 = · · ·+di(ci,1
~
βi,1+· · ·+ci,ni
~
βi,ni
)+· · · = · · ·+dici,1·~
βi,1+· · ·+dici,ni
·~
βi,ni
+· · ·
and the linear independence of B1
_
· · ·
_
Bk gives that each coefficient dici,j is
zero. Now, ~
wi is a nonzero vector, so at least one of the ci,j’s is zero, and thus
di is zero. This holds for each di, and therefore the linear relationship is trivial.
Finally, for (3) =⇒ (1), assume that, among nonzero vectors from different
Wi’s, any linear relationship is trivial. Consider two decompositions of a vector
~
v = ~
w1 + · · · + ~
wk and ~
v = ~
u1 + · · · + ~
uk in order to show that the two are the
same. We have
~
0 = (~
w1 + · · · + ~
wk) − (~
u1 + · · · + ~
uk) = (~
w1 − ~
u1) + · · · + (~
wk − ~
uk)
which violates the assumption unless each ~
wi − ~
ui is the zero vector. Hence,
decompositions are unique. QED
4.9 Definition A collection of subspaces {W1, . . . , Wk} is independent if no
nonzero vector from any Wi is a linear combination of vectors from the other
subspaces W1, . . . , Wi−1, Wi+1, . . . , Wk.
4.10 Definition A vector space V is the direct sum (or internal direct sum)
of its subspaces W1, . . . , Wk if V = W1 + W2 + · · · + Wk and the collection
{W1, . . . , Wk} is independent. We write V = W1 ⊕ W2 ⊕ . . . ⊕ Wk.
4.11 Example The benchmark model fits: R3
= x-axis ⊕ y-axis ⊕ z-axis.

4.12 Example The space of 2×2 matrices is this direct sum.
{
µ
a 0
0 d
¶
¯
¯ a, d ∈ R} ⊕ {
µ
0 b
0 0
¶
¯
¯ b ∈ R} ⊕ {
µ
0 0
c 0
¶
¯
¯ c ∈ R}
It is the direct sum of subspaces in many other ways as well; direct sum decom-
positions are not unique.
4.13 Corollary The dimension of a direct sum is the sum of the dimensions
of its summands.
Proof. In Lemma 4.8, the number of basis vectors in the concatenation equals
the sum of the number of vectors in the subbases that make up the concatena-
tion. QED
The special case of two subspaces is worth mentioning separately.
4.14 Definition When a vector space is the direct sum of two of its subspaces,
then they are said to be complements.
4.15 Lemma A vector space V is the direct sum of two of its subspaces W1
and W2 if and only if it is the sum of the two V = W1 +W2 and their intersection
is trivial W1 ∩ W2 = {~
0 }.
Proof. Suppose first that V = W1 ⊕ W2. By definition, V is the sum of the
two. To show that the two have a trivial intersection, let ~
v be a vector from
W1 ∩ W2 and consider the equation ~
v = ~
v. On the left side of that equation
is a member of W1, and on the right side is a linear combination of members
(actually, of only one member) of W2. But the independence of the spaces then
implies that ~
v = ~
0, as desired.
For the other direction, suppose that V is the sum of two spaces with a
trivial intersection. To show that V is a direct sum of the two, we need only
show that the spaces are independent — no nonzero member of the first is
expressible as a linear combination of members of the second, and vice versa.
This is true because any relationship ~
w1 = c1 ~
w2,1 + · · · + dk ~
w2,k (with ~
w1 ∈ W1
and ~
w2,j ∈ W2 for all j) shows that the vector on the left is also in W2, since
the right side is a combination of members of W2. The intersection of these two
spaces is trivial, so ~
w1 = ~
0. The same argument works for any ~
w2. QED
4.16 Example In the space R2
, the x-axis and the y-axis are complements,
that is, R2
= x-axis ⊕ y-axis. A space can have more than one pair of comple-
mentary subspaces; another pair here are the subspaces consisting of the lines
y = x and y = 2x.
4.17 Example In the space F = {a cos θ + b sin θ
¯
¯ a, b ∈ R}, the subspaces
W1 = {a cos θ
¯
¯ a ∈ R} and W2 = {b sin θ
¯
¯ b ∈ R} are complements. In addition
to the fact that a space like F can have more than one pair of complementary
subspaces, inside of the space a single subspace like W1 can have more than one
complement — another complement of W1 is W3 = {b sin θ + b cos θ
¯
¯ b ∈ R}.

4.18 Example In R3
, the xy-plane and the yz-planes are not complements,
which is the point of the discussion following Example 4.4. One complement of
the xy-plane is the z-axis. A complement of the yz-plane is the line through
(1, 1, 1).
4.19 Example Following Lemma 4.15, here is a natural question: is the simple
sum V = W1 + · · · + Wk also a direct sum if and only if the intersection of the
subspaces is trivial? The answer is that if there are more than two subspaces
then having a trivial intersection is not enough to guarantee unique decompo-
sition (i.e., is not enough to ensure that the spaces are independent). In R3
, let
W1 be the x-axis, let W2 be the y-axis, and let W3 be this.
W3 = {


q
q
r


¯
¯ q, r ∈ R}
The check that R3
= W1 + W2 + W3 is easy. The intersection W1 ∩ W2 ∩ W3 is
trivial, but decompositions aren’t unique.


x
y
z

 =


0
0
0

 +


0
y − x
0

 +


x
x
z

 =


x − y
0
0

 +


0
0
0

 +


y
y
z


(This example also shows that this requirement is also not enough: that all
pairwise intersections of the subspaces be trivial. See Exercise 30.)
In this subsection we have seen two ways to regard a space as built up from
component parts. Both are useful; in particular, in this book the direct sum
definition is needed to do the Jordan Form construction in the fifth chapter.
Exercises
X 4.20 Decide if R2
is the direct sum of each W1 and W2.
(a) W1 = {
µ
x
0
¶
¯
¯ x ∈ R}, W2 = {
µ
x
x
¶
¯
¯ x ∈ R}
(b) W1 = {
µ
s
s
¶
¯
¯ s ∈ R}, W2 = {
µ
s
1.1s
¶
¯
¯ s ∈ R}
(c) W1 = R2
, W2 = {~
0}
(d) W1 = W2 = {
µ
t
t
¶
¯
¯ t ∈ R}
(e) W1 = {
µ
1
0
¶
+
µ
x
0
¶
¯
¯ x ∈ R}, W2 = {
µ
−1
0
¶
+
µ
0
y
¶
¯
¯ y ∈ R}
X 4.21 Show that R3
is the direct sum of the xy-plane with each of these.
(a) the z-axis
(b) the line
{
Ã
z
z
z
!
¯
¯ z ∈ R}

4.22 Is P2 the direct sum of {a + bx2
¯
¯ a, b ∈ R} and {cx
¯
¯ c ∈ R}?
X 4.23 In Pn, the even polynomials are the members of this set
E = {p ∈ Pn
¯
¯ p(−x) = p(x) for all x}
and the odd polynomials are the members of this set.
O = {p ∈ Pn
¯
¯ p(−x) = −p(x) for all x}
Show that these are complementary subspaces.
4.24 Which of these subspaces of R3
W1: the x-axis, W2: the y-axis, W3: the z-axis,
W4: the plane x + y + z = 0, W5: the yz-plane
can be combined to
(a) sum to R3
? (b) direct sum to R3
?
X 4.25 Show that Pn = {a0
¯
¯ a0 ∈ R} ⊕ . . . ⊕ {anxn
¯
¯ an ∈ R}.
4.26 What is W1 + W2 if W1 ⊆ W2?
4.27 Does Example 4.5 generalize? That is, is this true or false: if a vector space V
has a basis h~
β1, . . . , ~
βni then it is the direct sum of the spans of the one-dimensional
subspaces V = [{~
β1}] ⊕ . . . ⊕ [{~
βn}]?
4.28 Can R4
be decomposed as a direct sum in two different ways? Can R1
?
4.29 This exercise makes the notation of writing ‘+’ between sets more natural.
Prove that, where W1, . . . , Wk are subspaces of a vector space,
W1 + · · · + Wk = {~
w1 + ~
w2 + · · · + ~
wk
¯
¯ ~
w1 ∈ W1, . . . , ~
wk ∈ Wk},
and so the sum of subspaces is the subspace of all sums.
4.30 (Refer to Example 4.19. This exercise shows that the requirement that pari-
wise intersections be trivial is genuinely stronger than the requirement only that
the intersection of all of the subspaces be trivial.) Give a vector space and three
subspaces W1, W2, and W3 such that the space is the sum of the subspaces, the
intersection of all three subspaces W1 ∩ W2 ∩ W3 is trivial, but the pairwise inter-
sections W1 ∩ W2, W1 ∩ W3, and W2 ∩ W3 are nontrivial.
X 4.31 Prove that if V = W1 ⊕ . . . ⊕ Wk then Wi ∩ Wj is trivial whenever i 6= j. This
shows that the first half of the proof of Lemma 4.15 extends to the case of more
than two subspaces. (Example 4.19 shows that this implication does not reverse;
the other half does not extend.)
4.32 Recall that no linearly independent set contains the zero vector. Can an
independent set of subspaces contain the trivial subspace?
X 4.33 Does every subspace have a complement?
X 4.34 Let W1, W2 be subspaces of a vector space.
(a) Assume that the set S1 spans W1, and that the set S2 spans W2. Can S1 ∪S2
span W1 + W2? Must it?
(b) Assume that S1 is a linearly independent subset of W1 and that S2 is a
linearly independent subset of W2. Can S1 ∪S2 be a linearly independent subset
of W1 + W2? Must it?
4.35 When a vector space is decomposed as a direct sum, the dimensions of the
subspaces add to the dimension of the space. The situation with a space that is
given as the sum of its subspaces is not as simple. This exercise considers the
two-subspace special case.
(a) For these subspaces of M2
×
2 find W1 ∩ W2, dim(W1 ∩ W2), W1 + W2, and
dim(W1 + W2).
W1 = {
µ
0 0
c d
¶
¯
¯ c, d ∈ R} W2 = {
µ
0 b
c 0
¶
¯
¯ b, c ∈ R}

(b) Suppose that U and W are subspaces of a vector space. Suppose that the
sequence h~
β1, . . . , ~
βki is a basis for U ∩ W. Finally, suppose that the prior
sequence has been expanded to give a sequence h~
µ1, . . . , ~
µj, ~
β1, . . . , ~
βki that is a
basis for U, and a sequence h~
β1, . . . , ~
βk, ~
ω1, . . . , ~
ωpi that is a basis for W. Prove
that this sequence
h~
µ1, . . . , ~
µj, ~
β1, . . . , ~
βk, ~
ω1, . . . , ~
ωpi
is a basis for for the sum U + W.
(c) Conclude that dim(U + W) = dim(U) + dim(W) − dim(U ∩ W).
(d) Let W1 and W2 be eight-dimensional subspaces of a ten-dimensional space.
List all values possible for dim(W1 ∩ W2).
4.36 Let V = W1 ⊕ . . . ⊕ Wk and for each index i suppose that Si is a linearly
independent subset of Wi. Prove that the union of the Si’s is linearly independent.
4.37 A matrix is symmetric if for each pair of indices i and j, the i, j entry equals
the j, i entry. A matrix is antisymmetric if each i, j entry is the negative of the j, i
entry.
(a) Give a symmetric 2×2 matrix and an antisymmetric 2×2 matrix. (Remark.
For the second one, be careful about the entries on the diagional.)
(b) What is the relationship between a square symmetric matrix and its trans-
pose? Between a square antisymmetric matrix and its transpose?
(c) Show that Mn
×
n is the direct sum of the space of symmetric matrices and
the space of antisymmetric matrices.
4.38 Let W1, W2, W3 be subspaces of a vector space. Prove that (W1 ∩W2)+(W1 ∩
W3) ⊆ W1 ∩ (W2 + W3). Does the inclusion reverse?
4.39 The example of the x-axis and the y-axis in R2
shows that W1 ⊕W2 = V does
not imply that W1 ∪ W2 = V . Can W1 ⊕ W2 = V and W1 ∪ W2 = V happen?
X 4.40 Our model for complementary subspaces, the x-axis and the y-axis in R2
,
has one property not used here. Where U is a subspace of Rn
we define the
orthocomplement of U to be
U⊥
= {~
v ∈ Rn
¯
¯ ~
v ~
u = 0 for all ~
u ∈ U}
(read “U perp”).
(a) Find the orthocomplement of the x-axis in R2
.
(b) Find the orthocomplement of the x-axis in R3
.
(c) Find the orthocomplement of the xy-plane in R3
.
(d) Show that the orthocomplement of a subspace is a subspace.
(e) Show that if W is the orthocomplement of U then U is the orthocomplement
of W.
(f) Prove that a subspace and its orthocomplement have a trivial intersection.
(g) Conclude that for any n and subspace U ⊆ Rn
we have that Rn
= U ⊕ U⊥
.
(h) Show that dim(U) + dim(U⊥
) equals the dimension of the enclosing space.
X 4.41 Consider Corollary 4.13. Does it work both ways — that is, supposing that
V = W1 + · · · + Wk, is V = W1 ⊕ . . . ⊕ Wk if and only if dim(V ) = dim(W1) +
· · · + dim(Wk)?
4.42 We know that if V = W1 ⊕ W2 then there is a basis for V that splits into a
basis for W1 and a basis for W2. Can we make the stronger statement that every
basis for V splits into a basis for W1 and a basis for W2?
4.43 We can ask about the algebra of the ‘+’ operation.
(a) Is it commutative; is W1 + W2 = W2 + W1?
(b) Is it associative; is (W1 + W2) + W3 = W1 + (W2 + W3)?

(c) Let W be a subspace of some vector space. Show that W + W = W.
(d) Must there be an identity element, a subspace I such that I +W = W +I =
W for all subspaces W?
(e) Does left-cancelation hold: if W1 + W2 = W1 + W3 then W2 = W3? Right
cancelation?
4.44 Consider the algebraic properties of the direct sum operation.
(a) Does direct sum commute: does V = W1 ⊕ W2 imply that V = W2 ⊕ W1?
(b) Prove that direct sum is associative: (W1 ⊕ W2) ⊕ W3 = W1 ⊕ (W2 ⊕ W3).
(c) Show that R3
is the direct sum of the three axes (the relevance here is that by
the previous item, we needn’t specify which two of the threee axes are combined
first).
(d) Does the direct sum operation left-cancel: does W1 ⊕ W2 = W1 ⊕ W3 imply
W2 = W3? Does it right-cancel?
(e) There is an identity element with respect to this operation. Find it.
(f) Do some, or all, subspaces have inverses with respect to this operation: is
there a subspace W of some vector space such that there is a subspace U with
the property that U ⊕ W equals the identity element from the prior item?

Topic: Fields 141
Topic: Fields
Linear combinations involving only fractions or only integers are much easier
for computations than combinations involving real numbers, because computing
with irrational numbers is awkward. Could other number systems, like the
rationals or the integers, work in the place of R in the definition of a vector
space?
Yes and no. If we take “work” to mean that the results of this chapter
remain true then an analysis of which properties of the reals we have used in
this chapter gives the following list of conditions an algebraic system needs in
order to “work” in the place of R.
Definition. A field is a set F with two operations ‘+’ and ‘·’ such that
(1) for any a, b ∈ F the result of a + b is in F and
• a + b = b + a
• if c ∈ F then a + (b + c) = (a + b) + c
(2) for any a, b ∈ F the result of a · b is in F and
• a · b = b · a
• if c ∈ F then a · (b · c) = (a · b) · c
(3) if a, b, c ∈ F then a · (b + c) = a · b + a · c
(4) there is an element 0 ∈ F such that
• if a ∈ F then a + 0 = a
• for each a ∈ F there is an element −a ∈ F such that (−a) + a = 0
(5) there is an element 1 ∈ F such that
• if a ∈ F then a · 1 = a
• for each non-0 element a ∈ F there is an element a−1
∈ F such that
a−1
· a = 1.
The number system comsisting of the set of real numbers along with the
usual addition and multiplication operation is a field, naturally. Another field is
the set of rational numbers with its usual addition and multiplication operations.
An example of an algebraic structure that is not a field is the integer number
system—it fails the final condition.
Some examples are surprising. The set {0, 1} under these operations:
+ 0 1
0 0 1
1 1 0
· 0 1
0 0 0
1 0 1
is a field (see Exercise 4).

We could develop Linear Algebra as the theory of vector spaces with scalars
from an arbitrary field, instead of sticking to taking the scalars only from R. In
that case, almost all of the statements in this book would carry over by replacing
‘R’ with ‘F’, and thus by taking coefficients, vector entries, and matrix entries
to be elements of F. (This says “almost all” because statements involving
distances or angles are exceptions.) Here are some examples; each applies to a
vector space V over a field F.
∗ For any ~
v ∈ V and a ∈ F, (i) 0 · ~
v = ~
0, and (ii) −1 · ~
v + ~
v = ~
0, and
(iii) a ·~
0 = ~
0.
∗ The span (the set of linear combinations) of a subset of V is a subspace
of V .
∗ Any subset of a linearly independent set is also linearly independent.
∗ In a finite-dimensional vector space, any two bases have the same number
of elements.
(Even statements that don’t explicitly mention F use field properties in their
proof.)
We won’t develop vector spaces in this more general setting because the
additional abstraction can be a distraction. The ideas we want to bring out
already appear when we stick to the reals.
The only exception is in Chapter Five. In that chapter we must factor
polynomials, so we will switch to considering vector spaces over the field of
complex numbers. We will discuss this more, including a brief review of complex
arithmetic, when we get there.
Exercises
1 Show that the real numbers form a field.
2 Prove that these are fields:
(a) the rational numbers (b) the complex numbers.
3 Give an example that shows that the integer number system is not a field.
4 Consider the set {0, 1} subject to the operations given above. Show that it is a
field.
5 Come up with suitable operations to make the set {0, 1, 2} a field.

Topic: Crystals 143
Topic: Crystals
Everyone has noticed that table salt comes in little cubes.
Remarkably, the explanation for the cubical external shape is the simplest one
possible: the internal shape, the way the atoms lie, is also cubical. The internal
structure is pictured below. Salt is sodium cloride, and the small spheres shown
are sodium while the big ones are cloride. (To simplify the view, only the
sodiums and clorides on the front, top, and right are shown.)
The specks of salt that we see when we spread a little out on the table consist of
many repetitions of this fundamental unit. That is, these cubes of atoms stack
up to make the larger cubical structure that we see. A solid, such as table salt,
with a regular internal structure is a crystal.
We can restrict our attention to the front face. There, we have this pattern
repeated many times.
The distance between the corners of this cell is about 3.34 Ångstroms (an
Ångstrom is 10−10
meters). Obviously that unit is unwieldly for describing
points in the crystal lattice. Instead, the thing to do is to take as a unit the
length of each side of the square. That is, we naturally adopt this basis.
h
µ
3.34
0
¶
,
µ
0
3.34
¶
i

Then we can describe, say, the corner in the upper right of the picture above as
3~
β1 + 2~
β2.
Another crystal from everyday experience is pencil lead. It is graphite,
formed from carbon atoms arranged in this shape.
This is a single plane of graphite. A piece of graphite consists of many of these
planes layered in a stack. (The chemical bonds between the planes are much
weaker than the bonds inside the planes, which explains why graphite writes—
it can be sheared so that the planes slide off and are left on the paper.) A
convienent unit of length can be made by decomposing the hexagonal ring into
three regions that are rotations of this unit cell.
A natural basis then would consist of the vectors that form the sides of that
unit cell. The distance along the bottom and slant is 1.42 Ångstroms, so this
h
µ
1.42
0
¶
,
µ
1.23
.71
¶
i
is a good basis.
The selection of convienent bases extends to three dimensions. Another
familiar crystal formed from carbon is diamond. Like table salt, it is built from
cubes, but the structure inside each cube is more complicated than salt’s. In
addition to carbons at each corner,
there are carbons in the middle of each face.

Topic: Crystals 145
(To show the added face carbons clearly, the corner carbons have been reduced
to dots.) There are also four more carbons inside the cube, two that are a
quarter of the way up from the bottom and two that are a quarter of the way
down from the top.
(As before, carbons shown earlier have been reduced here to dots.) The dis-
tance along any edge of the cube is 2.18 Ångstroms. Thus, a natural basis for
describing the locations of the carbons, and the bonds between them, is this.
h


2.18
0
0

 ,


0
2.18
0

 ,


0
0
2.18

i
Even the few examples given here show that the structures of crystals is com-
plicated enough that some organized system to give the locations of the atoms,
and how they are chemically bound, is needed. One tool for that organization
is a convienent basis. This application of bases is simple, but it shows a context
where the idea arises naturally. The work in this chapter just takes this simple
idea and develops it.
Exercises
1 How many fundamental regions are there in one face of a speck of salt? (With a
ruler, we can estimate that face is a square that is 0.1 cm on a side.)
2 In the graphite picture, imagine that we are interested in a point 5.67 Ångstroms
up and 3.14 Ångstroms over from the origin.
(a) Express that point in terms of the basis given for graphite.
(b) How many hexagonal shapes away is this point from the origin?
(c) Express that point in terms of a second basis, where the first basis vector is
the same, but the second is perpendicular to the first (going up the plane) and
of the same length.
3 Give the locations of the atoms in the diamond cube both in terms of the basis,
and in Ångstroms.
4 This illustrates how the dimensions of a unit cell could be computed from the
shape in which a substance crystalizes ([Ebbing], p. 462).
(a) Recall that there are 6.022×1023
atoms in a mole (this is Avagadro’s number).
From that, and the fact that platinum has a mass of 195.08 grams per mole,
calculate the mass of each atom.
(b) Platinum crystalizes in a face-centered cubic lattice with atoms at each lattice
point, that is, it looks like the middle picture given above for the diamond crystal.
Find the number of platinums per unit cell (hint: sum the fractions of platinums
that are inside of a single cell).
(c) From that, find the mass of a unit cell.

(d) Platinum crystal has a density of 21.45 grams per cubic centimeter. From
this, and the mass of a unit cell, calculate the volume of a unit cell.
(e) Find the length of each edge.
(f) Describe a natural three-dimensional basis.

Topic: Voting Paradoxes 147
Topic: Voting Paradoxes
Imagine that a Political Science class studying the American presidential pro-
cess holds a mock election. Members of the class are asked to rank, from most
preferred to least preferred, the nominees from the Democratic Party, the Re-
publican Party, and the Third Party, and this is the result (> means ‘is preferred
to’).
preference order
number with
that preference
Democrat > Republican > Third 5
Democrat > Third > Republican 4
Republican > Democrat > Third 2
Republican > Third > Democrat 8
Third > Democrat > Republican 8
Third > Republican > Democrat 2
total 29
What is the preference of the group as a whole?
Overall, the group prefers the Democrat to the Republican (by five votes;
seventeen voters ranked the Democrat above the Republican versus twelve the
other way). And, overall, the group prefers the Republican to the Third’s
nominee (by one vote; fifteen to fourteen). But, strangely enough, the group
also prefers the Third to the Democrat (by seven votes; eighteen to eleven).
Democrat
5 voters
Third
1 voter
Republican
7 voters
This is an example of a voting paradox, specifically, a majority cycle.
Voting paradoxes are studied in part because of their implications for practi-
cal politics. For instance, the instructor can manipulate the class into choosing
the Democrat as the overall winner by first asking the class to choose between
the Republican and the Third, and then asking the class to choose between the
winner of that contest (the Republican) and the Democrat. By similar manipu-
lations, any of the other two candidates can be made to come out as the winner.
(In this Topic we will stick to three-candidate elections, but similar results apply
to larger elections.)
Voting paradoxes are also studied simply because they are mathematically
interesting. One interesting aspect is that the group’s overall majority cycle
occurs despite that each single voters’s preference list is rational—in a straight-
line order. That is, the majority cycle seems to arise in the aggregate, without
being present in the elements of that aggregate, the preference lists. Recently,

however, linear algebra has been used [Zwicker] to argue that a tendency toward
cyclic preference is actually present in each voter’s list, and that it surfaces when
there is more adding of the tendency than cancelling.
For this argument, abbreviating the choices as D, R, and T, we can describe
how a voter with preference order D > R > T contributes to the above cycle
D 1 voter
T
1 voter
R
−1 voter
(the negative sign is here because the arrow describes T as preferred to D, but
this voter likes them the other way). The descriptions for the other preference
lists are in the table on page 150. Now, to conduct the election, we linearly
combine these descriptions; for instance, the Political Science mock election
5 ·
D 1 voter
T
1 voter
R
−1 voter
+ 4 ·
D 1 voter
T
−1 voter
R
−1 voter
+ · · · + 2 ·
D −1 voter
T
−1 voter
R
1 voter
yields the circular group preference shown earlier.
Of course, taking linear combinations is linear algebra. The above cycle no-
tation is suggestive but inconvienent, so we temporarily switch to using column
vectors by starting at the D and taking the numbers from the cycle in coun-
terclockwise order. Thus, the mock election and a single D > R > T vote are
represented in this way.


7
1
5

 and


−1
1
1


We will decompose vote vectors into two parts, one cyclic and the other acyclic.
For the first part, we say that a vector is purely cyclic if it is in this subspace
of R3
.
C = {


k
k
k


¯
¯ k ∈ R} = {k ·


1
1
1


¯
¯ k ∈ R}
For the second part, consider the subspace (see Exercise 6) of vectors that are
perpendicular to all of the vectors in C.
C⊥
= {


c1
c2
c3


¯
¯


c1
c2
c3




k
k
k

 = 0 for all k ∈ R}
= {


c1
c2
c3


¯
¯ c1 + c2 + c3 = 0}
= {c2


−1
1
0

 + c3


−1
0
1


¯
¯ c2, c3 ∈ R}

(Read that aloud as “C perp”.) Consideration of those two has led to this basis
of R3
.
h


1
1
1

 ,


−1
1
0

 ,


−1
0
1

i
We can represent votes with respect to this basis, and thereby decompose them
into a cyclic part and an acyclic part. (Note for readers who have covered the
optional section: that is, the space is the direct sum of C and C⊥
.)
For example, consider the D > R > T voter discussed above. The represen-
tation in terms of the basis is easily found,
c1 − c2 − c3 = −1
c1 + c2 = 1
c1 + c3 = 1
−ρ1+ρ2
−→
−ρ1+ρ3
(−1/2)ρ2+ρ3
−→
c1 − c2 − c3 = −1
2c2 + c3 = 2
(3/2)c3 = 1
so that c1 = 1/3, c2 = 2/3, and c3 = 2/3. Then


−1
1
1

 =
1
3
·


1
1
1

 +
2
3
·


−1
1
0

 +
2
3
·


−1
0
1

 =


1/3
1/3
1/3

 +


−4/3
2/3
2/3


gives the desired decomposition into a cyclic part and and an acyclic part.
D
1
T
1
R
−1
=
D 1/3
T
1/3
R
1/3
+
D 2/3
T
2/3
R
−4/3
Thus, this D > R > T voter’s rational preference list can indeed be seen to
have a cyclic part.
The T > R > D voter is opposite to the one just considered in that the ‘>’
symbols are reversed. This voter’s decomposition
D
−1
T
−1
R
1
=
D −1/3
T
−1/3
R
−1/3
+
D −2/3
T
−2/3
R
4/3
shows that these opposite preferences have decompositions that are opposite.
We say that the first voter has positive spin since the cycle part is with the
direction we have chosen for the arrows, while the second voter’s spin is negative.
The fact that that these opposite voters cancel each other is reflected in the
fact that their vote vectors add to zero. This suggests an alternate way to tally
an election. We could first cancel as many opposite preference lists as possible,
and then determine the outcome by adding the remaining lists.
The rows of the table below contain the three pairs of opposite preference
lists, and the columns group those pairs by spin. For instance, the first row
contains the two voters just considered.

positive spin negative spin
Democrat > Republican > Third
D 1
T
1
R
−1
=
D 1/3
T
1/3
R
1/3
+
D 2/3
T
2/3
R
−4/3
Third > Republican > Democrat
D −1
T
−1
R
1
=
D −1/3
T
−1/3
R
−1/3
+
D −2/3
T
−2/3
R
4/3
Republican > Third > Democrat
D −1
T
1
R
1
=
D 1/3
T
1/3
R
1/3
+
D −4/3
T
2/3
R
2/3
Democrat > Third > Republican
D 1
T
−1
R
−1
=
D −1/3
T
−1/3
R
−1/3
+
D 4/3
T
−2/3
R
−2/3
Third > Democrat > Republican
D 1
T
−1
R
1
=
D 1/3
T
1/3
R
1/3
+
D 2/3
T
−4/3
R
2/3
Republican > Democrat > Third
D −1
T
1
R
−1
=
D −1/3
T
−1/3
R
−1/3
+
D −2/3
T
4/3
R
−2/3
If we conduct the election as just described then after the cancellation of as many
opposite pairs of voters as possible, there will be left three sets of preference
lists, one set from the first row, one set from the second row, and one set from
the third row. We will finish by proving that a voting paradox can happen
only if the spins of these three sets are in the same direction. That is, for a
voting paradox to occur, the three remaining sets must all come from the left
of the table or all come from the right (see Exercise 3). This shows that there
is some connection between the majority cycle and the decomposition that we
are using—a voting paradox can happen only when the tendencies toward cyclic
preference reinforce each other.
For the proof, assume that opposite preference orders have been cancelled,
and we are left with one set of preference lists from each of the three rows.
Consider the sum of these three (here, a, b, and c could be positive, negative,
or zero).
D
a
T
a
R
−a
+
D
−b
T
b
R
b
+
D
c
T
−c
R
c
=
D
a − b + c
T
a + b − c
R
−a + b + c
A voting paradox occurs when the three numbers on the right, a − b + c and
a + b − c and −a + b + c, are all nonnegative or all nonpositive. On the left,
at least two of the three numbers, a and b and c, are both nonnegative or both
nonpositive. We can assume that they are a and b. That makes four cases: the
cycle is nonnegative and a and b are nonnegative, the cycle is nonpositive and
a and b are nonpositive, etc. We will do only the first case, since the second is
similar and the other two are also easy.
So assume that the cycle is nonnegative and that a and b are nonnegative.
The conditions 0 ≤ a − b + c and 0 ≤ −a + b + c add to give that 0 ≤ 2c, which
implies that c is also nonnegative, as desired. That ends the proof.
This result only says that having all three spin in the same direction is a
necessary condition for a majority cycle. It is not also a sufficient condition; see
Exercise 4.

Voting theory and associated topics are the subject of current research. The
are many surprising and intriguing results, most notably the one produced by
K. Arrow [Arrow], who won the Nobel Prize in part for this work, showing, es-
sentially, that no voting system is entirely fair. For more information, some good
introductory articles are [Gardner, 1970], [Gardner, 1974], [Gardner, 1980], and
[Neimi & Riker]. A quite readable recent book is [Taylor]. The material of this
Topic is largely drawn from [Zwicker]. (Author’s Note: I would like to thank
Professor Zwicker for his kind and illuminating discussions.)
Exercises
1 Here is a reasonable way in which a voter could have a cyclic preference. Suppose
that this voter ranks each candidate on each of three criteria.
(a) Draw up a table with the rows labelled ‘Democrat’, ‘Republican’, and ‘Third’,
and the columns labelled ‘character’, ‘experience’, and ‘policies’. Inside each
column, rank some candidate as most preferred, rank another as in the middle,
and rank the remaining one as least preferred.
(b) In this ranking, is the Democrat preferred to the Republican in (at least) two
out of three criteria, or vice versa? Is the Republican preferred to the Third?
(c) Does the table that was just constructed have a cyclic preference order? If
not, make one that does.
So it is possible for a voter to have a cyclic preference among candidates. The
paradox described above, however, is that even if each voter has a straight-line
preference list, there can still be a cyclic group preference.
2 Compute the values in the table of decompositions.
3 Do the cancellations of opposite preference orders for the Political Science class’s
mock election. Are all the remaining preferences from the left three rows of the
table or from the right?
4 The necessary condition that is proved above—a voting paradox can happen only
if all three preference lists remaining after cancellation have the same spin—is not
also sufficient.
(a) Continuing the positive cycle case considered in the proof, use the two in-
equalities 0 ≤ a − b + c and 0 ≤ −a + b + c to show that |a − b| ≤ c.
(b) Also show that c ≤ a + b, and hence that |a − b| ≤ c ≤ a + b.
(c) Give an example of a vote where there is a majority cycle, and addition of
one more voter with the same spin causes the cycle to go away.
(d) Can the opposite happen; can addition of one voter with a “wrong” spin
cause a cycle to appear?
(e) Give a condition that is both necessary and sufficient to get a majority cycle.
5 A one-voter election cannot have a majority cycle because of the requirement
that we’ve imposed that the voter’s list must be rational.
(a) Show that a two-voter election may have a majority cycle. (We consider the
group preference a majority cycle if all three group totals are nonnegative or if
all three are nonpositive—that is, we allow some zero’s in the group preference.)
(b) Show that for any number of voters greater than one, there is an election
involving that many voters that results in a majority cycle.
6 Let U be a subspace of R3
. Prove that the set U⊥
= {~
v
¯
¯ ~
v ~
u = 0 for all ~
u ∈ U}
of vectors that are perpendicular to each vector in U is also a subspace of R3
.

Topic: Dimensional Analysis
“You can’t add apples and oranges,” the old saying goes. It reflects the com-
mon experience that in applications the numbers are associated with units, and
keeping track of the units is worthwhile. Everyone is familiar with calculations
such as this one that use the units as a check.
60
sec
min
· 60
min
hr
· 24
hr
day
· 365
day
year
= 31 536 000
sec
year
However, the idea of paying attention to how the quantities are measured can
be pushed beyond bookkeeping. It can be used to draw conclusions about the
nature of relationships among physical quantities.
Consider this equation expressing a relationship: dist = 16 · (time)2
. If
distance is taken in feet and time in seconds then this is a true statement about
the motion of a falling body. But this equation is a correct description only in
the foot-second unit system. In the yard-second unit system it is not the case
that d = 16t2
. To get a complete equation—one that holds irrespective of the
size of the units—we will make the 16 a dimensional constant.
dist = 16
ft
sec2
· (time)2
Now, the equation holds in any units system, e.g., in yards and seconds we have
this.
dist in yd = 16
(1/3) yd
sec2
· (time in sec)2
=
16
3
yd
sec2
· (time in sec)2
The results below hold for complete equations.
Dimensional analysis can be applied to many areas, but we shall stick to
Newtonian dynamics. In the light of the prior paragraph, we shall work outside
of any particular unit system, and instead say that all quantities are measured
in combinations of (some units of) length L, mass M, and time T. Thus, for
instance, the dimensional formula of velocity is L/T and that of density is
M/L3
. We shall prefer to write those by including even the dimensions with a
zero exponent, e.g., as L1
M0
T−1
and L−3
M1
T0
.
In this terminology, the saying “You can’t add apples to oranges” becomes
the advice to have all of the terms in an equation have the same dimensional
formula. Such an equation is dimensionally homogeneous. An example is this
version of the falling body equation: d − gt2
= 0 where the dimensional formula
of d is L1
M0
T0
, that of g is L1
M0
T−2
, and that of t is L0
M0
T1
(g is the
dimensional constant expressed above in units of ft/sec2
). The gt2
term works
out as L1
M0
T−2
(L0
M0
T1
)2
= L1
M0
T0
, and so it has the same dimensional
formula as the d term.
Quantities with dimensional formula L0
M0
T0
, are said to be dimensionless.
An example of such a quantity is the measure of an angle. An angle measured
in radians is the ratio of the subtended arc to the radius.

Topic: Dimensional Analysis 153
rad
arc
θ
This is the ratio of a length to a length L1
M0
T0
/L1
M0
T0
and thus angles have
the dimensional formula L0
M0
T0
.
Paying attention to the dimensional formulas of the physical quantities will
help us to see which relationships are possible or impossible among the quanti-
ties. For instance, suppose that we want to give the period of a pendulum as
some formula p = · · · involving the other relevant physical quantities, length of
the string, etc. (see the table on page 154). The period is expressed in units of
time—it has dimensional formula L0
M0
T1
—and so the quantities on the other
side of the equation must have their dimensional formulas combine in such a
way that the L’s and M’s cancel and only a T is left. For instance, in that table,
the only quantities involving L are the length of the string and the acceleration
due to gravity. For these L’s to cancel, the quantities must enter the equation
in ratio, e.g., as (`/g)2
or as cos(`/g), or as (`/g)−1
. In this way, simply from
consideration of the dimensional formulas, we know that that the period can be
written as a function of `/g; the formula cannot possibly involve, say, `3
and
g−2
because the dimensional formulas wouldn’t cancel their L’s.
To do dimensional analysis systematically, we need two results (for proofs,
see [Bridgman], Chapter II and IV). First, each equation relating physical quan-
tities that we shall see involves a sum of terms, where each term has the form
mp1
1 mp2
2 . . . mpk
k
for numbers m1, . . . , mk that measure the quantities.
Next, observe that an easy way to construct a dimensionally homogeneous
expression is by taking a product of dimensionless quantities, or by adding
such dimensionless terms. The second result, Buckingham’s Theorem, is that
any complete relationship among quantities with dimensional formulas can be
algebraically manipulated into a form where there is some function f such that
f(Π1, . . . , Πn) = 0
for a complete set {Π1, . . . , Πn} of dimensionless products. (We shall see what
makes a set of dimensionless products ‘complete’ in the examples below.) We
usually want to express one of the quantities, m1 for instance, in terms of the
others, and for that we will assume that the above equality can be rewritten
m1 = m−p2
2 . . . m−pk
k · ˆ
f(Π2, . . . , Πn)
where Π1 = m1mp2
2 . . . mpk
k is dimensionless and the products Π2, . . . , Πn don’t
involve m1 (as with f, here ˆ
f is just some function, this time of n−1 arguments).
Thus, Buckingham’s Theorem says that to investigate the complete relationships
that are possible, we can look into the dimensionless products that are possible.
For that we will use the material of this chapter.

The classic example is a pendulum. An investigator trying to determine
the formula for its period might conjecture that these are the relevant physical
quantities.
quantity
dimensional
formula
period p L0
M0
T1
length of string ` L1
M0
T0
mass of bob m L0
M1
T0
acceleration due to gravity g L1
M0
T−2
arc of swing θ L0
M0
T0
To find which combinations of the powers in pp1
`p2
mp3
gp4
θp5
yield dimensionless
products, consider this equation.
(L0
M0
T1
)p1
(L1
M0
T0
)p2
(L0
M1
T0
)p3
(L1
M0
T−2
)p4
(L0
M0
T0
)p5
= L0
M0
T0
It gives three conditions on the powers.
p2 + p4 = 0
p3 = 0
p1 − 2p4 = 0
Note that p3 is 0—the mass of the bob does not affect the period. The system’s
solution space can be described in this way (p1 is taken as one of the parameters
in order to express the period in terms of the other quantities).
{






p1
p2
p3
p4
p5






=






1
−1/2
0
1/2
0






p1 +






0
0
0
0
1






p5
¯
¯ p1, p5 ∈ R}
Here is the linear algebra. The set of dimensionless products is the set of
products pp1
`p2
mp3
ap4
θp5
subject to the conditions in the above linear system.
This forms a vector space under the ‘+’ addition operation of multiplying two
such products and the ‘·’ scalar multiplication operation of raising such a prod-
uct to the power of the scalar (see Exercise 5). The term ‘complete set of
dimensionless products’ in Buckingham’s Theorem means a basis for this vector
space.
We can get a basis by first taking p1 = 1 and p5 = 0, and then taking p1 = 0
and p5 = 1. The associated dimensionless products are Π1 = p`−1/2
g1/2
and
Π2 = θ. The set {Π1, Π2} is complete, so we have
p = `1/2
g−1/2
· ˆ
f(θ)
=
p
`/g · ˆ
f(θ)

where ˆ
f is a function that we cannot determine from this analysis (by other
means we know that for small angles it is approximately the constant function
ˆ
f(θ) = 2π).
Thus, analysis of the relationships that are possible between the quantities
with the given dimensional formulas has given us a fair amount of information: a
pendulum’s period does not depend on the mass of the bob, and it rises with
the square root of the length of the string.
For the next example we try to determine the period of revolution of two
bodies in space orbiting each other under mutual gravitational attraction. An
experienced investigator could expect that these are the relevant quantities.
quantity
dimensional
formula
period of revolution p L0
M0
T1
mean radius of separation r L1
M0
T0
mass of the first m1 L0
M1
T0
mass of the second m2 L0
M1
T0
gravitational constant G L3
M−1
T−2
To get the complete set of dimensionless products we consider the equation
(L0
M0
T1
)p1
(L1
M0
T0
)p2
(L0
M1
T0
)p3
(L0
M1
T0
)p4
(L3
M−1
T−2
)p5
= L0
M0
T0
which gives rise to these relationships among the powers
p2 + 3p5 = 0
p3 + p4 − p5 = 0
p1 − 2p5 = 0
with the solution space
{






1
−3/2
1/2
0
1/2






p1 +






0
0
−1
1
0






p4
¯
¯ p1, p4 ∈ R}
(p1 is taken as a parameter so that we can state the period as a function of the
other quantities). As with the pendulum example, the linear algebra here is
that the set of dimensionless products of these quantities forms a vector space,
and we want to produce a basis for that space, a ‘complete’ set of dimensionless
products. One such set, gotten from setting p1 = 1 and p4 = 0, and also
setting p1 = 0 and p4 = 1 is {Π1 = pr−3/2
m
1/2
1 G1/2
, Π2 = m−1
1 m2}. With
that, Buckingham’s Theorem says that any complete relationship among these
quantities must be stateable this form.
p = r3/2
m
−1/2
1 G−1/2
· ˆ
f(m−1
1 m2)
=
r3/2
√
Gm1
· ˆ
f(m2/m1)

Remark. An especially interesting application of the above formula occurs
when when the two bodies are a planet and the sun. The mass of the sun m1
is much larger than that of the planet m2. Thus the argument to ˆ
f is approxi-
mately 0, and we can wonder if this part of the formula remains approximately
constant as m2 varies. One way to see that it does is this. The sun’s mass is
much larger than the planet’s mass and so the mutual rotation is approximately
about the sun’s center. If we vary the planet’s mass m2 by a factor of x then the
force of attraction is multiplied by x, and x times the force acting on x times
the mass results in the same acceleration, about the same center. Hence, the
orbit will be the same, and so its period will be the same, and thus the right side
of the above equation also remains unchanged (approximately). Therefore, for
m2’s much smaller than m1, the value of ˆ
f(m2/m1) is approximately constant
as m2 varies. This result is Kepler’s Third Law: the square of the period of a
planet is proportional to the cube of the mean radius of its orbit about the sun.
In the final example, we will see that sometimes dimensional analysis alone
suffices to essentially determine the entire formula. One of the earliest applica-
tions of the technique was to give the formula for the speed of a wave in deep
water. Lord Raleigh put these down as the relevant quantities.
quantity
dimensional
formula
velocity of the wave v L1
M0
T−1
density of the water d L−3
M1
T0
M0
T−2
wavelength λ L1
M0
T0
Considering
(L1
M0
T−1
)p1
(L−3
M1
T0
)p2
(L1
M0
T−2
)p3
(L1
M0
T0
)p4
= L0
M0
T0
gives this system
p1 − 3p2 + p3 + p4 = 0
p2 = 0
−p1 − 2p3 = 0
with this solution space
{




1
0
−1/2
−1/2



 p1
¯
¯ p1 ∈ R}
(as in the pendulum example, one of the quantities d turns out not to be in-
volved in the relationship). There is thus one dimensionless product, Π1 =
vg−1/2
λ−1/2
, and we have that v is
√
λg times a constant ( ˆ
f is constant since
it is a function of no arguments).
As those three examples show, analysis of the relationships possible among
quantities of the given dimensional formulas can bring us far toward expressing

the relationship among the quantities. For further reading, the classic refer-
ence is [Bridgman]—this brief book is a delight to read. Another source is
[Giordano, Wells, Wilde]. A description of how dimensional analysis fits into
the process of mathematical modeling is [Giordano, Jaye, Weir].
Exercises
1 [de Mestre] Consider a projectile, launched with initial velocity v0, at an angle
θ. An investigation of this motion might start with the guess that these are the
relevant quantities.
quantity
dimensional
formula
horizontal position x L1
M0
T0
vertical position y L1
M0
T0
initial speed v0 L1
M0
T−1
angle of launch θ L0
M0
T0
M0
T−2
time t L0
M0
T1
(a) Show that {gt/v0, gx/v2
0, gy/v2
0, θ} is a complete set of dimensionless prod-
ucts. (Hint. This can be done by finding the appropriate free variables in the
linear system that arises, but there is a shortcut that uses the properties of a
basis.)
(b) These two equations of motion for projectiles are familiar: x = v0 cos(θ)t and
y = v0 sin(θ)t−(g/2)t2
. Algebraic manipulate each to rewrite it as a relationship
among the dimensionless products of the prior item.
2 [Einstein] conjectured that the infrared characteristic frequencies of a solid may
be determined by the same forces between atoms as determine the solid’s ordanary
elastic behavior. The relevant quantities are
quantity
dimensional
formula
characteristic frequency ν L0
M0
T−1
compressibility k L1
M−1
T2
number of atoms per cubic cm N L−3
M0
T0
mass of an atom m L0
M1
T0
Show that there is one dimensionless product. Conclude that, in any complete
relationship among quantities with these dimensional formulas, k is a constant
times ν−2
N−1/3
m−1
. This conclusion played an important role in the early study
of quantum phenomena.
3 [Giordano, Wells, Wilde] Consider the torque produced by an engine. Torque
has dimensional formula L2
M1
T−2
. We may first guess that it depends on the
engine’s rotation rate (with dimensional formula L0
M0
T−1
), and the volume of
air displaced (with dimensional formula L3
M0
T0
).
(a) Try to find a complete set of dimensionless products. What goes wrong?
(b) Adjust the guess by adding the density of the air (with dimensional formula
L−3
M1
T0
). Now find a complete set of dimensionless products.
4 [Tilley] Dominoes falling make a wave. We may conjecture that the wave speed v
depends on the the spacing d between the dominoes, the height h of each domino,
and the acceleration due to gravity g.
(a) Find the dimensional formula for each of the four quantities.

(b) Show that {Π1 = h/d, Π2 = dg/v2
} is a complete set of dimensionless prod-
ucts.
(c) Show that if h/d is fixed then the propagation speed is proportional to the
square root of d.
5 Prove that the dimensionless products form a vector space under the ~
+ operation
of multiplying two such products and the ~
· operation of raising such the product
to the power of the scalar. (The vector arrows are a precaution against confusion.)
That is, prove that, for any particular homogeneous system, this set of products
of powers of m1, . . . , mk
{mp1
1 . . . m
pk
k
¯
¯ p1, . . . , pk satisfy the system}
is a vector space under:
mp1
1 . . . m
pk
k
~
+mq1
1 . . . m
qk
k = mp1+q1
1 . . . m
pk+qk
k
and
r~
·(mp1
1 . . . m
pk
k ) = mrp1
1 . . . m
rpk
k
(assume that all variables represent real numbers).
6 The advice about apples and oranges is not right. Consider the familiar equations
for a circle C = 2πr and A = πr2
.
(a) Check that C and A have different dimensional formulas.
(b) Produce an equation that is not dimensionally homogeneous (i.e., it adds
apples and oranges) but is nonetheless true of any circle.
(c) The prior item asks for an equation that is complete but not dimensionally
homogeneous. Produce an equation that is dimensionally homogeneous but not
complete.
(Just because the old saying isn’t strictly right, doesn’t keep it from being a useful
strategy. Dimensional homogeneity is often used as a check on the plausibility
of equations used in models. For an argument that any complete equation can
easily be made dimensionally homogeneous, see [Bridgman], Chapter I, especially
page 15.)

Chapter 3
Maps Between Spaces
3.I Isomorphisms
In the examples following the definition of a vector space we developed the
intuition that some spaces are “the same” as others. For instance, the space
of two-tall column vectors and the space of two-wide row vectors are not equal
because their elements—column vectors and row vectors—are not equal, but we
have the idea that these spaces differ only in how their elements appear. We
will now make this idea precise.
This section illustrates a common aspect of a mathematical investigation.
With the help of some examples, we’ve gotten an idea. We will next give a formal
definition, and then we will produce some results backing our contention that
the definition captures the idea. We’ve seen this happen already, for instance, in
the first section of the Vector Space chapter. There, the study of linear systems
led us to consider collections closed under linear combinations. We defined such
a collection as a vector space, and we followed it with some supporting results.
Of course, that definition wasn’t an end point, instead it led to new insights
such as the idea of a basis. Here too, after producing a definition, and supporting
it, we will get two (pleasant) surprises. First, we will find that the definition
applies to some unforeseen, and interesting, cases. Second, the study of the
definition will lead to new ideas. In this way, our investigation will build a
momentum.
3.I.1 Definition and Examples
We start with two examples that suggest the right definition.
1.1 Example Consider the example mentioned above, the space of two-wide
row vectors and the space of two-tall column vectors. They are “the same” in
that if we associate the vectors that have the same components, e.g.,
¡
1 2
¢
←→
µ
1
2
¶
159

160 Chapter 3. Maps Between Spaces
then this correspondence preserves the operations, for instance this addition
¡
1 2
¢
+
¡
3 4
¢
=
¡
4 6
¢
←→
µ
1
2
¶
+
µ
3
4
¶
=
µ
4
6
¶
and this scalar multiplication.
5 ·
¡
1 2
¢
=
¡
5 10
¢
←→ 5 ·
µ
1
2
¶
=
µ
5
10
¶
More generally stated, under the correspondence
¡
a0 a1
¢
←→
µ
a0
a1
¶
both operations are preserved:
¡
a0 a1
¢
+
¡
b0 b1
¢
=
¡
a0 + b0 a1 + b1
¢
←→
µ
a0
a1
¶
+
µ
b0
b1
¶
=
µ
a0 + b0
a1 + b1
¶
and
r ·
¡
a0 a1
¢
=
¡
ra0 ra1
¢
←→ r ·
µ
a0
a1
¶
=
µ
ra0
ra1
¶
(all of the variables are real numbers).
1.2 Example Another two spaces we can think of as “the same” are P2, the
space of quadratic polynomials, and R3
. A natural correspondence is this.
a0 + a1x + a2x2
←→


a0
a1
a2

 (e.g., 1 + 2x + 3x2
←→


1
2
3

)
The structure is preserved: corresponding elements add in a corresponding way
a0 + a1x + a2x2
+ b0 + b1x + b2x2
(a0 + b0) + (a1 + b1)x + (a2 + b2)x2
←→


a0
a1
a2

 +


b0
b1
b2

 =


a0 + b0
a1 + b1
a2 + b2


and scalar multiplication corresponds also.
r · (a0 + a1x + a2x2
) = (ra0) + (ra1)x + (ra2)x2
←→ r ·


a0
a1
a2

 =


ra0
ra1
ra2



Section I. Isomorphisms 161
1.3 Definition An isomorphism between two vector spaces V and W is a map
f : V → W that
(1) is a correspondence: f is one-to-one and onto;∗
(2) preserves structure: if ~
v1,~
v2 ∈ V then
f(~
v1 + ~
v2) = f(~
v1) + f(~
v2)
and if ~
v ∈ V and r ∈ R then
f(r~
v) = r f(~
v)
(we write V ∼
= W, read “V is isomorphic to W”, when such a map exists).
(“Morphism” means map, so “isomorphism” means a map expressing sameness.)
1.4 Example The vector space G = {c1 cos θ + c2 sin θ
¯
¯ c1, c2 ∈ R} of func-
tions of θ is isomorphic to the vector space R2
under this map.
c1 cos θ + c2 sin θ
f
7−→
µ
c1
c2
¶
We will check this by going through the conditions in the definition.
We will first verify condition (1), that the map is a correspondence between
the sets underlying the spaces.
To establish that f is one-to-one, we must prove that f(~
a) = f(~
b) only when
~
a = ~
b. If
f(a1 cos θ + a2 sin θ) = f(b1 cos θ + b2 sin θ)
then, by the definition of f,
µ
a1
a2
¶
=
µ
b1
b2
¶
from which we can conclude that a1 = b1 and a2 = b2 because column vectors are
equal only when they have equal components. We’ve proved that f(~
a) = f(~
b)
implies that ~
a = ~
b, which shows that f is one-to-one.
To check that f is onto we must check that any member of the codomain R2
mapped to. But that’s clear—any
µ
x
y
¶
∈ R2
is the image, under f, of this member of the domain: x cos θ + y sin θ ∈ G.
Next we will verify condition (2), that f preserves structure.
∗More information on one-to-one and onto maps is in the appendix.

This computation shows that f preserves addition.
f
¡
(a1 cos θ + a2 sin θ) + (b1 cos θ + b2 sin θ)
¢
= f
¡
(a1 + b1) cos θ + (a2 + b2) sin θ
¢
=
µ
a1 + b1
a2 + b2
¶
=
µ
a1
a2
¶
+
µ
b1
b2
¶
= f(a1 cos θ + a2 sin θ) + f(b1 cos θ + b2 sin θ)
A similar computation shows that f preserves scalar multiplication.
f
¡
r · (a1 cos θ + a2 sin θ)
¢
= f( ra1 cos θ + ra2 sin θ )
=
µ
ra1
ra2
¶
= r ·
µ
a1
a2
¶
= r · f(a1 cos θ + a2 sin θ)
With that, conditions (1) and (2) are verified, so we know that f is an
isomorphism, and we can say that the spaces are isomorphic G ∼
= R2
.
1.5 Example Let V be the space {c1x + c2y + c3z
¯
¯ c1, c2, c3 ∈ R} of linear
combinations of three variables x, y, and z, under the natural addition and
scalar multiplication operations. Then V is isomorphic to P2, the space of
quadratic polynomials.
To show this we will produce an isomorphism map. There is more than one
possibility; for instance, here are four.
c1x + c2y + c3z
f1
7−→ c1 + c2x + c3x2
f2
7−→ c2 + c3x + c1x2
f3
7−→ −c1 − c2x − c3x2
f4
7−→ c1 + (c1 + c2)x + (c1 + c3)x2
Although the first map is the more natural correspondence, below we shall
verify that the second one is an isomorphism, to underline that there are many
isomorphisms other than the obvious one that just carries the coefficients over
(showing that f1 is an isomorphism is Exercise 12).
To show that f2 is one-to-one, we will prove that if f2(c1x + c2y + c3z) =
f2(d1x + d2y + d3z) then c1x + c2y + c3z = d1x + d2y + d3z. The assumption
that f2(c1x+c2y +c3z) = f2(d1x+d2y +d3z) gives, by the definition of f2, that
c2 + c3x + c1x2
= d2 + d3 + d1x2
. Equal polynomials have equal coefficients, so
c2 = d2, c3 = d3, and c1 = d1. Thus f2(c1x + c2y + c3z) = f2(d1x + d2y + d3z)
implies that c1x + c2y + c3z = d1x + d2y + d3z and therefore f2 is one-to-one.

The map f2 is onto because any member a+bx+cx2
of the codomain is the
image of some member of the domain, namely it is the image of cx + ay + bz.
(For instance, 2 + 3x − 4x2
is f2(−4x + 2y + 3z).)
The computations for structure preservation for this map are like those in
the prior example. This map preserves addition
f2
¡
(c1x + c2y + c3z) + (d1x + d2y + d3z)
¢
= f2
¡
(c1 + d1)x + (c2 + d2)y + (c3 + d3)z
¢
= (c2 + d2) + (c3 + d3)x + (c1 + d1)x2
= (c2 + c3x + c1x2
) + (d2 + d3x + d1x2
)
= f2(c1x + c2y + c3z) + f2(d1x + d2y + d3z)
and scalar multiplication.
f2
¡
r · (c1x + c2y + c3z)
¢
= f2(rc1x + rc2y + rc3z)
= rc2 + rc3x + rc1x2
= r · (c2 + c3x + c1x2
)
= r · f2(c1x + c2y + c3z)
Thus f2 is an isomorphism and we write V ∼
= P2.
We are sometimes interested in an isomorphism of a space with itself, called
an automorphism. The identity map is easily seen to be an automorphism. The
next example shows that there are others.
1.6 Example Consider the space P5 of polynomials of degree 5 or less and the
map f that sends a polynomial p(x) to p(x − 1). For instance, under this map
x2
7→ (x−1)2
= x2
−2x+1 and x3
+2x 7→ (x−1)3
+2(x−1) = x3
−3x2
+5x−3.
This map is an automorphism of this space, the check is Exercise 21.
This isomorphism of P5 with itself does more than just tell us that the space
is “the same” as itself. It gives us some insight into the space’s structure. For
instance, below is shown a family of parabolas, graphs of members of P5. Each
has a vertex at y = −1, and the left-most one has zeroes at −2.25 and −1.75,
the next one has zeroes at −1.25 and −0.75, etc.
p0(x) p1(x)

Geometrically, the substitution of x − 1 for x in any function’s argument shifts
its graph to the right by one. In the case of the above picture, f(p0) = p1, and
more generally, f’s action is to shift all of the parabolas to the right by one.
Observe, though, that the picture before f is applied is the same as the picture
after f is applied, because while each parabola moves to the right, another one
comes in from the left to take its place. This also holds true for cubics, etc.
So the automorphism f gives us the insight that P5 has a certain horizontal-
homogeneity—the space looks the same near x = 1 as near x = 0.
1.7 Example A dilation map ds : R2
→ R2
that multiplies all vectors by a
nonzero scalar s is an automorphism of R2
.
~
u
~
v
d1.5
7−→
d1.5(~
u)
d1.5(~
v)
A rotation or turning map tθ : R2
→ R2
that rotates all vectors through an angle
θ is an automorphism.
~
v
tπ/3
7−→
tπ/3(~
v)
A third type of automorphism of R2
is a map f` : R2
→ R2
that flips or reflects
all vectors over a line ` through the origin.
~
v
f`
7−→
f`(~
v)
See Exercise 29.
As described in the preamble to this section, we will next produce some
results supporting the contention that the definition of isomorphism above cap-
tures our intuition of vector spaces being the same.
Of course the definition itself is persuasive: a vector space consists of two
components, a set and some structure, and the definition simply requires that
the sets correspond and that the structures correspond also. Also persuasive are
the examples above. In particular, Example 1.1, which gives an isomorphism
between the space of two-wide row vectors and the space of two-tall column
vectors, dramatizes our intuition that isomorphic spaces are the same in all

relevant respects. Sometimes people say, where V ∼
= W, that “W is just V
painted green”—any differences are merely cosmetic.
Further support for the definition, in case it is needed, is provided by the
following results that, taken together, suggest that all the things of interest
in a vector space correspond under an isomorphism. Since we studied vector
spaces to study linear combinations, “of interest” means “pertaining to linear
combinations”. Not of interest is the way that the vectors are typographically
laid out (or their color!).
As an example, although the definition of isomorphism doesn’t explicitly say
that the zero vectors must correspond, it is a consequence of that definition.
1.8 Lemma An isomorphism maps a zero vector to a zero vector.
Proof. Where f : V → W is an isomorphism, fix any ~
v ∈ V . Then f(~
0V ) =
f(0 · ~
v) = 0 · f(~
v) = ~
0W . QED
The definition of isomorphism requires that sums of two vectors correspond
and that so do scalar multiples. We can extend that to say that all linear
combinations correspond.
1.9 Lemma For any map f : V → W between vector spaces the statements
(1) f preserves structure
f(~
v1 + ~
v2) = f(~
v1) + f(~
v2) and f(c~
v) = c f(~
v)
(2) f preserves linear combinations of two vectors
f(c1~
v1 + c2~
v2) = c1f(~
v1) + c2f(~
v2)
(3) f preserves linear combinations of any finite number of vectors
f(c1~
v1 + · · · + cn~
vn) = c1f(~
v1) + · · · + cnf(~
vn)
are equivalent.
Proof. Since the implications (3) =⇒ (2) and (2) =⇒ (1) are clear, we need
only show that (1) =⇒ (3). Assume statement (1). We will prove statement (3)
by induction on the number of summands n.
The one-summand base case, that f(c~
v) = c f(~
v), is covered by state-
ment (1).
For the inductive step assume that statement (3) holds whenever there are k
or fewer summands, that is, whenever n = 1, or n = 2, . . . , or n = k. Consider
the k + 1-summand case. The first half of statement (1) gives
f(c1~
v1 + · · · + ck~
vk + ck+1~
vk+1) = f(c1~
v1 + · · · + ck~
vk) + f(ck+1~
vk+1)

by breaking the sum along the final +. Then the inductive hypothesis lets us
break up the sum of the k things.
= f(c1~
v1) + · · · + f(ck~
vk) + f(ck+1~
vk+1)
Finally, the second half of statement (1) gives
= c1 f(~
v1) + · · · + ck f(~
vk) + ck+1 f(~
vk+1)
when applied k + 1 times. QED
In addition to adding to the intuition that the definition of isomorphism
does indeed preserve things of interest in a vector space, that lemma’s second
item is an especially handy way of checking that a map preserves structure.
We close with a summary. We have defined the isomorphism relation ‘∼
=’
between vector spaces. We have argued that it is the right way to split the
collection of vector spaces into cases because it preserves the features of interest
in a vector space—in particular, it preserves linear combinations. The material
in this section augments the chapter on Vector Spaces. There, after giving the
definition of a vector space, we informally looked at what different things can
happen. We have now said precisely what we mean by ‘different’, and by ‘the
same’, and so we have precisely classified the vector spaces.
Exercises
X 1.10 Verify, using Example 1.4 as a model, that the two correspondences given
before the definition are isomorphisms.
(a) Example 1.1 (b) Example 1.2
X 1.11 For the map f : P1 → R2
given by
a + bx
f
7−→
µ
a − b
b
¶
Find the image of each of these elements of the domain.
(a) 3 − 2x (b) 2 + 2x (c) x
Show that this map is an isomorphism.
1.12 Show that the natural map f1 from Example 1.5 is an isomorphism.
X 1.13 Decide whether each map is an isomorphism (of course, if it is an isomorphism
then prove it and if it isn’t then state a condition that it fails to satisfy).
(a) f : M2
×
2 → R given by
µ
a b
c d
¶
7→ ad − bc
(b) f : M2
×
2 → R4
given by
µ
a b
c d
¶
7→



a + b + c + d
a + b + c
a + b
a



(c) f : M2
×
2 → P3 given by
µ
a b
c d
¶
7→ c + (d + c)x + (b + a)x2
+ ax3

(d) f : M2
×
2 → P3 given by
µ
a b
c d
¶
7→ c + (d + c)x + (b + a + 1)x2
+ ax3
1.14 Show that the map f : R1
→ R1
given by f(x) = x3
is one-to-one and onto.
Is it an isomorphism?
X 1.15 Refer to Example 1.1. Produce two more isomorphisms (of course, that they
satisfy the conditions in the definition of isomorphism must be verified).
1.16 Refer to Example 1.2. Produce two more isomorphisms (and verify that they
satisfy the conditions).
X 1.17 Show that, although R2
is not itself a subspace of R3
, it is isomorphic to the
xy-plane subspace of R3
.
1.18 Find two isomorphisms between R16
and M4
×
4.
X 1.19 For what k is Mm
×
n isomorphic to Rk
?
1.20 For what k is Pk isomorphic to Rn
?
1.21 Prove that the map in Example 1.6, from P5 to P5 given by p(x) 7→ p(x − 1),
is a vector space isomorphism.
1.22 Why, in Lemma 1.8, must there be a ~
v ∈ V ? That is, why must V be
nonempty?
1.23 Are any two trivial spaces isomorphic?
1.24 In the proof of Lemma 1.9, what about the zero-summands case (that is, if n
is zero)?
1.25 Show that any isomorphism f : P0 → R1
has the form a 7→ ka for some nonzero
real number k.
X 1.26 These prove that isomorphism is an equivalence relation.
(a) Show that the identity map id: V → V is an isomorphism. Thus, any vector
space is isomorphic to itself.
(b) Show that if f : V → W is an isomorphism then so is its inverse f−1
: W → V .
Thus, if V is isomorphic to W then also W is isomorphic to V .
(c) Show that a composition of isomorphisms is an isomorphism: if f : V → W is
an isomorphism and g : W → U is an isomorphism then so also is g ◦ f : V → U.
Thus, if V is isomorphic to W and W is isomorphic to U, then also V is isomor-
phic to U.
1.27 Suppose that f : V → W preserves structure. Show that f is one-to-one if and
only if the unique member of V mapped by f to ~
0W is ~
0V .
1.28 Suppose that f : V → W is an isomorphism. Prove that the set {~
v1, . . . ,~
vk} ⊆
V is linearly dependent if and only if the set of images {f(~
v1), . . . , f(~
vk)} ⊆ W is
linearly dependent.
X 1.29 Show that each type of map from Example 1.7 is an automorphism.
(a) Dilation ds by a nonzero scalar s.
(b) Rotation tθ through an angle θ.
(c) Reflection f` over a line through the origin.
Hint. For the second and third items, polar coordinates are useful.
1.30 Produce an automorphism of P2 other than the identity map, and other than
a shift map p(x) 7→ p(x − k).
1.31 (a) Show that a function f : R1
→ R1
is an automorphism if and only if it
has the form x 7→ kx for some k 6= 0.
(b) Let f be an automorphism of R1
such that f(3) = 7. Find f(−2).

(c) Show that a function f : R2
→ R2
is an automorphism if and only if it has
the form µ
x
y
¶
7→
µ
ax + by
cx + dy
¶
for some a, b, c, d ∈ R with ad − bc 6= 0. Hint. Exercises in prior subsections
have shown that µ
b
d
¶
is not a multiple of
µ
a
c
¶
if and only if ad − bc 6= 0.
(d) Let f be an automorphism of R2
with
f(
µ
1
3
¶
) =
µ
2
−1
¶
and f(
µ
1
4
¶
) =
µ
0
1
¶
.
Find
f(
µ
0
−1
¶
).
1.32 Refer to Lemma 1.8 and Lemma 1.9. Find two more things preserved by
isomorphism.
1.33 We show that isomorphisms can be tailored to fit in that, sometimes, given
vectors in the domain and in the range we can produce an isomorphism associating
those vectors.
(a) Let B = h~
β1, ~
β2, ~
β3i be a basis for P2 so that any ~
p ∈ P2 has a unique
representation as ~
p = c1
~
β1 + c2
~
β2 + c3
~
β3, which we denote in this way.
RepB(~
p) =
Ã
c1
c2
c3
!
Show that the RepB(·) operation is a function from P2 to R3
(this entails showing
that with every domain vector ~
v ∈ P2 there is an associated image vector in R3
,
and further, that with every domain vector ~
v ∈ P2 there is at most one associated
image vector).
(b) Show that this RepB(·) function is one-to-one and onto.
(c) Show that it preserves structure.
(d) Produce an isomorphism from P2 to R3
that fits these specifications.
x + x2
7→
Ã
1
0
0
!
and 1 − x 7→
Ã
0
1
0
!
1.34 Prove that a space is n-dimensional if and only if it is isomorphic to Rn
.
Hint. Fix a basis B for the space and consider the map sending a vector over to
its representation with respect to B.
1.35 (Requires the subsection on Combining Subspaces, which is optional.) Let U
and W be vector spaces. Define a new vector space, consisting of the set U × W =
{(~
u, ~
w)
¯
¯ ~
u ∈ U and ~
w ∈ W} along with these operations.
(~
u1, ~
w1) + (~
u2, ~
w2) = (~
u1 + ~
u2, ~
w1 + ~
w2) and r · (~
u, ~
w) = (r~
u, r ~
w)
This is a vector space, the external direct sum of U and W).
(a) Check that it is a vector space.
(b) Find a basis for, and the dimension of, the external direct sum P2 × R2
.
(c) What is the relationship among dim(U), dim(W), and dim(U × W)?

(d) Suppose that U and W are subspaces of a vector space V such that V =
U ⊕ W. Show that the map f : U × W → V given by
(~
u, ~
w)
f
7−→ ~
u + ~
w
is an isomorphism. Thus if the internal direct sum is defined then the internal
and external direct sums are isomorphic.
3.I.2 Dimension Characterizes Isomorphism
In the prior subsection, after stating the definition of an isomorphism, we
gave some results supporting the intuition that such a map describes spaces
as “the same”. Here we will formalize this intuition. While two spaces that
are isomorphic are not equal, we think of them as almost equal—as equivalent.
In this subsection we shall show that the relationship ‘is isomorphic to’ is an
equivalence relation.∗
2.1 Theorem Isomorphism is an equivalence relation between vector spaces.
Proof. We must prove that this relation has the three properties of being sym-
metric, reflexive, and transitive. For each of the three we will use item (2) of
Lemma 1.9 and show that the map preserves structure by showing that the it
preserves linear combinations of two members of the domain.
To check reflexivity, that any space is isomorphic to itself, consider the iden-
tity map. It is clearly one-to-one and onto. The calculation showing that it
preserves linear combinations is easy.
id(c1 · ~
v1 + c2 · ~
v2) = c1~
v1 + c2~
v2 = c1 · id(~
v1) + c2 · id(~
v2)
To check symmetry, that if V is isomorphic to W via some map f : V → W
then there is an isomorphism going the other way, consider the inverse map
f−1
: W → V . As stated in the appendix, the inverse of the correspondence
f is also a correspondence, so we need only check that the inverse preserves
linear combinations. Assume that ~
w1 = f(~
v1), i.e., that f−1
(~
w1) = ~
v1, and also
assume that ~
w2 = f(~
v2).
f−1
(c1 · ~
w1 + c2 · ~
w2) = f−1
¡
c1 · f(~
v1) + c2 · f(~
v2)
¢
= f−1
( f
¡
c1~
v1 + c2~
v2)
¢
= c1~
v1 + c2~
v2
= c1 · f−1
(~
w1) + c2 · f−1
(~
w2)
Finally, to check transitivity, that if V is isomorphic to W via some map f
and if W is isomorphic to U via some map g then also V is isomorphic to U,
consider the composition map g ◦ f : V → U. As stated in the appendix, the
∗ More information on equivalence relations is in the appendix.

composition of two correspondences is a correspondence, so we need only check
that the composition preserves linear combinations.
g ◦ f
¡
c1 · ~
v1 + c2 · ~
v2
¢
= g
¡
f(c1 · ~
v1 + c2 · ~
v2)
¢
= g
¡
c1 · f(~
v1) + c2 · f(~
v2)
¢
= c1 · g
¡
f(~
v1)) + c2 · g(f(~
v2)
¢
= c1 · g ◦ f (~
v1) + c2 · g ◦ f (~
v2)
Thus g ◦ f : V → U is an isomorphism. QED
As a consequence of that result, we know that the universe of vector spaces
is partitioned into classes: every space is in one and only one isomorphism class.
Finite-dimensional
vector spaces:
%
$
Ã
!
¿
À
. . .
.V
W .
V ∼
= W
The next result gives a simple criteria describing which spaces are in each class.
2.2 Theorem Vector spaces are isomorphic if and only if they have the same
dimension.
This theorem follows from the next two lemmas.
2.3 Lemma If spaces are isomorphic then they have the same dimension.
Proof. We shall show that an isomorphism of two spaces gives a correspondence
between their bases. That is, where f : V → W is an isomorphism and a basis
for the domain V is B = h~
β1, . . . , ~
βni, then the image set D = hf(~
β1), . . . , f(~
βn)i
is a basis for the codomain W. (The other half of the correspondence—that for
any basis of W the inverse image is a basis for V —follows on recalling that if
f is an isomorphism then f−1
is also an isomorphism, and applying the prior
sentence to f−1
.)
To see that D spans W, fix a ~
w ∈ W, use the fact that f is onto and so there
is a ~
v ∈ V with ~
w = f(~
v), and expand ~
v as a combination of basis vectors.
~
w = f(~
v) = f(v1
~
β1 + · · · + vn
~
βn) = v1 · f(~
β1) + · · · + vn · f(~
βn)
For linear independence of D, if
~
0W = c1f(~
β1) + · · · + cnf(~
βn) = f(c1
~
β1 + · · · + cn
~
βn)
then, since f is one-to-one and so the only vector sent to ~
0W is ~
0V , we have
that ~
0V = c1
~
β1 + · · · + cn
~
βn, implying that all the c’s are zero. QED

2.4 Lemma If spaces have the same dimension then they are isomorphic.
Proof. To show that any two spaces of dimension n are isomorphic, we can
simply show that any one is isomorphic to Rn
. Then we will have shown that
they are isomorphic to each other, by the transitivity of isomorphism (which
was established in Theorem 2.1).
Let V be an n-dimensional space. Fix a basis B = h~
β1, . . . , ~
βni for the
domain V and consider as a function the representation of the members of that
domain with respect to the basis.
~
v = v1
~
β1 + · · · + vn
~
βn
RepB
7−→



v1
.
.
.
vn



(This is well-defined since every ~
v has one and only one such representation—see
Remark 2.5 below.∗
)
This function is one-to-one because if
RepB(u1
~
β1 + · · · + un
~
βn) = RepB(v1
~
β1 + · · · + vn
~
βn)
then



u1
.
.
.
un


 =



v1
.
.
.
vn



and so u1 = v1, . . . , un = vn, and therefore the original arguments u1
~
β1 + · · · +
un
~
βn and v1
~
β1 + · · · + vn
~
βn are equal.
This function is onto; any n-tall vector
~
w =



w1
.
.
.
wn



is the image of some ~
v ∈ V , namely ~
w = RepB(v1
~
β1 + · · · + vn
~
βn).
Finally, this function preserves structure.
RepB(r · ~
u + s · ~
v) = RepB( (ru1 + sv1)~
β1 + · · · + (run + svn)~
βn )
=



ru1 + sv1
.
.
.
run + svn



= r ·



u1
.
.
.
un


 + s ·



v1
.
.
.
vn



= r · RepB(~
u) + s · RepB(~
v)
∗ More information on well-definedness is in the appendix.

Thus the function is an isomorphism, and we can say that any n-dimensional
space is isomorphic to the n-dimensional space Rn
. Consequently, as noted at
the start, any two spaces with the same dimension are isomorphic. QED
2.5 Remark The parenthetical comment in that proof about the role played
by the ‘one and only one representation’ result requires some explanation. We
need to show that each vector in the domain is associated by RepB with one
and only one vector in the codomain.
A contrasting example, where an association doesn’t have this property, is
illuminating. Consider this subset of P2, which is not a basis.
A = {1 + 0x + 0x2
, 0 + 1x + 0x2
, 0 + 0x + 1x2
, 1 + 1x + 2x2
}
Call those four polynomials ~
α1, . . . , ~
α4. If, mimicing above proof, we try to
write the members of P2 as ~
p = c1~
α1 + c2~
α2 + c3~
α3 + c4~
α4, and associate ~
p
with the four-tall vector with components c1, . . . , c4 then there is a problem.
For, consider ~
p(x) = 1 + x + x2
. The set A spans the space P2, so there is at
least one four-tall vector associated with ~
p. But A is not linearly independent
so vectors do not have unique decompositions. In this case, both
~
p(x) = 1~
α1 + 1~
α2 + 1~
α3 + 0~
α4 and ~
p(x) = 0~
α1 + 0~
α2 − 1~
α3 + 1~
α4
and so there is more than one four-tall vector associated with ~
p.




1
1
1
0



 and




0
0
−1
1




If we are trying to think of this association as a function then the problem is
that, for instance, with input ~
p the association does not have a well-defined
output value.
Any map whose definition appears possibly ambiguous must be checked to
see that it is well-defined. For the above proof that check is Exercise 19.
That ends the proof of Theorem 2.2. We say that the isomorphism classes
are characterized by dimension because we can describe each class simply by
giving the number that is the dimension of all of the spaces in that class.
This subsection’s results give us a collection of representatives of the isomor-
phism classes.∗
2.6 Corollary A finite-dimensional vector space is isomorphic to one and only
one of the Rn
.
2.7 Remark The proofs above pack many ideas into a small space. Through
the rest of this chapter we’ll consider these ideas again, and fill them out. For a
taste of this, we will close this section by indicating how we can expand on the
proof of Lemma 2.4.
∗ More information on equivalence class representatives is in the appendix.

2.8 Example The space M2×2 of 2×2 matrices is isomorphic to R4
. With this
basis for the domain
B = h
µ
1 0
0 0
¶
,
µ
0 1
0 0
¶
,
µ
0 0
1 0
¶
,
µ
0 0
0 1
¶
i
the isomorphism given in the lemma, the representation map, simply carries the
entries over.
µ
a b
c d
¶
f1
7−→




a
b
c
d




One way to understand the map f1 is this: we fix the basis B for the domain
and the basis E4 for the codomain, and associate ~
β1 with ~
e1, and ~
β2 with ~
e2,
etc. We then extend this association to all of the vectors in two spaces.
µ
a b
c d
¶
= a~
β1 + b~
β2 + c~
β3 + d~
β4
f1
7−→ a~
e1 + b~
e2 + c~
e3 + d~
e4 =




a
b
c
d




We say that the map has been extended linearly from the bases to the spaces.
We can do the same thing with different bases, for instance, taking this basis
for the domain.
A = h
µ
2 0
0 0
¶
,
µ
0 2
0 0
¶
,
µ
0 0
2 0
¶
,
µ
0 0
0 2
¶
i
Associating corresponding members of A and E4, and extending linearly,
µ
a b
c d
¶
= (a/2)~
α1 + (b/2)~
α2 + (c/2)~
α3 + (d/2)~
α4
f2
7−→ (a/2)~
e1 + (b/2)~
e2 + (c/2)~
e3 + (d/2)~
e4 =




a/2
b/2
c/2
d/2




gives rise to an isomorphism that is different than f1.
We can also change the basis for the codomain. Starting with these bases,
B and D = h




1
0
0
0



 ,




0
1
0
0



 ,




0
0
0
1



 ,




0
0
1
0



i

associating ~
β1 with ~
δ1, etc., and then linearly extending that correspondence to
all of the two spaces
a~
β1 + b~
β2 + c~
β3 + d~
β4 =
µ
a b
c d
¶
f3
7−→ a~
δ1 + b~
δ2 + c~
δ3 + d~
δ4 =




a
b
d
c




gives still another isomorphism.
So there is a connection between the maps between spaces and bases for
those spaces. We will explore that connection in later sections.
We now finish this section with a summary.
Recall that in the first chapter, we defined two matrices as row equivalent
if they can be derived from each other by elementary row operations (this was
the meaning of same-ness that was of interest there). We showed that is an
equivalence relation and so the collection of matrices is partitioned into classes,
where all the matrices that are row equivalent fall together into a single class.
Then, for insight into which matrices are in each class, we gave representatives
for the classes, the reduced echelon form matrices.
In this section, except that the appropriate notion of same-ness here is vector
space isomorphism, we have followed much the same outline. First we defined
isomorphism, saw some examples, and established some basic properties. Then
we showed that it is an equivalence relation, and now we have a set of class
representatives, the real vector spaces R1
, R2
, etc.
Finite-dimensional
vector spaces:
%
$
Ã
!
¿
À
. . .
.V
W .
?R0
?R1
?R2
?R4
?R3
One representative
per class
As before, the list of representatives helps us to understand the partition. It is
simply a classification of spaces by dimension.
In the second chapter, with the definition of vector spaces, we seemed to
have opened up our studies to many examples of new structures besides the
familiar Rn
’s. We now know that isn’t the case. Any finite-dimensional vector
space is actually “the same” as a real space. We are thus considering exactly
the structures that we need to consider.
In the next section, and in the rest of the chapter, we will fill out the work
that we have done here. In particular, in the next section we will consider maps
that preserve structure, but are not necessarily correspondences.
Exercises
X 2.9 Decide if the spaces are isomorphic.

(a) R2
, R4
(b) P5, R5
(c) M2
×
3, R6
(d) P5, M2
×
3 (e) M2
×
k, Ck
X 2.10 Consider the isomorphism RepB(·): P1 → R2
where B = h1, 1 + xi. Find the
image of each of these elements of the domain.
(a) 3 − 2x; (b) 2 + 2x; (c) x
X 2.11 Show that if m 6= n then Rm
6∼
= Rn
.
X 2.12 Is Mm
×
n
∼
= Mn
×
m?
X 2.13 Are any two planes through the origin in R3
isomorphic?
2.14 Find a set of equivalence class representatives other than the set of Rn
’s.
2.15 True or false: between any n-dimensional space and Rn
there is exactly one
isomorphism.
2.16 Can a vector space be isomorphic to one of its (proper) subspaces?
X 2.17 This subsection shows that for any isomorphism, the inverse map is also an iso-
morphism. This subsection also shows that for a fixed basis B of an n-dimensional
vector space V , the map RepB : V → Rn
is an isomorphism. Find the inverse of
this map.
X 2.18 Prove these facts about matrices.
(a) The row space of a matrix is isomorphic to the column space of its transpose.
(b) The row space of a matrix is isomorphic to its column space.
2.19 Show that the function from Theorem 2.2 is well-defined.
2.20 Is the proof of Theorem 2.2 valid when n = 0?
2.21 For each, decide if it is a set of isomorphism class representatives.
(a) {Ck
¯
¯ k ∈ N} (b) {Pk
¯
¯ k ∈ {−1, 0, 1, . . . }} (c) {Mm
×
n
¯
¯ m, n ∈ N}
2.22 Let f be a correspondence between vector spaces V and W (that is, a map
that is one-to-one and onto). Show that the spaces V and W are isomorphic via f
if and only if there are bases B ⊂ V and D ⊂ W such that corresponding vectors
have the same coordinates: RepB(~
v) = RepD(f(~
v)).
2.23 Consider the isomorphism RepB : P3 → R4
.
(a) Vectors in a real space are orthogonal if and only if their dot product is zero.
Give a definition of orthogonality for polynomials.
(b) The derivative of a member of P3 is in P3. Give a definition of the derivative
of a vector in R4
.
X 2.24 Does every correspondence between bases, when extended to the spaces, give
an isomorphism?
2.25 (Requires the subsection on Combining Subspaces, which is optional.) Suppose
that V = V1 ⊕V2 and that V is isomorphic to the space U under the map f. Show
that U = f(V1) ⊕ f(U2).
2.26 Show that this is not a well-defined function from the rational numbers to the
integers: with each fraction, associate the value of its numerator.

3.II Homomorphisms
The definition of isomorphism has two conditions. In this section we will con-
sider the second one, that the map must preserve the algebraic structure of the
space. We will focus on this condition by studying maps that are required only
to preserve structure; that is, maps that are not required to be correspondences.
Experience shows that this kind of map is tremendously useful in the study
of vector spaces. For one thing, as we shall see in the second subsection below,
while isomorphisms describe how spaces are the same, these maps describe how
spaces can be thought of as alike.
3.II.1 Definition
1.1 Definition A function between vector spaces h: V → W that preserves
the operations of addition
if ~
v1,~
v2 ∈ V then h(~
v1 + ~
v2) = h(~
v1) + h(~
v2)
and scalar multiplication
if ~
v ∈ V and r ∈ R then h(r · ~
v) = r · h(~
v)
is a homomorphism or linear map.
1.2 Example The projection map π: R3
→ R2


x
y
z

 π
7−→
µ
x
y
¶
is a homomorphism. It preserves addition
π(


x1
y1
z1

+


x2
y2
z2

) = π(


x1 + x2
y1 + y2
z1 + z2

) =
µ
x1 + x2
y1 + y2
¶
= π(


x1
y1
z1

) + π(


x2
y2
z2

)
and it preserves scalar multiplication.
π(r ·


x1
y1
z1

) = π(


rx1
ry1
rz1

) =
µ
rx1
ry1
¶
= r · π(


x1
y1
z1

)
Note that this map is not an isomorphism, since it is not one-to-one. For
instance, both ~
0 and ~
e3 in R3
are mapped to the zero vector in R2
.
1.3 Example The domain and codomain can be other than spaces of column
vectors. Both of these maps are homomorphisms.

Section II. Homomorphisms 177
(1) f1 : P2 → P3 given by
a0 + a1x + a2x2
7→ a0x + (a1/2)x2
+ (a2/3)x3
(2) f2 : M2×2 → R given by
µ
a b
c d
¶
7→ a + d
The verifications are straightforward.
1.4 Example Between any two spaces there is a zero homomorphism, sending
every vector in the domain to the zero vector in the codomain.
1.5 Example These two suggest why the term ‘linear map’ is used.
(1) The map g: R3
→ R given by


x
y
z

 g
7−→ 3x + 2y − 4.5z
is linear (i.e., is a homomorphism). In contrast, the map ĝ: R3
→ R given
by


x
y
z

 ĝ
7−→ 3x + 2y − 4.5z + 1
is not linear; for instance,
ĝ(


0
0
0

 +


1
0
0

) = 4 while ĝ(


0
0
0

) + ĝ(


1
0
0

) = 5
(to show that a map is not linear we need only produce one example of a
linear combination that is not preserved).
(2) The first of these two maps t1, t2 : R3
→ R2
is linear while the second is
not.


x
y
z

 t1
7−→
µ
5x − 2y
x + y
¶
and


x
y
z

 t2
7−→
µ
5x − 2y
xy
¶
(Finding an example that the second fails to preserve structure is easy.)
What distinguishes the homomorphisms is that the coordinate functions are
linear combinations of the arguments. See also Exercise 22.

Obviously, any isomorphism is a homomorphism—an isomorphism is a ho-
momorphism that is also a correspondence. So, one way to think of the ‘homo-
morphism’ idea is that it is a generalization of ‘isomorphism’, motivated by the
observation that many of the properties of isomorphisms have only to do with
the map respecting structure and not to do with it being a correspondence. As
examples, these two results from the prior section do not use one-to-one-ness or
onto-ness in their proof, and therefore apply to any homomorphism.
1.6 Lemma A homomorphism sends a zero vector to a zero vector.
1.7 Lemma Each of these is a necessary and sufficient condition for f : V → W
to be a homomorphism.
(1) for any c1, c2 ∈ R and ~
v1,~
v2 ∈ V ,
f(c1 · ~
v1 + c2 · ~
v2) = c1 · f(~
v1) + c2 · f(~
v2)
(2) for any c1, . . . , cn ∈ R and ~
v1, . . . ,~
vn ∈ V ,
f(c1 · ~
v1 + · · · + cn · ~
vn) = c1 · f(~
v1) + · · · + cn · f(~
vn)
This lemma simplifies the check that a function is linear since we can combine
the check that addition is preserved with the one that scalar multiplication is
preserved and since we need only check that combinations of two vectors are
preserved.
1.8 Example The map f : R2
→ R4
given by
µ
x
y
¶
f
7−→




x/2
0
x + y
3y




satisfies that check




r1(x1/2) + r2(x2/2)
0
r1(x1 + y1) + r2(x2 + y2)
r1(3y1) + r2(3y2)



 = r1




x1/2
0
x1 + y1
3y1



 + r2




x2/2
0
x2 + y2
3y2




and so it is a homomorphism.
(Sometimes, such as with Lemma 1.15 below, it is less awkward to check preser-
vation of addition and preservation of scalar multiplication separately, but this
is purely a matter of taste.)
However, some of the results that we have seen for isomorphisms fail to hold
for homomorphisms in general. An isomorphism between spaces gives a corre-
spondence between their bases, but a homomorphisms need not; Example 1.2
shows this and another example is the zero map between any two nontrivial
spaces. Instead, a weaker but still very useful result holds.

1.9 Theorem A homomorphism is determined by its action on a basis. That
is, if h~
β1, . . . , ~
βni is a basis of a vector space V and ~
w1, . . . , ~
wn are (perhaps
not distinct) elements of a vector space W then there exists a homomorphism
from V to W sending ~
β1 to ~
w1, . . . , and ~
βn to ~
wn, and that homomorphism is
unique.
Proof. We define the map h: V → W by associating ~
β1 with ~
w1, etc., and then
extending linearly to all of the domain. That is, where ~
v = c1
~
β1 + · · · + cn
~
βn,
let h(~
v) be c1 ~
w1 + · · · + cn ~
wn. This is well-defined because, with respect to the
basis, the representation of each domain vector ~
v is unique.
This map is a homomorphism since it preserves linear combinations; where
~
v1 = c1
~
β1 + · · · + cn
~
βn and ~
v2 = d1
~
β1 + · · · + dn
~
βn, we have this.
h(r1~
v1 + r2~
v2) = h((r1c1 + r2d1)~
β1 + · · · + (r1cn + r2dn)~
βn)
= (r1c1 + r2d1)~
w1 + · · · + (r1cn + r2dn)~
wn
= r1h(~
v1) + r2h(~
v2)
And, this map is unique since if ĥ: V → W is another homomorphism such
that ĥ(~
βi) = ~
wi for each i then h and ĥ agree on all of the vectors in the domain.
ĥ(~
v) = ĥ(c1
~
β1 + · · · + cn
~
βn)
= c1ĥ(~
β1) + · · · + cnĥ(~
βn)
= c1 ~
w1 + · · · + cn ~
wn
= h(~
v)
Thus, h and ĥ are the same map. QED
1.10 Example This result says that we can construct homomorphisms by fix-
ing a basis for the domain and specifying where the map sends those basis vec-
tors. For instance, if we specify a map h: R2
→ R2
that acts on the standard
basis E2 in this way
h(
µ
1
0
¶
) =
µ
−1
1
¶
and h(
µ
0
1
¶
) =
µ
−4
4
¶
then the action of h on any other member of the domain is also specified. For
instance, the value of h on this argument
h(
µ
3
−2
¶
) = h(3 ·
µ
1
0
¶
− 2 ·
µ
0
1
¶
) = 3 · h(
µ
1
0
¶
) − 2 · h(
µ
0
1
¶
) =
µ
5
−5
¶
is a direct consiquence of the value of h on the basis vectors. (Later in this
chapter we shall develop a scheme, using matrices, that is a convienent way to
do computations like this one.)
Just as isomorphisms of a space with itself are useful and interesting, so too
are homomorphisms of a space with itself.

1.11 Definition A linear map from a space into itself t: V → V is a linear
transformation.
In this book we use ‘linear transformation’ only in the case where the codomain
equals the domain, but it is also widely used as a general synonym for ‘homo-
morphism’.
1.12 Example The map on R2
that projects all vectors down to the x-axis
µ
x
y
¶
7→
µ
x
0
¶
is a linear transformation.
1.13 Example The derivative map d/dx: Pn → Pn
a0 + a1x + · · · + anxn d/dx
7−→ a1 + 2a2x + 3a3x2
+ · · · + nanxn−1
is a linear transformation by this result from calculus: d(c1f + c2g)/dx =
c1 (df/dx) + c2 (dg/dx).
1.14 Example The matrix transpose map
µ
a b
c d
¶
7→
µ
a c
b d
¶
is a linear transformation of M2×2. Note that this transformation is one-to-one
and onto, and so in fact is an automorphism.
We finish this subsection about maps by recalling that we can linearly com-
bine maps. For instance, for these maps from R2
to itself
µ
x
y
¶
f
7−→
µ
2x
3x − 2y
¶
and
µ
x
y
¶
g
7−→
µ
0
5x
¶
we can take the linear combination 5f − 2g to get this.
µ
x
y
¶
5f−2g
7−→
µ
10x
5x − 10y
¶
1.15 Lemma For vector spaces V and W, the set of linear functions from V
to W is itself a vector space, a subspace of the space of all functions from V to
W. It is denoted L(V, W).
Proof. This set is non-empty because it contains the zero homomorphism. So
to show that it is a subspace we need only check that it is closed under linear
combinations. Let f, g: V → W be linear. Then their sum is linear
(f + g)(c1~
v1 + c2~
v2) = c1f(~
v1) + c2f(~
v2) + c1g(~
v1) + c2g(~
v2)
= c1
¡
f + g
¢
(~
v1) + c2
¡
f + g
¢
(~
v2)

and any scalar multiple is also linear.
(r · f)(c1~
v1 + c2~
v2) = r(c1f(~
v1) + c2f(~
v2))
= c1(r · f)(~
v1) + c2(r · f)(~
v2)
Hence L(V, W) is a subspace. QED
We started this section by isolating the structure preservation property of
isomorphisms. That is, we defined homomorphisms as a generalization of iso-
morphisms. Some of the properties that we studied for isomorphisms carried
over unchanged, while others were adapted to this more general setting.
It would be a mistake, though, to view this new notion of homomorphism as
derived from or somehow secondary to that of isomorphism. In the rest of this
chapter we shall work mostly with homomorphisms, partly because any state-
ment made about homomorphisms is automatically true about isomorphisms,
but more because, while the isomorphism concept is perhaps more natural, ex-
perience shows that the homomorphism concept is actually more fruitful and
more central to further progress.
Exercises
X 1.16 Decide if each h: R3
→ R2
is linear.
(a) h(
Ã
x
y
z
!
) =
µ
x
x + y + z
¶
(b) h(
Ã
x
y
z
!
) =
µ
0
0
¶
(c) h(
Ã
x
y
z
!
) =
µ
1
1
¶
(d) h(
Ã
x
y
z
!
) =
µ
2x + y
3y − 4z
¶
X 1.17 Decide if each map h: M2
×
2 → R is linear.
(a) h(
µ
a b
c d
¶
) = a + d
(b) h(
µ
a b
c d
¶
) = ad − bc
(c) h(
µ
a b
c d
¶
) = 2a + 3b + c − d
(d) h(
µ
a b
c d
¶
) = a2
+ b2
X 1.18 Show that these two maps are homomorphisms.
(a) d/dx: P3 → P2 given by a0 + a1x + a2x2
+ a3x3
maps to a1 + 2a2x + 3a3x2
(b)
R
: P2 → P3 given by b0 + b1x + b2x2
maps to b0x + (b1/2)x2
+ (b2/3)x3
Are these maps inverse to each other?
1.19 Is (perpendicular) projection from R3
to the xz-plane a homomorphism? Pro-
jection to the yz-plane? To the x-axis? The y-axis? The z-axis? Projection to the
origin?
1.20 Show that, while the maps from Example 1.3 preserve linear operations, they
are not isomorphisms.
1.21 Is an identity map a linear transformation?

X 1.22 Stating that a function is ‘linear’ is different than stating that its graph is a
line.
(a) The function f1 : R → R given by f1(x) = 2x − 1 has a graph that is a line.
Show that it is not a linear function.
(b) The function f2 : R2
→ R given by
µ
x
y
¶
7→ x + 2y
does not have a graph that is a line. Show that it is a linear function.
X 1.23 Part of the definition of a linear function is that it respects addition. Does a
linear function respect subtraction?
1.24 Assume that h is a linear transformation of V and that h~
β1, . . . , ~
βni is a basis
of V . Prove each statement.
(a) If h(~
βi) = ~
0 for each basis vector then h is the zero map.
(b) If h(~
βi) = ~
βi for each basis vector then h is the identity map.
(c) If there is a scalar r such that h(~
βi) = r · ~
βi for each basis vector then
h(~
v) = r · ~
v for all vectors in V .
X 1.25 Consider the vector space R+
where vector addition and scalar multiplication
are not the ones inherited from R but rather are these: a + b is the product of
a and b, and r · a is the r-th power of a. (This was shown to be a vector space
in an earlier exercise.) Verify that the natural logarithm map ln: R+
→ R is a
homomorphism between these two spaces. Is it an isomorphism?
X 1.26 Consider this transformation of R2
.
µ
x
y
¶
7→
µ
x/2
y/3
¶
Find the image under this map of this ellipse.
{
µ
x
y
¶
¯
¯ (x2
/4) + (y2
/9) = 1}
X 1.27 Imagine a rope wound around the earth’s equator so that it fits snugly (sup-
pose that the earth is a sphere). How much extra rope must be added to raise the
circle to a constant six feet off the ground?
X 1.28 Verify that this map h: R3
→ R
Ã
x
y
z
!
7→
Ã
x
y
z
! Ã
3
−1
−1
!
= 3x − y − z
is linear. Generalize.
1.29 Show that every homomorphism from R1
to R1
acts via multiplication by a
scalar. Conclude that every nontrivial linear transformation of R1
is an isomor-
phism. Is that true for transformations of R2
? Rn
?
1.30 (a) Show that for any scalars a1,1, . . . , am,n this map h: Rn
→ Rm
is a ho-
momorphism.



x1
.
.
.
xn


 7→



a1,1x1 + · · · + a1,nxn
.
.
.
am,1x1 + · · · + am,nxn




(b) Show that for each i, the i-th derivative operator di
/dxi
is a linear trans-
formation of Pn. Conclude that for any scalars ck, . . . , c0 this map is a linear
transformation of that space.
f 7→
dk
dxk
f + ck−1
dk−1
dxk−1
f + · · · + c1
d
dx
f + c0f
1.31 Lemma 1.15 shows that a sum of linear functions is linear and that a scalar
multiple of a linear function is linear. Show also that a composition of linear
functions is linear.
X 1.32 Where f : V → W is linear, suppose that f(~
v1) = ~
w1, . . . , f(~
vn) = ~
wn for
some vectors ~
w1, . . . , ~
wn from W.
(a) If the set of ~
w ’s is independent, must the set of ~
v’s also be independent?
(b) If the set of ~
v ’s is independent, must the set of ~
w ’s also be independent?
(c) If the set of ~
w ’s spans W, must the set of ~
v ’s span V ?
(d) If the set of ~
v ’s spans V , must the set of ~
w ’s span W?
1.33 Generalize Example 1.14 by proving that the matrix transpose map is linear.
What is the domain and codomain?
1.34 (a) Where ~
u,~
v ∈ Rn
, the line segment connecting them is defined to be
the set ` = {t · ~
u + (1 − t) · ~
v
¯
¯ t ∈ [0..1]}. Show that the image, under a homo-
morphism h, of the segment between ~
u and ~
v is the segment between h(~
u) and
h(~
v).
(b) A subset of Rn
is convex if, for any two points in that set, the line segment
joining them lies entirely in that set. (The inside of a sphere is convex while the
skin of a sphere is not.) Prove that linear maps from Rn
to Rm
preserve the
property of set convexity.
X 1.35 Let h: Rn
→ Rm
be a homomorphism.
(a) Show that the image under h of a line in Rn
is a (possibly degenerate) line
in Rn
.
(b) What happens to a k-dimensional linear surface?
1.36 Prove that the restriction of a homomorphism to a subspace of its domain is
another homomorphism.
1.37 Assume that h: V → W is linear.
(a) Show that the rangespace of this map {h(~
v)
¯
¯ ~
v ∈ V } is a subspace of the
codomain W.
(b) Show that the nullspace of this map {~
v ∈ V
¯
¯ h(~
v) = ~
0W } is a subspace of
the domain V .
(c) Show that if U is a subspace of the domain V then its image {h(~
u)
¯
¯ ~
u ∈ U}
is a subspace of the codomain W. This generalizes the first item.
(d) Generalize the second item.
1.38 Consider the set of isomorphisms from a vector space to itself. Is this a
subspace of the space L(V, V ) of homomorphisms from the space to itself?
1.39 Does Theorem 1.9 need that h~
β1, . . . , ~
βni is a basis? That is, can we still get
a well-defined and unique homomorphism if we drop either the condition that the
set of ~
β’s be linearly independent, or the condition that it span the domain?
1.40 Let V be a vector space and assume that the maps f1, f2 : V → R1
are lin-
ear.
(a) Define a map F : V → R2
whose component functions are the given linear
ones.
~
v 7→
µ
f1(~
v)
f2(~
v)
¶

Show that F is linear.
(b) Does the converse hold—is any linear map from V to R2
made up of two
linear component maps to R1
?
(c) Generalize.
3.II.2 Rangespace and Nullspace
The difference between homomorphisms and isomorphisms is that while both
kinds of map preserve structure, homomorphisms needn’t be onto and needn’t
be one-to-one. Put another way, homomorphisms are a more general kind of
map; they are subject to fewer conditions than isomorphisms. In this subsection,
we will look at what can happen with homomorphisms that the extra conditions
rule out happening with isomorphisms.
We first consider the effect of dropping the onto requirement. Of course,
any function is onto some set, its range. The next result says that when the
function is a homomorphism, then this set is a vector space.
2.1 Lemma Under a homomorphism, the image of any subspace of the domain
is a subspace of the codomain. In particular, the image of the entire space, the
range of the homomorphism, is a subspace of the codomain.
Proof. Let h: V → W be linear and let S be a subspace of the domain V .
The image h(S) is nonempty because S is nonempty. Thus, to show that h(S)
is a subspace of the codomain W, we need only show that it is closed under
linear combinations of two vectors. If h(~
s1) and h(~
s2) are members of h(S) then
c1 · h(~
s1) + c2 · h(~
s2) = h(c1 ·~
s1) + h(c2 ·~
s2) = h(c1 ·~
s1 + c2 ·~
s2) is also a member
of h(S) because it is the image of c1 · ~
s1 + c2 · ~
s2 from S. QED
2.2 Definition The rangespace of h: V → W is
R(h) = {h(~
v)
¯
¯ ~
v ∈ V }
sometimes denoted h(V ). The dimension of the rangespace is the map’s rank.
(We shall soon see the connection between the rank of a map and the rank of a
matrix.)
2.3 Example Recall that the derivative map d/dx: P3 → P3 given by a0 +
a1x + a2x2
+ a3x3
7→ a1 + 2a2x + 3a3x2
is linear. The rangespace R(d/dx) is
the set of quadratic polynomials {r + sx + tx2
¯
¯ r, s, t ∈ R}. Thus, the rank of
this map is three.
2.4 Example With this homomorphism h: M2×2 → P3
µ
a b
c d
¶
7→ (a + b + 2d) + 0x + cx2
+ cx3

an image vector in the range can have any constant term, must have an x
coefficient of zero, and must have the same coefficient of x2
as of x3
. That is,
the rangespace is R(h) = {r + 0x + sx2
+ sx3
¯
¯ r, s ∈ R} and so the rank is two.
The prior result shows that, in passing from the definition of isomorphism to
the more general definition of homomorphism, omitting the ‘onto’ requirement
doesn’t make an essential difference. Any homomorphism is onto its rangespace.
However, omitting the ‘one-to-one’ condition does make a difference. A
homomorphism may have many elements of the domain map to a single element
in the range. The general picture is below. There is a homomorphism and its
domain, codomain, and range. The homomorphism is many-to-one, and two
elements of the range are shown that are each the image of more than one
member of the domain.
domain V
.
.
codomain W
)
R(h)
(Recall that for a map h: V → W, the set of elements of the domain that are
mapped to ~
w in the codomain {~
v ∈ V
¯
¯ h(~
v) = ~
w} is the inverse image of ~
w. It
is denoted h−1
(~
w); this notation is used even if h has no inverse function, that
is, even if h is not one-to-one.)
2.5 Example Consider the projection π: R3
→ R2


x
y
z

 π
7−→
µ
x
y
¶
which is a homomorphism but is not one-to-one. Picturing R2
as the xy-plane
inside of R3
allows us to see π(~
v) as the “shadow” of ~
v in the plane. In these
terms, the preservation of addition property says that
~
v1 above (x1, y1) plus ~
v2 above (x2, y2) equals ~
v1 + ~
v2 above (x1 + x2, y1 + y2).
Briefly, the shadow of a sum equals the sum of the shadows. (Preservation of
scalar multiplication has a similar interpretation.)

This description of the projection in terms of shadows is is memorable, but
strictly speaking, R2
isn’t equal to the xy-plane inside of R3
(it is composed of
two-tall vectors, not three-tall vectors). Separating the two spaces by sliding
R2
over to the right gives an instance of the general diagram above.
~
w1
~
w2
~
w1 + ~
w2
The vectors that map to ~
w1 on the right have endpoints that lie in a vertical
line on the left. One such vector is shown, in gray. Call any such member
of the inverse image of ~
w1 a “~
w1 vector”. Similarly, there is a vertical line of
“~
w2 vectors”, and a vertical line of “~
w1 + ~
w2 vectors”.
We are interested in π because it is a homomorphism. In terms of the
picture, this means that the classes add; any ~
w1 vector plus any ~
w2 vector
equals a ~
w1 + ~
w2 vector, simply because if π(~
v1) = ~
w1 and π(~
v2) = ~
w2 then
π(~
v1 + ~
v2) = π(~
v1) + π(~
v2) = ~
w1 + ~
w2. (A similar statement holds about the
classes under scalar multiplication.) Thus, although the two spaces R3
and R2
are not isomorphic, π describes a way in which they are alike: vectors in R3
add
like the associated vectors in R2
—vectors add as their shadows add.
2.6 Example A homomorphism can be used to express an analogy between
spaces that is more subtle than the prior one. For instance, this map from R2
to R1
is a homomorphism.
µ
x
y
¶
h
7−→ x + y
Fix two numbers a and b in the range R. Then the preservation of addition
condition says this for two vectors ~
u and ~
v from the domain.
if h(
µ
u1
u2
¶
) = a and h(
µ
v1
v2
¶
) = b then h(
µ
u1 + v1
u2 + v2
¶
) = a + b
As in the prior example, we illustrate by showing the class of vectors in the
domain that map to a, the class of vectors that map to b, and the class of
vectors that map to a + b. Vectors that map to a have components that add
to a, so a vector is in the inverse image h−1
(a) if its endpoint lies on the line
x+y = a. We can call these the “a vectors”. Similarly, we have the “b vectors”,
etc. Now the addition preservation statement becomes this.

(u1, u2)
(v1, v2)
(u1 + v1, u2 + v2)
an a vector plus a b vector equals an a + b vector
Restated, if an a vector is added to a b vector then the result is mapped by h to
the real number a+b. Briefly, the image of a sum is the sum of the images. Even
more briefly, h(~
u + ~
v) = h(~
u) + h(~
v). (The preservation of scalar multiplication
condition has a similar restatement.)
2.7 Example Inverse images can be structures other than lines. For the linear
map h: R3
→ R2


x
y
z

 7→
µ
x
x
¶
the inverse image sets are planes perpendicular to the x-axis.
2.8 Remark We won’t describe how every homomorphism that we will use in
this book is an analogy, both because the formal sense we make of “alike in this
way . . . ” is ‘a homomorphism exists such that . . . ’, and because many vector
spaces are hard to draw (e.g., a space of polynomials). Nonetheless, the idea
that a homomorphism between two spaces expresses how the domain’s vectors
fall into classes that act like the the range’s vectors, is a good way to view
homomorphisms.
We derive two insights from examples 2.5, 2.6, and 2.7.
First, in all three, each inverse image shown is a linear surface. In particular,
the inverse image of the range’s zero vector is a line or plane through the origin—
a subspace of the domain. The next result shows that this insight extends to
any vector space, not just spaces of column vectors (which are the only spaces
where the term ‘linear surface’ is defined).
2.9 Lemma For any homomorphism, the inverse image of a subspace of the
range is a subspace of the domain. In particular, the inverse image of the trivial
subspace of the range is a subspace of the domain.
Proof. Let h: V → W be a homomorphism and let S be a subspace of the
range of h. Consider {~
v ∈ V
¯
¯ h(~
v) ∈ S}, the inverse image of S. It is nonempty

because it contains ~
0V , as S contains ~
0W . To show that it is closed under
combinations, let ~
v1 and ~
v2 be elements of the inverse image, so that h(~
v1) and
h(~
v2) are members of S. Then c1~
v1 + c2~
v2 is also in the inverse image because
under h it is sent h(c1~
v1 +c2~
v2) = c1h(~
v1)+c2h(~
v2) to a member of the subspace
S. QED
2.10 Definition The nullspace or kernel of a linear map h: V → W is
N (h) = {~
v ∈ V
¯
¯ h(~
v) = ~
0W } = h−1
(~
0W ).
The dimension of the nullspace is the map’s nullity.
2.11 Example The map from Example 2.3 has this nullspace N (d/dx) =
{a0 + 0x + 0x2
+ 0x3
¯
¯ a0 ∈ R}.
2.12 Example The map from Example 2.4 has this nullspace.
N (h) = {
µ
a b
0 −(a + b)/2
¶
¯
¯ a, b ∈ R}
Now for the second insight from the above pictures. In Example 2.5, each
of the vertical lines is squashed down to a single point—π, in passing from the
domain to the range, takes all of these one-dimensional vertical lines and “zeroes
them out”, leaving the range one dimension smaller than the domain. Similarly,
in Example 2.6, the two-dimensional domain is mapped to a one-dimensional
range by breaking the domain into lines (here, they are diagonal lines), and
compressing each of those lines to a single member of the range. Finally, in
Example 2.7, the domain breaks into planes which get “zeroed out”, and so the
map starts with a three-dimensional domain but ends with a one-dimensional
range—this map “subtracts” two from the dimension. (Notice that, in this
third example, the codomain is two-dimensional but the range of the map is
only one-dimensional, and it is the dimension of the range that is of interest.)
2.13 Theorem A linear map’s rank plus its nullity equals the dimension of its
domain.
Proof. Let h: V → W be linear and let BN = h~
β1, . . . , ~
βki be a basis for
the nullspace. Extend that to a basis BV = h~
β1, . . . , ~
βk, ~
βk+1, . . . , ~
βni for the
entire domain. We shall show that BR = hh(~
βk+1), . . . , h(~
βn)i is a basis for the
rangespace. Then counting the size of these bases gives the result.
To see that BR is linearly independent, consider the equation ck+1h(~
βk+1)+
· · · + cnh(~
βn) = ~
0W . This gives that h(ck+1
~
βk+1 + · · · + cn
~
βn) = ~
0W and so
ck+1
~
βk+1 +· · ·+cn
~
βn is in the nullspace of h. As BN is a basis for this nullspace,
there are scalars c1, . . . , ck ∈ R satisfying this relationship.
c1
~
β1 + · · · + ck
~
βk = ck+1
~
βk+1 + · · · + cn
~
βn
But BV is a basis for V so each scalar equals zero. Therefore BR is linearly
independent.

To show that BR spans the rangespace, consider h(~
v) ∈ R(h) and write ~
v
as a linear combination ~
v = c1
~
β1 + · · · + cn
~
βn of members of BV . This gives
h(~
v) = h(c1
~
β1+· · ·+cn
~
βn) = c1h(~
β1)+· · ·+ckh(~
βk)+ck+1h(~
βk+1)+· · ·+cnh(~
βn)
and since ~
β1, . . . , ~
βk are in the nullspace, we have that h(~
v) = ~
0 + · · · + ~
0 +
ck+1h(~
βk+1) + · · · + cnh(~
βn). Thus, h(~
v) is a linear combination of members of
BR, and so BR spans the space. QED
2.14 Example Where h: R3
→ R4
is


x
y
z

 h
7−→




x
0
y
0




we have that the rangespace and nullspace are
R(h) = {




a
0
b
0




¯
¯ a, b ∈ R} and N (h) = {


0
0
z


¯
¯ z ∈ R}
and so the rank of h is two while the nullity is one.
2.15 Example If t: R → R is the linear transformation x 7→ −4x, then the
range is R(t) = R1
, and so the rank of t is one and the nullity is zero.
2.16 Corollary The rank of a linear map is less than or equal to the dimension
of the domain. Equality holds if and only if the nullity of the map is zero.
We know that there an isomorphism exists between two spaces if and only
if their dimensions are equal. Here we see that for a homomorphism to exist,
the dimension of the range must be less than or equal to the dimension of the
domain. For instance, there is no homomorphism from R2
onto R3
—there are
many homomorphisms from R2
into R3
, but none has a range that is all of
three-space.
The rangespace of a linear map can be of dimension strictly less than the
dimension of the domain (an example is that the derivative transformation on
P3 has a domain of dimension four but a range of dimension three). Thus, under
a homomorphism, linearly independent sets in the domain may map to linearly
dependent sets in the range (for instance, the derivative sends {1, x, x2
, x3
} to
{0, 1, 2x, 3x2
}). That is, under a homomorphism, independence may be lost. In
contrast, dependence is preserved.
2.17 Lemma Under a linear map, the image of a linearly dependent set is
linearly dependent.
Proof. Suppose that c1~
v1 + · · · + cn~
vn = ~
0V , with some ci nonzero. Then,
because h(c1~
v1 +· · ·+cn~
vn) = c1h(~
v1)+· · ·+cnh(~
vn) and because h(~
0V ) = ~
0W ,
we have that c1h(~
v1) + · · · + cnh(~
vn) = ~
0W with some nonzero ci. QED

When is independence not lost? One obvious sufficient condition is when
the homomorphism is an isomorphism (this condition is also necessary; see
Exercise 34.) We finish our comparison of homomorphisms and isomorphisms by
observing that a one-to-one homomorphism is an isomorphism from its domain
onto its range.
2.18 Definition A linear map that is one-to-one is nonsingular.
(In the next section we will see the connection between this use of ‘nonsingular’
for maps and its familiar use for matrices.)
2.19 Example This nonsingular homomorphism ι: R2
→ R3
µ
x
y
¶
ι
7−→


x
y
0


gives the obvious correspondence between R2
and the xy-plane inside of R3
.
We will close this section by adapting some results about isomorphisms to
this setting.
2.20 Theorem In an n-dimensional vector space V , then these
(1) h is nonsingular, that is, one-to-one
(2) h has a linear inverse
(3) N (h) = {~
0 }, that is, nullity(h) = 0
(4) rank(h) = n
(5) if h~
β1, . . . , ~
βni is a basis for V then hh(~
β1), . . . , h(~
βn)i is a basis for R(h)
are equivalent statements about a linear map h: V → W.
Proof. We will first show that (1) ⇐⇒ (2). We will then show that (1) =⇒
(3) =⇒ (4) =⇒ (5) =⇒ (2).
For (1) =⇒ (2), suppose that the linear map h is one-to-one, and so has an
inverse. The domain of that inverse is the range of h and so a linear combina-
tion of two members of that domain has the form c1h(~
v1) + c2h(~
v2). On that
combination, the inverse h−1
gives this.
h−1
(c1h(~
v1) + c2h(~
v2)) = h−1
(h(c1~
v1 + c2~
v2))
= h−1
◦ h (c1~
v1 + c2~
v2)
= c1~
v1 + c2~
v2
= c1h−1
◦ h (~
v1)c2h−1
◦ h (~
v2)
= c1 · h−1
(h(~
v1)) + c2 · h−1
(h(~
v2))
Thus the inverse of a one-to-one linear map is automatically linear. But this also
gives the (1) =⇒ (2) implication, because the inverse itself must be one-to-one.
Of the remaining implications, (1) =⇒ (3) holds because any homomor-
phism maps ~
0V to ~
0W , but a one-to-one map sends at most one member of V
to ~
0W .

Next, (3) =⇒ (4) is true since rank plus nullity equals the dimension of the
domain.
For (4) =⇒ (5), to show that hh(~
β1), . . . , h(~
βn)i is a basis for the rangespace
we need only show that it is a spanning set, because by assumption the range
has dimension n. Consider h(~
v) ∈ R(h). Expressing ~
v as a linear combination
of basis elements produces h(~
v) = h(c1
~
β1 + c2
~
β2 + · · · + cn
~
βn), which gives that
h(~
v) = c1h(~
β1) + · · · + cnh(~
βn), as desired.
Finally, for the (5) =⇒ (2) implication, assume that h~
β1, . . . , ~
βni is a basis
for V so that hh(~
β1), . . . , h(~
βn)i is a basis for R(h). Then every ~
w ∈ R(h) a the
unique representation ~
w = c1h(~
β1) + · · · + cnh(~
βn). Define a map from R(h) to
V by
~
w 7→ c1
~
β1 + c2
~
β2 + · · · + cn
~
βn
(uniqueness of the representation makes this well-defined). Checking that it is
linear and that it is the inverse of h are easy. QED
We’ve now seen that a linear map shows how the structure of the domain is
like that of the range. Such a map can be thought to organize the domain space
into inverse images of points in the range. In the special case that the map is
one-to-one, each inverse image is a single point and the map is an isomorphism
between the domain and the range.
Exercises
X 2.21 Let h: P3 → P4 be given by p(x) 7→ x · p(x). Which of these are in the
nullspace? Which are in the rangespace?
(a) x3
(b) 0 (c) 7 (d) 12x − 0.5x3
(e) 1 + 3x2
− x3
X 2.22 Find the nullspace, nullity, rangespace, and rank of each map.
(a) h: R2
→ P3 given by
µ
a
b
¶
7→ a + ax + ax2
(b) h: M2
×
2 → R given by
µ
a b
c d
¶
7→ a + d
(c) h: M2
×
2 → P2 given by
µ
a b
c d
¶
7→ a + b + c + dx2
(d) the zero map Z : R3
→ R4
X 2.23 Find the nullity of each map.

(a) h: R5
→ R8
of rank five (b) h: P3 → P3 of rank one
(c) h: R6
→ R3
, an onto map (d) h: M3
×
3 → M3
×
3, onto
X 2.24 What is the nullspace of the differentiation transformation d/dx: Pn → Pn?
What is the nullspace of the second derivative, as a transformation of Pn? The
k-th derivative?
2.25 Example 2.5 restates the first condition in the definition of homomorphism as
‘the shadow of a sum is the sum of the shadows’. Restate the second condition in
the same style.
2.26 For the homomorphism h: P3 → P3 given by h(a0 + a1x + a2x2
+ a3x3
) =
a0 + (a0 + a1)x + (a2 + a3)x3
find these.
(a) N (h) (b) h−1
(2 − x3
) (c) h−1
(1 + x2
)
X 2.27 For the map f : R2
→ R given by
f(
µ
x
y
¶
) = 2x + y
sketch these inverse image sets: f−1
(−3), f−1
(0), and f−1
(1).
X 2.28 Each of these transformations of P3 is nonsingular. Find the inverse function
of each.
(a) a0 + a1x + a2x2
+ a3x3
7→ a0 + a1x + 2a2x2
+ 3a3x3
(b) a0 + a1x + a2x2
+ a3x3
7→ a0 + a2x + a1x2
+ a3x3
(c) a0 + a1x + a2x2
+ a3x3
7→ a1 + a2x + a3x2
+ a0x3
(d) a0+a1x+a2x2
+a3x3
7→ a0+(a0+a1)x+(a0 +a1+a2)x2
+(a0+a1+a2+a3)x3
2.29 Describe the nullspace and rangespace of a transformation given by ~
v 7→ 2~
v.
2.30 List all pairs (rank(h), nullity(h)) that are possible for linear maps from R5
to R3
.
2.31 Does the differentiation map d/dx: Pn → Pn have an inverse?
X 2.32 Find the nullity of the map h: Pn → R given by
a0 + a1x + · · · + anxn
7→
Z x=1
x=0
a0 + a1x + · · · + anxn
dx.
2.33 (a) Prove that a homomorphism is onto if and only if its rank equals the
dimension of its codomain.
(b) Conclude that a homomorphism between vector spaces with the same di-
mension is one-to-one if and only if it is onto.
2.34 Show that a linear map is nonsingular if and only if it preserves linear inde-
pendence.
2.35 Corollary 2.16 says that for there to be an onto homomorphism from a vector
space V to a vector space W, it is necessary that the dimension of W be less
than or equal to the dimension of V . Prove that this condition is also sufficient;
use Theorem 1.9 to show that if the dimension of W is less than or equal to the
dimension of V , then there is a homomorphism from V to W that is onto.
2.36 Let h: V → R be a homomorphism, but not the zero homomorphism. Prove
that if h~
β1, . . . , ~
βni is a basis for the nullspace and if ~
v ∈ V is not in the nullspace
then h~
v, ~
β1, . . . , ~
βni is a basis for the entire domain V .
X 2.37 Recall that the nullspace is a subset of the domain and the rangespace is a
subset of the codomain. Are they necessarily distinct? Is there a homomorphism
that has a nontrivial intersection of its nullspace and its rangespace?

2.38 Prove that the image of a span equals the span of the images. That is, where
h: V → W is linear, prove that if S is a subset of V then h([S]) equals [h(S)]. This
generalizes Lemma 2.1 since it shows that if U is any subspace of V then its image
{h(~
u)
¯
¯ ~
u ∈ U} is a subspace of W, because the span of the set U is U.
X 2.39 (a) Prove that for any linear map h: V → W and any ~
w ∈ W, the set
h−1
(~
w) has the form
{~
v + ~
n
¯
¯ ~
n ∈ N (h)}
for ~
v ∈ V with h(~
v) = ~
w (if h is not onto then this set may be empty). Such a
set is a coset of N (h) and is denoted ~
v + N (h).
(b) Consider the map t: R2
→ R2
given by
µ
x
y
¶
t
7−→
µ
ax + by
cx + dy
¶
for some scalars a, b, c, and d. Prove that t is linear.
(c) Conclude from the prior two items that for any linear system of the form
ax + by = e
cx + dy = f
the solution set can be written (the vectors are members of R2
)
{~
p + ~
h
¯
¯ ~
h satisfies the associated homogeneous system}
where ~
p is a particular solution of that linear system (if there is no particular
solution then the above set is empty).
(d) Show that this map h: Rn
→ Rm
is linear



x1
.
.
.
xn


 7→



a1,1x1 + · · · + a1,nxn
.
.
.
am,1x1 + · · · + am,nxn



for any scalars a1,1, . . . , am,n. Extend the conclusion made in the prior item.
(e) Show that the k-th derivative map is a linear transformation of Pn for each
k. Prove that this map is a linear transformation of that space
f 7→
dk
dxk
f + ck−1
dk−1
dxk−1
f + · · · + c1
d
dx
f + c0f
for any scalars ck, . . . , c0. Draw a conclusion as above.
2.40 Prove that for any transformation t: V → V that is rank one, the map given
by composing the operator with itself t ◦ t: V → V satisfies t ◦ t = r · t for some
real number r.
2.41 Show that for any space V of dimension n, the dual space
L(V, R) = {h: V → R
¯
¯ h is linear}
is isomorphic to Rn
. It is often denoted V ∗
. Conclude that V ∗ ∼
= V .
2.42 Show that any linear map is the sum of maps of rank one.
2.43 Is ‘is homomorphic to’ an equivalence relation? (Hint: the difficulty is to
decide on an appropriate meaning for the quoted phrase.)
2.44 Show that the rangespaces and nullspaces of powers of linear maps t: V → V
form descending
V ⊇ R(t) ⊇ R(t2
) ⊇ . . .
and ascending
{~
0} ⊆ N (t) ⊆ N (t2
) ⊆ . . .
chains. Also show that if k is such that R(tk
) = R(tk+1
) then all following
rangespaces are equal: R(tk
) = R(tk+1
) = R(tk+2
) . . . . Similarly, if N (tk
) =
N (tk+1
) then N (tk
) = N (tk+1
) = N (tk+2
) = . . . .

3.III Computing Linear Maps
The prior section shows that a linear map is determined by its action on a basis.
In fact, the equation
h(~
v) = h(c1 · ~
β1 + · · · + cn · ~
βn) = c1 · h(~
β1) + · · · + cn · h(~
βn)
shows that, if we know the value of the map on the vectors in a basis, then we
can compute the value of the map on any vector ~
v at all just by finding the c’s
to express ~
v with respect to the basis.
This section gives the scheme that computes, from the representation of a
vector in the domain RepB(~
v), the representation of that vector’s image in the
codomain RepD(h(~
v)), using the representations of h(~
β1), . . . , h(~
βn).
3.III.1 Representing Linear Maps with Matrices
1.1 Example Consider a map h with domain R2
and codomain R3
(fixing
B = h
µ
2
0
¶
,
µ
1
4
¶
i and D = h


1
0
0

 ,


0
−2
0

 ,


1
0
1

i
as the bases for these spaces) that is determined by this action on the vectors
in the domain’s basis.
µ
2
0
¶
h
7−→


1
1
1


µ
1
4
¶
h
7−→


1
2
0


To compute the action of this map on any vector at all from the domain, we
first express h(~
β1) and h(~
β2) with respect to the codomain’s basis:


1
1
1

 = 0


1
0
0

 −
1
2


0
−2
0

 + 1


1
0
1

 so RepD(h(~
β1)) =


0
−1/2
1


D
and


1
2
0

 = 1


1
0
0

 − 1


0
−2
0

 + 0


1
0
1

 so RepD(h(~
β2)) =


1
−1
0


D

Section III. Computing Linear Maps 195
(these are easy to check). Then, as described in the preamble, for any member
~
v of the domain, we can express the image h(~
v) in terms of the h(~
β)’s.
h(~
v) = h(c1 ·
µ
2
0
¶
+ c2 ·
µ
1
4
¶
)
= c1 · h(
µ
2
0
¶
) + c2 · h(
µ
1
4
¶
)
= c1 · (0


1
0
0

−
1
2


0
−2
0

+ 1


1
0
1

) + c2 · (1


1
0
0

− 1


0
−2
0

+ 0


1
0
1

)
= (0c1 + 1c2) ·


1
0
0

 + (−
1
2
c1 − 1c2) ·


0
−2
0

 + (1c1 + 0c2) ·


1
0
1


Thus,
with RepB(~
v) =
µ
c1
c2
¶
then RepD( h(~
v) ) =


0c1 + 1c2
−(1/2)c1 − 1c2
1c1 + 0c2

.
For instance,
with RepB(
µ
4
8
¶
) =
µ
1
2
¶
B
then RepD( h(
µ
4
8
¶
) ) =


2
−5/2
1

.
This is a formula that computes how h acts on any argument.
We will express computations like the one above with a matrix notation.


0 1
−1/2 −1
1 0


B,D
µ
c1
c2
¶
B
=


0c1 + 1c2
(−1/2)c1 − 1c2
1c1 + 0c2


D
In the middle is the argument ~
v to the map, represented with respect to the
domain’s basis B by a column vector with components c1 and c2. On the right
is the value h(~
v) of the map on that argument, represented with respect to the
codomain’s basis D by a column vector with components 0c1 + 1c2, etc. The
matrix on the left is the new thing. It consists of the coefficients from the vector
on the right, 0 and 1 from the first row, −1/2 and −1 from the second row, and
1 and 0 from the third row.
This notation simply breaks the parts from the right, the coefficients and the
c’s, out separately on the left, into a vector that represents the map’s argument
and a matrix that we will take to represent the map itself.

1.2 Definition Suppose that V and W are vector spaces of dimensions n and
m with bases B and D, and that h: V → W is a linear map. If
RepD(h(~
β1)) =





h1,1
h2,1
.
.
.
hm,1





D
, . . . , RepD(h(~
βn)) =





h1,n
h2,n
.
.
.
hm,n





D
then
RepB,D(h) =





h1,1 h1,2 . . . h1,n
h2,1 h2,2 . . . h2,n
.
.
.
hm,1 hm,2 . . . hm,n





B,D
is the matrix representation of h with respect to B, D.
Briefly, the vectors representing the h(~
β)’s are adjoined to make the matrix
representing the map.
RepB,D(h) =




.
.
.
.
.
.
RepD( h(~
β1) ) · · · RepD( h(~
βn) )
.
.
.
.
.
.




Observe that the number of columns of the matrix is the dimension of the
domain of the map, and the number of rows is the dimension of the codomain.
1.3 Example If h: R3
→ P1 is given by


a1
a2
a3

 h
7−→ (2a1 + a2) + (−a3)x
then where
B = h


0
0
1

 ,


0
2
0

 ,


2
0
0

i and D = h1 + x, −1 + xi
the action of h on B is given by


0
0
1

 h
7−→ −x


0
2
0

 h
7−→ 2


2
0
0

 h
7−→ 4
and a simple calculation gives
RepD(−x) =
µ
−1/2
−1/2
¶
D
RepD(2) =
µ
1
−1
¶
D
RepD(4) =
µ
2
−2
¶
D

showing that this is the matrix representing h with respect to the bases.
RepB,D(h) =
µ
−1/2 1 2
−1/2 −1 −2
¶
B,D
We will use lower case letters for a map, upper case for the matrix, and
lower case again for the entries of the matrix. Thus for the map h, the matrix
representing it is H, with entries hi,j.
1.4 Theorem Assume that V and W are vector spaces of dimensions m and
n with bases B and D, and that h: V → W is a linear map. If h is represented
by
RepB,D(h) =





h1,1 h1,2 . . . h1,n
h2,1 h2,2 . . . h2,n
.
.
.
hm,1 hm,2 . . . hm,n





B,D
and ~
v ∈ V is represented by
RepB(~
v) =





c1
c2
.
.
.
cn





B
then the representation of the image of ~
v is this.
RepD( h(~
v) ) =





h1,1c1 + h1,2c2 + · · · + h1,ncn
h2,1c1 + h2,2c2 + · · · + h2,ncn
.
.
.
hm,1c1 + hm,2c2 + · · · + hm,ncn





D
Proof. Exercise 28. QED
We will think of the matrix RepB,D(h) and the vector RepB(~
v) as combining
to make the vector RepD(h(~
v)).
1.5 Definition The matrix-vector product of a m×n matrix and a n×1 vector
is this.





a1,1 a1,2 . . . a1,n
a2,1 a2,2 . . . a2,n
.
.
.
am,1 am,2 . . . am,n








c1
.
.
.
cn


 =





a1,1c1 + a1,2c2 + · · · + a1,ncn
a2,1c1 + a2,2c2 + · · · + a2,ncn
.
.
.
am,1c1 + am,2c2 + · · · + am,ncn






The point of Definition 1.2 is to generalize Example 1.1, that is, the point
of the definition is Theorem 1.4, that the matrix describes how to get from
the representation of a domain vector with respect to the domain’s basis to
the representation of its image in the codomain with respect to the codomain’s
basis. With Definition 1.5, we can restate this as: application of a linear map is
represented by the matrix-vector product of the map’s representative and the
vector’s representative.
1.6 Example With the matrix from Example 1.3 we can calculate where that
map sends this vector.
~
v =


4
1
0


This vector is represented, with respect to the domain basis B, by
RepB(~
v) =


0
1/2
2


B
and so this is the representation of the value h(~
v) with respect to the codomain
basis D.
RepD(h(~
v)) =
µ
−1/2 1 2
−1/2 −1 −2
¶
B,D


0
1/2
2


B
=
µ
(−1/2) · 0 + 1 · (1/2) + 2 · 2
(−1/2) · 0 − 1 · (1/2) − 2 · 2
¶
D
=
µ
9/2
−9/2
¶
D
To find h(~
v) itself, not its representation, take (9/2)(1+x)−(9/2)(−1+x) = 9.
1.7 Example Let π: R3
→ R2
be projection onto the xy-plane. To give a
matrix representing this map, we first fix bases.
B = h


1
0
0

 ,


1
1
0

 ,


−1
0
1

i D = h
µ
2
1
¶
,
µ
1
1
¶
i
For each vector in the domain’s basis, we find its image under the map.


1
0
0

 π
7−→
µ
1
0
¶


1
1
0

 π
7−→
µ
1
1
¶


−1
0
1

 π
7−→
µ
−1
0
¶
Then we find the representation of each image with respect to the codomain’s
basis
RepD(
µ
1
0
¶
) =
µ
1
−1
¶
RepD(
µ
1
1
¶
) =
µ
0
1
¶
RepD(
µ
−1
0
¶
) =
µ
−1
1
¶

(these are easily checked). Finally, adjoining these representations gives the
matrix representing π with respect to B, D.
RepB,D(π) =
µ
1 0 −1
−1 1 1
¶
B,D
We can illustrate Theorem 1.4 by computing the matrix-vector product repre-
senting the following statement about the projection map.
π(


2
2
1

) =
µ
2
2
¶
Representing this vector from the domain with respect to the domain’s basis
RepB(


2
2
1

) =


1
2
1


B
gives this matrix-vector product.
RepD( π(


2
1
1

) ) =
µ
1 0 −1
−1 1 1
¶
B,D


1
2
1


B
=
µ
0
2
¶
D
Expanding this representation into a linear combination of vectors from D
0 ·
µ
2
1
¶
+ 2 ·
µ
1
1
¶
=
µ
2
2
¶
checks that the map’s action is indeed reflected in the operation of the matrix.
(We will sometimes compress these three displayed equations into one


2
2
1

 =


1
2
1


B
h
7−→
H
µ
0
2
¶
D
=
µ
2
2
¶
in the course of a calculation.)
We now have two ways to compute the effect of projection, the straight-
forward formula that drops each three-tall vector’s third component to make
a two-tall vector, and the above formula that uses representations and matrix-
vector multiplication. Compared to the first way, the second way might seem
complicated. However, it has advantages. The next example shows that giving
a formula for some maps is simplified by this new scheme.
1.8 Example To represent a rotation map tθ : R2
→ R2
that turns all vectors
in the plane counterclockwise through an angle θ

~
v
tθ
7−→
tθ(~
v)
we start by fixing bases. Using E2 both as a domain basis and as a codomain
basis is natural, Now, we find the image under the map of each vector in the
domain’s basis.
µ
1
0
¶
tθ
7−→
µ
cos θ
sin θ
¶ µ
0
1
¶
tθ
7−→
µ
− sin θ
cos θ
¶
Then we represent these images with respect to the codomain’s basis. Because
this basis is E2, vectors are represented by themselves. Finally, adjoining the
representations gives the matrix representing the map.
RepE2,E2
(tθ) =
µ
cos θ − sin θ
sin θ cos θ
¶
The advantage of this scheme is that just by knowing how to represent the image
of the two basis vectors, we get a formula that tells us the image of any vector
at all; here a vector rotated by θ = π/6.
µ
3
−2
¶
tπ/6
7−→
µ√
3/2 −1/2
1/2
√
3/2
¶ µ
3
−2
¶
≈
µ
3.598
−0.232
¶
(Again, we are using the fact that, with respect to E2, vectors represent them-
selves.)
We have already seen the addition and scalar multiplication operations of
matrices and the dot product operation of vectors. Matrix-vector multiplication
is a new operation in the arithmetic of vectors and matrices. Nothing in Defi-
nition 1.5 requires us to view it in terms of representations. We can get some
insight into this operation by turning away from what is being represented, and
instead focusing on how the entries combine.
1.9 Example In the definition the width of the matrix equals the height of
the vector. Hence, the first product below is defined while the second is not.
µ
1 0 0
4 3 1
¶


1
0
2

 =
µ
1
6
¶ µ
1 0 0
4 3 1
¶ µ
1
0
¶
One reason that this product is not defined is purely formal: the definition re-
quires that the sizes match, and these sizes don’t match. (Behind the formality,
though, we have a reason why it is left undefined—the matrix represents a map
with a three-dimensional domain while the vector represents a member of a
two-dimensional space.)

A good way to view a matrix-vector product is as the dot products of the
rows of the matrix with the column vector.




.
.
.
ai,1 ai,2 . . . ai,n
.
.
.









c1
c2
.
.
.
cn





=




.
.
.
ai,1c1 + ai,2c2 + . . . + ai,ncn
.
.
.




Looked at in this row-by-row way, this new operation generalizes dot product.
Matrix-vector product can also be viewed column-by-column.





h1,1 h1,2 . . . h1,n
h2,1 h2,2 . . . h2,n
.
.
.
hm,1 hm,2 . . . hm,n










c1
c2
.
.
.
cn





=





h1,1c1 + h1,2c2 + · · · + h1,ncn
h2,1c1 + h2,2c2 + · · · + h2,ncn
.
.
.
hm,1c1 + hm,2c2 + · · · + hm,ncn





= c1





h1,1
h2,1
.
.
.
hm,1





+ · · · + cn





h1,n
h2,n
.
.
.
hm,n





1.10 Example
µ
1 0 −1
2 0 3
¶


2
−1
1

 = 2
µ
1
2
¶
− 1
µ
0
0
¶
+ 1
µ
−1
3
¶
=
µ
1
7
¶
The result has the columns of the matrix weighted by the entries of the
vector. This way of looking at it brings us back to the objective stated at the
start of this section, to compute h(c1
~
β1 +· · ·+cn
~
βn) as c1h(~
β1)+· · ·+cnh(~
βn).
We began this section by noting that the equality of these two enables us
to compute the action of h on any argument knowing only h(~
β1), . . . , h(~
βn).
We have developed this into a scheme to compute the action of the map by
taking the matrix-vector product of the matrix representing the map and the
vector representing the argument. In this way, any linear map is represented
with respect to some bases by a matrix. In the next subsection, we will show
the converse, that any matrix represents a linear map.
Exercises
X 1.11 Multiply, where it is defined, the matrix
Ã
1 3 1
0 −1 2
1 1 0
!
by each vector.
(a)
Ã
2
1
0
!
(b)
µ
−2
−2
¶
(c)
Ã
0
0
0
!
1.12 Perform, if possible, each matrix-vector multiplication.

(a)
µ
2 1
3 −1/2
¶ µ
4
2
¶
(b)
µ
1 1 0
−2 1 0
¶ Ã
1
3
1
!
(c)
µ
1 1
−2 1
¶ Ã
1
3
1
!
X 1.13 Solve this matrix equation.
Ã
2 1 1
0 1 3
1 −1 2
! Ã
x
y
z
!
=
Ã
8
4
4
!
X 1.14 For a homomorphism from P2 to P3 that sends
1 7→ 1 + x, x 7→ 1 + 2x, and x2
7→ x − x3
where does 1 − 3x + 2x2
go?
X 1.15 Assume that h: R2
→ R3
is determined by this action.
µ
1
0
¶
7→
Ã
2
2
0
! µ
0
1
¶
7→
Ã
0
1
−1
!
Using the standard bases, find
(a) the matrix representing this map;
(b) a general formula for h(~
v).
X 1.16 Let d/dx: P3 → P3 be the derivative transformation.
(a) Represent d/dx with respect to B, B where B = h1, x, x2
, x3
i.
(b) Represent d/dx with respect to B, D where D = h1, 2x, 3x2
, 4x3
i.
X 1.17 Represent each linear map with respect to each pair of bases.
(a) d/dx: Pn → Pn with respect to B, B where B = h1, x, . . . , xn
i, given by
a0 + a1x + a2x2
+ · · · + anxn
7→ a1 + 2a2x + · · · + nanxn−1
(b)
R
: Pn → Pn+1 with respect to Bn, Bn+1 where Bi = h1, x, . . . , xi
i, given by
a0 + a1x + a2x2
+ · · · + anxn
7→ a0x +
a1
2
x2
+ · · · +
an
n + 1
xn+1
(c)
R 1
0
: Pn → R with respect to B, E1 where B = h1, x, . . . , xn
i and E1 = h1i,
given by
a0 + a1x + a2x2
+ · · · + anxn
7→ a0 +
a1
2
+ · · · +
an
n + 1
(d) eval3 : Pn → R with respect to B, E1 where B = h1, x, . . . , xn
i and E1 = h1i,
given by
a0 + a1x + a2x2
+ · · · + anxn
7→ a0 + a1 · 3 + a2 · 32
+ · · · + an · 3n
(e) slide−1 : Pn → Pn with respect to B, B where B = h1, x, . . . , xn
i, given by
a0 + a1x + a2x2
+ · · · + anxn
7→ a0 + a1 · (x + 1) + · · · + an · (x + 1)n
1.18 Represent the identity map on any nontrivial space with respect to B, B,
where B is any basis.
1.19 Represent, with respect to the natural basis, the transpose transformation on
the space M2
×
2 of 2×2 matrices.
1.20 Assume that B = h~
β1, ~
β2, ~
β3, ~
β4i is a basis for a vector space. Represent with
respect to B, B the transformation that is determined by each.
(a) ~
β1 7→ ~
β2, ~
β2 7→ ~
β3, ~
β3 7→ ~
β4, ~
β4 7→ ~
0
(b) ~
β1 7→ ~
β2, ~
β2 7→ ~
0, ~
β3 7→ ~
β4, ~
β4 7→ ~
0

(c) ~
β1 7→ ~
β2, ~
β2 7→ ~
β3, ~
β3 7→ ~
0, ~
β4 7→ ~
0
1.21 Example 1.8 shows how to represent the rotation transformation of the plane
with respect to the standard basis. Express these other transformations also with
respect to the standard basis.
(a) the dilation map ds, which multiplies all vectors by the same scalar s
(b) the reflection map f`, which reflects all all vectors across a line ` through the
origin
X 1.22 Consider a linear transformation of R2
determined by these two.
µ
1
1
¶
7→
µ
2
0
¶ µ
1
0
¶
7→
µ
−1
0
¶
(a) Represent this transformation with respect to the standard bases.
(b) Where does the transformation send this vector?
µ
0
5
¶
(c) Represent this transformation with respect to these bases.
B = h
µ
1
−1
¶
,
µ
1
1
¶
i D = h
µ
2
2
¶
,
µ
−1
1
¶
i
(d) Using B from the prior item, represent the transformation with respect to
B, B.
1.23 Suppose that h: V → W is nonsingular so that by Theorem 2.20, for any
basis B = h~
β1, . . . , ~
βni ⊂ V the image h(B) = hh(~
β1), . . . , h(~
βn)i is a basis for
W.
(a) Represent the map h with respect to B, h(B).
(b) For a member ~
v of the domain, where the representation of ~
v has components
c1, . . . , cn, represent the image vector h(~
v) with respect to the image basis h(B).
1.24 Give a formula for the product of a matrix and ~
ei, the column vector that is
all zeroes except for a single one in the i-th position.
X 1.25 For each vector space of functions of one real variable, represent the derivative
transformation with respect to B, B.
(a) {a cos x + b sin x
¯
¯ a, b ∈ R}, B = hcos x, sin xi
(b) {aex
+ be2x
¯
¯ a, b ∈ R}, B = hex
, e2x
i
(c) {a + bx + cex
+ dxe2x
¯
¯ a, b, c, d ∈ R}, B = h1, x, ex
, xex
i
1.26 Find the range of the linear transformation of R2
represented with respect to
the standard bases by each matrix.
(a)
µ
1 0
0 0
¶
(b)
µ
0 0
3 2
¶
(c) a matrix of the form
µ
a b
2a 2b
¶
X 1.27 Can one matrix represent two different linear maps? That is, can RepB,D(h) =
RepB̂,D̂(ĥ)?
1.28 Prove Theorem 1.4.
X 1.29 Example 1.8 shows how to represent rotation of all vectors in the plane through
an angle θ about the origin, with respect to the standard bases.
(a) Rotation of all vectors in three-space through an angle θ about the x-axis is a
transformation of R3
. Represent it with respect to the standard bases. Arrange
the rotation so that to someone whose feet are at the origin and whose head is
at (1, 0, 0), the movement appears clockwise.

(b) Repeat the prior item, only rotate about the y-axis instead. (Put the person’s
head at ~
e2.)
(c) Repeat, about the z-axis.
(d) Extend the prior item to R4
. (Hint: ‘rotate about the z-axis’ can be restated
as ‘rotate parallel to the xy-plane’.)
1.30 (Schur’s Triangularization Lemma)
(a) Let U be a subspace of V and fix bases BU ⊆ BV . What is the relationship
between the representation of a vector from U with respect to BU and the
representation of that vector (viewed as a member of V ) with respect to BV ?
(b) What about maps?
(c) Fix a basis B = h~
β1, . . . , ~
βni for V and observe that the spans
[{~
0}] = {~
0} ⊂ [{~
β1}] ⊂ [{~
β1, ~
β2}] ⊂ · · · ⊂ [B] = V
form a strictly increasing chain of subspaces. Show that for any linear map
h: V → W there is a chain W0 = {~
0} ⊆ W1 ⊆ · · · ⊆ Wm = W of subspaces of
W such that
h([{~
β1, . . . , ~
βi}]) ⊂ Wi
for each i.
(d) Conclude that for every linear map h: V → W there are bases B, D so the
matrix representing h with respect to B, D is upper-triangular (that is, each
entry hi,j with i > j is zero).
(e) Is an upper-triangular representation unique?
3.III.2 Any Matrix Represents a Linear Map
The prior subsection shows that the action of a linear map h is described by
a matrix H, with respect to appropriate bases, in this way.
~
v =



v1
.
.
.
vn



B
h
7−→
H



h1,1v1 + · · · + h1,nvn
.
.
.
hm,1v1 + · · · + hm,nvn



D
= h(~
v)
In this subsection, we will show the converse, that each matrix represents a
linear map.
Recall that, in the definition of the matrix representation of a linear map,
the number of columns of the matrix is the dimension of the map’s domain and
the number of rows of the matrix is the dimension of the map’s codomain. Thus,
for instance, a 2×3 matrix cannot represent a map from R5
to R4
. The next
result says that, beyond this restriction on the dimensions, there are no other
limitations: the 2×3 matrix represents a map from any three-dimensional space
to any two-dimensional space.
2.1 Theorem Any matrix represents a homomorphism between vector spaces
of appropriate dimensions, with respect to any pair of bases.

Proof. For the matrix
H =





h1,1 h1,2 . . . h1,n
h2,1 h2,2 . . . h2,n
.
.
.
hm,1 hm,2 . . . hm,n





fix any n-dimensional domain space V and any m-dimensional codomain space
W. Also fix bases B = h~
β1, . . . , ~
βni and D = h~
δ1, . . . ,~
δmi for those spaces.
Define a function h: V → W by: where ~
v in the domain is represented as
RepB(~
v) =



v1
.
.
.
vn



B
then its image h(~
v) is the member the codomain represented by
RepD( h(~
v) ) =



h1,1v1 + · · · + h1,nvn
.
.
.
hm,1v1 + · · · + hm,nvn



D
that is, h(~
v) = h(v1
~
β1 + · · · + vn
~
βn) is defined to be (h1,1v1 + · · · + h1,nvn) ·~
δ1 +
· · · + (hm,1v1 + · · · + hm,nvn) ·~
δm. (This is well-defined by the uniqueness of the
representation RepB(~
v).)
Observe that h has simply been defined to make it the map that is repre-
sented with respect to B, D by the matrix H. So to finish, we need only check
that h is linear. If ~
v, ~
u ∈ V are such that
RepB(~
v) =



v1
.
.
.
vn


 and RepB(~
u) =



u1
.
.
.
un



and c, d ∈ R then the calculation
h(c~
v + d~
u) =
¡
h1,1(cv1 + du1) + · · · + h1,n(cvn + dun)
¢
· ~
δ1 +
· · · +
¡
hm,1(cv1 + du1) + · · · + hm,n(cvn + dun)
¢
· ~
δm
= c · h(~
v) + d · h(~
u)
provides this verification. QED
2.2 Example Which map the matrix represents depends on which bases are
used. If
H =
µ
1 0
0 0
¶
, B1 = D1 = h
µ
1
0
¶
,
µ
0
1
¶
i, and B2 = D2 = h
µ
0
1
¶
,
µ
1
0
¶
i,

then h1 : R2
→ R2
represented by H with respect to B1, D1 maps
µ
c1
c2
¶
=
µ
c1
c2
¶
B1
7→
µ
c1
0
¶
D1
=
µ
c1
0
¶
while h2 : R2
→ R2
represented by H with respect to B2, D2 is this map.
µ
c1
c2
¶
=
µ
c2
c1
¶
B2
7→
µ
c2
0
¶
D2
=
µ
0
c2
¶
These two are different. The first is projection onto the x axis, while the second
is projection onto the y axis.
So not only is any linear map described by a matrix but any matrix describes
a linear map. This means that we can, when convenient, handle linear maps
entirely as matrices, simply doing the computations, without have to worry that
a matrix of interest does not represent a linear map on some pair of spaces of
interest. (In practice, when we are working with a matrix but no spaces or
bases have been specified, we will often take the domain and codomain to be Rn
and Rm
and use the standard bases. In this case, because the representation
is transparent—the representation with respect to the standard basis of ~
v is
~
v—the column space of the matrix equals the range of the map. Consequently,
the column space of H is often denoted by R(H).)
With the theorem, we have characterized linear maps as those maps that act
in this matrix way. Each linear map is described by a matrix and each matrix
describes a linear map. We finish this section by illustrating how a matrix can
be used to tell things about its maps.
2.3 Theorem The rank of a matrix equals the rank of any map that it repre-
sents.
Proof. Suppose that the matrix H is m×n. Fix domain and codomain spaces
V and W of dimension n and m, with bases B = h~
β1, . . . , ~
βni and D. Then H
represents some linear map h between those spaces with respect to these bases
whose rangespace
{h(~
v)
¯
¯ ~
v ∈ V } = {h(c1
~
β1 + · · · + cn
~
βn)
¯
¯ c1, . . . , cn ∈ R}
= {c1h(~
β1) + · · · + cnh(~
βn)
¯
¯ c1, . . . , cn ∈ R}
is the span [{h(~
β1), . . . , h(~
βn)}]. The rank of h is the dimension of this range-
space.
The rank of the matrix is its column rank (or its row rank; the two are
equal). This is the dimension of the column space of the matrix, which is the
span of the set of column vectors [{RepD(h(~
β1)), . . . , RepD(h(~
βn))}].
To see that the two spans have the same dimension, recall that a represen-
tation with respect to a basis gives an isomorphism RepD : W → Rm
. Under
this isomorphism, there is a linear relationship among members of the range-
space if and only if the same relationship holds in the column space, e.g, ~
0 =

c1h(~
β1)+· · ·+cnh(~
βn) if and only if ~
0 = c1RepD(h(~
β1))+· · ·+cnRepD(h(~
βn)).
Hence, a subset of the rangespace is linearly independent if and only if the cor-
responding subset of the column space is linearly independent. This means that
the size of the largest linearly independent subset of the rangespace equals the
size of the largest linearly independent subset of the column space, and so the
two spaces have the same dimension. QED
2.4 Example Any map represented by




1 2 2
1 2 1
0 0 3
0 0 2




must, by definition, be from a three-dimensional domain to a four-dimensional
codomain. In addition, because the rank of this matrix is two (we can spot this
by eye or get it with Gauss’ method), any map represented by this matrix has
a two-dimensional rangespace.
2.5 Corollary Let h be a linear map represented by a matrix H. Then h
is onto if and only if the rank of H equals the number of its rows, and h is
one-to-one if and only if the rank of H equals the number of its columns.
Proof. For the first half, the dimension of the rangespace of h is the rank of
h, which equals the rank of H by the theorem. Since the dimension of the
codomain of h is the number of columns in H, if the rank of H equals the
number of columns, then the dimension of the rangespace equals the dimension
of the codomain. But a subspace with the same dimension as its superspace
must equal that superspace (a basis for the rangespace is a linearly independent
subset of the codomain, whose size is equal to the dimension of the codomain,
and so this set is a basis for the codomain).
For the second half, a linear map is one-to-one if and only if it is an isomor-
phism between its domain and its range, that is, if and only if its domain has the
same dimension as its range. But the number of columns in h is the dimension
of h’s domain, and by the theorem the rank of H equals the dimension of h’s
range. QED
The above results end any confusion caused by our use of the word ‘rank’ to
mean apparently different things when applied to matrices and when applied to
maps. We can also justify the dual use of ‘nonsingular’. We’ve defined a matrix
to be nonsingular if it is square and is the matrix of coefficients of a linear system
with a unique solution, and we’ve defined a linear map to be nonsingular if it is
one-to-one.
2.6 Corollary A square matrix represents nonsingular maps if and only if it
is a nonsingular matrix. Thus, a matrix represents an isomorphism if and only
if it is square and nonsingular.
Proof. Immediate from the prior result. QED

2.7 Example Any map from R2
to P1 represented with respect to any pair of
bases by
µ
1 2
0 3
¶
is nonsingular because this matrix has rank two.
2.8 Example Any map g: V → W represented by
µ
1 2
3 6
¶
is not nonsingular because this matrix is not nonsingular.
We’ve now seen that the relationship between maps and matrices goes both
ways: fixing bases, any linear map is represented by a matrix and any matrix
describes a linear map. That is, by fixing spaces and bases we get a correspon-
dence between maps and matrices. In the rest of this chapter we will explore
this correspondence. For instance, we’ve defined for linear maps the operations
of addition and scalar multiplication and we shall see what the corresponding
matrix operations are. We shall also see the matrix operation that represent
the map operation of composition. And, we shall see how to find the matrix
that represents a map’s inverse.
Exercises
X 2.9 Decide if the vector is in the column space of the matrix.
(a)
µ
2 1
2 5
¶
,
µ
1
−3
¶
(b)
µ
4 −8
2 −4
¶
,
µ
0
1
¶
(c)
Ã
1 −1 1
1 1 −1
−1 −1 1
!
,
Ã
2
0
0
!
X 2.10 Decide if each vector lies in the range of the map from R3
to R2
represented
with respect to the standard bases by the matrix.
(a)
µ
1 1 3
0 1 4
¶
,
µ
1
3
¶
(b)
µ
2 0 3
4 0 6
¶
,
µ
1
1
¶
X 2.11 Consider this matrix, representing a transformation of R2
, and these bases for
that space.
1
2
·
µ
1 1
−1 1
¶
B = h
µ
0
1
¶
,
µ
1
0
¶
i D = h
µ
1
1
¶
,
µ
1
−1
¶
i
(a) To what vector in the codomain is the first member of B mapped?
(b) The second member?
(c) Where is a general vector from the domain (a vector with components x and
y) mapped? That is, what transformation of R2
is represented with respect to
B, D by this matrix?
2.12 What transformation of F = {a cos θ + b sin θ
¯
¯ a, b ∈ R} is represented with
respect to B = hcos θ − sin θ, sin θi and D = hcos θ + sin θ, cos θi by this matrix?
µ
0 0
1 0
¶

X 2.13 Decide if 1 + 2x is in the range of the map from R3
to P2 represented with
respect to E3 and h1, 1 + x2
, xi by this matrix.
Ã
1 3 0
0 1 0
1 0 1
!
2.14 Example 2.8 gives a matrix that is nonsingular, and is therefore associated
with maps that are nonsingular.
(a) Find the set of column vectors representing the members of the nullspace of
any map represented by this matrix.
(b) Find the nullity of any such map.
(c) Find the set of column vectors representing the members of the rangespace
of any map represented by this matrix.
(d) Find the rank of any such map.
(e) Check that rank plus nullity equals the dimension of the domain.
X 2.15 Because the rank of a matrix equals the rank of any map it represents, if
one matrix represents two different maps H = RepB,D(h) = RepB̂,D̂(ĥ) (where
h, ĥ: V → W) then the dimension of the rangespace of h equals the dimension of
the rangespace of ĥ. Must these equal-dimensioned rangespaces actually be the
same?
X 2.16 Let V be an n-dimensional space with bases B and D. Consider a map that
sends, for ~
v ∈ V , the column vector representing ~
v with respect to B to the column
vector representing ~
v with respect to D. Show that is a linear transformation of
Rn
.
2.17 Example 2.2 shows that changing the pair of bases can change the map that
a matrix represents, even though the domain and codomain remain the same.
Could the map ever not change? Is there a matrix H, vector spaces V and W,
and associated pairs of bases B1, D1 and B2, D2 (with B1 6= B2 or D1 6= D2 or
both) such that the map represented by H with respect to B1, D1 equals the map
represented by H with respect to B2, D2?
X 2.18 A square matrix is a diagonal matrix if it is all zeroes except possibly for the
entries on its upper-left to lower-right diagonal—its 1, 1 entry, its 2, 2 entry, etc.
Show that a linear map is an isomorphism if there are bases such that, with respect
to those bases, the map is represented by a diagonal matrix with no zeroes on the
diagonal.
2.19 Describe geometrically the action on R2
of the map represented with respect
to the standard bases E2, E2 by this matrix.
µ
3 0
0 2
¶
Do the same for these. µ
1 0
0 0
¶ µ
0 1
1 0
¶ µ
1 3
0 1
¶
2.20 The fact that for any linear map the rank plus the nullity equals the dimension
of the domain shows that a necessary condition for the existence of a homomor-
phism between two spaces, onto the second space, is that there be no gain in
dimension. That is, where h: V → W is onto, the dimension of W must be less
than or equal to the dimension of V .
(a) Show that this (strong) converse holds: no gain in dimension implies that

there is a homomorphism and, further, any matrix with the correct size and
correct rank represents such a map.
(b) Are there bases for R3
such that this matrix
H =
Ã
1 0 0
2 0 0
0 1 0
!
represents a map from R3
to R3
whose range is the xy plane subspace of R3
?
2.21 Let V be an n-dimensional space and suppose that ~
x ∈ Rn
. Fix a basis
B for V and consider the map h~
x : V → R given ~
v 7→ ~
x RepB(~
v) by the dot
product.
(a) Show that this map is linear.
(b) Show that for any linear map g : V → R there is an ~
x ∈ Rn
such that g = h~
x.
(c) In the prior item we fixed the basis and varied the ~
x to get all possible linear
maps. Can we get all possible linear maps by fixing an ~
x and varying the basis?
2.22 Let V, W, X be vector spaces with bases B, C, D.
(a) Suppose that h: V → W is represented with respect to B, C by the matrix
H. Give the matrix representing the scalar multiple rh (where r ∈ R) with
respect to B, C by expressing it in terms of H.
(b) Suppose that h, g : V → W are represented with respect to B, C by H and
G. Give the matrix representing h + g with respect to B, C by expressing it in
terms of H and G.
(c) Suppose that h: V → W is represented with respect to B, C by H and
g : W → X is represented with respect to C, D by G. Give the matrix repre-
senting g ◦ h with respect to B, D by expressing it in terms of H and G.

Section IV. Matrix Operations 211
3.IV Matrix Operations
The prior section shows how matrices represent linear maps. A good strategy, on
seeing a new idea, is to explore how it interacts with some already-established
ideas. In the first subsection we will ask how the representation of the sum
of two maps f + g related to the representations F and G of the two maps,
and how the representation of a scalar product r · h of a map is related to the
representation H of that map. In later subsections we will see how to represent
map composition and map inverse.
3.IV.1 Sums and Scalar Products
Recall that for two maps f and g with the same domain and codomain, the
map sum f + g has the natural definition.
~
v
f+g
7−→ f(~
v) + g(~
v)
The easiest way to see how the representations of the maps combine to represent
the map sum is with an example.
1.1 Example Suppose that f, g: R2
→ R3
are represented with respect to the
bases B and D by these matrices.
F = RepB,D(f) =


1 3
2 0
1 0


B,D
G = RepB,D(g) =


0 0
−1 −2
2 4


B,D
Then, for any ~
v ∈ V represented with respect to B, computation of the repre-
sentation of f(~
v) + g(~
v)


1 3
2 0
1 0


µ
v1
v2
¶
+


0 0
−1 −2
2 4


µ
v1
v2
¶
=


1v1 + 3v2
2v1 + 0v2
1v1 + 0v2

 +


0v1 + 0v2
−1v1 − 2v2
2v1 + 4v2


gives this representation of f + g (~
v).


(1 + 0)v1 + (3 + 0)v2
(2 − 1)v1 + (0 − 2)v2
(1 + 2)v1 + (0 + 4)v2

 =


1v1 + 3v2
1v1 − 2v2
3v1 + 4v2


Thus, the action of f + g is described by this matrix-vector product.


1 3
1 −2
3 4


B,D
µ
v1
v2
¶
B
=


1v1 + 3v2
1v1 − 2v2
3v1 + 4v2


D
This matrix is the entry-by-entry sum of original matrices, e.g., the 1, 1 entry
of RepB,D(f + g) is the sum of the 1, 1 entry of F and the 1, 1 entry of G.

Representing a scalar multiple of a map works the same way.
1.2 Example If t is a transformation represented by
RepB,D(t) =
µ
1 0
1 1
¶
B,D
so that ~
v =
µ
v1
v2
¶
B
7→
µ
v1
v1 + v2
¶
D
= t(~
v)
then the scalar multiple map 5t acts in this way.
~
v =
µ
v1
v2
¶
B
7−→
µ
5v1
5v1 + 5v2
¶
D
= 5 · t(~
v)
Therefore, this is the matrix representing 5t.
RepB,D(5t) =
µ
5 0
5 5
¶
B,D
1.3 Definition The sum of two same-sized matrices is their entry-by-entry
sum. The scalar multiple of a matrix is the result of entry-by-entry scalar
multiplication.
1.4 Remark These extend the vector addition and scalar multiplication oper-
ations that we defined in the first chapter.
1.5 Theorem Let h, g: V → W be linear maps represented with respect to
bases B, D by the matrices H and G, and let r be a scalar. Then the map
h + g: V → W is represented with respect to B, D by H + G, and the map
r · h: V → W is represented with respect to B, D by rH.
Proof. Exercise 8; generalize the examples above. QED
A notable special case of scalar multiplication is multiplication by zero. For
any map 0 · h is the zero homomorphism and for any matrix 0 · H is the zero
matrix.
1.6 Example The zero map from any three-dimensional space to any two-
dimensional space is represented by the 2×3 zero matrix
Z =
µ
0 0 0
0 0 0
¶
no matter which domain and codomain bases are used.
Exercises
X 1.7 Perform the indicated operations, if defined.
(a)
µ
5 −1 2
6 1 1
¶
+
µ
2 1 4
3 0 5
¶
(b) 6 ·
µ
2 −1 −1
1 2 3
¶

(c)
µ
2 1
0 3
¶
+
µ
2 1
0 3
¶
(d) 4
µ
1 2
3 −1
¶
+ 5
µ
−1 4
−2 1
¶
(e) 3
µ
2 1
3 0
¶
+ 2
µ
1 1 4
3 0 5
¶
1.8 Prove Theorem 1.5.
(a) Prove that matrix addition represents addition of linear maps.
(b) Prove that matrix scalar multiplication represents scalar multiplication of
linear maps.
X 1.9 Prove each, where the operations are defined, where G, H, and J are matrices,
where Z is the zero matrix, and where r and s are scalars.
(a) Matrix addition is commutative G + H = H + G.
(b) Matrix addition is associative G + (H + J) = (G + H) + J.
(c) The zero matrix is an additive identity G + Z = G.
(d) 0 · G = Z
(e) (r + s)G = rG + sG
(f) Matrices have an additive inverse G + (−1) · G = Z.
(g) r(G + H) = rG + rH
(h) (rs)G = r(sG)
1.10 Fix domain and codomain spaces. In general, one matrix can represent many
different maps with respect to different bases. However, prove that a zero matrix
represents only a zero map. Are there other such matrices?
X 1.11 Let V and W be vector spaces of dimensions n and m. Show that the space
L(V, W) of linear maps from V to W is isomorphic to Mm
×
n.
X 1.12 Show that it follows from the prior questions that for any six transformations
t1, . . . , t6 : R2
→ R2
there are scalars c1, . . . , c6 ∈ R such that c1t1 + · · · + c6t6 is
the zero map. (Hint: this is a bit of a misleading question.)
1.13 The trace of a square matrix is the sum of the entries on the main diagonal
(the 1, 1 entry plus the 2, 2 entry, etc.; we will see the significance of the trace in
Chapter Five). Show that trace(H + G) = trace(H) + trace(G). Is there a similar
result for scalar multiplication?
1.14 Recall that the transpose of a matrix M is another matrix, whose i, j entry is
the j, i entry of M. Verifiy these identities.
(a) (G + H)trans
= Gtrans
+ Htrans
(b) (r · H)trans
= r · Htrans
X 1.15 A square matrix is symmetric if each i, j entry equals the j, i entry, that is, if
the matrix equals its transpose.
(a) Prove that for any H, the matrix H + Htrans
is symmetric. Does every
symmetric matrix have this form?
(b) Prove that the set of n×n symmetric matrices is a subspace of Mn
×
n.
X 1.16 (a) How does matrix rank interact with scalar multiplication—can a scalar
product of a rank n matrix have rank less than n? Greater?
(b) How does matrix rank interact with matrix addition—can a sum of rank n
matrices have rank less than n? Greater?

3.IV.2 Matrix Multiplication
After representing addition and scalar multiplication of linear maps in the
prior subsection, the natural next map operation to consider is composition.
2.1 Lemma A composition of linear maps is linear.
Proof. (This argument has appeared earlier, as part of the proof that isomor-
phism is an equivalence relation between spaces.) Let h: V → W and g: W → U
be linear. The natural calculation
g ◦ h
¡
c1 · ~
v1 + c2 · ~
v2
¢
= g
¡
h(c1 · ~
v1 + c2 · ~
v2)
¢
= g
¡
c1 · h(~
v1) + c2 · h(~
v2)
¢
= c1 · g
¡
h(~
v1)) + c2 · g(h(~
v2)
¢
= c1 · (g ◦ h)(~
v1) + c2 · (g ◦ h)(~
v2)
shows that g ◦ h: V → U preserves linear combinations. QED
To see how the representation of the composite arises out of the representations
of the two compositors, consider an example.
2.2 Example Let h: R4
→ R2
and g: R2
→ R3
, fix bases B ⊂ R4
, C ⊂ R2
,
D ⊂ R3
, and let these be the representations.
H = RepB,C(h) =
µ
4 6 8 2
5 7 9 3
¶
B,C
G = RepC,D(g) =


1 1
0 1
1 0


C,D
To represent the composition g ◦ h: R4
→ R3
we fix a ~
v, represent h of ~
v, and
then represent g of that. The representation of h(~
v) is the product of h’s matrix
and ~
v’s vector.
RepC( h(~
v) ) =
µ
4 6 8 2
5 7 9 3
¶
B,C




v1
v2
v3
v4




B
=
µ
4v1 + 6v2 + 8v3 + 2v4
5v1 + 7v2 + 9v3 + 3v4
¶
C
The representation of g( h(~
v) ) is the product of g’s matrix and h(~
v)’s vector.
RepD( g(h(~
v)) )=


1 1
0 1
1 0


C,D
µ
4v1 + 6v2 + 8v3 + 2v4
5v1 + 7v2 + 9v3 + 3v4
¶
C
=


1 · (4v1 + 6v2 + 8v3 + 2v4) + 1 · (5v1 + 7v2 + 9v3 + 3v4)
0 · (4v1 + 6v2 + 8v3 + 2v4) + 1 · (5v1 + 7v2 + 9v3 + 3v4)
1 · (4v1 + 6v2 + 8v3 + 2v4) + 0 · (5v1 + 7v2 + 9v3 + 3v4)


D
Distributing and regrouping on the v’s gives
=


(1 · 4 + 1 · 5)v1 + (1 · 6 + 1 · 7)v2 + (1 · 8 + 1 · 9)v3 + (1 · 2 + 1 · 3)v4
(0 · 4 + 1 · 5)v1 + (0 · 6 + 1 · 7)v2 + (0 · 8 + 1 · 9)v3 + (0 · 2 + 1 · 3)v4
(1 · 4 + 0 · 5)v1 + (1 · 6 + 0 · 7)v2 + (1 · 8 + 0 · 9)v3 + (1 · 2 + 0 · 3)v4


D

which we recognizing as the result of this matrix-vector product.
=


1 · 4 + 1 · 5 1 · 6 + 1 · 7 1 · 8 + 1 · 9 1 · 2 + 1 · 3
0 · 4 + 1 · 5 0 · 6 + 1 · 7 0 · 8 + 1 · 9 0 · 2 + 1 · 3
1 · 4 + 0 · 5 1 · 6 + 0 · 7 1 · 8 + 0 · 9 1 · 2 + 0 · 3


B,D




v1
v2
v3
v4




D
Thus, the matrix representing g◦h has the rows of G combined with the columns
of H.
2.3 Definition The matrix-multiplicative product of the m×r matrix G and
the r×n matrix H is the m×n matrix P, where
pi,j = gi,1h1,j + gi,2h2,j + · · · + gi,rhr,j
that is, the i, j-th entry of the product is the dot product of the i-th row and
the j-th column.
GH =




.
.
.
gi,1 gi,2 . . . gi,r
.
.
.









h1,j
. . . h2,j . . .
.
.
.
hr,j





=




.
.
.
. . . pi,j . . .
.
.
.




2.4 Example The matrices from Example 2.2 combine in this way.


1 · 4 + 1 · 5 1 · 6 + 1 · 7 1 · 8 + 1 · 9 1 · 2 + 1 · 3
0 · 4 + 1 · 5 0 · 6 + 1 · 7 0 · 8 + 1 · 9 0 · 2 + 1 · 3
1 · 4 + 0 · 5 1 · 6 + 0 · 7 1 · 8 + 0 · 9 1 · 2 + 0 · 3

 =


9 13 17 5
5 7 9 3
4 6 8 2


2.5 Example


2 0
4 6
8 2


µ
1 3
5 7
¶
=


2 · 1 + 0 · 5 2 · 3 + 0 · 7
4 · 1 + 6 · 5 4 · 3 + 6 · 7
8 · 1 + 2 · 5 8 · 3 + 2 · 7

 =


2 6
34 54
18 38


2.6 Theorem A composition of linear maps is represented by the matrix prod-
uct of the representatives.
Proof. (This argument parallels Example 2.2.) Let h: V → W and g: W → X
be represented by H and G with respect to bases B ⊂ V , C ⊂ W, and D ⊂ X,
of sizes n, r, and m. For any ~
v ∈ V , the k-th component of RepC( h(~
v) ) is
hk,1v1 + · · · + hk,nvn
and so the i-th component of RepD( g ◦ h (~
v) ) is this.
gi,1 · (h1,1v1 + · · · + h1,nvn) + gi,2 · (h2,1v1 + · · · + h2,nvn)
+ · · · + gi,r · (hr,1v1 + · · · + hr,nvn)

Distribute and regroup on the v’s.
= (gi,1h1,1 + gi,2h2,1 + · · · + gi,rhr,1) · v1
+ · · · + (gi,1h1,n + gi,2h2,n + · · · + gi,rhr,n) · vn
Finish by recognizing that the coefficient of each vj
gi,1h1,j + gi,2h2,j + · · · + gi,rhr,j
matches the definition of the i, j entry of the product GH. QED
The theorem is an example of a result that supports a definition. We can
picture what the definition and theorem together say with this arrow diagram
(‘w.r.t.’ abbreviates ‘with respect to’).
Vw.r.t. B
g◦h
−→
GH
Xw.r.t. D
%
h
H
Ww.r.t. C
&
g
G
Above the arrows, the maps show that the two ways of going from V to X,
straight over via the composition or else by way of W, have the same effect
~
v
g◦h
7−→ g(h(~
v)) ~
v
h
7−→ h(~
v)
g
7−→ g(h(~
v))
(this is just the definition of composition). Below the arrows, the matrices
indicate that the product does the same thing—multiplying GH into the column
vector RepB(~
v) has the same effect as multiplying the column first by H and
then multiplying the result by G.
RepB,D(g ◦ h) = GH = RepC,D(g) RepB,C(h)
The definition of the matrix-matrix product operation does not restrict us
to view it as a representation of a linear map composition. We can get insight
into this operation by studying it as a mechanical procedure. The striking thing
is the way that rows and columns combine.
One aspect of that combination is that the sizes of the matrices involved is
significant. Briefly, m×r times r×n equals m×n.
2.7 Example This product is not defined
µ
−1 2 0
0 10 1.1
¶ µ
0 0
0 2
¶
because the number of columns on the left does not equal the number of rows
on the right.

In terms of the underlying maps, the fact that the sizes must match up reflects
the fact that matrix multiplication is defined only when a corresponding function
composition
dimension n space
h
−→ dimension r space
g
−→ dimension m space
is possible.
2.8 Remark The order in which these things are written can be confusing. In
the ‘m×r times r×n equals m×n’ equation, the number written first m is the
dimension of g’s codomain and is thus the number that appears last in the map
dimension description above. The explanation is that while f is done first and
then g is applied, that composition is written g ◦ f, from the notation ‘g(f(~
v))’.
(Some people try to lessen confusion by reading ‘g ◦ f’ aloud as “g following
f”.) That order then carries over to matrices: g ◦ f is represented by GF.
Another aspect of the way that rows and columns combine in the matrix
product operation is that in the definition of the i, j entry
pi,j = gi, 1
h 1 ,j
+ gi, 2
h 2 ,j
+ · · · + gi, r h r ,j
the boxed subscripts on the g’s are column indicators while those on the h’s
indicate rows. That is, summation takes place over the columns of G but over
the rows of H; left is treated differently than right, so GH may be unequal to
HG. Matrix multiplication is not commutative.
2.9 Example Matrix multiplication hardly ever commutes. Test that by mul-
tiplying randomly chosen matrices both ways.
µ
1 2
3 4
¶ µ
5 6
7 8
¶
=
µ
19 22
43 50
¶ µ
5 6
7 8
¶ µ
1 2
3 4
¶
=
µ
23 34
31 46
¶
2.10 Example Commutativity can fail more dramatically:
µ
5 6
7 8
¶ µ
1 2 0
3 4 0
¶
=
µ
23 34 0
31 46 0
¶
while
µ
1 2 0
3 4 0
¶ µ
5 6
7 8
¶
isn’t even defined.
2.11 Remark The fact that matrix multiplication is not commutative may
be puzzling at first sight, perhaps just because most algebraic operations in
elementary mathematics are commutative. But on further reflection, it isn’t
so surprising. After all, matrix multiplication represents function composition,
which is not commutative—if f(x) = 2x and g(x) = x+1 then g ◦f(x) = 2x+1
while f ◦ g(x) = 2(x + 1) = 2x + 2. True, this g is not linear and we might
have hoped that linear functions commute, but this perspective shows that the
failure of commutativity for matrix multiplication fits into a larger context.

Except for the lack of commutativity, matrix multiplication is algebraically
well-behaved. Below are some nice properties and more are in Exercise 23 and
Exercise 24.
2.12 Theorem If F, G, and H are matrices, and the matrix products are
defined, then the product is associative (FG)H = F(GH) and distributes over
matrix addition F(G + H) = FG + FH and (G + H)F = GF + HF.
Proof. Associativity holds because matrix multiplication represents function
composition, which is associative: the maps (f ◦ g) ◦ h and f ◦ (g ◦ h) are equal
as both send ~
v to f(g(h(~
v))).
Distributivity is similar. For instance, the first one goes f ◦ (g + h) (~
v) =
f
¡
(g + h)(~
v)
¢
= f
¡
g(~
v) + h(~
v)
¢
= f(g(~
v)) + f(h(~
v)) = f ◦ g(~
v) + f ◦ h(~
v) (the
third equality uses the linearity of f). QED
2.13 Remark We could alternatively prove that result by slogging through
the indices. For example, associativity goes: the i, j-th entry of (FG)H is
(fi,1g1,1 + fi,2g2,1 + · · · + fi,rgr,1)h1,j
+ (fi,1g1,2 + fi,2g2,2 + · · · + fi,rgr,2)h2,j
.
.
.
+ (fi,1g1,s + fi,2g2,s + · · · + fi,rgr,s)hs,j
(where F, G, and H are m×r, r×s, and s×n matrices), distribute
fi,1g1,1h1,j + fi,2g2,1h1,j + · · · + fi,rgr,1h1,j
+ fi,1g1,2h2,j + fi,2g2,2h2,j + · · · + fi,rgr,2h2,j
.
.
.
+ fi,1g1,shs,j + fi,2g2,shs,j + · · · + fi,rgr,shs,j
and regroup around the f’s
fi,1(g1,1h1,j + g1,2h2,j + · · · + g1,shs,j)
+ fi,2(g2,1h1,j + g2,2h2,j + · · · + g2,shs,j)
.
.
.
+ fi,r(gr,1h1,j + gr,2h2,j + · · · + gr,shs,j)
to get the i, j entry of F(GH).
Contrast these two ways of verifying associativity, the one in the proof and
the one just above. The argument just above is hard to understand in the sense
that, while the calculations are easy to check, the arithmetic seems unconnected
to any idea (it also essentially repeats the proof of Theorem 2.6 and so is ineffi-
cient). The argument in the proof is shorter, clearer, and says why this property
“really” holds. This illustrates the comments made in the preamble to the chap-
ter on vector spaces—at least some of the time an argument from higher-level
constructs is clearer.

We have now seen how the representation of the composition of two linear
maps is derived from the representations of the two maps. We have called
the combination the product of the two matrices. This operation is extremely
important. Before we go on to study how to represent the inverse of a linear
map, we will explore it some more in the next subsection.
Exercises
X 2.14 Compute, or state ‘not defined’.
(a)
µ
3 1
−4 2
¶ µ
0 5
0 0.5
¶
(b)
µ
1 1 −1
4 0 3
¶ Ã
2 −1 −1
3 1 1
3 1 1
!
(c)
µ
2 −7
7 4
¶ Ã
1 0 5
−1 1 1
3 8 4
!
(d)
µ
5 2
3 1
¶ µ
−1 2
3 −5
¶
X 2.15 Where
A =
µ
1 −1
2 0
¶
B =
µ
5 2
4 4
¶
C =
µ
−2 3
−4 1
¶
compute or state ‘not defined’.
(a) AB (b) (AB)C (c) BC (d) A(BC)
2.16 Which products are defined?
(a) 3×2 times 2×3 (b) 2×3 times 3×2 (c) 2×2 times 3×3
(d) 3×3 times 2×2
X 2.17 Give the size of the product or state ‘not defined’.
(a) a 2×3 matrix times a 3×1 matrix
(b) a 1×12 matrix times a 12×1 matrix
(c) a 2×3 matrix times a 2×1 matrix
(d) a 2×2 matrix times a 2×2 matrix
X 2.18 Find the system of equations resulting from starting with
h1,1x1 + h1,2x2 + h1,3x3 = d1
h2,1x1 + h2,2x2 + h2,3x3 = d2
and making this change of variable (i.e., substitution).
x1 = g1,1y1 + g1,2y2
x2 = g2,1y1 + g2,2y2
x3 = g3,1y1 + g3,2y2
2.19 As Definition 2.3 points out, the matrix product operation generalizes the dot
product. Is the dot product of a 1×n row vector and a n×1 column vector the
same as their matrix-multiplicative product?
X 2.20 Represent the derivative map on Pn with respect to B, B where B is the
natural basis h1, x, . . . , xn
i. Show that the product of this matrix with itself is
defined; what the map does it represent?
2.21 Show that composition of linear transformations on R1
is commutative. Is
this true for any one-dimensional space?
2.22 Why is matrix multiplication not defined as entry-wise multiplication? That
would be easier, and commutative too.
X 2.23 (a) Prove that Hp
Hq
= Hp+q
and (Hp
)q
= Hpq
for positive integers p, q.
(b) Prove that (rH)p
= rp
· Hp
for any positive integer p and scalar r ∈ R.

X 2.24 (a) How does matrix multiplication interact with scalar multiplication: is
r(GH) = (rG)H? Is G(rH) = r(GH)?
(b) How does matrix multiplication interact with linear combinations: is F(rG+
sH) = r(FG) + s(FH)? Is (rF + sG)H = rFH + sGH?
2.25 We can ask how the matrix product operation interacts with the transpose
operation.
(a) Show that (GH)trans
= Htrans
Gtrans
.
(b) A square matrix is symmetric if each i, j entry equals the j, i entry, that is,
if the matrix equals its own transpose. Show that the matrices HHtrans
and
Htrans
H are symmetric.
X 2.26 Rotation of vectors in R3
about an axis is a linear map. Show that linear
maps do not commute by showing geometrically that rotations do not commute.
2.27 In the proof of Theorem 2.12 some maps are used. What are the domains and
codomains?
2.28 How does matrix rank interact with matrix multiplication?
(a) Can the product of rank n matrices have rank less than n? Greater?
(b) Show that the rank of the product of two matrices is less than or equal to
the minimum of the rank of each factor.
2.29 Is ‘commutes with’ an equivalence relation among n×n matrices?
X 2.30 (This will be used in the Matrix Inverses exercises.) Here is another property
of matrix multiplication that might be puzzling at first sight.
(a) Prove that the composition of the projections πx, πy : R3
→ R3
onto the x
and y axes is the zero map despite that neither one is itself the zero map.
(b) Prove that the composition of the derivatives d2
/dx2
, d3
/dx3
: P4 → P4 is
the zero map despite that neither is the zero map.
(c) Give a matrix equation representing the first fact.
(d) Give a matrix equation representing the second.
When two things multiply to give zero despite that neither is zero, each is said to
be a zero divisor.
2.31 Show that, for square matrices, (S + T)(S − T) need not equal S2
− T2
.
X 2.32 Represent the identity transformation id: V → V with respect to B, B for any
basis B. This is the identity matrix I. Show that this matrix plays the role in
matrix multiplication that the number 1 plays in real number multiplication: HI =
IH = H (for all matrices H for which the product is defined).
2.33 In real number algebra, quadratic equations have at most two solutions. That
is not so with matrix algebra. Show that the 2×2 matrix equation T2
= I has
more than two solutions, where I is the identity matrix (this matrix has ones in
its 1, 1 and 2, 2 entries and zeroes elsewhere; see Exercise 32).
2.34 (a) Prove that for any 2×2 matrix T there are scalars c0, . . . , c4 such that the
combination c4T4
+c3T3
+c2T2
+c1T +I is the zero matrix (where I is the 2×2
identity matrix, with ones in its 1, 1 and 2, 2 entries and zeroes elsewhere; see
Exercise 32).
(b) Let p(x) be a polynomial p(x) = cnxn
+ · · · + c1x + c0. If T is a square
matrix we define p(T) to be the matrix cnTn
+ · · · + c1T + I (where I is the
appropriately-sized identity matrix). Prove that for any square matrix there is
a polynomial such that p(T) is the zero matrix.
(c) The minimal polynomial m(x) of a square matrix is the polynomial of least
degree, and with leading coefficient 1, such that m(T) is the zero matrix. Find

the minimal polynomial of this matrix.
µ√
3/2 −1/2
1/2
√
3/2
¶
(This is the representation with respect to E2, E2, the standard basis, of a rotation
through π/6 radians counterclockwise.)
2.35 The infinite-dimensional space P of all finite-degree polynomials gives a mem-
orable example of the non-commutativity of linear maps. Let d/dx: P → P be the
usual derivative and let s: P → P be the shift map.
a0 + a1x + · · · + anxn s
7−→ 0 + a0x + a1x2
+ · · · + anxn+1
Show that the two maps don’t commute d/dx ◦ s 6= s ◦ d/dx; in fact, not only is
(d/dx ◦ s) − (s ◦ d/dx) not the zero map, it is the identity map.
2.36 Recall the notation for the sum of the sequence of numbers a1, a2, . . . , an.
n
X
i=1
ai = a1 + a2 + · · · + an
In this notation, the i, j entry of the product of G and H is this.
pi,j =
r
X
k=1
gi,khk,j
Using this notation,
(a) reprove that matrix multiplication is associative;
(b) reprove Theorem 2.6.
3.IV.3 Mechanics of Matrix Multiplication
In this subsection we consider matrix multiplication as a mechanical process,
putting aside for the moment any implications about the underlying maps. As
described earlier, the striking thing about matrix multiplication is the way rows
and columns combine. The i, j entry of the matrix product is the dot product
of the row i of the left matrix with column j of the right one. For instance, here
is a second row and a third column combining to make a 2, 3 entry.



1 1
0 1
1 0



Ã
4
5
6
7
8
9
2
3
!
=


9 13 17 5
5 7 9 3
4 6 8 2


We can view this as the left matrix acting by multiplying its rows, one at a time,
into the columns of the right matrix. Of course, another perspective is that the
right matrix uses its columns to act on the left matrix’s rows. Below, we will
examine actions from the left and from the right for some simple matrices.
The first case, the action of a zero matrix, is very easy.

3.1 Example Multiplying by an appropriately-sized zero matrix from the left
or from the right
µ
0 0
0 0
¶ µ
1 3 2
−1 1 −1
¶
=
µ
0 0 0
0 0 0
¶ µ
2 3
1 4
¶ µ
0 0
0 0
¶
=
µ
0 0
0 0
¶
results in a zero matrix.
After zero matrices, the matrices whose actions are easiest to understand
are the ones with a single nonzero entry.
3.2 Definition A matrix with all zeroes except for a one in the i, j entry is an
i, j unit matrix.
3.3 Example This is the 1, 2 unit matrix with three rows and two columns,
multiplying from the left.


0 1
0 0
0 0


µ
5 6
7 8
¶
=


7 8
0 0
0 0


Acting from the left, an i, j unit matrix copies row j of the multiplicand into
row i of the result. From the right an i, j unit matrix copies column i of the
multiplicand into column j of the result.


1 2 3
4 5 6
7 8 9




0 1
0 0
0 0

 =


0 1
0 4
0 7


3.4 Example Rescaling these matrices simply rescales the result. This is the
action from the left of the matrix that is twice the one in the prior example.


0 2
0 0
0 0


µ
5 6
7 8
¶
=


14 16
0 0
0 0


And this is the action of the matrix that is minus three times the one from the
prior example.


1 2 3
4 5 6
7 8 9




0 −3
0 0
0 0

 =


0 −3
0 −12
0 −21


Next in complication are matrices with two nonzero entries. There are two
cases. If a left-multiplier has entries in different rows then their actions don’t
interact.

3.5 Example


1 0 0
0 0 2
0 0 0




1 2 3
4 5 6
7 8 9

 = (


1 0 0
0 0 0
0 0 0

 +


0 0 0
0 0 2
0 0 0

)


1 2 3
4 5 6
7 8 9


=


1 2 3
0 0 0
0 0 0

 +


0 0 0
14 16 18
0 0 0


=


1 2 3
14 16 18
0 0 0


But if the left-multiplier’s nonzero entries are in the same row then that row of
the result is a combination.
3.6 Example


1 0 2
0 0 0
0 0 0




1 2 3
4 5 6
7 8 9

 = (


1 0 0
0 0 0
0 0 0

 +


0 0 2
0 0 0
0 0 0

)


1 2 3
4 5 6
7 8 9


=


1 2 3
0 0 0
0 0 0

 +


14 16 18
0 0 0
0 0 0


=


15 18 21
0 0 0
0 0 0


Right-multiplication acts in the same way, with columns.
These observations about matrices that are mostly zeroes extend to arbitrary
matrices.
3.7 Lemma In a product of two matrices G and H, the columns of GH are
formed by taking G times the columns of H
G ·




.
.
.
.
.
.
~
h1 · · · ~
hn
.
.
.
.
.
.



 =




.
.
.
.
.
.
G · ~
h1 · · · G · ~
hn
.
.
.
.
.
.




and the rows of GH are formed by taking the rows of G times H




· · · ~
g1 · · ·
.
.
.
· · · ~
gr · · ·



 · H =




· · · ~
g1 · H · · ·
.
.
.
· · · ~
gr · H · · ·




(ignoring the extra parentheses).

Proof. We will exhibit the 2×2 case, and leave the general case as an exercise.
GH =
µ
g1,1 g1,2
g2,1 g2,2
¶ µ
h1,1 h1,2
h2,1 h2,2
¶
=
µ
g1,1h1,1 + g1,2h2,1 g1,1h1,2 + g1,2h2,2
g2,1h1,1 + g2,2h2,1 g2,1h1,2 + g2,2h2,2
¶
The right side of the first equation in the result
µ
G
µ
h1,1
h2,1
¶
G
µ
h1,2
h2,2
¶¶
=
µµ
g1,1h1,1 + g1,2h2,1
g2,1h1,1 + g2,2h2,1
¶ µ
g1,1h1,2 + g1,2h2,2
g2,1h1,2 + g2,2h2,2
¶¶
is indeed the same as the right side of GH, except for the extra parentheses (the
ones marking the columns as column vectors). The other equation is similarly
easy to recognize. QED
An application of those observations is that there is a matrix that just copies
out the rows and columns.
3.8 Definition The main diagonal (or principle diagonal or diagonal) of a
square matrix goes from the upper left to the lower right.
3.9 Definition An identity matrix is square and has with all entries zero ex-
cept for ones in the main diagonal.
In×n =





1 0 . . . 0
0 1 . . . 0
.
.
.
0 0 . . . 1





3.10 Example The 3×3 identity leaves its multiplicand unchanged both from
the left


1 0 0
0 1 0
0 0 1




2 3 6
1 3 8
−7 1 0

 =


2 3 6
1 3 8
−7 1 0


and from the right.


2 3 6
1 3 8
−7 1 0




1 0 0
0 1 0
0 0 1

 =


2 3 6
1 3 8
−7 1 0


3.11 Example So does the 2×2 identity matrix.




1 −2
0 −2
1 −1
4 3




µ
1 0
0 1
¶
=




1 −2
0 −2
1 −1
4 3





In short, an identity matrix is the identity element of the set of n×n matrices,
with respect to the operation of matrix multiplication.
We next see two ways to generalize the identity matrix.
The first is that if the ones are relaxed to arbitrary reals, the resulting matrix
will rescale whole rows or columns.
3.12 Definition A diagonal matrix is square and has zeros off the main diag-
onal.





a1,1 0 . . . 0
0 a2,2 . . . 0
.
.
.
0 0 . . . an,n





3.13 Example From the left, the action of multiplication by a diagonal matrix
is to rescales the rows.
µ
2 0
0 −1
¶ µ
2 1 4 −1
−1 3 4 4
¶
=
µ
4 2 8 −2
1 −3 −4 −4
¶
From the right such a matrix rescales the columns.
µ
1 2 1
2 2 2
¶


3 0 0
0 2 0
0 0 −2

 =
µ
3 4 −2
6 4 −4
¶
The second generalization of identity matrices is that we can put a single one
in each row and column in ways other than putting them down the diagonal.
3.14 Definition A permutation matrix is square and is all zeros except for a
single one in each row and column.
3.15 Example From the left these matrices permute rows.


0 0 1
1 0 0
0 1 0




1 2 3
4 5 6
7 8 9

 =


7 8 9
1 2 3
4 5 6


From the right they permute columns.


1 2 3
4 5 6
7 8 9




0 0 1
1 0 0
0 1 0

 =


2 3 1
5 6 4
8 9 7


We finish this subsection by applying these observations to get matrices that
perform Gauss’ method and Gauss-Jordan reduction.

3.16 Example We have seen how to produce a matrix that will rescale rows.
Multiplying by this diagonal matrix rescales the second row of the other by a
factor of three.


1 0 0
0 3 0
0 0 1




0 2 1 1
0 1/3 1 −1
1 0 2 0

 =


0 2 1 1
0 1 3 −3
1 0 2 0


We have seen how to produce a matrix that will swap rows. Multiplying by this
permutation matrix swaps the first and third rows.


0 0 1
0 1 0
1 0 0




0 2 1 1
0 1 3 −3
1 0 2 0

 =


1 0 2 0
0 1 3 −3
0 2 1 1


To see how to perform a pivot, we observe something about those two ex-
amples. The matrix that rescales the second row by a factor of three arises in
this way from the identity.


1 0 0
0 1 0
0 0 1

 3ρ2
−→


1 0 0
0 3 0
0 0 1


Similarly, the matrix that swaps first and third rows arises in this way.


1 0 0
0 1 0
0 0 1

 ρ1↔ρ3
−→


0 0 1
0 1 0
1 0 0


3.17 Example The 3×3 matrix that arises as


1 0 0
0 1 0
0 0 1

 −2ρ2+ρ3
−→


1 0 0
0 1 0
0 −2 1


will, when it acts from the left, perform the pivot operation −2ρ2 + ρ3.


1 0 0
0 1 0
0 −2 1




1 0 2 0
0 1 3 −3
0 2 1 1

 =


1 0 2 0
0 1 3 −3
0 0 −5 7


3.18 Definition The elementary reduction matrices are obtained from identity
matrices with one Gaussian operation. We denote them:
(1) I
kρi
−→ Mi(k) for k 6= 0;
(2) I
ρi↔ρj
−→ Pi,j for i 6= j;
(3) I
kρi+ρj
−→ Ci,j(k) for i 6= j.

3.19 Lemma Gaussian reduction can be done through matrix multiplication.
(1) If H
kρi
−→ G then Mi(k)H = G.
(2) If H
ρi↔ρj
−→ G then Pi,jH = G.
(3) If H
kρi+ρj
−→ G then Ci,j(k)H = G.
Proof. Clear. QED
3.20 Example This is the first system, from the first chapter, on which we
performed Gauss’ method.
3x3 = 9
x1 + 5x2 − 2x3 = 2
(1/3)x1 + 2x2 = 3
It can be reduced with matrix multiplication. Swap the first and third rows,


0 0 1
0 1 0
1 0 0




0 0 3 9
1 5 −2 2
1/3 2 0 3

 =


1/3 2 0 3
1 5 −2 2
0 0 3 9


triple the first row,


3 0 0
0 1 0
0 0 1




1/3 2 0 3
1 5 −2 2
0 0 3 9

 =


1 6 0 9
1 5 −2 2
0 0 3 9


and then add −1 times the first row to the second.


1 0 0
−1 1 0
0 0 1




1 6 0 9
1 5 −2 2
0 0 3 9

 =


1 6 0 9
0 −1 −2 −7
0 0 3 9


Now back substitution will give the solution.
3.21 Example Gauss-Jordan reduction works the same way. For the matrix
ending the prior example, first adjust the leading entries


1 0 0
0 −1 0
0 0 1/3




1 6 0 9
0 −1 −2 −7
0 0 3 9

 =


1 6 0 9
0 1 2 7
0 0 1 3


and to finish, clear the third column and then the second column.


1 −6 0
0 1 0
0 0 1




1 0 0
0 1 −2
0 0 1




1 6 0 9
0 1 2 7
0 0 1 3

 =


1 0 0 3
0 1 0 1
0 0 1 3



We have observed the following result, which we shall use in the next sub-
section.
3.22 Corollary For any matrix H there are elementary reduction matrices
R1, . . . , Rr such that Rr · Rr−1 · · · R1 · H is in reduced echelon form.
Until now we have taken the point of view that our primary objects of study
are vector spaces and the maps between them, and have adopted matrices only
for computational convenience. This subsection show that this point of view
isn’t the whole story. Matrix theory is a fascinating and fruitful area.
In the rest of this book we shall continue to focus on maps as the primary
objects, but we will be pragmatic—if the matrix point of view gives some clearer
idea then we shall use it.
Exercises
X 3.23 Predict the result of each multiplication by an elementary reduction matrix,
and then check by multiplying it out.
(a)
µ
3 0
0 0
¶ µ
1 2
3 4
¶
(b)
µ
4 0
0 2
¶ µ
1 2
3 4
¶
(c)
µ
1 0
−2 1
¶ µ
1 2
3 4
¶
(d)
µ
1 2
3 4
¶ µ
1 −1
0 1
¶
(e)
µ
1 2
3 4
¶ µ
0 1
1 0
¶
X 3.24 The need to take linear combinations of rows and columns in tables of numbers
arises often in practice. For instance, this is a map of part of Vermont and New
York.
In part because of Lake Champlain,
there are no roads connecting some
pairs of towns. For instance, there
is no way to go from Winooski to
Grand Isle without going through
Colchester. (Of course, many other
roads and towns have been left off
to simplify the graph. From top to
bottom of this map is about forty
miles.)
Burlington
Colchester
Grand Isle
Swanton
Winooski
(a) The incidence matrix of a map is the square matrix whose i, j entry is the
number of roads from city i to city j. Produce the incidence matrix of this map
(take the cities in alphabetical order).
(b) A matrix is symmetric if it equals its transpose. Show that an incidence
matrix is symmetric. (These are all two-way streets. Vermont doesn’t have
many one-way streets.)
(c) What is the significance of the square of the incidence matrix? The cube?

X 3.25 The need to take linear combinations of rows and columns in tables of numbers
arises often in practice. For instance, this table gives the number of hours of each
type done by each worker, and the associated pay rates. Use matrices to compute
the wages due.
regular overtime
Alan 40 12
Betty 35 6
Catherine 40 18
Donald 28 0
wage
regular $25.00
overtime $45.00
3.26 Find the product of this matrix with its transpose.
µ
cos θ − sin θ
sin θ cos θ
¶
X 3.27 Prove that the diagonal matrices form a subspace of Mn
×
n. What is its
dimension?
3.28 Does the identity matrix represent the identity map if the bases are unequal?
3.29 Show that every multiple of the identity commutes with every square matrix.
Are there other matrices that commute with all square matrices?
3.30 Prove or disprove: nonsingular matrices commute.
X 3.31 Show that the product of a permutation matrix and its transpose is an identity
matrix.
3.32 Show that if the first and second rows of G are equal then so are the first and
second rows of GH. Generalize.
3.33 Describe the product of two diagonal matrices.
3.34 Write µ
1 0
−3 3
¶
as the product of two elementary reduction matrices.
X 3.35 Show that if G has a row of zeros then GH (if defined) has a row of zeros.
Does that work for columns?
3.36 Show that the set of unit matrices forms a basis for Mn
×
m.
3.37 Find the formula for the n-th power of this matrix.
µ
1 1
1 0
¶
X 3.38 The trace of a square matrix is the sum of the entries on its diagonal (its
significance appears in Chapter Five). Show that trace(GH) = trace(HG).
X 3.39 A square matrix is upper triangular if its only nonzero entries lie above, or
on, the diagonal. Show that the product of two upper triangular matrices is upper
triangular. Does this hold for lower triangular also?
3.40 A square matrix is a Markov matrix if each entry is between zero and one
and the sum along each row is one. Prove that a product of Markov matrices is
Markov.
X 3.41 Give an example of two matrices of the same rank with squares of differing
rank.
3.42 Combine the two generalizations of the identity matrix, the one allowing en-
tires to be other than ones, and the one allowing the single one in each row and
column to be off the diagonal. What is the action of this type of matrix?

3.43 On a computer multiplications are more costly than additions, so people are
interested in reducing the number of multiplications used to compute a matrix
product.
(a) How many real number multiplications are needed in formula we gave for the
product of a m×r matrix and a r×n matrix?
(b) Matrix multiplication is associative, so all associations yield the same result.
The cost in number of multiplications, however, varies. Find the association
requiring the fewest real number multiplications to compute the matrix product
of a 5×10 matrix, a 10×20 matrix, a 20×5 matrix, and a 5×1 matrix.
(c) (Very hard.) Find a way to multiply two 2×2 matrices using only seven
multiplications instead of the eight suggested by the naive approach.
3.44 [Putnam, 1990, A-5] If A and B are square matrices of the same size such
that ABAB = 0, does it follow that BABA = 0?
3.45 [Am. Math. Mon., Dec. 1966] Demonstrate these four assertions to get an al-
ternate proof that column rank equals row rank.
(a) ~
y · ~
y = ~
0 iff ~
y = ~
0.
(b) A~
x = ~
0 iff Atrans
A~
x = ~
0.
(c) dim(R(A)) = dim(R(Atrans
A)).
(d) col rank(A) = col rank(Atrans
) = row rank(A).
3.46 [Ackerson] Prove (where A is an n×n matrix and so defines a transformation
of any n-dimensional space V with respect to B, B where B is a basis) dim(R(A)∩
N (A)) = dim(R(A)) − dim(R(A2
)). Conclude
(a) N (A) ⊂ R(A) iff dim(N (A)) = dim(R(A)) − dim(R(A2
));
(b) R(A) ⊆ N (A) iff A2
= 0;
(c) R(A) = N (A) iff A2
= 0 and dim(N (A)) = dim(R(A)) ;
(d) dim(R(A) ∩ N (A)) = 0 iff dim(R(A)) = dim(R(A2
)) ;
(e) (Requires the Direct Sum subsection, which is optional.) V = R(A) ⊕ N (A)
iff dim(R(A)) = dim(R(A2
)).
3.IV.4 Inverses
We now consider how to represent the inverse of a linear map.
We start by recalling some facts about function inverses.∗
Some functions
have no inverse, or have an inverse on one side only.
4.1 Example Where π: R3
→ R2
is the projection map


x
y
z

 7→
µ
x
y
¶
and η: R2
→ R3
is the embedding
µ
x
y
¶
7→


x
y
0


∗ More information on function inverses is in the appendix.

the composition π ◦ η is the identity map on R2
.
π ◦ η :
µ
x
y
¶
η
7−→


x
y
0

 π
7−→
µ
x
y
¶
We say π is a left inverse map of η or, what is the same thing, that η is a right
inverse map of π. However, composition in the other order η◦π doesn’t give the
identity map—here is a vector that is not sent to itself under this composition.
η ◦ π :


0
0
1

 π
7−→
µ
0
0
¶
η
7−→


0
0
0


In fact, the projection π has no left inverse at all. For, if f were to be a left
inverse of π then we would have
f ◦ π :


x
y
z

 π
7−→
µ
x
y
¶
f
7−→


x
y
z


for all of the infinitely many z’s. But no function f can send a single argument
to more than one value.
(An example of a function with no inverse on either side, is the zero transfor-
mation on R2
.) Some functions have a two-sided inverse map, another function
that is the inverse of the first, both from the left and from the right. For in-
stance, the map given by ~
v 7→ 2 · ~
v has the two-sided inverse ~
v 7→ (1/2) · ~
v. In
this subsection we will focus on two-sided inverses. The appendix shows that a
function (linear or not) has a two-sided inverse if and only if it is both one-to-one
and onto. The appendix also shows that if a function f has a two-sided inverse
then it is unique, and so it is called ‘the’ inverse, and is denoted f−1
. So our
purpose in this subsection is, where a linear map h has an inverse, to find the
relationship between RepB,D(h) and RepD,B(h−1
).
4.2 Definition A matrix G is a left inverse matrix of the matrix H if GH is
the identity matrix. It is a right inverse matrix if HG is the identity. A matrix
H with a two-sided inverse is an invertible matrix. That two-sided inverse is
called the inverse matrix and is denoted H−1
.
Because of the correspondence between linear maps and matrices, statements
about map inverses translate into statements about matrix inverses.
4.3 Lemma If a matrix has both a left inverse and a right inverse then the
two are equal.
4.4 Theorem A matrix is invertible if and only if it is nonsingular.

Proof. (For both results.) Given a matrix H, fix spaces of appropriate dimen-
sion for the domain and codomain. Fix bases for these spaces. With respect to
these bases, H represents a map h. The statements are true about the map and
therefore they are true about the matrix. QED
4.5 Lemma A product of invertible matrices is invertible—if G and H are
invertible and if GH is defined then GH is invertible and (GH)−1
= H−1
G−1
.
Proof. (This is just like the prior proof except that it requires two maps.) Fix
appropriate spaces and bases and consider the represented maps h and g. Note
that h−1
g−1
is a two-sided map inverse of gh since (h−1
g−1
)(gh) = h−1
(id)h =
h−1
h = id and (gh)(h−1
g−1
) = g(id)g−1
= gg−1
= id. This equality is reflected
in the matrices representing the maps, as required. QED
Here is the arrow diagram giving the relationship between map inverses and
matrix inverses. It is a special case of the diagram for function composition and
matrix multiplication.
Vw.r.t. B
id
−→
I
Vw.r.t. B
%
h
H
Ww.r.t. D
&h−1
H−1
Beyond its place in our general program of seeing how to represent map
operations, another reason for our interest in inverses comes from solving linear
systems. A linear system is equivalent to a matrix equation, as here.
x1 + x2 = 3
2x1 − x2 = 2
⇐⇒
µ
1 1
2 −1
¶ µ
x1
x2
¶
=
µ
3
2
¶
(∗)
By fixing spaces and bases (e.g., R2
, R2
and E2, E2), we take the matrix H to
represent some map h. Then solving the system is the same as asking: what
domain vector ~
x is mapped by h to the result ~
d ? If we could invert h then we
could solve the system by multiplying RepD,B(h−1
) · RepD(~
d) to get RepB(~
x).
4.6 Example We can find a left inverse for the matrix just given
µ
m n
p q
¶ µ
1 1
2 −1
¶
=
µ
1 0
0 1
¶
by using Gauss’ method to solve the resulting linear system.
m + 2n = 1
m − n = 0
p + 2q = 0
p − q = 1
Answer: m = 1/3, n = 1/3, p = 2/3, and q = −1/3. This matrix is actually
the two-sided inverse of H, as can easily be checked. With it we can solve the
system (∗) above by applying the inverse.
µ
x
y
¶
=
µ
1/3 1/3
2/3 −1/3
¶ µ
3
2
¶
=
µ
5/3
4/3
¶

4.7 Remark Why solve systems this way, when Gauss’ method takes less
arithmetic (this assertion can be made precise by counting the number of arith-
metic operations, as computer algorithm designers do)? Beyond its conceptual
appeal of fitting into our program of discovering how to represent the various
map operations, solving linear systems by using the matrix inverse has at least
two advantages.
First, once the work of finding an inverse has been done, solving a system
with the same coefficients but different constants is easy and fast: if we change
the entries on the right of the system (∗) then we get a related problem
µ
1 1
2 −1
¶ µ
x
y
¶
=
µ
5
1
¶
wtih a related solution method.
µ
x
y
¶
=
µ
1/3 1/3
2/3 −1/3
¶ µ
5
1
¶
=
µ
2
3
¶
In applications, solving many systems having the same matrix of coefficients is
common.
Another advantage of inverses is that we can explore a system’s sensitivity
to changes in the constants. For example, tweaking the 3 on the right of the
system (∗) to
µ
1 1
2 −1
¶ µ
x1
x2
¶
=
µ
3.01
2
¶
can be solved with the inverse.
µ
1/3 1/3
2/3 −1/3
¶ µ
3.01
2
¶
=
µ
(1/3)(3.01) + (1/3)(2)
(2/3)(3.01) − (1/3)(2)
¶
to show that x1 changes by 1/3 of the tweak while x2 moves by 2/3 of that
tweak. This sort of analysis is used, for example, to decide how accurately data
must be specified in a linear model to ensure that the solution has a desired
accuracy.
We finish by describing the computational procedure usually used to find
the inverse matrix.
4.8 Lemma A matrix is invertible if and only if it can be written as the prod-
uct of elementary reduction matrices. The inverse can be computed by applying
to the identity matrix the same row steps, in the same order, as are used to
Gauss-Jordan reduce the invertible matrix.
Proof. A matrix H is invertible if and only if it is nonsingular and thus Gauss-
Jordan reduces to the identity. By Corollary 3.22 this reduction can be done
with elementary matrices Rr · Rr−1 . . . R1 · H = I. This equation gives the two
halves of the result.

First, elementary matrices are invertible and their inverses are also elemen-
tary. Applying R−1
r to the left of both sides of that equation, then R−1
r−1, etc.,
gives H as the product of elementary matrices H = R−1
1 · · · R−1
r · I (the I is
here to cover the trivial r = 0 case).
Second, matrix inverses are unique and so comparison of the above equation
with H−1
H = I shows that H−1
= Rr · Rr−1 . . . R1 · I. Therefore, applying R1
to the identity, followed by R2, etc., yields the inverse of H. QED
4.9 Example To find the inverse of
µ
1 1
2 −1
¶
we do Gauss-Jordan reduction, meanwhile performing the same operations on
the identity. For clerical convenience we write the matrix and the identity side-
by-side, and do the reduction steps together.
µ
1 1 1 0
2 −1 0 1
¶
−2ρ1+ρ2
−→
µ
1 1 1 0
0 −3 −2 1
¶
−1/3ρ2
−→
µ
1 1 1 0
0 1 2/3 −1/3
¶
−ρ2+ρ1
−→
µ
1 0 1/3 1/3
0 1 2/3 −1/3
¶
This calculation has found the inverse.
µ
1 1
2 −1
¶−1
=
µ
1/3 1/3
2/3 −1/3
¶
4.10 Example This one happens to start with a row swap.


0 3 −1 1 0 0
1 0 1 0 1 0
1 −1 0 0 0 1

 ρ1↔ρ2
−→


1 0 1 0 1 0
0 3 −1 1 0 0
1 −1 0 0 0 1


−ρ1+ρ3
−→


1 0 1 0 1 0
0 3 −1 1 0 0
0 −1 −1 0 −1 1


.
.
.
−→


1 0 0 1/4 1/4 3/4
0 1 0 1/4 1/4 −1/4
0 0 1 −1/4 3/4 −3/4


4.11 Example A non-invertible matrix is detected by the fact that the left
half won’t reduce to the identity.
µ
1 1 1 0
2 2 0 1
¶
−2ρ1+ρ2
−→
µ
1 1 1 0
0 0 −2 1
¶

This procedure will find the inverse of a general n×n matrix. The 2×2 case
is handy.
4.12 Corollary The inverse for a 2×2 matrix exists and equals
µ
a b
c d
¶−1
=
1
ad − bc
µ
d −b
−c a
¶
if and only if ad − bc 6= 0.
Proof. This computation is Exercise 22. QED
We have seen here, as in the Mechanics of Matrix Multiplication subsection,
that we can exploit the correspondence between linear maps and matrices. So
we can fruitfully study both maps and matrices, translating back and forth to
whichever helps us the most.
Over the entire four subsections of this section we have developed an algebra
system for matrices. We can compare it with the familiar algebra system for
the real numbers. Here we are working not with numbers but with matrices.
We have matrix addition and subtraction operations, and they work in much
the same way as the real number operations, except that they only combine
same-sized matrices. We also have a matrix multiplication operation and an
operation inverse to multiplication. These are somewhat like the familiar real
number operations (associativity, and distributivity over addition, for example),
but there are differences (failure of commutativity, for example). And, we have
scalar multiplication, which is in some ways another extension of real number
multiplication. This matrix system provides an example that algebra systems
other than the elementary one can be interesting and useful.
Exercises
4.13 Supply the intermediate steps in Example 4.10.
X 4.14 Use Corollary 4.12 to decide if each matrix has an inverse.
(a)
µ
2 1
−1 1
¶
(b)
µ
0 4
1 −3
¶
(c)
µ
2 −3
−4 6
¶
X 4.15 For each invertible matrix in the prior problem, use Corollary 4.12 to find its
inverse.
X 4.16 Find the inverse, if it exists, by using the Gauss-Jordan method. Check the
answers for the 2×2 matrices with Corollary 4.12.
(a)
µ
3 1
0 2
¶
(b)
µ
2 1/2
3 1
¶
(c)
µ
2 −4
−1 2
¶
(d)
Ã
1 1 3
0 2 4
−1 1 0
!
(e)
Ã
0 1 5
0 −2 4
2 3 −2
!
(f)
Ã
2 2 3
1 −2 −3
4 −2 −3
!
X 4.17 What matrix has this one for its inverse?
µ
1 3
2 5
¶

4.18 How does the inverse operation interact with scalar multiplication and addi-
tion of matrices?
(a) What is the inverse of rH?
(b) Is (H + G)−1
= H−1
+ G−1
?
X 4.19 Is (Tk
)−1
= (T−1
)k
?
4.20 Is H−1
invertible?
4.21 For each real number θ let tθ : R2
→ R2
be represented with respect to the
standard bases by this matrix.
µ
cos θ − sin θ
sin θ cos θ
¶
Show that tθ1+θ2 = tθ1 · tθ2 . Show also that tθ
−1
= t−θ.
4.22 Do the calculations for the proof of Corollary 4.12.
4.23 Show that this matrix
H =
µ
1 0 1
0 1 0
¶
has infinitely many right inverses. Show also that it has no left inverse.
4.24 In Example 4.1, how many left inverses has η?
4.25 If a matrix has infinitely many right-inverses, can it have infinitely many
left-inverses? Must it have?
X 4.26 Assume that H is invertible and that HG is the zero matrix. Show that G is
a zero matrix.
4.27 Prove that if H is invertible then the inverse commutes with a matrix GH−1
=
H−1
G if and only if H itself commutes with that matrix GH = HG.
X 4.28 Show that if T is square and if T4
is the zero matrix then (I − T)−1
=
I + T + T2
+ T3
. Generalize.
X 4.29 Let D be diagonal. Describe D2
, D3
, . . . , etc. Describe D−1
, D−2
, . . . , etc.
Define D0
appropriately.
4.30 Prove that any matrix row-equivalent to an invertible matrix is also invert-
ible.
4.31 The first question below appeared as Exercise 28.
(a) Show that the rank of the product of two matrices is less than or equal to
the minimum of the rank of each.
(b) Show that if T and S are square then TS = I if and only if ST = I.
4.32 Show that the inverse of a permutation matrix is its transpose.
4.33 The first two parts of this question appeared as Exercise 25.
(a) Show that (GH)trans
= Htrans
Gtrans
.
(b) A square matrix is symmetric if each i, j entry equals the j, i entry (that is, if
the matrix equals its transpose). Show that the matrices HHtrans
and Htrans
H
are symmetric.
(c) Show that the inverse of the transpose is the transpose of the inverse.
(d) Show that the inverse of a symmetric matrix is symmetric.
X 4.34 The items starting this question appeared as Exercise 30.
(a) Prove that the composition of the projections πx, πy : R3
→ R3
is the zero
map despite that neither is the zero map.
(b) Prove that the composition of the derivatives d2
/dx2
, d3
/dx3
: P4 → P4 is
the zero map despite that neither map is the zero map.
(c) Give matrix equations representing each of the prior two items.

When two things multiply to give zero despite that neither is zero, each is said to
be a zero divisor. Prove that no zero divisor is invertible.
4.35 In real number algebra, there are exactly two numbers, 1 and −1, that are
their own multiplicative inverse. Does H2
= I have exactly two solutions for 2×2
matrices?
4.36 Is the relation ‘is a two-sided inverse of’ transitive? Reflexive? Symmetric?
4.37 [Am. Math. Mon., Nov. 1951] Prove: if the sum of the elements of a square
matrix is k, then the sum of the elements in each row of the inverse matrix is 1/k.

3.V Change of Basis
Representations, whether of vectors or of maps, vary with the bases. For in-
stance, with respect to the two bases E2 and
B = h
µ
1
1
¶
,
µ
1
−1
¶
i
for R2
, the vector ~
e1 has two different representations.
RepE2
(~
e1) =
µ
1
0
¶
RepB(~
e1) =
µ
1/2
1/2
¶
Similarly, with respect to E2, E2 and E2, B, the identity map has two different
representations.
RepE2,E2
(id) =
µ
1 0
0 1
¶
RepE2,B(id) =
µ
1/2 1/2
1/2 −1/2
¶
With our point of view that the objects of our studies are vectors and maps, in
fixing bases we are adopting a scheme of tags or names for these objects, that
are convienent for computation. We will now see how to translate among these
names—we will see exactly how representations vary as the bases vary.
3.V.1 Changing Representations of Vectors
In converting RepB(~
v) to RepD(~
v) the underlying vector ~
v doesn’t change.
Thus, this translation is accomplished by the identity map on the space, de-
scribed so that the domain space vectors are represented with respect to B and
the codomain space vectors are represented with respect to D.
Vw.r.t. B
id


y
Vw.r.t. D
(The diagram is vertical to fit with the ones in the next subsection.)
1.1 Definition The change of basis matrix for bases B, D ⊂ V is the repre-
sentation of the identity map id: V → V with respect to those bases.
RepB,D(id) =




.
.
.
.
.
.
RepD(~
β1) · · · RepD(~
βn)
.
.
.
.
.
.





Section V. Change of Basis 239
1.2 Lemma Left-multiplication by the change of basis matrix for B, D con-
verts a representation with respect to B to one with respect to D. Conversly, if
left-multiplication by a matrix changes bases M · RepB(~
v) = RepD(~
v) then M
is a change of basis matrix.
Proof. For the first sentence, for each ~
v, as matrix-vector multiplication repre-
sents a map application, RepB,D(id) · RepB(~
v) = RepD( id(~
v) ) = RepD(~
v). For
the second sentence, with respect to B, D the matrix M represents some linear
map, whose action is ~
v 7→ ~
v, and is therefore the identity map. QED
1.3 Example With these bases for R2
,
B = h
µ
2
1
¶
,
µ
1
0
¶
i D = h
µ
−1
1
¶
,
µ
1
1
¶
i
because
RepD( id(
µ
2
1
¶
)) =
µ
−1/2
3/2
¶
D
RepD( id(
µ
1
0
¶
)) =
µ
−1/2
1/2
¶
D
the change of basis matrix is this.
RepB,D(id) =
µ
−1/2 −1/2
3/2 1/2
¶
We can see this matrix at work by finding the two representations of ~
e2
RepB(
µ
0
1
¶
) =
µ
1
−2
¶
RepD(
µ
0
1
¶
) =
µ
1/2
1/2
¶
and checking that the conversion goes as expected.
µ
−1/2 −1/2
3/2 1/2
¶ µ
1
−2
¶
=
µ
1/2
1/2
¶
We finish this subsection by recognizing that the change of basis matrices
are familiar.
1.4 Lemma A matrix changes bases if and only if it is nonsingular.
Proof. For one direction, if left-multiplication by a matrix changes bases then
the matrix represents an invertible function, simply because the function is
inverted by changing the bases back. Such a matrix is itself invertible, and so
nonsingular.
To finish, we will show that any nonsingular matrix M performs a change of
basis operation from any given starting basis B to some ending basis. Because
the matrix is nonsingular, it will Gauss-Jordan reduce to the identity, so there
are elementatry reduction matrices such that Rr · · · R1 · M = I. Elementary
matrices are invertible and their inverses are also elementary, so multiplying
from the left first by Rr
−1
, then by Rr−1
−1
, etc., gives M as a product of

elementary matrices M = R1
−1
· · · Rr
−1
. Thus, we will be done if we show
that elementary matrices change a given basis to another basis, for then Rr
−1
changes B to some other basis Br, and Rr−1
−1
changes Br to some Br−1,
. . . , and the net effect is that M changes B to B1. We will prove this about
elementary matrices by covering the three types as separate cases.
Applying a row-multiplication matrix
Mi(k)








c1
.
.
.
ci
.
.
.
cn








=








c1
.
.
.
kci
.
.
.
cn








changes a representation with respect to h~
β1, . . . , ~
βi, . . . , ~
βni to one with respect
to h~
β1, . . . , (1/k)~
βi, . . . , ~
βni in this way.
~
v = c1 · ~
β1 + · · · + ci · ~
βi + · · · + cn · ~
βn
7→ c1 · ~
β1 + · · · + kci · (1/k)~
βi + · · · + cn · ~
βn = ~
v
Similarly, left-multiplication by a row-swap matrix Pi,j changes a representation
with respect to the basis h~
β1, . . . , ~
βi, . . . , ~
βj, . . . , ~
βni into one with respect to the
basis h~
β1, . . . , ~
βj, . . . , ~
βi, . . . , ~
βni in this way.
~
v = c1 · ~
β1 + · · · + ci · ~
βi + · · · + cj
~
βj + · · · + cn · ~
βn
7→ c1 · ~
β1 + · · · + cj · ~
βj + · · · + ci · ~
βi + · · · + cn · ~
βn = ~
v
And, a representation with respect to h~
β1, . . . , ~
βi, . . . , ~
βj, . . . , ~
βni changes via
left-multiplication by a row-combination matrix Ci,j(k) into a representation
with respect to h~
β1, . . . , ~
βi − k~
βj, . . . , ~
βj, . . . , ~
βni
~
v = c1 · ~
β1 + · · · + ci · ~
βi + cj
~
βj + · · · + cn · ~
βn
7→ c1 · ~
β1 + · · · + ci · (~
βi − k~
βj) + · · · + (kci + cj) · ~
βj + · · · + cn · ~
βn = ~
v
(the definition of reduction matrices specifies that i 6= k and k 6= 0 and so this
last one is a basis). QED
1.5 Corollary A matrix is nonsingular if and only if it represents the identity
map with respect to some pair of bases.
In the next subsection we will see how to translate among representations
of maps, that is, how to change RepB,D(h) to RepB̂,D̂(h). The above corollary
is a special case of this, where the domain and range are the same space, and
where the map is the identity map.

Exercises
X 1.6 In R2
, where
D = h
µ
2
1
¶
,
µ
−2
4
¶
i
find the change of basis matrices from D to E2 and from E2 to D. Multiply the
two.
X 1.7 Find the change of basis matrix for B, D ⊆ R2
.
(a) B = E2, D = h~
e2,~
e1i (b) B = E2, D = h
µ
1
2
¶
,
µ
1
4
¶
i
(c) B = h
µ
1
2
¶
,
µ
1
4
¶
i, D = E2 (d) B = h
µ
−1
1
¶
,
µ
2
2
¶
i, D = h
µ
0
4
¶
,
µ
1
3
¶
i
1.8 For the bases in Exercise 7, find the change of basis matrix in the other direction,
from D to B.
X 1.9 Find the change of basis matrix for each B, D ⊆ P2.
(a) B = h1, x, x2
i, D = hx2
, 1, xi (b) B = h1, x, x2
i, D = h1, 1+x, 1+x+x2
i
(c) B = h2, 2x, x2
i, D = h1 + x2
, 1 − x2
, x + x2
i
X 1.10 Decide if each changes bases on R2
. To what basis is E2 changed?
(a)
µ
5 0
0 4
¶
(b)
µ
2 1
3 1
¶
(c)
µ
−1 4
2 −8
¶
(d)
µ
1 −1
1 1
¶
1.11 Find bases such that this matrix represents the identity map with respect to
those bases. Ã
3 1 4
2 −1 1
0 0 4
!
1.12 Conside the vector space of real-valued functions with basis hsin(x), cos(x)i.
Show that h2 sin(x)+cos(x), 3 cos(x)i is also a basis for this space. Find the change
of basis matrix in each direction.
1.13 Where does this matrix µ
cos(2θ) sin(2θ)
sin(2θ) − cos(2θ)
¶
send the standard basis for R2
? Any other bases? Hint. Consider the inverse.
X 1.14 What is the change of basis matrix with respect to B, B?
1.15 Prove that a matrix changes bases if and only if it is invertible.
1.16 Finish the proof of Lemma 1.4.
X 1.17 Let H be a n×n nonsingular matrix. What basis of Rn
does H change to the
standard basis?
X 1.18 (a) In P3 with basis B = h1 + x, 1 − x, x2
+ x3
, x2
− x3
i we have this repre-
senatation.
RepB(1 − x + 3x2
− x3
) =



0
1
1
2



B
Find a basis D giving this different representation for the same polynomial.
RepD(1 − x + 3x2
− x3
) =



1
0
2
0



D

(b) State and prove that any nonzero vector representation can be changed to
any other.
Hint. The proof of Lemma 1.4 is constructive—it not only says the bases change,
it shows how they change.
1.19 Let V, W be vector spaces, and let B, B̂ be bases for V and D, D̂ be bases for
W. Where h: V → W is linear, find a formula relating RepB,D(h) to RepB̂,D̂(h).
X 1.20 Show that the columns of an n×n change of basis matrix form a basis for
Rn
. Do all bases appear in that way: can the vectors from any Rn
basis make the
columns of a change of basis matrix?
X 1.21 Find a matrix having this effect.
µ
1
3
¶
7→
µ
4
−1
¶
That is, find a M that left-multiplies the starting vector to yield the ending vector.
Is there a matrix having these two effects?
(a)
µ
1
3
¶
7→
µ
1
1
¶ µ
2
−1
¶
7→
µ
−1
−1
¶
(b)
µ
1
3
¶
7→
µ
1
1
¶ µ
2
6
¶
7→
µ
−1
−1
¶
Give a necessary and sufficient condition for there to be a matrix such that ~
v1 7→ ~
w1
and ~
v2 7→ ~
w2.
3.V.2 Changing Map Representations
The first subsection shows how to convert the representation of a vector with
respect to one basis to the representation of that same vector with respect to
another basis. Here we will see how to convert the representation of a map with
respect to one pair of bases to the representation of that map with respect to
a different pair. That is, we want the relationship between the matrices in this
arrow diagram.
Vw.r.t. B
h
−
−
−
−
→
H
Ww.r.t. D
id


y id


y
Vw.r.t. B̂
h
−
−
−
−
→
Ĥ
Ww.r.t. D̂
To move from the lower-left of this diagram to the lower-right we can either go
straight over, or else up to VB then over to WD and then down. Restated in
terms of the matrices, we can calculate Ĥ = RepB̂,D̂(h) either by simply using
B̂ and D̂, or else by first changing bases with RepB̂,B(id) then multiplying
by H = RepB,D(h) and then changing bases with RepD,D̂(id). This equation
summarizes.
Ĥ = RepD,D̂(id) · H · RepB̂,B(id) (∗)

(To compare this equation with the sentence before it, remember that the equa-
tion is read from right to left because function composition is read right to left
and matrix multiplication represent the composition.)
2.1 Example The matrix
T =
µ
cos(π/6) − sin(π/6)
sin(π/6) cos(π/6)
¶
=
µ√
3/2 −1/2
1/2
√
3/2
¶
represents, with respect to E2, E2, the transformation t: R2
→ R2
that rotates
vectors π/6 radians counterclockwise.
µ
1
3
¶
t
7−→
µ
(−3 +
√
3)/2
(1 + 3
√
3)/2
¶
We can translate that representation with respect to E2, E2 to one with respect
to
B̂ = h
µ
1
1
¶ µ
0
2
¶
i D̂ = h
µ
−1
0
¶ µ
2
3
¶
i
by using the arrow diagram and formula (∗).
R2
w.r.t. E2
t
−
−
−
−
→
T
R2
w.r.t. E2
id


y id


y
R2
w.r.t. B̂
t
−
−
−
−
→
T̂
R2
w.r.t. D̂
T̂ = RepE2,D̂(id) · T · RepB̂,E2
(id)
Note that RepE2,D̂(id) can be calculated as the matrix inverse of RepD̂,E2
(id).
RepB̂,D̂(t) =
µ
−1 2
0 3
¶−1 µ√
3/2 −1/2
1/2
√
3/2
¶ µ
1 0
1 2
¶
=
µ
(5 −
√
3)/6 (3 + 2
√
3)/3
(1 +
√
3)/6
√
3/3
¶
Although the new matrix is messier-appearing, the map that it represents is the
same. For instance, to replicate the effect of t in the picture, start with B̂,
RepB̂(
µ
1
3
¶
) =
µ
1
1
¶
B̂
apply T̂,
µ
(5 −
√
3)/6 (3 + 2
√
3)/3
(1 +
√
3)/6
√
3/3
¶
B̂,D̂
µ
1
1
¶
B̂
=
µ
(11 + 3
√
3)/6
(1 + 3
√
3)/6
¶
D̂

and check it against D̂
11 + 3
√
3
6
·
µ
−1
0
¶
+
1 + 3
√
3
6
·
µ
2
3
¶
=
µ
(−3 +
√
3)/2
(1 + 3
√
3)/2
¶
to see that it is the same result as above.
2.2 Example On R3
the map


x
y
z

 t
7−→


y + z
x + z
x + y


that is represented with respect to the standard basis in this way
RepE3,E3
(t) =


0 1 1
1 0 1
1 1 0


can also be represented with respect to another basis
if B = h


1
−1
0

 ,


1
1
−2

 ,


1
1
1

i then RepB,B(t) =


−1 0 0
0 −1 0
0 0 2


in a way that is simpler, in that the action of a diagonal matrix is easy to
understand.
Naturally, we usually prefer basis changes that make the representation eas-
ier to understand. When the representation with respect to equal starting and
ending bases is a diagonal matrix we say the map or matrix has been diagonal-
ized. In Chaper Five we shall see which maps and matrices are diagonalizable,
and where one is not, we shall see how to get a representation that is nearly
diagonal.
We finish this subsection by considering the easier case where representa-
tions are with respect to possibly different starting and ending bases. Recall
that the prior subsection shows that a matrix changes bases if and only if it
is nonsingular. That gives us another version of the above arrow diagram and
equation (∗).
2.3 Definition Same-sized matrices H and Ĥ are matrix equivalent if there
are nonsingular matrices P and Q such that Ĥ = PHQ.
2.4 Corollary Matrix equivalent matrices represent the same map, with re-
spect to appropriate pairs of bases.
Exercise 19 checks that matrix equivalence is an equivalence relation. Thus
it partitions the set of matrices into matrix equivalence classes.

All matrices: %
$
Ã
!
¿
À
. . .
.H
Ĥ .
H matrix equivalent
to Ĥ
We can get some insight into the classes by comparing matrix equivalence with
row equivalence (recall that matrices are row equivalent when they can be re-
duced to each other by row operations). In Ĥ = PHQ, the matrices P and
Q are nonsingular and thus each can be written as a product of elementary
reduction matrices (Lemma 4.8). Left-multiplication by the reduction matrices
making up P has the effect of performing row operations. Right-multiplication
by the reduction matrices making up Q performs column operations. Therefore,
matrix equivalence is a generalization of row equivalence—two matrices are row
equivalent if one can be converted to the other by a sequence of row reduction
steps, while two matrices are matrix equivalent if one can be converted to the
other by a sequence of row reduction steps followed by a sequence of column
reduction steps.
Thus, if matrices are row equivalent then they are also matrix equivalent
(since we can take Q to be the identity matrix and so perform no column
operations). The converse, however, does not hold.
2.5 Example These two
µ
1 0
0 0
¶ µ
1 1
0 0
¶
are matrix equivalent because the second can be reduced to the first by the
column operation of taking −1 times the first column and adding to the second.
They are not row equivalent because they have different reduced echelon forms
(in fact, both are already in reduced form).
We will close this section by finding a set of representatives for the matrix
equivalence classes.∗
2.6 Theorem Any m×n matrix of rank k is matrix equivalent to the m×n
matrix that is all zeros except that the first k diagonal entries are ones.












1 0 . . . 0 0 . . . 0
0 1 . . . 0 0 . . . 0
.
.
.
0 0 . . . 1 0 . . . 0
0 0 . . . 0 0 . . . 0
.
.
.
0 0 . . . 0 0 . . . 0












∗ More information on class representatives is in the appendix.

Sometimes this is described as a block partial-identity form.
µ
I Z
Z Z
¶
Proof. As discussed above, Gauss-Jordan reduce the given matrix and combine
all the reduction matrices used there to make P. Then use the leading entries to
do column reduction and finish by swapping columns to put the leading ones on
the diagonal. Combine the reduction matrices used for those column operations
into Q. QED
2.7 Example We illustrate the proof by finding the P and Q for this matrix.


1 2 1 −1
0 0 1 −1
2 4 2 −2


First Gauss-Jordan row-reduce.


1 −1 0
0 1 0
0 0 1




1 0 0
0 1 0
−2 0 1




1 2 1 −1
0 0 1 −1
2 4 2 −2

 =


1 2 0 0
0 0 1 −1
0 0 0 0


Then column-reduce, which involves right-multiplication.


1 2 0 0
0 0 1 −1
0 0 0 0






1 −2 0 0
0 1 0 0
0 0 1 0
0 0 0 1








1 0 0 0
0 1 0 0
0 0 1 1
0 0 0 1



 =


1 0 0 0
0 0 1 0
0 0 0 0


Finish by swapping columns.


1 0 0 0
0 0 1 0
0 0 0 0






1 0 0 0
0 0 1 0
0 1 0 0
0 0 0 1



 =


1 0 0 0
0 1 0 0
0 0 0 0


Finally, combine the left-multipliers together as P and the right-multipliers
together as Q to get the PHQ equation.


1 −1 0
0 1 0
−2 0 1




1 2 1 −1
0 0 1 −1
2 4 2 −2






1 0 −2 0
0 0 1 0
0 1 0 1
0 0 0 1



 =


1 0 0 0
0 1 0 0
0 0 0 0


2.8 Corollary Two same-sized matrices are matrix equivalent if and only if
they have the same rank. That is, the matrix equivalence classes are character-
ized by rank.
Proof. Two same-sized matrices with the same rank are equivalent to the same
block partial-identity matrix. QED

2.9 Example Now that we know that the block partial-identity matrices form
canonical representatives of the matrix-equivalence classes, we can see what the
classes look like and how many classes there are. Consider the 2×2 matrices.
There are only three possible ranks: zero, one, or two. Thus the 2×2 matrices
fall into three matrix-equivalence classes.
All 2×2
matrices:
%
?
¡0 0
0 0
¢
¤
¤
¤
¤
¤
¤
.
?
¡1 0
0 0
¢
.
. .
.
?
¡1 0
0 1
¢ the three
equivalence
classes
Each class just consists of all the 2×2 matrices with the same rank.
In this subsection we have seen how to change the representation of a map
with respect to a first pair of bases to one with respect to a second pair. That
led to a definition describing when matrices are equivalent in this way. Finally
we noted that, with the proper choice of (possibly different) starting and ending
bases, any map can be represented in block partial-identity form.
One of the nice things about this representation is that, in some sense, we
can completely understand the map when it is expressed in this way: if the
bases are B = h~
β1, . . . , ~
βni and D = h~
δ1, . . . ,~
δmi then the map sends
c1
~
β1 + · · · + ck
~
βk + ck+1
~
βk+1 + · · · + cn
~
βn 7−→ c1
~
δ1 + · · · + ck
~
δk +~
0 + · · · +~
0
where k is the map’s rank. Thus, we can understand any linear map as a kind
of projection.










c1
.
.
.
ck
ck+1
.
.
.
cn










B
7→










c1
.
.
.
ck
0
.
.
.
0










D
Of course, “understanding” a map expressed in this way requires that we un-
derstand the relationship between B and D. However, despite that difficulty,
this is a good classification of linear maps.
Exercises
X 2.10 Decide if these matrices are matrix equivalent.
(a)
µ
1 3 0
2 3 0
¶
,
µ
2 2 1
0 5 −1
¶
(b)
µ
0 3
1 1
¶
,
µ
4 0
0 5
¶

(c)
µ
1 3
2 6
¶
,
µ
1 3
2 −6
¶
X 2.11 Find the canonical representative of the matrix-equivalence class of each ma-
trix.
(a)
µ
2 1 0
4 2 0
¶
(b)
Ã
0 1 0 2
1 1 0 4
3 3 3 −1
!
2.12 Suppose that, with respect to
B = E2 D = h
µ
1
1
¶
,
µ
1
−1
¶
i
the transformation t: R2
→ R2
is represented by this matrix.
µ
1 2
3 4
¶
Use change of basis matrices to represent t with respect to each pair.
(a) B̂ = h
µ
0
1
¶
,
µ
1
1
¶
i, D̂ = h
µ
−1
0
¶
,
µ
2
1
¶
i
(b) B̂ = h
µ
1
2
¶
,
µ
1
0
¶
i, D̂ = h
µ
1
2
¶
,
µ
2
1
¶
i
X 2.13 What size are P and Q?
X 2.14 Use Theorem 2.6 to show that a square matrix is nonsingular if and only if it
is equivalent to an identity matrix.
X 2.15 Show that, where A is a nonsingular square matrix, if P and Q are nonsingular
square matrices such that PAQ = I then QP = A−1
.
X 2.16 Why does Theorem 2.6 not show that every matrix is diagonalizable (see
Example 2.2)?
2.17 Must matrix equivalent matrices have matrix equivalent transposes?
2.18 What happens in Theorem 2.6 if k = 0?
X 2.19 Show that matrix-equivalence is an equivalence relation.
X 2.20 Show that a zero matrix is alone in its matrix equivalence class. Are there
other matrices like that?
2.21 What are the matrix equivalence classes of matrices of transformations on R1
?
R3
?
2.22 How many matrix equivalence classes are there?
2.23 Are matrix equivalence classes closed under scalar multiplication? Addition?
2.24 Let t: Rn
→ Rn
represented by T with respect to En, En.
(a) Find RepB,B(t) in this specific case.
T =
µ
1 1
3 −1
¶
B = h
µ
1
2
¶
,
µ
−1
−1
¶
i
(b) Describe RepB,B(t) in the general case where B = h~
β1, . . . , ~
βni.
2.25 (a) Let V have bases B1 and B2 and suppose that W has the basis D. Where
h: V → W, find the formula that computes RepB2,D(h) from RepB1,D(h).
(b) Repeat the prior question with one basis for V and two bases for W.
2.26 (a) If two matrices are matrix-equivalent and invertible, must their inverses
be matrix-equivalent?
(b) If two matrices have matrix-equivalent inverses, must the two be matrix-
equivalent?

(c) If two matrices are square and matrix-equivalent, must their squares be
matrix-equivalent?
(d) If two matrices are square and have matrix-equivalent squares, must they be
matrix-equivalent?
X 2.27 Square matrices are similar if they represent the same transformation, but
each with respect to the same ending as starting basis. That is, RepB1,B1
(t) is
similar to RepB2,B2
(t).
(a) Give a definition of matrix similarity like that of Definition 2.3.
(b) Prove that similar matrices are matrix equivalent.
(c) Show that similarity is an equivalence relation.
(d) Show that if T is similar to T̂ then T2
is similar to T̂2
, the cubes are similar,
etc. Contrast with the prior exercise.
(e) Prove that there are matrix equivalent matrices that are not similar.

3.VI Projection
The prior section describes the matrix equivalence canonical form as expressing a
projection and so this section takes the natural next step of studying projections.
However, this section is optional; only the last two sections of Chapter Five
require this material. In addition, this section requires some optional material
from the subsection on length and angle measure in n-space.
We have described the projection π from R3
into its xy plane subspace as
a ‘shadow map’. This shows why, but it also shows that some shadows fall
upward.
µ
1
2
2
¶
µ
1
2
0
¶
µ
1
2
−1
¶
µ
1
2
0
¶
So perhaps a better description is: the projection of ~
v is the ~
p in the plane with
the property that someone standing on ~
p and looking straight up or down sees
~
v. In this section we will generalize this to other projections, both orthogonal
(i.e., ‘straight up and down’) and nonorthogonal.
3.VI.1 Orthogonal Projection Into a Line
We first consider orthogonal projection into a line. To orthogonally project
a vector ~
v into a line `, darken a point on the line if someone on that line and
looking straight up or down (from that person’s point of view) sees ~
v.
~
v
`
~
p
The picture shows someone who has walked out on the line until the tip of
~
v is straight overhead. That is, where the line is described as the span of
some nonzero vector ` = {c · ~
s
¯
¯ c ∈ R}, the person has walked out to find the
coefficient c~
p with the property that ~
v − c~
p · ~
s is orthogonal to c~
p · ~
s.

Section VI. Projection 251
~
v
c~
p~
s
~
v − c~
p~
s
We can solve for this coefficient by noting that because ~
v − c~
p~
s is orthogonal to
a scalar multiple of ~
s it must be orthogonal to ~
s itself, and then the consequent
fact that the dot product (~
v − c~
p~
s) ~
s is zero gives that c~
p = ~
v ~
s/~
s ~
s.
1.1 Definition The orthogonal projection of ~
v into the line spanned by a
nonzero ~
s is this vector.
proj[~
s ](~
v) =
~
v ~
s
~
s ~
s
· ~
s
Exercise 19 checks that the outcome of the calculation depends only on the line
and not on which vector ~
s happens to be used to describe that line.
1.2 Remark The wording of that definition says ‘spanned by ~
s ’ instead the
more formal ‘the span of the set {~
s }’. This casual first phrase is common.
1.3 Example In R2
, to orthogonally project into the line y = 2x, we first pick
a direction vector for this line. For instance,
~
s =
µ
1
2
¶
will do. With that, calculation of a projection is routine.
~
v =
³
2
3
´
µ
2
3
¶ µ
1
2
¶
µ
1
2
¶ µ
1
2
¶ ·
µ
1
2
¶
=
8
5
·
µ
1
2
¶
=
µ
8/5
16/5
¶
1.4 Example In R3
, the orthogonal projection of a general vector


x
y
z


into the y-axis is


x
y
z




0
1
0




0
1
0




0
1
0


·


0
1
0

 =


0
y
0


which matches our intuitive expectation.

The picture above with the stick figure walking out on the line until ~
v’s tip
is overhead is one way to think of the orthogonal projection of a vector into a
line. We finish this subsection with two other ways.
1.5 Example A railroad car left on an east-west track without its brake is
pushed by a wind blowing toward the northeast at fifteen miles per hour; what
speed will the car reach?
For the wind we use a vector of length 15 that points toward the northeast.
~
v =
µ
15
p
1/2
15
p
1/2
¶
The car can only be affected by the part of the wind blowing in the east-west
direction—the part of ~
v in the direction of the x-axis is this (the picture has the
same perspective as the railroad car picture above).
east
north
~
p
~
p =
µ
15
p
1/2
0
¶
So the car will reach a velocity of 15
p
1/2 miles per hour toward the east.
Thus, another way to think of the picture that precedes the definition is that
it shows ~
v as decomposed into two parts, the part with the line (here, the part
with the tracks, ~
p), and the part that is orthogonal to the line (shown here lying
on the north-south axis). These two are “not interacting” or “independent”, in
the sense that the east-west car is not at all affected by the north-south part
of the wind (see Exercise 11). So the orthogonal projection of ~
v into the line
spanned by ~
s can be thought of as the part of ~
v that lies in the direction of ~
s.
Finally, another useful way to think of the orthogonal projection is to have
the person stand not on the line, but on the vector that is to be projected to the
line. This person has a rope over the line and pulls it tight, naturally making
the rope orthogonal to the line.

That is, we can think of the projection ~
p as being the vector in the line that is
closest to ~
v (see Exercise 17).
1.6 Example A submarine is tracking a ship moving along the line y = 3x+2.
Torpedo range is one-half mile. Can the sub stay where it is, at the origin on
the chart below, or must it move to reach a place where the ship will pass within
range?
east
north
The formula for projection into a line does not immediately apply because the
line doesn’t pass through the origin, and so isn’t the span of any ~
s. To adjust
for this, we start by shifting the entire map down two units. Now the line is
y = 3x, which is a subspace, and we can project to get the point ~
p of closest
approach, the point on the line through the origin closest to
~
v =
µ
0
−2
¶
the sub’s shifted position.
~
p =
µ
0
−2
¶ µ
1
3
¶
µ
1
3
¶ µ
1
3
¶ ·
µ
1
3
¶
=
µ
−3/5
−9/5
¶
The distance between ~
v and ~
p is approximately 0.63 miles and so the sub must
move to get in range.
This subsection has developed a natural projection map, orthogonal projec-
tion into a line. As suggested by the examples, it is often called for in appli-
cations. The next subsection shows how the definition of orthogonal projection
into a line gives us a way to calculate especially convienent bases for vector
spaces, again something that is common in applications. The final subsection
completely generalizes projection, orthogonal or not, into any subspace at all.
Exercises
X 1.7 Project the first vector orthogonally into the line spanned by the second vec-
tor.
(a)
µ
2
1
¶
,
µ
3
−2
¶
(b)
µ
2
1
¶
,
µ
3
0
¶
(c)
Ã
1
1
4
!
,
Ã
1
2
−1
!
(d)
Ã
1
1
4
!
,
Ã
3
3
12
!
X 1.8 Project the vector orthogonally into the line.

(a)
Ã
2
−1
4
!
, {c
Ã
−3
1
−3
!
¯
¯ c ∈ R} (b)
µ
−1
−1
¶
, the line y = 3x
1.9 Although the development of Definition 1.1 is guided by the pictures, we are
not restricted to spaces that we can draw. In R4
project this vector into this line.
~
v =



1
2
1
3


 ` = {c ·



−1
1
−1
1



¯
¯ c ∈ R}
X 1.10 Definition 1.1 uses two vectors ~
s and ~
v. Consider the transformation of R2
resulting from fixing
~
s =
µ
3
1
¶
and projecting ~
v into the line that is the span of ~
s. Apply it to these vec-
tors.
(a)
µ
1
2
¶
(b)
µ
0
4
¶
Show that in general the projection tranformation is this.
µ
x1
x2
¶
7→
µ
(x1 + 3x2)/10
(3x1 + 9x2)/10
¶
Express the action of this transformation with a matrix.
1.11 Example 1.5 suggests that projection breaks ~
v into two parts, proj[~
s ](~
v ) and
~
v − proj[~
s ](~
v ), that are “not interacting”. Recall that the two are orthogonal.
Show that any two nonzero orthogonal vectors make up a linearly independent
set.
1.12 (a) What is the orthogonal projection of ~
v into a line if ~
v is a member of
that line?
(b) Show that if ~
v is not a member of the line then the set {~
v,~
v − proj[~
s ](~
v )} is
linearly independent.
1.13 Definition 1.1 requires that ~
s be nonzero. Why? What is the right definition
of the orthogonal projection of a vector into the (degenerate) line spanned by the
zero vector?
1.14 Are all vectors the projection of some other vector into some line?
X 1.15 Show that the projection of ~
v into the line spanned by ~
s has length equal to
the absolute value of the number ~
v ~
s divided by the length of the vector ~
s .
1.16 Find the formula for the distance from a point to a line.
1.17 Find the scalar c such that (cs1, cs2) is a minimum distance from the point
(v1, v2) by using calculus (i.e., consider the distance function, set the first derivative
equal to zero, and solve). Generalize to Rn
.
X 1.18 Prove that the orthogonal projection of a vector into a line is shorter than the
vector.
X 1.19 Show that the definition of orthogonal projection into a line does not depend
on the spanning vector: if ~
s is a nonzero multiple of ~
q then (~
v ~
s/~
s ~
s ) · ~
s equals
(~
v ~
q/~
q ~
q ) · ~
q.
X 1.20 Consider the function mapping to plane to itself that takes a vector to its
projection into the line y = x. These two each show that the map is linear, the

first one in a way that is bound to the coordinates (that is, it fixes a basis and
then computes) and the second in a way that is more conceptual.
(a) Produce a matrix that describes the function’s action.
(b) Show also that this map can be obtained by first rotating everything in the
plane π/4 radians clockwise, then projecting into the x-axis, and then rotating
π/4 radians counterclockwise.
1.21 For ~
a,~
b ∈ Rn
let ~
v1 be the projection of ~
a into the line spanned by ~
b, let ~
v2 be
the projection of ~
v1 into the line spanned by ~
a, let ~
v3 be the projection of ~
v2 into
the line spanned by ~
b, etc., back and forth between the spans of ~
a and ~
b. That is,
~
vi+1 is the projection of ~
vi into the span of ~
a if i + 1 is even, and into the span of ~
b
if i + 1 is odd. Must that sequence of vectors eventually settle down—must there
be a sufficiently large i such that ~
vi+2 equals ~
vi and ~
vi+3 equals ~
vi+1? If so, what
is the earliest such i?
3.VI.2 Gram-Schmidt Orthogonalization
This subsection is optional. It requires material from the prior, also optional,
subsection. The work done here will only be needed in the final two sections of
Chapter Five.
The prior subsection suggests that projecting into the line spanned by ~
s
decomposes a vector ~
v into two parts
~
v
proj[~
s ](~
v )
~
v − proj[~
s ](~
v )
~
v = proj[~
s ](~
v) +
¡
~
v − proj[~
s ](~
v)
¢
that are orthogonal and so are “not interacting”. We now make that phrase
precise.
2.1 Definition Vectors ~
v1, . . . ,~
vk ∈ Rn
are mutually orthogonal when any two
are orthogonal: if i 6= j then the dot product ~
vi ~
vj is zero.
2.2 Theorem If the vectors in a set {~
v1, . . . ,~
vk} ⊂ Rn
are mutually orthogonal
and nonzero then that set is linearly independent.
Proof. Consider a linear relationship c1~
v1 + c2~
v2 + · · · + ck~
vk = ~
0. If i ∈ [1..k]
then taking the dot product of ~
vi with both sides of the equation
~
vi (c1~
v1 + c2~
v2 + · · · + ck~
vk) = ~
vi
~
0
ci · (~
vi ~
vi) = 0
shows, since ~
vi is nonzero, that ci is zero. QED

2.3 Corollary If the vectors in a size k subset of an k dimensional space are
mutually orthogonal and nonzero then that set is a basis for the space.
Proof. Any linearly independent size k subset of an k dimensional space is a
basis. QED
Of course, the converse of Corollary 2.3 does not hold—not every basis of
every subspace of Rn
is made of mutually orthogonal vectors. However, we can
get the partial converse that for every subspace of Rn
there is at least one basis
consisting of mutually orthogonal vectors.
2.4 Example The members ~
β1 and ~
β2 of this basis for R2
are not orthogonal.
B = h
µ
4
2
¶
,
µ
1
3
¶
i ~
β1
~
β2
However, we can derive from B a new basis for the same space that does have
mutually orthogonal members. For the first member of the new basis we simply
use ~
β1.
~
κ1 =
µ
4
2
¶
For the second member of the new basis, we take away from ~
β2 its part in the
direction of ~
κ1,
~
κ2 =
µ
1
3
¶
− proj[~
κ1](
µ
1
3
¶
) =
µ
1
3
¶
−
µ
2
1
¶
=
µ
−1
2
¶
~
κ2
which leaves the part, ~
κ2 pictured above, of ~
β2 that is orthogonal to ~
κ1 (it is
orthogonal by the definition of the projection into the span of ~
κ1). Note that,
by the corollary, {~
κ1,~
κ2} is a basis for R2
.
2.5 Definition An orthogonal basis for a vector space is a basis of mutually
orthogonal vectors.
2.6 Example To turn this basis for R3
h


1
1
1

 ,


0
2
1

 ,


1
0
3

i
into an orthogonal basis, we take the first vector as it is given.
~
κ1 =


1
1
1



We get ~
κ2 by starting with the given second vector ~
β2 and subtracting away the
part of it in the direction of ~
κ1.
~
κ2 =


0
2
0

 − proj[~
κ1](


0
2
0

) =


0
2
0

 −


2/3
2/3
2/3

 =


−2/3
4/3
−2/3


Finally, we get ~
κ3 by taking the third given vector and subtracting the part of
it in the direction of ~
κ1, and also the part of it in the direction of ~
κ2.
~
κ3 =


1
0
3

 − proj[~
κ1](


1
0
3

) − proj[~
κ2](


1
0
3

) =


−1
0
1


Again the corollary gives that
h


1
1
1

 ,


−2/3
4/3
−2/3

 ,


−1
0
1

i
is a basis for the space.
The next result verifies that the process used in those examples works with
any basis for any subspace of an Rn
(we are restricted to Rn
only because we
have not given a definition of orthogonality for other vector spaces).
2.7 Theorem (Gram-Schmidt orthogonalization) If h~
β1, . . . ~
βki is a basis
for a subspace of Rn
then, where
~
κ1 = ~
β1
~
κ2 = ~
β2 − proj[~
κ1](~
β2)
~
κ3 = ~
β3 − proj[~
κ1](~
β3) − proj[~
κ2](~
β3)
.
.
.
~
κk = ~
βk − proj[~
κ1](~
βk) − · · · − proj[~
κk−1](~
βk)
the ~
κ ’s form an orthogonal basis for the same subspace.
Proof. We will use induction to check that each ~
κi is nonzero, is in the span of
h~
β1, . . . ~
βii and is orthogonal to all preceding vectors: ~
κ1 ~
κi = · · · = ~
κi−1 ~
κi = 0.
With those, and with Corollary 2.3, we will have that h~
κ1, . . .~
κki is a basis for
the same space as h~
β1, . . . ~
βki.
We shall cover the cases up to i = 3, which give the sense of the argument.
Completing the details is Exercise 23.
The i = 1 case is trivial—setting ~
κ1 equal to ~
β1 makes it a nonzero vector
since ~
β1 is a member of a basis, it is obviously in the desired span, and the
‘orthogonal to all preceding vectors’ condition is vacuously met.

For the i = 2 case, expand the definition of ~
κ2.
~
κ2 = ~
β2 − proj[~
κ1](~
β2) = ~
β2 −
~
β2 ~
κ1
~
κ1 ~
κ1
· ~
κ1 = ~
β2 −
~
β2 ~
κ1
~
κ1 ~
κ1
· ~
β1
This expansion shows that ~
κ2 is nonzero or else this would be a non-trivial linear
dependence among the ~
β’s (it is nontrivial because the coefficient of ~
β2 is 1) and
also shows that ~
κ2 is in the desired span. Finally, ~
κ2 is orthogonal to the only
preceding vector
~
κ1 ~
κ2 = ~
κ1 (~
β2 − proj[~
κ1](~
β2)) = 0
because this projection is orthogonal.
The i = 3 case is the same as the i = 2 case except for one detail. As in the
i = 2 case, expanding the definition
~
κ3 = ~
β3 −
~
β3 ~
κ1
~
κ1 ~
κ1
· ~
κ1 −
~
β3 ~
κ2
~
κ2 ~
κ2
· ~
κ2
= ~
β3 −
~
β3 ~
κ1
~
κ1 ~
κ1
· ~
β1 −
~
β3 ~
κ2
~
κ2 ~
κ2
·
¡
~
β2 −
~
β2 ~
κ1
~
κ1 ~
κ1
· ~
β1
¢
shows that ~
κ3 is nonzero and is in the span. A calculation shows that ~
κ3 is
orthogonal to the preceding vector ~
κ1.
~
κ1 ~
κ3 = ~
κ1
¡
~
β3 − proj[~
κ1](~
β3) − proj[~
κ2](~
β3)
¢
= ~
κ1
¡
~
β3 − proj[~
κ1](~
β3)
¢
− ~
κ1 proj[~
κ2](~
β3)
= 0
(Here’s the difference from the i = 2 case—the second line has two kinds of
terms. The first term is zero because this projection is orthogonal, as in the
i = 2 case. The second term is zero because ~
κ1 is orthogonal to ~
κ2 and so is
orthogonal to any vector in the line spanned by ~
κ2.) The check that ~
κ3 is also
orthogonal to the other preceding vector ~
κ2 is similar. QED
Beyond having the vectors in the basis be orthogonal, we can do more; we
can arrange for each vector to have length one by dividing each by its own length
(we can normalize the lengths).
2.8 Example Normalizing the length of each vector in the orthogonal basis of
Example 2.6 produces this orthonormal basis.
h


1/
√
3
1/
√
3
1/
√
3

 ,


−1/
√
6
2/
√
6
−1/
√
6

 ,


−1/
√
2
0
1/
√
2

i
Besides its intuitive appeal, and its analogy with the standard basis En for Rn
,
an orthonormal basis also simplifies some computations. See Exercise 17, for
example.
Exercises
2.9 Perform the Gram-Schmidt process on each of these bases for R2
.

(a) h
µ
1
1
¶
,
µ
2
1
¶
i (b) h
µ
0
1
¶
,
µ
−1
3
¶
i (c) h
µ
0
1
¶
,
µ
−1
0
¶
i
Then turn those orthogonal bases into orthonormal bases.
X 2.10 Perform the Gram-Schmidt process on each of these bases for R3
.
(a) h
Ã
2
2
2
!
,
Ã
1
0
−1
!
,
Ã
0
3
1
!
i (b) h
Ã
1
−1
0
!
,
Ã
0
1
0
!
,
Ã
2
3
1
!
i
Then turn those orthogonal bases into orthonormal bases.
X 2.11 Find an orthonormal basis for this subspace of R3
: the plane x − y + z = 0.
2.12 Find an orthonormal basis for this subspace of R4
.
{



x
y
z
w



¯
¯ x − y − z + w = 0 and x + z = 0}
2.13 Show that any linearly independent subset of Rn
can be orthogonalized with-
out changing its span.
X 2.14 What happens if we apply the Gram-Schmidt process to a basis that is already
orthogonal?
2.15 Let h~
κ1, . . . ,~
κki be a set of mutually orthogonal vectors in Rn
.
(a) Prove that for any ~
v in the space, the vector ~
v−(proj[~
κ1](~
v )+· · ·+proj[~
vk](~
v ))
is orthogonal to each of ~
κ1, . . . , ~
κk.
(b) Illustrate the prior item in R3
by using ~
e1 as ~
κ1, using ~
e2 as ~
κ2, and taking
~
v to have components 1, 2, and 3.
(c) Show that proj[~
κ1](~
v ) + · · · + proj[~
vk](~
v ) is the vector in the span of the set
of ~
κ’s that is closest to ~
v. Hint. To the illustration done for the prior part,
add a vector d1~
κ1 + d2~
κ2 and apply the Pythagorean Theorem to the resulting
triangle.
2.16 Find a vector in R3
that is orthogonal to both of these.
Ã
1
5
−1
! Ã
2
2
0
!
X 2.17 One advantage of orthogonal bases is that they simplify finding the represen-
tation of a vector with respect to that basis.
(a) For this vector and this non-orthogonal basis for R2
~
v =
µ
2
3
¶
B = h
µ
1
1
¶
,
µ
1
0
¶
i
first represent the vector with respect to the basis. Then project the vector into
the span of each basis vector [~
β1] and [~
β2].
(b) With this orthogonal basis for R2
K = h
µ
1
1
¶
,
µ
1
−1
¶
i
represent the same vector ~
v with respect to the basis. Then project the vector
into the span of each basis vector. Note that the coefficients in the representation
and the projection are the same.
(c) Let K = h~
κ1, . . . ,~
κki be an orthogonal basis for some subspace of Rn
. Prove
that for any ~
v in the subspace, the i-th component of the representation RepK (~
v )
is the scalar coefficient (~
v ~
κi)/(~
κi ~
κi) from proj[~
κi](~
v ).

(d) Prove that ~
v = proj[~
κ1](~
v ) + · · · + proj[~
κk](~
v ).
2.18 Bessel’s Inequality. Consider these orthonormal sets
B1 = {~
e1} B2 = {~
e1,~
e2} B3 = {~
e1,~
e2,~
e3} B4 = {~
e1,~
e2,~
e3,~
e4}
along with the vector ~
v ∈ R4
whose components are 4, 3, 2, and 1.
(a) Find the coefficient c1 for the projection of ~
v into the span of the vector in
B1. Check that k~
v k2
≥ |c1|2
.
(b) Find the coefficients c1 and c2 for the projection of ~
v into the spans of the
two vectors in B2. Check that k~
v k2
≥ |c1|2
+ |c2|2
.
(c) Find c1, c2, and c3 associated with the vectors in B3, and c1, c2, c3, and c4
for the vectors in B4. Check that k~
v k2
≥ |c1|2
+ · · · + |c3|2
and that k~
v k2
≥
|c1|2
+ · · · + |c4|2
.
Show that this holds in general: where {~
κ1, . . . ,~
κk} is an orthonormal set and ci is
coefficient of the projection of a vector ~
v from the space then k~
v k2
≥ |c1|2
+ · · · +
|ck|2
. Hint. One way is to look at the inequality 0 ≤ k~
v − (c1~
κ1 + · · · + ck~
κk)k2
and expand the c’s.
2.19 Prove or disprove: every vector in Rn
is in some orthogonal basis.
2.20 Show that the columns of an n×n matrix form an orthonormal set if and only
if the inverse of the matrix is its transpose. Produce such a matrix.
2.21 Does the proof of Theorem 2.2 fail to consider the possibility that the set of
vectors is empty (i.e., that k = 0)?
2.22 Theorem 2.7 describes a change of basis from any basis B = h~
β1, . . . , ~
βki to
one that is orthogonal K = h~
κ1, . . . ,~
κki. Consider the change of basis matrix
RepB,K (id).
(a) Prove that the matrix RepK,B(id) changing bases in the direction opposite
to that of the theorem has an upper triangular shape—all of its entries below
the main diagonal are zeros.
(b) Prove that the inverse of an upper triangular matrix is also upper triangular
(if the matrix is invertible, that is). This shows that the matrix RepB,K (id)
changing bases in the direction described in the theorem is upper triangular.
2.23 Complete the induction argument in the proof of Theorem 2.7.
3.VI.3 Projection Into a Subspace
This subsection, like the others in this section, is optional. It also requires
material from the optional earlier subsection on Direct Sums.
The prior subsections project a vector into a line by decomposing it into two
parts: the part in the line proj[~
s ](~
v ) and the rest ~
v − proj[~
s ](~
v ). To generalize
projection to arbitrary subspaces, we follow this idea.
3.1 Definition For any direct sum V = M ⊕ N and any ~
v ∈ V , the projection
of ~
v into M along N is
projM,N (~
v ) = ~
m
where ~
v = ~
m + ~
n with ~
m ∈ M, ~
n ∈ N.

This definition doesn’t involve a sense of ‘orthogonal’ so we can apply it to
spaces other than subspaces of an Rn
. (Definitions of orthogonality for other
spaces are perfectly possible, but we haven’t seen any in this book.)
3.2 Example The space M2×2 of 2×2 matrices is the direct sum of these two.
M = {
µ
a b
0 0
¶
¯
¯ a, b ∈ R} N = {
µ
0 0
c d
¶
¯
¯ c, d ∈ R}
To project
A =
µ
3 1
0 4
¶
into M along N, we first fix bases for the two subspaces.
BM = h
µ
1 0
0 0
¶
,
µ
0 1
0 0
¶
i BN = h
µ
0 0
1 0
¶
,
µ
0 0
0 1
¶
i
The concatenation of these
B = BM
_
BN = h
µ
1 0
0 0
¶
,
µ
0 1
0 0
¶
,
µ
0 0
1 0
¶
,
µ
0 0
0 1
¶
i
is a basis for the entire space, because the space is the direct sum, so we can
use it to represent A.
µ
3 1
0 4
¶
= 3 ·
µ
1 0
0 0
¶
+ 1 ·
µ
0 1
0 0
¶
+ 0 ·
µ
0 0
1 0
¶
+ 4 ·
µ
0 0
0 1
¶
Now the projection of A into M along N is found by keeping the M part of this
sum and dropping the N part.
projM,N (
µ
3 1
0 4
¶
) = 3 ·
µ
1 0
0 0
¶
+ 1 ·
µ
0 1
0 0
¶
=
µ
3 1
0 0
¶
3.3 Example Both subscripts on projM,N (~
v ) are significant. The first sub-
script M matters because the result of the projection is an ~
m ∈ M, and changing
this subspace would change the possible results. For an example showing that
the second subscript matters, fix this plane subspace of R3
and its basis
M = {


x
y
z


¯
¯ y − 2z = 0} BM = h


1
0
0

 ,


0
2
1

i
and compare the projections along two different subspaces.
N = {k


0
0
1


¯
¯ k ∈ R} N̂ = {k


0
1
−2


¯
¯ k ∈ R}

(Verification that R3
= M ⊕ N and R3
= M ⊕ N̂ is routine.) We will check
that these projections are different by checking that they have different effects
on this vector.
~
v =


2
2
5


For the first one we find a basis for N
BN = h


0
0
1

i
and represent ~
v with respect to the concatenation BM
_
BN .


2
2
5

 = 2 ·


1
0
0

 + 1 ·


0
2
1

 + 4 ·


0
0
1


The projection of ~
v into M along N is found by keeping the M part and dropping
the N part.
projM,N (~
v ) = 2 ·


1
0
0

 + 1 ·


0
2
1

 =


2
2
1


For the other subspace N̂, this basis is natural.
BN̂ = h


0
1
−2

i
Representing ~
v with respect to the concatenation


2
2
5

 = 2 ·


1
0
0

 + (9/5) ·


0
2
1

 − (8/5) ·


0
1
−2


and then keeping only the M part gives this.
projM,N̂ (~
v ) = 2 ·


1
0
0

 + (9/5) ·


0
2
1

 =


2
18/5
9/5


Therefore projection along different subspaces may yield different results.
These pictures compare the two maps. Both show that the projection is
indeed ‘into’ the plane and ‘along’ the line.

M
N M
N̂
Notice that the projection along N is not orthogonal—there are members of the
plane M that are not orthogonal to the dotted line. But the projection along
N̂ is orthogonal.
A natural question is: what is the relationship between the projection op-
eration defined above, and the operation of orthogonal projection into a line?
The second picture above suggests the answer—orthogonal projection into a
line is a special case of the projection defined above; it is just projection along
a subspace perpendicular to the line.
M
N
In addition to pointing out that projection along a subspace is a generalization,
this scheme shows how to define orthogonal projection into any subspace of Rn
,
of any dimension.
3.4 Definition The orthogonal complement of a subspace M of Rn
is
M⊥
= {~
v ∈ Rn
¯
¯ ~
v is perpendicular to all vectors in M}
(read “M perp”). The orthogonal projection projM (~
v ) of a vector is its projec-
tion into M along M⊥
.
3.5 Example In R3
, to find the orthogonal complement of the plane
P = {


x
y
z


¯
¯ 3x + 2y − z = 0}
we start with a basis for P.
B = h


1
0
3

 ,


0
1
2

i
Any ~
v perpendicular to every vector in B is perpendicular to every vector in the
span of B (the proof of this assertion is Exercise 19). Therefore, the subspace

P⊥
consists of the vectors that satisfy these two conditions.


1
0
3




v1
v2
v3

 = 0


0
1
2




v1
v2
v3

 = 0
We can express those conditions more compactly as a linear system.
P⊥
= {


v1
v2
v3


¯
¯
µ
1 0 3
0 1 2
¶


v1
v2
v3

 =
µ
0
0
¶
}
We are thus left with finding the nullspace of the map represented by the matrix,
that is, with calculating the solution set of a homogeneous linear system.
P⊥
= {


v1
v2
v3


¯
¯ v1 + 3v3 = 0
v2 + 2v3 = 0
} = {k


−3
−2
1


¯
¯ k ∈ R}
3.6 Example Where M is the xy-plane subspace of R3
, what is M⊥
? A
common first reaction is that M⊥
is the yz-plane, but that’s not right. Some
vectors from the yz-plane are not perpendicular to every vector in the xy-plane.
Ã
1
1
0
!
6⊥
Ã
0
3
2
!
cos θ =
1 · 0 + 1 · 3 + 0 · 2
√
1 + 1 + 0 ·
√
0 + 9 + 4
so θ ≈ 0.94 rad
Instead M⊥
is the z-axis, since proceeding as in the prior example and taking
the natural basis for the xy-plane gives this.
M⊥
= {


x
y
z


¯
¯
µ
1 0 0
0 1 0
¶


x
y
z

 =
µ
0
0
¶
} = {


x
y
z


¯
¯ x = 0 and y = 0}
The two examples that we’ve seen since Definition 3.4 illustrate the first
sentence in that definition. The next result justifies the second sentence.
3.7 Lemma Let M be a subspace of Rn
. The orthogonal complement of M is
also a subspace. The space is the direct sum of the two Rn
= M ⊕ M⊥
. And,
for any ~
v ∈ Rn
, the vector ~
v − projM (~
v ) is perpendicular to every vector in M.
Proof. First, the orthogonal complement M⊥
is a subspace of Rn
because, as
noted in the prior two examples, it is a nullspace.
Next, we can start with any basis BM = h~
µ1, . . . , ~
µki for M and expand it to
a basis for the entire space. Apply the Gram-Schmidt process to get an orthog-
onal basis K = h~
κ1, . . . ,~
κni for Rn
. This K is the concatenation of two bases
h~
κ1, . . . ,~
κki (with the same number of members as BM ) and h~
κk+1, . . . ,~
κni.
The first is a basis for M, so if we show that the second is a basis for M⊥
then
we will have that the entire space is the direct sum of the two subspaces.

Exercise 17 from the prior subsection proves this about any orthogonal ba-
sis: each vector ~
v in the space is the sum of its orthogonal projections onto the
lines spanned by the basis vectors.
~
v = proj[~
κ1](~
v ) + · · · + proj[~
κn](~
v ) (∗)
To check this, represent the vector ~
v = r1~
κ1 +· · ·+rn~
κn, apply ~
κi to both sides
~
v ~
κi = (r1~
κ1 + · · · + rn~
κn) ~
κi = r1 · 0 + · · · + ri · (~
κi ~
κi) + · · · + rn · 0, and
solve to get ri = (~
v ~
κi)/(~
κi ~
κi), as desired.
Since obviously any member of the span of h~
κk+1, . . . ,~
κni is orthogonal to
any vector in M, to show that this is a basis for M⊥
we need only show the
other containment—that any ~
w ∈ M⊥
is in the span of this basis. The prior
paragraph does this. On projections into basis vectors from M, any ~
w ∈ M⊥
gives proj[~
κ1](~
w ) = ~
0, . . . , proj[~
κk](~
w ) = ~
0 and therefore (∗) gives that ~
w is a
linear combination of ~
κk+1, . . . ,~
κn. Thus this is a basis for M⊥
and Rn
is the
direct sum of the two.
The final sentence is proved in much the same way. Write ~
v = proj[~
κ1](~
v ) +
· · · + proj[~
κn](~
v ). Then projM (~
v ) is gotten by keeping only the M part and
dropping the M⊥
part projM (~
v ) = proj[~
κk+1](~
v ) + · · · + proj[~
κk](~
v ). Therefore
~
v − projM (~
v ) consists of a linear combination of elements of M⊥
and so is
perpendicular to every vector in M. QED
We can find the orthogonal projection into a subspace by following the steps
of the proof, but the next result gives a convienent formula.
3.8 Theorem Let ~
v be a vector in Rn
and let M be a subspace of Rn
with basis h~
β1, . . . , ~
βki. If A is the matrix whose columns are the ~
β’s then
projM (~
v ) = c1
~
β1 + · · · + ck
~
βk where the coefficients ci are the entries of the
vector (Atrans
A)Atrans
· ~
v. That is, projM (~
v ) = A(Atrans
A)−1
Atrans
· ~
v.
Proof. The vector projM (~
v) is a member of M and so it is a linear combination
of basis vectors c1 · ~
β1 + · · · + ck · ~
βk. Since A’s columns are the ~
β’s, that can
be expressed as: there is a ~
c ∈ Rk
such that projM (~
v ) = A~
c (this is expressed
compactly with matrix multiplication as in Example 3.5 and 3.6). Because
~
v −projM (~
v ) is perpendicular to each member of the basis, we have this (again,
expressed compactly).
~
0 = Atrans
¡
~
v − A~
c
¢
= Atrans
~
v − Atrans
A~
c
Solving for ~
c (showing that Atrans
A is invertible is an exercise)
~
c =
¡
Atrans
A
¢−1
Atrans
· ~
v
gives the formula for the projection matrix as projM (~
v ) = A · ~
c. QED
3.9 Example To orthogonally project this vector into this subspace
~
v =


1
−1
1

 P = {


x
y
z


¯
¯ x + z = 0}

first make a matrix whose columns are a basis for the subspace
A =


0 1
1 0
0 −1


and then compute.
A
¡
Atrans
A
¢−1
Atrans
=


0 1
1 0
0 −1


µ
0 1
1/2 0
¶ µ
1 0 −1
0 1 0
¶
=


1/2 0 −1/2
0 1 0
−1/2 0 1/2


With the matrix, calculating the orthogonal projection of any vector into P is
easy.
projP (~
v) =


1/2 0 −1/2
0 1 0
−1/2 0 1/2




1
−1
1

 =


0
−1
0


Exercises
X 3.10 Project the vectors into M along N.
(a)
µ
3
−2
¶
, M = {
µ
x
y
¶
¯
¯ x + y = 0}, N = {
µ
x
y
¶
¯
¯ −x − 2y = 0}
(b)
µ
1
2
¶
, M = {
µ
x
y
¶
¯
¯ x − y = 0}, N = {
µ
x
y
¶
¯
¯ 2x + y = 0}
(c)
Ã
3
0
1
!
, M = {
Ã
x
y
z
!
¯
¯ x + y = 0}, N = {c ·
Ã
1
0
1
!
¯
¯ c ∈ R}
X 3.11 Find M⊥
.
(a) M = {
µ
x
y
¶
¯
¯ x + y = 0} (b) M = {
µ
x
y
¶
¯
¯ −2x + 3y = 0}
(c) M = {
µ
x
y
¶
¯
¯ x − y = 0} (d) M = {~
0 } (e) M = {
µ
x
y
¶
¯
¯ x = 0}
(f) M = {
Ã
x
y
z
!
¯
¯ −x + 3y + z = 0} (g) M = {
Ã
x
y
z
!
¯
¯ x = 0 and y + z = 0}
3.12 This subsection shows how to project orthogonally in two ways, the method of
Example 3.2 and 3.3, and the method of Theorem 3.8. To compare them, consider
the plane P specified by 3x + 2y − z = 0 in R3
.
(a) Find a basis for P.
(b) Find P⊥
and a basis for P⊥
.
(c) Represent this vector with respect to the concatenation of the two bases from
the prior item.
~
v =
Ã
1
1
2
!

(d) Find the orthogonal projection of ~
v into P by keeping only the P part from
the prior item.
(e) Check that against the result from applying Theorem 3.8.
X 3.13 We have three ways to find the orthogonal projection of a vector into a line,
the Definition 1.1 way from the first subsection of this section, the Example 3.2
and 3.3 way of representing the vector with respect to a basis for the space and
then keeping the M part, and the way of Theorem 3.8. For these cases, do all
three ways.
(a) ~
v =
µ
1
−3
¶
, M = {
µ
x
y
¶
¯
¯ x + y = 0}
(b) ~
v =
Ã
0
1
2
!
, M = {
Ã
x
y
z
!
¯
¯ x + z = 0 and y = 0}
3.14 Check that the operation of Definition 3.1 is well-defined. That is, in Exam-
ple 3.2 and 3.3, doesn’t the answer depend on the choice of bases?
3.15 What is the orthogonal projection into the trivial subspace?
3.16 What is the projection of ~
v into M along N if ~
v ∈ M?
3.17 Show that if M ⊆ Rn
is a subspace with orthonormal basis h~
κ1, . . . ,~
κni then
the orthogonal projection of ~
v into M is this.
(~
v ~
κ1) · ~
κ1 + · · · + (~
v ~
κn) · ~
κn
X 3.18 Prove that the map p: V → V is the projection into M along N if and only
if the map id − p is the projection into N along M. (Recall the definition of the
difference of two maps: (id − p) (~
v) = id(~
v) − p(~
v) = ~
v − p(~
v).)
X 3.19 Show that if a vector is perpendicular to every vector in a set then it is
perpendicular to every vector in the span of that set.
3.20 True or false: the intersection of a subspace and its orthogonal complement is
trivial.
3.21 Show that the dimensions of orthogonal complements add to the dimension
of the entire space.
X 3.22 Suppose that ~
v1,~
v2 ∈ Rn
are such that for all complements M, N ⊆ Rn
, the
projections of ~
v1 and ~
v2 into M along N are equal. Must ~
v1 equal ~
v2? (If so, what
if we relax the condition to: all orthogonal projections of the two are equal?)
X 3.23 Let M, N be subspaces of Rn
. The perp operator acts on subspaces; we can
ask how it interacts with other such operations.
(a) Show that two perps cancel: (M⊥
)⊥
= M.
(b) Prove that M ⊆ N implies that N⊥
⊆ M⊥
.
(c) Show that (M + N)⊥
= M⊥
∩ N⊥
.
X 3.24 The material in this subsection allows us to express a geometric relationship
that we have not yet seen between the rangespace and the nullspace of a linear
map.
(a) Represent f : R3
→ R given by
Ã
v1
v2
v3
!
7→ 1v1 + 2v2 + 3v3

with respect to the standard bases and show that
Ã
1
2
3
!
is a member of the perp of the nullspace. Prove that N (f)⊥
is equal to the
span of this vector.
(b) Generalize that to apply to any f : Rn
→ R.
(c) Represent f : R3
→ R2
Ã
v1
v2
v3
!
7→
µ
1v1 + 2v2 + 3v3
4v1 + 5v2 + 6v3
¶
with respect to the standard bases and show that
Ã
1
2
3
!
,
Ã
4
5
6
!
are both members of the perp of the nullspace. Prove that N (f)⊥
is the span
of these two. (Hint. See the third item of Exercise 23.)
(d) Generalize that to apply to any f : Rn
→ Rm
.
This, and related results, is called the Fundamental Theorem of Linear Algebra in
[Strang 93].
3.25 Define a projection to be a linear transformation t: V → V with the property
that repeating the projection does nothing more than does the projection alone: (t◦
t) (~
v) = t(~
v) for all ~
v ∈ V .
(a) Show that orthogonal projection into a line has that property.
(b) Show that projection along a subspace has that property.
(c) Show that for any such t there is a basis B = h~
β1, . . . , ~
βni for V such that
t(~
βi) =
½
~
βi i = 1, 2, . . . , r
~
0 i = r + 1, r + 2, . . . , n
where r is the rank of t.
(d) Conclude that every projection is a projection along a subspace.
(e) Also conclude that every projection has a representation
RepB,B(t) =
µ
I Z
Z Z
¶
in block partial-identity form.
3.26 A square matrix is symmetric if each i, j entry equals the j, i entry (i.e., if the
matrix equals its transpose). Show that the projection matrix A(Atrans
A)−1
Atrans
is symmetric. Hint. Find properties of transposes by looking in the index under
‘transpose’.

Topic: Line of Best Fit 269
Topic: Line of Best Fit
This Topic requires the formulas from the subsections on Orthogonal Projection
Into a Line, and Projection Into a Subspace.
Scientists are often presented with a system that has no solution and they
must find an answer anyway, that is, they must find a value that is as close as
possible to being an answer. An often-encountered example is in finding a line
that, as closely as possible, passes through experimental data.
For instance, suppose that we have a coin to flip, and want to know: is it
fair? This question means that a coin has some proportion m of heads to flips,
determined by how it is balanced beween the two sides, and we want to know
if m = 1/2. We can get experimental information about it by flipping the coin
many times. This is the result a penny experiment, including some intermediate
numbers.
number of flips 30 60 90
number of heads 16 34 51
Naturally, because of randomness, the exact proportion is not found with this
sample — indeed, there is no solution to this system.
30m = 16
60m = 34
90m = 51
That is, the vector of experimental data is not in the subspace of solutions.


16
34
51

 6∈ {m


30
60
90


¯
¯ m ∈ R}
However, as described above, we expect that there is an m that nearly works.
An orthogonal projection of the data vector into the line subspace gives our best
guess at m.


16
34
51




30
60
90




30
60
90




30
60
90


·


30
60
90

 =
7110
12600
·


30
60
90


The estimate (m = 7110/12600 ≈ 0.56) is higher than 1/2, but not by much, so
probably the penny is fair enough for flipping purposes.
The line with the slope m ≈ 0.56 is called the line of best fit for this data.
flips
heads
30 60 90
30
60
b
c
b
c
b
c

Minimizing the distance between the given vector and the vector used as the
right-hand side minimizes the total of these vertical lengths (these have been
distorted, exaggerated by a factor of ten, to make them more visible).
b
c
b
c
b
c
Because it involves minimizing this total distance, we say that the line has been
obtained through fitting by least-squares.
In the previous example, the line that we use, whose slope is our best guess
of the true ratio of heads to flips, must pass through (0, 0). We can also handle
cases where the line is not required to pass through the origin.
For example, the different denominations of U.S. money have different aver-
age times in circulation (the $2 bill is left off as a special case). How long should
we expect a $25 bill to last?
denomination 1 5 10 20 50 100
average life (years) 1.5 2 3 5 9 20
The plot (see below) looks roughly linear. It isn’t a perfect line, i.e., the linear
system with equations b + 1m = 1.5, . . . , b + 100m = 20 has no solution, but
we can again use orthogonal projection to find a best approximation. Consider
the matrix of coefficients of that linear system and also its vector of constants,
the experimentally-determined values.
A =








1 1
1 5
1 10
1 20
1 50
1 100








~
v =








1.5
2
3
5
9
20








The ending result in the subsection on Projection into a Subspace says that
coefficients b and m so that the linear combination of the columns of A is as
close as possible to the vector ~
v are the entries of (Atrans
A)−1
Atrans
· ~
v. Some
calculation gives an intercept of b = 1.05 and a slope of m = 0.18.
denom
avg life
10 30 50 70 90
5
15
b
c b
c
b
c
b
c
b
c
b
c
Plugging x = 25 into the equation of the line shows that such a bill should last
between five and six years.

We close with an example [Oakley & Baker] that cautions about overusing
least-squares fitting. These are the world record times for the men’s mile race
that were in force on January first of the given years. We want to project when
a 3:40 mile will be run.
year 1870 1880 1890 1900 1910 1920 1930
seconds 268.8 264.5 258.4 255.6 255.6 252.6 250.4
1940 1950 1960 1970 1980 1990
246.4 241.4 234.5 231.1 229.0 226.3
The plot below shows that the data is surprisingly linear. With this input
A =







1 1860
1 1870
.
.
.
.
.
.
1 1980
1 1990







~
v =







280.0
268.8
.
.
.
229.0
226.32







MAPLE gives b = 970.68 and m = −0.37 (rounded to two places).
year
secs
1870 1890 1910 1930 1950 1970 1990
230
250
270
290
b
c
b
c
b
c
b
c b
c b
c b
c b
c
b
c
b
c
b
c
b
c b
c b
c
When will a 220 second mile be run? Solving 220 = 970.68 − 0.37x gives an
estimate of the year 2027.
This example is amusing, but serves as a caution because the linearity of the
data will break down someday — the tool of fitting by orthogonal projection
should be applied judicioulsy.
Exercises
The calculations here are most practically done on a computer. In addition, some
of the problems require more data, available in your library, on the net, or in the
Answers to the Exercises.
1 Use least-squares to judge if the coin in this experiment is fair.
flips 8 16 24 32 40
heads 4 9 13 17 20
2 For the men’s mile record, rather than give each of the many records and its
exact date, we’ve “smoothed” the data somewhat by taking a periodic sample. Do
the longer calculation and compare the conclusions.
3 Find the line of best fit for the men’s 1500 meter run. How does the slope compare
with that for the men’s mile (the distances are close; a mile is about 1609 meters)?
4 Find the line of best fit for the records for women’s mile.
5 Do the lines of best fit for the men’s and women’s miles cross?

6 When the space shuttle Challenger exploded in 1986, one of the criticisms made of
NASA’s decision to launch was in the way the analysis of number of O-ring failures
versus temperature was made (of course, O-ring failure caused the explosion). Four
O-ring failures will cause the rocket to explode. NASA had data from 24 previous
flights.
temp ◦
F 53 75 57 58 63 70 70 66 67 67 67
failures 3 2 1 1 1 1 1 0 0 0 0
68 69 70 70 72 73 75 76 76 78 79 80 81
0 0 0 0 0 0 0 0 0 0 0 0 0
The temperature that day was forecast to be 31◦
F.
(a) NASA based the decision to launch partially on a chart showing only the
flights that had at least one O-ring failure. Find the line that best fits these
seven flights. On the basis of this data, predict the number of O-ring failures
when the temperature is 31, and when the number of failures will exceed four.
(b) Find the line that best fits all 24 flights. On the basis of this extra data,
predict the number of O-ring failures when the temperature is 31, and when the
number of failures will exceed four.
Which do you think is the more accurate method of predicting? (An excellent
discussion appears in [Dalal, et. al.].)
7 This table lists the average distance from the sun to each of the first seven planets,
using earth’s average as a unit.
Mercury Venus Earth Mars Jupiter Saturn Uranus
0.39 0.72 1.00 1.52 5.20 9.54 19.2
(a) Plot the number of the planet (Mercury is 1, etc.) versus the distance. Note
that it does not look like a line, and so finding the line of best fit is not fruitful.
(b) It does, however look like an exponential curve. Therefore, plot the number
of the planet versus the logarithm of the distance. Does this look like a line?
(c) The asteroid belt between Mars and Jupiter is thought to be what is left of a
planet that broke apart. Renumber so that Jupiter is 6, Saturn is 7, and Uranus
is 8, and plot against the log again. Does this look better?
(d) Use least squares on that data to predict the location of Neptune.
(e) Repeat to predict where Pluto is.
(f) Is the formula accurate for Neptune and Pluto?
This method was used to help discover Neptune (although the second item is mis-
leading about the history; actually, the discovery of Neptune in position 9 prompted
people to look for the “missing planet” in position 5). See [Gardner, 1970]
8 William Bennett has proposed an Index of Leading Cultural Indicators for the
US ([Bennett], in 1993). Among the statistics cited are the average daily hours
spent watching TV, and the average combined SAT scores.
1960 1965 1970 1975 1980 1985 1990 1992
TV 5:06 5:29 5:56 6:07 6:36 7:07 6:55 7:04
SAT 975 969 948 910 890 906 900 899
Suppose that a cause and effect relationship is proposed between the time spent
watching TV and the decline in SAT scores (in this article, Mr. Bennett does not
argue that there is a direct connection).
(a) Find the line of best fit relating the independent variable of average daily
TV hours to the dependent variable of SAT scores.

(b) Find the most recent estimate of the average daily TV hours (Bennett’s cites
Neilsen Media Research as the source of these estimates). Estimate the associ-
ated SAT score. How close is your estimate to the actual average? (Warning: a
change has been made recently in the SAT, so you should investigate whether
some adjustment needs to be made to the reported average to make a valid
comparison.)

Topic: Geometry of Linear Maps
The geometric effect of linear maps h: Rn
→ Rm
is appealing both for its sim-
plicity and for its usefulness.
Even just in the case of linear transformations of R1
, the geometry is quite
nice. The pictures below contrast two nonlinear maps with two linear maps.
Each picture shows the domain R1
on the left mapped to the codomain R1
on
the right (the usual cartesian view, with the codomain drawn perpendicular to
the domain, doesn’t make the point as well as this one). The first two show the
nonlinear functions f1(x) = ex
and f2(x) = x2
. Arrows trace out where each
map sends x = 0, x = 1, x = 2, x = −1, and x = −2. Note how these nonlinear
maps distort the domain in transforming it into the range. In the left picture,
for instance, the top three arrows show that f1(1) is much further from f1(2)
than it is from f1(0) — the map is spreading the domain out unevenly so that
in being carried over to the range, an interval from the domain near x = 2 is
spread apart more than is an interval near x = 0.
−5
0
5
−5
0
5
−5
0
5
−5
0
5
Contrast those with the linear maps h1(x) = 2x and h2(x) = −x.
−5
0
5
−5
0
5
−5
0
5
−5
0
5
These maps are nicer, more regular, in that for each map all of the domain is
spread out by the same factor.
Because the only transformations of R1
are multiplications by a scalar, these
pictures are possibly misleading by being too simple. In higher-dimensional
spaces more can happen. For instance, this linear transformation of R2
, which
rotates all vectors counterclockwise, is not a simple scalar multiplication.
³
x
y
´
7→
−→
³
x cos θ − y sin θ
x sin θ + y cos θ
´
θ

Topic: Geometry of Linear Maps 275
And neither is this transformation of R3
, which projects vectors into the xz-
plane.
µ
x
y
z
¶
7→
−→
µ
x
0
z
¶
But even in higher-dimensional spaces, the situation isn’t complicated. Of
course, any linear map h: Rn
→ Rm
can be represented with respect to, say,
the standard bases by a matrix H. Recall that any matrix H can be factored as
H = PBQ where P and Q are nonsingular and B is a partial-identity matrix.
And, recall that nonsingular matrices factor into elementary matrices, matrices
that are obtained from the identity matrix with one Gaussian step
I
kρi
−→ Mi(k) I
ρi↔ρj
−→ Pi,j I
kρi+ρj
−→ Ci,j(k)
(i 6= j, k 6= 0). Thus we have the factorization H = TnTn−1 . . . TjBTj−1 . . . T1
where the T’s are elementary. Geometrically, a partial-identity matrix acts as a
projection, as here. (That is, the map that this matrix represents with respect
to the standard bases is a projection. We say that this is the map induced by
the matrix.)


x
y
z


µ
1 0 0
0 1 0
0 0 0
¶
E3, E3
−→


x
y
0


Therefore, we will have a description of the geometric action of h if we just
describe the geometric actions of the three kinds of elementary matrices. The
pictures below sticks to the elementary transformations of R2
only, for ease of
drawing.
The action of a matrix of the form Mi(k) (that is, the action of the trans-
formation of R2
that is induced by this matrix) is to stretch vectors by a factor
of k along the i-th axis. This is a dilation. This map stretches by a factor of 3
along the x-axis.
³
x
y
´
7→
−→
³
3x
y
´
Note that if 0 ≤ k < 1 or if k < 0 then the i-th component goes the other way;
here, toward the left.
³
x
y
´
7→
−→
³
−2x
y
´
The action of a matrix of the form Pi,j (that is, of the transformation induced
by this matrix) is to interchange the i-th and j-th axes; in two dimensions there
is only the single case P1,2, which reflects vectors about the line y = x.

³
x
y
´
7→
−→
³
y
x
´
(In higher dimensions, permutations involving many axes can be decomposed
into a combination of swaps of pairs of axes—see Exercise 5.)
The remaining case is the action of matrices of the form Ci,j(k). Recall that,
for instance, C1,2(2) does this.
µ
x
y
¶
³
1 0
2 1
´
E2, E2
−→
µ
x
2x + y
¶
The picture
~
u
~
v
³
x
y
´
7→
−→
³
x
2x + y
´
h(~
u)
h(~
v)
shows that any Ci,j(k) affects vectors depending on their i-th component; in
this example, the vector ~
v with the larger first component is affected more—it
is pushed further vertically, since h(~
v) is 4 higher than ~
v while h(~
u) is only 2
higher than ~
u. Another way to see the action of this map is to see where it
sends the unit square.
~
u ~
v
~
w
³
x
y
´
7→
−→
³
x
2x + y
´
h(~
u)
h(~
v)
h( ~
w)
In this picture, vectors with a first component of 0, like ~
u, are not pushed
vertically at all but vectors with a positive first component are slid up. In
general, for any Ci,j(k), the sliding happens in such a way that vectors with the
same i-th component are slid by the same amount. Here, ~
v and ~
w are each slid
up by 2. The resulting shape, a rhombus, has the same base and height as the
square (and thus the same area) but the right angles are gone. Because of this
action, this kind of map is called a skew.
Recall that under a linear map, the image of a subspace is a subspace. Thus
a linear transformation maps lines through the origin to lines through the origin
(the dimension of the image space cannot be greater than the dimension of the
domain space, so a line can’t map onto, say, a plane). Note, however, that all
four sides of the above rhombus are straight, not just the two sides lying in lines
through the origin. A skew — in fact a linear map of any kind — maps any
line to a line. Exercise 6 asks for a proof of this. That is, linear transformations
respect the linear structures of a space. This is the reason for the assertion
made above that, even on higher-dimensional spaces, linear maps are “nice” or
“regular”.

To finish, we will consider a familiar application, in calculus. On the left
below is a picture, like the ones that started this Topic, of the action of the non-
linear function y(x) = x2
+ x. As described at that start, overall the geometric
effect of this map is irregular in that at different domain points it has different
effects (e.g., as the domain point x goes from 2 to −2, the associated range point
f(x) at first decreases, then pauses instantaneously, and then increases).
0
5
0
5
0
5
0
5
But in calculus we don’t focus on the map overall, we focus on the local effect
of the map. The picture on the right looks more closely at what this map does
near x = 1. At x = 1 the derivative is y0
(1) = 3, so that near x = 1 we have
that ∆y ≈ 3 · ∆x; in other words, (1.0012
+ 1.001) − (12
+ 1) ≈ 3 · (0.001). That
is, in a neighborhood of x = 1, this map carries the domain to the codomain by
stretching by a factor of 3 — it is, locally, approximately, a dilation. This shows
a small interval in the domain (x − ∆x .. x + ∆x) carried over to an interval in
the codomain (y − ∆y .. y + ∆y) that is three times as wide: ∆y ≈ 3 · ∆x.
x = 1
y = 2
(When the above picture is drawn in the traditional cartesian way then the prior
sentence is usually rephrased as: the derivative y0
(1) = 3 gives the slope of the
line tangent to the graph at the point (1, 2).)
Calculus considers the map that locally approximates the change ∆x 7→
3 · ∆x, instead of the actual change map ∆x 7→ y(1 + ∆x) − y(1), because the
local map is easy to work with. Specifically, if the input change is doubled, or
tripled, etc., then the resulting output change will double, or triple, etc.
3(r ∆x) = r (3∆x)
(for r ∈ R) and adding changes in input adds the resulting output changes.
3(∆x1 + ∆x2) = 3∆x1 + 3∆x2
In short, what’s easy to work with about ∆x 7→ 3 · ∆x is that it is linear.

This point of view makes clear an often-misunderstood, but very important,
result about derivatives: the derivative of the composition of two functions is
computed by using the Chain Rule for combining their derivatives. Recall that
(with suitable conditions on the two functions)
d (g ◦ f)
dx
(x) =
dg
dx
(f(x)) ·
df
dx
(x)
so that, for instance, the derivative of sin(x2
+3x) is cos(x2
+3x)·(2x+3). How
does this combination arise? From this picture of the action of the composition.
x
f(x)
g(f(x))
The first map f dilates the neighborhood of x by a factor of
df
dx
(x)
and the second map g dilates some more, this time dilating a neighborhood of
f(x) by a factor of
dg
dx
( f(x) )
and as a result, the composition dilates by the product of these two.
Extending from the calculus of one-variable functions to more variables starts
with taking the natural next step: for a function y: Rn
→ Rm
and a point
~
x ∈ Rn
, the derivative is defined to be the linear map h: Rn
→ Rm
best approx-
imating how y changes near y(~
x). Then, for instance, the geometric description
given earlier of transformations of R2
characterizes how these derivatives of
functions y: R2
→ R2
can act. (Another example of how the extension steps
are natural is that when there is a composition, the Chain Rule just involves
multiplying the matrices expressing those derivatives.)
Exercises
1 Let h: R2
→ R2
be the transformation that rotates vectors clockwise by π/4 ra-
dians.
(a) Find the matrix H representing h with respect to the standard bases. Use
Gauss’ method to reduce H to the identity.
(b) Translate the row reduction to to a matrix equation TjTj−1 · · · T1H = I (the
prior item shows both that H is similar to I, and that no column operations are
needed to derive I from H).
(c) Solve this matrix equation for H.

(d) Sketch the geometric effect matrix, that is, sketch how H is expressed as a
combination of dilations, flips, skews, and projections (the identity is a trivial
projection).
2 What combination of dilations, flips, skews, and projections produces a rotation
counterclockwise by 2π/3 radians?
3 What combination of dilations, flips, skews, and projections produces the map
h: R3
→ R3
represented with respect to the standard bases by this matrix?
Ã
1 2 1
3 6 0
1 2 2
!
4 Show that any linear transformation of R1
is the map that multiplies by a scalar
x 7→ kx.
5 Show that for any permutation (that is, reordering) p of the numbers 1, . . . , n,
the map




x1
x2
.
.
.
xn



 7→




xp(1)
xp(2)
.
.
.
xp(n)




can be accomplished with a composition of maps, each of which only swaps a single
pair of coordinates. Hint: it can be done by induction on n. (Remark: in the fourth
chapter we will show this and we will also show that the parity of the number of
swaps used is determined by p. That is, although a particular permutation could
be accomplished in two different ways with two different numbers of swaps, either
both ways use an even number of swaps, or both use an odd number.)
6 Show that linear maps preserve the linear structures of a space.
(a) Show that for any linear map from Rn
to Rm
, the image of any line is a line.
The image may be a degenerate line, that is, a single point.
(b) Show that the image of any linear surface is a linear surface. This generalizes
the result that under a linear map the image of a subspace is a subspace.
(c) Linear maps preserve other linear ideas. Show that linear maps preserve
“betweeness”: if the point B is between A and C then the image of B is between
the image of A and the image of C.
7 Use a picture like the one that appears in the discussion of the Chain Rule to
answer: if a function f : R → R has an inverse, what’s the relationship between how
the function —locally, approximately — dilates space, and how its inverse dilates
space (assuming, of course, that it has an inverse)?

Topic: Markov Chains
Here is a simple game. A player bets on coin tosses, a dollar each time, and the
game ends either when the player has no money left or is up to five dollars. If
the player starts with three dollars, what is the chance the game takes at least
five flips? Twenty five flips?
At any point in the game, this player has either $0, or $1, . . . , or $5. We
say that the player is the state s0, s1, . . . , or s5. A game consists of moves,
with, for instance, a player in state s3 having on the next flip a .5 chance of
moving to state s2 and a .5 chance of moving to s4. Once in either state s0 or
state s5, the player never leaves that state. Writing pi,n for the probability that
the player is in state si after n flips, this equation sumarizes.








1 .5 0 0 0 0
0 0 .5 0 0 0
0 .5 0 .5 0 0
0 0 .5 0 .5 0
0 0 0 .5 0 0
0 0 0 0 .5 1
















p0,n
p1,n
p2,n
p3,n
p4,n
p5,n








=








p0,n+1
p1,n+1
p2,n+1
p3,n+1
p4,n+1
p5,n+1








For instance, the probability of being in state s0 after flip n + 1 is p0,n+1 =
p0,n + 0.5 · p1,n. With the initial condition that the player starts with three
dollars, calculation gives this.
n = 0 n = 1 n = 2 n = 3 n = 4 · · · n = 24








0
0
0
1
0
0
















0
0
.5
0
.5
0
















0
.25
0
.5
0
.25
















.125
0
.375
0
.25
.25
















.125
.1875
0
.3125
0
.375








· · ·








.39600
.00276
0
.00447
0
.59676








For instance, after the fourth flip there is a probability of 0.50 that the game
is already over — the player either has no money left or has won five dollars.
As this computational exploration suggests, the game is not likely to go on for
long, with the player quickly ending in either state s0 or state s5. (Because a
player who enters either of these two states never leaves, they are said to be
absorbtive. An argument that involves taking the limit as n goes to infinity will
show that when the player starts with $3, there is a probability of 0.60 that the
player eventually ends with $5 and consequently a probability of 0.40 that the
player ends the game with $0. That argument is beyond the scope of this Topic,
however; here we will just look at a few computations for applications.)
This game is an example of a Markov chain, named for work by A.A. Markov
at the start of this century. The vectors of p’s are probability vectors. The
matrix is a transition matrix. A Markov chain is historyless in that, with a
fixed transition matrix, the next state depends only on the current state and
not on any states that came before. Thus a player, say, who starts in state s3,

Topic: Markov Chains 281
then goes to state s2, then to s1, and then to s2 has exactly the same chance
at this point of moving next to state s3 as does a player whose history was to
start in s3, then go to s4, then to s3, and then to s2.
Here is a Markov chain from sociology. A study ([Macdonald & Ridge],
p. 202) divided occupations in the United Kingdom into upper level (executives
and professionals), middle level (supervisors and skilled manual workers), and
lower level (unskilled). To determine the mobility across these levels in a gen-
eration, about two thousand men were asked, “At which level are you, and at
which level was your father when you were fourteen years old?” This equation
summarizes the results.


.60 .29 .16
.26 .37 .27
.14 .34 .57




pU,n
pM,n
pL,n

 =


pU,n+1
pM,n+1
pL,n+1


For instance, a child of a lower class worker has a .27 probability of growing up to
be middle class. Notice that the Markov model assumption about history seems
reasonable—we expect that while a parent’s occupation has a direct influence
on the occupation of the child, the grandparent’s occupation has no such direct
influence. With the initial distribution of the respondents’s fathers given below,
this table lists the distributions for the next five generations.
n = 0 n = 1 n = 2 n = 3 n = 4 n = 5


.12
.32
.56




.23
.34
.42




.29
.34
.37




.31
.34
.35




.32
.33
.34




.33
.33
.34


One more example, from a very important subject, indeed. The World Series
of American baseball is played between the team winning the American League
and the team winning the National League (we follow [Brunner] but see also
[Woodside]). The series is won by the first team to win four games. That means
that a series is in one of twenty-four states: 0-0 (no games won yet by either
team), 1-0 (one game won for the American League team and no games for the
National League team), etc. If we assume that there is a probability p that the
American League team wins each game then we have the following transition
matrix.











0 0 0 0 . . .
p 0 0 0 . . .
1 − p 0 0 0 . . .
0 p 0 0 . . .
0 1 − p p 0 . . .
0 0 1 − p 0 . . .
.
.
.
.
.
.
.
.
.
.
.
.






















p0-0,n
p1-0,n
p0-1,n
p2-0,n
p1-1,n
p0-2,n
.
.
.











=











p0-0,n+1
p1-0,n+1
p0-1,n+1
p2-0,n+1
p1-1,n+1
p0-2,n+1
.
.
.











An especially interesting special case is p = 0.50; this table lists the resulting
components of the n = 0 through n = 7 vectors. (The code to generate this
table in the computer algebra system Octave follows the exercises.)

n = 0 n = 1 n = 2 n = 3 n = 4 n = 5 n = 6 n = 7
0 − 0
1 − 0
0 − 1
2 − 0
1 − 1
0 − 2
3 − 0
2 − 1
1 − 2
0 − 3
4 − 0
3 − 1
2 − 2
1 − 3
0 − 4
4 − 1
3 − 2
2 − 3
1 − 4
4 − 2
3 − 3
2 − 4
4 − 3
3 − 4
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.5
0.5
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.25
0.5
0.25
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.125
0.375
0.375
0.125
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.0625
0.25
0.375
0.25
0.0625
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.0625
0
0
0
0.0625
0.125
0.3125
0.3125
0.125
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0.0625
0
0
0
0.0625
0.125
0
0
0.125
0.15625
0.3125
0.15625
0
0
0
0
0
0
0
0
0
0
0
0
0.0625
0
0
0
0.0625
0.125
0
0
0.125
0.15625
0
0.15625
0.15625
0.15625
Note that evenly-matched teams are likely to have a long series—there is a
probability of 0.625 that the series goes at least six games.
One reason for the inclusion of this Topic is that Markov chains are one
of the most widely-used applications of matrix operations. Another reason is
that it provides an example of the use of matrices where we do not consider
the significance of any of the maps represented by the matrices. For more on
Markov chains, there are many sources such as [Kemeny & Snell] and [Iosifescu].
Exercises
Most of these problems need enough computation that a computer should be used.
1 These questions refer to the coin-flipping game.
(a) Check the computations in the table at the end of the first paragraph.
(b) Consider the second row of the vector table. Note that this row has alter-
nating 0’s. Must p1,j be 0 when j is odd? Prove that it must be, or produce a
counterexample.
(c) Perform a computational experiment to estimate the chance that the player
ends at five dollars, starting with one dollar, two dollars, and four dollars.
2 ([Feller], p. 424) We consider throws of a die, and say the system is in state si if
the largest number yet appearing on the die was i.
(a) Give the transition matrix.
(b) Start the system in state s1, and run it for five throws. What is the vector
at the end?
3 There has been much interest in whether industries in the United States are
moving from the Northeast and North Central regions to the South and West,

motivated by the warmer climate, by lower wages, and by less unionization. Here is
the transition matrix for large firms in Electric and Electronic Equipment ([Kelton],
p. 43)
NE NC S W Z
NE
NC
S
W
Z
0.787
0
0
0
0.021
0
0.966
0.063
0
0.009
0
0.034
0.937
0.074
0.005
0.111
0
0
0.612
0.010
0.102
0
0
0.314
0.954
For example, a firm in the Northeast region will be in the West region next year
with probability 0.111. (The Z entry is a “birth-death” state. For instance, with
probability 0.102 a large Electric and Electronic Equipment firm from the North-
east will move out of this system next year: go out of business, move abroad, or
move to another category of firm. There is a 0.021 probability that a firm in the
National Census of Manufacturers will move into Electronics, or be created, or
move in from abroad, into the Northeast. Finally, with probability 0.954 a firm
out of the categories will stay out, according to this research.)
(a) Does the Markov model assumption of lack of history seem justified?
(b) Assume that the initial distribution is even, except that the value at Z is
0.9. Compute the vectors for n = 1 through n = 4.
(c) Suppose that the initial distribution is this.
NE NC S W Z
0.0000 0.6522 0.3478 0.0000 0.0000
Calculate the distributions for n = 1 through n = 4.
(d) Find the distribution for n = 50 and n = 51. Has the system settled down
to an equilibrium?
4 This model has been suggested for some kinds of learning ([Wickens], p. 41). The
learner starts in an undecided state sU . Eventually the learner has to decide to do
either response A (that is, end in state sA) or response B (ending in sB). However,
the learner doesn’t jump right from being undecided to being sure A is the correct
thing to do (or B). Instead, the learner spends some time in a “tentative-A”
state, or a “tentative-B” state, trying the response out (denoted here tA and tB).
Imagine that once the learner has decided, it is final, so once sA or sB is entered
it is never left. For the other state changes, imagine a transition is made with
probability p, in either direction.
(a) Construct the transition matrix.
(b) Take p = 0.25 and take the initial vector to be 1 at sU . Run this for five
steps. What is the chance of ending up at sA?
(c) Do the same for p = 0.20.
(d) Graph p versus the chance of ending at sA. Is there a threshold value for p,
above which the learner is almost sure not to take longer than five steps?
5 A certain town is in a certain country (this is a hypothetical problem). Each year
ten percent of the town dwellers move to other parts of the country. Each year
one percent of the people from elsewhere move to the town. Assume that there
are two states sT , living in town, and sC , living elsewhere.
(a) Construct the transistion matrix.
(b) Starting with an initial distribution sT = 0.3 and sC = 0.7, get the results
for the first ten years.
(c) Do the same for sT = 0.2.

(d) Are the two outcomes alike or different?
6 For the World Series application, use a computer to generate the seven vectors
for p = 0.55 and p = 0.6.
(a) What is the chance of the National League team winning it all, even though
they have only a probability of 0.45 or 0.40 of winning any one game?
(b) Graph the probability p against the chance that the American League team
wins it all. Is there a threshold value—a p above which the better team is
essentially ensured of winning?
(Some sample code is included below.)
7 A Markov matrix has each entry positive, and each columns sums to 1.
(a) Check that the three transistion matrices shown in this Topic meet these two
conditions. Must any transition matrix do so?
(b) Observe that if A~
v0 = ~
v1 and A~
v1 = ~
v2 then A2
is a transition matrix from
~
v0 to ~
v2. Show that a power of a Markov matrix is also a Markov matrix.
(c) Generalize the prior item by proving that the product of two appropriately-
sized Markov matrices is a Markov matrix.
Computer Code
This is the code for the computer algebra system Octave that was used
to generate the table of World Series outcomes. First, this script is kept
in the file markov.m. (The sharp character # marks the rest of a line as a
comment.)
# Octave script file to compute chance of World Series outcomes.
function w = markov(p,v)
q = 1-p;
A=[0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-0
p,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-0
q,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-1_
0,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-0
0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-1
0,0,q,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-2__
0,0,0,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-0
0,0,0,q,p,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-1
0,0,0,0,q,p, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-2_
0,0,0,0,0,q, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 0-3
0,0,0,0,0,0, p,0,0,0,1,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 4-0
0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 3-1__
0,0,0,0,0,0, 0,q,p,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 2-2
0,0,0,0,0,0, 0,0,q,p,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0; # 1-3
0,0,0,0,0,0, 0,0,0,q,0,0, 0,0,1,0,0,0, 0,0,0,0,0,0; # 0-4_
0,0,0,0,0,0, 0,0,0,0,0,p, 0,0,0,1,0,0, 0,0,0,0,0,0; # 4-1
0,0,0,0,0,0, 0,0,0,0,0,q, p,0,0,0,0,0, 0,0,0,0,0,0; # 3-2
0,0,0,0,0,0, 0,0,0,0,0,0, q,p,0,0,0,0, 0,0,0,0,0,0; # 2-3__
0,0,0,0,0,0, 0,0,0,0,0,0, 0,q,0,0,0,0, 1,0,0,0,0,0; # 1-4
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,p,0, 0,1,0,0,0,0; # 4-2
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,q,p, 0,0,0,0,0,0; # 3-3_
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,q, 0,0,0,1,0,0; # 2-4
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,p,0,1,0; # 4-3
0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,0,0,0,0, 0,0,q,0,0,1]; # 3-4

w = A * v;
endfunction
Then the Octave session was this.
> v0=[1;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0]
> p=.5
> v1=markov(p,v0)
> v2=markov(p,v1)
...
Translating to another computer algebra system should be easy—all have
commands similar to these.

Topic: Orthonormal Matrices
In The Elements, Euclid considers two figures to be the same if they have the
same size and shape. That is, the triangles below are not equal because they are
not the same set of points. But they are congruent—essentially indistinguishable
for Euclid’s purposes—because we can imagine picking up the plane up, sliding
it over and turning it a bit (although not bending it or stretching it), and then
putting it back down, to superimpose the first figure on the second.
P1
P2
P3
Q1
Q2
Q3
(Euclid never explicitly states this principle but he uses it often [Casey].) In
modern terms, “picking the plane up . . . ” means taking a map from the plane
to itself. We, and Euclid, are considering only certain transformations of the
plane, ones that may possibly slide or turn the plane but not bend or stretch
it. Accordingly, we define a function f : R2
→ R2
to be distance-preserving (or
a rigid motion, or isometry) if for all points P1, P2 ∈ R2
, the map satisfies the
condition that the distance from f(P1) to f(P2) equals the distance from P1 to
P2. We define a plane figure to be a set of points in the plane and we say that
two figures are congruent if there is a distance-preserving map from the plane
to itself that carries one figure onto the other.
Many statements from Euclidean geometry follow easily from these defini-
tions. Some are: (i) collinearity is invariant under any distance-preserving map
(that is, if P1, P2, and P3 are collinear then so are f(P1), f(P2), and f(P3)),
(ii) betweeness is invariant under any distance-preserving map (if P2 is between
P1 and P3 then so is f(P2) between f(P1) and f(P3)), (iii) the property of
being a triangle is invariant under any distance-preserving map (if a figure is a
triangle then the image of that figure is also a triangle), (iv) and the property of
being a circle is invariant under any distance-preserving map. In 1872, F. Klein
suggested that Euclidean geometry can be characterized as the study of prop-
erties that are invariant under distance-preserving maps. (This forms part of
Klein’s Erlanger Program, which proposes the organizing principle that each
kind of geometry—Euclidean, projective, etc.—can be described as the study
of the properties that are invariant under some group of transformations. The
word ‘group’ here means more than just ‘collection’, but that lies outside of our
scope.)
We can use linear algebra to characterize the distance-preserving maps of
the plane.
First, there are distance-preserving transformations of the plane that are not
linear. The obvious example is this translation.
µ
x
y
¶
7→
µ
x
y
¶
+
µ
1
0
¶
=
µ
x + 1
y
¶
However, this example turns out to be the only example, in the sense that if f
is distance-preserving and sends ~
0 to ~
v0 then the map ~
v 7→ f(~
v) − ~
v0 is linear.

Topic: Orthonormal Matrices 287
That will follow immediately from this statement: a map t that is distance-
preserving and sends ~
0 to itself is linear. To prove this statement, let
t(~
e1) =
µ
a
b
¶
t(~
e2) =
µ
c
d
¶
for some a, b, c, d ∈ R. Then to show that t is linear, it suffices to show that it
can be represented by a matrix, that is, that t acts in this way for all x, y ∈ R.
~
v =
µ
x
y
¶
t
7−→
µ
ax + cy
bx + dy
¶
(∗)
Recall that if we fix three non-collinear points then any point in the plane can
be described by giving its distance from those three. So any point ~
v in the
domain is determined by its distance from the three fixed points ~
0, ~
e1, and ~
e2.
Similarly, any point t(~
v) in the codomain is determined by its distance from the
three fixed points t(~
0), t(~
e1), and t(~
e2) (these three are not collinear because, as
mentioned above, collinearity is invariant and ~
0, ~
e1, and ~
e2 are not collinear).
In fact, because t is distance-preserving, we can say more: for the point ~
v in the
plane that is determined by being the distance d0 from ~
0, the distance d1 from
~
e1, and the distance d2 from ~
e2, its image t(~
v) must be the unique point in the
codomain that is determined by being d0 from t(~
0), d1 from t(~
e1), and d2 from
t(~
e2). Because of the uniqueness, checking that the action in (∗) works in the
d0, d1, and d2 cases
dist(
µ
x
y
¶
,~
0) = dist(t(
µ
x
y
¶
), t(~
0)) = dist(
µ
ax + cy
bx + dy
¶
,~
0)
(t is assumed to send ~
0 to itself)
dist(
µ
x
y
¶
,~
e1) = dist(t(
µ
x
y
¶
), t(~
e1)) = dist(
µ
ax + cy
bx + dy
¶
,
µ
a
b
¶
)
and
dist(
µ
x
y
¶
,~
e2) = dist(t(
µ
x
y
¶
), t(~
e2)) = dist(
µ
ax + cy
bx + dy
¶
,
µ
c
d
¶
)
suffices to show that (∗) describes t. Those checks are routine.
Thus, any distance-preserving f : R2
→ R2
can be written f(~
v) = t(~
v) + ~
v0
for some constant vector ~
v0 and linear map t that is distance-preserving.
Not every linear map is distance-preserving, for example, ~
v 7→ 2~
v does not
preserve distances. But there is a neat characterization: a linear transformation
t of the plane is distance-preserving if and only if both kt(~
e1)k = kt(~
e2)k = 1 and
t(~
e1) is orthogonal to t(~
e2). The ‘only if’ half of that statement is easy—because
t is distance-preserving it must preserve the lengths of vectors, and because t
is distance-preserving the Pythagorean theorem shows that it must preserve
orthogonality. For the ‘if’ half, it suffices to check that the map preserves

lengths of vectors, because then for all ~
p and ~
q the distance between the two is
preserved kt(~
p − ~
q )k = kt(~
p) − t(~
q )k = k~
p − ~
q k. For that check, let
~
v =
µ
x
y
¶
t(~
e1) =
µ
a
b
¶
t(~
e2) =
µ
c
d
¶
and, with the ‘if’ assumptions that a2
+ b2
= c2
+ d2
= 1 and ac + bd = 0 we
have this.
kt(~
v )k2
= (ax + cy)2
+ (bx + dy)2
= a2
x2
+ 2acxy + c2
y2
+ b2
x2
+ 2bdxy + d2
y2
= x2
(a2
+ b2
) + y2
(c2
+ d2
) + 2xy(ac + bd)
= x2
+ y2
= k~
v k2
One thing that is neat about this characterization is that we can easily
recognize matrices that represent such a map with respect to the standard bases.
Those matrices have that when the columns are written as vectors then they
are of length one and are mutually orthogonal. Such a matrix is called an
orthonormal matrix or orthogonal matrix (the second term is commonly used to
mean not just that the columns are orthogonal, but also that they have length
one).
We can use this insight to delimit the geometric actions possible in distance-
preserving maps. Because kt(~
v )k = k~
v k, any ~
v is mapped by t to lie somewhere
on the circle about the origin that has radius equal to the length of ~
v, and
so in particular ~
e1 and ~
e2 are mapped to vectors on the unit circle. What’s
more, because of the orthogonality restriction, once we fix the unit vector ~
e1 as
mapped to the vector with components a and b then there are only two places
where ~
e2 can go.
¡a
b
¢
¡−b
a
¢
¡a
b
¢
¡ b
−a
¢
Thus, only two types of maps are possible.
RepE2,E2
(t) =
µ
a −b
b a
¶
RepE2,E2
(t) =
µ
a b
b −a
¶
We can geometrically describe these two cases. Let θ be the angle between the
x-axis and the image of ~
e1, measured counterclockwise.
The first matrix above represents, with respect to the standard bases, a
rotation of the plane by θ radians.

¡a
b
¢
¡−b
a
¢
µ
x
y
¶
t
7−→
µ
x cos θ − y sin θ
x sin θ + y cos θ
¶
The second matrix above represents a reflection of the plane through the line
bisecting the angle between ~
e1 and t(~
e1). (This picture shows ~
e1 reflected up
into the first quadrant and ~
e2 reflected down into the fourth quadrant.)
¡a
b
¢
¡ b
−a
¢
µ
x
y
¶
t
7−→
µ
x cos θ + y sin θ
x sin θ − y cos θ
¶
Note that in this second case, the right angle from ~
e1 to ~
e2 has a counterclockwise
sense but the right angle between the images of these two has a clockwise sense,
so the sense gets reversed. Geometers speak of a distance-preserving map as
direct if it preserves sense and as opposite if it reverses sense.
So, we have characterized the Euclidean study of congruence into the consid-
eration of the properties that are invariant under combinations of (i) a rotation
followed by a translation (possibly the trivial translation), or (ii) a reflection
followed by a translation (a reflection followed by a non-trivial translation is a
glide reflection).
Another idea, besides congruence of figures, encountered in elementary ge-
ometry is that figures are similar if they are congruent after a change of scale.
These two triangles are similar since the second is the same shape as the first,
but 3/2-ths the size.
P1
P2
P3
Q1
Q2
Q3
From the above work, we have that figures are similar if there is an orthonormal
matrix T such that the points ~
q on one are derived from the points ~
p by ~
q =
(kT)~
v + ~
p0 for some nonzero real number k and constant vector ~
p0.
Although many of these ideas were first explored by Euclid, mathematics is
timeless and they are very much in use today. One application of rigid motions
is in computer graphics. We can, for example, take this top view of a cube
and animate it by putting together film frames of it rotating.

Frame 1: Frame 2: Frame 3:
−.2 radians −.4 radians −.6 radians
We could also make the cube appear to be coming closer to us by producing
film frames of it gradually enlarging.
Frame 1: Frame 2: Frame 3:
110 percent 120 percent 130 percent
In practice, computer graphics incorporate many interesting techniques from
linear algebra (see Exercise 4).
So the analysis above of distance-preserving maps is useful as well as inter-
esting. For instance, it shows that to include in graphics software all possible
rigid motions of the plane, we need only include a few cases. It is not possible
that we’ve somehow ovelooked some rigid motions.
A beautiful book that explores more in this area is [Weyl]. More on groups,
of transformations and otherwise, can be found in any book on Modern Algebra,
for instance [Birkhoff & MacLane]. More on Klein and the Erlanger Program is
in [Yaglom].
Exercises
1 Decide if each of these is an orthonormal matrix.
(a)
µ
1/
√
2 −1/
√
2
−1/
√
2 −1/
√
2
¶
(b)
µ
1/
√
3 −1/
√
3
−1/
√
3 −1/
√
3
¶
(c)
µ
1/
√
3 −
√
2/
√
3
−
√
2/
√
3 −1/
√
3
¶
2 Write down the formula for each of these distance-preserving maps.
(a) the map that rotates π/6 radians, and then translates by ~
e2
(b) the map that reflects about the line y = 2x
(c) the map that reflects about y = −2x and translates over 1 and up 1
3 (a) The proof that a map that is distance-preserving and sends the zero vec-
tor to itself incidentally shows that such a map is one-to-one and onto (the
point in the domain determined by d0, d1, and d2 corresponds to the point
in the codomain determined by those three numbers). Therefore any distance-
preserving map has an inverse. Show that the inverse is also distance-preserving.
(b) Using the definitions given in this Topic, prove that congruence is an equiv-
alence relation between plane figures.

4 In practice the matrix for the distance-preserving linear transformation and the
translation are often combined into one. Check that these two computations yield
the same first two components.
µ
a c
b d
¶ µ
x
y
¶
+
µ
e
f
¶ Ã
a c e
b d f
0 0 1
! Ã
x
y
1
!
(These are homogeneous coordinates; see the Topic on Projective Geometry).
5 (a) Verify that the properties described in the second paragraph of this Topic
as invariant under distance-preserving maps are indeed so.
(b) Give two more properties that are of interest in Euclidean geometry from
your experience in studying that subject that are also invariant under distance-
preserving maps.
(c) Give a property that is not of interest in Euclidean geometry and is not
invariant under distance-preserving maps.

Chapter 4
Determinants
In the first chapter of this book we considered linear systems and we picked out
the special case of systems with the same number of equations as unknowns,
those of the form T~
x = ~
b where T is a square matrix. We noted a distinction
between two classes of T’s. While such systems may have a unique solution or
no solutions or infinitely many solutions, if a particular T is associated with a
unique solution in any system, such as the homogeneous system ~
b = ~
0, then
T is associated with a unique solution for every ~
b. We call such a matrix of
coefficients ‘nonsingular’. The other kind of T, where every linear system for
which it is the matrix of coefficients has either no solution or infinitely many
solutions, we call ‘singular’.
Through the second and third chapters the value of this distinction has been
a theme. For instance, we now know that nonsingularity of an n×n matrix T
is equivalent to each of these:
• a system T~
x = ~
b has a solution, and that solution is unique;
• Gauss-Jordan reduction of T yields an identity matrix;
• the rows of T form a linearly independent set;
• the columns of T form a basis for Rn
;
• any map that T represents is an isomorphism;
• an inverse matrix T−1
exists.
So when we look at a particular square matrix, the question of whether it
is nonsingular is one of the first things that we ask. This chapter develops
a formula to determine this. (Since we will restrict the discussion to square
matrices, in this chapter we will usually simply say ‘matrix’ in place of ‘square
matrix’.)
More precisely, we will develop infinitely many formulas, one for 1×1 ma-
trices, one for 2×2 matrices, etc. Of course, these formulas are related — that
is, we will develop a family of formulas, a scheme that describes the formula for
each size.
293

294 Chapter 4. Determinants
4.I Definition
For 1×1 matrices, determining nonsingularity is trivial.
¡
a
¢
is nonsingular iff a 6= 0
The 2×2 formula came out in the course of developing the inverse.
µ
a b
c d
¶
is nonsingular iff ad − bc 6= 0
The 3×3 formula can be produced similarly (see Exercise 9).


a b c
d e f
g h i

 is nonsingular iff aei + bfg + cdh − hfa − idb − gec 6= 0
With these cases in mind, we posit a family of formulas, a, ad−bc, etc. For each
n the formula gives rise to a determinant function detn×n : Mn×n → R such that
an n×n matrix T is nonsingular if and only if detn×n(T) 6= 0. (We usually omit
the subscript because if T is n×n then ‘det(T)’ could only mean ‘detn×n(T)’.)
4.I.1 Exploration
This subsection is optional. It briefly describes how an investigator might
come to a good general definition, which is given in the next subsection.
The three cases above don’t show an evident pattern to use for the general
n×n formula. We may spot that the 1×1 term a has one letter, that the 2×2
terms ad and bc have two letters, and that the 3×3 terms aei, etc., have three
letters. We may also observe that in those terms there is a letter from each row
and column of the matrix, e.g., the letters in the cdh term


c
d
h


come one from each row and one from each column. But these observations
perhaps seem more puzzling than enlightening. For instance, we might wonder
why some of the terms are added while others are subtracted.
A good problem solving strategy is to see what properties a solution must
have and then search for something with those properties. So we shall start by
asking what properties we require of the formulas.
At this point, our primary way to decide whether a matrix is singular is
to do Gaussian reduction and then check whether the diagonal of resulting
echelon form matrix has any zeroes (that is, to check whether the product
down the diagonal is zero). So, we may expect that the proof that a formula

Section I. Definition 295
determines singularity will involve applying Gauss’ method to the matrix, to
show that in the end the product down the diagonal is zero if and only if the
determinant formula gives zero. This suggests our initial plan: we will look for
a family of functions with the property of being unaffected by row operations
and with the property that a determinant of an echelon form matrix is the
product of its diagonal entries. Under this plan, a proof that the functions
determine singularity would go, “Where T → · · · → T̂ is the Gaussian reduction,
the determinant of T equals the determinant of T̂ (because the determinant is
unchanged by row operations), which is the product down the diagonal, which
is zero if and only if the matrix is singular”. In the rest of this subsection we
will test this plan on the 2×2 and 3×3 determinants that we know. We will end
up modifying the “unaffected by row operations” part, but not by much.
The first step in checking the plan is to test whether the 2×2 and 3×3
formulas are unaffected by the row operation of pivoting: if
T
kρi+ρj
−→ T̂
then is det(T̂) = det(T)? This check of the 2×2 determinant after the kρ1 + ρ2
operation
det(
µ
a b
ka + c kb + d
¶
) = a(kb + d) − (ka + c)b = ad − bc
shows that it is indeed unchanged, and the other 2×2 pivot kρ2 + ρ1 gives the
same result. The 3×3 pivot kρ3 + ρ2 leaves the determinant unchanged
det(


a b c
kg + d kh + e ki + f
g h i

) = a(kh + e)i + b(ki + f)g + c(kg + d)h
− h(ki + f)a − i(kg + d)b − g(kh + e)c
= aei + bfg + cdh − hfa − idb − gec
as do the other 3×3 pivot operations.
So there seems to be promise in the plan. Of course, perhaps the 4×4
determinant formula is affected by pivoting. We are exploring a possibility here
and we do not yet have all the facts. Nonetheless, so far, so good.
The next step is to compare det(T̂) with det(T) for the operation
T
ρi↔ρj
−→ T̂
of swapping two rows. The 2×2 row swap ρ1 ↔ ρ2
det(
µ
c d
a b
¶
) = cb − ad
does not yield ad − bc. This ρ1 ↔ ρ3 swap inside of a 3×3 matrix
det(


g h i
d e f
a b c

) = gec + hfa + idb − bfg − cdh − aei

also does not give the same determinant as before the swap — again there is a
sign change. Trying a different 3×3 swap ρ1 ↔ ρ2
det(


d e f
a b c
g h i

) = dbi + ecg + fah − hcd − iae − gbf
also gives a change of sign.
Thus, row swaps appear to change the sign of a determinant. This mod-
ifies our plan, but does not wreck it. We intend to decide nonsingularity by
considering only whether the determinant is zero, not by considering its sign.
Therefore, instead of expecting determinants to be entirely unaffected by row
operations, will look for them to change sign on a swap.
To finish, we compare det(T̂) to det(T) for the operation
T
kρi
−→ T̂
of multiplying a row by a scalar k 6= 0. One of the 2×2 cases is
det(
µ
a b
kc kd
¶
) = a(kd) − (kc)b = k · (ad − bc)
and the other case has the same result. Here is one 3×3 case
det(


a b c
d e f
kg kh ki

) = ae(ki) + bf(kg) + cd(kh)
−(kh)fa − (ki)db − (kg)ec
= k · (aei + bfg + cdh − hfa − idb − gec)
and the other two are similar. These lead us to suspect that multiplying a row
by k multiplies the determinant by k. This fits with our modified plan because
we are asking only that the zeroness of the determinant be unchanged and we
are not focusing on the determinant’s sign or magnitude.
In summary, to develop the scheme for the formulas to compute determi-
nants, we look for determinant functions that remain unchanged under the
pivoting operation, that change sign on a row swap, and that rescale on the
rescaling of a row. In the next two subsections we will find that for each n such
a function exists and is unique.
For the next subsection, note that, as above, scalars come out of each row
without affecting other rows. For instance, in this equality
det(


3 3 9
2 1 1
5 10 −5

) = 3 · det(


1 1 3
2 1 1
5 10 −5

)
the 3 isn’t factored out of all three rows, only out of the top row. The determi-
nant acts on each row of independently of the other rows. When we want to use
this property of determinants, we shall write the determinant as a function of
the rows: ‘det(~
ρ1, ~
ρ2, . . . ~
ρn)’, instead of as ‘det(T)’ or ‘det(t1,1, . . . , tn,n)’. The
definition of the determinant that starts the next subsection is written in this
way.

Exercises
X 1.1 Evaluate the determinant of each.
(a)
µ
3 1
−1 1
¶
(b)
Ã
2 0 1
3 1 1
−1 0 1
!
(c)
Ã
4 0 1
0 0 1
1 3 −1
!
1.2 Evaluate the determinant of each.
(a)
µ
2 0
−1 3
¶
(b)
Ã
2 1 1
0 5 −2
1 −3 4
!
(c)
Ã
2 3 4
5 6 7
8 9 1
!
X 1.3 Verify that the determinant of an upper-triangular 3×3 matrix is the product
down the diagonal.
det(
Ã
a b c
0 e f
0 0 i
!
) = aei
Do lower-triangular matrices work the same way?
X 1.4 Use the determinant to decide if each is singular or nonsingular.
(a)
µ
2 1
3 1
¶
(b)
µ
0 1
1 −1
¶
(c)
µ
4 2
2 1
¶
1.5 Singular or nonsingular? Use the determinant to decide.
(a)
Ã
2 1 1
3 2 2
0 1 4
!
(b)
Ã
1 0 1
2 1 1
4 1 3
!
(c)
Ã
2 1 0
3 −2 0
1 0 0
!
X 1.6 Each pair of matrices differ by one row operation. Use this operation to compare
det(A) with det(B).
(a) A =
µ
1 2
2 3
¶
B =
µ
1 2
0 −1
¶
(b) A =
Ã
3 1 0
0 0 1
0 1 2
!
B =
Ã
3 1 0
0 1 2
0 0 1
!
(c) A =
Ã
1 −1 3
2 2 −6
1 0 4
!
B =
Ã
1 −1 3
1 1 −3
1 0 4
!
1.7 Show this.
det(
Ã
1 1 1
a b c
a2
b2
c2
!
) = (b − a)(c − a)(c − b)
X 1.8 Which real numbers x make this matrix singular?
µ
12 − x 4
−8 8 − x
¶
1.9 Do the Gaussian reduction to check the formula for 3×3 matrices stated in the
preamble to this section.
Ã
a b c
d e f
g h i
!
is nonsingular iff aei + bfg + cdh − hfa − idb − gec 6= 0

1.10 Show that the equation of a line in R2
thru (x1, y1) and (x2, y2) is expressed
by this determinant.
det(
Ã
x y 1
x1 y1 1
x2 y2 1
!
) = 0 x1 6= x2
X 1.11 Many people know this mnemonic for the determinant of a 3×3 matrix: first
repeat the first two columns and then sum the products on the forward diagonals
and subtract the products on the backward diagonals. That is, first write
Ã
h1,1 h1,2 h1,3 h1,1 h1,2
h2,1 h2,2 h2,3 h2,1 h2,2
h3,1 h3,2 h3,3 h3,1 h3,2
!
and then calculate this.
h1,1h2,2h3,3 + h1,2h2,3h3,1 + h1,3h2,1h3,2
−h3,1h2,2h1,3 − h3,2h2,3h1,1 − h3,3h2,1h1,2
(a) Check that this agrees with the formula given in the preamble to this section.
(b) Does it extend to other-sized determinants?
1.12 The cross product of the vectors
~
x =
Ã
x1
x2
x3
!
~
y =
Ã
y1
y2
y3
!
is the vector computed as this determinant.
~
x × ~
y = det(
Ã
~
e1 ~
e2 ~
e3
x1 x2 x3
y1 y2 y3
!
)
Note that the first row is composed of vectors, the vectors from the standard basis
for R3
. Show that the cross product of two vectors is perpendicular to each vector.
1.13 Prove that each statement holds for 2×2 matrices.
(a) The determinant of a product is the product of the determinants det(ST) =
det(S) · det(T).
(b) If T is invertible then the determinant of the inverse is the inverse of the
determinant det(T−1
) = ( det(T) )−1
.
Matrices T and T0
are similar if there is a nonsingular matrix P such that T0
=
PTP−1
. (This definition is in Chapter Five.) Show that similar 2×2 matrices have
the same determinant.
X 1.14 Prove that the area of this region in the plane

x1
y1

x2
y2

is equal to the value of this determinant.
det(
µ
x1 x2
y1 y2
¶
)
Compare with this.
det(
µ
x2 x1
y2 y1
¶
)

1.15 Prove that for 2×2 matrices, the determinant of a matrix equals the determi-
nant of its transpose. Does that also hold for 3×3 matrices?
X 1.16 Is the determinant function linear — is det(x·T +y·S) = x·det(T)+y·det(S)?
1.17 Show that if A is 3×3 then det(c · A) = c3
· det(A) for any scalar c.
1.18 Which real numbers θ make
µ
cos θ − sin θ
sin θ cos θ
¶
singular? Explain geometrically.
1.19 [Am. Math. Mon., Apr. 1955] If a third order determinant has elements 1, 2,
. . . , 9, what is the maximum value it may have?
4.I.2 Properties of Determinants
As described above, we want a formula to determine whether an n×n matrix
is nonsingular. We will not begin by stating such a formula. Instead, we will
begin by considering the function that such a formula calculates. We will define
the function by its properties, then prove that the function with these proper-
ties exist and is unique and also describe formulas that compute this function.
(Because we will show that the function exists and is unique, from the start we
will say ‘det(T)’ instead of ‘if there is a determinant function then det(T)’ and
‘the determinant’ instead of ‘any determinant’.)
2.1 Definition A n×n determinant is a function det: Mn×n → R such that
(1) det(~
ρ1, . . . , k · ~
ρi + ~
ρj, . . . , ~
ρn) = det(~
ρ1, . . . , ~
ρj, . . . , ~
ρn) for i 6= j
(2) det(~
ρ1, . . . , ~
ρj, . . . , ~
ρi, . . . , ~
ρn) = − det(~
ρ1, . . . , ~
ρi, . . . , ~
ρj, . . . , ~
ρn) for i 6= j
(3) det(~
ρ1, . . . , k~
ρi, . . . , ~
ρn) = k · det(~
ρ1, . . . , ~
ρi, . . . , ~
ρn) for k 6= 0
(4) det(I) = 1 where I is an identity matrix
(the ~
ρ ’s are the rows of the matrix). We often write |T| for det(T).
2.2 Remark Property (2) is redundant since
T
ρi+ρj
−→
−ρj +ρi
−→
ρi+ρj
−→
−ρi
−→ T̂
swaps rows i and j. It is listed only for convenience.
The first result shows that a function satisfying these conditions gives a
criteria for nonsingularity. (Its last sentence is that, in the context of the first
three conditions, (4) is equivalent to the condition that the determinant of an
echelon form matrix is the product down the diagonal.)

2.3 Lemma A matrix with two identical rows has a determinant of zero. A
matrix with a zero row has a determinant of zero. A matrix is nonsingular if
and only if its determinant is nonzero. The determinant of an echelon form
matrix is the product down its diagonal.
Proof. To verify the first sentence, swap the two equal rows. The sign of the
determinant changes, but the matrix is unchanged and so its determinant is
unchanged. Thus the determinant is zero.
The second sentence is clearly true if the matrix is 1×1. If it has at least
two rows then apply property (1) of the definition with the zero row as row j
and with k = 1.
det(. . . , ~
ρi, . . . ,~
0, . . . ) = det(. . . , ~
ρi, . . . , ~
ρi +~
0, . . . )
The first sentence of this lemma gives that the determinant is zero.
For the third sentence, where T → · · · → T̂ is the Gauss-Jordan reduction,
by the definition the determinant of T is zero if and only if the determinant of
T̂ is zero (although they could differ in sign or magnitude). A nonsingular T
Gauss-Jordan reduces to an identity matrix and so has a nonzero determinant.
A singular T reduces to a T̂ with a zero row; by the second sentence of this
lemma its determinant is zero.
Finally, for the fourth sentence, if an echelon form matrix is singular then it
has a zero on its diagonal, that is, the product down its diagonal is zero. The
third sentence says that if a matrix is singular then its determinant is zero. So
if the echelon form matrix is singular then its determinant equals the product
down its diagonal.
If an echelon form matrix is nonsingular then none of its diagonal entries is
zero so we can use property (3) of the definition to factor them out (again, the
vertical bars | · · · | indicate the determinant operation).
¯
¯
¯
¯
¯
¯
¯
¯
¯
t1,1 t1,2 t1,n
0 t2,2 t2,n
...
0 tn,n
¯
¯
¯
¯
¯
¯
¯
¯
¯
= t1,1 · t2,2 · · · tn,n ·
¯
¯
¯
¯
¯
¯
¯
¯
¯
1 t1,2/t1,1 t1,n/t1,1
0 1 t2,n/t2,2
...
0 1
¯
¯
¯
¯
¯
¯
¯
¯
¯
Next, the Jordan half of Gauss-Jordan elimination, using property (1) of the
definition, leaves the identity matrix.
= t1,1 · t2,2 · · · tn,n ·
¯
¯
¯
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
...
0 1
¯
¯
¯
¯
¯
¯
¯
¯
¯
= t1,1 · t2,2 · · · tn,n · 1
Therefore, if an echelon form matrix is nonsingular then its determinant is the
product down its diagonal. QED

That result gives us a way to compute the value of a determinant function on
a matrix. Do Gaussian reduction, keeping track of any changes of sign caused by
row swaps and any scalars that are factored out, and then finish by multiplying
down the diagonal of the echelon form result. This procedure takes the same
time as Gauss’ method and so is sufficiently fast to be practical on the size
matrices that we see in this book.
2.4 Example Doing 2×2 determinants
¯
¯
¯
¯
2 4
−1 3
¯
¯
¯
¯ =
¯
¯
¯
¯
2 4
0 5
¯
¯
¯
¯ = 10
with Gauss’ method won’t give a big savings because the 2×2 determinant
formula is so easy. However, a 3×3 determinant is usually easier to calculate
with Gauss’ method than with the formula given earlier.
¯
¯
¯
¯
¯
¯
2 2 6
4 4 3
0 −3 5
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
¯
¯
2 2 6
0 0 −9
0 −3 5
¯
¯
¯
¯
¯
¯
= −
¯
¯
¯
¯
¯
¯
2 2 6
0 −3 5
0 0 −9
¯
¯
¯
¯
¯
¯
= −54
2.5 Example Determinants of matrices any bigger than 3×3 are almost always
most quickly done with this Gauss’ method procedure.
¯
¯
¯
¯
¯
¯
¯
¯
1 0 1 3
0 1 1 4
0 0 0 5
0 1 0 1
¯
¯
¯
¯
¯
¯
¯
¯
= −
¯
¯
¯
¯
¯
¯
¯
¯
1 0 1 3
0 1 1 4
0 0 0 5
0 0 −1 −3
¯
¯
¯
¯
¯
¯
¯
¯
= −
¯
¯
¯
¯
¯
¯
¯
¯
1 0 1 3
0 1 1 4
0 0 −1 −3
0 0 0 5
¯
¯
¯
¯
¯
¯
¯
¯
= −(−5) = 5
The prior example illustrates an important point. Although we have not yet
found a 4×4 determinant formula, if one exists then we know what value it gives
to the matrix — if there is a function with properties (1)-(4) then on the above
matrix the function must return 5.
2.6 Lemma For each n, if there is an n×n determinant function then it is
unique.
Proof. For any n×n matrix we can perform Gauss’ method on the matrix,
keeping track of how the sign alternates on row swaps, and then multiply down
the diagonal of the echelon form result. By the definition and the lemma, all n×n
determinant functions must return this value on this matrix. Thus all n×n de-
terminant functions are equal, that is, there is only one input argument/output
value relationship satisfying the four conditions. QED
The ‘if there is an n×n determinant function’ emphasizes that, although we
can use Gauss’ method to compute the only value that a determinant function
could possibly return, we haven’t yet shown that such a determinant function
exists for all n. In the rest of the section we will produce determinant functions.
Exercises
For these, assume that an n×n determinant function exists for all n.
X 2.7 Use Gauss’ method to find each determinant.

(a)
¯
¯
¯
¯
¯
3 1 2
3 1 0
0 1 4
¯
¯
¯
¯
¯
(b)
¯
¯
¯
¯
¯
¯
¯
1 0 0 1
2 1 1 0
−1 0 1 0
1 1 1 0
¯
¯
¯
¯
¯
¯
¯
2.8 Use Gauss’ method to find each.
(a)
¯
¯
¯
¯
2 −1
−1 −1
¯
¯
¯
¯ (b)
¯
¯
¯
¯
¯
1 1 0
3 0 2
5 2 2
¯
¯
¯
¯
¯
2.9 For which values of k does this system have a unique solution?
x + z − w = 2
y − 2z = 3
x + kz = 4
z − w = 2
X 2.10 Express each of these in terms of |H|.
(a)
¯
¯
¯
¯
¯
h3,1 h3,2 h3,3
h2,1 h2,2 h2,3
h1,1 h1,2 h1,3
¯
¯
¯
¯
¯
(b)
¯
¯
¯
¯
¯
−h1,1 −h1,2 −h1,3
−2h2,1 −2h2,2 −2h2,3
−3h3,1 −3h3,2 −3h3,3
¯
¯
¯
¯
¯
(c)
¯
¯
¯
¯
¯
h1,1 + h3,1 h1,2 + h3,2 h1,3 + h3,3
h2,1 h2,2 h2,3
5h3,1 5h3,2 5h3,3
¯
¯
¯
¯
¯
X 2.11 Find the determinant of a diagonal matrix.
2.12 Describe the solution set of a homogeneous linear system if the determinant
of the matrix of coefficients is nonzero.
X 2.13 Show that this determinant is zero.
¯
¯
¯
¯
¯
y + z x + z x + y
x y z
1 1 1
¯
¯
¯
¯
¯
2.14 (a) Find the 1×1, 2×2, and 3×3 matrices with i, j entry given by (−1)i+j
.
(b) Find the determinant of the square matrix with i, j entry (−1)i+j
.
2.15 (a) Find the 1×1, 2×2, and 3×3 matrices with i, j entry given by i + j.
(b) Find the determinant of the square matrix with i, j entry i + j.
X 2.16 Show that determinant functions are not linear by giving a case where |A +
B| 6= |A| + |B|.
2.17 The second condition in the definition, that row swaps change the sign of a
determinant, is somewhat annoying. It means we have to keep track of the number
of swaps, to compute how the sign alternates. Can we get rid of it? Can we replace
it with the condition that row swaps leave the determinant unchanged? (If so then
we would need new 1×1, 2×2, and 3×3 formulas, but that would be a minor
matter.)
2.18 Prove that the determinant of any triangular matrix, upper or lower, is the
product down its diagonal.
2.19 Refer to the definition of elementary matrices in the Mechanics of Matrix
Multiplication subsection.
(a) What is the determinant of each kind of elementary matrix?

(b) Prove that if E is any elementary matrix then |ES| = |E||S| for any appro-
priately sized S.
(c) (This question doesn’t involve determinants.) Prove that if T is singular then
a product TS is also singular.
(d) Show that |TS| = |T||S|.
(e) Show that if T is nonsingular then |T−1
| = |T|−1
.
2.20 Prove that the determinant of a product is the product of the determinants
|TS| = |T| |S| in this way. Fix the n × n matrix S and consider the function
d: Mn
×
n → R given by T 7→ |TS|/|S|.
(a) Check that d satisfies property (1) in the definition of a determinant func-
tion.
(b) Check property (2).
(c) Check property (3).
(d) Check property (4).
(e) Conclude the determinant of a product is the product of the determinants.
2.21 A submatrix of a given matrix A is one that can be obtained by deleting some
of the rows and columns of A. Thus, the first matrix here is a submatrix of the
second.
µ
3 1
2 5
¶ Ã
3 4 1
0 9 −2
2 −1 5
!
Prove that for any square matrix, the rank of the matrix is r if and only if r is the
largest integer such that there is an r×r submatrix with a nonzero determinant.
X 2.22 Prove that a matrix with rational entries has a rational determinant.
2.23 [Am. Math. Mon., Feb. 1953] Find the element of likeness in (a) simplifying a
fraction, (b) powdering the nose, (c) building new steps on the church, (d) keeping
emeritus professors on campus, (e) putting B, C, D in the determinant
¯
¯
¯
¯
¯
¯
¯
1 a a2
a3
a3
1 a a2
B a3
1 a
C D a3
1
¯
¯
¯
¯
¯
¯
¯
.
4.I.3 The Permutation Expansion
The prior subsection defines a function to be a determinant if it satisfies four
conditions and shows that there is at most one n×n determinant function for
each n. What is left is to show that for each n such a function exists.
How could such a function not exist? After all, we have done computations
that start with a square matrix, follow the conditions, and end with a number.
The difficulty is that, as far as we know, the computation might not give a
well-defined result. To illustrate this possibility, suppose that we were to change
the second condition in the definition of determinant to be that the value of a
determinant does not change on a row swap. By Remark 2.2 we know that
this conflicts with the first and third conditions. Here is an instance of the

conflict: here are two Gauss’ method reductions of the same matrix, the first
without any row swap
µ
1 2
3 4
¶
−3ρ1+ρ2
−→
µ
1 2
0 −2
¶
and the second with a swap.
µ
1 2
3 4
¶
ρ1↔ρ2
−→
µ
3 4
1 2
¶
−(1/3)ρ1+ρ2
−→
µ
3 4
0 2/3
¶
Following Definition 2.1 gives that both calculations yield the determinant −2
since in the second one we keep track of the fact that the row swap changes
the sign of the result of multiplying down the diagonal. But if we follow the
supposition and change the second condition then the two calculations yield
different values, −2 and 2. That is, under the supposition the outcome would not
be well-defined — no function exists that satisfies the changed second condition
along with the other three.
Of course, observing that Definition 2.1 does the right thing in this one
instance is not enough; what we will do in the rest of this section is to show
that there is never a conflict. The natural way to try this would be to define
the determinant function with: “The value of the function is the result of doing
Gauss’ method, keeping track of row swaps, and finishing by multiplying down
the diagonal”. (Since Gauss’ method allows for some variation, such as a choice
of which row to use when swapping, we would have to fix an explicit algorithm.)
Then we would be done if we verified that this way of computing the determinant
satisfies the four properties. For instance, if T and T̂ are related by a row swap
then we would need to show that this algorithm returns determinants that are
negatives of each other. However, how to verify this is not evident. So the
development below will not proceed in this way. Instead, in this subsection we
will define a different way to compute the value of a determinant, a formula,
and we will use this way to prove that the conditions are satisfied.
The formula that we shall use is based on an insight gotten from property (2)
of the definition of determinants. This property shows that determinants are
not linear.
3.1 Example For this matrix det(2A) 6= 2 · det(A).
A =
µ
2 1
−1 3
¶
Instead, the scalar comes out of each of the two rows.
¯
¯
¯
¯
4 2
−2 6
¯
¯
¯
¯ = 2 ·
¯
¯
¯
¯
2 1
−2 6
¯
¯
¯
¯ = 4 ·
¯
¯
¯
¯
2 1
−1 3
¯
¯
¯
¯
Since scalars come out a row at a time, we might guess that determinants
are linear a row at a time.

3.2 Definition Let V be a vector space. A map f : V n
→ R is multilinear if
(1) f(~
ρ1, . . . ,~
v + ~
w, . . . , ~
ρn) = f(~
ρ1, . . . ,~
v, . . . , ~
ρn) + f(~
ρ1, . . . , ~
w, . . . , ~
ρn)
(2) f(~
ρ1, . . . , k~
v, . . . , ~
ρn) = k · f(~
ρ1, . . . ,~
v, . . . , ~
ρn)
for ~
v, ~
w ∈ V and k ∈ R.
3.3 Lemma Determinants are multilinear.
Proof. The definition of determinants gives property (2) (Lemma 2.3 following
that definition covers the k = 0 case) so we need only check property (1).
det(~
ρ1, . . . ,~
v + ~
w, . . . , ~
ρn) = det(~
ρ1, . . . ,~
v, . . . , ~
ρn) + det(~
ρ1, . . . , ~
w, . . . , ~
ρn)
If the set {~
ρ1, . . . , ~
ρi−1, ~
ρi+1, . . . , ~
ρn} is linearly dependent then all three matrices
are singular and so all three determinants are zero and the equality is trivial.
Therefore assume that the set is linearly independent. This set of n-wide row
vectors has n − 1 members, so we can make a basis by adding one more vector
h~
ρ1, . . . , ~
ρi−1, ~
β, ~
ρi+1, . . . , ~
ρni. Express ~
v and ~
w with respect to this basis
~
v = v1~
ρ1 + · · · + vi−1~
ρi−1 + vi
~
β + vi+1~
ρi+1 + · · · + vn~
ρn
~
w = w1~
ρ1 + · · · + wi−1~
ρi−1 + wi
~
β + wi+1~
ρi+1 + · · · + wn~
ρn
giving this.
~
v + ~
w = (v1 + w1)~
ρ1 + · · · + (vi + wi)~
β + · · · + (vn + wn)~
ρn
By the definition of determinant, the value of det(~
ρ1, . . . ,~
v + ~
w, . . . , ~
ρn) is un-
changed by the pivot operation of adding −(v1 + w1)~
ρ1 to ~
v + ~
w.
~
v + ~
w − (v1 + w1)~
ρ1 = (v2 + w2)~
ρ2 + · · · + (vi + wi)~
β + · · · + (vn + wn)~
ρn
Then, to the result, we can add −(v2 + w2)~
ρ2, etc. Thus
det(~
ρ1, . . . ,~
v + ~
w, . . . , ~
ρn)
= det(~
ρ1, . . . , (vi + wi) · ~
β, . . . , ~
ρn)
= (vi + wi) · det(~
ρ1, . . . , ~
β, . . . , ~
ρn)
= vi · det(~
ρ1, . . . , ~
β, . . . , ~
ρn) + wi · det(~
ρ1, . . . , ~
β, . . . , ~
ρn)
(using (2) for the second equality). To finish, bring vi and wi back inside in
front of ~
β and use pivoting again, this time to reconstruct the expressions of ~
v
and ~
w in terms of the basis, e.g., start with the pivot operations of adding v1~
ρ1
to vi
~
β and w1~
ρ1 to wi~
ρ1, etc. QED
Multilinearity allows us to expand a determinant into a sum of determinants,
each of which involves a simple matrix.

3.4 Example We can use multilinearity to split this determinant into two,
first breaking up the first row
¯
¯
¯
¯
2 1
4 3
¯
¯
¯
¯ =
¯
¯
¯
¯
2 0
4 3
¯
¯
¯
¯ +
¯
¯
¯
¯
0 1
4 3
¯
¯
¯
¯
and then separating each of those two, breaking along the second rows.
=
¯
¯
¯
¯
2 0
4 0
¯
¯
¯
¯ +
¯
¯
¯
¯
2 0
0 3
¯
¯
¯
¯ +
¯
¯
¯
¯
0 1
4 0
¯
¯
¯
¯ +
¯
¯
¯
¯
0 1
0 3
¯
¯
¯
¯
We are left with four determinants, such that in each row of each matrix there
is a single entry from the original matrix.
3.5 Example In the same way, a 3×3 determinant separates into a sum of
many simpler determinants. We start by splitting along the first row, producing
three determinants (the zero in the 1, 3 position is underlined to set it off visually
from the zeroes that appear in the splitting).
¯
¯
¯
¯
¯
¯
2 1 −1
4 3 0
2 1 5
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
¯
¯
2 0 0
4 3 0
2 1 5
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 1 0
4 3 0
2 1 5
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 0 −1
4 3 0
2 1 5
¯
¯
¯
¯
¯
¯
Each of these three will itself split in three along the second row. Each of
the resulting nine splits in three along the third row, resulting in twenty seven
determinants
=
¯
¯
¯
¯
¯
¯
2 0 0
4 0 0
2 0 0
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
2 0 0
4 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
2 0 0
4 0 0
0 0 5
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
2 0 0
0 3 0
2 0 0
¯
¯
¯
¯
¯
¯
+ · · · +
¯
¯
¯
¯
¯
¯
0 0 −1
0 0 0
0 0 5
¯
¯
¯
¯
¯
¯
such that each row contains a single entry from the starting matrix.
So an n×n determinant expands into a sum of nn
determinants where each
row of each summands contains a single entry from the starting matrix. How-
ever, many of these summand determinants are zero.
3.6 Example In each of these three matrices from the above expansion, two
of the rows have their entry from the starting matrix in the same column, e.g.,
in the first matrix, the 2 and the 4 both come from the first column.
¯
¯
¯
¯
¯
¯
2 0 0
4 0 0
0 1 0
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0 0 −1
0 3 0
0 0 5
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
0 1 0
0 0 0
0 0 5
¯
¯
¯
¯
¯
¯
Any such matrix is singular, because in each, one row is a multiple of the other
(or is a zero row). Thus, any such determinant is zero, by Lemma 2.3.

Therefore, the above expansion of the 3×3 determinant into the sum of the
twenty seven determinants simplifies to the sum of these six.
¯
¯
¯
¯
¯
¯
2 1 −1
4 3 0
2 1 5
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
¯
¯
2 0 0
0 3 0
0 0 5
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
2 0 0
0 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 1 0
4 0 0
0 0 5
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 1 0
0 0 0
2 0 0
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 0 −1
4 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+
¯
¯
¯
¯
¯
¯
0 0 −1
0 3 0
2 0 0
¯
¯
¯
¯
¯
¯
We can bring out the scalars.
= (2)(3)(5)
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ (2)(0)(1)
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯
+ (1)(4)(5)
¯
¯
¯
¯
¯
¯
0 1 0
1 0 0
0 0 1
¯
¯
¯
¯
¯
¯
+ (1)(0)(2)
¯
¯
¯
¯
¯
¯
0 1 0
0 0 1
1 0 0
¯
¯
¯
¯
¯
¯
+ (−1)(4)(1)
¯
¯
¯
¯
¯
¯
0 0 1
1 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+ (−1)(3)(2)
¯
¯
¯
¯
¯
¯
0 0 1
0 1 0
1 0 0
¯
¯
¯
¯
¯
¯
To finish, we evaluate those six determinants by row-swapping them to the
identity matrix, keeping track of the resulting sign changes.
= 30 · (+1) + 0 · (−1)
+ 20 · (−1) + 0 · (+1)
− 4 · (+1) − 6 · (−1) = 12
That example illustrates the key idea. We’ve applied multilinearity to a 3×3
determinant to get 33
separate determinants, each with one distinguished entry
per row. We can drop most of these new determinants because the matrices
are singular, with one row a multiple of another. We are left with the one-
entry-per-row determinants also having only one entry per column (one entry
from the original determinant, that is). And, since we can factor scalars out, we
can further reduce to only considering determinants of one-entry-per-row-and-
column matrices where the entries are ones.
These are permutation matrices. Thus, the determinant can be computed
in this three-step way (Step 1) for each permutation matrix, multiply together
the entries from the original matrix where that permutation matrix has ones,
(Step 2) multiply that by the determinant of the permutation matrix and
(Step 3) do that for all permutation matrices and sum the results together.

To state this as a formula, we introduce a notation for permutation matrices.
Let ιj be the row vector that is all zeroes except for a one in its j-th entry, so
that the four-wide ι2 is
¡
0 1 0 0
¢
. We can construct permutation matrices
by permuting — that is, scrambling — the numbers 1, 2, . . . , n, and using them
as indices on the ι’s. For instance, to get a 4×4 permutation matrix matrix, we
can scramble the numbers from 1 to 4 into this sequence h3, 2, 1, 4i and take the
corresponding row vector ι’s.




ι3
ι2
ι1
ι4



 =




0 0 1 0
0 1 0 0
1 0 0 0
0 0 0 1




3.7 Definition An n-permutation is a sequence consisting of an arrangement
of the numbers 1, 2, . . . , n.
3.8 Example The 2-permutations are φ1 = h1, 2i and φ2 = h2, 1i. These are
the associated permutation matrices.
Pφ1
=
µ
ι1
ι2
¶
=
µ
1 0
0 1
¶
Pφ2
=
µ
ι2
ι1
¶
=
µ
0 1
1 0
¶
We sometimes write permutations as functions, e.g., φ2(1) = 2, and φ2(2) = 1.
Then the rows of Pφ2
are ιφ2(1) = ι2 and ιφ2(2) = ι1.
The 3-permutations are φ1 = h1, 2, 3i, φ2 = h1, 3, 2i, φ3 = h2, 1, 3i, φ4 =
h2, 3, 1i, φ5 = h3, 1, 2i, and φ6 = h3, 2, 1i. Here are two of the associated permu-
tation matrices.
Pφ2
=


ι1
ι3
ι2

 =


1 0 0
0 0 1
0 1 0

 Pφ5
=


ι3
ι1
ι2

 =


0 0 1
1 0 0
0 1 0


For instance, the rows of Pφ5
are ιφ5(1) = ι3, ιφ5(2) = ι1, and ιφ5(3) = ι2.
3.9 Definition The permutation expansion for determinants is
¯
¯
¯
¯
¯
¯
¯
¯
¯
t1,1 t1,2 . . . t1,n
t2,1 t2,2 . . . t2,n
.
.
.
tn,1 tn,2 . . . tn,n
¯
¯
¯
¯
¯
¯
¯
¯
¯
= t1,φ1(1)t2,φ1(2) · · · tn,φ1(n)|Pφ1
|
+ t1,φ2(1)t2,φ2(2) · · · tn,φ2(n)|Pφ2
|
.
.
.
+ t1,φk(1)t2,φk(2) · · · tn,φk(n)|Pφk
|
where φ1, . . . , φk are all of the n-permutations.
This formula is often written in summation notation
|T| =
X
permutations φ
t1,φ(1)t2,φ(2) · · · tn,φ(n) |Pφ|

read aloud as “the sum, over all permutations φ, of terms having the form
t1,φ(1)t2,φ(2) · · · tn,φ(n)|Pφ|”. This phrase is just a restating of the three-step
process (Step 1) for each permutation matrix, compute t1,φ(1)t2,φ(2) · · · tn,φ(n)
(Step 2) multiply that by |Pφ| and (Step 3) sum all such terms together.
3.10 Example The familiar formula for the determinant of a 2×2 matrix can
be derived in this way.
¯
¯
¯
¯
t1,1 t1,2
t2,1 t2,2
¯
¯
¯
¯ = t1,1t2,2 · |Pφ1
| + t1,2t2,1 · |Pφ2
|
= t1,1t2,2 ·
¯
¯
¯
¯
1 0
0 1
¯
¯
¯
¯ + t1,2t2,1 ·
¯
¯
¯
¯
0 1
1 0
¯
¯
¯
¯
= t1,1t2,2 − t1,2t2,1
(the second permutation matrix takes one row swap to pass to the identity).
Similarly, the formula for the determinant of a 3×3 matrix is this.
¯
¯
¯
¯
¯
¯
t1,1 t1,2 t1,3
t2,1 t2,2 t2,3
t3,1 t3,2 t3,3
¯
¯
¯
¯
¯
¯
= t1,1t2,2t3,3 |Pφ1
| + t1,1t2,3t3,2 |Pφ2
| + t1,2t2,1t3,3 |Pφ3
|
+ t1,2t2,3t3,1 |Pφ4
| + t1,3t2,1t3,2 |Pφ5
| + t1,3t2,2t3,1 |Pφ6
|
= t1,1t2,2t3,3 − t1,1t2,3t3,2 − t1,2t2,1t3,3
+ t1,2t2,3t3,1 + t1,3t2,1t3,2 − t1,3t2,2t3,1
Computing a determinant by permutation expansion usually takes longer
than Gauss’ method. However, here we are not trying to do the computation
efficiently, we are instead trying to give a determinant formula that we can
prove to be well-defined. While the permutation expansion is impractical for
computations, it is useful in proofs. In particular, we can use it for the result
that we are after.
3.11 Theorem For each n there is a n×n determinant function.
The proof is deferred to the following subsection. Also there is the proof of
the next result (they share some features).
3.12 Theorem The determinant of a matrix equals the determinant of its
transpose.
The consequence of this theorem is that, while we have so far stated results
in terms of rows (e.g., determinants are multilinear in their rows, row swaps
change the signum, etc.), all of the results also hold in terms of columns. The
final result gives examples.
3.13 Corollary A matrix with two equal columns is singular. Column swaps
change the sign of a determinant. Determinants are multilinear in their columns.
Proof. For the first statement, transposing the matrix results in a matrix with
the same determinant, and with two equal rows, and hence a determinant of
zero. The other two are proved in the same way. QED

We finish with a summary (although the final subsection contains the un-
finished business of proving the two theorems). Determinant functions exist,
are unique, and we know how to compute them. As for what determinants are
about, perhaps these lines [Kemp] help make it memorable.
Determinant none,
Solution: lots or none.
Determinant some,
Solution: just one.
Exercises
These summarize the notation used in this book for the 2- and 3- permutations.
i 1 2
φ1(i) 1 2
φ2(i) 2 1
i 1 2 3
φ1(i) 1 2 3
φ2(i) 1 3 2
φ3(i) 2 1 3
φ4(i) 2 3 1
φ5(i) 3 1 2
φ6(i) 3 2 1
X 3.14 Compute the determinant by using the permutation expansion.
(a)
¯
¯
¯
¯
¯
1 2 3
4 5 6
7 8 9
¯
¯
¯
¯
¯
(b)
¯
¯
¯
¯
¯
2 2 1
3 −1 0
−2 0 5
¯
¯
¯
¯
¯
X 3.15 Compute these both with Gauss’ method and with the permutation expansion
formula.
(a)
¯
¯
¯
¯
2 1
3 1
¯
¯
¯
¯ (b)
¯
¯
¯
¯
¯
0 1 4
0 2 3
1 5 1
¯
¯
¯
¯
¯
X 3.16 Use the permutation expansion formula to derive the formula for 3×3 deter-
minants.
3.17 List all of the 4-permutations.
3.18 A permutation, regarded as a function from the set {1, .., n} to itself, is one-
to-one and onto. Therefore, each permutation has an inverse.
(a) Find the inverse of each 2-permutation.
(b) Find the inverse of each 3-permutation.
3.19 Prove that f is multilinear if and only if for all ~
v, ~
w ∈ V and k1, k2 ∈ R, this
holds.
f(~
ρ1, . . . , k1~
v1 + k2~
v2, . . . , ~
ρn) = k1f(~
ρ1, . . . ,~
v1, . . . , ~
ρn) + k2f(~
ρ1, . . . ,~
v2, . . . , ~
ρn)
3.20 Find the only nonzero term in the permutation expansion of this matrix.
¯
¯
¯
¯
¯
¯
¯
0 1 0 0
1 0 1 0
0 1 0 1
0 0 1 0
¯
¯
¯
¯
¯
¯
¯
Compute that determinant by finding the signum of the associated permutation.
3.21 How would determinants change if we changed property (4) of the definition
to read that |I| = 2?
3.22 Verify the second and third statements in Corollary 3.13.

X 3.23 Show that if an n×n matrix has a nonzero determinant then any column vector
~
v ∈ Rn
can be expressed as a linear combination of the columns of the matrix.
3.24 True or false: a matrix whose entries are only zeros or ones has a determinant
equal to zero, one, or negative one.
3.25 (a) Show that there are 120 terms in the permutation expansion formula of
a 5×5 matrix.
(b) How many are sure to be zero if the 1, 2 entry is zero?
3.26 How many n-permutations are there?
3.27 A matrix A is skew-symmetric if Atrans
= −A, as in this matrix.
A =
µ
0 3
−3 0
¶
Show that n×n skew-symmetric matrices with nonzero determinants exist only for
even n.
X 3.28 What is the smallest number of zeros, and the placement of those zeros, needed
to ensure that a 4×4 matrix has a determinant of zero?
X 3.29 If we have n data points (x1, y1), (x2, y2), . . . , (xn, yn) and want to find a
polynomial p(x) = an−1xn−1
+ an−2xn−2
+ · · · + a1x + a0 passing through those
points then we can plug in the points to get an n equation/n unknown linear
system. The matrix of coefficients for that system is called the Vandermonde
matrix. Prove that the determinant of the transpose of that matrix of coefficients
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
1 1 . . . 1
x1 x2 . . . xn
x1
2
x2
2
. . . xn
2
.
.
.
x1
n−1
x2
n−1
. . . xn
n−1
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
equals the product, over all indices i, j ∈ {1, . . . , n} with i j, of terms of the
form xj − xi. (This shows that the determinant is zero, and the linear system has
no solution, if and only if the xi’s in the data are not distinct.)
3.30 A matrix can be divided into blocks, as here,
Ã
1 2 0
3 4 0
0 0 −2
!
which shows four blocks, the square 2×2 and 1×1 ones in the upper left and lower
right, and the zero blocks in the upper right and lower left. Show that if a matrix
can be partitioned as
T =
µ
J Z2
Z1 K
¶
where J and K are square, and Z1 and Z2 are all zeroes, then |T| = |J| · |K|.
X 3.31 Prove that for any n×n matrix T there are at most n distinct reals r such
that the matrix T − rI has determinant zero (we shall use this result in Chapter
Five).
3.32 [Math. Mag., Jan. 1963, Q307] The nine positive digits can be arranged into
3×3 arrays in 9! ways. Find the sum of the determinants of these arrays.
3.33 [Math. Mag., Jan. 1963, Q237] Show that
¯
¯
¯
¯
¯
x − 2 x − 3 x − 4
x + 1 x − 1 x − 3
x − 4 x − 7 x − 10
¯
¯
¯
¯
¯
= 0.

3.34 [Am. Math. Mon., Jan. 1949] Let S be the sum of the integer elements of a
magic square of order three and let D be the value of the square considered as a
determinant. Show that D/S is an integer.
3.35 [Am. Math. Mon., Jun. 1931] Show that the determinant of the n2
elements
in the upper left corner of the Pascal triangle
1 1 1 1 . .
1 2 3 . .
1 3 . .
1 . .
.
.
has the value unity.
4.I.4 Determinants Exist
This subsection is optional. It consists of proofs of two results from the prior
subsection. These proofs involve the properties of permutations, which will not
be used later, except in the optional Jordan Canonical Form subsection.
The prior subsection attacks the problem of showing that for any size there
is a determinant function on the set of square matrices of that size by using
multilinearity to develop the permutation expansion.
¯
¯
¯
¯
¯
¯
¯
¯
¯
t1,1 t1,2 . . . t1,n
t2,1 t2,2 . . . t2,n
.
.
.
tn,1 tn,2 . . . tn,n
¯
¯
¯
¯
¯
¯
¯
¯
¯
= t1,φ1(1)t2,φ1(2) · · · tn,φ1(n)|Pφ1
|
+ t1,φ2(1)t2,φ2(2) · · · tn,φ2(n)|Pφ2
|
.
.
.
+ t1,φk(1)t2,φk(2) · · · tn,φk(n)|Pφk
|
=
X
permutations φ
t1,φ(1)t2,φ(2) · · · tn,φ(n) |Pφ|
This reduces the problem to showing that there is a determinant function on
the set of permutation matrices of that size.
Of course, a permutation matrix can be row-swapped to the identity matrix
and to calculate its determinant we can keep track of the number of row swaps.
However, the problem is still not solved. We still have not shown that the result
is well-defined. For instance, the determinant of
Pφ =




0 1 0 0
1 0 0 0
0 0 1 0
0 0 0 1





could be computed with one swap
Pφ
ρ1↔ρ2
−→




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1




or with three.
Pφ
ρ3↔ρ1
−→




0 0 1 0
1 0 0 0
0 1 0 0
0 0 0 1




ρ2↔ρ3
−→




0 0 1 0
0 1 0 0
1 0 0 0
0 0 0 1




ρ1↔ρ3
−→




1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1




Both reductions have an odd number of swaps so we figure that |Pφ| = −1
but how do we know that there isn’t some way to do it with an even number of
swaps? Corollary 4.6 below proves that there is no permutation matrix that can
be row-swapped to an identity matrix in two ways, one with an even number of
swaps and the other with an odd number of swaps.
4.1 Definition Two rows of a permutation matrix









.
.
.
ιk
.
.
.
ιj
.
.
.









such that k j are in an inversion of their natural order.
4.2 Example This permutation matrix


ι3
ι2
ι1

 =


0 0 1
0 1 0
1 0 0


has three inversions: ι3 precedes ι1, ι3 precedes ι2, and ι2 precedes ι1.
4.3 Lemma A row-swap in a permutation matrix changes the number of in-
versions from even to odd, or from odd to even.
Proof. Consider a swap of rows j and k, where k j. If the two rows are
adjacent
Pφ =






.
.
.
ιφ(j)
ιφ(k)
.
.
.






ρk↔ρj
−→






.
.
.
ιφ(k)
ιφ(j)
.
.
.







then the swap changes the total number of inversions by one — either removing
or producing one inversion, depending on whether φ(j) φ(k) or not, since
inversions involving rows not in this pair are not affected. Consequently, the
total number of inversions changes from odd to even or from even to odd.
If the rows are not adjacent then they can be swapped via a sequence of
adjacent swaps, first bringing row k up













.
.
.
ιφ(j)
ιφ(j+1)
ιφ(j+2)
.
.
.
ιφ(k)
.
.
.













ρk↔ρk−1
−→
ρk−1↔ρk−2
−→ . . .
ρj+1↔ρj
−→













.
.
.
ιφ(k)
ιφ(j)
ιφ(j+1)
.
.
.
ιφ(k−1)
.
.
.













and then bringing row j down.
ρj+1↔ρj+2
−→
ρj+2↔ρj+3
−→ . . .
ρk−1↔ρk
−→













.
.
.
ιφ(k)
ιφ(j+1)
ιφ(j+2)
.
.
.
ιφ(j)
.
.
.













Each of these adjacent swaps changes the number of inversions from odd to even
or from even to odd. There are an odd number (k − j) + (k − j − 1) of them.
The total change in the number of inversions is from even to odd or from odd
to even. QED
4.4 Definition The signum of a permutation sgn(φ) is +1 if the number of
inversions in Pφ is even, and is −1 if the number of inversions is odd.
4.5 Example With the subscripts from Example 3.8 for the 3-permutations,
sgn(φ1) = 1 while sgn(φ2) = −1.
4.6 Corollary If a permutation matrix has an odd number of inversions then
swapping it to the identity takes an odd number of swaps. If it has an even
number of inversions then swapping to the identity takes an even number of
swaps.
Proof. The identity matrix has zero inversions. To change an odd number to
zero requires an odd number of swaps, and to change an even number to zero
requires an even number of swaps. QED

We still have not shown that the permutation expansion is well-defined be-
cause we have not considered row operations on permutation matrices other than
row swaps. We will finesse this problem: we will define a function d: Mn×n → R
by altering the permutation expansion formula, replacing |Pφ| with sgn(φ)
d(T) =
X
permutations φ
t1,φ(1)t2,φ(2) . . . tn,φ(n) sgn(φ)
(this gives the same value as the permutation expansion because the prior result
shows that det(Pφ) = sgn(φ)). This formula’s advantage is that the number of
inversions is clearly well-defined — just count them. Therefore, we will show
that a determinant function exists for all sizes by showing that d is it, that is,
that d satisfies the four conditions.
4.7 Lemma The function d is a determinant. Hence determinants exist for
every n.
Proof. We’ll must check that it has the four properties from the definition.
Property (4) is easy; in
d(I) =
X
perms φ
ι1,φ(1)ι2,φ(2) · · · ιn,φ(n) sgn(φ)
all of the summands are zero except for the product down the diagonal, which
is one.
For property (3) consider d(T̂) where T
kρi
−→T̂.
X
perms φ
t̂1,φ(1) · · · t̂i,φ(i) · · · t̂n,φ(n) sgn(φ) =
X
φ
t1,φ(1) · · · kti,φ(i) · · · tn,φ(n) sgn(φ)
Factor the k out of each term to get the desired equality.
= k ·
X
φ
t1,φ(1) · · · ti,φ(i) · · · tn,φ(n) sgn(φ) = k · d(T)
For (2), let T
ρi↔ρj
−→ T̂.
d(T̂) =
X
perms φ
t̂1,φ(1) · · · t̂i,φ(i) · · · t̂j,φ(j) · · · t̂n,φ(n) sgn(φ)
To convert to unhatted t’s, for each φ consider the permutation σ that equals φ
except that the i-th and j-th numbers are interchanged, σ(i) = φ(j) and σ(j) =
φ(i). Replacing the φ in t̂1,φ(1) · · · t̂i,φ(i) · · · t̂j,φ(j) · · · t̂n,φ(n) with this σ gives
t1,σ(1) · · · tj,σ(j) · · · ti,σ(i) · · · tn,σ(n). Now sgn(φ) = − sgn(σ) (by Lemma 4.3)
and so we get
=
X
σ
t1,σ(1) · · · tj,σ(j) · · · ti,σ(i) · · · tn,σ(n) ·
¡
− sgn(σ)
¢
= −
X
σ
t1,σ(1) · · · tj,σ(j) · · · ti,σ(i) · · · tn,σ(n) · sgn(σ)

where the sum is over all permutations σ derived from another permutation φ
by a swap of the i-th and j-th numbers. But any permutation can be derived
from some other permutation by such a swap, in one and only one way, so this
summation is in fact a sum over all permutations, taken once and only once.
Thus d(T̂) = −d(T).
To do property (1) let T
kρi+ρj
−→ T̂ and consider
d(T̂) =
X
perms φ
t̂1,φ(1) · · · t̂i,φ(i) · · · t̂j,φ(j) · · · t̂n,φ(n) sgn(φ)
=
X
φ
t1,φ(1) · · · ti,φ(i) · · · (kti,φ(j) + tj,φ(j)) · · · tn,φ(n) sgn(φ)
(notice: that’s kti,φ(j), not ktj,φ(j)). Distribute, commute, and factor.
=
X
φ
£
t1,φ(1) · · · ti,φ(i) · · · kti,φ(j) · · · tn,φ(n) sgn(φ)
+ t1,φ(1) · · · ti,φ(i) · · · tj,φ(j) · · · tn,φ(n) sgn(φ)
¤
=
X
φ
t1,φ(1) · · · ti,φ(i) · · · kti,φ(j) · · · tn,φ(n) sgn(φ)
+
X
φ
t1,φ(1) · · · ti,φ(i) · · · tj,φ(j) · · · tn,φ(n) sgn(φ)
= k ·
X
φ
t1,φ(1) · · · ti,φ(i) · · · ti,φ(j) · · · tn,φ(n) sgn(φ)
+ d(T)
We finish by showing that the terms t1,φ(1) · · · ti,φ(i) · · · ti,φ(j) . . . tn,φ(n) sgn(φ)
add to zero. This sum represents d(S) where S is a matrix equal to T except
that row j of S is a copy of row i of T (because the factor is ti,φ(j), not tj,φ(j)).
Thus, S has two equal rows, rows i and j. Since we have already shown that d
changes sign on row swaps, as in Lemma 2.3 we conclude that d(S) = 0. QED
We have now shown that determinant functions exist for each size. We
already know that for each size there is at most one determinant. Therefore,
the permutation expansion computes the one and only determinant value of a
square matrix.
We end this subsection by proving the other result remaining from the prior
subsection, that the determinant of a matrix equals the determinant of its trans-
pose.
4.8 Example Writing out the permutation expansion of the general 3×3 matrix
and of its transpose, and comparing corresponding terms
¯
¯
¯
¯
¯
¯
a b c
d e f
g h i
¯
¯
¯
¯
¯
¯
= · · · + cdh ·
¯
¯
¯
¯
¯
¯
0 0 1
1 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+ · · ·

(terms with the same letters)
¯
¯
¯
¯
¯
¯
a d g
b e h
c f i
¯
¯
¯
¯
¯
¯
= · · · + dhc ·
¯
¯
¯
¯
¯
¯
0 1 0
0 0 1
1 0 0
¯
¯
¯
¯
¯
¯
+ · · ·
shows that the corresponding permutation matrices are transposes. That is,
there is a relationship between these corresponding permutations. Exercise 15
shows that they are inverses.
4.9 Theorem The determinant of a matrix equals the determinant of its trans-
pose.
Proof. Call the matrix T and denote the entries of Ttrans
with s’s so that
ti,j = sj,i. Substitution gives this
|T| =
X
perms φ
t1,φ(1) . . . tn,φ(n) sgn(φ) =
X
φ
sφ(1),1 . . . sφ(n),n sgn(φ)
and we can finish the argument by manipulating the expression on the right
to be recognizable as the determinant of the transpose. We have written all
permutation expansions (as in the middle expression above) with the row indices
ascending. To rewrite the expression on the right in this way, note that because
φ is a permutation, the row indices in the term on the right φ(1), . . . , φ(n) are
just the numbers 1, . . . , n, rearranged. We can thus commute to have these
ascend, giving s1,φ−1(1) · · · sn,φ−1(n) (if the column index is j and the row index
is φ(j) then, where the row index is i, the column index is φ−1
(i)). Substituting
on the right gives
=
X
φ−1
s1,φ−1(1) · · · sn,φ−1(n) sgn(φ−1
)
(Exercise 14 shows that sgn(φ−1
) = sgn(φ)). Since every permutation is the
inverse of another, a sum over all φ−1
is a sum over all permutations φ
=
X
perms σ
s1,σ(1) . . . sn,σ(n) sgn(σ) =
¯
¯Ttrans
¯
¯
as required. QED
Exercises
These summarize the notation used in this book for the 2- and 3- permutations.
i 1 2
φ1(i) 1 2
φ2(i) 2 1
i 1 2 3
φ1(i) 1 2 3
φ2(i) 1 3 2
φ3(i) 2 1 3
φ4(i) 2 3 1
φ5(i) 3 1 2
φ6(i) 3 2 1

4.10 Give the permutation expansion of a general 2×2 matrix and its transpose.
X 4.11 This problem appears also in the prior subsection.
(a) Find the inverse of each 2-permutation.
(b) Find the inverse of each 3-permutation.
X 4.12 (a) Find the signum of each 2-permutation.
(b) Find the signum of each 3-permutation.
4.13 What is the signum of the n-permutation φ = hn, n − 1, . . . , 2, 1i?
4.14 Prove these.
(a) Every permutation has an inverse.
(b) sgn(φ−1
) = sgn(φ)
(c) Every permutation is the inverse of another.
4.15 Prove that the matrix of the permutation inverse is the transpose of the matrix
of the permutation Pφ−1 = Pφ
trans
, for any permutation φ.
X 4.16 Show that a permutation matrix with m inversions can be row swapped to
the identity in m steps. Contrast this with Corollary 4.6.
X 4.17 For any permutation φ let g(φ) be the integer defined in this way.
g(φ) =
Y
ij
[φ(j) − φ(i)]
(This is the product, over all indices i and j with i j, of terms of the given
form.)
(a) Compute the value of g on all 2-permutations.
(b) Compute the value of g on all 3-permutations.
(c) Prove this.
sgn(φ) =
g(φ)
|g(φ)|
Many authors give this formula as the definition of the signum function.

Section II. Geometry of Determinants 319
4.II Geometry of Determinants
The prior section develops the determinant algebraically, by considering what
formulas satisfy certain properties. This section complements that with a geo-
metric approach. One advantage of this approach is that, while we have so far
only considered whether or not a determinant is zero, here we shall give a mean-
ing to the value of that determinant. (The prior section handles determinants
as functions of the rows, but in this section columns are more convenient. The
final result of the prior section says that we can make the switch.)
4.II.1 Determinants as Size Functions
This parallelogram picture

x1
y1

x2
y2

is familiar from the construction of the sum of the two vectors. One way to
compute the area that it encloses is to draw this rectangle and subtract the
area of each subregion.
y1
y2
x2 x1
A
B
C
D
E
F
area of parallelogram
= area of rectangle − area of A − area of B
− · · · − area of F
= (x1 + x2)(y1 + y2) − x2y1 − x1y1/2
− x2y2/2 − x2y2/2 − x1y1/2 − x2y1
= x1y2 − x2y1
The fact that the area equals the value of the determinant
¯
¯
¯
¯
x1 x2
y1 y2
¯
¯
¯
¯ = x1y2 − x2y1
is no coincidence. The properties in the definition of determinants make rea-
sonable postulates for a function that measures the size of the region enclosed
by the vectors in the matrix.
For instance, this shows the effect of multiplying one of the box-defining
vectors by a scalar (the scalar used is k = 1.4).
~
v
~
w
k~
v
~
w

Compared to the shaded region enclosed by ~
v and ~
w, the region formed by
k~
v and ~
w is bigger by a factor of k. This illustrates that size(k~
v, ~
w) = k ·
size(~
v, ~
w). Generalized, we expect of the size measure that size(. . . , k~
v, . . . ) =
k · size(. . . ,~
v, . . . ). Of course, this postulate is already familiar as one of the
properties in the defintion of determinants.
Another property of determinants is that they are unaffected by pivoting.
Here are before-pivoting and after-pivoting boxes (the scalar used is k = 0.35).
~
v
~
w
~
v
k~
v + ~
w
Although the region on the right, the box formed by v and k~
v + ~
w, is more
slanted than the shaded region, the two have the same base and the same height
and hence the same area. This illustrates that size(~
v, k~
v + ~
w) = size(~
v, ~
w).
Generalized, size(. . . ,~
v, . . . , ~
w, . . . ) = size(. . . ,~
v, . . . , k~
v + ~
w, . . . ), which is a
restatement of the determinant postulate.
Of course, this picture
~
e1
~
e2
shows that size(~
e1,~
e2) = 1, and we naturally extend that to any number of
dimensions size(~
e1, . . . ,~
en) = 1, which is a restatement of the property that the
determinant of the identity matrix is one.
With that, because property (2) of determinants is redundant (as remarked
right after the definition), we have that all of the properties of determinants are
reasonable to expect of a function that gives the size of boxes. We can now cite
the work done in the prior section to show that the determinant exists and is
unique to be assured that these postulates are consistent and sufficient (we do
not need any more postulates). That is, we’ve got an intuitive justification to
interpret det(~
v1, . . . ,~
vn) as the size of the box formed by the vectors. (Com-
ment. An even more basic approach, which also leads to the definition below,
is [Weston].)
1.1 Example The volume of this parallelepiped, which can be found by the
usual formula from high school geometry, is 12.
2
0
2

0
3
1

−1
0
1

¯
¯
¯
¯
¯
¯
2 0 −1
0 3 0
2 1 1
¯
¯
¯
¯
¯
¯
= 12

1.2 Remark Although property (2) of the definition of determinants is redun-
dant, it raises an important point. Consider these two.
~
u
~
v
~
u
~
v
¯
¯
¯
¯
4 1
2 3
¯
¯
¯
¯ = 10
¯
¯
¯
¯
1 4
3 2
¯
¯
¯
¯ = −10
The only difference between them is in the order in which the vectors are taken.
If we take ~
u first and then go to ~
v, follow the counterclockwise arc shown, then
the sign is positive. Following a clockwise arc gives a negative sign. The sign
returned by the size function reflects the ‘orientation’ or ‘sense’ of the box. (We
see the same thing if we picture the effect of scalar multiplication by a negative
scalar.)
Although it is both interesting and important, the idea of orientation turns
out to be tricky. It is not needed for the development below, and so we will pass
it by. (See Exercise 27.)
1.3 Definition The box (or parallelepiped) formed by h~
v1, . . . ,~
vni (where each
vector is from Rn
) includes all of the set {t1~
v1 + · · · + tn~
vn
¯
¯ t1, . . . , tn ∈ [0..1]}.
The volume of a box is the absolute value of the determinant of the matrix with
those vectors as columns.
1.4 Example Volume, because it is an absolute value, does not depend on
the order in which the vectors are given. The volume of the parallelepiped in
Exercise 1.1, can also be computed as the absolute value of this determinant.
¯
¯
¯
¯
¯
¯
0 2 0
3 0 3
1 2 1
¯
¯
¯
¯
¯
¯
= −12
The definition of volume gives a geometric interpretation to something in
the space, boxes made from vectors. The next result relates the geometry to
the functions that operate on spaces.
1.5 Theorem A transformation t: Rn
→ Rn
changes the size of all boxes by
the same factor, namely the size of the image of a box |t(S)| is |T| times the
size of the box |S|, where T is the matrix representing t with respect to the
standard basis. That is, for all n×n matrices, the determinant of a product is
the product of the determinants |TS| = |T| · |S|.
The two sentences state the same idea, first in map terms and then in matrix
terms. Although we tend to prefer a map point of view, the second sentence,
the matrix version, is more convienent for the proof and is also the way that
we shall use this result later. (Alternate proofs are given as Exercise 23 and
Exercise 28.)

Proof. The two statements are equivalent because |t(S)| = |TS|, as both give
the size of the box that is the image of the unit box En under the composition
t ◦ s (where s is the map represented by S with respect to the standard basis).
First consider the case that |T| = 0. A matrix has a zero determinant if and
only if it is not invertible. Observe that if TS is invertible, so that there is an
M such that (TS)M = I, then the associative property of matrix multiplication
T(SM) = I shows that T is also invertible (with inverse SM). Therefore, if T
is not invertible then neither is TS — if |T| = 0 then |TS| = 0, and the result
holds in this case.
Now consider the case that |T| 6= 0, that T is nonsingular. Recall that any
nonsingular matrix can be factored into a product of elementary matrices, so
that TS = E1E2 · · · ErS. In the rest of this argument, we will verify that if E
is an elementary matrix then |ES| = |E| · |S|. The result will follow because
then |TS| = |E1 · · · ErS| = |E1| · · · |Er| · |S| = |E1 · · · Er| · |S| = |T| · |S|.
If the elementary matrix E is Mi(k) then Mi(k)S equals S except that row i
has been multiplied by k. The third property of determinant functions then
gives that |Mi(k)S| = k · |S|. But |Mi(k)| = k, again by the third property
because Mi(k) is derived from the identity by multiplication of row i by k, and
so |ES| = |E| · |S| holds for E = Mi(k). The E = Pi,j = −1 and E = Ci,j(k)
checks are similar. QED
1.6 Example Application of the map t represented with respect to the stan-
dard bases by
µ
1 1
−2 0
¶
will double sizes of boxes, e.g., from this
~
w
~
v
¯
¯
¯
¯
2 1
1 2
¯
¯
¯
¯ = 3
to this
t( ~
w)
t(~
v) ¯
¯
¯
¯
3 3
−4 −2
¯
¯
¯
¯ = 6
1.7 Corollary If a matrix is invertible then the determinant of its inverse is
the inverse of its determinant |T−1
| = 1/|T|.
Proof. 1 = |I| = |TT−1
| = |T| · |T−1
| QED
Recall that determinants are not additive homomorphisms, det(A+B) need
not equal det(A) + det(B). The above theorem says, in contrast, that determi-
nants are multiplicative homomorphisms: det(AB) does equal det(A) · det(B).

Exercises
1.8 Find the volume of the region formed.
(a) h
µ
1
3
¶
,
µ
−1
4
¶
i
(b) h
Ã
2
1
0
!
,
Ã
3
−2
4
!
,
Ã
8
−3
8
!
i
(c) h



1
2
0
1


 ,



2
2
2
2


 ,



−1
3
0
5


 ,



0
1
0
7


i
X 1.9 Is
Ã
4
1
2
!
inside of the box formed by these three?
Ã
3
3
1
! Ã
2
6
1
! Ã
1
0
5
!
X 1.10 Find the volume of this region.
X 1.11 Suppose that |A| = 3. By what factor do these change volumes?
(a) A (b) A2
(c) A−2
X 1.12 By what factor does each transformation change the size of boxes?
(a)
µ
x
y
¶
7→
µ
2x
3y
¶
(b)
µ
x
y
¶
7→
µ
3x − y
−2x + y
¶
(c)
Ã
x
y
z
!
7→
Ã
x − y
x + y + z
y − 2z
!
1.13 What is the area of the image of the rectangle [2..4] × [2..5] under the action
of this matrix?
µ
2 3
4 −1
¶
1.14 If t: R3
→ R3
changes volumes by a factor of 7 and s: R3
→ R3
changes vol-
umes by a factor of 3/2 then by what factor will their composition changes vol-
umes?
1.15 In what way does the definition of a box differ from the defintion of a span?
X 1.16 Why doesn’t this picture contradict Theorem 1.5?
¡
2 1
0 1
¢
−→
area is 2 determinant is 2 area is 5
X 1.17 Does |TS| = |ST|? |T(SP)| = |(TS)P|?
1.18 (a) Suppose that |A| = 3 and that |B| = 2. Find |A2
· Btrans
· B−2
· Atrans
|.
(b) Assume that |A| = 0. Prove that |6A3
+ 5A2
+ 2A| = 0.

X 1.19 Let T be the matrix representing (with respect to the standard bases) the
map that rotates plane vectors counterclockwise thru θ radians. By what factor
does T change sizes?
X 1.20 Must a transformation t: R2
→ R2
that preserves areas also preserve lengths?
X 1.21 What is the volume of a parallelepiped in R3
bounded by a linearly dependent
set?
X 1.22 Find the area of the triangle in R3
with endpoints (1, 2, 1), (3, −1, 4), and
(2, 2, 2). (Area, not volume. The triangle defines a plane—what is the area of the
triangle in that plane?)
X 1.23 An alternate proof of Theorem 1.5 uses the definition of determinant func-
tions.
(a) Note that the vectors forming S make a linearly dependent set if and only if
|S| = 0, and check that the result holds in this case.
(b) For the |S| 6= 0 case, to show that |TS|/|S| = |T| for all transformations,
consider the function d: Mn
×
n → R given by T 7→ |TS|/|S|. Show that d has
the first property of a determinant.
(c) Show that d has the remaining three properties of a determinant function.
(d) Conclude that |TS| = |T| · |S|.
1.24 Give a non-identity matrix with the property that Atrans
= A−1
. Show that
if Atrans
= A−1
then |A| = ±1. Does the converse hold?
1.25 The algebraic property of determinants that factoring a scalar out of a single
row will multiply the determinant by that scalar shows that where H is 3×3, the
determinant of cH is c3
times the determinant of H. Explain this geometrically,
that is, using Theorem 1.5,
X 1.26 Matrices H and G are said to be similar if there is a nonsingular matrix P
such that H = P−1
GP (we will study this relation in Chapter Five). Show that
similar matrices have the same determinant.
1.27 We usually represent vectors in R2
with respect to the standard basis so
vectors in the first quadrant have both coordinates positive.
~
v
RepE2
(~
v) =
µ
+3
+2
¶
Moving counterclockwise around the origin, we cycle thru four regions:
· · · −→
³
+
+
´
−→
³
−
+
´
−→
³
−
−
´
−→
³
+
−
´
−→ · · · .
Using this basis
B = h
µ
0
1
¶
,
µ
−1
0
¶
i ~
β2
~
β1
gives the same counterclockwise cycle. We say these two bases have the same
orientation.
(a) Why do they give the same cycle?
(b) What other configurations of unit vectors on the axes give the same cycle?
(c) Find the determinants of the matrices formed from those (ordered) bases.
(d) What other counterclockwise cycles are possible, and what are the associated
determinants?
(e) What happens in R1
?
(f) What happens in R3
?
A fascinating general-audience discussion of orientations is in [Gardner].

1.28 This question uses material from the optional Determinant Functions Exist
subsection. Prove Theorem 1.5 by using the permutation expansion formula for
the determinant.
X 1.29 (a) Show that this gives the equation of a line in R2
thru (x2, y2) and (x3, y3).
¯
¯
¯
¯
¯
x x2 x3
y y2 y3
1 1 1
¯
¯
¯
¯
¯
= 0
(b) [Petersen] Prove that the area of a triangle with vertices (x1, y1), (x2, y2),
and (x3, y3) is
1
2
¯
¯
¯
¯
¯
x1 x2 x3
y1 y2 y3
1 1 1
¯
¯
¯
¯
¯
.
(c) [Math. Mag., Jan. 1973] Prove that the area of a triangle with vertices at
(x1, y1), (x2, y2), and (x3, y3) whose coordinates are integers has an area of N
or N/2 for some positive integer N.

4.III Other Formulas
(This section is optional. Later sections do not depend on this material.)
Determinants are a fount of interesting and amusing formulas. Here is one
that is often seen in calculus classes and used to compute determinants by hand.
4.III.1 Laplace’s Expansion
1.1 Example In this permutation expansion
¯
¯
¯
¯
¯
¯
t1,1 t1,2 t1,3
t2,1 t2,2 t2,3
t3,1 t3,2 t3,3
¯
¯
¯
¯
¯
¯
= t1,1t2,2t3,3
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t1,1t2,3t3,2
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯
+ t1,2t2,1t3,3
¯
¯
¯
¯
¯
¯
0 1 0
1 0 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t1,2t2,3t3,1
¯
¯
¯
¯
¯
¯
0 1 0
0 0 1
1 0 0
¯
¯
¯
¯
¯
¯
+ t1,3t2,1t3,2
¯
¯
¯
¯
¯
¯
0 0 1
1 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+ t1,3t2,2t3,1
¯
¯
¯
¯
¯
¯
0 0 1
0 1 0
1 0 0
¯
¯
¯
¯
¯
¯
we can, for instance, factor out the entries from the first row
= t1,1 ·

t2,2t3,3
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t2,3t3,2
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯


+ t1,2 ·

t2,1t3,3
¯
¯
¯
¯
¯
¯
0 1 0
1 0 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t2,3t3,1
¯
¯
¯
¯
¯
¯
0 1 0
0 0 1
1 0 0
¯
¯
¯
¯
¯
¯


+ t1,3 ·

t2,1t3,2
¯
¯
¯
¯
¯
¯
0 0 1
1 0 0
0 1 0
¯
¯
¯
¯
¯
¯
+ t2,2t3,1
¯
¯
¯
¯
¯
¯
0 0 1
0 1 0
1 0 0
¯
¯
¯
¯
¯
¯


and swap rows in the permutation matrices to get this.
= t1,1 ·

t2,2t3,3
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t2,3t3,2
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯


− t1,2 ·

t2,1t3,3
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t2,3t3,1
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯


+ t1,3 ·

t2,1t3,2
¯
¯
¯
¯
¯
¯
1 0 0
0 1 0
0 0 1
¯
¯
¯
¯
¯
¯
+ t2,2t3,1
¯
¯
¯
¯
¯
¯
1 0 0
0 0 1
0 1 0
¯
¯
¯
¯
¯
¯



Section III. Other Formulas 327
The point of the swapping (one swap to each of the permutation matrices on
the second line and two swaps to each on the third line) is that the three lines
simplify to three terms.
= t1,1 ·
¯
¯
¯
¯
t2,2 t2,3
t3,2 t3,3
¯
¯
¯
¯ − t1,2 ·
¯
¯
¯
¯
t2,1 t2,3
t3,1 t3,3
¯
¯
¯
¯ + t1,3 ·
¯
¯
¯
¯
t2,1 t2,2
t3,1 t3,2
¯
¯
¯
¯
The formula given in Theorem 1.5, which generalizes this example, is a recur-
rence — the determinant is expressed as a combination of determinants. This
formula isn’t circular because, as here, the determinant is expressed in terms of
determinants of matrices of smaller size.
1.2 Definition For any n×n matrix T, the (n − 1)×(n − 1) matrix formed by
deleting row i and column j of T is the i, j minor of T. The i, j cofactor Ti,j of
T is (−1)i+j
times the determinant of the i, j minor of T.
1.3 Example The 1, 2 cofactor of the matrix from Example 1.1 is the negative
of the second 2×2 determinant.
T1,2 = −1 ·
¯
¯
¯
¯
t2,1 t2,3
t3,1 t3,3
¯
¯
¯
¯
1.4 Example Where
T =


1 2 3
4 5 6
7 8 9


these are the 1, 2 and 2, 2 cofactors.
T1,2 = (−1)1+2
·
¯
¯
¯
¯
4 6
7 9
¯
¯
¯
¯ = 6 T2,2 = (−1)2+2
·
¯
¯
¯
¯
1 3
7 9
¯
¯
¯
¯ = −12
1.5 Theorem (Laplace Expansion of Determinants) Where T is an n×n
matrix, the determinant can be found by expanding by cofactors on row i or
column j.
|T| = ti,1 · Ti,1 + ti,2 · Ti,2 + · · · + ti,n · Ti,n
= t1,j · T1,j + t2,j · T2,j + · · · + tn,j · Tn,j
Proof. Exercise 27. QED
1.6 Example We can compute the determinant
|T| =
¯
¯
¯
¯
¯
¯
1 2 3
4 5 6
7 8 9
¯
¯
¯
¯
¯
¯

by expanding along the first row, as in Example 1.1.
|T| = 1 · (+1)
¯
¯
¯
¯
5 6
8 9
¯
¯
¯
¯ + 2 · (−1)
¯
¯
¯
¯
4 6
7 9
¯
¯
¯
¯ + 3 · (+1)
¯
¯
¯
¯
4 5
7 8
¯
¯
¯
¯ = −3 + 12 − 9 = 0
Alternatively, we can expand down the second column.
|T| = 2 · (−1)
¯
¯
¯
¯
4 6
7 9
¯
¯
¯
¯ + 5 · (+1)
¯
¯
¯
¯
1 3
7 9
¯
¯
¯
¯ + 8 · (−1)
¯
¯
¯
¯
1 3
4 6
¯
¯
¯
¯ = 12 − 60 + 48 = 0
1.7 Example A row or column with many zeroes suggests a Laplace expan-
sion.
¯
¯
¯
¯
¯
¯
1 5 0
2 1 1
3 −1 0
¯
¯
¯
¯
¯
¯
= 0 · (+1)
¯
¯
¯
¯
2 1
3 −1
¯
¯
¯
¯ + 1 · (−1)
¯
¯
¯
¯
1 5
3 −1
¯
¯
¯
¯ + 0 · (+1)
¯
¯
¯
¯
1 5
2 1
¯
¯
¯
¯ = 16
We finish by applying this result to derive a new formula for the inverse
of a matrix. With Theorem 1.5, the determinant of an n×n matrix T can
be calculated by taking linear combinations of entries from a row and their
associated cofactors.
ti,1 · Ti,1 + ti,2 · Ti,2 + · · · + ti,n · Ti,n = |T| (∗)
Recall that a matrix with two identical rows has a zero determinant. Thus, for
any matrix T, weighing the cofactors by entries from the “wrong” row — row k
with k 6= i — gives zero
ti,1 · Tk,1 + ti,2 · Tk,2 + · · · + ti,n · Tk,n = 0 (∗∗)
because it represents the expansion along the row k of a matrix with row i equal
to row k. This equation summarizes (∗) and (∗∗).





t1,1 t1,2 . . . t1,n
t2,1 t2,2 . . . t2,n
.
.
.
tn,1 tn,2 . . . tn,n










T1,1 T2,1 . . . Tn,1
T1,2 T2,2 . . . Tn,2
.
.
.
T1,n T2,n . . . Tn,n





=





|T| 0 . . . 0
0 |T| . . . 0
.
.
.
0 0 . . . |T|





Note that the order of the subscripts in the matrix of cofactors is opposite to
the order of subscripts in the other matrix; e.g., along the first row of the matrix
of cofactors the subscripts are 1, 1 then 2, 1, etc.
1.8 Definition The matrix adjoint to the square matrix T is
adj(T) =





T1,1 T2,1 . . . Tn,1
T1,2 T2,2 . . . Tn,2
.
.
.
T1,n T2,n . . . Tn,n





where Tj,i is the j, i cofactor.

Section III. Other Formulas 329
1.9 Theorem Where T is a square matrix, T · adj(T) = adj(T) · T = |T| · I.
Proof. Equations (∗) and (∗∗). QED
1.10 Example If
T =


1 0 4
2 1 −1
1 0 1


then the adjoint adj(T) is


T1,1 T2,1 T3,1
T1,2 T2,2 T3,2
T1,3 T2,3 T3,3

=









¯
¯
¯
¯
1 −1
0 1
¯
¯
¯
¯ −
¯
¯
¯
¯
0 4
0 1
¯
¯
¯
¯
¯
¯
¯
¯
0 4
1 −1
¯
¯
¯
¯
−
¯
¯
¯
¯
2 −1
1 1
¯
¯
¯
¯
¯
¯
¯
¯
1 4
1 1
¯
¯
¯
¯ −
¯
¯
¯
¯
1 4
2 −1
¯
¯
¯
¯
¯
¯
¯
¯
2 1
1 0
¯
¯
¯
¯ −
¯
¯
¯
¯
1 0
1 0
¯
¯
¯
¯
¯
¯
¯
¯
1 0
2 1
¯
¯
¯
¯









=


1 0 −4
−3 −3 9
−1 0 1


and taking the product with T gives the diagonal matrix |T| · I.


1 0 4
2 1 −1
1 0 1




1 0 −4
−3 −3 9
−1 0 1

 =


−3 0 0
0 −3 0
0 0 −3


1.11 Corollary If |T| 6= 0 then T−1
= (1/|T|) · adj(T).
1.12 Example The inverse of the matrix from Example 1.10 is (1/−3)·adj(T).
T−1
=


1/−3 0/−3 −4/−3
−3/−3 −3/−3 9/−3
−1/−3 0/−3 1/−3

 =


−1/3 0 4/3
1 1 −3
1/3 0 −1/3


The formulas from this section are often used for by-hand calculation and
are sometimes useful with special types of matrices. However, they are not the
best choice for computation with arbitrary matrices because they require more
arithmetic than, for instance, the Gauss-Jordan method.
Exercises
X 1.13 Find the cofactor.
T =
Ã
1 0 2
−1 1 3
0 2 −1
!
(a) T2,3 (b) T3,2 (c) T1,3
X 1.14 Find the determinant by expanding
¯
¯
¯
¯
¯
3 0 1
1 2 2
−1 3 0
¯
¯
¯
¯
¯

(a) on the first row (b) on the second row (c) on the third column.
1.15 Find the adjoint of the matrix in Example 1.6.
X 1.16 Find the matrix adjoint to each.
(a)
Ã
2 1 4
−1 0 2
1 0 1
!
(b)
µ
3 −1
2 4
¶
(c)
µ
1 1
5 0
¶
(d)
Ã
1 4 3
−1 0 3
1 8 9
!
X 1.17 Find the inverse of each matrix in the prior question with Theorem 1.9.
1.18 Find the matrix adjoint to this one.



2 1 0 0
1 2 1 0
0 1 2 1
0 0 1 2



X 1.19 Expand across the first row to derive the formula for the determinant of a 2×2
matrix.
X 1.20 Expand across the first row to derive the formula for the determinant of a 3×3
matrix.
X 1.21 (a) Give a formula for the adjoint of a 2×2 matrix.
(b) Use it to derive the formula for the inverse.
X 1.22 Can we compute a determinant by expanding down the diagonal?
1.23 Give a formula for the adjoint of a diagonal matrix.
X 1.24 Prove that the transpose of the adjoint is the adjoint of the transpose.
1.25 Prove or disprove: adj(adj(T)) = T.
1.26 A square matrix is upper triangular if each i, j entry is zero in the part above
the diagonal, that is, when i j.
(a) Must the adjoint of an upper triangular matrix be upper triangular? Lower
triangular?
(b) Prove that the inverse of a upper triangular matrix is upper triangular, if an
inverse exists.
1.27 This question requires material from the optional Determinants Exist subsec-
tion. Prove Theorem 1.5 by using the permutation expansion.
1.28 Prove that the determinant of a matrix equals the determinant of its transpose
using Laplace’s expansion and induction on the size of the matrix.
1.29 [Am. Math. Mon., Jun. 1949] Show that
Fn =
¯
¯
¯
¯
¯
¯
¯
¯
¯
1 −1 1 −1 1 −1 . . .
1 1 0 1 0 1 . . .
0 1 1 0 1 0 . . .
0 0 1 1 0 1 . . .
. . . . . . . . .
¯
¯
¯
¯
¯
¯
¯
¯
¯
where Fn is the n-th term of 1, 1, 2, 3, 5, . . . , x, y, x+y, . . . , the Fibonacci sequence,
and the determinant is of order n − 1.

Topic: Cramer’s Rule 331
Topic: Cramer’s Rule
We introduced determinant functions algebraically, looking for a formula to
decide whether a matrix is nonsingular, that is, whether a linear system has a
unique solution. Then we saw a geometric interpretation, that the determinant
function gives the size of the box with sides formed by the columns of the matrix.
Here we will see a nice formula that connects the two views.
Consider this system
x1 + 2x2 = 6
3x1 + x2 = 8
Rewriting in vector form
x1 ·
µ
1
3
¶
+ x2 ·
µ
2
1
¶
=
µ
6
8
¶
and picturing with parallelograms
2
1

1
3

x1 ·
1
3

+ x2 ·
2
1

gives a geometric interpretation of solving the linear system: by what factor x1
must we dilate the first vector, and by what factor x2 must we dilate the second
vector, to expand the small parallegram to fill the larger one?
Of course, we routinely find the answer with the algebraic manipulations of
Gauss’ method. Nonetheless, the geometry can give us some insights — compare
the sizes of these three shaded boxes.
2
1

1
3

2
1

x1 ·
1
3

2
1

6
8

The second box is formed from x1
¡1
3
¢
and
¡2
1
¢
, and one of the properties of the
size function (that is, the determinant function) is that the size of the second
box is therefore x1 times the size of the first box. Since the third box is formed
from x1
¡1
3
¢
+ x2
¡2
1
¢
and
¡2
1
¢
, and sizes are unchanged by side operations (that

is, the determinant is unchanged by adding x2 times the second column to the
first column), the size of the third box equals the size of the second box.
¯
¯
¯
¯
6 2
8 1
¯
¯
¯
¯ = x1 ·
¯
¯
¯
¯
1 2
3 1
¯
¯
¯
¯
Solving gives the value of one of the variables.
x1 =
¯
¯
¯
¯
6 2
8 1
¯
¯
¯
¯
¯
¯
¯
¯
1 2
3 1
¯
¯
¯
¯
=
−10
−5
= 2
The theorem that generalizes this example, Cramer’s Rule, is: if |A| 6= 0
then the system A~
x = ~
b has the unique solution xi = |Bi|/|A| where the matrix
Bi is formed from A by replacing column i with the vector ~
b. Exercise 3 asks
for a proof.
For instance, to solve this system for x2


1 0 4
2 1 −1
1 0 1




x1
x2
x3

 =


2
1
−1


we do this computation.
x2 =
¯
¯
¯
¯
¯
¯
1 2 4
2 1 −1
1 −1 1
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
1 0 4
2 1 −1
1 0 1
¯
¯
¯
¯
¯
¯
=
−18
−3
Cramer’s Rule, with practice, allows us to solve two equations/two unknown
systems by eye. It is also sometimes used for three equations/three unknowns
systems. But computing large determinants takes a long time so that solving
large systems by Cramer’s Rule is impractical.
Exercises
1 Use Cramer’s Rule to solve each for each of the variables.
(a)
x − y = 4
−x + 2y = −7
(b)
−2x + y = −2
x − 2y = −2
2 Use Cramer’s Rule to solve this system for z.
2x + y + z = 1
3x + z = 4
x − y − z = 2
3 Prove Cramer’s Rule.

Topic: Cramer’s Rule 333
4 Suppose that a linear system with as many equations as unknowns, and with
integer coefficients and constants, has a matrix of coefficients with determinant 1.
Prove that the entries in the solution are all integers. (Remark. This is often used
to invent linear systems for exercises. If an instructor makes the linear system with
this property then the solution is not some disagreeable fraction.)
5 Use Cramer’s Rule to give a formula for the solution of a two equation/two
unknown linear system.
6 Can Cramer’s Rule tell the difference between a system with no solutions and
one with infinitely many?

Topic: Speed of Calculating Determinants
The permutation expansion formula for computing determinants is useful for
proving theorems, but the method of using row operations is a much better for
finding the determinants of a large matrix. We can make this statement precise
by considering, as computer algorithm designers do, the number of arithmetic
operations that each method uses.
The speed of an algorithm is measured by finding how the time taken by
the computer grows as the size of its input data set grows. For instance, how
much longer will the algorithm take if we increase the size of the input data by
a factor of ten, say from a 1000 row matrix to a 10, 000 row matrix, or from
10, 000 to 100, 000? Does the time taken grow by a factor of ten or by a factor
of a hundred, or by a factor of a thousand? That is, is the time taken by the
algorithm proportional to the size of the data set, or to the square of that size,
or to the cube of that size, etc.?
Recall the permutation expansion formula for determinants.
¯
¯
¯
¯
¯
¯
¯
¯
¯
t1,1 t1,2 . . . t1,n
t2,1 t2,2 . . . t2,n
.
.
.
tn,1 tn,2 . . . tn,n
¯
¯
¯
¯
¯
¯
¯
¯
¯
=
X
permutations φ
t1,φ(1)t2,φ(2) · · · tn,φ(n) |Pφ|
= t1,φ1(1) · t2,φ1(2) · · · tn,φ1(n) |Pφ1
|
+ t1,φ2(1) · t2,φ2(2) · · · tn,φ2(n) |Pφ2
|
.
.
.
+ t1,φk(1) · t2,φk(2) · · · tn,φk(n) |Pφk
|
There are n! = n·(n−1)·(n−2) · · · 2·1 different n-permutations. For numbers
n of any size at all, this is a quite large number; for instance, even if n is only
10 then the expansion has 10! = 3, 628, 800 terms, all of which are obtained by
multiplying n entries together. This is a very large number of multiplications
(for instance, [Knuth] suggests 10! steps as a rough boundary for the limit
of practical calculation). The factorial function grows faster than the square
function. It grows faster than the cube function, the fourth power function,
or any polynomial function. (One way to see that the factorial function grows
faster than the square is to note that multiplying the first two factors in n!
gives n · (n − 1), which for large n is approximately n2
, and then multiplying
in more factors will make it even larger. The same argument works for the
cube function, etc.) So a computer that is programmed to use the permutation
expansion formula, and thus to perform a number of operations that is greater
than or equal to the factorial of the number of rows, would take times that grow
very quickly as the input data set grows.
In contrast, the time taken by the row reduction method does not grow so
fast. This fragment of row-reduction code is in the computer language FOR-
TRAN. The matrix is stored in the N ×N array A. For each ROW between 1
and N parts of the program not shown here have already found the pivot entry

Topic: Speed of Calculating Determinants 335
A(ROW, COL). Now the program pivots.
−PIVINV · ρROW + ρi
(This code fragment is for illustration only, and is incomplete. Nonetheless,
analysis of a finished versions, including all of the tests and subcases, is messier
but gives essentially the same result.)
PIVINV=1.0/A(ROW,COL)
DO 10 I=ROW+1, N
DO 20 J=I, N
A(I,J)=A(I,J)-PIVINV*A(ROW,J)
20 CONTINUE
10 CONTINUE
The outermost loop (not shown) runs through N − 1 rows. For each row, the
nested I and J loops shown perform arithmetic on the entries in A that are
below and to the right of the pivot entry. Assume that the pivot is found in
the expected place, that is, that COL = ROW . Then there are (N − ROW )2
entries below and to the right of the pivot. On average, ROW will be N/2.
Thus, we estimate that the arithmetic will be performed about (N/2)2
times,
that is, will run in a time proportional to the square of the number of equations.
Taking into account the outer loop that is not shown, we get the estimate that
the running time of the algorithm is proportional to the cube of the number of
equations.
Finding the fastest algorithm to compute the determinant is a topic of cur-
rent research. Algorithms are known that run in time between the second and
third power.
Speed estimates like these help us to understand how quickly or slowly an
algorithm will run. Algorithms that run in time proportional to the size of the
data set are fast, algorithms that run in time proportional to the square of the
size of the data set are less fast, but typically quite usable, and algorithms that
run in time proportional to the cube of the size of the data set are still reasonable
in speed. However, algorithms that run in time (greater than or equal to) the
factorial of the size of the data set are not practical.
There are other methods besides the two discussed here that are also used
for computation of determinants. Those lie outside of our scope. Nonetheless,
this contrast of the two methods for computing determinants makes the point
that although in principle they give the same answer, in practice the idea is to
select the one that is fast.
Exercises
Most of these problems presume access to a computer.
1 Computer systems generate random numbers (of course, these are only pseudo-
random, in that they are generated by an algorithm, but they pass a number of
reasonable statistical tests for randomness).
(a) Fill a 5×5 array with random numbers (say, in the range [0..1)). See if it is
singular. Repeat that experiment a few times. Are singular matrices frequent
or rare (in this sense)?

(b) Time your computer algebra system at finding the determinant of ten 5×5
arrays of random numbers. Find the average time per array. Repeat the prior
item for 15×15 arrays, 25×25 arrays, and 35×35 arrays. (Notice that, when an
array is singular, it can sometimes be found to be so quite quickly, for instance
if the first row equals the second. In the light of your answer to the first part,
do you expect that singular systems play a large role in your average?)
(c) Graph the input size versus the average time.
2 Compute the determinant of each of these by hand using the two methods dis-
cussed above.
(a)
¯
¯
¯
¯
2 1
5 −3
¯
¯
¯
¯ (b)
¯
¯
¯
¯
¯
3 1 1
−1 0 5
−1 2 −2
¯
¯
¯
¯
¯
(c)
¯
¯
¯
¯
¯
¯
¯
2 1 0 0
1 3 2 0
0 −1 −2 1
0 0 −2 1
¯
¯
¯
¯
¯
¯
¯
Count the number of multiplications and divisions used in each case, for each of
the methods. (On a computer, multiplications and divisions take much longer than
additions and subtractions, so algorithm designers worry about them more.)
3 What 10×10 array can you invent that takes your computer system the longest
to reduce? The shortest?
4 Write the rest of the FORTRAN program to do a straightforward implementation
of calculating determinants via Gauss’ method. (Don’t test for a zero pivot.)
Compare the speed of your code to that used in your computer algebra system.
5 The FORTRAN language specification requires that arrays be stored “by col-
umn”, that is, the entire first column is stored contiguously, then the second col-
umn, etc. Does the code fragment given take advantage of this, or can it be
rewritten to make it faster, by taking advantage of the fact that computer fetches
are faster from contiguous locations?

Topic: Projective Geometry 337
Topic: Projective Geometry
There are geometries other than the familiar Euclidean one. One such geometry
arose in art, where it was observed that what a viewer sees is not necessarily
what is there. This is Leonardo da Vinci’s masterpiece The Last Supper.
What is there in the room, for instance where the ceiling meets the left and
right walls, are lines that are parallel. However, what a viewer sees is lines
that, if extended, would intersect. The intersection point is called the vanishing
point. This aspect of perspective is also familiar as the image of a long stretch
of railroad tracks that appear to converge at the horizon.
To depict the room, da Vinci has adopted a model of how we see, of how we
project the three dimensional scene to a two dimensional image. This model is
only a first approximation — it does not take into account that our retina is
curved and our lens bends the light, that we have binocular vision, or that our
brain’s processing greatly affects what we see — but nonetheless it is interesting,
both artistically and mathematically.
The projection is not orthogonal, it is a central projection from a single
point, to the plane of the canvas.
A
B
C
(It is not an orthogonal projection since the line from the viewer to C is not
orthogonal to the image plane.) As the picture suggests, the operation of cen-
tral projection preserves some geometric properties — lines project to lines.
However, it fails to preserve some others — equal length segments can project
to segments of unequal length; the length of AB is greater than the length of
BC because the segment projected to AB is closer to the viewer and closer
things look bigger. The study of the effects of central projections is projective
geometry. We will see how linear algebra can be used in this study.

There are three cases of central projection. The first is the projection done
by a movie projector.
source S image I
projector P
We can think that each source point is “pushed” from the domain plane out-
ward to the image point in the codomain plane. This case of projection has a
somewhat different character than the second case, that of the artist “pulling”
the source back to the canvas.
image I source S
painter P
In the first case S is in the middle while in the second case I is in the middle.
One more configuration is possible, with P in the middle. An example of this
is when we use a pinhole to shine the image of a solar eclipse onto a piece of
paper.
pinhole P
image I
source S
We shall take each of the three to be a central projection by P of S to I.
To illustrate some of the geometric effects of these projections, consider again
the effect of railroad tracks that appear to converge to a point. We model this
with parallel lines in a domain plane S and a projection via a P to a codomain
plane I. (The dotted lines are parallel to S and I.)
S
I
P
All three projection cases appear here. The first picture below shows P acting
like a movie projector by pushing points from part of S out to image points on
the lower half of I. The middle picture shows P acting like the artist by pulling
points from another part of S back to image points in I. In the third picture, P
acts like the pinhole. This picture is the trickiest—the points that are projected
near to the vanishing point are the ones that are far out to the bottom left of
S. Points in S that are near to the vertical dotted line are sent high up on I.

P P P
There are two awkward things about this situation. The first is that neither of
the two points in the domain nearest to the vertical dotted line (see below) has
an image because a projection from those two is along the dotted line that is
parallel to the codomain plane (we sometimes say that these two are projected
“to infinity”). The second awkward thing is that the vanishing point in I isn’t
the image of any point from S because a projection to this point would be along
the dotted line that is parallel to the domain plane (we sometimes say that the
vanishing point is the image of a projection “from infinity”).
For a better model, put the projector P at the origin. Imagine that P is
covered by a glass hemispheric dome. As P looks outward, anything in the line
of vision is projected to the same spot on the dome. This includes things on
the line between P and the dome, as in the case of projection by the movie
projector. It includes things on the line further from P than the dome, as in
the case of projection by the painter. It also includes things on the line that lie
behind P, as in the case of projection by a pinhole.
` = {k ·
µ
1
2
3
¶
¯
¯ k ∈ R}
From this perspective P, all of the spots on the line are seen as the same point.
Accordingly, for any nonzero vector ~
v ∈ R3
, we define the associated point v
in the projective plane to be the set {k~
v
¯
¯ k ∈ R and k 6= 0} of nonzero vectors
lying on the same line through the origin as ~
v. To describe a projective point
we can give any representative member of the line, so that the projective point
shown above can be represented in any of these three ways.


1
2
3




1/3
2/3
1




−2
−4
−6



Each of these is a homogeneous coordinate vector for v.
This picture, and the above definition that arises from it, clarifies the de-
scription of central projection but there is something awkward about the dome
model: what if the viewer looks down? If we draw P’s line of sight so that
the part coming toward us, out of the page, goes down below the dome then
we can trace the line of sight backward, up past P and toward the part of the
hemisphere that is behind the page. So in the dome model, looking down gives
a projective point that is behind the viewer. Therefore, if the viewer in the
picture above drops the line of sight toward the bottom of the dome then the
projective point drops also and as the line of sight continues down past the
equator, the projective point suddenly shifts from the front of the dome to the
back of the dome. This discontinuity in the drawing means that we often have
to treat equatorial points as a separate case. That is, while the railroad track
discussion of central projection has three cases, the dome model has two.
We can do better than this. Consider a sphere centered at the origin. Any
line through the origin intersects the sphere in two spots, which are said to be
antipodal. Because we associate each line through the origin with a point in the
projective plane, we can draw such a point as a pair of antipodal spots on the
sphere. Below, the two antipodal spots are shown connected by a dotted line
to emphasize that they are not two different points, the pair of spots together
make one projective point.
While drawing a point as a pair of antipodal spots is not as natural as the one-
spot-per-point dome mode, on the other hand the awkwardness of the dome
model is gone, in that if as a line of view slides from north to south, no sudden
changes happen on the picture. This model of central projection is uniform —
the three cases are reduced to one.
So far we have described points in projective geometry. What about lines?
What a viewer at the origin sees as a line is shown below as a great circle, the
intersection of the model sphere with a plane through the origin.
(One of the projective points on this line is shown to bring out a subtlety.
Because two antipodal spots together make up a single projective point, the
great circle’s behind-the-paper part is the same set of projective points as its
in-front-of-the-paper part.) Just as we did with each projective point, we will
also describe a projective line with a triple of reals. For instance, the members

of this plane through the origin in R3
{


x
y
z


¯
¯ x + y − z = 0}
project to a line that we can described with the triple
¡
1 1 −1
¢
(we use row
vectors to typographically distinguish lines from points). In general, for any
nonzero three-wide row vector ~
L we define the associated line in the projective
plane, to be the set L = {k~
L
¯
¯ k ∈ R and k 6= 0} of nonzero multiples of ~
L.
The reason that this description of a line as a triple is convienent is that
in the projective plane, a point v and a line L are incident — the point lies
on the line, the line passes throught the point — if and only if a dot product
of their representatives v1L1 + v2L2 + v3L3 is zero (Exercise 4 shows that this
is independent of the choice of representatives ~
v and ~
L). For instance, the
projective point described above by the column vector with components 1, 2,
and 3 lies in the projective line described by
¡
1 1 −1
¢
, simply because any
vector in R3
whose components are in ratio 1 : 2 : 3 lies in the plane through the
origin whose equation is of the form 1k ·x+1k ·y −1k ·z = 0 for any nonzero k.
That is, the incidence formula is inherited from the three-space lines and planes
of which v and L are projections.
Thus, we can do analytic projective geometry. For instance, the projective
line L =
¡
1 1 −1
¢
has the equation 1v1 + 1v2 − 1v3 = 0, because points
incident on the line are characterized by having the property that their repre-
sentatives satisfy this equation. One difference from familiar Euclidean anlaytic
geometry is that in projective geometry we talk about the equation of a point.
For a fixed point like
v =


1
2
3


the property that characterizes lines through this point (that is, lines incident
on this point) is that the components of any representatives satisfy 1L1 +2L2 +
3L3 = 0 and so this is the equation of v.
This symmetry of the statements about lines and points brings up the Duality
Principle of projective geometry: in any true statement, interchanging ‘point’
with ‘line’ results in another true statement. For example, just as two distinct
points determine one and only one line, in the projective plane, two distinct
lines determine one and only one point. Here is a picture showing two lines that
cross in antipodal spots and thus cross at one projective point.
(∗)
Contrast this with Euclidean geometry, where two distinct lines may have a
unique intersection or may be parallel. In this way, projective geometry is
simpler, more uniform, than Euclidean geometry.

That simplicity is relevant because there is a relationship between the two
spaces: the projective plane can be viewed as an extension of the Euclidean
plane. Take the sphere model of the projective plane to be the unit sphere in
R3
and take Euclidean space to be the plane z = 1. This gives us a way of
viewing some points in projective space as corresponding to points in Euclidean
space, because all of the points on the plane are projections of antipodal spots
from the sphere.
(∗∗)
Note though that projective points on the equator don’t project up to the plane.
Instead, these project ‘out to infinity’. We can thus think of projective space
as consisting of the Euclidean plane with some extra points adjoined — the
Euclidean plane is embedded in the projective plane. These extra points, the
equatorial points, are the ideal points or points at infinity and the equator is the
ideal line or line at infinity (note that it is not a Euclidean line, it is a projective
line).
The advantage of the extension to the projective plane is that some of the
awkwardness of Euclidean geometry disappears. For instance, the projective
lines shown above in (∗) cross at antipodal spots, a single projective point, on
the sphere’s equator. If we put those lines into (∗∗) then they correspond to
Euclidean lines that are parallel. That is, in moving from the Euclidean plane to
the projective plane, we move from having two cases, that lines either intersect
or are parallel, to having only one case, that lines intersect (possibly at a point
at infinity).
The projective case is nicer in many ways than the Euclidean case but has
the problem that we don’t have the same experience or intuitions with it. That’s
one advantage of doing analytic geometry, where the equations can lead us to
the right conclusions. Analytic projective geometry uses linear algebra. For
instance, for three points of the projective plane t, u, and v, setting up the
equations for those points by fixing vectors representing each, shows that the
three are collinear — incident in a single line — if and only if the resulting three-
equation system has infinitely many row vector solutions representing that line.
That, in turn, holds if and only if this determinant is zero.
¯
¯
¯
¯
¯
¯
t1 u1 v1
t2 u2 v2
t3 u3 v3
¯
¯
¯
¯
¯
¯
Thus, three points in the projective plane are collinear if and only if any three
representative column vectors are linearly dependent. Similarly (and illustrating
the Duality Principle), three lines in the projective plane are incident on a
single point if and only if any three row vectors representing them are linearly
dependent.

The following result is more evidence of the ‘niceness’ of the geometry of the
projective plane, compared to the Euclidean case. These two triangles are said
to be in perspective from P because their corresponding vertices are collinear.
O
T1
U1 V1
T2
U2
V2
Desargue’s Theorem is that when the three pairs of corresponding sides — T1U1
and T2U2, T1V1 and T2V2, U1V1 and U2V2 — are extended, they intersect
and further, those three intersection points are collinear.
We will prove this theorem, using projective geometry. (These are drawn as
Euclidean figures because it is the more familiar image. To consider them as
projective figures, we can imagine that, although the line segments shown are
parts of great circles and so are curved, the model has such a large radius
compared to the size of the figures that the sides appear in this sketch to be
straight.)
For this proof, we need a preliminary lemma [Coxeter]: if W, X, Y , Z are
four points in the projective plane (no three of which are collinear) then there
is a basis B for R3
such that
RepB(~
w) =


1
0
0

 RepB(~
x) =


0
1
0

 RepB(~
y) =


0
0
1

 RepB(~
z) =


1
1
1


where ~
w, ~
x, ~
y, ~
z are homogeneous coordinate vectors for the projective points.
The proof is straightforward. Because W, X, Y are not on the same projective
line, any homogeneous coordinate vectors ~
w0, ~
x0, ~
y0 do not line on the same
plane through the origin in R3
and so form a spanning set for R3
. Thus any
homogeneous coordinate vector for Z can be written as a combination ~
z0 =
a · ~
w0 + b · ~
x0 + c · ~
y0. Then ~
w = a · ~
w0, ~
x = b · ~
x0, ~
y = c · ~
y0, and ~
z = ~
z0 will do,
for B = h~
w, ~
x, ~
yi.
Now, to prove of Desargue’s Theorem, use the lemma to fix homogeneous
coordinate vectors and a basis.
RepB(~
t1) =


1
0
0

 RepB(~
u1) =


0
1
0

 RepB(~
v1) =


0
0
1

 RepB(~
o) =


1
1
1



Because the projective point T2 is incident on the projective line OT1, any
homogeneous coordinate vector for T2 lies in the plane through the origin in R3
that is spanned by homogeneous coordinate vectors of O and T1:
RepB(~
t2) = a


1
1
1

 + b


1
0
0


for some scalars a and b. That is, the homogenous coordinate vectors of members
T2 of the line OT1 are of the form on the left below, and the forms for U2 and
V2 are similar.
RepB(~
t2) =


t2
1
1

 RepB(~
u2) =


1
u2
1

 RepB(~
v2) =


1
1
v2


The projective line T1U1 is the image of a plane through the origin in R3
. A
quick way to get its equation is to note that any vector in it is linearly dependent
on the vectors for T1 and U1 and so this determinant is zero.
¯
¯
¯
¯
¯
¯
1 0 x
0 1 y
0 0 z
¯
¯
¯
¯
¯
¯
= 0 =⇒ z = 0
The equation of the plane in R3
whose image is the projective line T2U2 is this.
¯
¯
¯
¯
¯
¯
t2 1 x
1 u2 y
1 1 z
¯
¯
¯
¯
¯
¯
= 0 =⇒ (1 − u2) · x + (1 − t2) · y + (t2u2 − 1) · z = 0
Finding the intersection of the two is routine.
T1U1 ∩ T2U2 =


t2 − 1
1 − u2
0


(This is, of course, the homogeneous coordinate vector of a projective point.)
The other two intersections are similar.
T1V1 ∩ T2V2 =


1 − t2
0
v2 − 1

 U1V1 ∩ U2V2 =


0
u2 − 1
1 − v2


The proof is finished by noting that these projective points are on one projective
line because the sum of the three homogeneous coordinate vectors is zero.
Every projective theorem has a translation to a Euclidean version, although
the Euclidean result is often messier to state and prove. Desargue’s theorem
illustrates this. In the translation to Euclidean space, the case where O lies on

the ideal line must be treated separately for then the lines T1T2, U1U2, and V1V2
are parallel.
The parenthetical remark following the statement of Desargue’s Theorem
suggests thinking of the Euclidean pictures as figures from projective geometry
for a model of very large radius. That is, just as a small area of the earth appears
flat to people living there, the projective plane is also ‘locally Euclidean’.
Although its local properties are the familiar Euclidean ones, there is a global
property of the projective plane that is quite different. The picture below shows
a projective point. At that point is drawn an xy-axis. There is something
interesting about the way this axis appears at the antipodal ends of the sphere.
In the northern hemisphere, where the axis are drawn in black, a right hand put
down with fingers on the x-axis will have the thumb point along the y-axis. But
the antipodal axis, drawn in gray, has just the opposite: a right hand placed with
its fingers on the x-axis will have the thumb point in the wrong way, instead,
a left hand comes out correct. Briefly, the projective plane is not orientable: in
this geometry, left and right handedness are not fixed properties of figures.
The sequence of pictures below dramatizes this non-orientability. They sketch a
trip around this space in the direction of the y part of the xy-axis. (This trip is
not halfway around, it is a circuit, because antipodal spots are not two points,
they are one point, and so the antipodal spots in the third picture below form
the same projective point as the antipodal spots in the picture above.)
=⇒ =⇒
At the end of the circuit, the x arrow from the xy-axis sticks out in the other
direction. Another example of the same effect is that a clockwise spiral, on tak-
ing the same trip, would switch to counterclockwise. (This orientation reversal
appeared earlier, in the pinhole/eclipse picture.)
This exhibition of the existence of a non-orientable space raises the question
of whether our space is orientable: is is possible for a right handed astronaut to
take a trip to Mars and return left handed? An excellent nontechnical reference
is [Gardner]. An intriguing science fiction story about orientation reversal is
[Clarke].
So projective geometry is mathematically interesting and rewarding, in addi-
tion to the natural way in which it arises in art. It is more than just a technical
device to shorten some proofs. For an overview, see [Courant Robbins]. The
approach we’ve taken here, the analytic approach, leads to quick theorems and
— most importantly for us — illustrates the power of linear algebra (see [Hanes],

[Ryan], and [Eggar]). But, note that the other possible approach, the synthetic
approach of deriving the results from an axiom system, is both extraordinarily
beautiful and is also the historical route of development. Two fine sources for
this approach are [Coxeter] or [Seidenberg]. An interesting, easy, application is
[Davies]
Exercises
1 What is the equation of this point?
Ã
1
0
0
!
2 (a) Find the line incident on these points in the projective plane.
Ã
1
2
3
!
,
Ã
4
5
6
!
(b) Find the point incident on both of these projective lines.
¡
1 2 3
¢
,
¡
4 5 6
¢
3 Find the formula for the line incident on two projective points. Find the formula
for the point incident on two projective lines.
4 Prove that the definition of incidence is independent of the choice of the rep-
resentatives of p and L. That is, if p1, p2, p3, and q1, q2, q3 are two triples of
homogeneous coordinates for p, and L1, L2, L3, and M1, M2, M3 are two triples
of homogeneous coordinates for L, prove that p1L1 + p2L2 + p3L3 = 0 if and only
if q1M1 + q2M2 + q3M3 = 0.
5 Give a drawing to show that central projection does not preserve circles, that a
circle may project to an ellipse. Can a (non-circular) ellipse project to a circle?
6 Give the formula for the correspondence between the non-equatorial part of the
antipodal modal of the projective plane, and the plane z = 1.
7 (Pappus’s Theorem) Assume that T0, U0, and V0 are collinear and that T1, U1,
and V1 are collinear. Consider these three points: (i) the intersection V2 of the lines
T0U1 and T1U0, (ii) the intersection U2 of the lines T0V1 and T1V0, and (iii) the
intersection T2 of U0V1 and U1V0.
(a) Draw a (Euclidean) picture.
(b) Apply the lemma used in Desargue’s Theorem to get simple homogeneous
coordinate vectors for the T’s and V0.
(c) Find the resulting homogeneous coordinate vectors for U’s (these must each
involve a parameter as, e.g., U0 could be anywhere on the T0V0 line).
(d) Find the resulting homogeneous coordinate vectors for V1. (Hint: it involves
two parameters.)
(e) Find the resulting homogeneous coordinate vectors for V2. (It also involves
two parameters.)
(f) Show that the product of the three parameters is 1.
(g) Verify that V2 is on the T2U2 line..

Chapter 5
Similarity
While studying matrix equivalence, we have shown that for any any homomor-
phism there are bases B and D such that the representation matrix has a block
partial-identity form.
RepB,D(h) =
µ
Identity Zero
Zero Zero
¶
This representation lets us think of the map as sending c1
~
β1 + · · · + cn
~
βn to
c1
~
δ1 + · · · + ck
~
δk +~
0 + · · · +~
0, where n is the dimension of the domain and k is
the dimension of the range. So, under this representation the action of the map
is easy to understand because most of the matrix entries are zero.
This chapter considers the special case where the domain and the codomain
are equal, that is, where the homomorphism is a transformation. In this case
we naturally ask to find a single basis B so that RepB,B(t) is as simple as
possible (we will take ‘simple’ to mean that it has many zeroes). A matrix
having the above block partial-identity form is not always possible here. But,
we will develop a form that comes close — the representation will be nearly
diagonal.
5.I Complex Vector Spaces
This chapter will require that we factor polynomials. Of course, many polyno-
mials do not factor over the real numbers. For instance, x2
+ 1 does not factor
into the product of two linear polynomials with real coefficients. For that rea-
son, we shall from now on take our scalars from the complex numbers. In this
chapter the c’s in c1~
v1 + c2~
v2 + · · · + cn~
vn will be complex numbers.
So we are shifting from studying vector spaces over the real numbers to
vector spaces over the complex numbers. As a consequence, in this chapter
vector and matrix entries are complex. (The real numbers are a subset of the
complex numbers, and a quick glance through this chapter shows that most of
347

348 Chapter 5. Similarity
the examples use only pure-real numbers. Nonetheless, the critical theorems
require the use of the complex number system to go through.) Therefore, the
first section of this chapter is a quick review of complex numbers.
The idea of taking scalars from a structure other than the real numbers is
an interesting one. However, in this book we are moving to this more general
context only for the pragmatic reason that we must do so in order to develop
the representation. We will not go into using other sets of scalars in more
detail because it could distract from our task. For the more general approach,
delightful presentations are in [Halmos] or [Hoffman Kunze].
5.I.1 Factoring and Complex Numbers; A Review
This subsection is a review only and we take the main results as known.
For proofs, see [Birkhoff MacLane] or [Ebbinghaus].
Just as integers have a division operation — e.g., ‘4 goes 5 times into 21
with remainder 1’ — so do polynomials.
1.1 Theorem (Division Theorem for Polynomials) Let c(x) be a poly-
nomial. If m(x) is a non-zero polynomial then there are quotient and remainder
polynomials q(x) and r(x) such that
c(x) = m(x) · q(x) + r(x)
where the degree of r(x) is strictly less than the degree of m(x).
In this book constant polynomials, including the zero polynomial, are said to
have degree 0. (This is not the standard definition, but it is convienent here.)
The point of the integer division statement ‘4 goes 5 times into 21 with
remainder 1’ is that the remainder is less than 4 — while 4 goes 5 times, it does
not go 6 times. In the same way, the point of the polynomial division statement
is its final clause.
1.2 Example If c(x) = 2x3
− 3x2
+ 4x and m(x) = x2
+ 1 then q(x) = 2x − 3
and r(x) = 2x + 3. Note that r(x) has a lower degree than m(x).
1.3 Corollary The remainder when c(x) is divided by x − λ is the constant
polynomial r(x) = c(λ).
Proof. The remainder must be a constant polynomial. because it is of degree
less than the divisor x − λ, To determine the constant, taking m(x) from the
theorem to be x − λ and substituting λ for x yields c(λ) = (λ − λ) · q(λ) +
r(x). QED
If a divisor m(x) goes into a dividend c(x) evenly, meaning that r(x) is the
zero polynomial, then m(x) is a factor of c(x). Any root of the factor (any
λ ∈ R such that m(λ) = 0) is a root of c(x) since c(λ) = m(λ) · q(λ) = 0. The
prior corollary immediately yields the following converse.

Section I. Complex Vector Spaces 349
1.4 Corollary If λ is a root of the polynomial c(x) then x − λ divides c(x)
evenly, that is, x − λ is a factor of c(x).
Finding the roots and factors of a high-degree polynomial can be hard. But
for second-degree polynomials we have the quadratic formula: the roots of ax2
+
bx + c are
λ1 =
−b +
√
b2 − 4ac
2a
λ2 =
−b −
√
b2 − 4ac
2a
(if the discriminant b2
−4ac is negative then the polynomial has no real number
roots). A polynomial that cannot be factored into two lower-degree polynomials
with real number coefficients is irreducible over the reals.
1.5 Theorem Any constant or linear polynomial is irreducible over the reals.
A quadratic polynomial is irreducible over the reals if and only if its discriminant
is negative. No cubic or higher-degree polynomial is irreducible over the reals.
1.6 Corollary Any polynomial with real coefficients can be factored into linear
and irreducible quadratic polynomials. That factorization is unique; any two
factorizations have the same powers of the same factors.
Note the analogy with the prime factorization of integers. In both cases, the
uniqueness clause is very useful.
1.7 Example Because of uniqueness we know, without multiplying them out,
that (x + 3)2
(x2
+ 1)3
does not equal (x + 3)4
(x2
+ x + 1)2
.
1.8 Example By uniqueness, if c(x) = m(x) · q(x) then where c(x) = (x −
3)2
(x + 2)3
and m(x) = (x − 3)(x + 2)2
, we know that q(x) = (x − 3)(x + 2).
While x2
+ 1 has no real roots and so doesn’t factor over the real numbers,
if we imagine a root — traditionally denoted i so that i2
+ 1 = 0 — then x2
+ 1
factors into a product of linears (x − i)(x + i).
So we adjoin this root i to the reals and close the new system with respect
to addition, multiplication, etc. (i.e., we also add 3 + i, and 2i, and 3 + 2i, etc.,
putting in all linear combinations of 1 and i). We then get a new structure, the
complex numbers, denoted C.
In C we can factor (obviously, at least some) quadratics that would be ir-
reducible if we were to stick to the real numbers. Surprisingly, in C we can
not only factor x2
+ 1 and its close relatives, we can factor any quadratic. Any
quadratic polynomial factors over the complex numbers.
ax2
+ bx + c = a ·
¡
x −
−b +
√
b2 − 4ac
2a
¢
·
¡
x −
−b −
√
b2 − 4ac
2a
¢
1.9 Example The second degree polynomial x2
+x+1 factors over the complex
numbers into the product of two first degree polynomials.
¡
x −
−1 +
√
−3
2
¢¡
x −
−1 −
√
−3
2
¢
=
¡
x − (−
1
2
+
√
3
2
i)
¢¡
x − (−
1
2
−
√
3
2
i)
¢

1.10 Corollary (Fundamental Theorem of Algebra) Polynomials with
complex coefficients factor into linear polynomials with complex coefficients.
The factorization is unique.
5.I.2 Complex Representations
Recall the definitions of the complex number operations.
(a + bi) + (c + di) = (a + c) + (b + d)i
(a + bi)(c + di) = ac + adi + bci + bd(−1) = (ac − bd) + (ad + bc)i
2.1 Example For instance, (1−2i) + (5+4i) = 6+2i and (2−3i)(4−0.5i) =
6.5 − 13i.
Handling scalar operations with those rules, all of the operations that we’ve
covered for real vector spaces carry over unchanged.
2.2 Example Matrix multiplication is the same, although the scalar arith-
metic involves more bookkeeping.
µ
1 + 1i 2 − 0i
i −2 + 3i
¶ µ
1 + 0i 1 − 0i
3i −i
¶
=
µ
(1 + 1i) · (1 + 0i) + (2 − 0i) · (3i) (1 + 1i) · (1 − 0i) + (2 − 0i) · (−i)
(i) · (1 + 0i) + (−2 + 3i) · (3i) (i) · (1 − 0i) + (−2 + 3i) · (−i)
¶
=
µ
1 + 7i 1 − 1i
−9 − 5i 3 + 3i
¶
Everything else from prior chapters that we can, we shall also carry over
unchanged. For instance, we shall call this
h





1 + 0i
0 + 0i
.
.
.
0 + 0i





, . . . ,





0 + 0i
0 + 0i
.
.
.
1 + 0i





i
the standard basis for Cn
as a vector space over C and again denote it En.

Section II. Similarity 351
5.II Similarity
5.II.1 Definition and Examples
We’ve defined H and Ĥ to be matrix-equivalent if there are nonsingular
matrices P and Q such that Ĥ = PHQ. That definition is motivated by this
diagram
Vw.r.t. B
h
−
−
−
−
→
H
Ww.r.t. D
id


y id


y
Vw.r.t. B̂
h
−
−
−
−
→
Ĥ
Ww.r.t. D̂
showing that H and Ĥ both represent h but with respect to different pairs of
bases. We now specialize that setup to the case where the codomain equals the
domain, and where the codomain’s basis equals the domain’s basis.
Vw.r.t. B
t
−
−
−
−
→ Vw.r.t. B
id


y id


y
Vw.r.t. D
t
−
−
−
−
→ Vw.r.t. D
To move from the lower left to the lower right we can either go straight over, or
up, over, and then down. In matrix terms,
RepD,D(t) = RepB,D(id) RepB,B(t)
¡
RepB,D(id)
¢−1
(recall that a representation of composition like this one reads right to left).
1.1 Definition The matrices T and S are similar if there is a nonsingular P
such that T = PSP−1
.
Since nonsingular matrices are square, the similar matrices T and S must be
square and of the same size.
1.2 Example With these two,
P =
µ
2 1
1 1
¶
S =
µ
2 −3
1 −1
¶
calculation gives that S is similar to this matrix.
T =
µ
0 −1
1 1
¶
1.3 Example The only matrix similar to the zero matrix is itself: PZP−1
=
PZ = Z. The only matrix similar to the identity matrix is itself: PIP−1
=
PP−1
= I.

Since matrix similarity is a special case of matrix equivalence, if two ma-
trices are similar then they are equivalent. What about the converse: must
matrix equivalent square matrices be similar? The answer is no. The prior
example shows that the similarity classes are different from the matrix equiv-
alence classes, because the matrix equivalence class of the identity consists of
all nonsingular matrices of that size. Thus, for instance, these two are matrix
equivalent but not similar.
T =
µ
1 0
0 1
¶
S =
µ
1 2
0 3
¶
So some matrix equivalence classes split into two or more similarity classes —
similarity gives a finer partition than does equivalence. This picture shows some
matrix equivalence classes subdivided into similarity classes.
All square matrices:
%
$
Ã
!
¿
À
. . .
.A
B.
A matrix-equivalent to,
but not similar to, B.
To understand the similarity relation we shall study the similarity classes.
We approach this question in the same way that we’ve studied both the row
equivalence and matrix equivalence relations, by finding a canonical form for
representatives∗
of the similarity classes, called Jordan form. With this canon-
ical form, we can decide if two matrices are similar by checking whether they
reduce to the same representative. We’ve also seen with both row equivalence
and matrix equivalence that a canonical form gives us insight into the ways in
which members of the same class are alike (e.g., two identically-sized matrices
are matrix equivalent if and only if they have the same rank).
Exercises
1.4 For
S =
µ
1 3
−2 −6
¶
T =
µ
0 0
−11/2 −5
¶
P =
µ
4 2
−3 2
¶
check that T = PSP−1
.
X 1.5 Example 1.3 shows that the only matrix similar to a zero matrix is itself and
that the only matrix similar to the identity is itself.
(a) Show that the 1×1 matrix (2), also, is similar only to itself.
(b) Is a matrix of the form cI for some scalar c similar only to itself?
(c) Is a diagonal matrix similar only to itself?
1.6 Show that these matrices are not similar.
Ã
1 0 4
1 1 3
2 1 7
! Ã
1 0 1
0 1 1
3 1 2
!
∗ More information on representatives is in the appendix.

1.7 Consider the transformation t: P2 → P2 described by x2
7→ x + 1, x 7→ x2
− 1,
and 1 7→ 3.
(a) Find T = RepB,B(t) where B = hx2
, x, 1i.
(b) Find S = RepD,D(t) where D = h1, 1 + x, 1 + x + x2
i.
(c) Find the matrix P such that T = PSP−1
.
X 1.8 Exhibit an nontrivial similarity relationship in this way: let t: C2
→ C2
act by
µ
1
2
¶
7→
µ
3
0
¶ µ
−1
1
¶
7→
µ
−1
2
¶
and pick two bases, and represent t with respect to then T = RepB,B(t) and
S = RepD,D(t). Then compute the P and P−1
to change bases from B to D and
back again.
1.9 Explain Example 1.3 in terms of maps.
X 1.10 Are there two matrices A and B that are similar while A2
and B2
are not
similar?
X 1.11 Prove that if two matrices are similar and one is invertible then so is the
other.
X 1.12 Show that similarity is an equivalence relation.
1.13 Consider a matrix representing, with respect to some B, B, reflection across
the x-axis in R2
. Consider also a matrix representing, with respect to some D, D,
reflection across the y-axis. Must they be similar?
1.14 Prove that similarity preserves determinants and rank. Does the converse
hold?
1.15 Is there a matrix equivalence class with only one matrix similarity class inside?
One with infinitely many similarity classes?
1.16 Can two different diagonal matrices be in the same similarity class?
X 1.17 Prove that if two matrices are similar then their k-th powers are similar when
k 0. What if k ≤ 0?
X 1.18 Let p(x) be the polynomial cnxn
+ · · · + c1x + c0. Show that if T is similar to
S then p(T) = cnTn
+ · · · + c1T + c0I is similar to p(S) = cnSn
+ · · · + c1S + c0I.
1.19 List all of the matrix equivalence classes of 1×1 matrices. Also list the sim-
ilarity classes, and describe which similarity classes are contained inside of each
matrix equivalence class.
1.20 Does similarity preserve sums?
1.21 Show that if T − λI and N are similar matrices then T and N + λI are also
similar.
5.II.2 Diagonalizability
The prior subsection defines the relation of similarity and shows that, al-
though similar matrices are necessarily matrix equivalent, the converse does not
hold. Some matrix-equivalence classes break into two or more similarity classes
(the nonsingular n×n matrices, for instance). This means that the canoni-
cal form for matrix equivalence, a block partial-identity, cannot be used as a

canonical form for matrix similarity because the partial-identities cannot be in
more than one similarity class, so there are similarity classes without one. This
picture illustrates. As earlier in this book, class representatives are shown with
stars.
Equivalence classes
subdivided into
similarity classes. %
$
Ã
!
¿
À
. . .
?
?
?
?
?
?
? ?
?
?
?
This finer partition needs
more representatives.
We are developing a canonical form for representatives of the similarity classes.
We naturally try to build on our previous work, meaning first that the partial
identity matrices should represent the similarity classes into which they fall,
and beyond that, that the representatives should be as simple as possible. The
simplest extension of the partial-identity form is a diagonal form.
2.1 Definition A transformation is diagonalizable if it has a diagonal repre-
sentation with respect to the same basis for the codomain as for the domain. A
diagonalizable matrix is one that is similar to a diagonal matrix: T is diagonal-
izable if there is a nonsingular P such that PTP−1
is diagonal.
µ
4 −2
1 1
¶
is diagonalizable.
µ
2 0
0 3
¶
=
µ
−1 2
1 −1
¶ µ
4 −2
1 1
¶ µ
−1 2
1 −1
¶−1
2.3 Example Not every matrix is diagonalizable. The square of
N =
µ
0 0
1 0
¶
is the zero matrix. Thus, for any map n that N represents (with respect to the
same basis for the domain as for the codomain), the composition n ◦ n is the
zero map. This implies that no such map n can be diagonally represented (with
respect to any B, B) because no power of a nonzero diagonal matrix is zero.
That is, there is no diagonal matrix in N’s similarity class.
That example shows that a diagonal form will not do for a canonical form —
we cannot find a diagonal matrix in each matrix similarity class. However, the
canonical form that we are developing has the property that if a matrix can
be diagonalized then the diagonal matrix is the canonical representative of the
similarity class. The next result characterizes which maps can be diagonalized.

2.4 Corollary A transformation t is diagonalizable if and only if there is a
basis B = h~
β1, . . . , ~
βni and scalars λ1, . . . , λn such that t(~
βi) = λi
~
βi for each i.
Proof. This follows from the definition by considering a diagonal representation
matrix.
RepB,B(t) =




.
.
.
.
.
.
RepB(t(~
β1)) · · · RepB(t(~
βn))
.
.
.
.
.
.



 =



λ1 0
.
.
.
...
.
.
.
0 λn



This representation is equivalent to the existence of a basis satisfying the stated
conditions simply by the definition of matrix representation. QED
2.5 Example To diagonalize
T =
µ
3 2
0 1
¶
we take it as the representation of a transformation with respect to the standard
basis T = RepE2,E2
(t) and we look for a basis B = h~
β1, ~
β2i such that
RepB,B(t) =
µ
λ1 0
0 λ2
¶
that is, such that t(~
β1) = λ1 and t(~
β2) = λ2.
µ
3 2
0 1
¶
~
β1 = λ1 · ~
β1
µ
3 2
0 1
¶
~
β2 = λ2 · ~
β2
We are looking for scalars x such that this equation
µ
3 2
0 1
¶ µ
b1
b2
¶
= x ·
µ
b1
b2
¶
has solutions b1 and b2, which are not both zero. Rewrite that as a linear system.
(3 − x) · b1 + 2 · b2 = 0
(1 − x) · b2 = 0
(∗)
In the bottom equation the two numbers multiply to give zero only if at least
one of them is zero so there are two possibilities, b2 = 0 and x = 1. In the b2 = 0
possibility, the first equation gives that either b1 = 0 or x = 3. Since the case
of both b1 = 0 and b2 = 0 is disallowed, we are left looking at the possibility of
x = 3. With it, the first equation in (∗) is 0 · b1 + 2 · b2 = 0 and so associated
with 3 are vectors with a second component of zero and a first component that
is free.
µ
3 2
0 1
¶ µ
b1
0
¶
= 3 ·
µ
b1
0
¶

That is, one solution to (∗) is λ1 = 3, and we have a first basis vector.
~
β1 =
µ
1
0
¶
In the x = 1 possibility, the first equation in (∗) is 2 · b1 + 2 · b2 = 0, and so
associated with 1 are vectors whose second component is the negative of their
first component.
µ
3 2
0 1
¶ µ
b1
−b1
¶
= 1 ·
µ
b1
−b1
¶
Thus, another solution is λ2 = 1 and a second basis vector is this.
~
β2 =
µ
1
−1
¶
To finish, drawing the similarity diagram
R2
w.r.t. E2
t
−
−
−
−
→
T
R2
w.r.t. E2
id


y id


y
R2
w.r.t. B
t
−
−
−
−
→
D
R2
w.r.t. B
and noting that the matrix RepB,E2
(id) is easy leads to this diagonalization.
µ
3 0
0 1
¶
=
µ
1 1
0 −1
¶−1 µ
3 2
0 1
¶ µ
1 1
0 −1
¶
In the next subsection, we will expand on that example by considering more
closely the property of Corollary 2.4. This includes seeing another way, the way
that we will routinely use, to find the λ’s.
Exercises
X 2.6 Repeat Example 2.5 for the matrix from Example 2.2.
2.7 Diagonalize these upper triangular matrices.
(a)
µ
−2 1
0 2
¶
(b)
µ
5 4
0 1
¶
X 2.8 What form do the powers of a diagonal matrix have?
2.9 Give two same-sized diagonal matrices that are not similar. Must any two
different diagonal matrices come from different similarity classes?
2.10 Give a nonsingular diagonal matrix. Can a diagonal matrix ever be singular?
X 2.11 Show that the inverse of a diagonal matrix is the diagonal of the the inverses,
if no element on that diagonal is zero. What happens when a diagonal entry is
zero?

2.12 The equation ending Example 2.5
µ
1 1
0 −1
¶−1 µ
3 2
0 1
¶ µ
1 1
0 −1
¶
=
µ
3 0
0 1
¶
is a bit jarring because for P we must take the first matrix, which is shown as an
inverse, and for P−1
we take the inverse of the first matrix, so that the two −1
powers cancel and this matrix is shown without a superscript −1.
(a) Check that this nicer-appearing equation holds.
µ
3 0
0 1
¶
=
µ
1 1
0 −1
¶ µ
3 2
0 1
¶ µ
1 1
0 −1
¶−1
(b) Is the previous item a coincidence? Or can we always switch the P and the
P−1
?
2.13 Show that the P used to diagonalize in Example 2.5 is not unique.
2.14 Find a formula for the powers of this matrix Hint: see Exercise 8.
µ
−3 1
−4 2
¶
X 2.15 Diagonalize these.
(a)
µ
1 1
0 0
¶
(b)
µ
0 1
1 0
¶
2.16 We can ask how diagonalization interacts with the matrix operations. Assume
that t, s: V → V are each diagonalizable. Is ct diagonalizable for all scalars c?
What about t + s? t ◦ s?
X 2.17 Show that matrices of this form are not diagonalizable.
µ
1 c
0 1
¶
c 6= 0
2.18 Show that each of these is diagonalizable.
(a)
µ
1 2
2 1
¶
(b)
µ
x y
y z
¶
x, y, z scalars
5.II.3 Eigenvalues and Eigenvectors
In this subsection we will focus on the property of Corollary 2.4.
3.1 Definition A transformation t: V → V has a scalar eigenvalue λ if there
is a nonzero eigenvector ~
ζ ∈ V such that t(~
ζ) = λ · ~
ζ.
(“Eigen” is German for “characteristic of” or “peculiar to”; some authors call
these characteristic values and vectors. No authors call them “peculiar”.)
3.2 Example The projection map


x
y
z

 π
7−→


x
y
0

 x, y, z ∈ C

has an eigenvalue of 1 associated with any eigenvector of the form


x
y
0


where x and y are non-0 scalars. On the other hand, 2 is not an eigenvalue of
π since no non-~
0 vector is doubled.
That example shows why the ‘non-~
0’ appears in the definition. Disallowing
~
0 as an eigenvector eliminates trivial eigenvalues.
3.3 Example The only transformation on the trivial space {~
0 } is ~
0 7→ ~
0. This
map has no eigenvalues because there are no non-~
0 vectors ~
v mapped to a scalar
multiple λ · ~
v of themselves.
3.4 Example Consider the homomorphism t: P1 → P1 given by c0 + c1x 7→
(c0 +c1)+(c0 +c1)x. The range of t is one-dimensional. Thus an application of
t to a vector in the range will simply rescale that vector: c + cx 7→ (2c) + (2c)x.
That is, t has an eigenvalue of 2 associated with eigenvectors of the form c + cx
where c 6= 0.
This map also has an eigenvalue of 0 associated with eigenvectors of the form
c − cx where c 6= 0.
3.5 Definition A square matrix T has a scalar eigenvalue λ associated with
the non-~
0 eigenvector ~
ζ if T ~
ζ = λ · ~
ζ.
3.6 Remark Although this extension from maps to matrices is obvious, there
is a point that must be made. Eigenvalues of a map are also the eigenvalues of
matrices representing that map, and so similar matrices have the same eigen-
values. But the eigenvectors are different — similar matrices need not have the
same eigenvectors.
For instance, consider again the transformation t: P1 → P1 given by c0 +
c1x 7→ (c0+c1)+(c0+c1)x. It has an eigenvalue of 2 associated with eigenvectors
of the form c + cx where c 6= 0. If we represent t with respect to B = h1 +
1x, 1 − 1xi
T = RepB,B(t) =
µ
2 0
0 0
¶
then 2 is an eigenvalue of T, associated with these eigenvectors.
{
µ
c0
c1
¶
¯
¯
µ
2 0
0 0
¶ µ
c0
c1
¶
=
µ
2c0
2c1
¶
} = {
µ
c0
0
¶
¯
¯ c0 ∈ C, c0 6= 0}
On the other hand, representing t with respect to D = h2 + 1x, 1 + 0xi gives
S = RepD,D(t) =
µ
3 0
−3 2
¶

and the eigenvectors of S associated with the eigenvalue 2 are these.
{
µ
c0
c1
¶
¯
¯
µ
3 0
−3 2
¶ µ
c0
c1
¶
=
µ
2c0
2c1
¶
} = {
µ
0
c1
¶
¯
¯ c1 ∈ C, c1 6= 0}
Thus similar matrices can have different eigenvectors.
Here is an informal description of what’s happening. The underlying trans-
formation doubles the eigenvectors ~
v 7→ 2·~
v. But when the matrix representing
the transformation is T = RepB,B(t) then it “assumes” that column vectors are
representations with respect to B. In contrast, S = RepD,D(t) “assumes” that
column vectors are representations with respect to D. So the vectors that get
doubled by each matrix look different.
The next example illustrates the basic tool for finding eigenvectors and eigen-
values.
3.7 Example What are the eigenvalues and eigenvectors of this matrix?
T =


1 2 1
2 0 −2
−1 2 3


To find the scalars x such that T ~
ζ = x~
ζ for non-~
0 eigenvectors ~
ζ, bring every-
thing to the left-hand side


1 2 1
2 0 −2
−1 2 3




z1
z2
z3

 − x


z1
z2
z3

 = ~
0
and factor (T −xI)~
ζ = ~
0. (Note that it says T −xI; the expression T −x doesn’t
make sense because T is a matrix while x is a scalar.) This homogeneous linear
system


1 − x 2 1
2 0 − x −2
−1 2 3 − x




z1
z2
z3

 =


0
0
0


has a non-~
0 solution if and only if the matrix is nonsingular. We can determine
when that happens.
0 = |T − xI|
=
¯
¯
¯
¯
¯
¯
1 − x 2 1
2 0 − x −2
−1 2 3 − x
¯
¯
¯
¯
¯
¯
= x3
− 4x2
+ 4x
= x(x − 2)2

The eigenvalues are λ1 = 0 and λ2 = 2. To find the associated eigenvectors,
plug in each eigenvalue:


1 − 0 2 1
2 0 − 0 −2
−1 2 3 − 0




z1
z2
z3

 =


0
0
0

 =⇒


z1
z2
z3

 =


a
0
a


for a scalar parameter a 6= 0 (a is non-0 because eigenvectors must be non-~
0).
In the same way,


2 − 1 2 1
2 2 − 0 −2
−1 2 2 − 3




z1
z2
z3

 =


0
0
0

 =⇒


z1
z2
z3

 =


b
−b
b


with b 6= 0.
3.8 Example If
S =
µ
π 1
0 3
¶
(here π is not a projection map, it is the number 3.14 . . .) then
¯
¯
¯
¯
µ
π − x 1
0 3 − x
¶¯
¯
¯
¯ = (x − π)(x − 3)
so S has eigenvalues of λ1 = π and λ2 = 3. To find associated eigenvectors, first
plug in λ1 for x:
µ
π − π 1
0 3 − π
¶ µ
z1
z2
¶
=
µ
0
0
¶
=⇒
µ
z1
z2
¶
=
µ
a
0
¶
for a scalar a 6= 0, and then plug in λ2:
µ
π − 3 1
0 3 − 3
¶ µ
z1
z2
¶
=
µ
0
0
¶
=⇒
µ
z1
z2
¶
=
µ
−(1/π)b
b
¶
where b 6= 0.
3.9 Definition The characteristic polynomial of a square matrix T is the de-
terminant of the matrix T − xI, where x is a variable. The characteristic
equation is |T − xI| = 0. The characteristic polynomial of a transformation t is
the polynomial of any RepB,B(t).
Exercise 30 checks that the characteristic polynomial of a transformation is
well-defined, that is, any choice of basis yields the same polynomial.
3.10 Lemma A linear transformation on a nontrivial vector space has at least
one eigenvalue.

Proof. Any root of the characteristic polynomial is an eigenvalue. Over the
complex numbers, any polynomial of degree one or greater has a root. (This is
the reason that in this chapter we’ve gone to scalars that are complex.) QED
Notice the familiar form of the sets of eigenvectors in the above examples.
3.11 Definition The eigenspace of a transformation t associated with the
eigenvalue λ is Vλ = {~
ζ
¯
¯ t(~
ζ ) = λ~
ζ } ∪ {~
0 }. The eigenspace of a matrix is
defined analogously.
3.12 Lemma An eigenspace is a subspace.
Proof. An eigenspace must be nonempty — for one thing it contains the zero
vector — and so we need only check closure. Take vectors ~
ζ1, . . . , ~
ζn from Vλ,
to show that any linear combination is in Vλ
t(c1
~
ζ1 + c2
~
ζ2 + · · · + cn
~
ζn) = c1t(~
ζ1) + · · · + cnt(~
ζn)
= c1λ~
ζ1 + · · · + cnλ~
ζn
= λ(c1
~
ζ1 + · · · + cn
~
ζn)
(the second equality holds even if any ~
ζi is ~
0 since t(~
0) = λ ·~
0 = ~
0). QED
3.13 Example In Example 3.8 the eigenspace associated with the eigenvalue
π and the eigenspace associated with the eigenvalue 3 are these.
Vπ = {
µ
a
0
¶
¯
¯ a ∈ R} V3 = {
µ
−b/π
b
¶
¯
¯ b ∈ R}
3.14 Example In Example 3.7, these are the eigenspaces associated with the
eigenvalues 0 and 2.
V0 = {


a
0
a


¯
¯ a ∈ R}, V2 = {


b
−b
b


¯
¯ b ∈ R}.
3.15 Remark The characteristic equation is 0 = x(x − 2)2
so in some sense 2
is an eigenvalue “twice”. However there are not “twice” as many eigenvectors,
in that the dimension of the eigenspace is one, not two. The next example shows
a case where a number, 1, is a double root of the characteristic equation and
the dimension of the associated eigenspace is two.
3.16 Example With respect to the standard bases, this matrix


1 0 0
0 1 0
0 0 0



represents projection.


x
y
z

 π
7−→


x
y
0

 x, y, z ∈ C
Its eigenspace associated with the eigenvalue 0 and its eigenspace associated
with the eigenvalue 1 are easy to find.
V0 = {


0
0
c3


¯
¯ c3 ∈ C} V1 = {


c1
c2
0


¯
¯ c1, c2 ∈ C}
By the lemma, if two eigenvectors ~
v1 and ~
v2 are associated with the same
eigenvalue then any linear combination of those two is also an eigenvector as-
sociated with that same eigenvalue. But, if two eigenvectors ~
v1 and ~
v2 are
associated with different eigenvalues then the sum ~
v1 + ~
v2 need not be related
to the eigenvalue of either one. In fact, just the opposite. If the eigenvalues are
different then the eigenvectors are not linearly related.
3.17 Theorem For any set of distinct eigenvalues of a map or matrix, a set of
associated eigenvectors, one per eigenvalue, is linearly independent.
Proof. We will use induction on the number of eigenvalues. If there is no eigen-
value or only one eigenvalue then the set of associated eigenvectors is empty or is
a singleton set with a non-~
0 member, and in either case is linearly independent.
For induction, assume that the theorem is true for any set of k distinct eigen-
values, suppose that λ1, . . . , λk+1 are distinct eigenvalues, and let ~
v1, . . . ,~
vk+1
be associated eigenvectors. If c1~
v1 + · · · + ck~
vk + ck+1~
vk+1 = ~
0 then after multi-
plying both sides of the displayed equation by λk+1, applying the map or matrix
to both sides of the displayed equation, and subtracting the first result from the
second, we have this.
c1(λk+1 − λ1)~
v1 + · · · + ck(λk+1 − λk)~
vk + ck+1(λk+1 − λk+1)~
vk+1 = ~
0
The induction hypothesis now applies: c1(λk+1 −λ1) = 0, . . . , ck(λk+1 −λk) = 0.
Thus, as all the eigenvalues are distinct, c1, . . . , ck are all 0. Finally, now ck+1
must be 0 because we are left with the equation ~
vk+1 6= ~
0. QED
3.18 Example The eigenvalues of


2 −2 2
0 1 1
−4 8 3


are distinct: λ1 = 1, λ2 = 2, and λ3 = 3. A set of associated eigenvectors like
{


2
1
0

 ,


9
4
4

 ,


2
1
2

}
is linearly independent.

3.19 Corollary An n×n matrix with n distinct eigenvalues is diagonalizable.
Proof. Form a basis of eigenvectors. Apply Corollary 2.4. QED
Exercises
3.20 For each, find the characteristic polynomial and the eigenvalues.
(a)
µ
10 −9
4 −2
¶
(b)
µ
1 2
4 3
¶
(c)
µ
0 3
7 0
¶
(d)
µ
0 0
0 0
¶
(e)
µ
1 0
0 1
¶
X 3.21 For each matrix, find the characteristic equation, and the eigenvalues and
associated eigenvectors.
(a)
µ
3 0
8 −1
¶
(b)
µ
3 2
−1 0
¶
3.22 Find the characteristic equation, and the eigenvalues and associated eigenvec-
tors for this matrix. Hint. The eigenvalues are complex.
µ
−2 −1
5 2
¶
3.23 Find the characteristic polynomial, the eigenvalues, and the associated eigen-
vectors of this matrix.
Ã
1 1 1
0 0 1
0 0 1
!
X 3.24 For each matrix, find the characteristic equation, and the eigenvalues and
associated eigenvectors.
(a)
Ã
3 −2 0
−2 3 0
0 0 5
!
(b)
Ã
0 1 0
0 0 1
4 −17 8
!
X 3.25 Let t: P2 → P2 be
a0 + a1x + a2x2
7→ (5a0 + 6a1 + 2a2) − (a1 + 8a2)x + (a0 − 2a2)x2
.
Find its eigenvalues and the associated eigenvectors.
3.26 Find the eigenvalues and eigenvectors of this map t: M2 → M2.
µ
a b
c d
¶
7→
µ
2c a + c
b − 2c d
¶
X 3.27 Find the eigenvalues and associated eigenvectors of the differentiation operator
d/dx: P3 → P3.
3.28 Prove that the eigenvalues of a triangular matrix (upper or lower triangular)
are the entries on the diagonal.
X 3.29 Find the formula for the characteristic polynomial of a 2×2 matrix.
3.30 Prove that the characteristic polynomial of a transformation is well-defined.
X 3.31 (a) Can any non-~
0 vector in any nontrivial vector space be a eigenvector?
That is, given a ~
v 6= ~
0 from a nontrivial V , is there a transformation t: V → V
and a scalar λ ∈ R such that t(~
v) = λ~
v?
(b) Given a scalar λ, can any non-~
0 vector in any nontrivial vector space be an
eigenvector associated with the eigenvalue λ?

X 3.32 Suppose that t: V → V and T = RepB,B(t). Prove that the eigenvectors of T
associated with λ are the non-~
0 vectors in the kernel of the map represented (with
respect to the same bases) by T − λI.
3.33 Prove that if a, . . . , d are all integers and a + b = c + d then
µ
a b
c d
¶
has integral eigenvalues, namely a + b and a − c.
X 3.34 Prove that if T is nonsingular and has eigenvalues λ1, . . . , λn then T−1
has
eigenvalues 1/λ1, . . . , 1/λn. Is the converse true?
X 3.35 Suppose that T is n×n and c, d are scalars.
(a) Prove that if T has the eigenvalue λ with an associated eigenvector ~
v then ~
v
is an eigenvector of cT + dI associated with eigenvalue cλ + d.
(b) Prove that if T is diagonalizable then so is cT + dI.
X 3.36 Show that λ is an eigenvalue of T if and only if the map represented by T −λI
is not an isomorphism.
3.37 [Strang 80]
(a) Show that if λ is an eigenvalue of A then λk
is an eigenvalue of Ak
.
(b) What is wrong with this proof generalizing that? “If λ is an eigenvalue of A
and µ is an eigenvalue for B, then λµ is an eigenvalue for AB, for, if A~
x = λ~
x
and B~
x = µ~
x then AB~
x = Aµ~
x = µA~
xµλ~
x”?
3.38 Do matrix-equivalent matrices have the same eigenvalues?
3.39 Show that a square matrix with real entries and an odd number of rows has
at least one real eigenvalue.
3.40 Diagonalize.
Ã
−1 2 2
2 2 2
−3 −6 −6
!
3.41 Suppose that P is a nonsingular n×n matrix. Show that the similarity trans-
formation map tP : Mn
×
n → Mn
×
n sending T 7→ PTP−1
is an isomorphism.
3.42 [Math. Mag., Nov. 1967] Show that if A is an n square matrix and each row
(column) sums to c then c is a characteristic root of A.

Section III. Nilpotence 365
5.III Nilpotence
The goal of this chapter is to show that every square matrix is similar to one
that is a sum of two kinds of simple matrices. The prior section focused on the
first kind, diagonal matrices. We now consider the other kind.
5.III.1 Self-Composition
This subsection is optional, although it is necessary for later material in this
section and in the next one.
A linear transformations t: V → V , because it has the same domain and
codomain, can be iterated.∗
That is, compositions of t with itself such as t2
= t◦t
and t3
= t ◦ t ◦ t are defined.
~
v
t(~
v )
t2
(~
v )
Note that this power notation for the linear transformation functions dovetails
with the notation that we’ve used earlier for their square matrix representations
because if RepB,B(t) = T then RepB,B(tj
) = Tj
.
1.1 Example For the derivative map d/dx: P3 → P3 given by
a + bx + cx2
+ dx3 d/dx
7−→ b + 2cx + 3dx2
the second power is the second derivative
a + bx + cx2
+ dx3 d2
/dx2
7−→ 2c + 6dx
the third power is the third derivative
a + bx + cx2
+ dx3 d3
/dx3
7−→ 6d
and any higher power is the zero map.
1.2 Example This transformation of the space of 2×2 matrices
µ
a b
c d
¶
t
7−→
µ
b a
d 0
¶
∗ More information on function interation is in the appendix.

has this second power
µ
a b
c d
¶
t2
7−→
µ
a b
0 0
¶
and this third power.
µ
a b
c d
¶
t3
7−→
µ
b a
0 0
¶
After that, t4
= t2
and t5
= t3
, etc.
These examples suggest that on iteration more and more zeros appear until
there is a settling down. The next result makes this precise.
1.3 Lemma For any transformation t: V → V , the rangespaces of the powers
form a descending chain
V ⊇ R(t) ⊇ R(t2
) ⊇ · · ·
and the nullspaces form an ascending chain.
{~
0 } ⊆ N (t) ⊆ N (t2
) ⊆ · · ·
Further, there is a k such that for powers less than k the subsets are proper (if
j k then R(tj
) ⊃ R(tj+1
) and N (tj
) ⊂ N (tj+1
)), while for powers greater
than k the sets are equal (if j ≥ k then R(tj
) = R(tj+1
) and N (tj
) = N (tj+1
)).
Proof. We will do the rangespace half and leave the rest for Exercise 13. Recall,
however, that for any map the dimension of its rangespace plus the dimension
of its nullspace equals the dimension of its domain. So if the rangespaces shrink
then the nullspaces must grow.
That the rangespaces form chains is clear because if ~
w ∈ R(tj+1
), so that
~
w = tj+1
(~
v), then ~
w = tj
( t(~
v) ) and so ~
w ∈ R(tj
). To verify the “further”
property, first observe that if any pair of rangespaces in the chain are equal
R(tk
) = R(tk+1
) then all subsequent ones are also equal R(tk+1
) = R(tk+2
),
etc. This is because if t: R(tk+1
) → R(tk+2
) is the same map, with the same
domain, as t: R(tk
) → R(tk+1
) and it therefore has the same range: R(tk+1
) =
R(tk+2
) (and induction shows that it holds for all higher powers). So if the
chain of rangespaces ever stops being strictly decreasing then it is stable from
that point onward.
But the chain must stop decreasing. Each rangespace is a subspace of the one
before it. For it to be a proper subspace it must be of strictly lower dimension
(see Exercise 11). These spaces are finite-dimensional and so the chain can fall
for only finitely-many steps, that is, the power k is at most the dimension of
V . QED
1.4 Example The derivative map a + bx + cx2
+ dx3 d/dx
7−→ b + 2cx + 3dx2
of
Example 1.1 has this chain of rangespaces
P3 ⊃ P2 ⊃ P1 ⊃ P0 ⊃ {~
0 } = {~
0 } = · · ·

and this chain of nullspaces.
{~
0 } ⊂ P0 ⊂ P1 ⊂ P2 ⊂ P3 = P3 = · · ·
1.5 Example The transformation π: C3
→ C3
projecting onto the first two
coordinates


c1
c2
c3

 π
7−→


c1
c2
0


has C3
⊃ R(π) = R(π2
) = · · · and {~
0 } ⊂ N (π) = N (π2
) = · · · .
1.6 Example Let t: P2 → P2 be the map c0 + c1x + c2x2
7→ 2c0 + c2x. As the
lemma describes, on iteration the rangespace shrinks
R(t0
) = P2 R(t) = {a + bx
¯
¯ a, b ∈ C} R(t2
) = {a
¯
¯ a ∈ C}
and then stabilizes R(t2
) = R(t3
) = · · · , while the nullspace grows
N (t0
) = {0} N (t) = {cx
¯
¯ c ∈ C} N (t2
) = {cx + d
¯
¯ c, d ∈ C}
and then stabilizes N (t2
) = N (t3
) = · · · .
This graph illustrates Lemma 1.3. The horizontal axis gives the power j
of a transformation. The vertical axis gives the dimension of the rangespace
of tj
as the distance above zero — and thus also shows the dimension of the
nullspace as the distance below the gray horizontal line, because the two add to
the dimension n of the domain.
0 1 2 j n
n
rank(tj
)
As sketched, on iteration the rank falls and with it the nullity grows until the
two reach a steady state. This state must be reached by the n-th iterate. The
steady state’s distance above zero is the dimension of the generalized rangespace
and its distance below n is the dimension of the generalized nullspace.
1.7 Definition Let t be a transformation on an n-dimensional space. The
generalized rangespace (or the closure of the rangespace) is R∞(t) = R(tn
) The
generalized nullspace (or the closure of the nullspace) is N∞(t) = N (tn
).

Exercises
1.8 Give the chains of rangespaces and nullspaces for the zero and identity trans-
formations.
1.9 For each map, give the chain of rangespaces and the chain of nullspaces, and
the generalized rangespace and the generalized nullspace.
(a) t0 : P2 → P2, a + bx + cx2
7→ b + cx2
(b) t1 : R2
→ R2
,
µ
a
b
¶
7→
µ
0
a
¶
(c) t2 : P2 → P2, a + bx + cx2
7→ b + cx + ax2
(d) t3 : R3
→ R3
,
Ã
a
b
c
!
7→
Ã
a
a
b
!
1.10 Prove that function composition is associative (t ◦ t) ◦ t = t ◦ (t ◦ t) and so we
can write t3
without specifying a grouping.
1.11 Check that a subspace must be of dimension less than or equal to the dimen-
sion of its superspace. Check that if the subspace is proper (the subspace does not
equal the superspace) then the dimension is strictly less. (This is used in the proof
of Lemma 1.3.)
1.12 Prove that the generalized rangespace R∞(t) is the entire space, and the
generalized nullspace N∞(t) is trivial, if the transformation t is nonsingular. Is
this ‘only if’ also?
1.13 Verify the nullspace half of Lemma 1.3.
1.14 Give an example of a transformation on a three dimensional space whose
range has dimension two. What is its nullspace? Iterate your example until the
rangespace and nullspace stabilize.
1.15 Show that the rangespace and nullspace of a linear transformation need not
be disjoint. Are they ever disjoint?
5.III.2 Strings
This subsection is optional, and requires material from the optional Direct Sum
subsection.
The prior subsection shows that as j increases, the dimensions of the R(tj
)’s
fall while the dimensions of the N (tj
)’s rise, in such a way that this rank and
nullity split the dimension of V . Can we say more; do the two split a basis —
is V = R(tj
) ⊕ N (tj
)?
The answer is yes for the smallest power j = 0 since V = R(t0
) ⊕ N (t0
) =
V ⊕ {~
0}. The answer is also yes at the other extreme.
2.1 Lemma Where t: V → V is a linear transformation, the space is the direct
sum V = R∞(t) ⊕ N∞(t). That is, both dim(V ) = dim(R∞(t)) + dim(N∞(t))
and R∞(t) ∩ N∞(t) = {~
0 }.

Proof. We will verify the second sentence, which is equivalent to the first. The
first clause, that the dimension n of the domain of tn
equals the rank of tn
plus
the nullity of tn
, holds for any transformation and so we need only verify the
second clause.
Assume that ~
v ∈ R∞(t) ∩ N∞(t) = R(tn
) ∩ N (tn
), to prove that ~
v is ~
0.
Because ~
v is in the nullspace, tn
(~
v) = ~
0. On the other hand, because R(tn
) =
R(tn+1
), the map t: R∞(t) → R∞(t) is a dimension-preserving homomorphism
and therefore is one-to-one. A composition of one-to-one maps is one-to-one,
and so tn
: R∞(t) → R∞(t) is one-to-one. But now — because only ~
0 is sent by
a one-to-one linear map to ~
0 — the fact that tn
(~
v) = ~
0 implies that ~
v = ~
0. QED
2.2 Note Technically we should distinguish the map t: V → V from the map
t: R∞(t) → R∞(t) because the domains or codomains might differ. The second
one is said to be the restriction∗
of t to R(tk
). We shall use later a point from
that proof about the restriction map, namely that it is nonsingular.
In contrast to the j = 0 and j = n cases, for intermediate powers the space
V might not be the direct sum of R(tj
) and N (tj
). The next example shows
that the two can have a nontrivial intersection.
2.3 Example Consider the transformation of C2
defined by this action on the
elements of the standard basis.
µ
1
0
¶
n
7−→
µ
0
1
¶ µ
0
1
¶
n
7−→
µ
0
0
¶
N = RepE2,E2
(n) =
µ
0 0
1 0
¶
The vector
~
e2 =
µ
0
1
¶
is in both the rangespace and nullspace. Another way to depict this map’s
action is with a string.
~
e1 7→ ~
e2 7→ ~
0
2.4 Example A map n̂: C4
→ C4
whose action on E4 is given by the string
~
e1 7→ ~
e2 7→ ~
e3 7→ ~
e4 7→ ~
0
has R(n̂)∩N (n̂) equal to the span [{~
e4}], has R(n̂2
)∩N (n̂2
) = [{~
e3,~
e4}], and
has R(n̂3
) ∩ N (n̂3
) = [{~
e4}]. The matrix representation is all zeros except for
some subdiagonal ones.
N̂ = RepE4,E4
(n̂) =




0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0




∗ More information on map restrictions is in the appendix.

2.5 Example Transformations can act via more than one string. A transfor-
mation t acting on a basis B = h~
β1, . . . , ~
β5i by
~
β1 7→ ~
β2 7→ ~
β3 7→ ~
0
~
β4 7→ ~
β5 7→ ~
0
is represented by a matrix that is all zeros except for blocks of subdiagonal ones
RepB,B(t) =






0 0 0 0 0
1 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 1 0






(the lines just visually organize the blocks).
In those three examples all vectors are eventually transformed to zero.
2.6 Definition A nilpotent transformation is one with a power that is the zero
map. A nilpotent matrix is one with a power that is the zero matrix. In either
case, the least such power is the index of nilpotency.
2.7 Example In Example 2.3 the index of nilpotency is two. In Example 2.4
it is four. In Example 2.5 it is three.
2.8 Example The differentiation map d/dx: P2 → P2 is nilpotent of index
three since the third derivative of any quadratic polynomial is zero. This map’s
action is described by the string x2
7→ 2x 7→ 2 7→ 0 and taking the basis
B = hx2
, 2x, 2i gives this representation.
RepB,B(d/dx) =


0 0 0
1 0 0
0 1 0


Not all nilpotent matrices are all zeros except for blocks of subdiagonal ones.
2.9 Example With the matrix N̂ from Example 2.4, and this four-vector basis
D = h




1
0
1
0



 ,




0
2
1
0



 ,




1
1
1
0



 ,




0
0
0
1



i
a change of basis operation produces this representation with respect to D, D.




1 0 1 0
0 2 1 0
1 1 1 0
0 0 0 1








0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0








1 0 1 0
0 2 1 0
1 1 1 0
0 0 0 1




−1
=




−1 0 1 0
−3 −2 5 0
−2 −1 3 0
2 1 −2 0





The new matrix is nilpotent; it’s fourth power is the zero matrix since
(PN̂P−1
)4
= PN̂P−1
· PN̂P−1
· PN̂P−1
· PN̂P−1
= PN̂4
P−1
and N̂4
is the zero matrix.
The goal of this subsection is Theorem 2.13, which shows that the prior
example is prototypical in that every nilpotent matrix is similar to one that is
all zeros except for blocks of subdiagonal ones.
2.10 Definition Let t be a nilpotent transformation on V . A t-string gener-
ated by ~
v ∈ V is a sequence h~
v, t(~
v), . . . , tk−1
(~
v)i. This sequence has length k.
A t-string basis is a basis that is a concatenation of t-strings.
2.11 Example In Example 2.5, the t-strings h~
β1, ~
β2, ~
β3i and h~
β4, ~
β5i, of length
three and two, can be concatenated to make a basis for the domain of t.
2.12 Lemma If a space has a t-string basis then the longest string in it has
length equal to the index of nilpotency of t.
Proof. Suppose not. Those strings cannot be longer; if the index is k then
tk
sends any vector — including those starting the string — to ~
0. So suppose
instead that there is a transformation t of index k on some space, such that
the space has a t-string basis where all of the strings are shorter than length
k. Because t has index k, there is a vector ~
v such that tk−1
(~
v) 6= ~
0. Represent
~
v as a linear combination of basis elements and apply tk−1
. We are supposing
that tk−1
sends each basis element to ~
0 but that it does not send ~
v to ~
0. That
is impossible. QED
We shall show that every nilpotent map has an associated string basis. Then
our goal theorem, that every nilpotent matrix is similar to one that is all zeros
except for blocks of subdiagonal ones, is immediate, as in Example 2.5.
Looking for a counterexample — a nilpotent map without an associated
string basis that is disjoint — will suggest the idea for the proof. Consider the
map t: C5
→ C5
with this action.
~
e1
~
e2
7→
7→
~
e3 7→ ~
0
~
e4 7→ ~
e5 7→ ~
0
RepE5,E5
(t) =






0 0 0 0 0
0 0 0 0 0
1 1 0 0 0
0 0 0 0 0
0 0 0 1 0






Even after ommitting the zero vector, these three strings aren’t disjoint, but
that doesn’t end hope of finding a t-string basis. It only means that E5 will not
do for the string basis.
To find a basis that will do, we first find the number and lengths of its
strings. Since t’s index of nilpotency is two, Lemma 2.12 says that at least one

string in the basis has length two. Thus the map must act on a string basis in
one of these two ways.
~
β1 7→ ~
β2 7→ ~
0
~
β3 7→ ~
β4 7→ ~
0
~
β5 7→ ~
0
~
β1 7→ ~
β2 7→ ~
0
~
β3 7→ ~
0
~
β4 7→ ~
0
~
β5 7→ ~
0
Now, the key point. A transformation with the left-hand action has a nullspace
of dimension three since that’s how many basis vectors are sent to zero. A
transformation with the right-hand action has a nullspace of dimension four.
Using the matrix representation above, calculation of t’s nullspace
N (t) = {






x
−x
z
0
r






¯
¯ x, z, r ∈ C}
shows that it is three-dimensional, meaning that we want the left-hand action.
To produce a string basis, first pick ~
β2 and ~
β4 from R(t) ∩ N (t)
~
β2 =






0
0
1
0
0






~
β4 =






0
0
0
0
1






(other choices are possible, just be sure that {~
β2, ~
β4} is linearly independent).
For ~
β5 pick a vector from N (t) that is not in the span of {~
β2, ~
β4}.
~
β5 =






1
−1
0
0
0






Finally, take ~
β1 and ~
β3 such that t(~
β1) = ~
β2 and t(~
β3) = ~
β4.
~
β1 =






0
1
0
0
0






~
β3 =






0
0
0
1
0







Now, with respect to B = h~
β1, . . . , ~
β5i, the matrix of t is as desired.
RepB,B(t) =






0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 0






2.13 Theorem Any nilpotent transformation t is associated with a t-string
basis. While the basis is not unique, the number and the length of the strings
is determined by t.
This illustrates the proof. Basis vectors are categorized into kind 1, kind 2, and
kind 3. They are also shown as squares or circles, according to whether they
are in the nullspace or not.
3 7→ 1 7→ · · · · · · 7→ 1 7→ 1 7→ ~
0
3 7→ 1 7→ · · · · · · 7→ 1 7→ 1 7→ ~
0
.
.
.
3 7→ 1 7→ · · · 7→ 1 7→ 1 7→ ~
0
2 7→ ~
0
.
.
.
2 7→ ~
0
Proof. Fix a vector space V ; we will argue by induction on the index of nilpo-
tency of t: V → V . If that index is 1 then t is the zero map and any basis is
a string basis ~
β1 7→ ~
0, . . . , ~
βn 7→ ~
0. For the inductive step, assume that the
theorem holds for any transformation with an index of nilpotency between 1
and k − 1 and consider the index k case.
First observe that the restriction to the rangespace t: R(t) → R(t) is also
nilpotent, of index k − 1. Apply the inductive hypothesis to get a string basis
for R(t), where the number and length of the strings is determined by t.
B = h~
β1, t(~
β1), . . . , th1
(~
β1)i
_
h~
β2, . . . , th2
(~
β2)i
_
· · ·
_
h~
βi, . . . , thi
(~
βi)i
(In the illustration these are the basis vectors of kind 1, so there are i strings
shown with this kind of basis vector.)
Second, note that taking the final nonzero vector in each string gives a basis
C = hth1
(~
β1), . . . , thi
(~
βi)i for R(t) ∩ N (t). (These are illustrated with 1’s in
squares.) For, a member of R(t) is mapped to zero if and only if it is a linear
combination of those basis vectors that are mapped to zero. Extend C to a
basis for all of N (t).
Ĉ = C
_
h~
ξ1, . . . , ~
ξpi

(The ~
ξ’s are the vectors of kind 2 so that Ĉ is the set of squares.) While many
choices are possible for the ~
ξ’s, their number p is determined by the map t as it
is the dimension of N (t) minus the dimension of R(t) ∩ N (t).
Finally, B
_
Ĉ is a basis for R(t)+N (t) because any sum of something in the
rangespace with something in the nullspace can be represented using elements
of B for the rangespace part and elements of Ĉ for the part from the nullspace.
Note that
dim
¡
R(t) + N (t)
¢
= dim(R(t)) + dim(N (t)) − dim(R(t) ∩ N (t))
= rank(t) + nullity(t) − i
= dim(V ) − i
and so B
_
Ĉ can be extended to a basis for all of V by the addition of i more
vectors. Specifically, remember that each of ~
β1, . . . , ~
βi is in R(t), and extend
B
_
Ĉ with vectors ~
v1, . . . ,~
vi such that t(~
v1) = ~
β1, . . . , t(~
vi) = ~
βi. (In the
illustration, these are the 3’s.) The check that linear independence is preserved
by this extension is Exercise 29. QED
2.14 Corollary Every nilpotent matrix is similar to a matrix that is all zeros
except for blocks of subdiagonal ones. That is, every nilpotent map is repre-
sented with respect to some basis by such a matrix.
This form is unique in the sense that if a nilpotent matrix is similar to two
such matrices then those two simply have their blocks ordered differently. Thus
this is a canonical form for the similarity classes of nilpotent matrices provided
that we order the blocks, say, from longest to shortest.
M =
µ
1 −1
1 −1
¶
has an index of nilpotency of two, as this calculation shows.
p Mp
N (Mp
)
1 M =
µ
1 −1
1 −1
¶
{
µ
x
x
¶
¯
¯ x ∈ C}
2 M2
=
µ
0 0
0 0
¶
C2
The calculation also describes how a map m represented by M must act on any
string basis. With one map application the nullspace has dimension one and so
one vector of the basis is sent to zero. On a second application, the nullspace
has dimension two and so the other basis vector is sent to zero. Thus, the action
of the map is ~
β1 7→ ~
β2 7→ ~
0 and the canonical form of the matrix is this.
µ
0 0
1 0
¶

We can exhibit such a m-string basis and the change of basis matrices wit-
nessing the matrix similarity. For the basis, take M to represent m with respect
to the standard bases, pick a ~
β2 ∈ N (m) and also pick a ~
β1 so that m(~
β1) = ~
β2.
~
β2 =
µ
1
1
¶
~
β1 =
µ
1
0
¶
(If we take M to be a representative with respect to some nonstandard bases
then this picking step is just more messy.) Recall the similarity diagram.
C2
w.r.t. E2
m
−
−
−
−
→
M
C2
w.r.t. E2
id


yP id


yP
C2
w.r.t. B
m
−
−
−
−
→ C2
w.r.t. B
The canonical form equals RepB,B(m) = PMP−1
, where
P−1
= RepB,E2
(id) =
µ
1 1
0 1
¶
P = (P−1
)−1
=
µ
1 −1
0 1
¶
and the verification of the matrix calculation is routine.
µ
1 −1
0 1
¶ µ
1 −1
1 −1
¶ µ
1 1
0 1
¶
=
µ
0 0
1 0
¶






0 0 0 0 0
1 0 0 0 0
−1 1 1 −1 1
0 1 0 0 0
1 0 −1 1 −1






is nilpotent. These calculations show the nullspaces growing.
p Np
N (Np
)
1






0 0 0 0 0
1 0 0 0 0
−1 1 1 −1 1
0 1 0 0 0
1 0 −1 1 −1






{






0
0
u − v
u
v






¯
¯ u, v ∈ C}
2






0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0






{






0
y
z
u
v






¯
¯ y, z, u, v ∈ C}
3 –zero matrix– C5

That table shows that any string basis must satisfy: the nullspace after one map
application has dimension two so two basis vectors are sent directly to zero,
the nullspace after the second application has dimension four so two additional
basis vectors are sent to zero by the second iteration, and the nullspace after
three applications is of dimension five so the final basis vector is sent to zero in
three hops.
~
β1 7→ ~
β2 7→ ~
β3 7→ ~
0
~
β4 7→ ~
β5 7→ ~
0
To produce such a basis, first pick two independent vectors from N (n)
~
β3 =






0
0
1
1
0






~
β5 =






0
0
0
1
1






then add ~
β2, ~
β4 ∈ N (n2
) such that n(~
β2) = ~
β3 and n(~
β4) = ~
β5
~
β2 =






0
1
0
0
0






~
β4 =






0
1
0
1
0






and finish by adding ~
β1 ∈ N (n3
) = C5
) such that n(~
β1) = ~
β2.
~
β1 =






1
0
1
0
0






Exercises
X 2.17 What is the index of nilpotency of the left-shift operator, here acting on the
space of triples of reals?
(x, y, z) 7→ (0, x, y)
X 2.18 For each string basis state the index of nilpotency and give the dimension of
the rangespace and nullspace of each iteration of the nilpotent map.
(a) ~
β1 7→ ~
β2 7→ ~
0
~
β3 7→ ~
β4 7→ ~
0
(b) ~
β1 7→ ~
β2 7→ ~
β3 7→ ~
0
~
β4 7→ ~
0
~
β5 7→ ~
0
~
β6 7→ ~
0
(c) ~
β1 7→ ~
β2 7→ ~
β3 7→ ~
0
Also give the canonical form of the matrix.
2.19 Decide which of these matrices are nilpotent.

(a)
µ
−2 4
−1 2
¶
(b)
µ
3 1
1 3
¶
(c)
Ã
−3 2 1
−3 2 1
−3 2 1
!
(d)
Ã
1 1 4
3 0 −1
5 2 7
!
(e)
Ã
45 −22 −19
33 −16 −14
69 −34 −29
!
X 2.20 Find the canonical form of this matrix.





0 1 1 0 1
0 0 1 1 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0





X 2.21 Consider the matrix from Example 2.16.
(a) Use the action of the map on the string basis to give the canonical form.
(b) Find the change of basis matrices that bring the matrix to canonical form.
(c) Use the answer in the prior item to check the answer in the first item.
X 2.22 Each of these matrices is nilpotent.
(a)
µ
1/2 −1/2
1/2 −1/2
¶
(b)
Ã
0 0 0
0 −1 1
0 −1 1
!
(c)
Ã
−1 1 −1
1 0 1
1 −1 1
!
Put each in canonical form.
2.23 Describe the effect of left or right multiplication by a matrix that is in the
canonical form for nilpotent matrices.
2.24 Is nilpotence invariant under similarity? That is, must a matrix similar to a
nilpotent matrix also be nilpotent? If so, with the same index?
X 2.25 Show that the only eigenvalue of a nilpotent matrix is zero.
2.26 Is there a nilpotent transformation of index three on a two-dimensional space?
2.27 In the proof of Theorem 2.13, why isn’t the proof’s base case that the index
of nilpotency is zero?
X 2.28 Let t: V → V be a linear transformation and suppose ~
v ∈ V is such that
tk
(~
v) = ~
0 but tk−1
(~
v) 6= ~
0. Consider the t-string h~
v, t(~
v), . . . , tk−1
(~
v)i.
(a) Prove that t is a transformation on the span of the set of vectors in the string,
that is, prove that t restricted to the span has a range that is a subset of the
span. We say that the span is a t-invariant subspace.
(b) Prove that the restriction is nilpotent.
(c) Prove that the t-string is linearly independent and so is a basis for its span.
(d) Represent the restriction map with respect to the t-string basis.
2.29 Finish the proof of Theorem 2.13.
2.30 Show that the terms ‘nilpotent transformation’ and ‘nilpotent matrix’, as
given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is
represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an
only if there is a basis such that the map’s representation with respect to that
basis is a nilpotent matrix, or that any representation is a nilpotent matrix?)
2.31 Let T be nilpotent of index four. How big can the rangespace of T3
be?
2.32 Recall that similar matrices have the same eigenvalues. Show that the converse
does not hold.
2.33 Prove a nilpotent matrix is similar to one that is all zeros except for blocks of
super-diagonal ones.

X 2.34 Prove that if a transformation has the same rangespace as nullspace. then the
dimension of its domain is even.
2.35 Prove that if two nilpotent matrices commute then their product and sum are
also nilpotent.
2.36 Consider the transformation of Mn
×
n given by tS(T) = ST − TS where S is
an n×n matrix. Prove that if S is nilpotent then so is tS.
2.37 Show that if N is nilpotent then I − N is invertible. Is that ‘only if’ also?

Section IV. Jordan Form 379
5.IV Jordan Form
This section uses material from three optional subsections: Direct Sum, Deter-
minants Exist, and Other Formulas for the Determinant.
The chapter on linear maps shows that every h: V → W can be represented
by a partial-identity matrix with respect to some bases B ⊂ V and D ⊂ W.
This chapter revisits this issue in the special case that the map is a linear
transformation t: V → V . Of course, the general result still applies but with
the codomain and domain equal we naturally ask about having the two bases
also be equal. That is, we want a canonical form to represent transformations
as RepB,B(t).
After a brief review section, we began by noting that a block partial identity
form matrix is not always obtainable in this B, B case. We therefore considered
the natural generalization, diagonal matrices, and showed that if its eigenvalues
are distinct then a map or matrix can be diagonalized. But we also gave an
example of a matrix that cannot be diagonalized and in the section prior to this
one we developed that example. We showed that a linear map is nilpotent —
if we take higher and higher powers of the map or matrix then we eventually
get the zero map or matrix — if and only if there is a basis on which it acts via
disjoint strings. That led to a canonical form for nilpotent matrices.
Now, this section concludes the chapter. We will show that the two cases
we’ve studied are exhaustive in that for any linear transformation there is a
basis such that the matrix representation RepB,B(t) is the sum of a diagonal
matrix and a nilpotent matrix in its canonical form.
5.IV.1 Polynomials of Maps and Matrices
Recall that the set of square matrices is a vector space under entry-by-entry
addition and scalar multiplication and that this space Mn×n has dimension n2
.
Thus, for any n×n matrix T the n2
+1-member set {I, T, T2
, . . . , Tn2
} is linearly
dependent and so there are scalars c0, . . . , cn2 such that cn2 Tn2
+· · ·+c1T +c0I
is the zero matrix.
1.1 Remark This observation is small but important. It says that every trans-
formation exhibits a generalized nilpotency: the powers of a square matrix can-
not climb forever without a “repeat”.
1.2 Example Rotation of plane vectors π/6 radians counterclockwise is rep-
resented with respect to the standard basis by
T =
µ√
3/2 −1/2
1/2
√
3/2
¶
and verifying that 0T4
+ 0T3
+ 1T2
− 2T − 1I equals the zero matrix is easy.

1.3 Definition For any polynomial f(x) = cnxn
+ · · · + c1x + c0, where t is a
linear transformation then f(t) is the transformation cntn
+ · · · + c1t + c0(id)
on the same space and where T is a square matrix then f(T) is the matrix
cnTn
+ · · · + c1T + c0I.
1.4 Remark If, for instance, f(x) = x − 3, then most authors write in the
identity matrix: f(T) = T − 3I. But most authors don’t write in the identity
map: f(t) = t − 3. In this book we shall also observe this convention.
Of course, if T = RepB,B(t) then f(T) = RepB,B(f(t)), which follows from
the relationships Tj
= RepB,B(tj
), and cT = RepB,B(ct), and T1 + T2 =
RepB,B(t1 + t2).
As Example 1.2 shows, there may be polynomials of degree smaller than n2
that zero the map or matrix.
1.5 Definition The minimal polynomial m(x) of a transformation t or a square
matrix T is the polynomial of least degree and with leading coefficient 1 such
that m(t) is the zero map or m(T) is the zero matrix.
A minimal polynomial always exists by the observation opening this subsec-
tion. A minimal polynomial is unique by the ‘with leading coefficient 1’ clause.
This is because if there are two polynomials m(x) and m̂(x) that are both of the
minimal degree to make the map or matrix zero (and thus are of equal degree),
and both have leading 1’s, then their difference m(x) − m̂(x) has a smaller de-
gree than either and still sends the map or matrix to zero. Thus m(x)−m̂(x) is
the zero polynomial and the two are equal. (The leading coefficient requirement
also prevents a minimal polynomial from being the zero polynomial.)
1.6 Example We can see that m(x) = x2
− 2x − 1 is minimal for the matrix
of Example 1.2 by computing the powers of T up to the power n2
= 4.
T2
=
µ
1/2 −
√
3/2
√
3/2 1/2
¶
T3
=
µ
0 −1
1 0
¶
T4
=
µ
−1/2 −
√
3/2
√
3/2 −1/2
¶
Next, put c4T4
+ c3T3
+ c2T2
+ c1T + c0I equal to the zero matrix
−(1/2)c4 + (1/2)c2 + (
√
3/2)c1 + c0 = 0
−(
√
3/2)c4 − c3 − (
√
3/2)c2 − (1/2)c1 = 0
(
√
3/2)c4 + c3 + (
√
3/2)c2 + (1/2)c1 = 0
−(1/2)c4 + (1/2)c2 + (
√
3/2)c1 + c0 = 0
and use Gauss’ method.
c4 − c2 −
√
3c1 − 2c0 = 0
c3 +
√
3c2 + 2c1 +
√
3c0 = 0
Setting c4, c3, and c2 to zero forces c1 and c0 to also come out as zero. To get
a leading one, the most we can do is to set c4 and c3 to zero. Thus the minimal
polynomial is quadratic.

Using the method of that example to find the minimal polynomial of a 3×3
matrix would mean doing Gaussian reduction on a system with nine equations
in ten unknowns. We shall develop an alternative. To begin, note that we can
break a polynomial of a map or a matrix into its components.
1.7 Lemma Suppose that the polynomial f(x) = cnxn
+ · · · + c1x + c0 factors
as k(x − λ1)q1
· · · (x − λ`)q`
. If t is a linear transformation then these two are
equal maps.
cntn
+ · · · + c1t + c0 = k · (t − λ1)q1
◦ · · · ◦ (t − λ`)q`
Consequently, if T is a square matrix then f(T) and k·(T −λ1I)q1
· · · (T −λ`I)q`
are equal matrices.
Proof. This argument is by induction on the degree of the polynomial. The
cases where the polynomial is of degree 0 and 1 are clear. The full induction
argument is Exercise 1.7 but the degree two case gives its sense.
A quadratic polynomial factors into two linear terms f(x) = k(x − λ1) · (x −
λ2) = k(x2
+ (λ1 + λ2)x + λ1λ2) (the roots λ1 and λ2 might be equal). We can
check that substituting t for x in the factored and unfactored versions gives the
same map.
¡
k · (t − λ1) ◦ (t − λ2)
¢
(~
v) =
¡
k · (t − λ1)
¢
(t(~
v) − λ2~
v)
= k ·
¡
t(t(~
v)) − t(λ2~
v) − λ1t(~
v) − λ1λ2~
v
¢
= k ·
¡
t ◦ t (~
v) − (λ1 + λ2)t(~
v) + λ1λ2~
v
¢
= k · (t2
− (λ1 + λ2)t + λ1λ2) (~
v)
The third equality holds because the scalar λ2 comes out of the second term, as
t is linear. QED
In particular, if a minimial polynomial m(x) for a transformation t factors
as m(x) = (x − λ1)q1
· · · (x − λ`)q`
then m(t) = (t − λ1)q1
◦ · · · ◦ (t − λ`)q`
is
the zero map. Since m(t) sends every vector to zero, at least one of the maps
t − λi sends some nonzero vectors to zero. So, too, in the matrix case — if m is
minimal for T then m(T) = (T −λ1I)q1
· · · (T −λ`I)q`
is the zero matrix and at
least one of the matrices T −λiI sends some nonzero vectors to zero. Rewording
both cases: at least some of the λi are eigenvalues. (See Exercise 29.)
Recall how we have earlier found eigenvalues. We have looked for λ such that
T~
v = λ~
v by considering the equation ~
0 = T~
v−x~
v = (T −xI)~
v and computing the
determinant of the matrix T − xI. That determinant is a polynomial in x, the
characteristic polynomial, whose roots are the eigenvalues. The major result
of this subsection, the next result, is that there is a connection between this
characteristic polynomial and the minimal polynomial. This results expands
on the prior paragraph’s insight that some roots of the minimal polynomial
are eigenvalues by asserting that every root of the minimal polynomial is an
eigenvalue and further that every eigenvalue is a root of the minimal polynomial
(this is because it says ‘1 ≤ qi’ and not just ‘0 ≤ qi’).

1.8 Theorem (Cayley-Hamilton) If the characteristic polynomial of a
transformation or square matrix factors into
k · (x − λ1)p1
(x − λ2)p2
· · · (x − λ`)p`
then its minimal polynomial factors into
(x − λ1)q1
(x − λ2)q2
· · · (x − λ`)q`
where 1 ≤ qi ≤ pi for each i between 1 and `.
The proof takes up the next three lemmas. Although they are stated only in
matrix terms, they apply equally well to maps. We give the matrix version only
because it is convenient for the first proof.
The first result is the key — some authors call it the Cayley-Hamilton Theo-
rem and call Theorem 1.8 above a corollary. For the proof, observe that a matrix
of polynomials can be thought of as a polynomial with matrix coefficients.
µ
2x2
+ 3x − 1 x2
+ 2
3x2
+ 4x + 1 4x2
+ x + 1
¶
=
µ
2 1
3 4
¶
x2
+
µ
3 0
4 1
¶
x +
µ
−1 2
1 1
¶
1.9 Lemma If T is a square matrix with characteristic polynomial c(x) then
c(T) is the zero matrix.
Proof. Let C be T − xI, the matrix whose determinant is the characteristic
polynomial c(x) = cnxn
+ · · · + c1x + c0.
C =





t1,1 − x t1,2 . . .
t2,1 t2,2 − x
.
.
.
...
tn,n − x





Recall that the product of the adjoint of a matrix with the matrix itself is the
determinant of that matrix times the identity.
c(x) · I = adj(C)C = adj(C)(T − xI) = adj(C)T − adj(C) · x (∗)
The entries of adj(C) are polynomials, each of degree at most n − 1 since the
minors of a matrix drop a row and column. Rewrite it, as suggested above, as
adj(C) = Cn−1xn−1
+ · · · + C1x + C0 where each Ci is a matrix of scalars. The
left and right ends of equation (∗) above give this.
cnIxn
+ cn−1Ixn−1
+ · · · + c1Ix + c0I = (Cn−1T)xn−1
+ · · · + (C1T)x + C0T
− Cn−1xn
− Cn−2xn−1
− · · · − C0x

Equate the coefficients of xn
, the coefficients of xn−1
, etc.
cnI = −Cn−1
cn−1I = −Cn−2 + Cn−1T
.
.
.
c1I = −C0 + C1T
c0I = C0T
Multiply (from the right) both sides of the first equation by Tn
, both sides
of the second equation by Tn−1
, etc. Add. The result on the left is cnTn
+
cn−1Tn−1
+ · · · + c0I, and the result on the right is the zero matrix. QED
We sometimes refer to that lemma by saying that a matrix or map satisfies
its characteristic polynomial.
1.10 Lemma Where f(x) is a polynomial, if f(T) is the zero matrix then f(x)
is divisible by the minimal polynomial of T. That is, any polynomial satisfied
by T is divisable by T’s minimal polynomial.
Proof. Let m(x) be minimal for T. The Division Theorem for Polynomials
gives f(x) = q(x)m(x) + r(x) where the degree of r is strictly less than the
degree of m. Plugging T in shows that r(T) is the zero matrix, because T
satisfies both f and m. That contradicts the minimality of m unless r is the
zero polynomial. QED
Combining the prior two lemmas gives that the minimal polynomial divides
the characteristic polynomial. Thus, any root of the minimal polynomial is
also a root of the characteristic polynomial. That is, so far we have that if
m(x) = (x − λ1)q1
. . . (x − λi)qi
then c(x) must has the form (x − λ1)p1
. . . (x −
λi)pi
(x − λi+1)pi+1
. . . (x − λ`)p`
where each qj is less than or equal to pj. The
proof of the Cayley-Hamilton Theorem is finished by showing that in fact the
characteristic polynomial has no extra roots λi+1, etc.
1.11 Lemma Each linear factor of the characteristic polynomial of a square
matrix is also a linear factor of the minimal polynomial.
Proof. Let T be a square matrix with minimal polynomial m(x) and assume
that x − λ is a factor of the characteristic polynomial of T, that is, assume that
λ is an eigenvalue of T. We must show that x − λ is a factor of m, that is, that
m(λ) = 0.
In general, where λ is associated with the eigenvector ~
v, for any polyno-
mial function f(x), application of the matrix f(T) to ~
v equals the result of
multiplying ~
v by the scalar f(λ). (For instance, if T has eigenvalue λ associ-
ated with the eigenvector ~
v and f(x) = x2
+ 2x + 3 then (T2
+ 2T + 3) (~
v) =
T2
(~
v) + 2T(~
v) + 3~
v = λ2
·~
v + 2λ ·~
v + 3 ·~
v = (λ2
+ 2λ + 3) ·~
v.) Now, as m(T) is
the zero matrix, ~
0 = m(T)(~
v) = m(λ) · ~
v and therefore m(λ) = 0. QED

1.12 Example We can use the Cayley-Hamilton Theorem to help find the
minimal polynomial of this matrix.
T =




2 0 0 1
1 2 0 2
0 0 2 −1
0 0 0 1




First, its characteristic polynomial c(x) = (x−1)(x−2)3
can be found with the
usual determinant. Now, the Cayley-Hamilton Theorem says that T’s minimal
polynomial is either (x − 1)(x − 2) or (x − 1)(x − 2)2
or (x − 1)(x − 2)3
. We can
decide among the choices just by computing:
(T − 1I)(T − 2I) =




1 0 0 1
1 1 0 2
0 0 1 −1
0 0 0 0








0 0 0 1
1 0 0 2
0 0 0 −1
0 0 0 −1



 =




0 0 0 0
1 0 0 1
0 0 0 0
0 0 0 0




and
(T − 1I)(T − 2I)2
=




0 0 0 0
1 0 0 1
0 0 0 0
0 0 0 0








0 0 0 1
1 0 0 2
0 0 0 −1
0 0 0 −1



 =




0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0




and so m(x) = (x − 1)(x − 2)2
.
Exercises
X 1.13 What are the possible minimal polynomials if a matrix has the given charac-
teristic polynomial?
(a) 8 · (x − 3)4
(b) (1/3) · (x + 1)3
(x − 4) (c) −1 · (x − 2)2
(x − 5)2
(d) 5 · (x + 3)2
(x − 1)(x − 2)2
What is the degree of each possibility?
X 1.14 Find the minimal polynomial of each matrix.
(a)
Ã
3 0 0
1 3 0
0 0 4
!
(b)
Ã
3 0 0
1 3 0
0 0 3
!
(c)
Ã
3 0 0
1 3 0
0 1 3
!
(d)
Ã
2 0 1
0 6 2
0 0 2
!
(e)
Ã
2 2 1
0 6 2
0 0 2
!
(f)





−1 4 0 0 0
0 3 0 0 0
0 −4 −1 0 0
3 −9 −4 2 −1
1 5 4 1 4





1.15 Find the minimal polynomial of this matrix.
Ã
0 1 0
0 0 1
1 0 0
!
X 1.16 What is the minimal polynomial of the differentiation operator d/dx on Pn?

X 1.17 Find the minimal polynomial of matrices of this form








λ 0 0 . . . 0
1 λ 0 0
0 1 λ
...
λ 0
0 0 . . . 1 λ








where the scalar λ is fixed (i.e., is not a variable).
1.18 What is the minimal polynomial of the transformation of Pn that sends p(x)
to p(x + 1)?
1.19 What is the minimal polynomial of the map π : C3
→ C3
projecting onto the
first two coordinates?
1.20 Find a 3×3 matrix whose minimal polynomial is x2
.
1.21 What is wrong with this claimed proof of Lemma 1.9: “if c(x) = |T −xI| then
c(T) = |T − TI| = 0”?
1.22 Verify Lemma 1.9 for 2×2 matrices by direct calculation.
X 1.23 Prove that the minimal polynomial of an n×n matrix has degree at most
n (not n2
as might be guessed from this subsection’s opening). Verify that this
maximum, n, can happen.
X 1.24 The only eigenvalue of a nilpotent map is zero. Show that the converse state-
ment holds.
1.25 What is the minimal polynomial of a zero map or matrix? Of an identity map
or matrix?
X 1.26 Interpret the minimal polynomial of Example 1.2 geometrically.
1.27 What is the minimal polynomial of a diagonal matrix?
X 1.28 A projection is any transformation t such that t2
= t. (For instance, the
transformation of the plane R2
projecting each vector onto its first coordinate will,
if done twice, result in the same value as if it is done just once.) What is the
minimal polynomial of a projection?
1.29 The first two items of this question are review.
(a) Prove that the composition of one-to-one maps is one-to-one.
(b) Prove that if a linear map is not one-to-one then at least one nonzero vector
from the domain is sent to the zero vector in the codomain.
(c) Verify the statement, excerpted here, that preceeds Theorem 1.8.
. . . if a minimial polynomial m(x) for a transformation t factors as
m(x) = (x − λ1)q1
· · · (x − λ`)q` then m(t) = (t − λ1)q1
◦ · · · ◦ (t − λ`)q`
is the zero map. Since m(t) sends every vector to zero, at least one
of the maps t − λi sends some nonzero vectors to zero. . . . Rewording
. . . : at least some of the λi are eigenvalues.
1.30 True or false: for a transformation on an n dimensional space, if the minimal
polynomial has degree n then the map is diagonalizable.
1.31 Let f(x) be a polynomial. Prove that if A and B are similar matrices then
f(A) is similar to f(B).
(a) Now show that similar matrices have the same characteristic polynomial.
(b) Show that similar matrices have the same minimal polynomial.

(c) Decide if these are similar.
µ
1 3
2 3
¶ µ
4 −1
1 1
¶
1.32 (a) Show that a matrix is invertible if and only if the constant term in its
minimal polynomial is not 0.
(b) Show that if a square matrix T is not invertible then there is a nonzero
matrix S such that ST and TS both equal the zero matrix.
X 1.33 (a) Finish the proof of Lemma 1.7.
(b) Give an example to show that the result does not hold if t is not linear.
5.IV.2 Jordan Canonical Form
This subsection moves from the canonical form for nilpotent matrices to the
one for all matrices.
We have shown that if a map is nilpotent then all of its eigenvalues are zero.
We can now prove the converse.
2.1 Lemma A linear transformation whose only eigenvalue is zero is nilpotent.
Proof. If a transformation t on an n-dimensional space has only the single
eigenvalue of zero then its characteristic polynomial is xn
. The Cayley-Hamilton
Theorem says that a map satisfies its characteristic polynimial so tn
is the zero
map. Thus t is nilpotent. QED
We have a canonical form for nilpotent matrices, that is, for each matrix
whose single eigenvalue is zero: each such matrix is similar to one that is all
zeroes except for blocks of subdiagonal ones. (To make this representation
unique we can fix some arrangement of the blocks, say, from longest to shortest.)
We next extend this to all single-eigenvalue matrices.
Observe that if t’s only eigenvalue is λ then t − λ’s only eigenvalue is 0
because t(~
v) = λ~
v if and only if (t − λ) (~
v) = 0 · ~
v. The natural way to extend
the results for nilpotent matrices is to represent t − λ in the canonical form N,
and try to use that to get a simple representation T for t. The next result says
that this try works.
2.2 Lemma If the matrices T − λI and N are similar then T and N + λI are
also similar, via the same change of basis matrices.
Proof. With N = P(T − λI)P−1
= PTP−1
− P(λI)P−1
we have N =
PTP−1
− PP−1
(λI) since the diagonal matrix λI commutes with anything,
and so N = PTP−1
− λI. Therefore N + λI = PTP−1
, as required. QED
2.3 Example The characteristic polynomial of
T =
µ
2 −1
1 4
¶

is (x − 3)2
and so T has only the single eigenvalue 3. Thus for
T − 3I =
µ
−1 −1
1 1
¶
the only eigenvalue is 0, and T − 3I is nilpotent. The null spaces are routine
to find; to ease this computation we take T to represent the transformation
t: C2
→ C2
with respect to the standard basis (we shall maintain this convention
for the rest of the chapter).
N (t − 3) = {
µ
−y
y
¶
¯
¯ y ∈ C} N ((t − 3)2
) = C2
The dimensions of these null spaces show that the action of an associated map
t − 3 on a string basis is ~
β1 7→ ~
β2 7→ ~
0. Thus, the canonical form for t − 3 with
one choice for a string basis is
RepB,B(t − 3) = N =
µ
0 0
1 0
¶
B = h
µ
1
1
¶
,
µ
−2
2
¶
i
and by Lemma 2.2, T is similar to this matrix.
Rept(B, B) = N + 3I =
µ
3 0
1 3
¶
We can produce the similarity computation. Recall from the Nilpotence
section how to find the change of basis matrices P and P−1
to express N as
P(T − 3I)P−1
. The similarity diagram
C2
w.r.t. E2
t−3
−
−
−
−
→
T −3I
C2
w.r.t. E2
id


yP id


yP
C2
w.r.t. B
t−3
−
−
−
−
→
N
C2
w.r.t. B
describes that to move from the lower left to the upper left we multiply by
P−1
=
¡
RepE2,B(id)
¢−1
= RepB,E2
(id) =
µ
1 −2
1 2
¶
and to move from the upper right to the lower right we multiply by this matrix.
P =
µ
1 −2
1 2
¶−1
=
µ
1/2 1/2
−1/4 1/4
¶
So the similarity is expressed by
µ
3 0
1 3
¶
=
µ
1/2 1/2
−1/4 1/4
¶ µ
2 −1
1 4
¶ µ
1 −2
1 2
¶
which is easily checked.

2.4 Example This matrix has characteristic polynomial (x − 4)4
T =




4 1 0 −1
0 3 0 1
0 0 4 0
1 0 0 5




and so has the single eigenvalue 4. The nullities of t − 4 are: the null space of
t − 4 has dimension two, the null space of (t − 4)2
has dimension three, and the
null space of (t − 4)3
has dimension four. Thus, t − 4 has the action on a string
basis of ~
β1 7→ ~
β2 7→ ~
β3 7→ ~
0 and ~
β4 7→ ~
0. This gives the canonical form N for
t − 4, which in turn gives the form for t.
N + 4I =




4 0 0 0
1 4 0 0
0 1 4 0
0 0 0 4




An array that is all zeroes, except for some number λ down the diagonal
and blocks of subdiagonal ones, is a Jordan block. We have shown that Jordan
block matrices are canonical representatives of the similarity classes of single-
eigenvalue matrices.
2.5 Example The 3×3 matrices whose only eigenvalue is 1/2 separate into
three similarity classes. The three classes have these canonical representatives.


1/2 0 0
0 1/2 0
0 0 1/2




1/2 0 0
1 1/2 0
0 0 1/2




1/2 0 0
1 1/2 0
0 1 1/2


In particular, this matrix


1/2 0 0
0 1/2 0
0 1 1/2


belongs to the similarity class represented by the middle one, because we have
adopted the convention of ordering the blocks of subdiagonal ones from the
longest block to the shortest.
We will now finish the program of this chapter by extending this work to
cover maps and matrices with multiple eigenvalues. The best possibility for
general maps and matrices would be if we could break them into a part involving
their first eigenvalue λ1 (which we represent using its Jordan block), a part with
λ2, etc.
This ideal is in fact what happens. For any transformation t: V → V , we
shall break the space V into the direct sum of a part on which t−λ1 is nilpotent,
plus a part on which t − λ2 is nilpotent, etc. More precisely, we shall take three

steps to get to this section’s major theorem and the third step shows that
V = N∞(t − λ1) ⊕ · · · ⊕ N∞(t − λ`) where λ1, . . . , λ` are t’s eigenvalues.
Suppose that t: V → V is a linear transformation. Note that the restriction∗
of t to a subspace M need not be a linear transformation on M because there may
be an ~
m ∈ M with t(~
m) 6∈ M. To ensure that the restriction of a transformation
to a ‘part’ of a space is a transformation on the partwe need the next condition.
2.6 Definition Let t: V → V be a transformation. A subspace M is t invari-
ant if whenever ~
m ∈ M then t(~
m) ∈ M (shorter: t(M) ⊆ M).
Two examples are that the generalized null space N∞(t) and the generalized
range space R∞(t) of any transformation t are invariant. For the generalized null
space, if ~
v ∈ N∞(t) then tn
(~
v) = ~
0 where n is the dimension of the underlying
space and so t(~
v) ∈ N∞(t) because tn
( t(~
v) ) is zero also. For the generalized
range space, if ~
v ∈ R∞(t) then ~
v = tn
(~
w) for some ~
w and then t(~
v) = tn+1
(~
w) =
tn
( t(~
w) ) shows that t(~
v) is also a member of R∞(t).
Thus the spaces N∞(t − λi) and R∞(t − λi) are t − λi invariant. Observe
also that t−λi is nilpotent on N∞(t−λi) because, simply, if ~
v has the property
that some power of t − λi maps it to zero — that is, if it is in the generalized
null space — then some power of t − λi maps it to zero. The generalized null
space N∞(t − λi) is a ‘part’ of the space on which the action of t − λi is easy
to understand.
The next result is the first of our three steps. It establishes that t−λj leaves
t − λi’s part unchanged.
2.7 Lemma A subspace is t invariant if and only if it is t−λ invariant for any
scalar λ. In particular, where λi is an eigenvalue of a linear transformation t,
then for any other eigenvalue λj, the spaces N∞(t − λi) and R∞(t − λi) are
t − λj invariant.
Proof. For the first sentence we check the two implications of the ‘if and only
if’ separately. One of them is easy: if the subspace is t − λ invariant for any λ
then taking λ = 0 shows that it is t invariant. For the other implication suppose
that the subspace is t invariant, so that if ~
m ∈ M then t(~
m) ∈ M, and let λ
be any scalar. The subspace M is closed under linear combinations and so if
t(~
m) ∈ M then t(~
m) − λ~
m ∈ M. Thus if ~
m ∈ M then (t − λ) (~
m) ∈ M, as
required.
The second sentence follows straight from the first. Because the two spaces
are t − λi invariant, they are therefore t invariant. From this, applying the first
sentence again, we conclude that they are also t − λj invariant. QED
The second step of the three that we will take to prove this section’s major
result makes use of an additional property of N∞(t − λi) and R∞(t − λi), that
they are complementary. Recall that if a space is the direct sum of two others
V = N ⊕ R then any vector ~
v in the space breaks into two parts ~
v = ~
n + ~
r
where ~
n ∈ N and ~
r ∈ R, and recall also that if BN and BR are bases for N
∗ More information on restrictions of functions is in the appendix.

and R then the concatenation BN
_
BR is linearly independent (and so the two
parts of ~
v do not “overlap”). The next result says that for any subspaces N
and R that are complementary as well as t invariant, the action of t on ~
v breaks
into the “non-overlapping” actions of t on ~
n and on ~
r.
2.8 Lemma Let t: V → V be a transformation and let N and R be t invariant
complementary subspaces of V . Then t can be represented by a matrix with
blocks of square submatrices T1 and T2
µ
T1 Z2
Z1 T2
¶
}dim(N )-many rows
}dim(R)-many rows
where Z1 and Z2 are blocks of zeroes.
Proof. Since the two subspaces are complementary, the concatenation of a basis
for N and a basis for R makes a basis B = h~
ν1, . . . ,~
νp, ~
µ1, . . . , ~
µqi for V . We
shall show that the matrix
RepB,B(t) =




.
.
.
.
.
.
RepB(t(~
ν1)) · · · RepB(t(~
µq))
.
.
.
.
.
.




has the desired form.
Any vector ~
v ∈ V is in N if and only if its final q components are zeroes
when it is represented with respect to B. As N is t invariant, each of the
vectors RepB(t(~
ν1)), . . . , RepB(t(~
νp)) has that form. Hence the lower left of
RepB,B(t) is all zeroes.
The argument for the upper right is similar. QED
To see that t has been decomposed into its action on the parts, observe
that the restrictions of t to the subspaces N and R are represented, with
respect to the obvious bases, by the matrices T1 and T2. So, with subspaces
that are invariant and complementary, we can split the problem of examining a
linear transformation into two lower-dimensional subproblems. The next result
illustrates this decomposition into blocks.
2.9 Lemma If T is a matrices with square submatrices T1 and T2
T =
µ
T1 Z2
Z1 T2
¶
where the Z’s are blocks of zeroes, then |T| = |T1| · |T2|.
Proof. Suppose that T is n×n, that T1 is p×p, and that T2 is q×q. In the
permutation formula for the determinant
|T| =
X
permutations φ
t1,φ(1)t2,φ(2) · · · tn,φ(n) sgn(φ)

each term comes from a rearrangement of the column numbers 1, . . . , n into a
new order φ(1), . . . , φ(n). The upper right block Z2 is all zeroes, so if a φ has at
least one of p + 1, . . . , n among its first p column numbers φ(1), . . . , φ(p) then
the term arising from φ is zero, e.g., if φ(1) = n then t1,φ(1)t2,φ(2) . . . tn,φ(n) =
0 · t2,φ(2) . . . tn,φ(n) = 0.
So the above formula reduces to a sum over all permutations with two
halves: any significant φ is the composition of a φ1 that rearranges only 1, . . . , p
and a φ2 that rearranges only p + 1, . . . , p + q. Now, the distributive law (and
the fact that the signum of a composition is the product of the signums) gives
that this
|T1| · |T2| =
µ X
perms φ1
of 1,...,p
t1,φ1(1) · · · tp,φ1(p) sgn(φ1)
¶
·
µ X
perms φ2
of p+1,...,p+q
tp+1,φ2(p+1) · · · tp+q,φ2(p+q) sgn(φ2)
¶
equals |T| =
P
significant φ t1,φ(1)t2,φ(2) · · · tn,φ(n) sgn(φ). QED
2.10 Example
¯
¯
¯
¯
¯
¯
¯
¯
2 0 0 0
1 2 0 0
0 0 3 0
0 0 0 3
¯
¯
¯
¯
¯
¯
¯
¯
=
¯
¯
¯
¯
2 0
1 2
¯
¯
¯
¯ ·
¯
¯
¯
¯
3 0
0 3
¯
¯
¯
¯ = 36
From Lemma 2.9 we conclude that if two subspaces are complementary and
t invariant then t is nonsingular if and only if its restrictions to both subspaces
are nonsingular.
Now for the promised third, final, step to the main result.
2.11 Lemma If a linear transformation t: V → V has the characteristic poly-
nomial (x − λ1)p1
. . . (x − λ`)p`
then (1) V = N∞(t − λ1) ⊕ · · · ⊕ N∞(t − λ`)
and (2) dim(N∞(t − λi)) = pi.
Proof. Because dim(V ) is the degree p1 + · · · + p` of the characteristic poly-
nomial, to establish statement (1) we need only show that statement (2) holds
and that N∞(t − λi) ∩ N∞(t − λj) is trivial whenever i 6= j.
For the latter, by Lemma 2.7, both N∞(t−λi) and N∞(t−λj) are t invariant.
Notice that an intersection of t invariant subspaces is t invariant and so the
restriction of t to N∞(t −λi)∩N∞(t −λj) is a linear transformation. But both
t − λi and t − λj are nilpotent on this subspace and so if t has any eigenvalues
on the intersection then its “only” eigenvalue is both λi and λj. That cannot
be, so this restriction has no eigenvalues: N∞(t − λi) ∩ N∞(t − λj) is trivial
(Lemma 3.10 shows that the only transformation without any eigenvalues is on
the trivial space).

To prove statement (2), fix the index i. Decompose V as N∞(t − λi) ⊕
R∞(t − λi) and apply Lemma 2.8.
T =
µ
T1 Z2
Z1 T2
¶
}dim( N∞(t − λi) )-many rows
}dim( R∞(t − λi) )-many rows
By Lemma 2.9, |T − xI| = |T1 − xI| · |T2 − xI|. By the uniqueness clause of the
Fundamental Theorem of Arithmetic, the determinants of the blocks have the
same factors as the characteristic polynomial |T1 −xI| = (x−λ1)q1
. . . (x−λ`)q`
and |T2 − xI| = (x − λ1)r1
. . . (x − λ`)r`
, and the sum of the powers of these
factors is the power of the factor in the characteristic polynomial: q1 + r1 = p1,
. . . , q` + r` = p`. Statement (2) will be proved if we will show that qi = pi and
that qj = 0 for all j 6= i, because then the degree of the polynomial |T1 − xI| —
which equals the dimension of the generalized null space — is as required.
For that, first, as the restriction of t − λi to N∞(t − λi) is nilpotent on that
space, the only eigenvalue of t on it is λi. Thus the characteristic equation of t
on N∞(t − λi) is |T1 − xI| = (x − λi)qi
. And thus qj = 0 for all j 6= i.
Now consider the restriction of t to R∞(t − λi). By Note II.2.2, the map
t − λi is nonsingular on R∞(t − λi) and so λi is not an eigenvalue of t on that
subspace. Therefore, x − λi is not a factor of |T2 − xI|, and so qi = pi. QED
Our major result just translates those steps into matrix terms.
2.12 Theorem Any square matrix is similar to one in Jordan form







Jλ1
–zeroes–
Jλ2
...
Jλ`−1
–zeroes– Jλ`







where each Jλ is the Jordan block associated with the eigenvalue λ of the orig-
inal matrix (that is, is all zeroes except for λ’s down the diagonal and some
subdiagonal ones).
Proof. Given an n×n matrix T, consider the linear map t: Cn
→ Cn
that it
represents with respect to the standard bases. Use the prior lemma to write
Cn
= N∞(t − λ1) ⊕ · · · ⊕ N∞(t − λ`) where λ1, . . . , λ` are the eigenvalues of t.
Because each N∞(t − λi) is t invariant, Lemma 2.8 and the prior lemma show
that t is represented by a matrix that is all zeroes except for square blocks along
the diagonal. To make those blocks into Jordan blocks, pick each Bλi
to be a
string basis for the action of t − λi on N∞(t − λi). QED
Jordan form is a canonical form for similarity classes of square matrices,
provided that we make it unique by arranging the Jordan blocks from least
eigenvalue to greatest and then arranging the subdiagonal 1 blocks inside each
Jordan block from longest to shortest.

2.13 Example This matrix has the characteristic polynomial (x − 2)2
(x − 6).
T =


2 0 1
0 6 2
0 0 2


We will handle the eigenvalues 2 and 6 separately.
Computation of the powers, and the null spaces and nullities, of T − 2I is
routine. (Recall from Example 2.3 the convention of taking T to represent a
transformation, here t: C3
→ C3
, with respect to the standard basis.)
power p (T − 2I)p
N ((t − 2)p
) nullity
1



0 0 1
0 4 2
0 0 0


 {



x
0
0



¯
¯ x ∈ C} 1
2



0 0 0
0 16 8
0 0 0


 {



x
−z/2
z



¯
¯ x, z ∈ C} 2
3



0 0 0
0 64 32
0 0 0


 –same– —
So the generalized null space N∞(t − 2) has dimension two. We’ve noted that
the restriction of t − 2 is nilpotent on this subspace. From the way that the
nullities grow we know that the action of t − 2 on a string basis ~
β1 7→ ~
β2 7→ ~
0.
Thus the restriction can be represented in the canonical form
N2 =
µ
0 0
1 0
¶
= RepB,B(t − 2) B2 = h


1
1
−2

 ,


−2
0
0

i
where many choices of basis are possible. Consequently, the action of the re-
striction of t to N∞(t − 2) is represented by this matrix.
J2 = N2 + 2I = RepB2,B2
(t) =
µ
2 0
1 2
¶
The second eigenvalue’s computations are easier. Because the power of x−6
in the characteristic polynomial is one, the restriction of t−6 to N∞(t−6) must
be nilpotent of index one. Its action on a string basis must be ~
β3 7→ ~
0 and since
it is the zero map, its canonical form N6 is the 1×1 zero matrix. Consequently,
the canonical form J6 for the action of t on N∞(t−6) is the 1×1 matrix with the
single entry 6. For the basis we can use any nonzero vector from the generalized
null space.
B6 = h


0
1
0

i

Taken together, these two give that the Jordan form of T is
RepB,B(t) =


2 0 0
1 2 0
0 0 6


where B is the concatenation of B2 and B6.
2.14 Example Contrast the prior example with
T =


2 2 1
0 6 2
0 0 2


which has the same characteristic polynomial (x − 2)2
(x − 6).
While the characteristic polynomial is the same,
power p (T − 2I)p
N ((t − 2)p
) nullity
1



0 2 1
0 4 2
0 0 0


 {



x
−z/2
z



¯
¯ x, z ∈ C} 2
2



0 8 4
0 16 8
0 0 0


 –same– —
here the action of t−2 is stable after only one application — the restriction of of
t−2 to N∞(t−2) is nilpotent of index only one. (So the contrast with the prior
example is that while the characteristic polynomial tells us to look at the action
of the t − 2 on its generalized null space, the characteristic polynomial does not
describe completely its action and we must do some computations to find, in
this example, that the minimal polynomial is (x − 2)(x − 6).) The restriction of
t − 2 to the generalized null space acts on a string basis as ~
β1 7→ ~
0 and ~
β2 7→ ~
0,
and we get this Jordan block associated with the eigenvalue 2.
J2 =
µ
2 0
0 2
¶
For the other eigenvalue, the arguments for the second eigenvalue of the
prior example apply again. The restriction of t − 6 to N∞(t − 6) is nilpotent
of index one (it can’t be of index less than one, and since x − 6 is a factor of
the characteristic polynomial to the power one it can’t be of index more than
one either). Thus t − 6’s canonical form N6 is the 1×1 zero matrix, and the
associated Jordan block J6 is the 1×1 matrix with entry 6.
Therefore, T is diagonalizable.
RepB,B(t) =


2 0 0
0 2 0
0 0 6

 B = B2
_
B6 = h


1
0
0

 ,


0
1
−2

 ,


3
4
0

i
(Checking that the third vector in B is in the nullspace of t − 6 is routine.)

2.15 Example A bit of computing with
T =






−1 4 0 0 0
0 3 0 0 0
0 −4 −1 0 0
3 −9 −4 2 −1
1 5 4 1 4






shows that its characteristic polynomial is (x − 3)3
(x + 1)2
. This table
power p (T − 3I)p
N ((t − 3)p
) nullity
1








−4 4 0 0 0
0 0 0 0 0
0 −4 −4 0 0
3 −9 −4 −1 −1
1 5 4 1 1








{








−(u + v)/2
−(u + v)/2
(u + v)/2
u
v








¯
¯ u, v ∈ C} 2
2








16 −16 0 0 0
0 0 0 0 0
0 16 16 0 0
−16 32 16 0 0
0 −16 −16 0 0








{








−z
−z
z
u
v








¯
¯ z, u, v ∈ C} 3
3








−64 64 0 0 0
0 0 0 0 0
0 −64 −64 0 0
64 −128 −64 0 0
0 64 64 0 0








–same– —
shows that the restriction of t − 3 to N∞(t − 3) acts on a string basis via the
two strings ~
β1 7→ ~
β2 7→ ~
0 and ~
β3 7→ ~
0.
A similar calculation for the other eigenvalue
power p (T + 1I)p
N ((t + 1)p
) nullity
1








0 4 0 0 0
0 4 0 0 0
0 −4 0 0 0
3 −9 −4 3 −1
1 5 4 1 5








{








−(u + v)
0
−v
u
v








¯
¯ u, v ∈ C} 2
2








0 16 0 0 0
0 16 0 0 0
0 −16 0 0 0
8 −40 −16 8 −8
8 24 16 8 24








–same– —

shows that the restriction of t + 1 to its generalized null space acts on a string
basis via the two separate strings ~
β4 7→ ~
0 and ~
β5 7→ ~
0.
Therefore T is similar to this Jordan form matrix.






−1 0 0 0 0
0 −1 0 0 0
0 0 3 0 0
0 0 1 3 0
0 0 0 0 3






We close with the statement that the subjects considered earlier in this
Chpater are indeed, in this sense, exhaustive.
2.16 Corollary Every square matrix is similar to the sum of a diagonal matrix
and a nilpotent matrix.
Exercises
2.17 Do the check for Example 2.3.
2.18 Each matrix is in Jordan form. State its characteristic polynomial and its
minimal polynomial.
(a)
µ
3 0
1 3
¶
(b)
µ
−1 0
0 −1
¶
(c)
Ã
2 0 0
1 2 0
0 0 −1/2
!
(d)
Ã
3 0 0
1 3 0
0 1 3
!
(e)



3 0 0 0
1 3 0 0
0 0 3 0
0 0 1 3


 (f)



4 0 0 0
1 4 0 0
0 0 −4 0
0 0 1 −4


 (g)
Ã
5 0 0
0 2 0
0 0 3
!
(h)



5 0 0 0
0 2 0 0
0 0 2 0
0 0 0 3


 (i)



5 0 0 0
0 2 0 0
0 1 2 0
0 0 0 3



X 2.19 Find the Jordan form from the given data.
(a) The matrix T is 5×5 with the single eigenvalue 3. The nullities of the powers
are: T − 3I has nullity two, (T − 3I)2
has nullity three, (T − 3I)3
has nullity
four, and (T − 3I)4
has nullity five.
(b) The matrix S is 5×5 with two eigenvalues. For the eigenvalue 2 the nullities
are: S − 2I has nullity two, and (S − 2I)2
has nullity four. For the eigenvalue
−1 the nullities are: S + 1I has nullity one.
2.20 Find the change of basis matrices for each example.
(a) Example 2.13 (b) Example 2.14 (c) Example 2.15
X 2.21 Find the Jordan form and a Jordan basis for each matrix.
(a)
µ
−10 4
−25 10
¶
(b)
µ
5 −4
9 −7
¶
(c)
Ã
4 0 0
2 1 3
5 0 4
!

(d)
Ã
5 4 3
−1 0 −3
1 −2 1
!
(e)
Ã
9 7 3
−9 −7 −4
4 4 4
!
(f)
Ã
2 2 −1
−1 −1 1
−1 −2 2
!
(g)



7 1 2 2
1 4 −1 −1
−2 1 5 −1
1 1 2 8



X 2.22 Find all possible Jordan forms of a transformation with characteristic poly-
nomial (x − 1)2
(x + 2)2
.
2.23 Find all possible Jordan forms of a transformation with characteristic poly-
nomial (x − 1)3
(x + 2).
X 2.24 Find all possible Jordan forms of a transformation with characteristic poly-
nomial (x − 2)3
(x + 1) and minimal polynomial (x − 2)2
(x + 1).
2.25 Find all possible Jordan forms of a transformation with characteristic poly-
nomial (x − 2)4
(x + 1) and minimal polynomial (x − 2)2
(x + 1).
X 2.26 Diagonalize these.
(a)
µ
1 1
0 0
¶
(b)
µ
0 1
1 0
¶
X 2.27 Find the Jordan matrix representing the differentiation operator on P3.
X 2.28 Decide if these two are similar.
µ
1 −1
4 −3
¶ µ
−1 0
1 −1
¶
2.29 Find the Jordan form of this matrix.
µ
0 −1
1 0
¶
Also give a Jordan basis.
2.30 How many similarity classes are there for 3×3 matrices whose only eigenvalues
are −3 and 4?
X 2.31 Prove that a matrix is diagonalizable if and only if its minimal polynomial
has only linear factors.
2.32 Give an example of a linear transformation on a vector space that has no
non-trivial invariant subspaces.
2.33 Show that a subspace is t − λ1 invariant if and only if it is t − λ2 invariant.
2.34 Prove or disprove: two n×n matrices are similar if and only if they have the
same characteristic and minimal polynomials.
2.35 The trace of a square matrix is the sum of its diagonal entries.
(a) Find the formula for the characteristic polynomial of a 2×2 matrix.
(b) Show that trace is invariant under similarity, and so we can sensibly speak
of the ‘trace of a map’. (Hint: see the prior item.)
(c) Is trace invariant under matrix equivalence?

(d) Show that the trace of a map is the sum of its eigenvalues (counting multi-
plicities).
(e) Show that the trace of a nilpotent map is zero. Does the converse hold?
2.36 To use Definition 2.6 to check whether a subspace is t invariant, we seemingly
have to check all of the infinitely many vectors in a (nontrivial) subspace to see if
they satisfy the condition. Prove that a subspace is t invariant if and only if its
subbasis has the property that for all of its elements, t(~
β) is in the subspace.
X 2.37 Is t invariance preserved under intersection? Under union? Complementation?
Sums of subspaces?
2.38 Give a way to order the Jordan blocks if some of the eigenvalues are complex
numbers. That is, suggest a reasonable ordering for the complex numbers.
2.39 Let Pj(R) be the vector space over the reals of degree j polynomials. Show
that if j ≤ k then Pj(R) is an invariant subspace of Pk(R) under the differentiation
operator. In P7(R), does any of P0(R), . . . , P6(R) have an invariant complement?
2.40 In Pn(R), the vector space (over the reals) of degree n polynomials,
E = {p(x) ∈ Pn(R)
¯
¯ p(−x) = p(x) for all x}
and
O = {p(x) ∈ Pn(R)
¯
¯ p(−x) = −p(x) for all x}
are the even and the odd polynomials; p(x) = x2
is even while p(x) = x3
is odd.
Show that they are subspaces. Are they complementary? Are they invariant under
the differentiation transformation?
2.41 Lemma 2.8 says that if M and N are invariant complements then t has a
representation in the given block form (with respect to the same ending as starting
basis, of course). Does the implication reverse?
2.42 A matrix S is the square root of another T if S2
= T. Show that any nonsin-
gular matrix has a square root.

Topic: Computing Eigenvalues—the Method of Powers 399
Topic: Computing Eigenvalues—the Method of
Powers
In practice, calculating eigenvalues and eigenvectors is a difficult problem. Find-
ing, and solving, the characteristic polynomial of the large matrices often en-
countered in applications is too slow and too hard. Other techniques, indirect
ones that avoid the characteristic polynomial, are used. Here we shall see such
a method that is suitable for large matrices that are ‘sparse’ (the great majority
of the entries are zero).
Suppose that the n×n matrix T has the n distinct eigenvalues λ1, λ2, . . . , λn.
Then Rn
has a basis that is composed of the associated eigenvectors h~
ζ1, . . . , ~
ζni.
For any ~
v ∈ Rn
, where ~
v = c1
~
ζ1 + · · · + cn
~
ζn, iterating T on ~
v gives these.
T~
v = c1λ1
~
ζ1 + c2λ2
~
ζ2 + · · · + cnλn
~
ζn
T2
~
v = c1λ2
1
~
ζ1 + c2λ2
2
~
ζ2 + · · · + cnλ2
n
~
ζn
T3
~
v = c1λ3
1
~
ζ1 + c2λ3
2
~
ζ2 + · · · + cnλ3
n
~
ζn
.
.
.
Tk
~
v = c1λk
1
~
ζ1 + c2λk
2
~
ζ2 + · · · + cnλk
n
~
ζn
If one of the eigenvaluse, say, λ1, has a larger absolute value than any of the
other eigenvalues then its term will dominate the above expression. Put another
way, dividing through by λk
1 gives this,
Tk
~
v
λk
1
= c1
~
ζ1 + c2
λk
2
λk
1
~
ζ2 + · · · + cn
λk
n
λk
1
~
ζn
and, because λ1 is assumed to have the largest absolute value, as k gets larger
the fractions go to zero. Thus, the entire expression goes to c1
~
ζ1.
That is (as long as c1 is not zero), as k increases, the vectors Tk
~
v will
tend toward the direction of the eigenvectors associated with the dominant
eigenvalue, and, consequently, the ratios of the lengths k Tk
~
v k/k Tk−1
~
v k will
tend toward that dominant eigenvalue.
For example (sample computer code for this follows the exercises), because
the matrix
T =
µ
3 0
8 −1
¶
is triangular, its eigenvalues are just the entries on the diagonal, 3 and −1.
Arbitrarily taking ~
v to have the components 1 and 1 gives
~
v T~
v T2
~
v · · · T9
~
v T10
~
v
µ
1
1
¶ µ
3
7
¶ µ
9
17
¶
· · ·
µ
19 683
39 367
¶ µ
59 049
118 097
¶
and the ratio between the lengths of the last two is 2.999 9.

Two implementation issues must be addressed. The first issue is that, instead
of finding the powers of T and applying them to ~
v, we will compute ~
v1 as T~
v and
then compute ~
v2 as T~
v1, etc. (i.e., we never separately calculate T2
, T3
, etc.).
These matrix-vector products can be done quickly even if T is large, provided
that it is sparse. The second issue is that, to avoid generating numbers that are
so large that they overflow our computer’s capability, we can normalize the ~
vi’s
at each step. For instance, we can divide each ~
vi by its length (other possibilities
are to divide it by its largest component, or simply by its first component). We
thus implement this method by generating
~
w0 = ~
v0/k~
v0k
~
v1 = T ~
w0
~
w1 = ~
v1/k~
v1k
~
v2 = T ~
w2
.
.
.
~
wk−1 = ~
vk−1/k~
vk−1k
~
vk = T ~
wk
until we are satisfied. Then the vector ~
vk is an approximation of an eigenvector,
and the approximation of the dominant eigenvalue is the ratio k~
vkk/k~
wk−1k =
k~
vkk.
One way we could be ‘satisfied’ is to iterate until our approximation of the
eigenvalue settles down. We could decide, for instance, to stop the iteration
process not after some fixed number of steps, but instead when k~
vkk differs
from k~
vk−1k by less than one percent, or when they agree up to the second
significant digit.
The rate of convergence is determined by the rate at which the powers of
kλ2/λ1k go to zero, where λ2 is the eigenvalue of second largest norm. If that
ratio is much less than one then convergence is fast, but if it is only slightly
less than one then convergence can be quite slow. Consequently, the method of
powers is not the most commonly used way of finding eigenvalues (although it
is the simplest one, which is why it is here as the illustration of the possibility of
computing eigenvalues without solving the characteristic polynomial). Instead,
there are a variety of methods that generally work by first replacing the given
matrix T with another that is similar to it and so has the same eigenvalues, but
is in some reduced form such as tridiagonal form: the only nonzero entries are
on the diagonal, or just above or below it. Then special techniques can be used
to find the eigenvalues. Once the eigenvalues are known, the eigenvectors of T
can be easily computed. These other methods are outside of our scope. A good
reference is [Goult, et al.]
Exercises
1 Use ten iterations to estimate the largest eigenvalue of these matrices, starting
from the vector with components 1 and 2. Compare the answer with the one
obtained by solving the characteristic equation.

Topic: Computing Eigenvalues—the Method of Powers 401
(a)
µ
1 5
0 4
¶
(b)
µ
3 2
−1 0
¶
2 Redo the prior exercise by iterating until k~
vkk − k~
vk−1k has absolute value less
than 0.01 At each step, normalize by dividing each vector by its length. How many
iterations are required? Are the answers significantly different?
3 Use ten iterations to estimate the largest eigenvalue of these matrices, starting
from the vector with components 1, 2, and 3. Compare the answer with the one
obtained by solving the characteristic equation.
(a)
Ã
4 0 1
−2 1 0
−2 0 1
!
(b)
Ã
−1 2 2
2 2 2
−3 −6 −6
!
4 Redo the prior exercise by iterating until k~
vkk − k~
vk−1k has absolute value less
than 0.01. At each step, normalize by dividing each vector by its length. How
many iterations does it take? Are the answers significantly different?
5 What happens if c1 = 0? That is, what happens if the initial vector does not to
have any component in the direction of the relevant eigenvector?
6 How can the method of powers be adopted to find the smallest eigenvalue?
Computer Code
This is the code for the computer algebra system Octave that was used
to do the calculation above.
T=[3, 0;
8, -1]
T=
3 0
8 -1
v0=[1; 2]
v0=
1
1
v1=T*v0
v1=
3
7
v2=T*v1
v2=
9
17
T9=T**9
T9=
19683 0
39368 -1
T10=T**10
T10=
59049 0
118096 1
v9=T9*v0
v9=
19683

39367
v10=T10*v0
v10=
59049
118096
norm(v10)/norm(v9)
ans=2.9999
(It has been lightly edited to remove blank lines, etc. Remark: we are ignor-
ing the power of Octave here; there are built-in functions to automatically
apply quite sophisticated methods to find eigenvalues and eigenvectors. In-
stead, we are using just the system as a calculator.)

Topic: Stable Populations 403
Topic: Stable Populations
Imagine a reserve park with animals from a species that we are trying to protect.
The park doesn’t have a fence and so animals cross the boundary, both from
the inside out and in the other direction. Every year, 10% of the animals from
inside of the park leave, and 1% of the animals from the outside find their way
in. We can ask if we can find a stable level of population for this park: is there a
population that, once established, will stay constant over time, with the number
of animals leaving equal to the number of animals entering?
To answer that question, we must first establish the equations. Let the year
n population in the park be pn and in the rest of the world be rn.
pn+1 = .90pn + .01rn
rn+1 = .10pn + .99rn
We can set this system up as a matrix equation (see the Markov Chain topic).
µ
pn+1
rn+1
¶
=
µ
.90 .01
.10 .99
¶ µ
pn
rn
¶
Now, “stable level” means that pn+1 = pn and rn+1 = rn, so that the matrix
equation ~
vn+1 = T~
vn becomes ~
v = T~
v. We are therefore looking for eigenvectors
for T that are associated with the eigenvalue 1. The equation (I − T)~
v = ~
0 is
µ
.10 .01
.10 .01
¶ µ
p
r
¶
=
µ
0
0
¶
which gives the eigenspace: vectors with the restriction that p = .1r. Coupled
with additional information, that the total world population of this species is is
p + r = 110 000, we find that the stable state is p = 10, 000 and r = 100, 000.
If we start with a park population of ten thousand animals, so that the rest of
the world has one hundred thousand, then every year ten percent (a thousand
animals) of those inside will leave the park, and every year one percent (a
thousand) of those from the rest of the world will enter the park. It is stable,
self-sustaining.
Now imagine that we are trying to gradually build up the total world pop-
ulation of this species. We can try, for instance, to have the world population
grow at a rate of 1% per year. In this case, we can take a “stable” state for
the park’s population to be that it also grows at 1% per year. The equation
~
vn+1 = 1.01 · ~
vn = T~
vn leads to ((1.01 · I) − T)~
v = ~
0, which gives this system.
µ
.11 .01
.10 .02
¶ µ
p
r
¶
=
µ
0
0
¶
The matrix is nonsingular, and so the only solution is p = 0 and r = 0. Thus,
there is no (usable) initial population that we can establish at the park and
expect that it will grow at the same rate as the rest of the world.

Knowing that an annual world population growth rate of 1% forces an un-
stable park population, we can ask which growth rates there are that would
allow an initial population for the park that will be self-sustaining. We consider
λ~
v = T~
v and solve for λ.
0 =
¯
¯
¯
¯
λ − .9 .01
.10 λ − .99
¯
¯
¯
¯ = (λ − .9)(λ − .99) − (.10)(.01) = λ2
− 1.89λ + .89
A shortcut to factoring that quadratic is our knowledge that λ = 1 is an eigen-
value of T, so the other eigenvalue is .89. Thus there are two ways to have a
stable park population (a population that grows at the same rate as the popu-
lation of the rest of the world, despite the leaky park boundaries): have a world
population that is does not grow or shrink, and have a world population that
shrinks by 11% every year.
So this is one meaning of eigenvalues and eigenvectors—they give a sta-
ble state for a system. If the eigenvalue is 1 then the system is static. If
the eigenvalue isn’t 1 then the system is either growing or shrinking, but in a
dynamically-stable way.
Exercises
1 What initial population for the park discussed above should be set up in the case
where world populations are allowed to decline by 11% every year?
2 What will happen to the population of the park in the event of a growth in world
population of 1% per year? Will it lag the world growth, or lead it? Assume
that the inital park population is ten thousand, and the world population is one
hunderd thousand, and calculate over a ten year span.
3 The park discussed above is partially fenced so that now, every year, only 5% of
the animals from inside of the park leave (still, about 1% of the animals from the
outside find their way in). Under what conditions can the park maintain a stable
population now?
4 Suppose that a species of bird only lives in Canada, the United States, or in
Mexico. Every year, 4% of the Canadian birds travel to the US, and 1% of them
travel to Mexico. Every year, 6% of the US birds travel to Canada, and 4%
go to Mexico. From Mexico, every year 10% travel to the US, and 0% go to
Canada.
(a) Give the transition matrix.
(b) Is there a way for the three countries to have constant populations?
(c) Find all stable situations.

Topic: Linear Recurrences 405
Topic: Linear Recurrences
In 1202 Leonardo of Pisa, also known as Fibonacci, posed this problem.
A certain man put a pair of rabbits in a place surrounded on all
sides by a wall. How many pairs of rabbits can be produced from
that pair in a year if it is supposed that every month each pair begets
a new pair which from the second month on becomes productive?
This moves past an elementary exponential growth model for population in-
crease to include the fact that there is an initial period where newborns are not
fertile. However, it retains other simplyfing assumptions, such as that there is
no gestation period and no mortality.
The number of newborn pairs that will appear in the upcoming month is
simply the number of pairs that were alive last month, since those will all be
fertile, having been alive for two months. The number of pairs alive next month
is the sum of the number alive last month and the number of newborns.
f(n + 1) = f(n) + f(n − 1) where f(0) = 1, f(1) = 1
The is an example of a recurrence relation (it is called that because the values
of f are calculated by looking at other, prior, values of f). From it, we can
easily answer Fibonacci’s twelve-month question.
month 0 1 2 3 4 5 6 7 8 9 10 11 12
pairs 1 1 2 3 5 8 13 21 34 55 89 144 233
The sequence of numbers defined by the above equation (of which the first few
are listed) is the Fibonacci sequence. The material of this chapter can be used
to give a formula with which we can can calculate f(n + 1) without having to
first find f(n), f(n − 1), etc.
For that, observe that the recurrence is a linear relationship and so we can
give a suitable matrix formulation of it.
µ
1 1
1 0
¶ µ
f(n)
f(n − 1)
¶
=
µ
f(n + 1)
f(n)
¶
where
µ
f(1)
f(0)
¶
=
µ
1
1
¶
Then, where we write T for the matrix and ~
vn for the vector with components
f(n+1) and f(n), we have that ~
vn = Tn
~
v0. The advantage of this matrix formu-
lation is that by diagonalizing T we get a fast way to compute its powers: where
T = PDP−1
we have Tn
= PDn
P−1
, and the n-th power of the diagonal
matrix D is the diagonal matrix whose entries that are the n-th powers of the
entries of D.
The characteristic equation of T is λ2
− λ − 1. The quadratic formula gives
its roots as (1 +
√
5)/2 and (1 −
√
5)/2. Diagonalizing gives this.
µ
1 1
1 0
¶
=
µ
1+
√
5
2
1−
√
5
2
1 1
¶ Ã
1+
√
5
2 0
0 1−
√
5
2
! Ã
1
√
5
−1−
√
5
2
√
5
−1
√
5
1+
√
5
2
√
5
!

Introducing the vectors and taking the n-th power, we have
µ
f(n + 1)
f(n)
¶
=
µ
1 1
1 0
¶n µ
f(1)
f(0)
¶
=
µ
1+
√
5
2
1−
√
5
2
1 1
¶ Ã
1+
√
5
2
n
0
0 1−
√
5
2
n
! Ã
1
√
5
−1−
√
5
2
√
5
−1
√
5
1+
√
5
2
√
5
! µ
f(1)
f(0)
¶
We can compute f(n) from the second component of that equation.
f(n) =
1
√
5
Ã
1 +
√
5
2
!n
−
Ã
1 −
√
5
2
!n#
Notice that f is dominated by its first term because (1 −
√
5)/2 is less than
one, so its powers go to zero. Although we have extended the elementary model
of population growth by adding a delay period before the onset of fertility, we
nonetheless still get an (asmyptotically) exponential function.
In general, a linear recurrence relation has the form
f(n + 1) = anf(n) + an−1f(n − 1) + · · · + an−kf(n − k)
(it is also called a difference equation). This recurrence relation is homogeneous
because there is no constant term; i.e, it can be put into the form 0 = −f(n +
1) + anf(n) + an−1f(n − 1) + · · · + an−kf(n − k). This is said to be a relation
of order k. The relation, along with the initial conditions f(0), . . . , f(k)
completely determine a sequence. For instance, the Fibonacci relation is of
order 2 and it, along with the two initial conditions f(0) = 1 and f(1) = 1,
determines the Fibonacci sequence simply because we can compute any f(n) by
first computing f(2), f(3), etc. In this Topic, we shall see how linear algebra
can be used to solve linear recurrence relations.
First, we define the vector space in which we are working. Let V be the set
of functions f from the natural numbers N = {0, 1, 2, . . . } to the real numbers.
(Below we shall have functions with domain {1, 2, . . . }, that is, without 0, but
it is not an important distinction.)
Putting the initial conditions aside for a moment, for any recurrence, we can
consider the subset S of V of solutions. For example, without initial conditions,
in addition to the function f given above, the Fibonacci relation is also solved by
the function g whose first few values are g(0) = 1, g(1) = 1, g(2) = 3, g(3) = 4,
and g(4) = 7.
The subset S is a subspace of V . It is nonempty because the zero function
is a solution. It is closed under addition since if f1 and f2 are solutions, then
an+1(f1 + f2)(n + 1) + · · · + an−k(f1 + f2)(n − k)
= (an+1f1(n + 1) + · · · + an−kf1(n − k))
+ (an+1f2(n + 1) + · · · + an−kf2(n − k))
= 0.

And, it is closed under scalar multiplication since
an+1(rf1)(n + 1) + · · · + an−k(rf1)(n − k)
= r(an+1f1(n + 1) + · · · + an−kf1(n − k))
= r · 0
= 0.
We can give the dimension of S. Consider this map from the set of functions S
to the set of vectors Rk
.
f 7→





f(0)
f(1)
.
.
.
f(k)





Exercise 3 shows that this map is linear. Because, as noted above, any solution
of the recurrence is uniquely determined by the k initial conditions, this map is
one-to-one and onto. Thus it is an isomorphism, and thus S has dimension k,
the order of the recurrence.
So (again, without any initial conditions), we can describe the set of solu-
tions of any linear homogeneous recurrence relation of degree k by taking linear
combinations of only k linearly independent functions. It remains to produce
those functions.
For that, we express the recurrence f(n + 1) = anf(n) + · · · + an−kf(n − k)
with a matrix equation.









an an−1 an−2 . . . an−k+1 an−k
1 0 0 . . . 0 0
0 1 0
0 0 1
.
.
.
.
.
.
...
.
.
.
0 0 0 . . . 1 0














f(n)
f(n − 1)
.
.
.
f(n − k)





=





f(n + 1)
f(n)
.
.
.
f(n − k + 1)





In trying to find the characteristic function of the matrix, we can see the pattern
in the 2×2 case
µ
an − λ an−1
1 −λ
¶
= λ2
− anλ − an−1
and 3×3 case.


an − λ an−1 an−2
1 −λ 0
0 1 −λ

 = −λ3
+ anλ2
+ an−1λ + an−2

Exercise 4 shows that the characteristic equation is this.
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
an − λ an−1 an−2 . . . an−k+1 an−k
1 −λ 0 . . . 0 0
0 1 −λ
0 0 1
.
.
.
.
.
.
...
.
.
.
0 0 0 . . . 1 −λ
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
= ±(−λk
+ anλk−1
+ an−1λk−2
+ · · · + an−k+1λ + an−k)
We call that the polynomial ‘associated’ with the recurrence relation. (We will
be finding the roots of this polynomial and so we can drop the ± as irrelevant.)
If −λk
+ anλk−1
+ an−1λk−2
+ · · · + an−k+1λ + an−k has no repeated roots
then the matrix is diagonalizable and we can, in theory, get a formula for f(n)
as in the Fibonacci case. But, because we know that the subspace of solutions
has dimension k, we do not need to do the diagonalization calculation, provided
that we can exhibit k linearly independent functions satisfying the relation.
Where r1, r2, . . . , rk are the distinct roots, consider the functions fr1
(n) =
rn
1 through frk
(n) = rn
k of powers of those roots. Exercise 5 shows that each is
a solution of the recurrence and that the k of them form a linearly independent
set. So, given the homogeneous linear recurrence f(n + 1) = anf(n) + · · · +
an−kf(n−k) (that is, 0 = −f(n+1)+anf(n)+· · ·+an−kf(n−k)) we consider
the associated equation 0 = −λk
+ anλk−1
+ · · · + an−k+1λ + an−k. We find its
roots r1, . . . , rk, and if those roots are distinct then any solution of the relation
has the form f(n) = c1rn
1 + c2rn
2 + · · · + ckrn
k for c1, . . . , cn ∈ R. (The case of
repeated roots is also easily done, but we won’t cover it here—see any text on
Discrete Mathematics.)
Now, given some initial conditions, so that we are interested in a particular
solution, we can solve for c1, . . . , cn. For instance, the polynomial associated
with the Fibonacci relation is −λ2
+ λ + 1, whose roots are (1 ±
√
5)/2 and so
any solution of the Fibonacci equation has the form f(n) = c1((1 +
√
5)/2)n
+
c2((1 −
√
5)/2)n
. Including the initial conditions for the cases n = 0 and n = 1
gives
c1 + c2 = 1
(1 +
√
5/2)c1 + (1 −
√
5/2)c2 = 1
which yields c1 = 1/
√
5 and c2 = −1/
√
5, as was calculated above.
We close by considering the nonhomogeneous case, where the relation has the
form f(n+1) = anf(n)+an−1f(n−1)+· · ·+an−kf(n−k)+b for some nonzero
b. As in the first chapter of this book, only a small adjustment is needed to make
the transition from the homogeneous case. This classic example illustrates.
In 1883, Edouard Lucas posed the following problem.
In the great temple at Benares, beneath the dome which marks
the center of the world, rests a brass plate in which are fixed three

diamond needles, each a cubit high and as thick as the body of a
bee. On one of these needles, at the creation, God placed sixty four
disks of pure gold, the largest disk resting on the brass plate, and
the others getting smaller and smaller up to the top one. This is the
Tower of Bramah. Day and night unceasingly the priests transfer
the disks from one diamond needle to another according to the fixed
and immutable laws of Bramah, which require that the priest on
duty must not move more than one disk at a time and that he must
place this disk on a needle so that there is no smaller disk below
it. When the sixty-four disks shall have been thus transferred from
the needle on which at the creation God placed them to one of the
other needles, tower, temple, and Brahmins alike will crumble into
dusk, and with a thunderclap the world will vanish. (Translation of
[De Parville] from [Ball Coxeter].)
How many disk moves will it take? Instead of tackling the sixty four disk
problem right away, we will consider the problem for smaller numbers of disks,
starting with three.
To begin, all three disks are on the same needle.
After moving the small disk to the far needle, the mid-sized disk to the middle
needle, and then moving the small disk to the middle needle we have this.
Now we can move the big disk over. Then, to finish, we repeat the process of
moving the smaller disks, this time so that they end up on the third needle, on
top of the big disk.
So the thing to see is that to move the very largest disk, the bottom disk,
at a minimum we must: first move the smaller disks to the middle needle, then
move the big one, and then move all the smaller ones from the middle needle to
the ending needle. Those three steps give us this recurence.
T(n + 1) = T(n) + 1 + T(n) = 2T(n) + 1 where T(1) = 1
We can easily get the first few values of T.
n 1 2 3 4 5 6 7 8 9 10
T(n) 1 3 7 15 31 63 127 255 511 1023

We recognize those as being simply one less than a power of two.
To derive this equation instead of just guessing at it, we write the original
relation as −1 = −T(n + 1) + 2T(n), consider the homogeneous relation 0 =
−T(n) + 2T(n − 1), get its associated polynomial −λ + 2, which obviously has
the single, unique, root of r1 = 2, and conclude that functions satisfying the
homogeneous relation take the form T(n) = c12n
.
That’s the homogeneous solution. Now we need a particular solution.
Because the nonhomogeneous relation −1 = −T(n+1)+2T(n) is so simple,
in a few minutes (or by remembering the table) we can spot the particular
solution T(n) = −1 (there are other particular solutions, but this one is easily
spotted). So we have that—without yet considering the initial condition—any
solution of T(n + 1) = 2T(n) + 1 is the sum of the homogeneous solution and
this particular solution: T(n) = c12n
− 1.
The initial condition T(1) = 1 now gives that c1 = 1, and we’ve gotten the
formula that generates the table: the n-disk Tower of Hanoi problem requires a
minimum of 2n
− 1 moves.
Finding a particular solution in more complicated cases is, naturally, more
complicated. A delightful and rewarding, but challenging, source on recur-
rence relations is [Graham, Knuth, Patashnik]., For more on the Tower of Hanoi,
[Ball Coxeter] or [Gardner 1957] are good starting points. So is [Hofstadter].
Some computer code for trying some recurrence relations follows the exercises.
Exercises
1 Solve each homogeneous linear recurrence relations.
(a) f(n + 1) = 5f(n) − 6f(n − 1)
(b) f(n + 1) = 4f(n − 1)
(c) f(n + 1) = 6f(n) + 7f(n − 1) + 6f(n − 2)
2 Give a formula for the relations of the prior exercise, with these initial condi-
tions.
(a) f(0) = 1, f(1) = 1
(b) f(0) = 0, f(1) = 1
(c) f(0) = 1, f(1) = 1, f(2) = 3.
3 Check that the isomorphism given betwween S and Rk
is a linear map. It is
argued above that this map is one-to-one. What is its inverse?
4 Show that the characteristic equation of the matrix is as stated, that is, is the
polynomial associated with the relation. (Hint: expanding down the final column,
and using induction will work.)
5 Given a homogeneous linear recurrence relation f(n + 1) = anf(n) + · · · +
an−kf(n−k), let r1, . . . , rk be the roots of the associated polynomial.
(a) Prove that each function fri (n) = rn
k satisfies the recurrence (without initial
conditions).
(b) Prove that no ri is 0.
(c) Prove that the set {fr1 , . . . , frk } is linearly independent.
6 (This refers to the value T(64) = 18, 446, 744, 073, 709, 551, 615 given in the com-
puter code below.) Transferring one disk per second, how many years would it
take the priests at the Tower of Hanoi to finish the job?

Computer Code
This code allows the generation of the first few values of a function de-
fined by a recurrence and initial conditions. It is in the Scheme dialect of
LISP (specifically, it was written for A. Jaffer’s free scheme interpreter SCM,
although it should run in any Scheme implementation).
First, the Tower of Hanoi code is a straightforward implementation of
the recurrence.
(define (tower-of-hanoi-moves n)
(if (= n 1)
1
(+ (* (tower-of-hanoi-moves (- n 1))
2)
1) ) )
(Note for readers unused to recursive code: to compute T(64), the computer
is told to compute 2∗T(63)−1, which requires, of course, computing T(63).
The computer puts the ‘times 2’ and the ‘plus 1’ aside for a moment to
do that. It computes T(63) by using this same piece of code (that’s what
‘recursive’ means), and to do that is told to compute 2 ∗ T(62) − 1. This
keeps up (the next step is to try to do T(62) while the other arithmetic is
held in waiting), until, after 63 steps, the computer tries to compute T(1).
It then returns T(1) = 1, which now means that the computation of T(2)
can proceed, etc., up until the original computation of T(64) finishes.)
The next routine calculates a table of the first few values. (Some language
notes: ’() is the empty list, that is, the empty sequence, and cons pushes
something onto the start of a list. Note that, in the last line, the procedure
proc is called on argument n.)
(define (first-few-outputs proc n)
(first-few-outputs-helper proc n ’()) )
;
(define (first-few-outputs-aux proc n lst)
(if ( n 1)
lst
(first-few-outputs-aux proc (- n 1) (cons (proc n) lst)) ) )
The session at the SCM prompt went like this.
(first-few-outputs tower-of-hanoi-moves 64)
Evaluation took 120 mSec
(1 3 7 15 31 63 127 255 511 1023 2047 4095 8191 16383 32767
65535 131071 262143 524287 1048575 2097151 4194303 8388607
16777215 33554431 67108863 134217727 268435455 536870911
1073741823 2147483647 4294967295 8589934591 17179869183
34359738367 68719476735 137438953471 274877906943 549755813887
1099511627775 2199023255551 4398046511103 8796093022207
17592186044415 35184372088831 70368744177663 140737488355327
281474976710655 562949953421311 1125899906842623
2251799813685247 4503599627370495 9007199254740991
18014398509481983 36028797018963967 72057594037927935
144115188075855871 288230376151711743 576460752303423487

1152921504606846975 2305843009213693951 4611686018427387903
9223372036854775807 18446744073709551615)
This is a list of T(1) through T(64). (The 120 mSec came on a 50 mHz ’486
running in an XTerm of XWindow under Linux. The session was edited to
put line breaks between numbers.)

Appendix
Introduction
Mathematics is made of arguments (reasoned discourse that is, not pottery
throwing). This section is a reference to the most used techniques. A reader
having trouble with, say, proof by contradiction, can turn here for an outline of
that method.
But this section gives only a sketch. For more, these are classics: Propo-
sitional Logic by Copi, Induction and Analogy in Mathematics by Pólya, and
Naive Set Theory by Halmos.
Propositions
The point at issue in an argument is the proposition. Mathematicians usually
write the point in full before the proof and label it either Theorem for major
points, Lemma for results chiefly used to prove others, or Corollary for points
that follow immediately from a prior result.
Propositions can be complex, with many subparts. The truth or falsity of
the entire proposition depends both on the truth value of the parts, and on the
words used to assemble the statement from its parts.
Not. For example, where P is a proposition, ‘it is not the case that P’ is true
provided P is false. Thus ‘n is not prime’ is true only when n is the product of
smaller integers.
We can picture the ‘not’ operation with a Venn diagram:
%
'$
P
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Where the box encloses all natural numbers, and inside the circle are the primes,
the dots are numbers satisfying ‘not P’.
To prove a ‘not P’ statement holds, show P is false.
A-1

A-2
And. Consider the statement form ‘P and Q’. For the statement to be true
both halves must hold: ‘7 is prime and so is 3’ is true, while ‘7 is prime and 3
is not’ is false.
Here is the Venn diagram for ‘P and Q’.
%
'$
%
'$
P Q
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To prove ‘P and Q’, prove each half holds.
Or. A ‘P or Q’ is true when either half holds: ‘7 is prime or 4 is prime’ is
true, while ‘7 is not prime or 4 is prime’ is false. We take ‘or’ inclusively so that
if both halves are true ‘7 is prime or 4 is not’ then the statement as a whole
is true. (In everyday speech sometimes ‘or’ is meant in an exclusive way: “Eat
your vegetables or no dessert” does not intend both halves to hold.)
The Venn diagram for ‘or’ includes all of both circles.
%
'$
%
'$
P Q
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To prove ‘P or Q’ show that at all times at least one half holds (perhaps
sometimes one half and sometimes the other, but at all times at least one).
Implication. A ‘P implies Q’ statement (perhaps instead phrased ‘if P then
Q’ or ‘P =⇒ Q’) is true unless P is true while Q is false. Thus ‘7 is prime
implies 4 is not’ is true (contrary to its use in casual speech, in mathematics ‘if
P then Q’ does not connote that P precedes Q or causes Q) while ‘7 is prime
implies 4 is also prime’ is false.
More subtly, ‘P =⇒ Q’ is always true when P is false: ‘4 is prime implies 7
is prime’ and ‘4 is prime implies 7 is not’ are both true statements, sometimes
said to be vacuously true. (We adopt this convention because we want ‘if a
number is a perfect square then it is not prime’ to be always true, for instance
when the number is 5 or when the number is 6.)
The diagram
%
'$
'

$
%
P Q
shows Q holds whenever P does (another phrasing is ‘P is sufficient to give Q’).
Notice again that if P does not hold, Q may or may not be in force.
There are two main ways to establish an implication. The first way is direct:
assume P is true and, using that assumption, prove Q. For instance, to show

A-3
‘if a number is divisible by 5 then twice that number is divisible by 10’, assume
the number is 5n and deduce that 2(5n) = 10n. The second way is indirect:
prove the contrapositive statement: ‘if Q is false then P is false’ (rephrased, ‘Q
can only be false when P is also false’). As an example, to show ‘if a number is
prime then it is not a perfect square’, argue that if it were a square p = n2
then
it could be factored p = n · n where n p and so wouldn’t be prime (of course,
p = 0 or p = 1 don’t give n p, but they are non-primes by definition)
Two comments about implication.
First, a ‘P =⇒ Q’ result can sometimes be improved by weakening P or
strengthening Q. Thus, ‘if a number is divisible by p2
then its square is also
divisible by p2
’ could be upgraded either by relaxing its hypothesis: ‘if a number
is divisible by p then its square is divisible by p2
’, or by tightening its conclusion:
‘if a number is divisible by p2
then its square is divisible by p4
’.
Second, after showing ‘P implies Q’, a good next step is to look into whether
there are cases where Q holds but P does not. The idea is to better understand
the relationship between P and Q, with an eye toward strengthening the propo-
sition.
Equivalence. An implication cannot be improved when not only does P
imply Q, but also Q implies P, ( some ways to say this are: ‘P if and only if
Q’, ‘P iff Q’, ‘P and Q are logically equivalent’, ‘P is necessary and sufficient
to give Q’, ‘P ⇐⇒ Q’). For example, ‘a number is divisible by a prime if and
only if that number squared is divisible by the prime squared’.
The picture here shows that P and Q hold in exactly the same cases.
%
'$
P Q
Although in simple cases a chain P ⇐⇒ R ⇐⇒ S ⇐⇒ · · · ⇐⇒ Q may
be practical, we often show P =⇒ Q and Q =⇒ P separately. For instance,
‘P ⇐⇒ Q’ can be shown with both of the contrapositives: ‘not Q implies not
P’ and ‘not P implies not Q’.
Quantifiers
Compare these two statements about natural numbers: ‘there is an x such
that x is divisible by x2
’ is true, while ‘for all numbers x, that x is divisible
by x2
’ is false. Clearly the prefixes deserve some study. We call the delimiting
prefixes quantifiers.
For all. The ‘for all’ prefix is called the universal quantifier and symbolized
∀.

A-4
Venn diagrams aren’t very helpful with quantifiers, but in a sense the box we
draw to border the diagram represents the universal quantifier since it dilineates
the universe of possible members.
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To show a statement holds in all cases, we must show it holds in each case.
Thus, to prove ‘every number divisible by p has its square divisible by p2
’, take a
single number of the form pn and square it: (pn)2
= p2
n2
. Hence the statement
holds for each number divisible by p. (This is a “typical element” proof. This
kind of argument requires that no properties are assumed for that element other
than those in the hypothesis—for instance, this is a wrong argument: “if n is
divisible by a prime, say 2, so that n = 2k then n2
= (2k)2
= 4k2
and the square
of the number is divisible by the square of the prime”. That is an argument
about the case p = 2, but it isn’t a proof for general p.)
There exists. The other common quantifier is ‘there exists’, symbolized ∃.
This quantifier is in some sense the opposite of ‘for all’. For instance, contrast
these two definitions of primality of an integer p: (1) for all n, if n is not 1 and
n is not p then n does not divide p, and (2) it is not the case that there exists
an n (with n 6= 1 and n 6= p) such that n divides p.
As noted above, Venn diagrams are not much help with quantifiers, but a
picture of ‘there is a number such that P’ would show both that there can be
more than one and that not all numbers need satisfy P.
%
'$
P .
.
.
.
An existence proposition can be proved by producing something satisfying
the property: one time, to settle the question of primality of 225
+ 1, Euler
produced the divisor 641. But there are proofs that show that something exists
without saying how to find it; Euclid’s argument, given in the next subsection,
shows there are infinitely many primes without naming them. In general, while
demonstrating existence is better than nothing, giving an example is better than
that, and an exhaustive list of all instances is great. Still, mathematicians take
what they can get.
Finally, along with “Are there any?” we often ask “How many?” That
is why the issue of uniqueness often arises in conjunction with questions of
existence. Many times the two arguments are simpler if separated, so note that
just as proving something exists does not show it is unique, neither does proving
something is unique show it exists.

A-5
Techniques of Proof
Induction. Many proofs are iterative, “Here’s why it’s true for 1, it then
follows for 2, from there to 3, and so on . . . ”. These are called proofs by
induction. Such a proof has two steps. In the base step the proposition is
established for some first number, often 0 or 1. Then in the inductive step we
assume the proposition holds for numbers up to some k then it holds for the
next number.
Examples explain it best.
We prove that 1 + 2 + 3 + · · · + n = n(n + 1)/2.
For the base step we must show the formula holds when n = 1. That’s
easy, the sum of the first 1 numbers equals 1(1 + 1)/2.
For the inductive step, assume the formula holds for 1, 2, . . . , k
1 = 1(1 + 1)/2
and 1 + 2 = 2(2 + 1)/2
and1 + 2 + 3 = 3(3 + 1)/2
.
.
.
and1 + · · · + k = k(k + 1)/2
and then show it therefore holds in the k + 1 case. That’s just algebra:
1 + 2 + · · · + k + (k + 1) =
k(k + 1)
2
+ (k + 1) =
(k + 1)(k + 2)
2
.
The idea of induction is simple. We’ve shown the proposition holds for 1,
and we’ve shown that if it holds for 1 then it holds for 2, and if it holds for 1
and 2 then it holds for 3, etc. Thus it holds for any natural number greater
than or equal to 1.
We prove that every integer greater than 1 is a product of primes.
The base step is easy, 2 is the product of a single prime.
For the inductive step assume each of 2, 3, . . . , k is a product of primes,
aiming to show k + 1 is also. There are two cases: (1) if k + 1 is not
divisible by a number smaller than itself then it is a prime and so the
product of primes, and (2) if k + 1 factors then each factor can be written
as a product of primes by the inductive hypothesis and so k + 1 can be
rewritten as a product of primes. That ends the proof.
(Remark. The Prime Factorization Theorem of Number Theory says that
not only does a factorization exist, but that it is unique. We’ve shown the
easy half.)
Two remarks about ‘next number’.
For one thing, while induction works on the integers, it’s no good on the
reals. There is no ‘next’ real.

A-6
The other thing is that we sometimes use induction to go down, say, from
10 to 9 to 8, etc., down to 0. This is OK—‘next number’ could mean ‘next
lowest number’. Of course, at the end we have not shown the fact for all natural
numbers, only for those less than or equal to 10.
Contradiction. Another technique of proof is to show something is true by
showing it can’t be false.
The classic example is Euclid’s, that there are infinitely many primes.
Suppose there are only finitely many primes p1, . . . , pk. Consider p1 ·
p2 . . . pk +1. None of the primes on this supposedly exhaustive list divides
that number evenly, each leaves a remainder of 1. But every number is
a product of primes so this can’t be. Thus there cannot be only finitely
many primes.
Every proof by contradiction has the same form—assume the proposition is
false and derive some contradiction to known facts.
Another example is this proof that
√
2 is not a rational number.
Suppose
√
2 = m/n. Then
2n2
= m2
.
Factor out the 2’s: n = 2kn
· n̂ and m = 2km
· m̂. Rewrite:
2 · (2kn
· n̂)2
= (2km
· m̂)2
.
The Prime Factorization Theorem says there must be the same number
of factors of 2 on both sides, but there are an odd number (1 + 2kn) on
the left and an even number (2km) on the right. That’s a contradiction
so a rational with a square of 2 cannot be.
Both these examples aimed to prove something doesn’t exist. A negative
proposition often suggests a proof by contradiction.
Sets, Functions, and Relations
Sets. The perfect squares less than 20, the roots of x5
− 3x3
+ 2, the primes—
all are collections. Mathematicians work with sets, collections that satisfy the
Principle of Extensionality stated below.
A set can be given as a listing between curly braces: {1, 4, 9, 16}, or, if that’s
unwieldy, by using set-builder notation: {x
¯
¯ x5
− 3x3
+ 2 = 0} (read “the set
of all x such that . . . ”). We name sets with capital roman letters: P =
{2, 3, 5, 7, 11, . . . } except for the set of real numbers, written R, and the set
of complex numbers, written C. To denote that something is an element (or
member) of a set we use ‘ ∈ ’, so 7 ∈ {3, 5, 7} while 8 6∈ {3, 5, 7}.

A-7
The Principle of Extensionality is that two sets with the same elements are
equal. Hence repeats collapse {7, 7} = {7} and order doesn’t matter {2, π} =
{π, 2}.
We use ‘⊂’ for the subset relationship: {2, π} ⊂ {2, π, 7} and ‘⊆’ for subset or
equality (if A is a subset of B but A 6= B then A is a proper subset of B). These
symbols may be flipped to signify the reverse relationship: {2, π, 5} ⊃ {2, 5}.
Because of Extensionality, to prove A = B just show they have the same
members. Usually we show mutual inclusion: both A ⊆ B and A ⊇ B.
Set operations. Venn diagrams are handy here. For instance, ‘x ∈ P’ can be
pictured
%
'$
P
.x
and ‘P ⊆ Q’ looks like this.
%
'$
'

$
%
P Q
Note this is also the diagram for implication. That’s because ‘P ⊆ Q’ means
x ∈ P =⇒ x ∈ Q.
In general, for every propositional logic operator there is an associated set
operator. For instance, the complement of P is Pcomp
= {x
¯
¯ not(x ∈ P)}
%
'$
P
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the union is P ∪ Q = {x
¯
¯ (x ∈ P) or (x ∈ Q)}
%
'$
%
'$
P Q
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and the intersection is P ∩ Q = {x
¯
¯ (x ∈ P) and (x ∈ Q)}.
%
'$
%
'$
P Q
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A-8
When two sets share no members their intersection is the empty set {},
symbolized ∅. Any set has the empty set for a subset by the ‘vacuously true’
property of the definition of implication.
Sequences. We shall also use collections where order does matter and where
repeats do not collapse. These are sequences, denoted with angle brackets:
h2, 3, 7i 6= h2, 7, 3i. A sequence of length 2 is sometimes called an ordered pair
and written with parentheses: (π, 3). We also sometimes say ‘ordered triple’,
‘ordered 4-tuple’, etc. The set of ordered n-tuples of elements of a set A is
denoted An
. Thus the set of pairs of reals is R2
.
Functions. Functions are used to express relationships; for instance, Galelio
rolled balls down inclined planes to find the function relating elapsed time to
distance covered.
We first see functions in elementary Algebra, where they are presented as
formulas (e.g., f(x) = 16x2
− 100), but progressing to more advanced Math-
ematics reveals more general functions—trigonometric ones, exponential and
logarithmic ones, and even constructs like absolute value that involve piecing
together parts—and we see that functions aren’t formulas, instead the key idea
is that a function associates with each input x a single output f(x).
Consequently, a function (or map) is defined to be a set of ordered pairs
(x, f(x) ) such that x suffices to determine f(x), that is x1 = x2 =⇒ f(x1) =
f(x2) (this requirement is referred to by saying a function is well-defined).∗
Each input x is one of the function’s arguments and each output f(x) is a
value. The set of all arguments is f’s domain and the set of output values is its
range. Often we don’t need to produce exactly the range, and we instead work
with a superset of the range, the codomain. The notation for a function f with
domain X and codomain Y is f : X → Y .
'

$
%
X
-
f
'

$
%
Y
¾ »
)
range of f
We sometimes instead use the notation x
f
7−→ 16x2
− 100, read ‘x maps under
f to 16x2
− 100’, or ‘16x2
− 100 is the image of x’.
Complicated maps, like x 7→ sin(1/x), can be thought of as combinations
of simple maps, for instance here applying the function g(y) = sin(y) to the
image of f(x) = 1/x. The composition of g: Y → Z with f : X → Y , denoted
g ◦ f : X → Z, is the map sending x ∈ X to g( f(x) ) ∈ Z. This definition only
makes sense if the range of f is a subset of the domain of g.
Observe that the identity map id: Y → Y, defined by id(y) = y, has the
property that for any f : X → Y , the composition id ◦ f is equal to the map f.
∗More on this is in the section on isomorphisms

A-9
So an identity map plays the same role with respect to function composition
that the number 0 plays in real number addition, or that the number 1 plays in
multiplication.
In line with that analogy, define a left inverse of a map f : X → Y to be a
function g: range(f) → X such that g ◦ f is the identity map on X. Of course,
a right inverse of f is a h: Y → X such that f ◦ h is the identity.
A map that is both a left and right inverse of f is called simply an inverse.
An inverse, if one exists, is unique because if both g1 and g2 are inverses of f
then g1(x) = g1 ◦ (f ◦ g2)(x) = (g1 ◦ f) ◦ g2(x) = g2(x) (the middle equality
comes from the associativity of function composition), so we often call it “the”
inverse, written f−1
. For instance, the inverse of the function f : R → R given
by f(x) = 2x − 3 is the function f−1
: R → R given by f−1
(x) = (x + 3)/2.
The superscript ‘f−1
’ notation for function inverse can be confusing — it
doesn’t mean 1/f(x). It is used because it fits into a larger scheme. Func-
tions that have the same codomain as domain can be iterated, so that where
f : X → X, we can consider the composition of f with itself: f ◦f, and f ◦f ◦f,
etc. Naturally enough, we write f ◦ f as f2
and f ◦ f ◦ f as f3
, etc. Note
that the familiar exponent rules for real numbers obviously hold: fi
◦ fj
= fi+j
and (fi
)j
= fi·j
. The relationship with the prior paragraph is that, where f is
invertible, writing f−1
for the inverse and f−2
for the inverse of f2
, etc., gives
that these familiar exponent rules continue to hold, once f0
is defined to be the
identity map.
If the codomain Y equals the range of f then we say that the function is onto.
A function has a right inverse if and only if it is onto (this is not hard to check).
If no two arguments share an image, if x1 6= x2 implies that f(x1) 6= f(x2),
then the function is one-to-one. A function has a left inverse if and only if it is
one-to-one (this is also not hard to check).
By the prior paragraph, a map has an inverse if and only if it is both onto
and one-to-one; such a function is a correspondence. It associates one and only
one element of the domain with each element of the range (for example, finite
sets must have the same number of elements to be matched up in this way).
Because a composition of one-to-one maps is one-to-one, and a composition of
onto maps is onto, a composition of correspondences is a correspondence.
We sometimes want to shrink the domain of a function. For instance, we
may take the function f : R → R given by f(x) = x2
and, in order to have an
inverse, limit input arguments to nonnegative reals ˆ
f : R+
→ R. Technically,
ˆ
f is a different function than f; we call it the restriction of f to the smaller
domain.
A final point on functions: neither x nor f(x) need be a number. As an
example, we can think of f(x, y) = x + y as a function that takes the ordered
pair (x, y) as its argument.
Relations. Some familiar operations are obviously functions: addition maps
(5, 3) to 8. But what of ‘’ or ‘=’? We here take the approach of rephrasing
‘3 5’ to ‘(3, 5) is in the relation ’. That is, define a binary relation on a set
A to be a set of ordered pairs of elements of A. For example, the relation is

A-10
the set {(a, b)
¯
¯ a b}; some elements of that set are (3, 5), (3, 7), and (1, 100).
Another binary relation on the natural numbers is equality; this relation is
formally written as the set {. . . , (−1, −1), (0, 0), (1, 1), . . . }.
Still another example is ‘closer than 10’, the set {(x, y)
¯
¯ |x − y| 10}. Some
members of that relation are (1, 10), (10, 1), and (42, 44). Neither (11, 1) nor
(1, 11) is a member.
Those examples illustrate the generality of the definition. All kinds of re-
lationships (e.g., ‘both numbers even’ or ‘first number is the second with the
digits reversed’) are covered under the definition.
Equivalence Relations. We shall need to say, formally, that two objects
are alike in some way. While these alike things aren’t identical, they are related
(e.g., two integers that ‘give the same remainder when divided by 2’).
A binary relation {(a, b), . . . } is an equivalence relation when it satisfies
(1) reflexivity: any object is related to itself;
(2) symmetry: if a is related to b then b is related to a;
(3) transitivity: if a is related to b and b is related to c then a is related to c.
(To see these conditions formalize being the same, read them again, replacing
‘is related to’ with ‘is like’.)
Some examples (on the integers): ‘=’ is an equivalence relation, ‘’ does
not satisfy symmetry, ‘same sign’ is a equivalence, while ‘nearer than 10’ fails
transitivity.
Partitions. In ‘same sign’ {(1, 3), (−5, −7), (−1, −1), . . . } there are two kinds
of pairs, the first with both numbers positive and the second with both negative.
So integers fall into exactly one of two classes, positive or negative.
A partition of a set S is a collection of subsets {S1, S2, . . . } such that every
element of S is in one and only one Si: (1) S1 ∪ S2 ∪ . . . = S and (2) if i is not
equal to j then Si ∩ Sj = ∅. Picture S being decomposed into distinct parts.
S:
%
$
Ã
!
¿
À
. . .
S1
S2
S3
S4
S5
Thus, the first paragraph says ‘same sign’ partitions the integers into the posi-
tives and the negatives.
¤
¤
¤
¤
¤
¤
.−2
.−3
.−4
.−1
. . .
.2
.0
.1
. . .

A-11
Similarly, the equivalence relation ‘=’ partitions the integers into one-element
sets.
. . .
¤
¤
¤
¤
¤
¤
.−1
¤
¤
¤
¤
¤
¤
.0
¤
¤
¤
¤
¤
¤
.1
¤
¤
¤
¤
¤
¤
.2
¤
¤
¤
¤
¤
¤
. . .
Before we show that equivalence relations always give rise to partitions,
we first illustrate the argument. Consider the relationship between two in-
tegers of ‘same parity’, the set {(−1, 3), (2, 4), (0, 0), . . . } (i.e., ‘give the same
remainder when divided by 2’). We want to say that the natural numbers
split into two pieces, the evens and the odds, and inside a piece each mem-
ber has the same parity as each other. So for each x we define the set of
numbers associated with it: Sx = {y
¯
¯ (x, y) ∈ ‘same parity’}. Some examples
are S1 = {. . . , −3, −1, 1, 3, . . . }, and S4 = {. . . , −2, 0, 2, 4, . . . }, and S−1 =
{. . . , −3, −1, 1, 3, . . . }. These are the parts, e.g., S1 is the odds.
Theorem. An equivalence relation induces a partition on the underlying set.
Proof. Call the set S and the relation R.
For each x ∈ S define Sx = {y
¯
¯ (x, y) ∈ R}. Observe that, as x is in Sx, the
union of all these sets is S.
All that remains is to show that distinct parts are disjoint: if Sx 6= Sy then
Sx ∩ Sy = ∅. To argue the contrapositive, assume Sx ∩ Sy 6= ∅, aiming to show
Sx = Sy. Let p be an element of the intersection, so that each of (x, p), (p, x),
(y, p), and (p, y) is in R. To show that Sx = Sy we show each is a subset of the
other.
Assume q is in Sx so (q, x) is in R. Use transitivity along with (x, p) ∈ R to
conclude that (q, p) is also an element of R. But (p, y) is in R, so another use
of transitivity gives that (q, y) is in R. Thus q is in Sy. Hence q ∈ Sx implies
q ∈ Sy, and so Sx ⊆ Sy.
The same argument in the other direction gives the other inclusion, and so
equality holds, completing the contrapositive. QED
We call each part of a partition an equivalence class (for our purposes ‘class’
means the same as ‘set’).
A last remark about classification. We often pick a single element to be the
representative.
One representative
from each class: %
$
Ã
!
¿
À
. . .
?
?
?
?
?

A-12
Usually when we pick representatives we have some natural scheme in mind. In
that case we call them the canonical class representatives.
For example, when considering the even and odd natural numbers,
¤
¤
¤
¤
¤
¤
.−1
.−3
.1
.3
. . .
.−2
.0
.2
. . .
we may pick 0 and 1 as canonical representatives because each is the smallest
nonnegative member if its class.
¤
¤
¤
¤
¤
¤
.−1
.−3
?1
.3
. . .
.−2
?0
.2
. . .
Another example is the simplest form of a fraction. We consider 3/5 and
9/15 to be equivalent fractions. That is, we partition symbols of the form ‘d/n’
where d and n are integers and n 6= 0 according to the relationship that ‘d1/n1’
is equivalent to ‘d2/n2’ if and only if ‘d1n2 = d2n1’.
All d/n with n 6= 0:
%
$
Ã
!
¿
À
. . .
.3
5
9
15 .
3
5
equivalent to 9
15
.
We usually use the reduced form symbols as representatives.
One representative
from each class: %
$
Ã
!
¿
À
. . .
? 1
2
? 2
3
? 3
5
? 2
1
? −8
7

Bibliography
[Abbot] Edwin Abbott, Flatland, Dover.
[Ackerson] R. H. Ackerson, A Note on Vector Spaces, American Mathematical
Monthly, volume 62 number 10 (Dec. 1955), p. 721.
[Agnew] Jeanne Agnew, Explorations in Number Theory, Brooks/Cole, 1972.
[Am. Math. Mon., Jun. 1931] C. A. Rupp (proposer), H. T. R. Aude (solver), problem
3468, American Mathematical Monthly, volume 37 number 6 (June-July 1931),
p. 355.
[Am. Math. Mon., Feb. 1933] V. F. Ivanoff (proposer), T. C. Esty (solver), problem
3529, American Mathematical Monthly, volume 39 number 2 (Feb. 1933),
p. 118.
[Am. Math. Mon., Jan. 1935] W. R. Ransom (proposer), Hansraj Gupta (solver), El-
ementary problem 105, American Mathematical Monthly, volume 42 number
1 (Jan. 1935), p. 47.
[Am. Math. Mon., Jan. 1949] C. W. Trigg (proposer), R. J. Walker (solver), Elemen-
tary problem 813, American Mathematical Monthly, volume 56 number 1 (Jan.
1949), p. 33.
[Am. Math. Mon., Jun. 1949] Don Walter (proposer), Alex Tytun (solver), Elemen-
tary problem 834, American Mathematical Monthly, volume 56 number 6
(June-July 1949), p. 409.
[Am. Math. Mon., Nov. 1951] Albert Wilansky, The Row-Sums of the Inverse Matrix,
American Mathematical Monthly, volume 58 number 9 (Nov. 1951), p. 614.
[Am. Math. Mon., Feb. 1953] Norman Anning (proposer), C. W. Trigg (solver), Ele-
mentary problem 1016, American Mathematical Monthly, volume 60 number
2 (Feb. 1953), p. 115.
[Am. Math. Mon., Apr. 1955] Vern Haggett (proposer), F. W. Saunders (solver), El-
ementary problem 1135, American Mathematical Monthly, volume 62 number
4 (Apr. 1955), p. 257.
[Am. Math. Mon., Jan. 1963] Underwood Dudley, Arnold Lebow (proposers), David
Rothman (solver), Elemantary problem 1151, American Mathematical
Monthly, volume 70 number 1 (Jan. 1963), p. 93.
[Am. Math. Mon., Dec. 1966] Hans Liebeck, A Proof of the Equality of Column Rank
and Row Rank of a Matrix American Mathematical Monthly, volume 73 num-
ber 10 (Dec. 1966), p. 1114.

[Anton] Howard Anton, Elementary Linear Algebra, John Wiley Sons, 1987.
[Arrow] Kenneth J. Arrow, Social Choice and Individual Values, Wiley, 1963.
[Ball Coxeter] W.W. Rouse Ball, Mathematical Recreations and Essays, revised by
H.S.M. Coxeter, MacMillan, 1962.
[Bennett] William J. Bennett, Quantifying America’s Decline, in Wall Street Journal,
15 March, 1993.
[Birkhoff MacLane] Garrett Birkhoff, Saunders MacLane, Survey of Modern Alge-
bra, third edition, Macmillan, 1965.
[Blass 1984] A. Blass, Existence of Bases Implies the Axiom of Choice, pp. 31-33, Ax-
iomatic Set Theory, J. E. Baumgartner, ed., American Mathematical Society,
Providence RI, 1984.
[Bridgman] P.W. Bridgman, Dimensional Analysis, Yale University Press, 1931.
[Casey] John Casey, The Elements of Euclid, Books I to VI and XI, ninth edition,
Hodges, Figgis, and Co., Dublin, 1890.
[Clark Coupe] David H. Clark, John D. Coupe, The Bangor Area Economy Its
Present and Future, report to the city of Bangor ME, Mar. 1967.
[Clarke] Arthur C. Clarke, Technical Error, Fantasy, December 1946, reprinted in
Great SF Stories 8 (1946), DAW Books, 1982.
[Con. Prob. 1955] The Contest Problem Book, 1955 number 38.
[Coxeter] H.S.M. Coxeter, Projective Geometry, second edition, Springer-Verlag,
1974.
[Courant Robbins] Richard Courant, Herbert Robbins, What is Mathematics?, Ox-
ford University Press, 1978.
[Cullen] Charles G. Cullen, Matrices and Linear Transformations, second edition,
Dover, 1990.
[Dalal, et. al.] Siddhartha R. Dalal, Edward B. Fowlkes, Bruce Hoadley, Lesson
Learned from Challenger: A Statistical Perspective, Stats: the Magazine for
Students of Statistics, Fall 1989, p. 3.
[Davies] Thomas D. Davies, New Evidence Places Peary at the Pole, National Geo-
graphic Magazine, vol. 177 no. 1 (Jan. 1990), p. 44.
[de Mestre] Neville de Mestre, The Mathematics of Projectiles in Sport, Cambridge
University Press, 1990.
[De Parville] De Parville, La Nature, Paris, 1884, part I, p. 285-286 (citation from
[Ball Coxeter]).
[Duncan] W.J. Duncan, Method of Dimensions, in Encyclopaedic Dictionary of
Physics Macmillan, 1962.
[Ebbing] Darrell D. Ebbing, General Chemistry, fourth edition, Houghton Mifflin,
1993.
[Ebbinghaus] H. D. Ebbinghaus, Numbers, Springer-Verlag, 1990.
[Einstein] A. Einstein, Annals of Physics, v. 35, 1911, p. 686.
[Eggar] M.H. Eggar, Pinhole Cameras, Perspecitve, and Projective Geometry, Amer-
ican Mathematicsl Monthly, August-September 1998, p. 618-630.

[Feller] William Feller, An Introduction to Probability Theory and Its Applications
(vol. 1, 3rd ed.), John Wiley, 1968.
[Finkbeiner] Daniel T. Finkbeiner III, Introduction to Matrices and Linear Transfor-
mations, third edition, W. H. Freeman and Company, 1978.
[Fraleigh Beauregard] Fraleigh Beauregard, Linear Algebra.
[Gardner] Martin Gardner, The New Ambidextrous Universe, third revised edition,
W. H. Freeman and Company, 1990.
[Gardner 1957] Martin Gardner, Mathematical Games: About the remarkable similar-
ity between the Icosian Game and the Tower of Hanoi, Scientific American,
May 1957, p. 150-154.
[Gardner, 1970] Martin Gardner, Mathematical Games, Some mathematical curiosi-
ties embedded in the solar system, Scientific American, April 1970, p. 108-112.
[Gardner, 1980] Martin Gardner, Mathematical Games, From counting votes to mak-
ing votes count: the mathematics of elections, Scientific American, October
1980.
[Gardner, 1974] Martin Gardner, Mathematical Games, On the paradoxical situations
that arise from nontransitive relations, Scientific American, October 1974.
[Giordano, Wells, Wilde] Frank R. Giordano, Michael E. Wells, Carroll O. Wilde,
Dimensional Analysis, UMAP Unit 526, in UMAP Modules, 1987, COMAP,
1987.
[Giordano, Jaye, Weir] Frank R. Giordano, Michael J. Jaye, Maurice D. Weir, The
Use of Dimensional Analysis in Mathematical Modeling, UMAP Unit 632, in
UMAP Modules, 1986, COMAP, 1986.
[Glazer] A.M. Glazer, The Structure of Crystals, Adam Hilger, 1987.
[Goult, et al.] R.J. Goult, R.F. Hoskins, J.A. Milner, M.J. Pratt, Computational
Methods in Linear Algebra, Wiley, 1975.
[Graham, Knuth, Patashnik] Ronald L. Graham, Donald E. Knuth, Oren Patashnik,
Concrete Mathematics, Addison-Wesley, 1988.
[Halmos] Paul P. Halmos, Finite Dimensional Vector Spaces, second edition, Van Nos-
trand, 1958.
[Hamming] Richard W. Hamming, Introduction to Applied Numerical Analysis, Hemi-
sphere Publishing, 1971.
[Hanes] Kit Hanes, Analytic Projective Geometry and its Applications, UMAP Unit
710, UMAP Modules, 1990, p. 111.
[Heath] T. Heath, Euclid’s Elements, volume 1, Dover, 1956.
[Hoffman Kunze] Kenneth Hoffman, Ray Kunze, Linear Algebra, second edition,
Prentice-Hall, 1971.
[Hofstadter] Douglas R. Hofstadter, Metamagical Themas: Questing for the Essence
of Mind and Pattern, Basic Books, 1985.
[Iosifescu] Marius Iofescu, Finite Markov Processes and Their Applications, John Wi-
ley, 1980.
[Kelton] Christina M.L. Kelton, Trends on the Relocation of U.S. Manufacturing, UMI
Research Press, 1983.

[Kemeny Snell] John G. Kemeny, J. Laurie Snell, Finite Markov Chains, D. Van
Nostrand, 1960.
[Kemp] Franklin Kemp Linear Equations, American Mathematical Monthly, volume
89 number 8 (Oct. 1982), p. 608.
[Knuth] Donald E. Knuth, The Art of Computer Programming, Addison Wesley, 1988.
[Lang] Serge Lang, Linear Algebra, Addison-Wesley, 1966.
[Leontief 1951] Wassily W. Leontief, Input-Output Economics, Scientific American,
volume 185 number 4 (Oct. 1951), p. 15.
[Leontief 1965] Wassily W. Leontief, The Structure of the U.S. Economy, Scientific
American, volume 212 number 4 (Apr. 1965), p. 25.
[Lieberman] David Lieberman, The Big Bucks Ad Battles Over TV’s Most Expensive
Minutes, TV Guide, Jan. 26 1991, p. 11.
[Macdonald Ridge] Kenneth Macdonald, John Ridge, Social Mobility, in British
Social Trends Since 1900, A.H. Halsey, Macmillian, 1988.
[Macmillan Dictionary] William D. Halsey, Macmillan, 1979.
[Math. Mag., Sept. 1952] Dewey Duncan (proposer), W. H. Quelch (solver), Mathe-
matics Magazine, volume 26 number 1 (Sept-Oct. 1952), p. 48.
[Math. Mag., Jan. 1957] M. S. Klamkin (proposer), Trickie T-27, Mathematics Mag-
azine, volume 30 number 3 (Jan-Feb. 1957), p. 173.
[Math. Mag., Jan. 1963, Q237] D. L. Silverman (proposer), C. W. Trigg (solver),
Quickie 237, Mathematics Magazine, volume 36 number 1 (Jan. 1963).
[Math. Mag., Jan. 1963, Q307] C. W. Trigg (proposer). Quickie 307, Mathematics
Magazine, volume 36 number 1 (Jan. 1963), p. 77.
[Math. Mag., Nov. 1967] Clarence C. Morrison (proposer), Quickie, Mathematics
Magazine, volume 40 number 4 (Nov. 1967), p. 232.
[Math. Mag., Jan. 1973] Marvin Bittinger (proposer), Quickie 578, Mathematics
Magazine, volume 46 number 5 (Jan. 1973), p. 286, 296.
[Munkres] James R. Munkres, Elementary Linear Algebra, Addison-Wesley, 1964.
[Neimi Riker] Richard G. Neimi, William H. Riker, The Choice of Voting Systems,
Scientific American, June 1976, p. 21-27.
[Nering] Evar D. Nering, Linear Algebra and Matrix Theory, second edition, John
Wiley, 1970.
[Niven Zuckerman] I. Niven, H. Zuckerman, An Introduction to the Theory of Num-
bers, third edition, John Wiley, 1972.
[Oakley Baker] Cletus O. Oakley, Justine C. Baker, Least Squares and the 3 : 40
Mile, Mathematics Teacher, Apr. 1977.
[Ohanian] Hans O’Hanian, Physics, volume one, W. W. Norton, 1985.
[Onan] Onan, Linear Algebra.
[Petersen] G. M. Petersen, Area of a Triangle, American Mathematical Monthly, vol-
ume 62 number 4 (Apr. 1955), p. 249.
[Polya] G. Polya, Patterns of Plausible Inference,

[Programmer’s Ref.] Microsoft Programmers Reference.
[Putnam, 1988, A-6] William Lowell Putnam Mathematical Competition, Problem A-
6, 1988, solution: American Mathematical Monthly, volume 96 number 8 (Oct.
1989), p. 688.
[Putnam, 1990, A-5] William Lowell Putnam Mathematical Competition, Problem A-
5, 1990.
[Rice] John R. Rice, Numerical Mathods, Software, and Analysis, second edition,
Academic Press, 1993.
[Rucker] Rudy Rucker, Infinity and the Mind, Birkhauser, 1982.
[Ryan] Patrick J. Ryan, Euclidean and Non-Euclidean Geometry: an Analytic Ap-
proach, Cambridge University Press, 1986.
[Seidenberg] A. Seidenberg, Lectures in Projective Geometry, Van Nostrand, 1962.
[Smith] Larry Smith, Linear Algebra, second edition, Springer-Verlag, 1984.
[Strang 93] Gilbert Strang The Fundamental Theorem of Linear Algebra, American
Mathematical Monthly, Nov. 1993, p. 848-855.
[Strang 80] Gilbert Strang, Linear Algebra and its Applications, second edition, Har-
court Brace Jovanovich, 1980.
[Tan] Tan, Finite Mathematics.
[Taylor] Alan D. Taylor, Mathematics and Politics: Strategy, Voting, Power, and
Proof, Springer-Verlag, 1995.
[Tilley] Burt Tilley, private communication, 1996.
[Trono] Tony Trono, compiler, University of Vermont Mathematics Department High
School Prize Examinations 1958-1991, mimeographed printing, 1991.
[USSR Olympiad no. 174] The USSR Mathematics Olympiad, number 174.
[Weston] J. D. Weston, Volume in Vector Spaces, American Mathematical Monthly,
volume 66 number 7 (Aug./Sept. 1959), p. 575-577.
[Weyl] Hermann Weyl, Symmetry, Princeton University Press, 1952.
[Wickens] Thomas D. Wickens, Models for Behavior, W.H. Freeman, 1982.
[Wilkinson 1965] Wilkinson, 1965
[Wohascum no. 2] The Wohascum County Problem Book problem number 2.
[Wohascum no. 47] The Wohascum County Problem Book problem number 47.
[Yaglom] I.M. Yaglom, Felix Klein and Sophus Lie: Evolution of the Idea of Symmetry
in the Nineteenth Century, translated by Sergei Sossinsky, Birkhäuser, 1988.
[Zwicker] William S. Zwicker, The Voters’ Paradox, Spin, and the Borda Count, Math-
ematical Social Sciences, vol. 22 (1991), p. 187-227.

Index
accuracy
of Gauss’ method, 67–71
rounding error, 68
addition
vector, 80
additive inverse, 80
adjoint matrix, 328
angle, 42
antipodal, 340
antisymmetric matrix, 139
arrow diagram, 216, 232, 238, 242, 351
augmented matrix, 14
automorphism, 163
dilation, 164
reflection, 164
rotation, 164
back-substitution, 5
basis, 113–124
change of, 238
definition, 113
orthogonal, 256
orthogonalization, 257
orthonormal, 258
standard, 114, 350
standard over the complex num-
bers, 350
string, 371
best fit line, 269
block matrix, 311
box, 321
orientation, 321
sense, 321
volume, 321
C language, 67
canonical form
for matrix equivalence, 245
for nilpotent matrices, 374
for row equivalence, 57
for similarity, 392
canonical representative, A-12
Cauchy-Schwartz Inequality, 41
Cayley-Hamilton theorem, 382
central projection, 337
change of basis, 238–249
characteristic
vectors, values, 357
characteristic equation, 360
characteristic polynomial, 360
characterized, 172
characterizes, 246
Chemistry problem, 1, 9
chemistry problem, 22
circuits
parallel, 73
series, 73
series-parallel, 74
closure, 95
of nullspace, 367
of rangespace, 367
codomain, A-8
cofactor, 327
column, 13
rank, 126
vector, 15
column rank
full, 131
column space, 126
complementary subspaces, 136
orthogonal, 263
complex numbers
vector space over, 91
component, 15
composition
self, 365
computer algebra systems, 61–62
concatenation, 134
conditioning number, 70
congruent figures, 286

congruent plane figures, 286
contradiction, A-6
convex set, 183
coordinates
homogeneous, 340
with respect to a basis, 116
correspondence, 161, A-9
coset, 193
Cramer’s rule, 331–333
cross product, 298
crystals, 143–146
diamond, 144
graphite, 144
salt, 143
unit cell, 144
da Vinci, Leonardo, 337
determinant, 294, 299–318
cofactor, 327
Cramer’s rule, 332
definition, 299
exists, 309, 315
Laplace expansion, 327
minor, 327
permutation expansion, 308, 312,
334
diagonal matrix, 209, 225
diagonalizable, 354–357
difference equation, 406
homogeneous, 406
dilation, 164, 275
representing, 203
dimension, 121
direct sum, 131
), 140
definition, 135
external, 168
of two subspaces, 136
direction vector, 35
distance-preserving, 286
division theorem, 348
dot product, 39
dual space, 193
echelon form, 5
free variable, 12
leading variable, 5
reduced, 46
eigenspace, 361
eigenvalue, eigenvector
of a matrix, 358
of a transformation, 357
elementary
matrix, 226, 275
elementary reduction operations, 4
pivoting, 4
rescaling, 4
swapping, 4
elementary row operations, 4
entry, 13
equivalence
class, A-11
canonical representative, A-12
representative, A-11
equivalence relation, A-10, A-11
isomorphism, 169
matrix equivalence, 244
matrix similarity, 351
row equivalence, 50
equivalent statements, A-3
Erlanger Program, 286
Euclid, 286
even functions, 99, 138
even polynomials, 398
external direct sum, 168
Fibonacci sequence, 405
field, 141–142
definition, 141
finite-dimensional vector space, 119
flat, 36
form, 55
free variable, 12
full column rank, 131
full row rank, 131
function, A-8
inverse image, 185
codomain, A-8
composition, 215, A-8
correspondence, A-9
domain, A-8
even, 99
identity, A-8
inverse, 231, A-9
left inverse, 231
multilinear, 305
odd, 99
one-to-one, A-9
onto, A-9
range, A-8

restriction, A-9
right inverse, 231
structure preserving, 161, 165
seehomomorphism, 176
two-sided inverse, 231
well-defined, A-8
zero, 177
Fundamental Theorem
of Linear Algebra, 268
Gauss’ method, 2
accuracy, 67–71
back-substitution, 5
elementary operations, 4
Gauss-Jordan, 46
Gauss-Jordan, 46
generalized nullspace, 367
generalized rangespace, 367
Geometry of Linear Maps, 274–279
Gram-Schmidt process, 255–260
homogeneous coordinate vector, 340
homogeneous coordinates, 291
homogeneous equation, 21
homomorphism, 176
composition, 215
matrix representing, 194–204
nonsingular, 190, 207
nullity, 188
nullspace, 188
rangespace, 184
rank, 206
zero, 177
ideal line, 342
ideal point, 342
identity
function, A-8
matrix, 224
ill-conditioned, 68
implication, A-2
improper subspace, 92
incidence matrix, 228
index
of nilpotency, 370
induced map, 275
induction, 23, A-5
inner product, 39
Input-Output Analysis, 63–66
internal direct sum, 135
invariant
subspace, 377
invariant subspace
definition, 389
inverse, 231, A-9
additive, 80
exists, 231
left, 231, A-9
matrix, 329
right, 231, A-9
two-sided, A-9
inverse function, 231
inverse image, 185
inversion, 313
isometry, 286
isomorphism, 159–175
characterized by dimension, 172
definition, 161
of a space with itself, 163
Jordan block, 388
Jordan form, 379–398
represents similarity classes, 392
kernel, 188
Kirchhoff’s Laws, 73
Klein, F., 286
Laplace expansion, 326–330
computes determinant, 327
leading variable, 5
least squares, 269–273
length, 39
Leontief, W., 63
line
best fit, 269
in projective plane, 341
line at infinity, 342
line of best fit, 269–273
linear
transpose operation, 131
linear combination, 52
Linear Combination Lemma, 52
linear equation, 2
coefficients, 2
constant, 2
homogeneous, 21
inconsistent systems, 269
satisfied by a vector, 15
solution of, 2

Gauss’ method, 3
Gauss-Jordan, 46
solutions of
Cramer’s rule, 332
system of, 2
linear map
dilation, 275
reflection, 289
rotation, 274, 288
seehomomorphism, 176
skew, 276
trace, 397
linear recurrence, 406
linear recurrences, 405–412
linear relationship, 103
linear surface, 36
linear transformation
seetransformation, 180
linearly dependent, 103
linearly independent, 103
LINPACK, 61
map
distance-preserving, 286
extended linearly, 173
induced, 275
self composition, 365
Maple, 61
Markov chains, 280–285
Markov matrix, 284
Mathematica, 61
mathematical induction, 23
MATLAB, 61
matrix, 13
adjoint, 328
antisymmetric, 139
augmented, 14
block, 246, 311
change of basis, 238
cofactor, 327
column, 13
column space, 126
conditioning number, 70
determinant, 294, 299
diagonal, 209, 225
diagonalizable, 354
diagonalized, 244
elementary reduction, 226, 275
entry, 13
equivalent, 244
identity, 220, 224
incidence, 228
induced map, 275
inverse, 329
main diagonal, 224
Markov, 229, 284
matrix-vector product, 197
minimal polynomial, 220, 380
minor, 327
multiplication, 215
nilpotent, 370
orthogonal, 288
orthonormal, 286–291
permutation, 225
rank, 206
representation, 196
row, 13
row equivalence, 50
row rank, 124
row space, 124
scalar multiple, 212
similar, 324
similarity, 351
singular, 27
skew-symmetric, 311
submatrix, 303
sum, 212
symmetric, 118, 139, 213, 220, 228,
268
trace, 213, 229, 397
transpose, 19, 126, 213
triangular, 204, 229, 330
unit, 222
Vandermonde, 311
matrix equivalence, 242–249
canonical form, 245
definition, 244
matrix:form, 55
mean
arithmetic, 44
geometric, 44
method of powers, 399–402
minimal polynomial, 220, 380
minor, 327
morphism, 161
multilinear, 305
multiplication
matrix-matrix, 215

matrix-vector, 197
mutual inclusion, A-7
natural representative, A-12
networks, 72–78
Kirchhoff’s Laws, 73
nilpotent, 368–378
canonical form for, 374
definition, 370
matrix, 370
transformation, 370
nilpotentcy
index, 370
homomorphism, 190
matrix, 27
normalize, 258
nullity, 188
nullspace, 188
closure of, 367
generalized, 367
Octave, 61
odd functions, 99, 138
order
of a recurrence, 406
orientation, 321, 324
orthogonal, 42
basis, 256
complement, 263
mutually, 255
projection, 263
orthogonal matrix, 288
orthogonalization, 257
orthonormal basis, 258
orthonormal matrix, 286–291
parallelepiped, 321
parallelogram rule, 35
parameter, 13
partial pivoting, 69
partition, A-10–A-12
matrix equivalence classes, 244, 247
row equivalence classes, 50
partitions
into isomorphism classes, 170
permutation, 308
inversions, 313
matrix, 225
signum, 314
permutation expansion, 308, 312, 334
perp, 263
perpendicular, 42
perspective
triangles, 343
Physics problem, 1
pivoting
full, 69
pivoting on rows, 4
plane figure, 286
congruence, 286
point
at infinity, 342
in projective plane, 339
polynomial
even, 398
minimal, 380
of map, matrix, 379
polynomials
division theorem, 348
populations, stable, 403–404
powers, method of, 399–402
preserves structure, 176
projection, 176, 185, 250, 268, 385
along a subspace, 260
central, 337
vanishing point, 337
into a line, 251
into a subspace, 260
orthogonal, 251, 263
Projective Geometry, 337–346
projective geometry
Duality Principle, 341
projective plane
ideal line, 342
ideal point, 342
lines, 341
proof techniques
induction, 23
proper subspace, 92
rangespace, 184
closure of, 367
generalized, 367
rank, 128, 206
column, 126
of a homomorphism, 184, 188
recurrence, 327, 406
homogeneous, 406
initial conditions, 406

reduced echelon form, 46
reflection, 289
glide, 289
reflection (or flip) about a line, 164
relation, A-9
equivalence, A-10
relationship
linear, 103
representation
of a matrix, 196
of a vector, 116
representative, A-11
canonical, A-12
for row equivalence classes, 57
of matrix equivalence classes, 245
of similarity classes, 392
rescaling rows, 4
restriction, A-9
rigid motion, 286
rotation, 274, 288
rotation (or turning), 164
represented, 199
row, 13
rank, 124
vector, 15
row equivalence, 50
row rank
full, 131
row space, 124
scalar, 80
scalar multiple
matrix, 212
vector, 15, 34, 80
scalar product, 39
Schwartz Inequality, 41
SciLab, 61
self composition
of maps, 365
sense, 321
sequence, A-8
concatenation, 134
sets, A-6
dependent, independent, 103
empty, 105
mutual inclusion, A-7
proper subset, A-7
span of, 95
subset, A-7
sgn
seesignum, 314
signum, 314
similar, 298, 324
canonical form, 392
similar matrices, 351
similarity, 351–364
similarity transformation, 364
singular
matrix, 27
size, 319, 321
skew, 276
skew-symmetric, 311
span, 95
of a singleton, 99
spin, 149
square root, 398
stable populations, 403–404
standard basis, 114
Statics problem, 5
string, 371
basis, 371
of basis vectors, 369
structure
preservation, 176
submatrix, 303
subspace, 91–101
closed, 93
complementary, 136
definition, 91
direct sum, 135
improper, 92
independence, 135
invariant, 389
orthocomplement, 139
proper, 92
sum, 132
sum
of matrices, 212
of subspaces, 132
vector, 15, 34, 80
summation notation
for permutation expansion, 308
swapping rows, 4
symmetric matrix, 118, 139, 213, 220
system of linear equations, 2
Gauss’ method, 2
solving, 2
trace, 213, 229, 397
transformation

composed with itself, 365
diagonalizable, 354
eigenspace, 361
eigenvalue, eigenvector, 357
Jordan form for, 392
minimal polynomial, 380
nilpotent, 370
canonical representative, 374
projection, 385
size change, 321
transpose, 19, 126
determinant, 309, 317
interaction with sum and scalar
multiplication, 213
Triangle Inequality, 40
triangular matrix, 229
Triangularization, 204
trivial space, 84, 114
turning map, 164
unit matrix, 222
Vandermonde matrix, 311
vanishing point, 337
vector, 15, 33
angle, 42
canonical position, 33
column, 15
component, 15
cross product, 298
direction, 35
dot product, 39
free, 33
homogeneous coordinate, 340
length, 39
orthogonal, 42
representation of, 116, 238
row, 15
satisfies an equation, 15
scalar multiple, 15, 34, 80
sum, 15, 34, 35, 80
unit, 43
zero, 22, 80
vector space, 80–101
basis, 113
closure, 80
complex scalars, 91
definition, 80
dimension, 121
dual, 193
finite dimensional, 119
homomorphism, 176
isomorphism, 161
map, 176
over complex numbers, 347
subspace, 91
trivial, 84, 114
volume, 321
voting paradox, 147
majority cycle, 147
rational preference order, 147
voting paradoxes, 147–151
spin, 149
well-defined, A-8
Wheatstone bridge, 75
zero
divisor, 220
zero divison, 237
zero divisor, 220
zero homomorphism, 177
zero vector, 22, 80

Linear Algebra_ Theory_Jim Hefferon

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Linear Algebra_ Theory_Jim Hefferon