mmtoro/doc/multilineal/linearbrown... · preface for the past two years, i have been teaching a...

A Second Course in Linear Algebra


WILLIAM C. BROWN Michigan State University East Lansing, Michigan

A Wiley-lnterscience Publication

JOHN WILEY & SONS

New York • Chichester • Brisbane • Toronto • Singapore

Copyright © 1988 by John Wiley & Sons, Inc. All rights reserved. Published simultaneously in Canada.

Reproduction or translation of any part of this work beyond that permitted by Section 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons, Inc.

Library of Congress Cataloging-in-Publication Data

Brown, William C. (William Clough), 1943-A second course in linear algebra.

"A Wiley-Interscience publication." Bibliography: p. Includes index. 1. Algebras, Linear.

QA184.B765 1987 ISBN 0-471-62602-3

I. Title. 512'.5

Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

87-23117

To Linda

Preface

For the past two years, I have been teaching a first-year graduate-level course in linear algebra and analysis. My basic aim in this course has been to prepare students for graduate-level work. This book consists mainly of the linear algebra in my lectures. The topics presented here are those that I feel are most important for students intending to do advanced work in such areas as algebra, analysis, topology, and applied mathematics.

Normally, a student interested in mathematics, engineering, or the physical sciences will take a one-term course in linear algebra, usually at the junior level. In such a course, a student will first be exposed to the theory of matrices, vector spaces, determinants, and linear transformations. Often, this is the first place where a student is required to do a mathematical proof. It has been my experience that students who have had only one such linear algebra course in their undergraduate training are ill prepared to do advanced-level work. I have written this book specifically for those students who will need more linear algebra than is normally covered in a one-term junior-level course.

This text is aimed at seniors and beginning iiraduate students who have had at least one course in linear algebra. The text has been designed for a onequarter or semester course at the senior or first-year graduate level. It is assumed that the reader is familiar with such animals as functions, matrices, determinants, and elementary set theory. The presentation of the material in this text is deliberately formal, consisting mainly of theorems and proofs, very much in the spirit of a graduate-level course.

The reader will nQte that many familiar ideas are discussed in Chapter I. I urge the reader not to skip this chapter. The topics are familiar, but my approach, as well as the notation I use; is more sophisticated than a junior-level

vii

viii PREFACE

treatment. The material discussed in Chapters II-V is usually only touched upon (if at all) in a one-term course. I urge the reader to study these chapters carefully.

Having written five chapters for this book, I obviously feel that the reader should study all five parts of the text. However, time considerations often demand that a student or instructor do less. A shorter but adequate course could consist of Chapter I, Sections 1-6, Chapter II, Sections 1 and 2, and Chapters III and V. If the reader is willing to accept a few facts about extending scalars, then Chapters III, IV, and V can be read with no reference to Chapter II. Hence, a still shorter course could consist of Chapter I, Sections 1-6 and Chapters III and V.

It is my firm belief that any second course in linear algebra ought to contain material on tensor products and their functorial properties. For this reason, I urge the reader to follow the first version of a short course if time does not permit a complete reading of the text. It is also my firm belief that the basic linear algebra needed to understand normed linear vector spaces and real inner product spaces should not be divorced from the intrinsic topology and analysis involved. I have therefore presented the material in Chapter IV and the first half of Chapter V in the same spirit as many analysis texts on the subject. My original lecture notes on normed linear vector spaces and (real) inner product spaces were based on Loomis and Sternberg's classic text Advanced Calculus. Although I have made many changes in my notes for this book, I would still like to take this opportunity to acknowledge my debt to these authors and their fine text for my current presentation of this material.

One final word about notation is in order here. All important definitions are clearly displayed in the text with a number. Notation for specific ideas (e.g., 1\1 for the set of natural numbers) is introduced in the main body of the text as needed. Once a particular notation is introduced, it will be used (with only a few exceptions) with the same meaning throughout the rest of the text. A glossary of notation has been provided at the back of the book for the reader's convenience.

East Lansing, Michigan September 1987

WILLIAM C. BROWN

Contents

Chapter I. Linear Algebra

1. Definitions and Examples of Vector Spaces 2. ·Bases and Dimension 3. Linear Transformations 4. Products and Direct Sums 5. Quotient Spaces and the Isomorphism Theorems 6. Duals and Adjoints 7. Symmetric Bilinear Forms

Chapter II. Multilinear Algebra

1. Multilinear Maps and Tensor Products 2. Functorial Properties of Tensor Products 3. Alternating Maps and Exterior Powers 4. Symmetric Maps and Symmetric Powers

Chapter Ill. Canonical Forms of Matrices

1. Preliminaries on Fields 2. Minimal and Characteristic Polynomials 3. Eigenvalues and Eigenvectors 4. The Jordan Canonical Form 5. The Real Jordan Canonical Form 6. The Rational Canonical Form

1

1 8

17 30 38 46 53

59

59 68 83 94

98

98 105 117 132 141 159

hr

X CONTENTS

Chapter IV. Normed Linear Vector Spaces 171

1. Basic Definitions and Examples 171 2. Product Norms and Equivalence 180 3. Sequential Compactness and the Equivalence of

Norms 186 4. Banach Spaces 200

Chapter V. Inner Product Spaces

1. Real Inner Product Spaces 2. Self-adjoint Transformations 3. Complex Inner Product Spaces 4. Normal Operators

Glossary of Notation

References

Subject Index

206

206 221 236 243

254

259

261


Chapter I

Linear Algebra

1. DEFINITIONS AND EXAMPLES OF VECTOR SPACES

In this book, the symbol F will denote an arbitrary field. A field is defined as follows:

Definition 1.1: A nonempty set F together with two functions (x, y)-+ x + y and (x, y) -+ xy from F x F to F is called a field if the following nine axioms are satisfied:

Fl. X + y = y + X for all X, y E F. F2. x + (y + z) = (x + y) + z for all x, y, z e F. F3. There exists a unique· element 0 e F such that x + 0 = x for all x e F. F4. For every xeF, there exists a unique element -xeF such that

x + (-x) = 0. FS. xy = yx for all x, yeF. F6. x(yz) = (xy)z for all x, y, z e F.

F7. There exists a unique element 1 :/= 0 in F such that xl = x for all xeF. F8. For every x :/= 0 in F, there exists a unique yeF such that xy = 1. F9. x(y + z) = xy + xz for all x, y, zeF.

Strictly speaking a field is an ordered triple (F, (x, y) -+ x + y, (x, y)-+ xy) satisfying axioms Fl-F9 above. The map from F x F-+ F given by (x, y) -+ x + y is called addition, and the map (x, y) -+ xy is called multiplication. When referring to some field (F, (x, y)-+ x + y, (x, y)-+ xy), references to addition and multiplication are dropped from the notation, and the letter F is used to

2 LINEAR ALGEBRA

denote both the set and the two maps satisfying axioms F1-F9. Although this procedure is somewhat ambiguous, it causes no confusion in concrete situations. In our first example below, we introduce some notation that we shall use throughout the rest of this book.

Example 1.2: We shall let iQ denote the set of rational numbers, IR, the set of real numbers, and C, the set of complex numbers. With the usual addition and multiplication, iQ, IR, and C are all fields with iQ ~ IR ~ C. 0

The fields in Example 1.1 are all infinite in the sense that the cardinal number attached to the underlying set in question is infinite. Finite fields are very important in linear algebra as well. Much of coding theory is done over finite algebraic extensions of the field IFP described in Example 1.3 below.

Example 1.3: Let 7L denote the set of integers with the usual addition x + y and multiplication xy inherited from iQ. Let p be a positive prime in 7L and set IFP = {0,1, ... , p -1}. IFP becomes a (finite) field if we define additi?n E9 .and multiplication · modulo p. Thus, for elements x, y e IF P' there extst umque integers k, z e 7L such that x + y = kp + z with z e IF p· We define x E9 y to be z. Similarly, x·y = w where xy = k'p +wand 0 ~ w < p. .

The reader can easily check that (IFP, Ee, ·)satisfies axioms Fl-F9. Thus,IFP ts a finite field of cardinality p. 0

Except for some results in Section 7, the definitions and theorems in Chapter I are completely independent of the field F. Hence, we shall assume that F is an arbitrary field and study vector spaces over F.

Definition 1.4: A vector space V over F is a nonempty set together with two functions, (IX, p)-+ a:+ p from V x V to V (called addition) and (x, IX)-+ xa: from F x V to V (called scalar multiplication), which satisfy the following axioms:

Vl. a.+ p = p +a for all a., pev. V2. a:+ (p + y) =(IX+ p) + y for all a., p, yeV. V3. There exists an element 0 e V such that 0 + a. = a. for all a e V. V4. For every IXEV, there exists a peV such that a+ P = 0. V5. (xy)a: = x(ya.) for all x, y e F, and IX e V.

V6. x(IX + p) = xa. + xP for all xeF, and IX, PeV. V7. (x + y)IX = xa. + y(l. for all x, y e F, and IX e V. V8. 1a: = a: for all a: e V.

As with fields, we should make the comment that a vector space over F is really a triple (V, (a, p) -+a: + p, (x, a:) -+ xa.) consisting of a no?en;tpty s~t V together with two functions from V x V to V and F x V to .v satisfymg ~xtoms Vl-V8. There may be many different ways to endow a gtven set V Wtth the

DEFINITIONS AND EXAMPLES OF VECTOR SPACES 3

structure of a vector space over F. Nevertheless, we shall drop any reference to addition and scalar multiplication when no confusion can arise and just use the notation V to indicate a given vector space over F.

IfV is a vector space over F, then the elements ofV will be called vectors and the elements ofF scalars. We assume the reader is familiar with the elementary arithmetic in V, and, thus, we shall use freely such expressions as -a:, a: - p, and a1 + · · · + 1X0 when dealing with vectors in V. Let us review some well-known examples of vector spaces.

Example 1.5: Let N = {1, 2, 3, ... } denote the set of natural numbers. For each n eN, we have the vector space F 0 = {(x1, ... , x.J I x1 e F} consisting of all ntuples of elements from F. Vector addition and scalar multiplication are defined componentwise by (x1, ... ,x.J + (Yt•· .. ,y.J = (xl + Yt>· .. ,xn + y.J and x(x1 , ••. , x.J = (xx1 , .•• , xx.J. In particular, when n = 1, we see F itself is a vector space over F. 0

If A and B are two sets, let us denote the set of functions from A to B by BA. Thus, BA = {f: A-+ B 1 f is a function}. In Example 1.5, P can be viewed as the set of functions from {1, 2, ... , n} to F. Thus, IX= (x1 , ... , x.JeFn is identified with the function g.. e F 1

• .. •• nl given by g..(i) = x1 fori = 1, ... , n. These remarks suggest the following generalization of Example 1.5.

Example 1.6: Let V be a vector space over F and A an arbitrary set. Then the set yA consisting of all functions from A to V becomes a vector space over F when we define addition and scalar multiplication pointwise. Thus, iff,geVA, f + g is the function from A to V defined by(f + g)(a) = f(a) + g(a) for all a eA. For xeF and feVA, xf is de.fined by (xf)(a) = x(f(a)). 0

If A is a finite set of cardinality n in Example 1.6, then we shall shorten our notation for the vector space yA and simply write vn. In particular, if V = F, then yn = F 0 and we recover the example in 1.5.

Example 1.7: We shall denote the set of m x n matrices (a11) with coefficients aiJeF by Mmxn(F). The usual addition of matrices (a1J)+ (b1J) = (aiJ + biJ) and scalar multiplication x(aiJ) = (xa1J make Mm x 0 (F) a vector space over F. 0

Note that our choice of notation implies that Fn and M 1 x 0(F) are the same

vector space. Although we now have two different notations for the same vector space, this redundancy is useful and will cause no confusion in the sequel.

Example 1.8: We shall let F[X] denote the set of all polynomials in an indeterminate X over F. Thus, a typical element in F[X] is a finite sum of the form a0 X0 +a0 _ 1xn-l+ .. ·+a0 • Here neNu{O}, and a0 , ... ,a

0EF. The

usual notions of adding two polynomials and multiplying a polynomial by a

4 LINEAR ALGEBRA

constant, which the reader is familiar with from the elementary calculus, make sense over any field F. These operations give F[X] the structure of a vector space over F. D

Many interesting examples of vector spaces come from analysis. Here are some typical examples.

Example 1.9: Let I be an interval (closed, open, or half open) in IR. We shall let C(I) denote the set of all continuous, real valued functions on I. If keN, we shall let Ck(I) denote those f e C(I) that are k-times differentiable on the interior of I. Then C(I) 2 C1(I) 2 C2(I) 2 · · ·. These sets are all vector spaces over IR when endowed with the usual pointwise addition (f + g)(x) = f(x) + g(x), x e I, and scalar multiplication (yf)(x) = y(f(x)). D

Example 1.10: Let A = [a1, b 1] x · · · x [~. bJ s;;; IR0 be a closed rectangle. We shall let Bl(A) denote the set of all real valued functions on A that are Riemann integrable. Clearly Bl(A) is a vector space over IR when addition and scalar multiplication are defined as in Example 1.9. D

We conclude our list of examples with a vector space, which we shall study carefully in Chapter III.

Example 1.11: Consider the following system of linear differential equations:

Here f1 , ••. , fn e C1(I), where I is some open interval in IR. fj denotes the derivative off1, and the au are scalars in IR. Set A= (a1J)e M0 xn(IR). A is called the .· matrix of the system. If B is any matrix, we shall let Bt denote the transpose of B. Set f = (f1 , ••• , f0)l, We may think off as a function from {1, ... , n} to C 1(I), that is, f e C 1(I)n. With this notation, our system of differential equations becomes f' = Af. The set of solutions to our system is V = {f e C 1(1)0 If' = Af}. Clearly, V is a vector space over IR if we define addition and scalar multiplication componentwise as in Example 1.9. D

Now suppose Vis a vector space over F. One rich source of vector spaces associated with V is the set of subspaces of V. Recall the following definition:

Definition 1.12: A nonempty subset W of V is a subspace of V if W is a vector space under the same vector addition and scalar multiplication as for V.

Thus, a subset W of V is a subspace if W is closed under the operations of V. For example, C([a, b]), Ck([a, b]), IR[X], and Bl([a, b]) are all subspaces of IJlla,bJ.

DEFINITIONS AND EXAMPLES OF VECTOR SPACES 5

If we have a collection f/' = {W11ieA} ofsubspaces ofV, then there are some obvious ways offorming new subspaces from f/'. We gather these constructions together in. the following example:

Example 1.13: Let f/' = {Wd ieA} be an indexed collection of subspaces of V. In what follows, the indexing set A of f/' can be finite or infinite. Certainly the intersection, nleA WI> Of the SUbspaceS in f/' is a SUbspace Of V. The set of all :finite sums of vectors from UteA W1 is also a subspace ofV. We shall denote this subspace by LteA W1• Thus, LteA W, = {LteAa:tla:,eW, for all ieA}. Here and throughout the rest of this book, if A is infinite, then the notation LteA a:1 means that all a:1 are zero except possibly for finitely many i eA. If A is finite, then without any loss of generality, we can assume A = { 1, ... , n} for some n e ~1. (If A= </J, then Ve<1 w, = (0).) We shall then write LieA w, = wl + ... + Wn.'

If f/' has the property that for every i, j e !J. there exists a k e !J. such that w, u wj s;;; w k• then clearly UteA w, is a subspace of v. D

In general, the union of two subspaces of V is not a subspace of V. In fact, if W 1 and W 2 are subspaces of V, then W 1 u W 2 is a subspace if and only if W 1 s;;; W 2 or W 2 s;;; W 1• This fact is easy to prove and is left as an exercise. In our first theorem, we discuss one more important fact about unions.

Theorem 1.14: Let V be a vector space over an infinite field F. Then V cannot be the union of a finite number of proper subspaces.

Proof Suppose W 1 , ••• , W n are proper subspaces of V such that V = W 1 u · · · u Wn·. We shall show that this equation is impossible. We remind the reader that a subspace W of V is proper if W =F V. Thus, V - W =F cf> for a proper subspace W of V.

We may assume without loss of generality that W 1 ¢ W 2 u · · · u W n· Let a.eW1 - Ui= 2 W1• Let PeV- W 1• Since F is infinite, and neither a: nor Pis zero, !J. = {a: + xp 1 x e F} is an infinite subset of V. Since there are only finitely many subspaces W" there exists a j e { 1, ... , n} such that !J. n W J is infinite.

Suppose j e { 2, ... , n}. Then there exist two nonzero scalars x, x' e F such that x =F x', and a:+ xp, a.+ x'PeWJ. Since WJ is a subspace, (x'- x)a: = x'(a: + xp)- x(a: + x'P)eWJ. Since x'- x:).!: 0, we conclude a:eWJ. But this is contrary to our choice of a:¢ W 2 u · · · u W n. Thus, j = 1.

Now if j = 1, then again there exist two nonzero scalars x, x' e F such that x =F x', and a.+ xp, a:+ x'PeW1• Then (x- x1P =(a:+ xp)- (a:+ x'p)eW1•

Since x - x' =F 0, peW 1• This is impossible since P was chosen in V - W 1• We conclude that Y cannot be equal to the union of W 1 , ••• , W n· This completes the proof of Theorem 1.14. D

IfF is finite, then·Theorem 1.14 is false in general. For example, let V = (IF2)2

•

Then V = W 1 uW2 uW3, where W 1 = {(0,0), (1, 1)}, W 2 = {(0, 0),(0, 1)}, and w 3 = {(0, 0), (1, 0)}.

6 LINEAR ALGEBRA

Any subset S of a vector space V determines a subspace L(S) = n{W I W a subspace ofV, W 2 S}. We shall call L(S) the linear span of S. Clearly, L(S) is the smallest subspace of V containing S. Thus, in Example 1.13, for instance,

L(UieA WJ = Lie4 WI. Let &l(V) denote the set of all subsets of V. If 9'(V) denotes the set of all

subspaces of V, then 9'(V) £ &'(V), and we have a natural function L: &'(V) _.. 9'(V), which sends a subset S e &'(V) to its linear span L(S) e 9'(V). Clearly, L is a surjective map whose restriction to .9'(V) is the identity. We conclude this section with a list of the more important properties of the function L(·).

Theorem 1.15: The function L: &'(V) -+ 9'(V) satisfies the following poperties:

(a) For S e&'(V), L(S) is the subspace of V consisting of all finite linear combinations of vectors from S. Thus,

(b) If S1 £ S2, then L(S1) £ L(S2). (c) If ex e L(S), then there exists a finite subset S' £ S such that ex e L(S').

(d) S £ L(S) for all S E &'(V).

(e) For every Se&'(V), L(L(S)) = L(S).

(f) If fJ e L(S u {ex}) and fJ ¢ L(S), then ex E L(S u { fJ} ). Here IX, fJ E V and Se&'(V).

Proof: Properties (a)-( e) follow directly from the definition of the linear span. We prove (f). If fJ e L(S u { o:}) - L(S), then fJ is a finite linear combination of vectors from S u {ex}. Furthermore, ex must occur with a nonzero coefficie~t in any such linear combination. Otherwise, fJ e L(S). Thus, there extst vectors cx1 , ••• , cxn E S and nonzero scalars x1, ... , Xn, Xn + 1 E F such that fJ = x 1 IX 1 + .. · + XnOCn + Xn + 1 a. Since Xn + 1 =f: 0, we can writ~ ~ as a linear

· · d N I -t fJ - 1 combmat10n of fJ an oc1, ... , ocn. arne y, IX= Xn+l - Xn+1X11X1-.. ·- x,;;1xnocn. Thus, cxeL(S u {fJ}). D ·

EXERCISES FOR SECTION 1

(1) Complete the details in Example 1.3 and argue (IF P' Ea, ·) is a field.

(2) Let IR(X) = {f(x)fg(x)jf, ge!R[X] and g =f: 0} denote the set of ~a.tional functions on !fit Show that IR(X) is a field under the usual defimtton of addition f/g + hfk = (kf + gh)/gk and multiplication (f/g)(hfk) = th/gk IR(X) is called the field of rational functions over IR. Does F(X) make sense for any field F?

EXERCISES FOR SECTION 1 7

(3) Set F ={a+ bj='Sja, beQ}. Show that F is a subfield of C, that is, F is a field under complex addition and multiplication. Show that

{a + bJ=S"I a, b integers} is not a subfield of C.

( 4) Let I be an open interval in !fit Let a E I. Let V a = { f e IR11 f has a derivative at a}. Show that Ya is a subspace of IR1•

(5) The vector space IRN is just the set of all sequences { a1} = (at> a2, a3

, ••• )

with a1 E IR. What are vector addition and scalar multiplication here?

(6) Show that the following sets are subspaces of IRN: (a) W 1 = { { a1} E IRN llimi .... co a1 = 0}. (b) W2 = {{a1}e1RNI{a1} is a bounded sequence}.

(c) W 3 = {{a1}e!RNIL~ 1 a~ < oo}.

(7) Let (a1 , ••• , aJeFn- (0). Show that {(x1 , ••• , xJePILi'= 1 a1x1 = 0} is a proper subspace of P.

(8) Identify all subspaces of IR2. Find two subspaces W 1 and W 2 of IR2 such that W 1 u W 2 is not a subspace.

(9) Let V be a vector space over F. Suppose W 1 and W 2 are subspaces of V. Show that W 1 u W 2 is a subspace of V if and only if W 1 £ W 2 or W2 £W1.

(10) Consider the following subsets of IR[X]:

(a) W 1 = {fe!R[X] lf(O) = 0}.

(b) W 2 = {fe IR[X] 12f(O) = f(1)}.

(c) W 3 = {fe IR[X] I the degree off~ n}.

(d) w4 = {fe!R[X] lf(t) = f(1- t) for all te!R}.

In which of these cases is W1 a subspace of IR[X]?

(11) Let K, L, and M be subspaces of a vector space V. Suppose K ~ L. Prove Dedekind's modular law: K n (L + M) = L + (K n M).

(12) Let V = IR3• Show that o1 = (1, 0, 0) is not in the linear span of a, {J, andy

where ex = (1, 1, 1), fJ = (0, 1, -1), and y = (1, 0, 2) .

(13) If S1 and S2 are subsets of a vector space V, show that L(S1 u S2) = L(S1) + L(Sa).

(14) Let S be any subset of IR[X] £ !Rill. Show that e'" ¢ L(S).

{15) Let tX1 = (a11'llt2)eF2 fori= 1, 2. Show that F 2 = L({cx1, cx2}) if and only if the determinant of the 2 x 2 matrix M = (a1J) is nonzero. Generalize this result to Fn.

(16) Generalize Example 1.8 to n + 1 variables X0 , ••• , Xn. The resulting vector space over F is called the ring of polynomials in n + 1 variables (over F). It is denoted F[X0 , ••• , XJ. Show that this vector space is spanned by all monomials Xg'0 , ••• , X:• as (m0 , ••• , mJe(N u {O})n+ 1•

8 LINEAR ALGEBRA

(17) A polynomial f e F[X0, ... , XJ is said to be homogeneous of degree d iff is a finite linear combination of monomials Xg'O, ... , X:'" of degree d (i.e., m0 + · · · + mn = d). Show that the set of homogeneous polynomials of degree d is a subspace of F[X0 , ••• , Xn]. Show that any polynomial f can be written uniquely as a finite sum of homogeneous polynomials.

(18) Let V = {AeMnxn(F)IA =A'}. Show that Vis a subspace of Mnxn(F). V is the subspace of symmetric matrices of Mn x n(F).

(19) Let W = {A e Mn,. 0 (F) I A' = -A}. Show that W is a subspace of Mn x 0 (F). W is the subspace of all skew-symmetric matrices in Mn x n(F).

(20) Let W be a subspace of V, and let a, p e V. Set A = a: + Wand B = P + W. Show that A= B or AnB = 0.

2. BASES AND DIMENSION

Before proceeding with the main results of this section, let us recall a few facts from set theory. If A is any set, we shall denote the cardinality of A by !AI. Thus, A is a finite set if and only if !AI < oo. If A is not finite, we shall write !AI = oo. The only fact from cardinal arithmetic that we shall need in this section is the following:

2.1: Let A and B be sets, and suppose IAI = oo. If for each x e A, we have some finite set Ax!;;;;; B, then !AI~ IUxeAAxl·

A proof of 2.1 can be found in any standard text in set theory (e.g., [1]), and, consequently, we omit it.

A relation R on a set A is any subset of the. crossed product A x A. Suppose R is a relation on a set A. lfx, yeA and (x, y) e R, then we shall say x relates toy and write x ""' y. Thus, x ""' y <=> (x, y) e R We shall use the notation (A, ""') to indicate the composite notion of a set A and a relation R £;;;; A x A. This notation is a bit ambiguous since the symbol "' has no reference to R in it. However, the use of ""' will always be clear from the context. In fact, the only relation R we shall systematically exploit in this section is the inclusion relation £;;;; among subsets of BI'(V) [V some vector space over a field F].

. A set A is said to be partially ordered if A has a relation R £;;;; A x A such that (1) x ""'x for all xeA, (2) if x""' y, andy"' x, then x = y, and (3) if x""' y, and y ""' z, then x ""' z. A typical example of a partially ordered set is BI'(V) together with the relation A ""' B if and only if A £;;;; B. If (A, "') is a partially ordered set, and A1 £;;;; A, then we say A1 is totally ordered if for any two elements x, y eAt> we have at least one of the relations x - y or y ""' x. If (A, ""') is a partially ordered set, and A 1 £;;;; A, then an element x e A is called an upper bound for A 1 if y ""'x for all yeA1. Finally, an element xe(A, -)is a maximal element of A if x - y implies x = y.

BASES AND DIMENSION 9

We say a partially ordered set (A, ""') is inductive if every totally ordered subset of A has an upper bound in A. The crucial point about inductive sets is the following result, which is called Zorn's lemma:

2.2: If a partially ordered set (A, "') is inductive, then a maximal element of A exists.

We shall not give a proof of Zorn's lemma here. The interested reader may consult (3, p. 33] for more details. ·

Now suppose Vis an arbitrary vector space over a field F. LetS be a subset of v.

Definition 2.3: S is linearly dependent over F if there exists a finite subset {a:1, ... ,a:n} £;;;; S and nonzero scalars x1, ... ,xneF such that x1a:1 + · · · + XnCXn = 0. S is linearly independent (over F) if S is not linearly dependent.

Thus, if S is linearly independent, then whenever Lf=1 x1a:1 = 0 with {a:t> · .. , a:n} £;;;; S and {x1, ... , xn} £;;;; F, then x1 = .. · Xn = 0. Note that our definition implies the empty set cp is fuiearly independent over F. When considering questions of dependence, we shall drop the words "over F" whenever F is clear from the context. It should be obvious, however, that if more than one field is involved, a given set S could be dependent over one field and independent over another. The following example makes this clear.

Example 2.4: Suppose V = IR, the field of real numbers. Let F 1 = Q, and F2 =A~. Then V is a vector space over both F 1 and F 2. Let S = {a:1 = 1,

!X2 = ,J2}. Sis a set of two vectors in V. Using the fact that every integer factors uniquely into a product of primes, one sees easily that S is independent over F 1. But, clearly S is dependent over F 2 since (.j2)cx1 + ( -1)cx2 = 0. D

Definition 2.5: A subset S of V is called a basis of V if S is linearly independent over F and L(S) = V.

If S is a basis Of a vector space V, then every nonzero Vector a: E V can be Written uniquely in the form CX = X1 0:1 + · · · + Xn!Xn, where { a1~ ... , a:n} !;;;;; S and x~, · .. , Xn are nonzero scalars in F. Every vector space h.as a basis. In fact, any gJ.Ven linearly independent subset S of V can be expanded to a basis.

!Jteorem 2.6: Let V be a vector space over F, and suppose S is a linearly Independent subset of V. Then there exists a basis B of V such that B ;;2 s.

\

Proof: Let f/ denote .the set of all independent subsets of V that contain s. Thus, 9' = {A e BI'(V) I A ;;2 S and A is linearly independent over F}. We note that 9' ::P cp since Se9'. We partially order f/ by inclusion Thus for A A e9'

• ' h 2 '

10 LINEAR ALGEBRA

At - A2 if and only if At s;;; A2. The fact that (9', s;;;) is a partially ordered set is clear.

Suppose f!r = { A11 i e !:J.} is an indexed collection of elements from 9' that form a totally ordered subset of 9'. We show f!r has an upper bound. Set A= UteAA,. aearly, Ae~(V). s s;;; A, and A, s;;; A for all ie!:J.. If A fails to be linearly independent, then there exists a finite subset { IX1, .•• , 1X0 } s;;; A and nonzero scalars x1 , ••• , X0 e F such that Xt1X1 + · · · + X0 1Xn = 0. Since f!r is totally ordered, there exists an index i0 e !:J. such that {IX!> ••• , 1X0 } s;;; A10• But then A,o is dependent, which is impossible since A~ae9'. We conclude that A is linearly independent, and, consequently, Ae9'. Thus, f!r has an upper bound A in 9'.

Since f!r was arbitrary, we can now conclude that (9', s;;;) is an inductive set. Applying 2.2, we see that 9' has a maximal element B. Since Be 9', B ;;;! S and B is linearly independent. We claim that B is in fact a basis of V. To prove this assertion, we need only argue L(B) = V. Suppose L(B)-:/: V. Then there exists a vector IX e V - L(B). Since a:¢ L(B), the set B u {a:} is clearly linearly independent. But then B u {a:} e 9', and B u {IX} is strictly larger than B. This is contrary to the maximality of Bin 9'. Thus, L(B) = V, and B is a basis of V containing s. 0

Let us ·look at a few concrete examples of bases before continuing.

Example 2.7: The empty set ¢ is a basis for the zero subspace (0) of any vector space V. If we regard a field F as a vector space over itself, then any nonzero element a: ofF forms a basis of F. 0

Example 2.8: Suppose V = F0, n eN. For each i = 1, ... , n, let Dt =

(0, ... , 1, ... , 0). Thus, 151 is then-tuple whose entries are all zero except for a 1 in the ith position. Set §={D1 , ••• ,D0 }. Since (xl, ... ,xJ=xtDl+···+ x t5 we see t5 is a basis of P. We shall call t5 the canonical (standard) basis of n D' - _

F0• 0

Example 2.9: Let V = Mm x u(F). For any i = 1, ... , m, and j = 1, ... , n, let e1J denote the m x n matrix whose entries are all zero except for a 1 in the (i j)th position. Since ( a1J) = Lt,J 9-tJeiJ, we see B = { e1j 11 ~ i ~ m, 1 ~ j ~ n} is a basis for V. The elements e1J in B are called the matrix units of Mm x n(F). 0

Example 2.10: Let V = F[X]. Let B denote the set of all monic monomials in X. Thus, B = {1 = X0, X, X2, ... }. Clearly, B is a basis of F[X]. 0

A specific basis for the vector space Ck(l) in Example 1.9 is hard to write down. However, since IR[X] s;;; Ck(I), Theorem 2.6 guarantees that one basis of Ck(l) contains the monomials 1, X, X2, ....

Theorem 2.6 says that any linearly independent subset of V can be expanded to a basis of V. There is a companion result, which we shall need in Section 3. Namely, if some subset S of V spans V, then S contains a basis of V.

L


Theorem 2.11: Let V be a vector space over F, and suppose V = L(S). Then S contains a basis of V.

Proof: IfS =¢or {0}, then V = (0). In this case,¢ is a basis ofV contained inS. So, we can sume S contains a nonzero vector a:. Let 9' = {A s;;; S 1 A linearly independent over F}. Clearly, {a:} e9'. Partially order [I' by inclusion. If ff ={Ad ie!:J.} is a totally ordered subset of 9', then U 1s4A1 is an upper bound for f!r in 9'. Thus, (tl', s;;;) is inductive. Applying 2.2, we see that 9' has a maximal element B.

We claim B is a basis for V. Since Betl', B s;;; Sand B is linearly independent over F. If L(B) = V, then B is a basis of V, and the proof is complete. Suppose L(B) -:/: V. Then S ¢ L(B), for otherwise V = L(S) s;;; L(L(B)) = L(B). Hence there exists a vector PeS - L(B). Oearly, B u { P} is linearly independent ovet F. Thus, B u {P} etl'. But P¢L(B) implies P¢B. Hence, B u {P} is strictly larger than B in 9'. Since B is maximal, this is a contradiction. Therefore, L(B) = V and our proof is complete. 0

A given vector space V has many different bases. For example, ~ = {(0, ... , x, ... , 0) +t5tli = 1, ... , n} is clearly a basis for P for any x-:/: -1 in F. What all bases ofV have in common is their cardinality. We prove this fact in our next theorem.

Theorem 2.12: Let V be a vector space over F, and suppose B1 and B2 are two bases of V. Then IB11 = IB2I·

Proof: We divide this proof into two cases.

CASE 1: Suppose V has a basis B that is finite. In this case, we shall argue IB11 = IBI = IB2I· Suppose B = {at, ... , a:n}· It

clearly suffices to show !Btl= n. We suppose IB11-:/: nand derive a contradiction. There are two possibilities to consider here. Either IB11 = m < n or IB11 > n. Let us first suppose B1 = {Pt, ... , Pm} with m < n. Since Pte L(B), /11 = x1a:1 + · · · + X 0 CX0 • At least one x1 here is nonzero since /11 -:/: 0. Relabeling the a:1 if need be, we can assume x1 -:/: 0. Since B is linearly independent over F, we conclude that /11 eL({a:2, ... , a:n} u {1Xl})- L({a:2, ... ,an}). It now follows from Theorem 1.15(f) that a:1 e L({/11, a:2, ... , CX0 }). Since {a:l, ... , a:n} is linearly independent over F, and PI¢L({a:2, ... ,a:0 }), we see that {Pt> a:2, ... , a:0 } is linearly independent over F. Since a:1 e L({P1, a:2, ... , 1X0 }),

V = L({/11, a:2, ... , CX0 })." Thus, {/11, a:2, ... , a:n} is a basis of V. Now we ·can repeat this argument m times. We get after possibly some

permutation of the a:1 that {P1 , ... ,pm,a:m+t•· .. ,a:0 } is a basis of V. But {p1 , ... ,/1m} is ~lready a basis of V. Thus, a:m+ 1 eL({/11 , ... ,pm}). This implies { p 1 , •.. , Pmo a:m + 1 , •.• , a:0 } is linearly dependent which is a contradiction. Thus, IB1I cannot be less than n.

12' LINEAR ALGEBRA

Now suppose !Btl > n (!Btl could be infinite here). By an argument similar to that given above, we can exchange n vectors ofBt with ext, ... , exn. Thus, we construct a basis of V of the form B u S, where S is some nonempty subset of Bt. But B is already a basis of V. Since S :F t/J, BuS must then be linearly dependent. This is impossible. Thus, if V has a basis consisting of n vectors, then any basis of V has cardinality n, and the proof of the theorem is complete in case 1.

CASE 2: Suppose no basis of V is finite. In this case, both B1 and B2 are infinite sets. Let exeB1• Since B2 is a basis

of V there exists a unique, finite subset A~~. s;;; B2 such that ex e L(AJ and a¢ r.{A 1 for any proper subset A' of All.' Thus, we have a well-defined function l{J: B1 ~ &J(B2) given by lfJ(cx) = A~~.. Since B1 is infinite, we may apply .2.1 and conclude that IB11 ;:a: IU~~.eB, A~~.l· Since ex e L(AJ for all ex e B1, Y_ = L(Ufi.EB, AJ. Thus U~~.eB, A~~. is a subset of B2 tha~ spans all of V. Since B2 ~s a basis of V, we conclude UII.GB, A~~. = B2. In particular, !Btl ;:a: IB2I· Reversmg the roles ofB1 and B2 gives IB2 1 ;:a: !Btl· This completes the proof of Theorem 2.12. 0

We shall call the common cardinality of any basis ofV the dimension ofV. We shall write dim V for the dimension of V. If we want to stress what field we are over, then we shall use the notation dimF V for the dimension of the F-vect~r space V. Thus, dim V = IBI, where B is any basis of V when the base field F IS

understood. Let us check the dimensions of some of our previous examples. In Example

2 4 dim y = 1 and dimF (V) = 1~1. the cardinality of R In Example 2.7, • J Fz ' I • (F) dim~O) = 0. In Example 2.8, dim F0 = n. In Example 2.9, dtm Mm" n = mn. In Example 2.10, dim V = IN I, the cardinality of N.

If the dimension of a vector space Vis infinite, as in Examples 2.4 and 2.10, we shall usually make no attempt to distinguish which cardin.al nun;tber gives dim V. Instead, we shall merely write dim V = oo. IfV has a firut~ bas~s {ext, ... , exn}, we shall call V a finite-dimensional vector space and wnte dtm V < oo, or, more precisely, dim V = n < oo. Thus, for example, dimR Ck(l) = oo, wherea.:s dimR ~n = n < oo. In our next theorem, we gather together some of the mor~ elementary facts about dim V.

Theorem 2.13: Let V be a vector space over F.

(a) If W is a subspace of V, then dim W ~ dim V. (b) If y is finite dimensional ~nd W is a subspace of V such that

dim W = dim V, then W = V. (c) If w is a subspace of V, then there exists a subspace W' of V such that

W + W' = V and W n W' = (0). (d) If y is finite dimensional and W t ~nd W 2 a~e subspaces of V, then

dim(W 1 + W 2) + dim(W t n W 2) = dtm W t + dtm W 2·


proof: It follows from Theorem 2.6 that any basis of a subspace W of V can be enlarged to a basis of V. This immediately proves (a) and (b). Suppose W is a subspace of V. Let B be a basis of W. By Theorem 2.6, there exists a basis C of V such that B s;;; C. Let W' = L(C - B). Since C = B u (C - B), y:::: L(C) = L(B) + L(C- B)= W + W'. Since Cis linearly independent and B n (C - B) = t/J, L(B) n L(C - B) = (0). Thus, W n W' = (0), and the proof of (c) is complete.

To prove (d), let B0 ={ext, ... , exn} be a basis ofW 1 n W 2• IfW1 n W2 = (0), then we take B0 to be the empty set t/J. We can enlarge B0 to a basis B1 = {cx1, ... , ex0 , Pt> ... , Pm} of W t· We can also enlarge B0 to a basis B2 = {cxl,. .. , exn, 1'1>'"' ')lp} ofW2. Thus, dim wt n w2 = n, dim wt = n + m, and dim w2 = n + p. We claim that B ={ext, ... ' exn, Pt .... ' Pm. 'Yt· ... ' 1'p} is a basis of W 1 + W 2. Oearly L(B) = W 1 + W 2• We need only argue B is lineatly independent. Suppose Ir=t x,a, + Ll"=t y,p, + Lf=t Zj')l, = 0 for some x1, y1,

z1 e F. Then Lf= t Zj')l1 e W t n W 2 = L({ ex1, ... , exn}). Thus, Lf= t Zj')l, = Ir= 1 w,ex, for some w1 e F. Since B2 is a basis of W 2, we conclude that Zt = · · · = Zp = 0. Since B1 is a basis ofW1o x1 = ··· = X0 = y1 = .. :Ym = 0. In particular, B is linearly independent. Thus, dim(W 1 + W 2) = IBI = n + m + p, and the proof of (d) follows. 0

A few comments about Theorem 2.13 are in order here. Part (d) is true whether V is finite dimesional or not. The proof is the same as that given above when dim(W t + W 2) < oo. If dim(W 1 + W 2) = oo, then either W 1 or W 2 is an infinite-dimensional subspace with the same dimension as W t + W 2. Thus, the result is still true but rather uninteresting.

If Vis not finite dimensional, then (b) is false in general. A simple example illustrates this point.

Example 2.14: Let V = F[X], and let W be the subspace of V consisting of all even polynomials. Thus, W = {L a1X

21 1 ~ e F}. A basis of W is clearly all even powers of X. Thus, dim V = dim W, but W :F V. 0

The subspace W' of V constructed in part (c) of Theorem 2.0 is called a complement ofW. Note that W' is not in general unique. For ~xample, ifV = ~2

and W = L((1, 0)), then any subspace of the form L((a, b)) with b :F 0 is a complement of W.

Finally, part (d) of Theorem 2.13 has a simple extension to finitely many subspaces W h ••• , W k of V. We record this extension as a corollary.

Corollary 2.15~ Let V be a finite-dimensional vector space of dimension n. Suppose W1 ,.~ .• Wk are subspaces of V. For each i= 1, ... ,k, set f1 = n- dimW1• Then

(a) dim(W1 n · · · ~ WJ = n- }J=t f, + "'J;;;f {n- dim((Wt n · .. n Wj) + WJ+t)}.

14 LINEAR ALGEBRA

(b) dim(W1 n · · · n Wk) ~ n -!J= 1 f1• (c) dim(W1n···nWJ=n-}:J= 1 f1 if and only if for all i=1, ... ,k,

W; + <nJ'fl WJ) = V.

Proof: Part (a) follows from Theorem 2.13 (d) by induction. Parts (b) and (c) are easy consequences of(a). We leave the technical details for an exercise at the end of this section. D

Before closing this section, let us develop some useful notation concerning bases. Suppose V is a finite-dimensional·vector space over F. If~ = { IX1, ••• , cxn} is a basis of V, then we have a natural function [·]!! V-+ Mnx 1(F) defined as follows.

Definition 2.16: If ~ = { cx1, ... , cxn} is a basis of V, then [PJf! =

(x1, ••• , xJ'e Mn x 1(F) if and only if L:r=t X;IX; = p.

Since ~ is a basis of V, the representation of a given vector P as a linear combination of cx1 , ••• , 1X

0 is unique. Thus, Definition 2.16 is unambiguous. The

function [·Ja:: V-+ Mnx 1(F) is clearly bijective and preserves vector addition and scalar multiplication. Consequently, [xp + yo]!!= x[PJf! + y[o]f! for all x, Y e F and p, (j e V. The column vector [PJa: is often called the ~ skeleton of p.

Suppose ~ = {a 1 , ••• , 1X0

} and ~ = { o 1, ... , 00 } are two bases of V. Then there is a simple relationship between the ~ and ~ skeletons of a given vector p. Let M(~, ~) denote the n x n matrix whose columns are defined by the following equation:

2.17: M(~, ~) = ([o1Ja: I·· ·I [c5J~

·In equation 2.17, the ith column of M(~, ~) is the n x 1 matrix [oJf!. Multiplication by M(~, ~) induces a map from Mnx 1(F) to Mnx 1(F) that connects the ~ and ~ skeletons. Namely:

Theorem 2.18: M(~, ~)[p]~ = [PJf! for all P e V.

Proof: Let us denote the ith column of any matrix M by Col;(M). Then for each i = 1, ... , n, we have M(~, ~)[ 01].4 = M(~, ~)(0, ... , 0, 1, 0, ... , 0)' = Col1(M(~, ~)) = [oJa:· Thus, the theorem is correct for Pe~.

Now we have already noted that [·]~and [·]f! preserve vector addition and scalar multiplication. So does multiplication by M(~, ~) as a map on Mn x 1 (F). Since any p e V is a linear combination of the vectors in ~. we conclude that M(~, ~)[p]~ = [PJf! for every P e V. D

The matrix M(~, ~) defined in 2.17 is called the change of basis matrix (between~ and~). It is often convenient to think of Theorem 2.18 in terms ofthe


following commutative diagram:

2.19:

By a diagram, we shall mean a collection of vector spaces and maps (represented by arrows) between these spaces. A diagram is said to be commutative if any two sequences of maps (i.e., composites of functions in the diagram) that originate at the same space and end at the same space are equal. Thus, 2.19 is commutative if and only if the two paths from V to Mn x 1 (F), clockwise and counterclockwise, are the same maps. This is precisely what Theorem 2.18 says.

Most of the maps or functions that we shall encounter in the diagrams in this book will be linear transformations. We take up the formal study of linear transformations in Section 3.


(1) Let Vn = {f(X)eF[X] I degree f ~ n}. Show that each Vn is a finitedimensional subspace of F[X] of dimension n + 1. Since F[X] = U:"= 1 Vn, observe that Theorem 1.14 is false when the word "finite" is taken out of the theorem.

(2) Let V n be as in Exercise 1 with F = IF 2• Find a basis of V 5 containing 1 + x and x2 + x + 1.

(3) Show that any set of nonzero polynomials in F[X], no two of which have the same degree, is linearly independent over F.

(4) Let V ={(at> a 2, ••• )e ~1\ill a1 = 0 for all i sufficiently large}. Show that Vis an infinite-dimensional subspace of ~N. Find a basis for V.

(5) Prove Theorem 2.13(d) when dim(W1 + W 2) = oo.

(6) Prove Corollary 2.15.

(7) Find the dimension of the subspace V = L({1X, p, y, c5}) s;; ~4, where a= (1, 2, 1, 0), p = ( -1, 1, -4, 3), y = (2, 3, 3, -1), and o = (0, 1, -1, 1).

(8) Compute the following dimensions: (a) dimR(C).

(b) dimo(~). {c) dim0 (F), where F is the field given in Exercise 3 of Section 1.

16 LINEAR ALGEBRA

(9) Suppose V is an n-dimensional vector space over the finite field IF p· Argue that V is a finite set and find lVI.

(10) Suppose V is a vector space over a field F for which lVI > 2. Show that V has more than one basis.

(11) Let F be a subfield of the field F'. This means that the operations of addition and multiplication on F' when restricted to F make F a field.

(a) Show that F' is a vector space over F.

(b) Suppose dim~F1 = n. Let V be an m-dimensional vector space over F'. Show that V is an ron-dimensional vector space over F.

(12) Show that dim(V") = n dim(V).

(13) Return to the space v n in Exercise 1. Let pt(X) = LJ~O ajlxj fori = 1' ... ' r. Set A= (a1JeM<n+l)x.(F). Show that the dimension of L({p1 , ••• , Pr}) is precisely the rank of A.

(14) Show that the dimension of the subspace of homogeneous polynomials of degree din F[X0 , ••• , XJ is the binomial coefficient (n~d).

(15) Find the dimensions of the vector spaces in Exercises 18 and19 of Section 1.

(16) Let AeMmxn(F). Set CS(A) = {AXIXeMnxl(F)}. CS(A) is called the column space of A. Set NS(A) = {X e Mn x 1 (F) I AX = 0}. NS(A) is called the null space of A. Show that CS(A) is a subspace ofMmx 1(F}, and NS(A) is a subspace of M 0 x 1 (F). Show that dim(CS(A)) + dim(NS(A)) = n.

(17) With the same notation as in Exercise 16, show the linear system AX= B has a solution if and only if dim(CS(A)) = dim(CS(A I B)). Here BeMmx 1(F}, and (AlB) is the m x (n + 1) augmented matrix obtained from A by adjoining the column B.

(18) Suppose V and W are two vector spaces over a field F such that lVI = IWI. Is dim V = dim W?

(19) Consider the set W of 2 x 2 matrices of the form

and the set Y of 2 x 2 matrir..es of the form

Show that Wand Yare subspaces of M 2 xiF) and compute the numbers dim(W), dim(Y), dim(W + Y), and dim(W n Y).

LINEAR TRANSFORMATIONS 17

3. LINEAR TRANSFORMATIONS

Let V and W be vector spaces over a field F.

Definition 3.1: A function T: V -+ W is called a linear transformation (linear map, homomorphism) ifT(xa+ yfJ) = xT(a) + yT{/J) for all x, yeF and a, PeV.

Before we state any general theorems about linear transformations, let us consider a few examples.

Example 3.2: The map that sends every vector in V to 0 e W is clearly a linear map. We shall call this map the zero map and denote it by 0. If T: V-+ W and S: W -+ Z are linear trap.sformations, then clearly the composite map ST: V -+ Z is a linear transformation. 0

Example 3.3: If V is finite dimensional with basis ~ = {a 1 , ... , a0 }, then [ · ],.: V -+ M

0 x 1 (F) is a linear transformation that is bijective. 0 -

Example 3.4: Taking the transpose, A-+ A', is clearly a linear map from Mm x o(F} -+ Mo x m(F}. 0

Example 3.5: Suppose V = Mm x n(F) and A E Mm x m(F). Then multiplication by A (necessarily on the left) induces a linear transformation T A: V-+ V given by TA(B) = AB for all BeY. 0

Example 3.3 and 3.5 show that the commutative diagram in 2.19 consists of linear transformations.

Example 3,6: Suppose V = Ck(I) with k ~ 2. Then ordinary differentiation f-+ f' is a linear transformation from Ck(I) to ck- 1(!). 0

Example 3.7: Suppose V = F[X]. We can formally define a derivative f-+ f' on V as follows: If f(X) = Lf = 0 BjXi, then f'(X) = Lf = 1 ia1X

1-

1• The reader can easily check that this map, which is called the canonical derivative on F[X], is a linear transformation. 0

Example 3.8: Suppose V =~(A) as in Example 1.10. Then T(f) ~ JAfisa linear transformation from V to IR. D

We shall encounter many more examples of linear transformations as we proceed. At this point, let us introduce a name for the collection of all linear transformations from V toW.

Definition 3.9: Let V and W be vector spaces over F. The set of all linear transformations from V toW will be denoted by Hom~V, W).

18 LINEAR ALGEBRA

When the base field F is clear from the context, we shall often write Hom(V, W) instead of Homp(V, W). Thus, Hom(V, W) is the subset of the vector space wv (Example 1.6) consisting of all linear transformations from V to W. If T, S e Hom(V, W) and x, y e F, then the function xT + yS e wv is in fact a linear transformation. For if a, be F and a, p e V, then (xT + yS)(aa + bp) = xT(aa + bp) + yS(aa + bp) = xaT(a) + xbT(p) + yaS(a) + ybS(p) = a(xT(a) + yS(a)) + b(xT(p) + yS(p)) = a(xT + yS)(a) + b(xT + yS)(p). Therefore, xT + ySeHom(V, W). We have proved the following theorem:

Theorem 3.10: Hom(V, W) is a subspace of wv. 0

Since any T e Hom(V, W) has the property that T(O) = 0, we see that Hom(V, W) is always a proper subspace of wv whenever W :f. (0).

At this point, it is convenient to introduce the following terminology.

Definition 3.11: Let T e Hom(V, W). Then,

(a) ker T = {aeVIT(a) = 0}. (b) lm T = {T(a)eWI aeV}. (c) Tis injective (monomorphism, 1 - 1) if ker T =(0).

(d) Tis surjective (epimorphism, onto) if Im T = W. (e) Tis bijective (isomorphism) if Tis both injective and surjective. (f) We say V and W are isomorphic and write V ~ W if there exists an

isomorphism T e Hom(V, W).

The set ker T is called the kernel ofT and is clearly a subspace of V. Im T is called the image ofT and is a subspace of W. Before proceeding further, let us give a couple of important examples of isomorphisms between vector spaces.

Example 3.12: Mnx 1(F)~ M 1 xn(F) via the transpose A- A'. We have already mentioned that F:n = M 1 xn(F). Thus, all three of the vector spaces Mnx 1(F), M 1 xiF), and Pare isomorphic to each other. 0

Example 3.13: Suppose V is a finite-dimensional vector space over F. Then every basis ~ = {a1 , ••• , a

0} of V deterinines a linear transformation T(!!):

V- P given by T(~)(p) = (x1 , •.. , xJ ifand only iQJ= 1 xiai = p. T(~) is just the composite of the coordinate map [·] .. : V- Mn x 1(F) and the transpose Mnx

1(F)- M1 xn(F) = P. Since both of-these maps are isomorphisms, we see

T(~) is an isomorphism. 0

It is often notationally convenient to switch back and forth from column vectors to row vectors. For this reason, we give a formal name to the isomorphism T(~) introduced in Example 3.13.


Definition 3.14: Let V be a finite-dimensional vector space over F. If a is a basis ofV, then(·)!: V- Pis the linear transformation defined by (P) .. = ([PJJ1 for all peV. - -

Thus, (p)! = T(~)(p) for all PeV. We can now state the following theorem, whose proof is given in Example 3.13:

Theorem 3.15: Let V be a finite-dimensional vector space over F and suppose dim V = n. Then every basis~ ofV determines an isomorphism(·)!: V-+ F0 • 0

We now have two isomorphisms [·Ji V- Mnx 1(F) and(·).,: V- F 0 for every choice of basis ~ of a (finite-dimensional) vector space V. We -shall be careful to distinguish between these two maps although they only differ by an isomorphism from Mnx 1(F) to M1 x0 (F). Notationally, Fn is easier to write than Mnx 1(F), and so most of our subsequent theorems will be written using the map(·).,. With this in mind, let us reinterpret the commutative diagram given in 2.19.-

If A is any n x n matrix with coefficients in F, then A induces a linear transformation SA: pn- P given by the following equation:

3.16:

Using the notation in Example 3.5, we see SA is the linear transformation that makes the following diagram commutative:

3.17:

M;.x 1(Ij~') ]Mnx l(F)

(·)' (·)'

Fn ___ .::,SA~-~Fn

The vertical arrows in 3.17 are isomorphisms. Clearly, TA is an isomorphism if ~nd only if A is invertible. Thus, SA is an isomorphism if and only if A is mvertible.

We shall replace the notation SA (or TJ with A1 (or A) and simply write

A' Fn·_ --~fR or

i .

I dirrNow_ suppose ~ and Q a~e tw~ bases of a finite-dimensional vector space v of

1

mellSlon n. Ir we combtne diagrams 2.19 and 3.17, we have the following

_ _j

20 LINEAR ALGEBRA

commutative diagram:

3.18:

Since (·)! = ([·Jl and (·)g =([·]g)\ we get the following corollary to Theorem 3.15:

Corollary 3.19: Suppose V is a finite-dimensional vector space of dimension n over F. If~ and ~ are two bases of V, then

3.20:

('l«/Fjn y/ M(~,g)'

~yn is a commutative diagram of isomorphisms.

Proof: We have already noted from 3.18 that 3.20 is commutative. Both(·)! and (')., are isomorphisms from Theorem 3.15. We need only argue M(§, ~) is an invertible matrix. Then the map M(~, ~)1 = SM@,!J: Fn - P is an isomorphism.

Now change of basis matrices M(§, ~)are always invertible. This follows from Theorem 2.18. For any {JeV, we have M(~. §)M(§, ~)({J]~ = M(~ §)[fJJ! = [fJ]~. This equation easily implies M(~ ~)M(§, ~) = In, the n x n identity matrix. 0

In our next theorem, we shall need an isomorphic description of the vector space yn introduced in Example 1.6.

Example 3.21: In this example, we construct a vector space isomorphic to vn. Let V be a vector space over F, and let n e 1'1. Consider the Cartesian product V x ... x V (n times)= {(cx1 , ... ,cxJia1eV}. Clearly, V x ... x Vis a vector space over F when we define vector addition and scalar multiplication by (cx1, ... , cxJ + ({11 , • .. , fJJ = (a1 + {11 , ... , an+ fJJ, and x(cx1, ... , an)= (xcx1, ... , x!XJ.


Suppose A is any finite set with IAI = n. We can without any loss of generality assume A= {1, ... , n}. Then VA= vn. There is a natural isomorphism T:V x ··· x V- yn given by T((cx1, ... ,cxJ) = fevn, where f(i) = cx1 for all i == 1, ... , n. The fact that T is an isomorphism is an easy exercise, which we leave to the reader. 0

Henceforth, we shall identify the vector spaces V x · · · x V (n times), yn and yA with !AI = nand write just yn to represent any one of these spaces. Using this notation, we have the following theorem:

Theorem 3.22: Let V and W be vector spaces over F, and suppose Vis finite dimensional. Let dim V = n.

(a) If~= {ext> ... , CX0 } is a basis of V, then for every ({J1, ... , fJn)E Wn, there exists a unique T e Hom(V, W) such that T(cx1) = {11 for i = 1, ... , n.

(b) Every basis ~ of V determines an isomorphism 'P(g): Hom(V, W'r-, wn.

Proof: (a). Let ~={cx1 , ... ,an} be a basis for V. Then (·) .. eHom(V,Fn) is an isomorphism. Let {J = ({Jl, ... , {J,) E Wn. The n=-tuple {J determines a linear transformation L11 eHom(F", W) given by L11((x1 , ••• , xJ) = Lf=t x1{J1• The fact that L11 is a linear transformation is obvious. Set T = L11(·)!. Then TeHom(V, W) and T(aJ = {J1 for all i = 1, ... , n. The fact that T is the only linear transformation from V to W for which T(cx1) = {11 fori = 1, ... , n is an easy exercise left to the reader.

(b) Fix a basis g = { cx1, ... , an} of V. Define 'P(g): Hom(V, W) - wn by 'P(~)(T) = (T(a1), ... , T(aJ). The fact that 'P(~) is a linear transformation is obvious. We can define an inverse map x: WU- Hom(V, W) by x((fJ1 , ... , fJJ) = Lp(·)cz. Here fJ = (fJ1 , ... ,fJJ. Hence, 'P(g) is an isomorphism. 0 -

Theorem 3.22(a) implies that a given linear transformation T e Hom(V, W) is completely determined by its values on a basis ~ of V. This remark is true whether V is finite dimensional or infinite dimensional. To define a linear transformation T from V to W, we need only define t · on some basis B = {a1lieA} ofVand then extend the definition ofT linearly to all ofL(B) = V. Thus, ifT(aJ = {J1 for all ieA, then T(L1e4 x1cxJ is defined to be Lte4 x1{J1• These remarks provide a proof of the following generalization of Theorem 3.22(a):

3.23: Let V and W be vector spaces over F and suppose B = { a11 i e A} is a basis of V. If {fJ11 i e A} is any subset of W, then there exists a unique T e Hom(V, W)

. such that T(aJ = fJ1 for all i eA. 0

I du:;~w suppose V and W are both finite-<timensional vector spaces over F. Let L_ =n and dimW=m. If ~={a1 , ••• ,cxn} is a basis of V and

22 LINEAR ALGEBRA

~ = {P1, ... , Pm} is a basis of W, then the pair (g, ~) determines a linear transformation r(g, ~): Hom(V, W) -+ Mm X n(F) defined by the following equation:

3.24:

r(g, ~)(T) = ([T(a1)J~ I·· ·I [T(aJJp)

In equation 3.24, T e Hom{V, W), and r(g, ~)(T) is the m x n matrix whose ith column is the m x 1 matrix [T(aJ]f!. If T 1, T 2 e Hom(V, W) and x, y e F, then

r(g, ~)(xT 1 + yT 2) = ([{xT 1 + yT 2)(a1)]p I·· ·I [{xT 1 + yT 2)(aJJ2)

= (x[T1(a1)]f! + y[T2(al)]l!l···lx[T1(aJ]l! + y[T2(aJ]l!)

= x([T 1(a1)J2 I·· ·I [T t(aJJl!) + y([T 2(at)]t! I·· ·I [T 2(ao)]l!)

= xr(g, ~)(T 1) + yr(g, ~)(T 2)

Thus r(g, ~ is indeed a linear transformation from Hom(V, W) to Mm X n(F). Suppose T e ker r(g, p). Then r(g, p)(T) = 0. In particular, [T(aJ]p = 0 for all

i = 1, ... , n. But then- 0 = [T(aJ]~-= (T(a1))f!, and Theorem ns implies T(aJ = 0. Thus, T = 0, and we conclude that r(g, ~ is an injective linear transformation.

r(g, p) is surjective as well. To see this, let A = ( x11) e Mm x 0 (F). Let ')11 = 2;: 1 XJ1P1 for i = 1, ... , n. Then {Y1, ... , Yn} £;; W, and [/'Jp = (x11 , ••• , xmil' = Col1(A) for all i = 1, ... , n. It follows from Theorem 3.22 fhat there exists a (necessarily unique) TeHom(V, W) such that T(aJ = y1 for i = 1, ... , n. Thus, r(g, p)(T) = A and r(g, p) is surjective. We have, now proved the first statement in the following theorem:

Theorem 3.25: Let V and W be finite-dimensional vector spaces over F of dimensions nand m, respectively. Let g be a basis ofV and p a basis ofW. Then the map r(g, ~: Hom{V, W)-+ Mmxn(F) defined by equation 3.24 is an isomorphism. For every T e Hom(V, W), the following diagram is commutative:

3.26:

v T w

(•\ 1 r(g,f!.'(f)'

1 , .• pn Fm

Proof: We need only argue that the diagram in 3.26 is commutative. Using the


same notation as in 3.17, we have the following diagram:

3:Z7:

Since all the maps in 3.27 are linear and the bottom square commutes, we need only check [·]f!T = r(g, ~[·]!!on a basis of V. Then the top square of 3.27 is commutative, and the commutativity of 3.26 follows. For any a1 e g, we have ([·]pT)(aJ = [T(cxJ]f! = Col1(r(g, ~) = r(g, ~0, ... , 1, ... , 0)' =

n~~[aa!. o

r(g, p)(T) is called the matrix representation of the linear transformation T with res{>ect to the bases g and p. Since the vertical arrows in 3.26 and r(g, p) are isomorphisms, V, W, Hom(V;W), and T are often identified with P~ P, Mm x 0 {F), and A = r(g, ~(T). Thus, the distinction between a linear transformation and a matrix is often blurred in the literature.

The matrix representation r(g, p)(T) ofT of course depends on the particular bases g and p chosen. It is an easy matter to keep track of how r(g, p)(T) changes with g and ~· -

Theorem 3.28: Let V and W be finite-dimensional vector spaces over F of dimensions nand m, respectively. Suppose g and g' are two bases ofV and P and ~~ two bases of W. Then for every T e Hom{V, W), we have ... -

3.29:

i r(g', ~')(T) = M{~, ~1r(g, ~M(g, gr1

I Proof: Before proving equation 3.29, we note that M{~, ~1 (and M(~ g1) is the I Ill x m (and n x n) change of basis matrix given in equation 2.17. We have 1 already noted that change of bases matrices are invertible and consequently all I th\terms in equation. 3.29 make sense. Lo see that 3.29 is in fact a valid equation, we merely combine the

_j

24 LINEAR ALGEBRA

commutative diagrams 2.19 and 3.27. Consider the following diagram:

3.30:

Mnx1(F) r<g,l_)(T)

/ •• ,(F)

( ~ Q)

[·J, ) T w (?) M(~, ~') M(~ g') /(v

@) \[·]~: Mnx 1(F) Mmx 1(F)

rc~·. ()(f)

The diagram 3.30 is made up off our parts, which we have labeled CD, @, Q) , and @). By Theorem 2.18, diagrams CD and (?) are commutative. By Theorem 3.25, diagrams Q) and @) are commutative. It follows that the entire diagram 3.30 is commutative. In particular, M(p, P')r(f!:, P)(T) = r(f!:',P')(T)M(f!:, 1!:1· Solving this equation for r(f!:', ~m gives 3.29. 0 - -

Recall that two m x n matrices A, Be Mm x nCF) are said to be equivalent if there exist invertible matrices P e Mm x m(F) and Q e Mn x o(F) such that A = PBQ. Equation 3.29 says that a given matrix representation r(f!:, P)(T) ofT relative to a pair of bases (f!:, P) changes to an equivalent matrix when we replace (f!:, ~ by new bases (g', ~l This leads to the following question: What is the simplest representation of a given linear transformation T? If we set A = r(~ P)(T), then we are asking, What is the simplest matrix B equivalent to A? -

Recalling a few facts from elementary matrix theory gives us an easy answer to that question. Any invertible matrix P is a product, P = E, · · · Eto of elementary matrices E1, ... , E,. PA = E,(· .. (E1A)·") is them x n matrix obtained from .A by preforming the elementary row operations on A represented by E1, ... , E,. Similarly (PA)Q is the m x n matrix obtained from PA by preforming a finite number of elementary column operations on PA. Let us denote the rank of any m x n matrix A (i.e., the number of linearly independent rows or columns of A) by rk(A). If rk(A) = s, then we can clearly find invertible matrices P and Q such that

PAQ=(+I ~) Here our notation

(~I~) means PAQ will have the s x s identity matrix, I., in its upper left-hand corner and zeros everywhere else.


If we apply these remarks to our situation in Theorem 3.28, we get the following corollary:

Corollary 3.31: Let V and W be finite-dimensional vector spaces over F of dimensions n and m, respectively. Let 1!: and ~ be bases of V and W. Let T e IIom(V, W), and set A = r(~ ~)(T). If rk(A) = s, then there exist bases g' and ~' of V and W, respectively, such that

3.32:

r(g'. ~')(T) = ( ~ I+) D

There is another representation problem that naturally arises when considering Theorem 3.28. Suppose V = W. If g is a basis ofV, then any T e Hom(V, V) is represented in terms of g by an n x n matrix A = r(f!:, g)(T). If we change 1!: to a new basis g' of V, then the representation ofT changes to B = r(g', g1(T). Equation 3.29 implies that B = PAP - 1, where P = M(g, 1!:1· Recall that two n x n matrices A and B are similar if there exists an invertible n x n matrix P such that B = PAP-1. Thus, different representations of the same T e Hom(V, V) with respect to different bases of V are similar matrices.

Now we can ask, What is the simplest representation ofT? If we choose any basis g of V and set A = r(f!:, g)(T), then our question becomes, What is the simplest matrix B similar to A? That question is not so easy to answer as the previous equivalence problem. We shall present some solutions to this question in Chapter III of this book.

Theorem 3.25 implies that dim Hom(V, W) =(dim V)(dim W) when V and W are finite dimensional. In our next theorem, we gather together some miscellaneous facts about linear transformations and the dim(·) function.

Theorem 3.33: Let V and W be vector spaces over F and suppose TeHom(V, W). Then

(a) If T is surjective, dim V ;;;: dim W.

(b) If dim V = dim W < oo, then Tis an isomorphism if and only if either Tis injective or T is surjective.

(c) dim(Im T) + dim(ker T) = dim V.

Proof: (a) follows immediately from Theorem 2.11. In (b), if T is an isomorphism, then T is. both injective and surjective. Suppose T is injective, and n ==dim V =dim W. Let g = {a1, ... , an} be a basis of V. Since T. is injective, Tg == {T(a1), ... , T(aJ} is a linearly independent set in W. Then dim W = n implies Tf!: is a basis ofW. In particular, W = L(Tg) = T(L(g)) = T(V). Thus, Tis surjective, and hence, an isomorphism.

Suppose T is surjective. If g = { a 1 , ... , an} is a basis of V, then

26 LINEAR ALGEBRA

W = TM = T(L(~)) = L(T~). By Theorem 2.11, T~ contains a basis ofW. Since dim W = n, Tg is a basis of W. Now let aeker T. Write ct = L x1ct1• Then 0 = L x1T(ctJ. Since Tg is a basis of W, x1 = · · · = x.. = 0. Thus, ct = 0 and T is injective. This competes the proof of (b).

We prove (c) in the case that dim V = n < oo. The infinite-dimensional case is left as an exercise at the end of this section. Let g = { ct 1 , ... , tXr} be a basis of ker T. We take r = 0, and g = ¢ if T is injective. By Theorem 2.6, we can expand g to a basis 4 = {a1 , ... , a., P1 , •.. , P.} ofV. Here r + s = n. We complete the proof of (c) by arguing that TP = {T(P1), .•• , T(P.)} is a basis of Im T.

Suppose be 1m T. Then - {J = T(y) for some y e V. Since V = L(4), y = X1tX1 + ··· + XrtXr + Y1P1 + ··· + YsPs for some X~o y1eF. Applying T to this equation, gives fJ E L(Tp). Thus, TP spans Im T.

Suppose L:=lYIT(PJ = 0 for some y,eF. Then L:=1YIPtekerT. Thus, L:= 1 y1p, = Lf= 1 x1ct1 for some x1 e F. Since 4 is a basis of V, we conclude that x1 = ··· = Xr = y1 = · · · = Ys = 0. In particular, {T(P1), ••• , T(p.)} is linearly independent. Thus, TP is a basis of 1m T, and the proof of (c) is complete. 0

We finish this section with a generalization of Theorem 3.33(c). We shall need the following definition.

Definition 3.34: By a chain complex C = { (V~o dJ I i E Z} of vector spaces over F, we shall mean an infinite sequence {V1} of vector spaces Vi> one for each integer i E Z, together with a sequence { d1} of linear transformations, d1 e Hom(V1, V1_ tl for each i E Z, such that d1 + 1 d1 = 0 for all i e Z.

We usually draw a chain complex as an infinite sequence of spaces and maps as follows:

3.35:

H a chain complex C has only finitely many nonzero terms, then we can change notation and write C as

3.36:

It is understood here that all other vector spaces and maps not explicitly appearing in 3.36 are zero.

Definition 3.37: A chain complex

c: · ·· ..... vl+l


is said to be exact if 1m d1 + 1 = ker d1 for every i e Z.

Let us consider an important example.

Example 3.38: Let V and W be vector space over F, and let T e Hom(V, W). Then

C: 0--+ ker T --~ V __ T---) 1m T --~ 0

is an exact chain complex. Here i denotes the inclusion of ker T into V. 0

We can generalize Example 3.38 slightly as follows:

Definition 3.39: By a short exact sequence, we shall mean an exact chain complex C of the following form:

3.40:

Thus, the example in 3.38 is a short exact sequence with V 2 = ker T, d2 = i, V 1 = V, etc. Oearly, a chain complex C of the form depicted in 3.40 is a short exact sequence if and only if d 2 is injective, d1 is surjective, and Im d2 = ker d1. Theorem 3.33(c) implies that if C is a short exact sequence, then dim V 2 - dim V 1 + dim V0 = 0. We can now prove the following generalization of this result:

Theorem 3.41: Suppose

is an exact chain complex. Then Lf=o ( -1)1 dim V1 = 0.

Proof: The chain complex C can be decomposed into the following short exact sequences

J

28 LINEAR ALGEBRA

If we now apply Theorem 3.33(c) to each C1 and add the results, we get Lf=o( -1)1dim V1 = 0. D


(1) Let V and W be vector spaces over F.

(a) Show that the Cartesian product V x W = {(a, p) I a e V, PeW} is a vector space under componentwise addition and scalar multiplication.

(b) Compute dim(V x W) when V and W are finite dimensional.

(c) Suppose T: V-+ W is a function. Show that T eHom(V, W) if and only if the graph GT = {(a, T(a)) e V x WI a e V} of T is a subspace of VxW.

(2) Let TeHom(V, W) and SeHom(W, V). Prove the following statements:

(a) If ST is surjective, then Sis surjective.

(b) If ST is injective, then Tis injective.

(c) If ST = Iv (the identity map on V) and TS = Iw, then T is an isomorphism.

(d) IfV and W have the same finite dimension n, then ST = Iv implies Tis an isomorphism. Similarly, TS = lw implies T is an isomorphism.

(3) Show that Exercise 2(d) is false in general. (Hint: Let V = W be the vector space in Exercise 4 of Section 2.)

(4) Show that P~ F~n = m.

(5) Let T e Hom(V, V). If T is not injective, show there exists a nonzero SeHom(V, V) with TS = 0. If T is not surjective, show there exists a nonzero S e Hom(V, V) such that ST = 0.

(6) In the proof of Corollary 3.19, we claimed that M(g, §)M(§, g)[PJ., = [PJ, for all p e V implies M(g, §)M(§, g) = In. Give a proof of this facf -

(7) When considering diagram 3.17 we claimed T A is an isomorphism if and only if A is an invertible matrix. Give a proof of this fact.

(8) Show that Theorem 3.33(c) is correct for any vector spaces V and W. Some knowledge of cardinal arithmetic is needed for this exercise.

(9) Let T e Hom(V, V). Show that T 2 = 0 if and only if there exist two subspaces M and N of V such that

(a) M + N = V.

(b) M n N = (0). (c) T(N) = 0.

(d) T(M) ~ N.


(10) Let Te Hom(V, V) be an involution, that is, T 2 = Iv. Show that there exists · two subspaces M and N of V such that

(a) M + N= V. (b) M n N = (0).

(c) T(a) = a for every a eM.

(d) T(a) = -a for every aeN.

In Exercise 10,we assume 2 :1= 0 in F. IfF= IF2, are there subspaces M and N satisfying (a)-(d)? ·

(11) Let T E Homp(V, V). If f(X) = RaXn + ' .. + a1 X + a0 E F[X], tllen f(T) =RaT"+ ... + a 1T + a0IveHom(V, V). Show that dimF V = m < oo implies there exists a nonzero polynomial f(X) e F[X] such that f(T) = 0. .,

(12) If S, T e Homp(V, F) such that S(a) = 0 implies T(a) = 0, prove that T =· xS for some xeF.

(13) Let W be a subspace of V with m =dim W ~dim V = n < oo. Let Z = {T e Hom(V, V) I T(a) = 0 for all a e W}. Show that Z is a subspace of Hom(V, V) and compute its dimension.

(14) Suppose V is a finite-dimensional vector space over F, and .let S, T e Homp(V, V). If ST = Iv, show there exists a polynomial f(X) e F[X] such that S = f(T).

(15) Use two appropriate diagrams as in 3.27 to prove the following theorem: Let V, W, Z be finite-dimensional vector spaces of dimensions n, m, and p, respectively. Let g, f}, and 1 be bases of V, W, and Z. If T e Hom(V, W) and s E Hom(W, Z), then r(g, ¥XST) = r(f}, ¥)(S)r(g, f})(T).

(16) Suppose

c: ... ~ vl+ 1 ~+1 vi d,

"'-+ Yt dt V0 -+0

and

C': ... -+ Vi+t di+t v; dl ... -+ V't di V0-+0

are chain complexes with C exact. Let T 0 e HomF(V 0 , V0). Show that there exists T1eHomp(Vb V;) such that T1_ 1d1 = diT1 for all i = 1, .... The collection of linear transformations {T1} is called a chain map from C to C'.

(17) Suppose C = {(V .. dJ 1 i e Z} and C' = {(VI, d;) 1 i e Z} are two chain complexes. LetT = {T1hez be a chain map from C to C'. Thus, T1: V1 -+ v;, and T1_ 1d 1 = djT1 for all ieZ. For each ieZ, set VI= vi-1 X v; = {(cx,fl)laeVI-1• peV;}. Define a map df': VI-+ V'f.-t by

J

30 LINEAR ALGEBRA

dl'(cx, fJ) = ( -d1- 1(cx), T1- 1(cx) + di(P).) Show that C" = {(Vi', di11 ieZ} is a chain complex. The complex C" is called the mapping cone ofT.

(18) Use The~rem 3.33(c) to give another proof of Exercise 16 in Section 2.

(19) Find a T e HomR(C, C) that is not C-linear.

(20) Let V be a finite-dimensional vector space over F. Suppose T e HomF(V, V) such that dim(lm(T2)) = dim(Im(T)). Show that Im(T) n ker(T) = {0}.

(21) The special case of equation 3.29 where V = W, ~ = ~. and g' = ~~ is very important. Write out all the matrices and verify equation 3.29 in the following example: T: IR3 - IR3 is the linear transformation given by T(l51) = 2151 + 2152 + 153, T(l52) = 151 + 3152 - 153, and T(153) = - .5t + 2152. Let ~ = { cx1, cx2, cx3}, where cx1 = (1, 2, 1), cx2 = (1, 0, -1) and cx3 = (0, 1, -1 ). Compute r(~, ~)(T), r(g, g)(T), and the change of bases matrices in 3.29.

(22) Let V be a finite-dimensional vector space over F. Suppose TS = ST for every SeHorn~, V). Show that T = xlv for some xeF.

(23) Let A, Be Mn x n(F) with at least one of these matrices nonsingular. Show that AB and BA are similar. Does this remain true if both A and Bare singular?

4. PRODUCTS AND DIRECT SUMS

Let {VilieA} be a collection of vector spaces over a common field F. bur indexing set A may be finite or infinite. We define the product nleA vi of the vi as follows:

Definition 4.1: n.eA v. = {f: A- UieA Vilfis a function with f(i)eVi for all ieA}.

We can give the set nleA VI the structure of a vector space (over F) by defining addition and scalar multiplication pointwise. Thus, iff, g e nleA V., then f + g is defined by (f + g)(i) = f(i) + g(i). If f e OleA VI and X e F, then xf is defined by (xf)(i) = x(f(i)). The fact that nleA v. is a vector space with these operations is straightforward. Henceforth, the symbol TiieA V1 will denote the vector space whose underlying set is given in 4.1 and whose vector operations are pointwise addition and scalar multiplication.

Suppose V = fl1eA V1 is a product. It is sometimes convenient to identify a given vector fe V with its set of values {f(i) I ieA}. f(i) is called the ith coordinate off, and we think off as the "A-tuple" (f(i))1eA· Addition and scalar multiplication in V are given in temis of A-tuples as follows: (f(i))1e4 + (g(i))ieA = (f(i) + g(i))ieA> and x(f(i))1eA = (xf(i))1eA. This particular viewpoint is especially fruitful. when IAI = n < oo. In this case, we can assume A= {1, 2, ... , n}. Each feV ts then identified with the n-tuple (f(l), ... , f(n)). When IAI = n, we shall use the

PRODUCTS AND DIRECT SUMS 31

notation V 1 X • . • X V n instead of nleA Vi. Thus, the examples given in 1.5, 3.21, and Exercise 1 of Section 3 are all special cases of finite products. Example 1.5 is a product in which every Vi is the same vector space V.

If V = TiieA Vb then for every pair of indices (p, q) e A x A, there exist linear transformations n:PeHom(V, VP), and OqeHom(Vq, V) defined as follows:

Definition 4.2:

(a) n:P: V- VP is given by n:p(f) = f(p) for all feY.

(b) Oq: V q - V is given by

Oq(cx)(i) = {~ if i = q

if i¥=q

In Definition 4.2(b), cxeVq. Oq(cx) is that function in V whose only nonzero value is ex taken on at i = q. The fact that n:P and Oq are linear transformations is obvious. Our next theorem lists some of the interesting properties these two sets of maps have.

Theorem 4.3: Let v = n.eA v •. Then

(a) nPOP = Iv •• the identity map on VP, for all peA.

(b) npOq = 0 for.all p ¥= q in A.

(c) If A is finite, LpeA OPn:P = Iv, the identity map on V. (d) 7tp is surjective and op is injective for all peA.

(e) Let W be a second vector space over F. A function T: W- Vis a linear transformation if and only if n:P T e Hom(W, V p) for all peA.

(f) The vector space V together with the set { nP I peA} of linear transformations satisfies the following universal mapping property: Suppose W is any vector space over F and {TPeHom(W, VP)I peA} a set of linear transformations. Then there exists a unique T e Hom(W, V) such that for every peA the following diagram is commutative:

4.4:

vp

i Proof: (a), (b), and (c) follow immediately from the definitions. n:P is surjective I and OP is injective sinee n:POP = Iv •. Thus, (d) is clear. As for (e), we need only l . argue that Tis linear provided .. Tis linear for all peA. Let a, peW and x, y e F.

32 LINEAR ALGEBRA

Then for every peL\, we have 7tp{T(xa + yP)) = 7tpT(xcx + yP) = X7tpT(cx) + y1tP T(P) = 7tp(xT(cx) + yT(P)). Now it is clear from our definitions that two functions f, g e V are equal if and only if 7tp(f) = 1tP(g) for all peL.\. Consequently, T(xa + yp) = xT(cx) + yT(P), and T is linear.

Finally, we come to the proof of (f). We shall have no. use for this fact in this text. We mention this result only because in general category theory products are defined as the unique object satisfying the universal mapping property given in (f). The map T: W-+ V making 4.4 commute is given byT(a) = (T1(cx))164. We leave the details for an exercise at the end of this section. D

The map 1tP:V-+ VP in Definition 4.2(a) is called the pth projection or pth coordinate map of V. the map Oq: V q -+Vis often called the qth injection of V q into V. These maps can be used to analyze linear transformations to and from products. We begin first with the case where IL.\1 < oo.

Theorem 4.5: Suppose V = V 1 x · · · x V n is a. finite product of vector spaces, and let W be another vector space. IfT1 e Hom(W, V J fori = 1, ... , n, then there exists a unique T e Hom(W, V 1 x · · · x V J such that 1t1T = T1 for all i=1, ... ,n.

Proof: Set T = Lf= 1 81T 1 and apply Theorem 4.3. D

As an immediate corollary to Theorem 4.5, we get the following result:

Corollary 4.6: If IL.\1 = n < oo, then Hom(W,nie4 VI)~ nle4Hom(W, VJ.

Proof: Define a map 'P: Hom(W, V 1 x .. · x V J-+ Hom(W, V 1) x .. · x Hom(W, V J by 'P(T) = (1t1 T, ... , 1tn T). One easily checks that 'P is an

injective, linear transformation. Theorem 4.5 implies 'P is surjective. D

We have a similar result for products in the first slot of Hom.

Theorem 4.7: Suppose v = v1 X ••• X vn is a finite product of vector spaces, and let W be another vector space. IfT1 e Hom(Vb W) fori = 1, ... , n, then there exists a unique T e Hom(V 1 x · · · x V n• W) such that T81 = T1 for all i = 1, ... ,n.

Proof· Set T = :Li = 1 T11t1 and apply Theorem 4.3. D

Corollary 4.8: If IL.\1 = n < oo, then Hom(U164 Vt> W) ~ U 1e4 Hom(V~o W).

Proof: Define a map 'P: Hom(U1e4 Vb W)-+ U1e4 Hom(Vt> W) by 'P(T) = (T81, ... , TOJ. Again the reader can easily check that 'P is an injective, linear transformation. Theorem 4. 7 implies that 'P is surjective. D

I i


Suppose V = V 1 x · · · x V n is a finite product of vector spaces over F. Let B1 be a basis of Vb i = 1, ... , n. We can think of the vectors in V as n-tuples (er:

1, ••• , aJ with cx1 eV1. For any i and aeV~o 81(cx) = (0, ... , a, 0, ... , 0). Thus,

8 (er:) is the n-tuple of V that is zero everywhere except for an a in the ith slot. since (}I is injective, (JI: VI~ (}I(V J. In particular, (JI(BJ is a basis of the subspace (J.(V J. Since Ot(BJ n L(U J1'I OJ(BJ)) = (0), B = UleA 81(BJ is a linearly independe~t set. Clearly, V = Li=1 81(VJ. Consequently, B is a basis ofV. We have now proved the following theorem:

Theorem 4.9: Let V · = V 1 x · · · x V n be a finite product of vector spaces. If B1 is a basis of V1, i = 1, ... , n, then B = Ui= 1 81(BJ is a basis of V. In particular, if each V1 is finite dimensional, then so is V. In this case, we have

dim v = Li=l dim VI. D

At this point, let us say a few words about our last three theorems when IL.\1 = oo. Corollary 4.6 is true for any indexing set L\. The map 'P(T) = (1t1T)164 is an injective, linear transformation as before. We cannot use Theorem 4.5 to conclude 'Pis surjective, since L164 81T1 makes no sense when IL.\1 = oo. However, we can argue directly that 'P is surjective. Let (T J 164 e U 1e4 Hom(W, V J. Define TeHom(W,Uie4 VJ by T(oc} = (T1(cx})164. Clearly 'P(T) = (TJ154. Thus, we have the following generalization of 4.6:

4.10: For any indexing set L.\, Hom(W, UteA VJ~ Ute4 Hom(W, VJ.

In general, Corollary 48 is false when IL.\1 = oo. For example, if W = F and V 1 = F for all i e L\, then the reader can easily see that 1HomF(Uie4 F, F)l > 1n164 Fl when L\ is infinite. Since HomF(F, F)~ F, we see that Hom(U164 F, F) cannot be isomorphic to U 164 Hom(F,F).

If v = nle4 vi with IL.\1 = 00 and Bt is a basis of VI> then Ute4 (JI(BJ is a linearly independent subset of V. But in general, V :1: }2164 Ot(V J. For a concrete example, consider V = IRF\1 in Exercise 5 of Section 1. Thus, U 1e4 91(BJ is not in general a basis for V. In particular, Theorem 4.9 is false when IL.\1 = oo.

Let us again suppose v = nle4 VI with L\ an arbitrary set. There is an important subspace of V that we wish to study.

Definition 4.11: Let EBie4 VI= {feUie4 Vdf(i) = 0 exCept possibly for finitely many iei.\}.

Clearly EB 164 V1 is a subspace of V under pointwise addition and scalar multiplication. In terms of !\-tuples, the vector f = (cxJ1e4 lies in EB leA V1 if and only if there exists some finite subset L.\0 (possibly empty) of L\ such that ext = 0 for all iei.\- L\0 • Ifli.\1 < oo, then E9 1e4 V1 =UreA V1. Ifli.\1 = oo, then EB1eA V1 is usually a proper suospace of V. Consider the following example:

34 UN EAR ALGEBRA

Example 4.12: Let F = IR, t1 = N, and V1 = IR for all i eA. Then the N-tuple (l),.,N, that is, the function f: N -+ U ieN IR given by f(i) = 1 for all i e N, is a vector in v = nieN IR but not a vector in EBleN R D

The vector space EeleA V1 is called the direct sum of the V1• It is also called the subdirect product of the V1 and written UleA V1• In this text, we shall consistently use the notation EE) 1eA V1 to indicate the direct sum of the V1• If llil = n < oo, then we can assume t1 = {1, 2, ... , n}. In this case we shall write V 1 EJ) • • • EJ) V or EB~=1 v, instead of EBieA v,. Thus, v1 Ei) •.• Ei) Vn, E9~=1 vi, v.l X ••• X vn: nleA VI> and EeieA VI are all the same space when llil = n < 00.

Since EeleA v; = LieA O,(V J, our comments after 4.10 imply the following theorem:

Theorem 4.13: Suppose V = EB leA V1 is the direct sum of vector spaces V1• Let B1

be a basis of VI. Then B = uleA O,(BJ is a basis of v. D

The subspace EB teA V1 constructed in Definition 4.11 is sometimes called the external direct sum of the V1 because the vector spaces {V11 i e t1} a priori have no relationship to each other. We finish this section with a construction that is often called an internal direct sum.

Suppose Vis a vector space over F. Let {V1 1 ieA} be a collection ofsubspaces of V. Here our indexing set t1 may be finite or infinite. We can construct the (external) direct sum E9teA V1 of the V1 as in Definition 4.11 and consider the natural linear transformation S: EB teA V1 -+ V given by S((aJ1eJ = LteA a1• Since (aJ1eA e E9teA Vi> only finitely many of the a1 are nonzero. Therefore, LieA a1 is a well defined finite sum in V. Thus, S is well defined and clearly linear.

Definition 4.14: Let {Vd ietl} be a collection of subspaces of V. We say these subspaces are independent if the linear transformation S: E9taA V1 -+ V defined above is injective.

Note Im S = LieA V,. Thus, the subspaces VI> i e tl, are indepedent if and only if E&.aA V1 ~ LiaA V1 viaS. A simple example of independent subspaces is provided by Theorem 2.13(c).

Example 4.15: Let V be a vector space over F and W a subspace of V. Let W' be any complement ofW. Then W, W' are independent. The direct sum W EJ) W' is just the product W x W', and S: W x W' -+ W + W' is given by S((a, /3)) = a + fl. If (a, f3) e ker S, then a + f3 = 0. But W n W' = 0. Therefore, a = - f3 e W n W' implies a = f3 = 0. Thus, S is injective, and W, W' are independent. D

In our next theorem, we collect a few simple facts about independent subs paces.


Theorem 4.16: Let {V,Iieli} be a collection of subspaces of V. Then the following statements are equivalent:

(a) The VI> i e A are independent.

(b) Every vector ae LieA V1 can be written uniquely in the form a= Li a1 with a1eV1 for all ieA. eA

(b') If LieA a, = 0 with a1 e VI> then a1 = 0 for all i e tl.

(c) For every j e t1, V1 n (L11 J V J = (0).

Proof: In statements (b) and (b'), LiEA a1 means a1 = 0 for all but possibly finitely many ieA. It is obvious that (b) and (b') are equivalent. So, we argue

oo-~-~ . Suppose the V1 are independent. If LtaA a1 = 0 with a1 e V1 for all i e tl, then

S((aJ1eJ = LiEA a1 = 0. Since S is injective, we conclude that a1 = 0 for all i e tl. Thus, (a) implies (b'). Similarly, (b') implies (a).

Suppose we assume (b'). Fix j eA. Let a e V1 n (L11 J V J. Then a = a1 for some a1eV1, and a= LieA-{j}at for some a1eV1• As usual, all the cx1 here are zero except possibly for finitely many indices i ':/: j. Thus, 0 = LteA-{j} a1 + ( -l)a1•

(b') then implies a1 = a1 = 0 for all ietl- {j}. In particular, a= 0, and (c) is established.

Suppose we assume (c). Let LieAa1 = 0 with a1eV1• If every a1 = 0, there is nothing to prove. Suppose some al> sa:¥ a1, is not zero. Then a1 = - Lt'f' J a1 e V1 n (L11 1 V J implies V1 n (L11 1 V J ':/: 0. This is contrary to our assumption. Thus, (c) implies (b'), and our proof is complete. 0

If {V1lieA} is a .collection of independent subspaces of V such that LieA V1 = V, then we say V is the internal direct sum of the V1• In this case, V ~ EB leA V1 viaS, and we often just identify V with EB leA V1• If llil = n < oo, we shall simply write V = V 1 EJ) • • • EJ) V n when V is an internal direct sum of subspaces v 1> ... ' v n•

The reader will note that there is no difference in notation between an external direct sum and an internal direct sum. This deliberate ambiguity will cause no real confusion in the future.

Finally, suppose V = V 1 EJ) • • • EJ) V n is an internal direct sum of independent su~spaces V1> ... , Vn. Then by Theorem 4.16(b), eyery vector aeV can be wntten uniquely in the form a= a 1 +···+!Xu with cx1eV1• Thus, the map ~~: v.-+ V, which sends a to !XJ, is a well-defined function. Theorem 4.16(b) unpbes that each P1 is a linear transformation such that 1m P1 = V1. We give formal names to these maps P 1, ... , P n.

Definition 4.17: Let V = V 1 E9 · · · E9 Vn be the internal direct sum of independent subspaces V1 , ... , VI!-. For each i = 1, ... , n, the linear transformation P1

defined above is called the ith projection map of V relative to the decomposition v1 ffi · ·· E9 vn.

36 LINEAR ALGEBRA

Our next theorem is an immediate consequence of Theorem 4.16(b).

Theorem 4.18: Let V = V 1 EB · · · EB V n be an internal direct sum of the independent subspaces V 1 , ••• , V n. Suppose P 1 , .•• , P n e Hom(V, V) are the associated projection maps. Then

(a) P1PJ = 0 if i =I: j. (b) P1P1 = P1. (c) 2:f= 1 P1 = Iv. the identity map on V. 0

Theorem 4.18 says that every internal direct sum decomposition V = V 1 EB · · · EB V n determines a set {P t> ••• , P .. } of pairwise orthogonal [4.18(a)] idempotents [4.18(b)] whose sum is lv [4.18(c)] in the algebra of endomorphisms G(V) = Homp(V, V). Let us take this opportunity to define some of the words in our last sentence.

Definition 4.19: By an associative algebra A over F, we shall mean a vector space (A, (ex, fJ) -+ex + p, (x, ex)-+ xa) over F together with a second function (ex, fJ) -+ a.fJ from A x A to A satisfying the following axioms:

Al. a.({Jy) = (exp)y for all ex, p, yeA. A2. ex(p + y) = a.p + a.y for all a., p, yEA.

A3. (fJ + y)a. = fJa + ya for all ex, p, yeA. A4. x(a.fJ) = (xa.)fJ = a.(xfJ) for all ex, peA, x e F. AS. There exists an element 1 e A such that lex = ttl = a. for all a eA.

We have seen several examples of (associative) algebras in this book already. Any field F is an associative alebra over F. Mnxn(F) and F[X] with the usual multiplication of matrices or polynomials is an algebra over F. IfV is any vector space over F, then Hom~V, V) becomes an (associative) algebra over F when we define the product of two linear transformations T 1 and T 2 to be their composite T 1 T 2• Clearly axioms Al-AS are satisfied. Here 1 is the identity map from V to V. Linear transformations from V to V are called endomorphisms of V. The algebra G(V) = Homp(V, V) is called the algebra of endomorphisms of V.

Suppose A is any algebra over F. An element ex e A is idempotent if excx = a.. In F or F[X], for example, the only idempotents are 0 and 1. In M., x .,(F), e11 , •.• , e .... are all idempotents different from 0 or 1. Idempotents { a. 1 , •.• , ex .. } in an algebra A are said to be pairwise orthogonal if cx1exJ = 0 whenever i =1: j. Thus, { e11 , ••• , e .. .,} is a set of pairwise orthogonal idempotents in M .. x .,(F).

Theorem 4.18 says that every internal direct sum decomposition V = V 1 EB · · · EB V n determines a set of pairwise orthogonal idempotents whose sum is 1 in G(V). Our last theorem of this section is the converse of this result.

Theorem 4.20: Let V be a vector space over F, and suppose {P 1 , ••• , P .. } is a set of pairwise orthogonal idempotents in H(V) such that P 1 + · · · + P" = 1. Let VI= ImPI. Then v = vl EB"·EBV ...


Proof: We must show V = V1 + ... + v .. and VJ n (LI'f'J V1) = 0. Let cxeV and set ex1 = P1(cx). Then ex = l(cx) = (P 1 + .. · + P nXex) = P 1 (a.) + · .. + P .. (a.) = al + '"O:n. Since a.lelmPI =VI> we conclude v £; vl + ... + v ... Thus, V=V1+"·+V ...

Fix J, and suppose .<5eyJn(L17'JVJ. Then <5=PJ(fJ)=2:17'JP1({J1) for some {JeV and fJ1eV (1 =I:J). Then o = P1({J) = P1P1(fJ) = PJ(L17'1P1({J1)) = LI7'JPJPI(PJ = 0. Thus, VJ n (LI'f'J VJ = (0), and the proof is complete. 0


(1) Let B = { <511 i e d} be a basis ofV. Show that Vis the internal direct sum of {Fo1lied}.

(2) Show Hom~E91eA VI, W)~ nleAHomp(VIt W).

(3) Give a careful proof of Theorem 4.3(f).

(4) Let V = V 1 x · · · x V "' and for each i = 1, ... , n, set T1 = 8111:1• Show that {T 1 ,. .. , T .. } is a set of pairwise orthogonal idempotents in 8(V) whose sum is 1.

(5) Let V = V 1 x · · · x V n. Show that V has a collection of subspaces {W~> .. ·• W~} such that V1~ W1 fori= l, ... ,n and V = E9f= 1 W1•

(6) Give a combined version of Corollaries 4.6 and 4.8 by showing directly that I/I:Homp(V1 X ... X v.,, wl X ... X W.J-+ nr=1Thm=l Hom(Vh Wj) given by J/!(T) = (xJTOJ1=1, ... ,n,J= 1, ... ,m is an isomorphism.

(7) Suppose V = V 1 EB · · · EB V.,. Let T e Hom(V, V) such that T(V J £; V1 for all i = 1, ... , n. Find a basis g of V such that

( (M1 . r g. g)(T) = o . . .

where M1 describes the action ofT on V1·

(8) If X, Y, Z are subspaces of V such that X EB Y = X EB Z = V, is Y = Z? Is y~ Z? ·.

(9) Find three subspaces V 1> V 2, V 3 of V = F[X] such that V = V 1 EB V 2 EB V 3·

(10) If V = V 1 + V 2, show that there exists a subspace W of V such that W £; V 2 'and V = V 1 EB W.

(11) Let A be an algebra over F. A linear transformation T e Hom~A, A) is called an algebra homomorphism if T(cxfJ) = T(a.)T(/1) for all ex, {JeA. Exhibit a nontrivial algebra homomorphism on the algebras F[X] and Mnxn(F).

38 LINEAR ALGEBRA

(12) Suppose V is a vector space over F. Let S: V~ V be an isomorphism of V. Show that the map T ~ S - 1TS is an algebra homomorphism of G(V) which is one to one and onto.

(13) Let F be a field. Show that the vector space V = F (over F) is not the direct sum of any two proper subspaces.

(14) An algebra A over F is said to be commutative if rx{J = {Jrx for all a., peA. Suppose V is a vector space over F such that dim~ > 1. Show that G(V) is not commutative.

(15) Suppose V is a vector space over F. Let TeG(V) be idempotent. Show V = ker(T) Ee Im(T).

(16) Let V be a vector space over F, and let TeG(V). If T 3 = T, show that V = V 0 Ee V 1 Ee V 2 where the V1 are subspaces of V with the following properties: rx e V 0 => T(a.) = 0, a. e V 1 => T(rx) = ex, and ex e V 2 => T(ex) = -ex. In this exercise, assume 2 =/:: 0 in F.

(17) Suppose V is a finite-dimensional vector space over F. If T eG(V) is nonzero, show there exists an S e G(V) such that ST is a nonzero idempotent of G(V).

(18) Suppose T e G(V) is not zero and not an isomorphism of V. Prove there is an S e G(V) such that ST = 0, but TS =/:: 0.

(19) Suppose V is a finite-dimensional vector space over F with subspaces W 1 , ... , W". Suppose V=W1 +···+Wk, and dim(V)=L~= 1 dim(WJ. Show that V = W 1 Ee · .. Ee Wk.

5. QUOTIENT SPACES AND THE ISOMORPHISM THEOREMS

In this section, we develop the notion of a quotient space of V. In order to do that, we need to consider equivalence relations. Suppose A is a nonempty set and R s;;; A x A is a relation on A. The reader will recall from Section 2 that we used the notation x "'"' y to mean (x, y) e R The relation "'"' is called an equivalence relation if the following conditions are satisfied:

5.1: (a) x"'"' x for all xeA. (b) If x"'"' y, then y"'"' x for all x, yeA. (c) Ifx"'"' y andy"'"' z, then x "'"'z for all x, y, zeA.

A relation satisfying 5.1(a) is called reflexive. If 5.1(b) is satisfied, the relation is said to be symmetric. A relation satisfying 5.1(c) is said to be transitive. Thus, an equivalence relation is a reflexive, symmetric relation that is transitive.

Example 5.2: Let A = 71.., and suppose p is a positive prime. Define a relation = (congruence mod p) on A by x = y if and only if pIx - y. The reader can easily check that = is an equivalence relation on 71... 0

r I

QUOTIENT SPACES AND THE ISOMORPHISM THEOREMS 39

The equivalence relation introduced in Example 5.2 is called a congruence, and we shall borrow the symbol = to indicate a general equivalence relation. Thus, if R s;;; A x A is an equivalence relation on A and (x, y) e R, then we shall write x = y. We shall be careful in the rest of this text to use the symbol s only when dealing with an equivalence relation.

Now suppose = is an equivalence relation on a set A. For each x e A, we set x = {yeAiy = x}. xis a subset of A containing x .. xis called the equivalence class of x. The function from A to f?l'(A) given by x --+ x satisfies the following properties:

5.3: (a) xex.

(b) x = y if and only if x = y. (c) For any x, yeA, either x = y or iny = rjJ.

(d) A = UxeA i.

The proofs of the statements in 5.3 are all easy consequences of the definitions. If we examine Example 5.2 again, we see 7L is the disjoint union of the p equivalence classes 0, I, ... , p- 1. It follows from 5.3(c) and 5.3(d) that any equivalence relation on a set A divides A into a disjoint union of equivalence ~asses. The reader probably has noted that the equivalence classes {0, 1, ... , p - 1} of 7L inherit an addition and multiplication from 7L and form the field IF P discussed in Example 1.3. This is a common phenomenon in algebra. The set of equivalence classes on a set A often inherits some algebraic operations from A itself. This type of inheritance of algebraic structure is particularly fruitful in the study of vector spaces.

Let V be a vector space over a field F, and suppose W is a subspace ofV. The subspace W determines an equivalence relation = on V defined as follows:

5.4:

rx=P if rx-{JeW

Let ~s. check that the relation = defined in 5.4 is reflexive, symmetric, and transttive. Clearly, ex = ex. If ex = p, then ex - peW .. Since W is a subspace, f1 -.a. e ~· Therefore, P = a.. Suppose ex = p and p = y. Then ex - p, p - yeW. ~gam, smce W is a subspace, ex- y = (ex- p) + (p- y) e W, and, thus ex = y. So, Indeed = is an equivalence relation on V. The reader should realize that the equivalence relation = depends on the subspace W. We have deliberately suppressed any reference to W in the symbol = to simplify notation. This will cause no confusion in the sequel. ·

D 6ni. e . tion 5.5: Let '?I be a subspace of V, and let = denote the equivalence :elation defined in 5.4. If ex e V, then the equivalence class of ex will be denoted by IX. The set of all equivalence classes {~I a e V} will be denoted by V fW.

40 UN EAR ALGEBRA

Thus, ii = {PeVIP =a} and VjW = {iilaeV}. Note that the elements in V jW are subsets of V. Hence V fW consists of a collection of elements from t?PM.

Definition 5.6: If W is a subspace of V and a e V, then the subset a + W = {a + 1 I 1 e W} is called a coset of W.

Clearly, pea + W if and only if a - peW. Thus, the coset a + W is the same set as the equivalence class ii of a under =. So, V fW is the set of all cosets of W. In particular, the equivalence class ii of a has a nice geometric interpretation. ii = a + W is the translate of the subspace W through the vector a.

Let us pause for a second and discuss the other names that some of these objects have. A coset a+ W is also called an affine subspace or flat ofV. We shall not use the word "fiat" again in this text, but we want to introduce formally the set of affine subspaces of V.

Definition 5.7: The set of an affine subspaces of v will be denoted dM.

Thus, A e dM if and only if A = a + W for some subspace W £ V and some a e V. Note that an affine subspace A = a + W is not a subspace of V unless a = 0. Thus, we must be careful to use the word "affine" when considering elements in dM. Since dM consists of all cosets of all subspaces of V, V fW £ dM £ t?PM and these inclusions are usually strict.

The set V fW is called the quotient of V by W and is read "V mod W". We shall see shortly that V fW inherits a vector space structure from V. Before discussing this point, we gather together some of the more useful properties of affine subspaces in general.

Theorem 5.8: Let V be a vector space over F, and let dM denote the set of all affine subspaces of V. ·

(a) If {AdieA} is an indexed collection of affine subspaces in dM, then either nie4 AI = 4> or nie4 Ate dM.

(b) If A, BedM, then A+ BedM. (c) If AedM and xeF, then xAedM. (d) If AedM and TeHom(V, V'), then T(A)ed(V'). (e) If A' e d(V') and T e Hom(V, V'), then T- 1(A') is either empty or an affine

subspace of V.

Proof: The proofs of (b)-(e) are ·an straightforward. In (e), T- 1(A') = {aeVIT(a)eA'}. We give a proof of (a) only. Suppose. A1 = tX1 + W1 for each i eA. Here W1 is a subspace of V and a1 a vector m V. Suppose nle4Ai 'I= <f>. Let pe nie4AI. Then for each ie.6., p =IX!+ 'YI with f'IEWI. But then p + WI = IX! + w" and nieA AI = nleA (p + w J.

We claim that n 1e4(P + W J = p + (nlsA W J. Clearly, p + (nie4 W J £ ·


nieA(p + WJ, so let IXE nle4(p +WI). Then, for i 'I= j, IX= p +~I= p + ~j with ~1 eW1 and ~1eW1 • But then ~~=~1 and aeP+(W1nW1). Thus, nieA(p + WJ £ p + <nleA WJ. Therefore, n 1e4(p + WJ = p + (n1e4 WJ. Since P + (n1e4 W J e dM, the proof of (a) is complete. 0

We can generalize Theorem 5.8(d)' one step further by introducing the concept of an affine map between two vector spaces. If a e V, then by translation through tX, we shall mean the function s .. : V --+ V given by S .. (p) = a + p. Any coset a + W is just S..(W) for the translations ... Note that when a 'I= 0, s .. is not a linear transformation.

Definition 5.9: Let V and V' be two vector spaces over a field F. A function f: V--+ V' is called an affine transformation iff= s .. T for some Te HomF{V, V') and some aeV'. The set of all affine transformations from V to V' will be denoted AffFfV, V').

Oearly, HomFfV, V') £ Aft'F(V, V') £ (V')v. Theorem 5.8(d) can be restated as follows:

Theorem 5.10: If AedM and feAffFfV, V'), then f(A)ed(V'). 0

Let us now return to the special subset V /W of dM. The cosets of W can be given the structure of a vector space. We first define a binary operation -i- on V jW by the following formula:

5.11:

In equation 5.11, a and p are vectors in V and ii and 7J are their corresponding equivalence classes. ii -i- fJ is defined to be the equivalence class that contains a + p. We note that our definition of ii -i- fJ depends only on the equivalence classes ii and 7J and not on the particular elements aeii and PeP (used to form the right-hand side of 5.11). To see this, suppose a 1 eii and.p1 e]J. Then tXt- a and P1 - Pare in W. Therefore, (at·+ P1)- (a+ p)eW and a1 + P1 =a+ p. Thus, -i-: V fW x V fW --+ V fW is a well-defined function .. The reader can easily check that (V jW, -i-) satisfies axioms Vl-V 4 of Definition 1.4. 0 is the zero element of V /W, and -a is the inverse of ii under -i-. The function -i- is called addition on V /W, and, henceforth, we shall simply write + for this operation. Thus, ii + 7J = a + p defines the operation of vector addition on V fW.

We can define scalar multiplication on VfW by the following formula:

5.12:

Xii =XIX

42 LINEAR ALGEBRA

~equation 5.12, xeF~nd aeVfW. Again we observe that if a 1 ea., then

XIX 1 = XIX. Thus (x, a) -+ XIX is a well-defined function from F x V fW to V fW. The reader can easily check that scalar multiplication satisfies axioms V5-VS in Definition 1.4. Thus, (V /W, (a, lJ) -+a + lJ, (x, a) -+ xa) is a vector space over F. We shall refer to this vector space in the future as simply VfW.

Equations 5.11 and 5.12 imply that the natural map IT:V-+ VfW given by IT(IX) = a. is a linear transformation. Clearly, IT is surjective and has kernel W. Thus, if i: W-+ V denotes the inclusion of W into V, then we have the following short exact sequence:

5.13:

0-+ W--'--~ v-.::.::n~ V fW-+ 0

ln particular, Theorem 3.33 implies the following theorem:

Theorem 5.14: Suppose Vis a finite-dimensional vector space over F and W a subspace of V. Then dim V = dim W + dim V /W. 0

We shall finish this section on quotients with three theorems that are collectively known as the isomorphism theorems. These theorems appear in various forms all over mathematics and are very useful.

Theorem 5.15 (First Isomorphism Theorem): Let T E Homp(V, V'), and suppose W is a subspace of V for which T(W) = 0. Let IT: V -+ V /W be the natural map. Then there exists a unique T E Homp(V fW, V1 such that the following diagram commutes:

5.16:

T

v~~· V/W

Proof We define T by T(a) = T(IX). Again, we remaind the reader that a is a subset of V containing IX. To ensure that our definition of T makes sense, we must argue that T(1X1) = T(IX) for any IX 1 e iX. If 1X 1 E iX, then IX1 - IX E W. Since T is zero on W, we get T(1X1) = T(a). Thus, our definition of T(iX) depends only on the coset iX and not on any particular representative of iX. Since T(xiX + ylJ) = T(xa + y[:J) = T(xa + y[:J) = xT(a) + yT({J) = xT(iX) + y'f{lJ), we


see TeHom(VjW, V1. TIT(a) = T(a) = T(a) and so 5.16 commutes. Only the uniqueness of T remains to be proved.

If T' e Hom(V jW, V') is another map for which T'IT = T, then T = T' on Im n. But IT is surjective. Therefore, T = T'. 0

Corollary 5.17: Suppose TeHomp(V, V'). Then ImT~ Vjker T.

Proof We can view T as a surjective, linear transformation from V to 1m T. Applying Theorem 5.15, we get a unique linear transformation T: V jker T -+ Im T for which the following diagram is commutative:

5.18:

V T 1m T

~/ Vjker T

In 5.18, IT is the natural map from V to Vjker T. We claim Tis an isomorphism. Since TIT = T and T: V-+ 1m T is surjective, T is surjective. Suppose iX e ker T. Then_ T(a) = Trr(1X) = T(a) = 0. Thus, aekerT. But, then IT{1X) = 0. Thus, a= 0, and T is injective. 0

The second isomophism theorem deals with multiple quotients. Suppose W is a subspace of V and consider the natural projection IT: V-+ V fW. If W' is a subspace of V containing W, then IT{W1 is a subspace of V [W. Hence, we can form the quotient space (V /W)(fi(W'). By Corollary 5.17, IT{W1 is isomorphic to W'/W. Thus we may rewrite (V/W)(fi(W1 as (V/W)/{W'/W).

Theorem 5.19 (Second Isomorphism Theorem): Suppose W !;;;; W' are subspaces ofV. Then (V/W)/{W'/W)~ V/W'.

!'ro?f Let IT: V-+ V /W and IT': V fW-+ (V /W)(fi(W') be the natural proJec~tons. Set T = IT'll: V -+ (V /W)(fi(W'). Since IT and IT'. are both surjective, T ts a surjective, linear transformation. Clearly, W' !;;;; ker T. Let a e ker T. Then 0 = IT'IT(1X). Thus, a= IT(1X)eiT(W'). Let [:JeW' such that IT([:J) = IT(a). Then IT([:J- a)= 0. Thus, [:J- IXEkeriT = W!;;;; W'. In particular, IXEW'. We have now proved that kerT = W'. Applying Corollary 5.17, we have {V!W)(fi(W') =1m T~ VjkerT = VjW'. 0

The third isomorphism theorem deals with sums and quotients.

Theorem 5.20 (Third Isomorphism Theorem): Suppose W and W' are subspaces ofV. Then (W + W')/W~ W'/{W n W1.

44 LINEAR ALGEBRA

Proof: Let IT: W + W' -+ (W + W')/W be the natural projection. The inclusion map ofW' into W + W' when composed with I1 gives us a linear transformation T: W' -+ (W + W1/W· Since the kernel of I1 is W, ker T = W n W'. We claim T is surjective. To see this, consider a typical element ye(W + W1/W· y is a coset ofW of the form y = t5 + W with fJeW + W'. Thus, {J =ex+ f3 with exeW and peW'. But ex + W = W. So, y = t5 + W = (/3 + ex) + W = f3 + W. In particular, T(p) = p + W = y, and T is surjective. By Corollary 5.17, (W + W')/W = 1m T~ W'fker T = W'/W n W'. 0

We close this section with a typical application of the isomorphism theorems. Suppose V is an internal direct sum of subs paces V 1 , ••• , V n. Thus, V = V

1 Ea · .. Ea V n· Since v, n (LJ1'' VJ) = (0), Theorem 5.20 implies

V(V1 = (V

1 + LHt V1)(V1 ~ (LJor 1 V1)/(V1 n (LJr 1 VJ)) = (L; 71 Vj)/(0) = V 1 E9 · ..

E9 V1 E9 · • • E9 V n· Here the little hat(~) above V1 means V1 is not present in this

sum.


(1) Suppose f e HomF(V, F). Iff =I= 0, show V fker f~ F.

(2) Let T e Hom(V, V) and suppose T(ex) = ex for all ex e W, a subspace of V.

(a) Show that T induces a map S e Hom(V /W, V /W). (b) If Sis the identity map on V /W, show that R = T - Iv has the property

that R2 = 0. (c) Conversely, suppose T = Iv + R with ReHom(V, V) and R

2 = 0.

Show that there exists a subspace W ofV such that Tis the identity on W and the induced map S is the identity on V fW.

(3) A subspace W of V is said to have finite codimension n if dim V /W = n. If W has finite codimension, we write codim W < oo. Show that if W 1 and W 2 have finite codimension in V, then so does W 1 n W 2. Show codim(W 1 n W 2) ~ codim W 1 + codim W 2•

(4) In Exercise 3, suppose Vis finite dimensional and codim W 1 = codim W 2· Show that dim(W 1/W 1 n W 2) = dim(W ~ 1 n W 2).

(5) LetT e Hom(V, V'), and suppose Tis surjective. Set K = ker T. Show there exists a one-to-one, inclusion-preserving correspondence between the subspaces of V' and the subspaces of V containing K.

(6) LetT E Hom(V, V1, and let K = ker T. Show that all vectors ofV t_hat have the same image under T belong to the same coset of V jK.

(7) Suppose W is a finite-dimensional subspace of V such that V /W is finite dimensional. Show V must be finite dimensional.


(8) ~t V be ~ finite-dimensional vector space. If W is a subspace with dim W = d.Im V- 1, then the cosets of W are called hyperplanes in v. Sup.P_ose SIS an affine subspace of V and H = ex + W is a hyperplane. Show that if H n S = 0, then S ~ p + W for some p e V.

(9) ~ S = .a + We d(V), we define dim S = dim W. Suppose y is finite dimens.Ional, H a hyperplane in V, and S e d(V). Show that s n H =1= 0 => dim(S n H) = dim S - 1. Assume S ¢. H.

(10) tiLet S e d(V) .with dimS = m - 1. Show that S = {Ll"= 1 x,ad Ll"= 1 x1 = 1} or some choice of m vectors a 1 , ... , am e V.

(11) Suppose C = {(V" dJiieZ} is a chain complex. For each ieZ, set H1(C) = ker dJim d1+ 1• H1(C) is called the ith homology of C. (a) Show that C is exact if and only if H1(C) = 0 for all i e z. (b) Let C = {(V" dJiieZ} and C' = {(Vj, d~lieZ} be two chain

complexes. Show that any chain map T = {T1}iez: C -+ C induces a linear transformation 'f1:Ht(C)-+ H1(C1 such that 'f1(a+Imd1+ 1)= T1(a) + Imdi+1·

(c) Suppose C: 0 -+ Vn d. V ~ dl V 0 n-1~··· 1-+

is a ~te chain complex. Show that L(- 1)1 dim H,(C) = L( -1)' dim V,. Here each V1 is assumed finite dimensional.

(12) Suppose V is an n-dimensonal vector space over F and W W are b

' 1, ... , k

~u spaces of codimension e1 = n - dim(W J. Let S1 = a1 + w, for I= 1, ... , k. IfS1 n ... nSk = </J, show dim(W1 n ... n WJ > n- L~= 1 e1•

(13) Use Exercise 12 to' prove the following assertion: Let S1 = a 1 + w 1 and S2 = ex2 + W 2 be two cosets of dimension k [i.e., dim(W J = k]. Show that ~ 1 and S2 are parallel (i.e., W 1 = W 2) if and only if S 1 and S are contained m a coset of dimension k + 1, and have empty intersectio~.

U4) I~ IR3, ~how ~at ~he ~ntersection of two nonparallel planes (i.e., cosets of

dimen~Ion 2) IS a line (I.e., a coset of dimension 1). The same problem makes sense many three-dimensional vector space V.

(15) Let S1o S2, and S3 be planes in IR3 such that S1 n S2 n S3 = <jJ, but no two S1 are parallel. Show that the lines S1 n S2, S1 n S3 and S2 n S3 are parallel.

(16) Let f(X). = xn + au-lxn-l+···+aoEIR[X]. Let w = {pflpe!R[X]}. Show t~a~ Yf IS a subspace of IR[X]. Show that dim(IR[X]/W) = n. (Hint: Use the diVISion alg~rithm in IR[X].)

(17) In Thea 5 15 'fT · · · · -. rem . , I IS suiJective and W = kerT, then Tis an isomorph-Ism [prove!]. In .Particular, S = (f) -l is a well-defined map from V' to V /W. Show that the process of indefinite integration is an example of such a mapS.

46 LINEAR ALGEBRA

6. DUALS AND ADJOINTS

Let V be a vector space over F.

Definition 6.1: V* = Hom~V, F) is called the dual of V.

lfV is a finite-dimensional vector space over F, then it follows fro~ Theor:~ 3.25 that V* is finite dimensional with dim V* =dim V. We record th1s fact Wtt

a different proof in 6.2.

fieorem 6.2: Let V be finite dimensional. Then dim V* =dim V.

_ { } be a basis of V. For each i = 1, ... , n, define an Proof tLe! ~V-;. bayl ::; :..:_ann(·) Here (·) . V -+ p is the isomorphism determined elemen !Xi e "'I - I "'' "'' • h d' t f

h b · nd 1t • Fn -+ F-is the natural projection onto the tt coor ma e o by t e asts a a 1· * . . b Fn. Thus, if;= x1a1 + ... + x0 a0 EV, then aJ IS gtven Y

6.3:

af(x1oc1 + .. · + XnaJ = Xt

We claim that~·= {aT, ... , a!'} is a basis ofV*. Suppose L~=l Y1ai = 0.:_ Let · e {1, ... , n}. Then equation 6.3 implies?= <l:r=t Y1a;}{aJ) = Ll= 1 Ytal(ocJ- Y1· J _. .. = 0 and oc* is linearly mdependent over F. . . Thus, Yt V* th~n T = rr:l T(ocJar. This last equation follows tmme~ately fro~ T6~3. Thus, L(~*) = V*, and g* is a basis of V*. In particular,

dimV* = 1~*1 = n = dimV. 0

Th b · * - {N* ""*} constructed in 6.3 is called the dual basis of ~· e astsa - ""t•·"•""n d' Thus every b~is a of a finite-dimensional vector space V has a corre~pon t~g dual basis a* ofV*. Furthermore, V ~ V* under the linear map T, which sen s

- . * * every a eoc to the corresponding !Xi E~ · . . . d'fti t If y \s -not finite dimensional over F, then the situation 1s .qu1te t ~re;·

Theorem 6.2 is false when dim V = oo. If dim V = oo, t~en dtm V* > dtm · Instead of proving that fact, we shall content ourselves Wtth an example.

I 6 4 Let V - ffi «> F that is V is the direct sum of the vector spaces Examp e • : - Wt=l ' ' f s t' n 4 that V*""' {V - Flie N} It follows from Exercise 2 o ec to I- • ""noo H (F F)~ n~lF. From Theorem 4.13, we

HomF(ffii'=t.F, F) =1,...

11A=l .

0~: c~unti~g e~ercise will convince the reader know that dtm V = •~ · strop h d. V* dtm' (fl"" F) is strictly larger than 1~'~1· 0 t at tm = t=l

. xt result we need the following definition: Before stating our ne '

DUALS AND ADJOINTS 47

Definition 6.5: Let V, V', and W be vector spaces over F, and let ro: V x V'-+ W be a function. We call ro a bilinear map if for all aeV, ro{a, ·)eHomF(V', W) and for all PeV', ro(·, P)eHom~, W).

Thus, a function ro: V x V' -+ W is a bilinear map if and only if w(xa1 + ya2, P) = xro(al> p) + yro(a2, p), and ro{oc, xP1 + yp~ = xro(a, {3

1) +

yw(a, {32) for all x, y e F, a, oct> a2 e V and p, P 1, P2 e V'. If V is any vector space over F, there is a natural bilinear map ro: V x V* -+ F given by

6.6:

w(a, T) = T(a)

In equation 6.6, a e V and T e V*. The fact that ro is a bilinear map is obvious .. ro determines a natural, injective, linear transformation 1/1: V -+ V** in the following way. If aeV, set 1/!(a) = w(a, ·). Thus, for any TeV*, 1/l(a)(T) = w(a, T) = T(a). If x, yeF, a, PeV and TeV*, then 1/!(xa + yp)(T) = ro(xa + yp, T) = xro(a, T) + yro(p, T) = (xl/l(a) + yl/l(p))(T). Consequently, 1/1 e Hom~, V**). To see that 1/1 is injective, we need to generalize equation 6.3. Suppose g = { a11 i e ~} is a basis of V (finite or infinite). Then for every i e ~ we can define a dual transformation ~Xt* e V* as follows: For each nonzero vector a e V, there exists a unique finite subset ~(a) = {jit ... 'jn uk E ~} of~ such that a = xhah + ... + xJ.aj .. Here XJI' ... ' xi. are all nonzer<~ scalars in F. We then define ~Xt*(oc) = x1k if i = jk for some k = 1' ... ' n. If 1¢ ~(a), we set ar(a) = 0. If a = 0, we of course define tXt*( a)= 0. Clearly arE V*, and

!Xt*(aJ = {~ if i = j if i=Fj

No.wifaekerl/1, then T(a) = Ofor all TeV*. In particular, af(a) = Ofor all ie~. This clearly implies a = 0, and, thus, 1/1 is injective.

We note in passing that the set ~· = {~Xt* 1 ie~} £:;; V*, wJ:rich we have just ~onstructed above, is clearly linearly independent over F. If dim V < oo, this is ~ust the dual basis ofV* coming from g. If dim V = oo, then~· does not span V*, ~d, therefore, cannot be called a dual basis. At any rate, we have proved the

st Part of the following theorem:

!~~orern 6.7: Let V be a vector space over F and suppose ro: V x V* -+ F is the t/JIInear map given in· equation 6.6. Then the map 1/1: V-+ V** given by

(o:) == co(a, ·) is an injective linear transformation. If dim V < oo, then 1/1 is a natural isomorphism.

48 UN EAR ALGEBRA

Proof Only the last sentence in Theorem 6. 7 remains to be proved. If dim V < oo, then Theorem 6.2 implies dim V =dim V** < oo. Since 1/1 is injective, our result follows from Theorem 3.33(b). 0

The word "natural" in Theorem 6.7 has a precise meaning in category theory, but here we mean only that the isomorphism l/f: V ~ V** is independ~nt of any choice of bases in V and V**. The word "natural" when applied to. an isomorphism l/f: V-+ V** also means ce~ain diagrams mu~t be commutative. See Exercise 4 at the end of this section for more det~s. We h~d ~oted previously that when dim V < oo, then~~. V*. Th~ type of Isomorphism IS pot natural, since it is constructed by first pickmg a basis (! = { cx1, ... , cxn} of V and then mapping cx1 to ext in V*. .

The bilinear map w: V x V* -+ F can also be used to set up certain correspondences between &i'(V) and &i'(V*).

Definition 6.8: If A is any subset of V, let A .L = { P E V* I w( ex, P) = 0 for all ex E A}.

Thus A .L is precisely the set of all vectors in V* that vanish on A. It is easy to see that A .Lis in fact a subspace of V*. We have a similar definition for subsets of

V*.

Definition 6.9: If A is a subset of V*, Let A .L = {ex E V I ro( ex, p) = 0 for all P E A}· '

Thus, if A £;; V*, then A .L is the set of all vectors in V that are zero under the maps in A. Oearly, A.L is a subspace of V for any A£;; V*.

Theorem 6.10: Let A and B be subsets of V (or V*).

(a) A£;; B implies A.L 2 B.L.

(b) L(Al = A.L. (c) (Au B).L = A.L n B.L.

(d) A £;; Au. (e) If W is a subspace of a finite-dimensional vector space V, then

dimV = dimW + dimW.L.

Proof (a)-(d) are straightforward, and we leave their proofs .as exercises. w_e prove (e). Let { ex 1 , ••• , cxm} be a basis of W. We extend this set to a basts

{ } f V Thus dim W = m and dim V = n. Let g = CXt' •. • 'CXm, exm+ 1> • .. ' exn 0 • ' . g* ={ex!, ... , ex:} be a dual basis of g. We complete the proof of (e) by argmng , that {ex!+ 1' •.• ' ex:} is a basis of w.L. .

If m + 1 ~ j ~ n, then cxf(exJ = 0 for i = 1, ... , m. In. parti~ular, a*+

1, ••• , ex: E w.L. Since {a!+ 1 , ••• , ex:} £;; ~·. {ex!+ 1 , ... , ex:} is linearly mde-

;ndent over F. We must show L({oc!+ 1 , ••• ,cx:})=W.L. Let P.ew ·Then p = Lf=

1 c,cx;. Since cx1 , ••• , am E W, we have for any J = 1, ... , Ill,

0 = p(a~ = (Lf=1 c,cxt)(a1) = 2J= 1 c1cxt(a~ = c1•

L({oc!+l•····cx:}). D

DUALS AND ADJOINTS

Thus,

49

If T e HomF<V, W), then T determines a linear transformation T*eHomp{W*, V*), which we call the adjoint ofT.

Definition 6.11: Let TeHomF<V, W). Then T*eHom~W*, V*) is the linear transformation defined by T*(f) = IT for all feW*.

Since the composite V -+TW -+rF of the linear transformations f and Tis again a linear map from V to F, we see T*(f)eV*. Ifx, yeF and f1of2eW*, then T*(xft + yf2) = (xf1 + yf2)T = x(f1 T) + y(f2 T) = xT*(f1) + yT*(f2). Thus, T* is a linear transformation from W* to V*.

Theorem 6.12: Let V and W be vector spaces over F. The map T-+ T* from HomF<V,W)-+ Homp{W*,V*) is an injective transformation. If V and W are finite dimensional, then this map is an isomorphism.

Proof Let x: Hom(V, W)-+ Hom(W*, V*) be defined by x(T) = T*. Our comments above imply x is a well-defined function. Suppose x,y e F, T1, T2eHom(V, W), and feW*. Then x(xT1 + yT2)(f) = (xT1 + yT2)*(f) = f(xT 1 + yT 2) = x(IT 1) + y(IT 2) = xTf(f) + yT~(f) = (xTf + yT~)(f) = (xx(T 1) + yx(T 2))(f). Thus, x(xT 1 + yT 2) = xx(T 1) + yx(T 2), and x is ~ linear transformation.

Suppose Teker X· Then for every feW*, 0 = x(T)(f) = T*(f) =IT. Now if we follow the same argument given in the proof of Theorem 6. 7, we know that if pis a nonzero vector in W, then there exists an feW* such that f(p) =1: 0. Thus, IT = 0 for all feW* implies Im T = (0). Therefore, T = 0, and x is injective.

Now suppose V and Ware finite dimensional. Then Theorems 6.2 and 3.25 imply dim{HomF<V, W)} = dim{Homp(W*, V*)}. Since x is injective, Theorem 3.33(b) implies xis an isomorphism. 0

We note in passing that forming the adjoint of a product is the product of the adjoints in the opposite order. More specifically, suppose T e HomF<V, W) and SeHomp{W, Z). Then STeHomF<V, Z). If feZ*, .. then (ST)*(f) = f(ST) = (fS)T = T*(fS) = T*(S*(f)) = T*S*(f). Thus, we get equation 6.13:

6.13:

(ST)* = T*S*

The connection between adjoints and Theorem 6.10 is easily described.

Theorem 6.14: Let T e HomF<V, W). Then

(a) (Im T*).L = ker·T. (b) ker T* = (Im T).L.

50 LINEAR ALGEBRA

Proof (a) Let cxe(lm T*)1., and suppose ro:V x V*-+ F is the bilinear map defined in equation 6.6. Then ro(cx, 1m T*) = 0. Thus, for all feW*, 0 = ro(~X. T*(f)) = ro(IX, fT) = IT( ex) = f(T(1X)). But we have seen that f(T(cx)) = 0 for all feW* implies T(cx) = 0. Thus, ex e kerT. Conversely, if cxekerT, then 0 = f(T(cx)) = ro(cx, T*(f)) and cxe(lm T*)l..

(b) Suppose fekerT*. Then 0 = T*(f) =fT. In particular, f(T(cx)) = 0 for all cxeV. Therefore, 0 = ro(T(cx), f) and fe(lm T).L. Thus, kerT* s;;; (1m T).L. The steps in this proof are easily reversed and so (1m T).L s;;; ker T*. 0

Theorem 6.14 has an interesting corollary. IfTeHom~, W), let us define the rank ofT, rk{T}, to be dim(lm T). Thus, rk{T} = dim(Im T). Then we have the following:

Corollary 6.15: Let V and W be finite-dimensional vector spaces over F, and let TeHomp{V, W). Then rk{T} = rk{T*}.

Proof The following integers are all equal:

rk {T} = dim(lm T) = dim V - dim(ker T)

=dim V- dim{(Im T*).L}

=dim V*- dim{(Im T*).L}

= dim(lmT*)

= rk{T*} 0

[Theorem 3.33(c)]

[Theorem 6.14]

[Theorem 6.2]

[Theorem 6.10(e)]

Corollary 6.15 has a familiar interpretation when we switch to matrices. If I! is any basis of V and !} any basis of W, then Theorem 3.25 implies rk{T} = rk(r(l!, P)(T)). Let A = r{l!, p)(T). In Theorem 6.16 below, we shall show that the matrix representation ofT*: W* -+ V* with respect toP* and I!* is given by the transpose of A. Thus, r(p*, I!*)(T*) = N. In particular,-Corollary 6.15 is the familiar statement that a matrix A and its transpose A' have the same rank.

Theorem 6.16: Let V and W be finite-dimensional vector spaces over F. Suppose I! and P are bases of V and W, respectively. Let g* and P* be the corresponding dual bases in V* and W*. Then for all T e Homp{V, w), we have

6.17:

r(~*, g*)(T*) = (r(g, !})(T))'

Proof Suppose g = {IX 1 , ... , 1Xn} and ~ = { P 1 , .•• , Pm}. Set A = r(g, p)(T). Then A = (aiJ) e Mm x u(F), and from 3.24, we have T(cxJ) = LF=

1 a,Jp, for all

j = 1, ... , n.

I

i I

I i 1....


r(f}*, I!*)(T*) is the n x m matrix that makes the following diagram commute:

The transpose of A is the n x m matrix A' = (bpq), where bpq = aqp for all p=l, ... ,n, and q=l, ... ,m.lt follows from 3.24 that r(~*,g*)(T*)=A' provided that the following equation is true:

6.18:

for all q = 1, ... , m

Fix q = 1, ... , m. To show that T*(P:) and L:= 1 bpqa: are the same vector in V*, it suffices to show that these two maps agree on the basis g of V. For any r = 1, ... , n, (T*(&))(cxr) = P:(T(cx,)) = P:<LF= 1 auPJ = LF= 1 8-j,P:<PJ = aqr· On the other hand, (L..:= 1 bpqCX:)(a,) = L:= 1 bpqCX:(a:,) = brq = aqr· Thus, equation 6.18 is established, and the proof of Theorem 6.16 is complete. 0


(1) Prove (a)-(d) in Theorem 6.10.

(2) Let V and W be finite-dimensional vector spaces over F with bases g and f}, respectively. Suppose T Hom~, W). Show that rk{T} = rk(r(l!. /})(T)).

(3) Let 0 :f:: pev and feV*- (0). Define T: V-+ V by T(cx) = f(cx)p. A function defined in this way is called a dyad.

(a) Show T e Hom(V, V) such that dim(lm T) = 1.

(b) If S e Hom(V, V) such that dim(lm S) = 1, show that S is a dyad. (c) If Tis a dyad on V, show that T* is a dyad on V*.- ·

(4) Let V and W be finite-dimensional vector spaces over F. Let 1/Jv: V-+ V** and 1/Jw: W-+ W** be the isomorphisms given in Theorem 6.7. Show that for every T e Hom(V, W) the following diagram is commutative:

T

1/Jvvj Wj 1/Jw T** V** __ ___::. __ -+ W**

62 LINEAR ALGEBRA

(5) Let A= {f1o ... ,fn} s;;; V*. Show A.t = nf= 1 ker f1•

(6) Let A = {f1 , •.• , fn} s;;; V* and suppose g e V* such that g vanishes on A .L.

Show geL(A). [Hint: First assume dim(V) < oo; then use Exercise 3 of Section 5 for the general case.]

(7) Let V and W be finite-dimensional vector spaces over F, and let co: V x W --. F be an arbitrary bilinear map. Let T: V --. W* and S: W--. V* be defined from co as follows: T(ex)(/1) =co( ex, /1) and s(p)(ex} = co( ex, p). Show that S = T* if we identify W with W** via 1/Jw.

(8) Show that (V fW)*;:;. W .L.

(9) Let V be a finite-dimensional vector space over F. Let W = V Ei3 V*. Show that the map (ex, /1) --. (p, ex) is an isomorphism between W and W*.

(10) If S T o-. v--=--~w--~z-.o

is a short exact sequence of vector spaces over F, show that

T• s• 0--. Z*----+ W*--~ V*--. 0

is exact.

(11) Let {Wd i e Z} be a sequence of vector spaces over F. Suppose for each ieZ, we have a linear transformation e1eHomp(Wb W1+1). Then D = {(WbeJjieZ} is called a cochain complex ifel+ 1e1 = 0 for all ieZ. D is said to be exact if Im e1 = ker ei+ 1 for all i e Z.

(a) If C = {(C~o d1)1ieZ} is a chain complex, show that C* = {(C~, e1 = dt+ 1)jieZ} is a cochain complex.

(b) If Cis exact, show that C* is also exact.

(12) Prove that(V 1 Ei3 • • • Ei3 V J* ':!:. Vf EB · · · Ei3 v:. (13) Let V be a finite-dimensional vector space over F with basis ~ =

{exl> ... ,ex0 }. Define T:P-.V by T(:x1 , ••• ,xJ=~)= 1 x1ex1 • Show that T*(f) = (f)«* for all f e V*. Here you will need to identify (P}* with Fn in a natural way.

(14) Let {z1}~0 be a sequence of complex numbers. Define a map T: C[X] -t C by T(L,:=oakXk} = L~=oakzk. Show that Te(C[X])*. Show that every T e (C[X]}* is given by such a sequence.

(15) Let V = IR[X]. Which of the following functions on V are elements in y•: (a) T(p) = JA p(X)dX. (b) T(p) = JA p(X)2 dX.

r

. (c) T(p} = JA X2p(X} dX. (d) T(p) = dp/dX.

(e) T(p) = dp/dXIx=o·

SYMMETRIC BILINEAR FORMS 53

(16) Suppose F is a finite field (e.g., IFp). Let V be a vector space over F of dimension n. For every m ~ n, show the number of subspaces of V of dimension m is precisely the same as the number of subspaces of V of dimension n - m. ·

(17) An important linear functional on Mnxn(F) is the trace map Tr: Mnxn(F) -t F defined by Tr(A) = Lf= 1 au where A= (a1J). Show that Tr() e(Mn xn(F))*.

(18) In Exercise 17, show Tr(AB) = Tr(BA) for all A, BeMnxn(F).

(19) Let m, ne ~. Let f1o ... .fm e(F")*. Define T: F" -.Fm by T(a) = (f1(ex), ... , fm(ex)). Show that T e Homp(F", Fm). Show that every TeHomp(P, Fm) is given in this way for some f1 , ••. , fm.

(20) Let V be a finite-dimensional vector space over C. Suppose ex1, ... , exn are distinct, nonzero vectors in V. Show there exists a T e V* such that T(exJ #: 0 for all k = 1, ... , n.

7. SYMMETRIC BILINEAR FORMS

In this last section of Chapter I, we discuss symmetric bilinear forms on a vector space V. Unlike the first six sections, the nature of the base field F is important here. In our main theorems, we shall assume Vis a finite-dimensional vector space over the reals R ·

Let V be a vector space over an arbitrary field F.

Definition 7.1: By a bilinear form co on V, we shall mean any bilinear map w: V x V--. F. We say co is symmetric if co( ex, P) = co(p, a) for all ex, p e V.

Example 7.2: The standard example to keep in mind here is the form w((xlt · · ·, xJ, (y1, ... , yJ) = Lf=- 1 XtYt· Clearly, co is a symmetric, bilinear form on F0

• 0

Suppose co is a bilinear form on a finite-dimensional vect~r space V. Then for everyb· { } · aSis ~ = ex1, ... , an of V, we can define an n x n matrix ~(co, ~)e Mnxn(F).whose (i,j)th entry is given by {M(co, ~)h,J = co(exb aJ). In terms fthe.usual coordmate map[·]!!: V -t Mnx 1(F), co is then given by the following

equatton:

7.3:

54 LINEAR ALGEBRA

Clearly, ru is symmetric if and only if M(ru, ~)is a symmetric matrix.

Definition 7.4: Suppose ru is a bilinear form on Y. The function q: Y-+ F defined by q(e) = ru<e. e) is called the quadratic form associated with ru.

If Y is finite dimensional with basis ~ = {a 1 , ... , an}, then equation 7.3 implies q@ = [eJ~M(ru, ~)[e),.= L~J=l aux1xJ. Here (x1 , ••• , xJ1 = [eJ~ and (aiJ) = M(ru, ~). Thus, q(e> is -a quadratic homogeneous polynomial in the coordinates x1, ... , Xn of e. That fact explains why q is called a quadratic form on Y. In Example 7.2, for instance, q((xt, ... , xJ) = Lf=1 x~.

At this point, a natural question arises. Suppose ru is a symmetric, bilinear form on a finite-dimensional vector space Y. Can we choose a basis ~ of Y so that the representation of ruin equation 7.3 is as simple as possible? What would the corresponding quadratic form q look like in this representation? We shall give answers to both of these questions when F = R For a more general treatment, we refer the reader to [2].

For the rest of this section, we assume Y is a finite-dimensional vector space over R Let ru be a symmetric, bilinear form on Y.

Definition 7.5: A basis ~ = {a 1 , ... , an} of Y is said to be ru-orthonormal if

(a) ru(at> aJ) = 0 whenever i :1= j, and (b) ru(ai> aJ E { -1, 0, 1} for all i = 1, ... , n.

In Example 7.2, for instance, the canonical basis ~ = {c5, = (0,. ·., 1 , ... , 0) 1 i = 1, ... , n} is an ru-orthonormal basis of !Rn. Our first theorem in this section guarantees ru-orthonormal bases exist.

Theorem 7.6: Let Y be a finite-dimensional vector space over 1R and suppose OJ

is a symmetric, bilinear form on Y. Then Y has an (I)-Orthonormal basis.

Proof: We proceed via induction on n =dim Y. If Y = (0), then the result is trivial. So, suppose n = 1. Then any nonzero vector of Y is a basis of Y. If ru(a, a)= 0 for every aE Y, then any nonzero vector of Y is an ru-orthonormal basis. Suppose there exists a {3 E Y such that ru({J, {3) :1= 0. Then c == lru({J, {3)1- 112 is a positive scalar in IR, and { c{J} is an ru-orthonormal. basis. of V. Thus, we have established the result for all vector spaces of dtmensiOn 1 over IR.

Suppose n > 1, and we have proved the theorem for any vector space over IR ' of dimension less than n. Since ru is symmetric, we have

7.7:

for all e, r[EY

r SYMMETRIC BILINEAR FORMS 56

In equation 7.7, q is the quadratic form associated with ru. Now if OJ( a, a) = q(a) = 0 for all a E Y, then 7. 7 implies ru is identically zero. In this case, any basis of Y is an ru-orthonormal basis. Thus, we can assume there exists a nonzero vector {3 E Y such that ru({J, {3) :1= 0. As in the case n = 1, we can then adjust {3 by a scalar multiple if need be and find an an =F 0 in Y such that m(an, aJ E { -1, 1 }.

Next define a linear transformation fEY* by f(e) = ru(an,e). Since f(aJ = ru(an, aJ :1= 0, f is a nonzero map. Set N =kerf. Since f :1= 0, and dimR IR = 1, f is surjective. Thus, Corollary 5.17 implies Y fN ~ IR. In particular, Theorem 5.14 implies dim N = dim Y - 1. ru when restricted to N is clearly a symmetric bilinear form. Hence our induction hypothesis implies N has an ru-orthonormal basis {at> ... ,an-1}.

We claim ~={at> ... , an_ 1, an} is an ru-orthonormal basis of Y. Since f(aJ :1= 0, Cln ¢ N. In particular, ~ is linearly independent over IR. · Since dimR (V) = n, ~ is a basis of Y. Conditions (a) and (b) of Definitions 7.5 are satisfied for { a1 , ••• , an _1} since this set is an (I)-Orthonormal basis of N. Since N = kerf, ru(a~o aJ = 0 for i = 1, ... , n - 1. Thus, ~ is an (I)-Orthonormal basis of V and the proof of Theorem 7.6 is complete. D

The existence of ru-orthonormal bases of Y answers our first question about representing ru. Suppose~= {at> ... , an} is an ru-orthonormal basis ofY. Then the matrix M{ru, ~) is just an n x n diagonal matrix, diag(q(a1),: .. , q(an)), with q(aJ=ru(abaJE{-1,0,1}. If e. r[EY with [eJ .. =(xl, ... ,Xn)1 and ['7] .. = (y1, ... , Yn)\ then equation 7.3 implies ru(e, rt)~ Lf= 1 x1y1q(aJ. By reordering the elements of ~ if need be, we can assume ~ = { a 1 ,

... , ap} u {ap+ 1, ... , ap+m} u {ap+m+ 1, ... , ap+m+r}, where

7.8:

q(aJ= H for i = 1, ... , p

for i = p + 1, ... , p + m

for i = p + m + 1, ... , p + m + r

The vector spaceY then decomposes into the direct sum Y = Y _1 EB Y0 EB Y1, Where y_1=L({ap+l•···•IXp+m}), Yo=L({ap+m+l•···•ap+~+r}), and Y1= L({ IX1, ... , ap}). · · ;

Our quadratic form q is positive on Y 1 - (0), zero on Y 0, and negative on V -1- (0). For example, suppose {JEY _1 - (0). Then {3 = x1o:p+l + ... ~xm~+m for some x1 , ... ,XmEF. Thus, q({J)=ru({J,{J)=l:'f= 1 ~q(ap+J· Since {3 :1= 0, some x1 is nonzero. Since q( aP + 1) = - 1 for all i = 1 , ... , m, we see ~{J)<Q .

The subspaces Y -1> Y 0, and Y 1 are pairwise ru-orthogonal in the sense that w(V" V1) = 0 whenever i,j E { -1, 0, 1} and i :1= j. Thus, any ru-orthonormal basis ~ of V decomposes Y 'into a direct sum Y = V _1 EB Y0 EB Y 1 of pairwise ruorthogonal subspaces Y1. The sign of the associated quadratic form q is constant

56 LINEAR ALGEBRA

on each V1 - (0). An important fact here is that the dimensions of these three subspaces, p, m, and r, depend only on ro and not on the particular roorthonormal basis ~ chosen.

Lemma 7.9: Suppose P = {P11 ••• , Po} is a second ro-orthonormal basis of V, and let V = W _ 1 Ea W~ Ea W 1 be the corresponding decomposition of V. Then dim wj =dim vj for j = -1,0, 1.

Proof: W _ 1 is the subspace of V spanned by those P1 for which q(pJ = -1. Let cxeW - 1 r'\(V0 + V1). If cx =I= 0, then q(cx) < 0 since cxeW _ 1• But cxeV0 +V1 implies q(cx) ;;::: 0, which is impossible. Thus, ex = 0. So, W _ 1 r'\ (V 0 + V 1) = (0). By expanding the basis of W _ 1 if need be, we can then construct a subspace P of V such that W _ 1 £;; P, and PEa (V 0 + V 1) = V. Thus, from Theorem 4.9, we have dim(W _1) ~ dim P = dim V - dim V 0 - dim V 1 = dim(V _ 1). Therefore, dim(W _ 1) ~dim V _ 1• Reversing the roles of the W1 and V1 in this proof gives dim(V _ 1) ~ dim(W _ 1). Thus, dim(W _ 1) = dim(V _ 1). A similar proof shows dim(W 1) = dim(V 1). Then dim(W 0) = dim(V0) by Theorem 4.9. This completes the proof of Lemma 7.9. D

Let us agree when discussing ro-orthonormal bases ~ of V always to order the basis elements cx1eg according to equation 7.8. Then Lemma 7.9 implies that the integers p, m, and r do not depend on g but only on ro. In particular, the following definition makes sense.

Definition 7.10: p- miscalled the signature of q. p +miscalled the rank of q.

We have now proved the following theorem:

Theorem 7.11: Let ro be a symmetric, bilinear form on a finite-dimensional vector space V over R Then there exists integers m and p such that if g = {a1 , ... , ex,} is any ro-orthonormal basis ofV and [eJ.. = (x11 ••• , X 0 )

1, then q(e> = :Lr=l x~- rr~.:"+1 x~. o -

A quadratic form q, associated with some symmetric bilinear form ro on V, is said to be definite if q@ = 0 implies e = 0. For instance, in Example 7.2, q((x1 , ••• , xJ) = Lf= 1 Xf is definite when F =RIfF= C, then q is not definite

since, for example, q((l, ~. 0, ... , 0)) = 0. If q is a definite quadratic form on a finite-dimensional vector space V over IR,

then Theorem 7.11 implies q(e) > 0 for all e E V - (0) or q@ < 0 for all e e V - (0). In general, we say a quadratic form q is positive definite if q( e) > 0 for all eev- (0). We say q is negative definite if q(e) < 0 for all eev- (0).

Definition 7.12: Let V be a vector space over R A symmetric, bilinear form ro on V whose associated quadratic form is positive definite is called an inner product on V.

r

L


Note in our definition that we do not require that V be finite dimensional. We finish this section with a few examples of inner products.

Example 7.13: Let V = ~n, and define ro as in Example 7.2. D

Example 7.14: Let V = E9'f=1 ~ and define ro by ro((xh x2 , ••• ),

(y1, y2, ••• )) = L~ 1 x1y1. Since both sequences {x1} and {y1} are eventually zero, w is well defined and is clearly an inner product on V. · D

Example 7.15: Let V = C([a, b]). Define ro(f, g) = J: f(x)g(x) dx. Clearly, ro is an inner product on V. D

We shall come back to the study of inner products in Chapter V.


(1) In our proof of Lemma 7.9, we used the following fact: If Wand W' are subspaces of V such that W r'\ W' = (0), then there exists a complement of W' that contains W. Give a proof of this fact.

(2) Let V = Mm x 0{F), and let C e Mm x m(F). Define a map ro: V x V -+ F by the

formula ro(A, B) = Tr(A'CB). Show that ro is a bilinear form. Is ro symmetric?

(3) Let V = M0

x 0{F). Define a map ro: V x V -+ F by ro(A, B~ = n Tr(AB)

- Tr(A) Tr(B). Show that ro is a bilinear form. Is ro symmetnc?

(4) Exhibit a bilinear form on ~n that is not symmetric.

(5) Find a symmetric bilinear form on en whose associated quadratic form is positive definite.

(6) Describe explicitly all symmetric bilinear forms on ~3 •

(7) Describe explicitly all skew-symmetric bilinear froms on ~3 • A bilinear form ro is skew-symmetric if ro(a, p) = -ro(P, cx).

(8) Let ro: V x V -+ F be a bilinear form on a finite dimensional vector space V. Show that the following conditions are equivalent:

(a) {aeVjro(a, p) = 0 for all peV} = (0). (b) {aeVjro(p, a)= 0 for all PeV} = (0).

(c) ~(m, g) is nonsingular for any basis ~ of V. We say ro is nondegenerate if ro satisfies the conditions listed above.

(9) Suppose ca: V x V -+ F is a nondegenerate, bilinear form on a finitedimensional. vector space V. Let W be a subspace of V. Set wl. = {aeVI ro(cx, p) = 0 for all PeW}. Show that v = w Ea wl..

68 LINEAR ALGEBRA

(10) With the same hypotheses as in Exercise 9, suppose f e V*. Prove that there exists an rxeV such that f({f) = w(cx,p) for all pev.

(11) Suppose co: V x V--+ F is a bilinear form on V. Let W 1 and W 2 be subspaces of V. Show that (W1 + W2).L = Wf n wt. If co is nondegenerate, prove that (W1 n W2).L = Wf + Wf.

(12) Let co be a nondegenerate, bilinear form on a finite-dimensional vector space V. Let co' be any bilinear form on V. Show there exists a unique T e HomF"fV, V) such that co'(a, {f) = w(T(rx), {f) for all rx, p e V. Show that co' is nondegenerate if and only if T is bijective.

(13) With the same hypotheses as in Exercise 12, show that for every T e HomFfV, V) there exists a unique T' e HomFfV, V) such that w(T(rx), {f) = w(rx, T'(P)) for all rx, p e V.

(14) Let Bil(V) denote the set of all bilinear forms on the vector space V. Define addition in Bil(V) by (co + co')(rx, P) = co(rx, p) + co'(cx, p), and scalar multiplication by (xco)(cx, {f) = xco(cx, {f). Prove that Bil(V) is a vector space over F with these definitions. What is the dimension of Bil(V) when Vis finite dimensional?

(15) Find an co-orthonormal basis for !R2 when co is given by w((xh y1), (x2,Y2)) = X1Y2 + X2Y1·

(16) Argue that w(f, g)= J:f(x)g(x)dx is an inner product on C([a, b]).

(17) Let V = {p(X)e!R[X] ldeg(p) ~ 5}. Suppose co:V x V--+ !R is given by co(f, g) = J5 f(x)g(x) dx. Find an co-orthonormal basis of V. ·

(18) Let V be the subspace of C([ -n, n]) spanned by the functions 1, sin(x), cos(x), sin(2x), cos(2x), ..• , sin(nx), cos(nx). Find an co-orthonormal basis of V where co is the inner product given in Exercise 16.

Chapter II

Multilinear Algebra

1. MULTILINEAR MAPS AND TENSOR PRODUCTS

In Chapter I, we dealt mainly with functions of one variable between vector spaces. Those functions were linear in that variable and were called linear transformations. In this chapter, we examine functions of several variables between vector spaces. If such a function is linear in each of its variables, then the function is called a multilinear mapping. Along with any theory of multilinear maps comes a sequence of universal mapping problems whose solutions are the fundamental ideas in multilinear algebra. In this and the next few sections, we shall give a careful explanation of the principal constructions of the subject matter. Applications of the ideas discussed here will abound throughout the rest of the book.

Let us first give a careful definition of a multilinear mapping. As usual, F will denote an arbitrary field. SuppOse V 1 , ••• , V n and V are vector spaces over F. Let ¢: V 1 X • • • X V n --+ V be a function from the finite product V 1 X • • • X V n to V. We had seen in Section 4 of Chapter I that a typical vector in V 1 x · · · x V n is ann-tuple (rx1 , ••• , rxJ with rx1 e V1• Thus, we can think of¢ as a function of then Variable vectors rx1 , ••• , rxu.

~efinition 1.1: A function ¢: V 1 x · · · x V n --+ V is called a multilinear mapping tf for each i = 1, ... , n, we have ·

(a) ¢(rxt> ... ,.rx1 + rxj, ... , rxJ = ¢(rx1, ... , rxh .. ·• rxJ + ¢(rx1 , ... , rxj, ... , rxJ, and

(b) ¢(rxl> ... , xrx~o ... , rxJ = x¢(rx1 , ... , rx~o ... , rxJ.

mmtoro/doc/multilineal/linearbrown... · preface for the past two years, i have been teaching a...

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