Galois Theory and Advanced Linear Algebra

Rajnikant Sinha Galois Theory and Advanced Linear Algebra Galois Theory and Advanced Linear Algebra Rajnikant Sinha...

Author: Rajnikant Sinha

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Rajnikant Sinha

Galois Theory and Advanced Linear Algebra


Rajnikant Sinha


123

Rajnikant Sinha Samne Ghat Varanasi, Uttar Pradesh, India

ISBN 978-981-13-9848-3 ISBN 978-981-13-9849-0 https://doi.org/10.1007/978-981-13-9849-0

(eBook)

© Springer Nature Singapore Pte Ltd. 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

Evariste Galois (25 October 1811–31 May 1832) was a great French mathematician. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a problem standing for 350 years. The famous ancient problem of “trisecting an angle by using solely straightedge and compass” was later solved by using the fundamental theorem of Galois theory. Many students are overwhelmed to learn this. They take keen interest in learning the theory Galois had discovered. Unfortunately, there is no literature which can lead to the fundamental theorem without going through painful learning process. Further, there is no right kind of book on linear algebra that can provide good theoretical foundation needed for later applications in Riemannian geometry, quantum mechanics, etc. These voids prompted me to write this book. This book is meant to be an introduction to abstract algebra. The reader of this book is assumed to have some prior exposure to elementary properties of groups, rings, fields, and vector spaces. At times, we shall assume familiarity with inner product space of finite dimension, and linear transformations. For the readers who have only learned a minimum of abstract algebra will also find this book friendly. A nodding acquaintance with elementary properties of positive integers is beneficial. Most of the material usually taught in an abstract algebra course are presented in this text. However, some results appear for the first time in a textbook form. The ordering of the topics as well as the approach we have taken sometimes deviate from the standard path, simply because of pedagogical reasons. Aside from the usual approach, we sometimes have also developed a more elementary approach that uses standard calculation techniques. Wherever required, we also have supplied abundantly “second layer of proof” (that is, proof within proof), so that comprehensibility of the proof gets enhanced. In some named theorems, rarely we need a third layer of proof. In the first part of Chap. 1, we have developed necessary field theory. Using these theorems, we have tried to prove the fundamental theorem of Galois groups. This is the theorem for which Galois became immortal. In Chap. 2, some wonderful applications of Galois theory are presented. Solution to the ancient famous problem of “trisection of a given angle by ruler and compass” is given here with enough v

vi

Preface

detailed proof. In Chap. 3, we have supplied the proofs of many celebrated theorems by using linear transformation tools as well as matrix methods. These are beautiful areas of mathematics in itself. Their applications to quantum mechanics, manifold theory, etc., are well known. In Chap. 4, we have dealt with some amazing, but forgotten, results of yesteryears. It is known that “signature” of a quadratic form is invariant, but it is difficult to find an accessible proof to it. We have tried to supply a proof which could be an effortless reading. Finally, on a personal note, I would like to thank my lovely wife, Bina, for her patient endurance and constant encouragement. Uttar Pradesh, India

Rajnikant Sinha

Contents

1 Galois Theory I . . . . . . . . . . 1.1 Euclidean Rings . . . . . . 1.2 Polynomial Rings . . . . . 1.3 The Eisenstein Criterion 1.4 Roots of Polynomials . . 1.5 Splitting Fields . . . . . . . Exercises . . . . . . . . . . . . . . .

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1 1 13 31 44 65 89

2 Galois Theory II . . . . . . . . . . . . . . 2.1 Simple Extensions . . . . . . . . . 2.2 Galois Groups . . . . . . . . . . . . 2.3 Applications of Galois Theory . 2.4 Solvability By Radicals . . . . . . Exercises . . . . . . . . . . . . . . . . . . . .

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91 91 103 129 152 165

3 Linear Transformations . . . . . . . . . 3.1 Eigenvalues . . . . . . . . . . . . . . . 3.2 Canonical Forms . . . . . . . . . . . 3.3 The Cayley–Hamilton Theorem . Exercises . . . . . . . . . . . . . . . . . . . . .

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167 167 187 223 252

4 Sylvester’s Law of Inertia . . . . . . . . . . . . 4.1 Positive Definite Matrices . . . . . . . . . 4.2 Sylvester’s Law . . . . . . . . . . . . . . . . 4.3 Application to Riemannian Geometry Exercises . . . . . . . . . . . . . . . . . . . . . . . . .

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255 255 290 323 347

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vii

About the Author

Rajnikant Sinha is a former Professor of Mathematics at Magadh University, Bodh Gaya, India. A passionate mathematician, Prof. Sinha has published numerous interesting research findings in international journals, and has authored three textbooks with Springer Nature: Smooth Manifolds, Real and Complex Analysis: Volume 1, and Real and Complex Analysis: Volume 2; and a contributed volume on Solutions to Weatherburn’s Elementary Vector Analysis: With Applications to Geometry and Mechanics with another publisher. His research focuses on topological vector spaces, differential geometry and manifolds.

ix

Chapter 1

Galois Theory I

Roughly, a field is a commutative ring in which division by every nonzero element is allowed. In algebra, fields play a central role. Results about fields find important applications in the theory of numbers. The theory of fields comprises the subject matter of the theory of equations. Here, we shall deal lightly with the field of algebraic numbers. Our main emphasis will be on aspects of field theory that concern the roots of polynomials. The beautiful ideas, due to the brilliant French mathematician Évariste Galois (1811–1832), served as an inspiration for the development of abstract algebra. We shall prove the fundamental theorem of Galois theory.

1.1

Euclidean Rings

1.1.1 Definition Let R be an integral domain. Suppose that for every nonzero member a of R, d ðaÞ is a nonnegative integer. If 1. for every nonzero a; b 2 R, d ðaÞ d ðabÞ, 2. for every nonzero a; b 2 R, there exist q; r 2 R such that a ¼ qb þ r, and either r ¼ 0 or d ðr Þ\d ðbÞ, then we say that R is a Euclidean ring. 1.1.2 Theorem Let R be a Euclidean ring. Let A be an ideal of R. Then there exists a0 2 A such that a0 R ¼ A. Proof If A ¼ f0g, then 0 serves the purpose of a0 . So we consider the case A 6¼ f0g. It follows that there exists a nonzero member a of A, and hence fd ð xÞ : x 2 A and x 6¼ 0g is a nonempty set of nonnegative integers. Hence, minfd ð xÞ : x 2 A and x 6¼ 0g exists. It follows that there exists a nonzero member a0 of A such that © Springer Nature Singapore Pte Ltd. 2020 R. Sinha, Galois Theory and Advanced Linear Algebra, https://doi.org/10.1007/978-981-13-9849-0_1

1

2

1 Galois Theory I

d ða0 Þ ¼ minfd ð xÞ : x 2 A and x 6¼ 0g:

ðÞ

It remains to show that A ¼ a0 R. For this purpose, take an arbitrary a 2 A. If a ¼ 0, then a ¼ 0 ¼ a0 0 2 a0 R. Thus if a ¼ 0, then a 2 a0 R. Now we consider the case a 6¼ 0. Since a; a0 are nonzero members of R, and R is a Euclidean ring, there exist q; r 2 R such that a ¼ qa0 þ r and either r ¼ 0 or d ðr Þ\d ðbÞ. Since a; a0 are members of the ideal A of R, we have r ¼ ða qa0 Þ 2 A ; |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} and hence r 2 A. Now, if r 6¼ 0, then from ðÞ, d ða0 Þ d ðr Þ. Further, either r ¼ 0 or d ðr Þ\d ða0 Þ. This shows that ¼ ffl0} ; a qa0 ¼ r|fflffl{zffl and hence a ¼ qa0 ¼ a0 q 2 a0 R. Thus in all cases, a 2 A ) a 2 a0 R. Hence A a0 R. It remains to show that a0 R A. For this purpose, let us take an arbitrary b 2 R. We have to show that a0 b 2 A. Since a0 is a member of the ideal A of R, and b 2 R, we have a0 b 2 A: ■ Definition Let R be an integral domain. Let a 2 R. It is clear that aR is an ideal of R. The ideal aR of R is denoted by ðaÞ. Definition Let R be an integral domain. If 1. R has a unit element, 2. every ideal of R is of the form ðaÞ, then we say that R is a principal ideal ring. 1.1.3 Theorem Let R be a Euclidean ring. Then R has a unit element. Proof Since R is an ideal of the Euclidean ring R, by 1.1.2, there exists a0 2 R such that a0 R ¼ R. Since a0 R ¼ R and a0 2 R, we have a0 2 a0 R, and hence there exists e 2 R such that a0 ¼ a0 e. It suffices to show that e functions as a unit element in R. To this end, let us take an arbitrary b 2 R. We have to show that be ¼ b. Since b 2 R ¼ a0 R, there exists c 2 R such that b ¼ a0 c. Now, LHS ¼ be ¼ ða0 cÞe ¼ ðca0 Þe ¼ cða0 eÞ ¼ ca0 ¼ a0 c ¼ b ¼ RHS; where LHS and RHS are the left- and right-hand sides of the equality to be proved. ■ 1.1.4 Theorem Let R be a Euclidean ring. Then R is a principal ideal ring. Proof By 1.1.3, R has a unit element. Next, by 1.1.2, every ideal of R is of the form ðaÞ. It follows, by the definition of principal ideal ring, that R is a principal ideal ring. ■

1.1 Euclidean Rings

3

1.1.5 Theorem Let R be a Euclidean ring. Let a; b 2 R. Then the greatest common divisor ða; bÞ of a and b exists in R, in the sense that 1. 2. 3. 4.

ða; bÞ 2 R, ða; bÞja, ða; bÞjb, ðcja and cjbÞ ) cjða; bÞ. Further, there exist s; t 2 R such that ða; bÞ ¼ as þ bt:

Proof Since R is a Euclidean ring, by 1.1.3, R has a unit element, say e. Let A fax þ by : x; y 2 Rg: Clearly A is an ideal of the Euclidean ring R. Now, by 1.1.2, there exists f 2 R such that f ¼ fe 2 fR ¼ A ¼ fax þ by : x; y 2 Rg; |fflfflffl{zfflfflffl} and hence f 2 fax þ by : x; y 2 Rg. It follows that there exist s; t 2 R such that f ¼ as þ bt: Since fR ¼ fax þ by : x; y 2 Rg 3 ðae þ b0Þ ¼ a, we have a 2 fR, and hence f ja. Similarly, f jb. Next suppose that cja and cjb. It remains to show that cjðas þ btÞ. This is clearly true. ■ Definition Let R be a commutative ring with unit element 1. Let a 2 R. If there exists b 2 R such that ab ¼ 1, then we say that a is a unit in R. 1.1.6 Theorem Let R be an integral domain with unit element 1. Let a; b be nonzero members of R. Suppose that ajb and bja. Then there exists u 2 R such that au ¼ b, and u is a unit in R. Proof Since ajb, there exists u 2 R such that au ¼ b. Similarly, there exists v 2 R such that bv ¼ a. It follows that aðuvÞ ¼ ðauÞv ¼ a ¼ a1; |fflfflfflfflfflffl{zfflfflfflfflfflffl} and hence aðuvÞ ¼ a1. Now, since a is a nonzero member of the integral domain R, we have uv ¼ 1, and hence u is a unit in R. ■ Definition Let R be a commutative ring with unit element 1. Let a; b 2 R. If there exists a unit u in R such that au ¼ b, then we say that a and b are associates, and we denote this relationship by a b.

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1 Galois Theory I

It is clear that is an equivalence relation over R. Hence R is partitioned by into equivalence classes. 1.1.7 Theorem Let R be a Euclidean ring. Let a; b 2 R. Let c be a greatest common divisor of a and b. Let d be a greatest common divisor of a and b. Then c d. Proof Since c is a greatest common divisor of a and b, we have ðcja and cjbÞ. Now, since d is a greatest common divisor of a and b, we have cjd. Similarly, djc. Since cjd and djc, by 1.1.6 there exists a unit u 2 R such that cu ¼ d, and hence c d. ■ 1.1.8 Theorem Let R be a Euclidean ring with unit element 1. Let a; b be nonzero members of R. Let b be a nonunit. Then d ðaÞ\d ðabÞ: Proof Suppose to the contrary that d ðaÞ ¼ d ðabÞ. We seek a contradiction. Since aR is an ideal of R, by 1.1.2 there exists b 2 R such that ðabÞR ¼ aR. Now, since a ¼ a1 2 aR ¼ ðabÞR, we have a 2 ðabÞR, and hence there exists a nonzero member c of R such that a ¼ ðabÞc. Hence a1 ¼ a ¼ aðbcÞ : |fflfflfflfflfflffl{zfflfflfflfflfflffl} Since a1 ¼ aðbcÞ and a is a nonzero member of the integral domain R, we have 1 ¼ bc, and hence b is a unit. This is a contradiction. ■ Definition Let R be a Euclidean ring with unit element 1. Let p be a nonzero member of R that is not a unit. If ða; b 2 R and p ¼ abÞ ) ða is a unit or b is a unitÞ; then we say that p is a prime element of R. 1.1.9 Theorem Let R be a Euclidean ring with unit element 1. Let a be a nonzero member of R. Suppose that d ðaÞ ¼ d ð1Þ. Then a is a unit. Proof Suppose to the contrary that a is a nonunit. We seek a contradiction. Here 1; a are nonzero members of R. Also, a is not a unit. So by 1.1.8, d ð1Þ\d ð1aÞ ¼ d ðaÞ; |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} and hence d ð1Þ\d ðaÞ. Thus d ðaÞ 6¼ d ð1Þ. This is a contradiction.

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1.1.10 Problem Let R be a Euclidean ring with unit element 1. Let a be a nonzero member of R. Let a be a unit. Then d ðaÞ ¼ d ð1Þ.

1.1 Euclidean Rings

5

Proof Suppose to the contrary that d ðaÞ 6¼ d ð1Þ. We seek a contradiction. Since a is a unit, there exists a nonzero member b of R such that ab ¼ 1. Now, since R is a Euclidean ring, we have d ðaÞ d ðabÞ ¼ d ð1Þ d ð1aÞ ¼ d ðaÞ; |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} and hence d ðaÞ ¼ d ð1Þ. This is a contradiction.

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1.1.11 Problem Let R be a Euclidean ring with unit element 1. Then for every nonzero member a of R, either a is a unit or a can be expressed as a product of finitely many prime elements of R. Proof (Induction on d ðaÞÞ: Let us first consider the case that a is a nonzero member of R and d ðaÞ ¼ 0. Since 0 d ð1Þ d ð1aÞ ¼ d ðaÞ ¼ 0; we have d ð1Þ ¼ d ðaÞ. Now by 1.1.9, a is a unit. Thus the statement “either a is a unit or a can be expressed as a product of finitely many prime elements of R” holds in this case. Next suppose that the statement “either a is a unit or a can be expressed as a product of finitely many prime elements of R” holds for all a in R for which d ðaÞ n. Next suppose that b is a nonzero member of R for which d ðbÞ ¼ n þ 1. We have to show that the statement “either b is a unit or b can be expressed as a product of finitely many prime elements of R” holds. Case I: b is a prime element of R. In this case, the statement “b can be expressed as a product of finitely many prime elements of R” holds, and hence the statement “either b is a unit or b can be expressed as a product of finitely many prime elements of R” holds. Case II: b is not a prime element of R. Subcase I: b is a unit. In this subcase, the statement “either b is a unit or b can be expressed as a product of finitely many prime elements of R” holds. Subcase II: b is not a unit. Here b is not a prime element of R, so by the definition of prime element, there exist c; e 2 R such that 1. b ¼ ce, 2. c is not a unit, 3. e is not a unit. Since b is a nonzero member of R and b ¼ ce, it follows that c; e are nonzero members of R. Since c is not a unit, by 1.1.8,

6

1 Galois Theory I

d ðeÞ\d ðecÞ ¼ d ðceÞ ¼ d ðbÞ ¼ n þ 1; |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} and hence d ðeÞ\n þ 1. Now, since d ðeÞ in an integer, we have d ðeÞ n. Similarly, d ðcÞ n. Since d ðcÞ n and c is not a unit, by hypothesis, c can be expressed as a product of finitely many prime elements of R. Similarly, e can be expressed as a product of finitely many prime elements of R. It follows that ce can be expressed as a product of finitely many prime elements of R. Now, since b ¼ ce, b can be expressed as a product of finitely many prime elements of R. ■ Definition Let R be a Euclidean ring with unit element 1. Let a; b be nonzero members of R (By 1.1.5, a greatest common divisor of a and b exists in R.). If there exists a unit u in R such that u is a greatest common divisor of a and b, then we say that a and b are relatively prime. 1.1.12 Problem Let R be a Euclidean ring with unit element 1. Let a; b; c be any nonzero elements of R. Suppose that ajbc. Let a and b be relatively prime. Then ajc. Proof Since a and b are relatively prime, there exists a unit u in R such that u is a greatest common divisor of a and b. Now, by 1.1.5, there exist s; t 2 R such that u ¼ as þ bt: Since u is a unit in R, there exists v in R such that uv ¼ 1. It follows that asv þ btv ¼ ðas þ btÞv ¼ 1; |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} and hence acsv þ bctv ¼ asvc þ btvc ¼ ðasv þ btvÞc ¼ 1c ¼ c: |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus acsv þ ðbcÞtv ¼ c. Since ajbc, there exists a nonzero member k of R such that ak ¼ bc. It follows that aðcsv þ ktvÞ ¼ acsv þ ðak Þtv ¼ c; |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence al ¼ c, where l ðcsv þ ktvÞ 2 R. Thus ajc.

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1.1.13 Problem Let R be a Euclidean ring with unit element 1. Let a; p be any nonzero elements of R. Suppose that p is a prime element of R. Then either pja or (p and a are relatively prime). Proof By 1.1.5, there exists e 2 R such that ejp, eja, and ððcjp and cjaÞ ) cjeÞ, that is, e is a greatest common divisor of p and a.

1.1 Euclidean Rings

7

Since ejp, there exists k 2 R such that ek ¼ p. Now, since p is a prime element of R, either e is a unit or k is a unit. Case I: k is a unit. It follows that there exists l 2 R such that kl ¼ 1. Now, since ek ¼ p, we have e ¼ e1 ¼ eðklÞ ¼ ðek Þl ¼ pl ; |fflfflfflfflfflffl{zfflfflfflfflfflffl} and hence e ¼ pl. Now, since eja, we have ðplÞja, and hence there exists m 2 R such that pðlmÞ ¼ ðplÞm ¼ a. Thus pn ¼ a, where n lm 2 R. Hence pja. Thus |fflfflfflfflfflffl{zfflfflfflfflfflffl} the statement “either pja or (p and a are relatively prime)” holds. Case II: e is a unit. Since e is a greatest common divisor of p and a, and e is a unit, p and a are relatively prime, and hence the statement “either pja or (p and a are relatively prime)” holds. ■ 1.1.14 Problem Let R be a Euclidean ring with unit element 1. Let p be a nonzero element of R. Suppose that p is a prime element of R. Then ða; b 2 R and pjðabÞÞ ) ðpja or pjbÞ: Proof Let us take arbitrary nonzero members a and b of R such that pjðabÞ. It suffices to show that pja or pjb. Suppose to the contrary that p-a and p-b. We have to arrive at a contradiction. Since p is a prime element of R, by 1.1.13 we have pja or (p and a are relatively prime). Now, since p-a, p and a are relatively prime. Next, since pjðabÞ and p and a are relatively prime, by 1.1.12, we have pjb. This is a contradiction. ■ 1.1.15 Theorem Let R be a Euclidean ring with unit element 1. Let a be a nonzero element of R. Suppose that a is not a unit in R (By 1.1.11, a can be expressed as a product of finitely many prime elements of R.). Let a ¼ p1 p2 . . .pm ; where each pi ði ¼ 1; 2; . . .; mÞ is a prime element of R. Let a ¼ p01 p02 . . .p0n ; where each p0j ðj ¼ 1; 2; . . .; nÞ is a prime element of R. Then 1. each pi is an associate of some p0j , 2. each p0j is an associate of some pi , 3. n ¼ m. This theorem is known as the unique factorization theorem.

8

1 Galois Theory I

Proof Since p1 jðp1 p2 . . .pm Þ and p1 p2 . . .pm ¼ a ¼ p01 p02 . . .p0n , we have p1 j p01 p02 . . .p0n . Now, since p1 is a prime element of R, by 1.1.14 we have p1 jp0j for some j 2 f1; 2; . . .; ng. Here for some j 2 f1; 2; . . .; ng, we have p1 jp0j , and hence there exists a nonzero k 2 R such that p1 k ¼ p0j . Now, since p0j is a prime element of R, either p1 is a unit or k is a unit. Since p1 is a prime element of R, p1 is not a unit. It follows that k is a unit. Now, since p1 k ¼ p0j , we have p1 p0j , where j 2 f1; 2; . . .; ng. Similarly, p2 p0k , where k 2 f1; 2; . . .; ng, etc. Thus, each pi is an associate of some p0j . Similarly, each p0j is an associate of some pi . This proves (1) and (2). For (3): Suppose to the contrary that m\n. We seek a contradiction. Since each p1 is an associate of some p0j and p1 p2 . . .pm ¼ p01 p02 . . .p0n ; we get an equality of the form p2 p3 . . .pm ¼ up01 p02 . . .p0j1 p0j þ 1 . . .p0n ; where u is a unit. Next, since p2 jðp2 p3 . . .pm Þ, and p2 p3 . . .pm ¼ up01 p02 . . . p0j1 p0j þ 1 . . .p0n , we have p2 j up01 p02 . . .p0j1 p0j þ 1 . . .p0n . Now, since p2 is a prime element of R, by 1.1.14 we have p2 ju or p2 jp0k for some k 2 ðf1; 2; . . .; ng f jgÞ. We claim that p2 -u. Suppose to the contrary that p2 ju. We seek a contradiction. Since u is a unit, there exists a nonzero v 2 R such that uv ¼ 1. Since p2 ju, we have p2 jðuvÞ, and hence p2 j1. This shows that p2 is a unit. Since p2 is a prime element of R, p2 is not a unit. This is a contradiction. Thus our claim is true, that is, p2 -u. It follows that p2 jp0k for some k 2 ðf1; 2; . . .; ng f jgÞ. Similarly, p3 jp0l for some l 2 ðf1; 2; . . .; ng fj; kgÞ, etc. On repeating this argument m times, we get an equality of the form 1 ¼ w0 p0m þ 1 p0m þ 2 . . .p0n ; where w0 is a unit. It follows that p0m þ 1 is a unit, and hence p0m þ 1 is not a prime element. This is a contradiction. ■ 1.1.16 Problem Let R be a Euclidean ring with unit element 1. Let p be a nonzero element of R. Suppose that p is not a unit in R. Let p be a prime element of R. Then the ideal ðpÞ is maximal. Proof If ðpÞ ¼ R, then it is clear that ðpÞ is a maximal ideal. So we consider the case that the ideal ðpÞ is a proper subset of R. We have to show that ðpÞ is maximal. Suppose to the contrary that there exists an ideal U of R such that ðpÞ is a proper subset of U, and U is a proper subset of R. We seek a contradiction.

1.1 Euclidean Rings

9

By 1.1.4, R is a principal ideal ring. It follows that there exists a 2 U such that ðaÞ ¼ U. Thus ðpÞ is a proper subset of ðaÞ, and ðaÞ is a proper subset of R. Since ðaÞ is a proper subset of R, a is not a unit. Since ðpÞ is a subset of ðaÞ and p 2 ðpÞ, we have p 2 ðaÞ. Hence there exists a nonzero u 2 R such that p ¼ au. Now, since p is a prime element of R, a is a unit or u is a unit. Next since a is not a unit, u is a unit. Since u is a unit and p ¼ au, we have ðpÞ ¼ ðaÞ. This contradicts the fact that ðpÞ is a proper subset of ðaÞ. ■ 1.1.17 Problem Let R be a Euclidean ring with unit element 1. Let p be a nonzero element of R. Suppose that p is not a unit in R. Let the ideal ðpÞ be maximal. Then p is a prime element of R. Proof Suppose to the contrary that there exist nonzero a; b in R such that p ¼ ab, and neither a nor b is a unit. We seek a contradiction. Since p ¼ ab, we have ðpÞ ðaÞ. Since ðpÞ ðaÞ and ðpÞ is maximal, either ðpÞ ¼ ðaÞ or ðaÞ ¼ R. Since a is not a unit, we have ðaÞ 6¼ R. It follows that ðpÞ ¼ ðaÞ, and hence there exists u 2 R such that a ¼ pu. Now, since p ¼ ab, we have p1 ¼ p ¼ ðpuÞb ¼ pðubÞ; |fflfflfflfflfflffl{zfflfflfflfflfflffl} and hence p1 ¼ pðubÞ. Next, since p is a nonzero element of R, we have 1 ¼ ub, and hence b is a unit. This is a contradiction. ■ 1.1.18 Notation The collection of all complex numbers a þ ib, where a; b are integers, is denoted by J ½i. Its members are called Gaussian integers. It is easy to see that J ½i is an integral domain with unit element 1ð¼1 þ i0 2 J ½iÞ. For every nonzero a þ ib 2 J ½i, by d ða þ ibÞ we shall mean the positive integer a2 þ b2 . 1.1.19 Note Observe that for every nonzero ða þ ibÞ; ðe þ if Þ 2 J ½i, we have 0\ða2 þ b2 Þ, and 1 ðe2 þ f 2 Þ, and hence d ða þ ibÞ ¼ a2 þ b2 1 a2 þ b2 e2 þ f 2 ¼ ja þ ibj2 je þ if j2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðja þ ibjje þ if jÞ2 ¼ jða þ ibÞðe þ if Þj2 ¼ d ðða þ ibÞðe þ if ÞÞ: Thus for every nonzero ða þ ibÞ; ðe þ if Þ 2 J ½i, we have d ða þ ibÞ d ðða þ ibÞðe þ if ÞÞ: 1.1.20 Note Let ða þ ibÞ be any nonzero member of J ½i, where a and b are integers. Let x be any positive integer. By the divisibility property of integers, there exist two integers q1 ; r1 such that a ¼ q1 x þ r1 , where 2x r1 2x.

10

1 Galois Theory I

Examples: Let x ¼ 7 and a ¼ 42. Since 42 ¼ 6 7 þ 0, we can take q1 ¼ 6 and r1 ¼ 0 2 72 ; 72 . Let x ¼ 7 and a ¼ 47. Since 47 ¼ 6 7 þ 5 ¼ ð6 þ 1Þ 7 þ ð2Þ, we can take q1 ¼ 7 and r1 ¼ 2 2 72 ; 72 . Let x ¼ 6 and a ¼ 45. Since 45 ¼ 7 6 þ 3, we can take q1 ¼ 7 and r1 ¼ 3 2 62 ; 62 . Similarly, there exist two integers q2 ; r2 such that b ¼ q2 x þ r2 , where 2x r2 2x. It follows that ða þ ibÞ ¼ ðq1 x þ r1 Þ þ iðq2 x þ r2 Þ ¼ ðq1 þ iq2 Þx þ ðr1 þ ir2 Þ; and hence ða þ ibÞ ¼ qx þ r; where q ðq1 þ iq2 Þ 2 J ½i, and r ðr1 þ ir2 Þ 2 J ½i. Suppose that r 6¼ 0. Since jr1 j 2x and jr2 j 2x, we have d ðr Þ ¼ d ðr1 þ ir2 Þ ¼ ðr1 Þ2 þ ðr2 Þ2

x2 \x2 ¼ d ðx þ i0Þ ¼ d ð xÞ: 2

1.1.21 Conclusion Let ða þ ibÞ be any nonzero member of J ½i, where a and b are integers. Let x be any positive integer. Then there exist q; r 2 J ½i such that ða þ ibÞ ¼ qx þ r and ðeither r ¼ 0 or d ðr Þ\d ð xÞÞ. 1.1.22 Note Let ða þ ibÞ be any nonzero member of J ½i, where a and b are integers. Let ðe þ if Þ be any nonzero member of J ½i, where e and f are integers. Since ðe þ if Þ is a nonzero member of J ½i, e2 þ f 2 is a positive integer. Now, by 1.1.20, there exist ðq1 þ iq2 Þ; ðr1 þ ir2 Þ 2 J ½i such that 1. ða þ ibÞðe if Þ ¼ ðq1 þ iq2 Þðe2 þ f 2 Þ þ ðr1 þ ir2 Þ, 2. ðeither r1 þ ir2 ¼ 0 or d ðr1 þ ir2 Þ\d ðe2 þ f 2 ÞÞ. Case I: r1 þ ir2 ¼ 0. From item 1 above, ða þ ibÞðe if Þ ¼ ðq1 þ iq2 Þ e2 þ f 2 : Next, since e2 þ f 2 is a positive integer, we have ða þ ibÞ ¼ ðq1 þ iq2 Þðe þ if Þ þ ð0 þ i0Þ. Thus the statement “there exist q; r 2 J ½i such that ða þ ibÞ ¼ qðe þ if Þ þ r and ðeither r ¼ 0 or d ðr Þ\d ðe þ if ÞÞ” holds. Case II: r1 þ ir2 6¼ 0. It follows from (2) that d ðr1 þ ir2 Þ\d e2 þ f 2 ;

1.1 Euclidean Rings

11

and hence from (1), d ða þ ibÞðe if Þ ðq1 þ iq2 Þ e2 þ f 2 \d e2 þ f 2 ; that is, ða þ ibÞðe if Þ ðq1 þ iq2 Þ e2 þ f 2 2 \ e2 þ f 2 2 ; that is,

2 e2 þ f 2 jða þ ibÞ ðq1 þ iq2 Þðe þ if Þj2 \ e2 þ f 2 ;

that is, jða þ ibÞ ðq1 þ iq2 Þðe þ if Þj2 \ e2 þ f 2 :

ð Þ

Let us put s1 þ is2 ðða þ ibÞ ðq1 þ iq2 Þðe þ if ÞÞ 2 J ½i; where s1 ; s2 are integers. Here ða þ ibÞ ¼ ðq1 þ iq2 Þðe þ if Þ þ ðs1 þ is2 Þ; and from ðÞ, js1 þ is2 j2 \ e2 þ f 2 : It follows that either s1 þ is2 ¼ 0 or d ðs1 þ is2 Þ\d ðe þ if Þ. Thus the statement “there exist q; r 2 J ½i such that ða þ ibÞ ¼ qðe þ if Þ þ r, and ðeither r ¼ 0 or d ðr Þ\d ðe þ if ÞÞ” holds. 1.1.23 Conclusion For every nonzero a; b 2 J ½i, there exist q; r 2 J ½i such that a ¼ qb þ r and ðeither r ¼ 0 or d ðr Þ\d ðbÞÞ. Also, we have seen that for every nonzero a; b 2 J ½i, we have d ðaÞ d ðabÞ. Hence J ½i is a Euclidean ring with unit element 1. 1.1.24 Problem It is clear that the collection Z of all integers is an integral domain with unit element 1. For every nonzero integer a, by d ðaÞ we shall mean the absolute value jaj of a. Observe that for every nonzero a; b 2 Z, d ðaÞ ¼ jaj jajjbj ¼ jabj ¼ d ðabÞ; so for every nonzero a; b 2 Z, d ðaÞ d ðabÞ.

12

1 Galois Theory I

Next, let us take arbitrary nonzero a; b 2 Z. By the divisibility property of integers, there exist q; r 2 Z such that a ¼ qb þ r and 0 r\jbj. Since 0 r\jbj, we have ðeither r ¼ 0 or 0\r\jbjÞ, and hence ðeither r ¼ 0 or jr j\jbjÞ. Thus, for every nonzero a; b 2 Z, there exist q; r 2 Z such that a ¼ qb þ r and ðeither r ¼ 0 or d ðr Þ\d ðbÞÞ. This shows that Z is a Euclidean ring. Further, Z is a subring of the Euclidean ring J ½i. Proof Since for every integer a, we have a ¼ ða þ i0Þ 2 J ½i, it follows that Z is a subset of J ½i. Also, Z is itself an integral domain. Since Z is a subset of J ½i and J ½i is a Euclidean ring, we have d ðaÞ d ðabÞ for every nonzero a; b 2 Z. Next, let us take arbitrary nonzero a; b 2 Z. By the divisibility property of integers, there exist q; r 2 Z such that a ¼ qb þ r and 0 r\jbj. Since 0 r\jbj, we have ðeither r ¼ 0 or 0\r\jbjÞ, and hence either r ¼ 0 or jr j2 \jbj2 . Thus,

for every nonzero a; b 2 Z, there exist q; r 2 Z such that a ¼ qb þ r and ðeither r ¼ 0 or d ðr Þ\d ðbÞÞ. This shows that Z is itself a Euclidean ring. Thus Z is a subring of the Euclidean ring J ½i. ■ 1.1.25 Problem Let pð2 ZÞ be a prime number, in the sense that 1\p and ðajp ) ða ¼ 1 or 1 or p or pÞÞ. Let a; b; c be any integers satisfying 1. c is relatively prime to p, 2. cp ¼ ða2 þ b2 Þ. Then p is not a prime element of the Euclidean ring J ½i. Proof Suppose to the contrary that p is a prime element of the Euclidean ring J ½i. We seek a contradiction. From (2), pjða þ ibÞða ibÞ. It follows, by 1.1.14, that pjða þ ibÞ or pjða ibÞ. Case I: pjða þ ibÞ. It follows that there exist integers e; f such that pðe þ if Þ ¼ ða þ ibÞ, and hence ja þ ibj2 ¼ jpðe þ if Þj2 . Thus p2 ðe2 þ f 2 Þ ¼ ða2 þ b2 Þ. Now, from (2), p2 ðe2 þ f 2 Þ ¼ cp, and hence pðe2 þ f 2 Þ ¼ c. This shows that pjc, and hence c is not relatively prime to p. This contradicts (1). Case II: pjða ibÞ. This case is similar to Case I.

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1.1.26 Problem Let pð2 ZÞ be a prime number. Let a; b; c be any integers satisfying 1. c is relatively prime to p, 2. cp ¼ ða2 þ b2 Þ. Then there exist integers e and f such that p ¼ e2 þ f 2 . Proof By 1.1.25, p is not a prime element of the Euclidean ring J ½i. Hence there exist integers e; f ; g; h such that

1.1 Euclidean Rings

13

p ¼ ðe þ if Þðg þ ihÞ; and neither ðe þ if Þ is a unit in J ½i nor ðg þ ihÞ is a unit in J ½i. Since ðe þ if Þ is not a unit in J ½i, we have e2 þ f 2 6¼ 1. Proof Suppose to the contrary that e2 þ f 2 ¼ 1. We seek a contradiction. Since e2 þ f 2 ¼ 1, we have ðe þ if Þðe if Þ ¼ 1, and hence ðe þ if Þ is a unit in J ½i. This is a contradiction. ■ Similarly, g2 þ h2 6¼ 1. Since p ¼ ðe þ if Þðg þ ihÞ, we have p2 ¼ e 2 þ f 2 g2 þ h2 ;

ð Þ

and hence ðe2 þ f 2 Þjp2 . Now, since p is a prime number, we have ðe2 þ f 2 Þ ¼ 1 or ðe2 þ f 2 Þ ¼ p or ðe2 þ f 2 Þ ¼ p2 . And since e2 þ f 2 6¼ 1, we have ðe2 þ f 2 Þ ¼ p or ð e2 þ f 2 Þ ¼ p2 . If ðe2 þ f 2 Þ ¼ p2 , then from ðÞ, we have p2 ¼ p2 ðg2 þ h2 Þ, and hence 2 g þ h2 ¼ 1. This is a contradiction. So ðe2 þ f 2 Þ 6¼ p2 . Since ðe2 þ f 2 Þ ¼ p or ðe2 þ f 2 Þ ¼ p2 , we have ðe2 þ f 2 Þ ¼ p. ■

1.2

Polynomial Rings

1.2.1 Note Let us observe that the quadratic congruence x2 1ðmod 8Þ has f1; 3; 5; 7g as a solution set. Thus the quadratic congruence x2 1ðmod 8Þ has four solutions. Definition Let n be an integer such that n 1. By uðnÞ we mean the number of positive integers m such that m n and (m; n are relatively prime). Here u : f1; 2; 3; . . .g ! f1; 2; 3; . . .g is called the Euler totient function. For example, uð1Þ ¼ 1; uð2Þ ¼ 1; uð3Þ ¼ 2; uð4Þ ¼ 2; uð5Þ ¼ 4; uð6Þ ¼ 2, etc. Definition Let m be an integer such that m 1. By a reduced residue system modulo m we mean a collection A of integers such that 1. the number of elements in A is uðmÞ, 2. no two members of A are congruent modulo m, 3. each member of A is relatively prime to m. Example: f1; 29g is a reduced residue system modulo 6.

1.2.2 Problem Let m be an integer such that m 1. Let a1 ; a2 ; . . .; auðmÞ be a reduced residue

system modulo m. Let k be a positive integer that is relatively prime to m. Then ka1 ; ka2 ; . . .; kauðmÞ is also a reduced residue system modulo m. Proof The proof is straightforward.

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14

1 Galois Theory I

1.2.3 Problem Let a; m be any integers such that a 1, and m 1. Suppose that a is relatively prime to m. Then auðmÞ 1ðmod mÞ:

Proof Let b1 ; b2 ; . . .; buðmÞ be a reduced residue system modulo m. Since a is

relatively prime to m, by 1.2.2, ab1 ; ab2 ; . . .; abuðmÞ is also a reduced residue system modulo m. Since a is relatively prime to m, we have ab1 b1 ðmod mÞ. Similarly, ab2 b2 ðmod mÞ, etc. This shows that

ðab1 Þðab2 Þ. . . abuðmÞ b1 b2 . . .buðmÞ ðmod mÞ;

and hence

auðmÞ

b1 b2 . . .buðmÞ

b1 b2 . . .buðmÞ ðmod mÞ:

Thus

auðmÞ b1 b2 . . .buðmÞ b1 b2 . . .buðmÞ m

is an integer, that is, b1 b2 . . .buðmÞ auðmÞ 1 m

is an integer. Since b1 ; b2 ; . . .; buðmÞ is a reduced residue system modulo m, each bi is relatively prime to m. Since each bi is relatively prime to m, and b1 b2 . . .buðmÞ auðmÞ 1 m is an integer, auðmÞ 1 m is an integer, and hence auðmÞ 1ðmod mÞ.

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1.2.4 Problem Let a be any integer such that a 1. Let p be a prime. Suppose that p does not divide a. Then ap1 1ðmod pÞ:

1.2 Polynomial Rings

15

Proof Since p is a prime and p does not divide a, a is relatively prime to p. Now, by 1.2.3, auð pÞ 1ðmod pÞ: Since p is a prime, we have uð pÞ ¼ p 1. Hence ap1 1ðmod pÞ: ■ 1.2.5 Theorem Let a be any integer such that a 1. Let p be a prime. Then ap aðmod pÞ: This result is known as the little Fermat theorem. Proof Case I: p does not divide a. Here, by 1.2.4, ap1 1ðmod pÞ, and hence ap1 a 1aðmod pÞ. Thus ap aðmod pÞ. Case II: p divides a. It follows that ap ðap1 1Þ is an integer, and hence a pa is an integer. This shows that ap aðmod pÞ. Hence in all cases, ap aðmod pÞ. ■ p

1.2.6 Problem Let a; b; m be any integers such that a 1; b 1, and m 1. Suppose that a is relatively prime to m. Then the polynomial congruence ax bðmod mÞ has a unique solution, namely, x auðmÞ1 bðmod mÞ. Proof Existence: We must show that a auðmÞ1 b bðmod mÞ; that is, auðmÞ b bðmod mÞ: Since a is relatively prime to m, by 1.2.3, auðmÞ 1ðmod mÞ, and hence auðmÞ b 1bðmod mÞ. Thus auðmÞ b bðmod mÞ. Uniqueness: Suppose that the polynomial congruence ax bðmod mÞ

16

1 Galois Theory I

has two solutions x1 and x2 , that is, ax1 bðmod mÞ : ax2 bðmod mÞ We have to show that x1 x2 ðmod mÞ. Since ax1 bðmod mÞ ; ax2 bðmod mÞ we have ax1 ax2 ðmod mÞ; and hence ax1 ax2 m 2 is an integer, and hence aðx1mx2 Þ is an integer. Since a is relatively prime to m, x1 x m is an integer, and hence

x1 x2 ðmod mÞ: ■ 1.2.7 Theorem Let p be a prime. Let f ð xÞ c0 þ c1 x þ þ cn xn be any polynomial in x with integer coefficients. Suppose that cn is a positive integer that is not divisible by p. Then the polynomial congruence f ð xÞ 0ðmod pÞ has at most n solutions. This result is due to Lagrange. Proof (Induction on nÞ: Let us take the case n ¼ 1. Thus f ð xÞ c0 þ c1 x. Here we have to solve the congruence ðc0 þ c1 xÞ 0ðmod pÞ; where c1 is not divisible by p. Since p is a prime and c1 is not divisible by p, c1 is relatively prime to p, and hence by 1.2.6, the polynomial congruence


17

c1 x c0 ðmod mÞ has a unique solution. Hence the statement “f ð xÞ 0ðmod pÞ has at most n solutions” holds for n ¼ 1. Now suppose that the statement “f ð xÞ 0ðmod pÞ has at most ðn 1Þ solutions” holds for all polynomials of degree ðn 1Þ. Also, suppose that the polynomial congruence f ð xÞ 0ðmod pÞ has ðn þ 1Þ noncongruent solutions, say x0 ; x1 ; . . .; xn . We seek a contradiction. Observe that f ð xÞ f ð x0 Þ ¼ c1 ð x x0 Þ þ c1 x2 ð x0 Þ 2 þ þ cn ð xn ð x0 Þ n Þ ¼ ðx x0 Þ c1 þ ð Þx þ þ cn xn1 ; so f ð xÞ f ðx0 Þ ¼ ðx x0 Þgð xÞ; where gð xÞ is a polynomial of degree n 1, with leading coefficient cn . Now, since cn is a positive integer that is not divisible by p, the leading coefficient of gð xÞ is a positive integer that is not divisible by p. It follows, by the induction hypothesis, that gð xÞ 0ðmod pÞ has at most ðn 1Þ noncongruent solutions. Since x0 ; x1 are solutions of f ð xÞ 0ðmod pÞ, we have f ðx0 Þ 0ðmod pÞ ; f ðx1 Þ 0ðmod pÞ and hence f ðx1 Þ f ðx0 Þðmod pÞ: This shows that ðf ðx1 Þ f ðx0 ÞÞ 0ðmod pÞ; that is, ððx1 x0 Þgðx1 ÞÞ 0ðmod pÞ:

18

1 Galois Theory I

Now, since, x1 ; x0 are noncongruent modulo p, we have gðx1 Þ 0ðmod pÞ, and hence x1 is a solution of gð xÞ 0ðmod pÞ. Similarly, x2 is a solution of gð xÞ 0ðmod pÞ; . . .; xn is a solution of gð xÞ 0ðmod pÞ. Thus gð xÞ 0ðmod pÞ has n noncongruent solutions. This is a contradiction. ■ 1.2.8 Problem Let p be a prime. Let f ð xÞ c0 þ c1 x þ þ cn xn be any polynomial in x. Suppose that the polynomial congruence f ð xÞ 0ðmod pÞ has more than n solutions. Then each ci is divisible by p. Proof Suppose to the contrary that there exists a largest positive integer k n such that ck is not divisible by p. We seek a contradiction. Here f ð xÞ ¼ c0 þ c1 x þ þ ck xk þ p ð Þxk þ 1 þ ð Þxk þ 2 þ þ ð Þxn : Suppose that the polynomial congruence f ð xÞ 0ðmod pÞ has ðn þ 1Þ noncongruent solutions x0 ; x1 ; . . .; xn . It follows that f ðx0 Þ 0ðmod pÞ; that is, c0 þ c1 x0 þ þ ck ðx0 Þk þ p ð Þðx0 Þk þ 1 þ ð Þðx0 Þk þ 2 þ þ ð Þðx0 Þn 0ðmod pÞ: Now, since p ð Þðx0 Þk þ 1 þ ð Þðx0 Þk þ 2 þ þ ð Þðx0 Þn 0ðmod pÞ; we have c0 þ c1 x0 þ þ ck ðx0 Þk þ p ð Þðx0 Þk þ 1 þ ð Þðx0 Þk þ 2 þ þ ð Þðx0 Þn p ð Þðx0 Þk þ 1 þ ð Þðx0 Þk þ 2 þ þ ð Þðx0 Þn ð0 0Þðmod pÞ;


19

and hence

c0 þ c1 x0 þ þ ck ðx0 Þk 0ðmod pÞ:

This shows that x0 is a solution of the polynomial congruence

c0 þ c1 x þ þ ck xk 0ðmod pÞ:

Similarly, x1 is a solution of the polynomial congruence

c0 þ c1 x þ þ ck xk 0ðmod pÞ; .. .

xn is a solution of the polynomial congruence c0 þ c1 x þ þ ck xk 0ðmod pÞ: Thus the number of solutions of the polynomial congruence

c0 þ c1 x þ þ ck xk 0ðmod pÞ

is strictly greater than n. By 1.2.7, the number of solutions of the polynomial congruence

c0 þ c1 x þ þ ck xk 0ðmod pÞ

is at most k. It follows that n\k. This contradicts k n.

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1.2.9 Problem Let p be a prime. Let f ð xÞ ðx 1Þðx 2Þ. . .ðx ðp 1ÞÞ xp1 1 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðp1Þfactors

Then each coefficient of the ðp 2Þth-degree polynomial f ð xÞ is divisible by p. Proof By 1.2.4, for every x 2 f1; 2; . . .; ðp 1Þg, we have xp1 1 0ðmod pÞ: It is clear that for every x 2 f1; 2; . . .; ðp 1Þg, we have ðx 1Þðx 2Þ. . .ðx ðp 1ÞÞ 0ðmod pÞ:

20

1 Galois Theory I

Hence for every x 2 f1; 2; . . .; ðp 1Þg, we have

ðx 1Þðx 2Þ. . .ðx ðp 1ÞÞ xp1 1 ð0 0Þðmod pÞ:

Thus for every x 2 f1; 2; . . .; ðp 1Þg, we have f ð xÞ 0ðmod pÞ: Thus the number of solutions of the polynomial congruence f ð xÞ 0ðmod pÞ is strictly greater than ðp 2Þ. Now, since f ð xÞ is a polynomial of degree ðp 2Þ, by 1.2.8, each coefficient of the polynomial f ð xÞ is divisible by p. ■ 1.2.10 Theorem Let p be a prime. Then ðp 1Þ! 1ðmod pÞ. This result is known as Wilson’s theorem. Proof If p ¼ 2, then ðp 1Þ! 1ðmod pÞ becomes 1! 1ðmod 2Þ. This is trivially true. So we shall consider only the case in which the prime p is odd. By 1.2.9, each coefficient of the ðp 2Þth-degree polynomial ðx 1Þðx 2Þ ðx ðp 1ÞÞ ðxp1 1Þ is divisible by p. Now, since the |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðp1Þ factors

constant term of the polynomial ðx 1Þðx 2Þ ðx ðp 1ÞÞ xp1 1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðp1Þ factors

is ð1Þp1 ðp 1Þ! þ 1 ¼ ð1Þodd1 ðp 1Þ! þ 1 ¼ ðp 1Þ! þ 1 ; ðp 1Þ! þ 1 is divisible by p, that is, ðp 1Þ! 1ðmod pÞ.

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1.2.11 Note Let p be a prime number of the form 4n þ 1. It follows that 12 ðp 1Þ is an integer 2. Put

1 1 ð p 1Þ ¼ ð p 1Þ ! : a 1 2 3 2 2 It follows that

1 1 1 a2 ¼ 1 2 3 ð p 1Þ ð p 1Þ ðp 1Þ 1 3 2 1 ; 2 2 2


21

and hence

1 1 1 1 2 3 ðp 1Þ ðp 1 Þ ðp 1Þ 1 3 2 1 ðmod pÞ: 2 2 2

a2

Thus

1 1 1 a2 1 2 3 ðp 1Þ ðp 1Þ ðp 3Þ 3 2 1 ðmod pÞ: 2 2 2 Since 1 2 ðp 1 2 ðp

1Þ 1 2 ðp þ 1Þðmod pÞ; 3Þ 1 2 ðp þ 3Þðmod pÞ; .. . 3 ðp 3Þðmod pÞ; 2 ðp 2Þðmod pÞ; 1 ðp 1Þðmod pÞ;

we have

1 1 ðp 1Þ ðp 3Þ 3 2 1 2 2

1 1 ð p þ 1Þ ðp þ 3Þ ððp 3ÞÞððp 2ÞÞððp 1ÞÞ ðmod pÞ; 2 2 that is,

1 1 ðp 1Þ ðp 3Þ 3 2 1 2 2

p1 1 1 2 ð1Þ ð p þ 1Þ ðp þ 3Þ ððp 3ÞÞððp 2ÞÞððp 1ÞÞ ðmod pÞ; 2 2

that is,

1 1 ð p 1Þ ðp 3 Þ 3 2 1 2 2

1 1 ð1Þeven ðp þ 1Þ ðp þ 3Þ ððp 3ÞÞððp 2ÞÞððp 1ÞÞ ðmod pÞ; 2 2

22

1 Galois Theory I

that is,

1 1 ðp 1Þ ð p 3Þ 3 2 1 2 2

1 1 ð p þ 1Þ ðp þ 3Þ ððp 3ÞÞððp 2ÞÞððp 1ÞÞ ðmod pÞ: 2 2 Now, since

1 1 1 ðp 1Þ ðp 1Þ ðp 3Þ 3 2 1 ðmod pÞ; 1 2 3 2 2 2

a2

we have

1 ð p 1Þ a 1 2 3 2

1 1 ð p þ 1Þ ðp þ 3Þ ððp 3ÞÞððp 2ÞÞððp 1ÞÞ ðmod pÞ; 2 2 2

that is, a2 ðp 1Þ!ðmod pÞ: By 1.2.10, ðp 1Þ! 1ðmod pÞ. Now, since a2 ðp 1Þ!ðmod pÞ, we have that a is a solution of the quadratic congruence a 1ðmod pÞ. This shows x2 1ðmod pÞ, that is, 12 ðp 1Þ ! is a solution of the quadratic congruence x2 1ðmod pÞ. 2

1.2.12 Conclusion Let p be a prime number of the form 4n þ1. Then there exists a solution of x2 1ðmod pÞ. One such solution is 12 ðp 1Þ !. 1.2.13 Theorem Let p be a prime number of the form 4n þ 1. Then there exist integers a and b such that p ¼ a2 þ b2 . This result is due to Fermat. Proof It is clear that 5 p and that 12 ðp 1Þ is an even integer. By 1.2.12, there

exists an integer x 2 0; 1; . . .; 12 ðp 1Þ; . . .; ðp 1Þ such that x2 1ðmod pÞ. It follows that there exists an integer c such that cp ¼ x2 þ 12 . It follows that there exists an integer y 2 12 ðp 1Þ; 12 ðp 1Þ þ 1; . . .0; . . .; 12 ðp 1Þg such that y2 1ðmod pÞ.

Proof Here x 2 0; 1; . . .; 12 ðp 1Þ; . . .; ðp 1Þ , so either x 2

1 0; 1; . . .; ðp 1Þ 2

or x 2

1 ðp 1Þ þ 1; . . .; ðp 1Þ : 2


23

Case I: x 2 0; 1; . . .; 12 ðp 1Þ . In this case, let us take x for y. Since x2 1ðmod pÞ, we have Since y2 1ðmod pÞ: 1 y ¼ x 2 0; 1; . . .; ðp 1Þ 2 1 1 1 ðp 1Þ; ðp 1Þ þ 1; . . .0; . . .; ðp 1Þ ; 2 2 2

we have y 2 12 ðp 1Þ; 12 ðp 1Þ þ 1; . . .0; . . .; 12 ðp 1Þ . Case II: x 2 Since

1

1Þ þ 1; . . .; ðp 1Þ . In this case, let us take ðp xÞ for y.

2 ðp

x2 1ðmod pÞ, Since

x2 þ 1 p

is an integer, and hence p 2x þ

x2 þ 1 p

is an integer.

y2 þ 1 ðp xÞ2 þ 1 x2 þ 1 ¼ ¼ p 2x þ ; p p p y2 þ 1 p

is an integer. Thus y2 1ðmod pÞ. It remains to show that y2

Since x 2

1

1 1 1 ðp 1Þ; ðp 1Þ þ 1; . . .0; . . .; ðp 1Þ : 2 2 2

2 ðp

1Þ þ 1; . . .; ðp 1Þ ; we have

y ¼ ðp xÞ 2 p 12 ðp 1Þ þ 1 ; . . .; p ðp 1Þ ¼ 12 ðp 1Þ; . . .; 1

12 ðp 1Þ; 12 ðp 1Þ þ 1; . . .0; . . .; 12 ðp 1Þ ; and hence y2

1 1 1 ðp 1Þ; ðp 1Þ þ 1; . . .0; . . .; ðp 1Þ : 2 2 2

So in all cases, there exists an integer y 2 12 ðp 1Þ; 12 ðp 1Þ þ 1; . . .0; . . .; 12 ðp 1Þg such that y2 1ðmod pÞ: ■ Since y2 1ðmod pÞ, there exists a positive integer e such that ep ¼ ðy2 þ 12 Þ. In view of 1.1.26, it suffices to show that e is relatively prime to p. Since

24

1 Galois Theory I

y 2 12 ðp 1Þ; 12 ðp 1Þ þ 1; . . .0; . . .; 12 ðp 1Þ , we have j yj 12 ðp 1Þ, and hence y2 14 ðp 1Þ2 . It follows that 2 1 y þ 12 ðp 1Þ2 þ 1; 4 and hence

1 1 1 2 1 1 2 2 2 ð p 1Þ þ 1 \ ðp 1Þ þ 1 e ¼ y þ1 p p 4 p1 4 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 1 1 1 1 p ¼ ð p 1Þ þ ðp 1Þ þ ¼ \p: 4 p1 4 4 4 Thus e is a positive integer strictly smaller than p, and since p is a prime, e must be relatively prime to p. ■ 1.2.14 Definition Let F be a field. Let F ½ x be the collection of all polynomials f ð xÞ in the “indeterminant” x having coefficients in F. We know that F ½ x is an integral domain with unit element 1. For every nonzero polynomial f ð xÞ, put d ðf ð xÞÞ degðf ð xÞÞ: It is known that 1. for every nonzero f ð xÞ; gð xÞ 2 F ½ x, degðf ð xÞÞ degðf ð xÞgð xÞÞ, 2. for every nonzero f ð xÞ; gð xÞ 2 F ½ x, there exist qð xÞ; r ð xÞ 2 F ½ x such that f ð xÞ ¼ qð xÞgð xÞ þ r ð xÞ, and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðgð xÞÞÞ. This shows that F ½ x is a Euclidean ring. Also, it is clear that ðÞ for every nonzero f ð xÞ; gð xÞ 2 F ½ x, degðf ð xÞgð xÞÞ ¼ degðf ð xÞÞ þ degðgð xÞÞ. 1.2.15 Problem F ½ x is a principal ideal ring. Proof Since F ½ x is a Euclidean ring, by 1.1.4, F ½ x is a principal ideal ring.

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Definition Let f ð xÞ be a nonzero member of the Euclidean ring F ½ x. If f ð xÞ is a unit or f ð xÞ is a prime element of F ½ x, then we say that f ð xÞ is irreducible over F. 1.2.16 Problem Let f ð xÞ be a nonzero member of the Euclidean ring F ½ x. Then f ð xÞ is a unit if and only if f ð xÞ is a constant. Proof Let f ð xÞ be a unit in the Euclidean ring F ½ x. We have to show that f ð xÞ is a constant. Since f ð xÞ is a unit in the Euclidean ring F ½ x, there exists gð xÞ 2 F ½ x such that f ð xÞgð xÞ ¼ 1. Now, since 1 6¼ 0, and F ½ x is an integral domain, it follows that f ð xÞ 6¼ 0 and gð xÞ 6¼ 0, and then that


25

0 deg ðf ð xÞÞ deg ðf ð xÞgð xÞÞ ¼ degð1Þ ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence degðf ð xÞÞ ¼ 0. Thus f ð xÞ is a constant. Conversely, let f ð xÞ be a constant. It follows that degðf ð xÞÞ ¼ 0, and hence f ð xÞ 2 F. Now, since f ð xÞ is a nonzero member of the field F, there exists a nonzero member b of F ð F ½ xÞ such that f ð xÞb ¼ 1. Thus f ð xÞ is a unit. ■ 1.2.17 Problem Let f ð xÞ be a nonzero member of the Euclidean ring F ½ x. Then f ð xÞ is irreducible if and only if ðgð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞÞ ) ðgð xÞ is a unit or hð xÞ is a unitÞ: Proof Let f ð xÞ be irreducible. We have to show that ðgð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞÞ ) ðgð xÞ is a unit or hð xÞ is a unitÞ: Since f ð xÞ is irreducible, f ð xÞ is a unit or f ð xÞ is a prime element of F ½ x. Case I: f ð xÞ is a unit. Suppose that gð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞ. We have to show that gð xÞ is a unit or hð xÞ is a unit. Suppose to the contrary that gð xÞ is not a unit and hð xÞ is not a unit. We seek a contradiction. Since f ð xÞ is a unit, by 1.2.16, f ð xÞ is a constant, and hence degðf ð xÞÞ ¼ 0. Since f ð xÞ is nonzero and f ð xÞ ¼ gð xÞhð xÞ, we have gð xÞ 6¼ 0. Now, since gð xÞ is not a unit, by 1.2.16, gð xÞ is a not constant, and hence 1 degðgð xÞÞ. Since f ð xÞ ¼ gð xÞhð xÞ, we have 1 degðgð xÞÞ degðgð xÞhð xÞÞ ¼ degðf ð xÞÞ ¼ 0: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} This is a contradiction. Case II: f ð xÞ is a prime element of F ½ x. Suppose that gð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞ. We have to show that gð xÞ is a unit or hð xÞ is a unit. Since f ð xÞ is a prime element of the Euclidean ring F ½ x, gð xÞ; hð xÞ 2 F ½ x; and f ð xÞ ¼ gð xÞhð xÞ, we have that either gð xÞ is a unit or hð xÞ is a unit. So in all cases, ðgð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞÞ ) ðgð xÞ is a unit or hð xÞ is a unitÞ: Conversely, suppose that ðgð xÞ; hð xÞ 2 F ½ x and f ð xÞ ¼ gð xÞhð xÞÞ ) ðgð xÞ is a unit or hð xÞ is a unitÞ:

ð Þ

26

1 Galois Theory I

We have to show that f ð xÞ is irreducible, that is, f ð xÞ is a unit or f ð xÞ is a prime element of F ½ x. Suppose to the contrary that f ð xÞ is not a unit and f ð xÞ is not a prime element of F ½ x. We seek a contradiction. Since f ð xÞ is not a unit and f ð xÞ is not a prime element of the Euclidean ring F ½ x, there exist gð xÞ; hð xÞ 2 F ½ x such that f ð xÞ ¼ gð xÞhð xÞ, gð xÞ is not a unit and hð xÞ is not a unit. This contradicts ðÞ. ■ 1.2.18 Problem Clearly, the polynomial 1 þ x2 is irreducible over the field R of all real numbers. Proof Suppose to the contrary that it is reducible. Then by 1.2.17, there exist gð xÞ; hð xÞ 2 R½ x such that 1 þ x2 ¼ gð xÞhð xÞ, gð xÞ is not a unit, and hð xÞ is not a unit. We seek a contradiction. Since gð xÞ is not a unit, by 1.2.16, gð xÞ is a constant, and hence degðgð xÞÞ 1. Similarly, degðhð xÞÞ 1. Next, degðgð xÞÞ þ degðhð xÞÞ ¼ degðgð xÞhð xÞÞ ¼ deg 1 þ x2 ¼ 2; so degðgð xÞÞ þ degðhð xÞÞ ¼ 2. Now, since degðgð xÞÞ 1 and degðhð xÞÞ 1, we have degðgð xÞÞ ¼ 1 and degðhð xÞÞ ¼ 1. So we can suppose that gð xÞ x þ a and hð xÞ x þ b, where a; b are real numbers. Thus 1 þ x2 ¼ ðx þ aÞðx þ bÞ ¼ ab þ ða þ bÞx þ x2 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence

ab ¼ 1 : aþb ¼ 0

This shows that 0 a2 ¼ 1. This is a contradiction.

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1.2.19 Problem Clearly, the polynomial 1 þ x2 is not irreducible over the field C of all complex numbers. Proof Observe that 1 þ x2 ¼ ðx þ iÞðx iÞ: Also, x þ i; x i are members of C½ x. Clearly, x þ i and x i are not units in C½ x. Thus by 1.2.17, 1 þ x2 is not irreducible in C½ x. ■ 1.2.20 Problem Let f ð xÞ be a nonzero member of the Euclidean ring F ½ x. Let f ð xÞ be a nonunit. Then f ð xÞ can be expressed as a product of finitely many irreducible polynomials of degree 1 in F ½ x.


27

Proof Since f ð xÞ is not a unit, by 1.2.16, f ð xÞ is not a constant, and hence degðf ð xÞÞ 1. Further, by 1.1.11, f ð xÞ can be expressed as a product of finitely many prime elements of F ½ x. Since a prime element of F ½ x is not a unit, by 1.2.16, the degree of a prime element of F ½ x is 1. Now, by the definition of irreducibility over F, f ð xÞ can be expressed as a product of finitely many irreducible polynomials of degree 1 in F ½ x. ■ 1.2.21 Theorem Let f ð xÞ be a nonzero member of the Euclidean ring F ½ x. Suppose that f ð xÞ is not a unit in F ½ x. (By 1.2.20, f ð xÞ can be expressed as a product of finitely many irreducible polynomials of degree 1 in F ½ x:Þ Let f ð xÞ ¼ p1 ð xÞp2 ð xÞ. . .pm ð xÞ; where each pi ð xÞði ¼ 1; 2; . . .; mÞ is an irreducible polynomial of degree 1 in F ½ x. Let f ð xÞ ¼ p01 ð xÞp02 ð xÞ. . .p0n ð xÞ; where each p0j ð xÞðj ¼ 1; 2; . . .; nÞ is an irreducible polynomial of degree 1 in F ½ x. Then 1. each pi ð xÞ is an associate of some p0j ð xÞ; 2. each p0j ð xÞ is an associate of some pi ð xÞ, 3. n ¼ m. This theorem is known as the unique factorization theorem of polynomials over F. Proof By 1.1.15, the proof is immediate.

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1.2.22 Problem Let pð xÞ be a nonzero member of the Euclidean ring F ½ x. Let pð xÞ be an irreducible polynomial of degree 1 in F ½ x. Then the ideal ðpð xÞÞ is “maximal” in the sense that (i) ðpð xÞÞ is a proper subset of F ½ x, (ii) if M is an ideal containing ðpð xÞÞ and M is a proper subset of F ½ x, then M ¼ ðpð xÞÞ. Proof By 1.2.16, pð xÞ is not a unit. Hence by the definition of irreducible polynomial, pð xÞ is a prime element of F ½ x. Now by 1.1.16, the ideal ðpð xÞÞ is maximal. ■ 1.2.23 Problem Let pð xÞ be a nonzero member of the Euclidean ring F ½ x. Suppose that degðpð xÞÞ 1. Let the ideal ðpð xÞÞ be maximal. Then pð xÞ is an irreducible polynomial in F ½ x. Proof By 1.2.16, pð xÞ is not a unit. Hence by the definition of irreducible polynomial, it suffices to show that pð xÞ is a prime element of F ½ x. Since the ideal ðpð xÞÞ is maximal, by 1.1.17, pð xÞ is a prime element of F ½ x. ■

28

1 Galois Theory I

1.2.24 Problem Let pð xÞ be a nonzero member of the Euclidean ring F ½ x. Suppose that degðpð xÞÞ 1. Let pð xÞ be irreducible over the field F. Then by 1.2.22, the ideal ðpð xÞÞ is maximal. Further, the quotient ring ðpFð½xxÞÞ is a field. Proof Since F ½ x is an integral domain with unit element 1, the quotient ring ðpFð½xxÞÞ is a commutative ring with unit element 1 þ ðpð xÞÞ. Next, let us take arbitrary nonzero elements f ð xÞ þ ðpð xÞÞ and gð xÞ þ ðpð xÞÞ of ðpFð½xxÞÞ, where f ð xÞ; gð xÞ 2 F ½ x. We have to show that f ð xÞgð xÞ þ ðpð xÞÞ is nonzero. Suppose to the contrary that f ð xÞgð xÞ 2 ðpð xÞÞ. We seek a contradiction. Since f ð xÞgð xÞ 2 ðpð xÞÞ, there exists hð xÞ 2 F ½ x such that f ð xÞgð xÞ ¼ pð xÞhð xÞ. Now, since pð xÞ is irreducible, by 1.2.21, pð xÞjf ð xÞ or pð xÞjgð xÞ. It follows that either f ð xÞ 2 ðpð xÞÞ or gð xÞ 2 ðpð xÞÞ. In other words, either f ð xÞ þ ðpð xÞÞ is the zero element of ðpFð½xxÞÞ or gð xÞ þ ðpð xÞÞ is the zero element of ðpFð½xxÞÞ. This is a contradiction. Thus we have shown that the product of nonzero elements of ðpFð½xxÞÞ is nonzero.

Next, let f ð xÞ þ ðpð xÞÞ be a nonzero element of ðpFð½xxÞÞ. It follows that f ð xÞ 62 ðpð xÞÞ, and f ð xÞ is a nonzero polynomial. Hence pð xÞ-f ð xÞ. By 1.1.5, there exists a greatest common divisor hð xÞ of pð xÞ and f ð xÞ in F ½ x. Further, there exist kð xÞ; lð xÞ 2 F ½ x such that hð xÞ ¼ kð xÞpð xÞ þ lð xÞf ð xÞ:

We claim that hð xÞ is a unit. Suppose to the contrary that hð xÞ is not a unit. We seek a contradiction. Since hð xÞ is a greatest common divisor of pð xÞ and f ð xÞ in F ½ x, we have hð xÞjpð xÞ and hð xÞjf ð xÞ. Since hð xÞjpð xÞ, there exists k ð xÞ 2 F ½ x such that pð xÞ ¼ hð xÞkð xÞ. Now, since pð xÞ is irreducible, by 1.2.17, hð xÞ is a unit or k ð xÞ is a unit. Since hð xÞ is not a unit, k ð xÞ is a unit. It follows from pð xÞ ¼ hð xÞ kð xÞ that pð xÞ and hð xÞ are associates. And since pð xÞ-f ð xÞ, we have hð xÞ-f ð xÞ. This is a contradiction. Thus our claim is true, that is, hð xÞ is a unit. It follows that there exists lð xÞ 2 F ½ x such that 1 ¼ hð xÞlð xÞ, and hence 1 ¼ ðkð xÞpð xÞ þ lð xÞf ð xÞÞlð xÞ ¼ ðkð xÞlð xÞÞpð xÞ þ ðlð xÞlð xÞÞf ð xÞ: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus 1 gð xÞf ð xÞ ¼ ðkð xÞlð xÞÞpð xÞ 2 ðpð xÞÞ; where gð xÞ kð xÞlð xÞ 2 F ½ x. Hence gð xÞ þ ðpð xÞÞ serves the purpose of the inverse element of f ð xÞ þ ðpð xÞÞ in ðpFð½xxÞÞ. Thus

F ½ x ðpðxÞÞ

is a field.

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29

1.2.25 Example The field of all rational numbers is denoted by Q. Observe that the polynomial x3 2 is a member of the Euclidean ring Q½ x. Also, x3 2 is irreducible in Q½ x. ½ x And hence by 1.2.24, the quotient ring ðxQ3 2 Þ is a field. Proof Suppose to the contrary that x3 2 ¼ gð xÞhð xÞ; where gð xÞ; hð xÞ 2 Q½ x, and neither gð xÞ nor hð xÞ is a unit. It follows that either degðgð xÞÞ ¼ 1 or degðhð xÞÞ ¼ 1. For definiteness, suppose that degðgð xÞÞ ¼ 1. Now we can suppose that gð xÞ x a; where a 2 Q. It follows that x3 2 ¼ ðx aÞhð xÞ; and hence a3 2 ¼ ða aÞhðaÞ ¼ 0hðaÞ ¼ 0: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus a3 ¼ 2. Since a 2 Q, there exist two integers r and s such that r 3 ¼ 2 s3 : Observe that in the prime factorization of r 3 , 23ðintegerÞ will occur, but in the prime factorization of ðr 3 ¼Þ 2 s3 , 23ðintegerÞ þ 1 will occur. But 23ðintegerÞ 6¼ 23ðintegerÞ þ 1 , which contradicts the uniqueness property of prime factorization of integers. ■ 1.2.26 Note Suppose that f ð xÞ þ ðx3 2Þ is a member of the field

Q ½ x ðx3 2Þ,

where

f ð xÞ 2 Q½ x. Let us denote the polynomial x 2 by gð xÞ. It follows that there exist qð xÞ; r ð xÞ 2 Q½ x such that f ð xÞ ¼ qð xÞgð xÞ þ r ð xÞ, and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðgð xÞÞ ¼ 3Þ. Hence we can suppose that 3

r ð x Þ a0 þ a1 x þ a2 x 2 ; where a0 ; a1 ; a2 2 Q. Thus f ð xÞ ¼ qð xÞgð xÞ þ a0 þ a1 x þ a2 x2 ;

30

1 Galois Theory I

and hence f ð xÞ þ x3 2 ¼ qð xÞgð xÞ þ a0 þ a1 x þ a2 x2 þ x3 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ qð xÞgð xÞ þ a0 þ a1 x þ a2 x2 þ ðgð xÞÞ ¼ a0 þ a1 x þ a2 x2 þ ðqð xÞgð xÞ þ ðgð xÞÞÞ: Thus f ð xÞ þ x3 2 ¼ a0 þ a1 x þ a2 x2 þ ðqð xÞgð xÞ þ ðgð xÞÞÞ: Clearly, qð xÞgð xÞ 2 ðgð xÞÞ. Now, since ðgð xÞÞ is an additive group, qð xÞgð xÞ þ ðgð xÞÞ ¼ ðgð xÞÞ. Thus f ð xÞ þ x3 2 ¼ a0 þ a1 x þ a2 x2 þ ðgð xÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ða0 þ ðgð xÞÞÞ þ ða1 þ ðgð xÞÞÞðx þ ðgð xÞÞÞ þ ða2 þ ðgð xÞÞÞðx þ ðgð xÞÞÞ2 ; or f ð xÞ þ x3 2 ¼ ða0 þ ðgð xÞÞÞ þ ða1 þ ðgð xÞÞÞt þ ða2 þ ðgð xÞÞÞt2 ; where t x þ ðgð xÞÞ. Next, t3 ¼ ðx þ ðgð xÞÞÞ3 ¼ x3 þ ðgð xÞÞ ¼ ðgð xÞ þ ðgð xÞÞÞ þ ð2 þ ðgð xÞÞÞ ¼ ð0 þ ðgð xÞÞÞ þ ð2 þ ðgð xÞÞÞ ¼ 2 þ ðgð xÞÞ; so t3 ¼ 2 þ ðgð xÞÞ: ½ x Notation Observe that the field ðxQ3 2 Þ can be thought of as a vector space over the field Q under the obvious definition of “scalar multiplication”: For every a 2 Q and for every f ð xÞ 2 Q½ x,

a f ð xÞ þ x3 2 a þ x3 2 f ð xÞ þ x3 2 ¼ af ð xÞ þ x3 2 : That is why for every a 2 Q, it is customary to denote a þ ðgð xÞÞ simply by a. ½ x 1.2.27 Conclusion All the elements of the field ðxQ3 2 Þ can be expressed “uniquely” as

a0 þ a 1 t þ a 2 t 2 ; where t x þ ðx3 2Þ, and a0 ; a1 ; a2 2 Q. Further, t3 2 ¼ 0.


31

Proof of uniqueness part To this end, suppose that a0 þ a1 t þ a2 t 2 ¼ b0 þ b1 t þ b2 t 2 ; where a0 ; a1 ; a2 ; b0 ; b1 ; b2 2 Q. We have to show that ai ¼ bi ði ¼ 0; 1; 2Þ. Since a0 þ a 1 t þ a 2 t 2 ¼ b 0 þ b 1 t þ b 2 t 2 ; we have ða0 b0 Þ þ ða1 b1 Þt þ ða2 b2 Þt2 ¼ 0; and hence ða0 b0 Þ þ ða1 b1 Þx þ ða2 b2 Þx2 2 x3 2 : Now, since the degree of each nonzero member of ðx3 2Þ is 3, ða0 b0 Þ þ ða1 b1 Þx þ ða2 b2 Þx2 is the zero polynomial, and hence ai ¼ bi ði ¼ 0; 1; 2Þ. ■

1.3

The Eisenstein Criterion

1.3.1 Definition The field of all integers is denoted by Z. Let a0 þ a1 x þ þ an xn be a member of the ring Z½ x, where each ai is an integer. If 1 is a greatest common divisor of a0 ; a1 ; . . .; an , then we say that a0 þ a1 x þ þ an xn is a primitive polynomial. 1.3.2 Problem Let a0 þ a1 x þ þ an xn and b0 þ b1 x þ þ bm xm be two primitive polynomials. Their product is c0 þ c1 x þ þ cn þ m xn þ m ; where 9 c 0 a0 b0 > > = c 1 a1 b0 þ a0 b1 c 2 a2 b0 þ a1 b1 þ a0 b2 > : > ; .. . Then c0 þ c1 x þ þ cn þ m xn þ m is primitive. Proof Suppose to the contrary that c0 þ c1 x þ þ cn þ m xn þ m is imprimitive. We seek a contradiction.

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1 Galois Theory I

Since c0 þ c1 x þ þ cn þ m xn þ m is not primitive, 1 is not a greatest common divisor of c0 ; c1 ; ; cn þ m , and hence there exists a prime number pð [ 1Þ such that pjci ði ¼ 0; 1; . . .; n þ mÞ. Since p [ 1, and 1 is a greatest common divisor of a0 ; a1 ; . . .; an , there exists j 2 f0; 1; . . .; ng such that p-aj and pjal ðl ¼ 0; 1; . . .; j 1Þ. Similarly, there exists k 2 f0; 1; . . .; mg such that p-bk and pjbl ðl ¼ 0; 1; . . .; k 1Þ. Hence a0 þ a1 x þ þ an xn is of the form pð Þ þ pð Þx þ pð Þx2 þ þ pð Þxj1 þ aj x j þ aj þ 1 xj þ 1 þ ; and b0 þ b1 x þ þ bm xm is of the form pð Þ þ pð Þx þ pð Þx2 þ þ pð Þxk1 þ bk xk þ bk þ 1 xk þ 1 þ : Since p is a prime number, p-aj , and p-bk , we have p-aj bk . Further, cj þ k ¼ aj þ k b0 þ aj þ k1 b1 þ þ a0 bj þ k ¼ aj þ k b0 þ aj þ k1 b1 þ þ aj þ 1 bk1 þ aj bk þ aj1 bk þ 1 þ þ a0 bj þ k ; so aj bk ¼ cj þ k aj þ k b0 þ aj þ k1 b1 þ þ aj þ 1 bk1 aj1 bk þ 1 þ þ a0 bj þ k ¼ cj þ k aj þ k pð Þ þ aj þ k1 pð Þ þ þ aj þ 1 pð Þ pð Þbk þ 1 þ þ pð Þbj þ k ¼ cj þ k pð Þ pð Þ ¼ cj þ k pð Þ;

and hence aj bk ¼ cj þ k pð Þ: Since pjcj þ k , we have pjaj bk . This is a contradiction.

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1.3.3 Theorem Let a0 þ a1 x þ þ an xn ð2 Z½ xÞ be a primitive polynomial. Suppose that a0 þ a1 x þ þ an xn ¼ ðr0 þ r1 x þ þ rm xm Þðs0 þ s1 x þ þ snm xnm Þ; where each ri is a rational number and each sj is a rational number. Then there exist two polynomials kð xÞ; lð xÞ 2 Z½ x such that a0 þ a1 x þ þ an xn ¼ kð xÞlð xÞ: This result is known as the Gauss’s lemma.

1.3 The Eisenstein Criterion

33

Proof By clearing denominators and taking out common factors, we can write ðr0 þ r1 x þ þ rm xm Þðs0 þ s1 x þ þ snm xnm Þ as ab kð xÞlð xÞ, where a; b are positive integers and kð xÞ; lð xÞ are primitive polynomials. Now, since a0 þ a1 x þ þ an xn ¼ ðr0 þ r1 x þ þ rm xm Þðs0 þ s1 x þ þ snm xnm Þ; we have a a0 þ a1 x þ þ an xn ¼ kð xÞlð xÞ; b or ba0 þ ba1 x þ þ pan xn ¼ akð xÞlð xÞ:

ð Þ

Since a0 þ a1 x þ þ an xn ð2 Z½ xÞ is a primitive polynomial, 1 is a greatest common divisor of a0 ; a1 ; . . .; an , and hence b is a greatest common divisor of ba0 ; ba1 ; . . .; ban . Now, from ðÞ, b is a greatest common divisor of all the coefficients of the various powers of x in akð xÞlð xÞ. Since kð xÞ; lð xÞ are primitive, by 1.3.2, kð xÞlð xÞ is primitive, and hence a is a greatest common divisor of all the coefficients of the various powers of x in akð xÞlð xÞ. Since a; b are positive integers and b is a greatest common divisor of all the coefficients of the various powers of x in akð xÞlð xÞ, we have a ¼ b. Since a ¼ b, by ðÞ, we have a0 þ a1 x þ þ an xn ¼ kð xÞlð xÞ: Also, kð xÞ; lð xÞ are primitive polynomials.

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1.3.4 Problem Let a0 þ a1 x þ þ an x ð2 Z½ xÞ be a primitive polynomial. Let pð 2Þ be a prime number. Suppose that pjai ði ¼ 0; 1; . . .; n 1Þ, p-an , and p2 -a0 . Then a0 þ a1 x þ þ an xn is irreducible over Q, that is, a0 þ a1 x þ þ an xn cannot be factored into two nontrivial polynomials with rational numbers as coefficients. n

Proof Suppose to the contrary that a0 þ a1 x þ þ an xn ¼ ðr0 þ r1 x þ þ rm xm Þðs0 þ s1 x þ þ snm xnm Þ; where each ri is a rational number and each sj is a rational number. We seek a contradiction. By 1.3.3, there exist two polynomials b0 þ b1 x þ þ bm xm ; c0 þ c1 x þ þ cnm xnm 2 Z½ x such that

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1 Galois Theory I

a0 þ a1 x þ þ an xn ¼ ðb0 þ b1 x þ þ bm xm Þðc0 þ c1 x þ þ cnm xnm Þ: Since pjai ði ¼ 0; 1; . . .; n 1Þ, the above equality takes the form p a00 þ a01 x þ þ a0n1 xn1 þ an xn ¼ ðb0 þ b1 x þ þ bm xm Þ ðc0 þ c1 x þ þ cnm xnm Þ

ðÞ:

Since p-an , p does not divide any greatest common divisor of a0 ; a1 ; . . .; an . Here pa00 ¼ b0 c0 , and p is a prime, so pjb0 or pjc0 . Since p2 -a0 and a0 ¼ pa00 ¼ b0 c0 , we have p2 -b0 c0 , and hence pjb0 and pjc0 cannot be true simultaneously. So for the sake of definiteness, suppose that pjb0 , and p-c0 . Now ðÞ takes the form p a00 þ a01 x þ þ a0n1 xn1 þ an xn ¼ ðpð Þ þ b1 x þ þ bm xm Þ ðc0 þ c1 x þ þ cnm xnm Þ: Since p-an , from ðÞ, we find that there exists j 2 f1; 2; . . .; mg such that 1. p-bj , 2. pjbk ðk ¼ 0; 1; . . .; j 1Þ. Now we can write p a00 þ a01 x þ þ a0n1 xn1 þ an xn ðpð Þ þ pð Þx þ þ pð Þxj1 þ bj x j þ þ bm xm ðc0 þ c1 x þ þ cnm xnm Þ: It follows that pa0j ¼ bj c0 þ pð Þc1 þ pð Þc2 þ þ pð Þcj : This shows that pjbj c0 . Now, since p is a prime number, either pjbj or pjc0 . This is a contradiction. ■ 1.3.5 Theorem Let a0 þ a1 x þ þ an xn ð2 Z½ xÞ be a polynomial. Let pð 2Þ be a prime number. Suppose that pjai ði ¼ 0; 1; . . .; n 1Þ, p-an and p2 -a0 . Then a0 þ a1 x þ þ an xn is irreducible over Q, that is, a0 þ a1 x þ þ an xn cannot be factored into two nontrivial polynomials with rational numbers as coefficients. This result is known as Eisenstein’s criterion. Proof Let d be the positive greatest common divisor of a0 ; a1 ; . . .; an . We can write a0 þ a1 x þ þ an xn ¼ d a00 þ a01 x þ þ a0n xn ;


35

where the positive greatest common divisor of a00 ; a01 ; . . .; a0n is 1. It follows that a00 þ a01 x þ þ a0n xn is a primitive polynomial. Since, p-an , p does not divide the positive greatest common divisor d of a0 ; a1 ; . . .; an . Since pja0 , and a0 ¼ da00 , we have pjda00 . Now, since p is a prime and p-d, we have pja00 . Since pja1 , and a1 ¼ da01 , we have pjda01 . Since p is a prime and p-d, we have pja01 . Similarly, pja02 , etc. Thus pja0i ði ¼ 0; 1; . . .; n 1Þ. Since p-an and an ¼ da0n , we have p-da0n . It follows that p-a0n . Since p2 -a0 , and a0 ¼ da00 , we have p2 -da00 . It follows that p2 -a00 . Now, by 1.3.4, a00 þ a01 x þ þ a0n xn is irreducible over Q, and hence 0 d a0 þ a01 x þ þ a0n xn is irreducible over Q. Since a0 þ a1 x þ þ an xn ¼ d a00 þ a01 x þ þ a0n xn ; a0 þ a1 x þ þ an xn is irreducible over Q.

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1.3.6 Definition Let R be a commutative ring with unit element 1. We know that R½x1 is a commutative ring with unit element 1. Since R½x1 is a commutative ring with unit element 1, ðR½x1 Þ½x2 is also a commutative ring with unit element 1. Here ðR½x1 Þ½x2 is denoted by R½x1 ; x2 . Observe that the elements of R½x1 ; x2 ð¼ ðR½x1 Þ½x2 Þ are of the form

a00 þ a10 x1 þ a20 ðx1 Þ2 þ þ a01 þ a11 x1 þ a21 ðx1 Þ2 þ x2 þ a02 þ a12 x1 þ a22 ðx1 Þ2 þ ðx2 Þ2 þ ;

that is, a00 þ ða10 x1 þ a01 x2 Þ þ a20 ðx1 Þ2 þ a11 x1 x2 þ a02 ðx2 Þ2 þ a30 ðx1 Þ3 þ a21 ðx1 Þ2 x2 þ a12 x1 ðx2 Þ2 þ a03 ðx2 Þ3 þ ; that is, X i þ j¼0

X X aij ðx1 Þi ðx2 Þ j þ aij ðx1 Þi ðx2 Þ j þ aij ðx1 Þi ðx2 Þ j þ : i þ j¼1

i þ j¼2

Thus each member of the ring R½x1 ; x2 is of the form X i þ j¼0

X X aij ðx1 Þi ðx2 Þ j þ aij ðx1 Þi ðx2 Þ j þ aij ðx1 Þi ðx2 Þ j þ : i þ j¼1

i þ j¼2

36

1 Galois Theory I

Definition Let R be a commutative ring with unit element 1. We know that R½x1 ; x2 is a commutative ring with unit element 1. Since R½x1 ; x2 is a commutative ring with unit element 1, ðR½x1 ; x2 Þ½x3 is also a commutative ring with unit element 1. Here ðR½x1 ; x2 Þ½x3 is denoted by R½x1 ; x2 ; x3 . Observe that the elements of R½x1 ; x2 ; x3 ð¼ ðR½x1 ; x2 Þ½x3 Þ are of the form

a000 þ ða100 x1 þ a010 x2 Þ þ a200 ðx1 Þ2 þ a110 x1 x2 þ a020 ðx2 Þ2 þ þ a001 þ ða101 x1 þ a011 x2 Þ þ a201 ðx1 Þ2 þ a111 x1 x2 þ a021 ðx2 Þ2 þ x3 þ a002 þ ða102 x1 þ a012 x2 Þ þ a202 ðx1 Þ2 þ a112 x1 x2 þ a022 ðx2 Þ2 þ ðx3 Þ2 þ ;

that is, a000 þ ða100 x1 þ a010 x2 þ a001 x3 Þ þ a200 ðx1 Þ2 þ a110 x1 x2 þ a020 ðx2 Þ2 þ a101 x1 x3 þ a011 x2 x3 þ a002 ðx3 Þ2 þ ; that is, X aijk ðx1 Þi ðx2 Þ j ðx3 Þk þ i þ j þ k¼0

þ

X

aijk ðx1 Þi ðx2 Þ j ðx3 Þk

i þ j þ k¼1

X aijk ðx1 Þi ðx2 Þ j ðx3 Þk þ :

i þ j þ k¼2

Similar definitions can be supplied for R½x1 ; x2 ; x3 ; x4 , etc. The commutative ring R½x1 ; . . .; xn with unit element 1 is called the ring of polynomials in n variables x1 ; . . .; xn over R. 1.3.7 Problem Let R be an integral domain. Then R½ x is an integral domain. And hence R½x1 ; . . .; xn is an integral domain. Proof Let a0 þ a1 x þ a2 x2 þ and b0 þ b1 x þ b2 x2 þ be any two nonzero members of R½ x. It suffices to show that their product ða0 b0 Þ þ ða1 b0 þ a0 b1 Þx þ ða2 b0 þ a1 b1 þ a0 b2 Þx2 þ is nonzero. Since a0 þ a1 x þ a2 x2 þ is nonzero, there exists j 2 f0; 1; 2; . . .g such that 1. aj is nonzero, 2. al ¼ 0ðl ¼ 0; 1; . . .; j 1Þ.


37

Similarly, there exists k 2 f0; 1; 2; . . .g such that 1′. bk is nonzero, 2′. bl ¼ 0ðl ¼ 0; 1; . . .; k 1Þ. It follows that a0 þ a1 x þ a2 x 2 þ ¼ aj x j þ aj þ 1 x j þ 1 þ and b0 þ b1 x þ b2 x 2 þ ¼ bk x k þ bk þ 1 x k þ 1 þ ; and hence the coefficient of xj þ k in

a0 þ a1 x þ a2 x 2 þ b0 þ b1 x þ b2 x 2 þ

is aj bk . It suffices to show that aj bk is nonzero. Since aj is a nonzero member of R, bk is a nonzero member of R, and R is an integral domain, it follows that aj bk is nonzero. ■ Definition Let F be a field. By 1.3.7, F ½x1 ; . . .; xn is an integral domain. Now we can construct its field of quotients. This field is denoted by F ðx1 ; . . .; xn Þ and is called the field of rational functions in x1 ; . . .; xn over F. The field F ðx1 ; . . .; xn Þ is important in algebraic geometry. Definition Let R be an integral domain with unit element 1. Let p be a nonzero member of R that is not a unit. If ða; b 2 R and p ¼ abÞ ) ða is a unit or b is a unitÞ; then we say that p is irreducible (or p is a prime element of RÞ. Definition Let R be an integral domain with unit element 1. If a. every nonzero member of R that is not a unit can be written as a product of finitely many irreducible elements of R, b. the decomposition in part (a) is unique up to the order and associates of the irreducible elements of R, then we say that R is a unique factorization domain. ðÞ Since every nonzero member of a field is a unit, every field is an example of a unique factorization domain. 1.3.8 Problem Let R be a unique factorization domain with unit element 1. Let a; b 2 R. Then clearly, a greatest common divisor of a and b exists in R. Definition Let R be a unique factorization domain with unit element 1. Let a; b be nonzero members of R (By 1.3.8, a greatest common divisor of a and b exists in R.).

38

1 Galois Theory I

If there exists a unit u in R such that u is a greatest common divisor of a and b, then we say that a and b are relatively prime. 1.3.9 Problem Let R be a unique factorization domain with unit element 1. Let a; b; c be any nonzero elements of R. Suppose that ajbc. Let a and b be relatively prime. Then clearly, ajc. 1.3.10 Problem Let R be a unique factorization domain with unit element 1. Let a; b; c be any nonzero elements of R. Suppose that a is an irreducible element. Suppose that ajbc. Then clearly, either ajb or ajc. Definition Let R be a unique factorization domain with unit element 1. Let a0 þ a1 x þ þ an xn be a member of the ring R½ x, where each ai is in R. If 1 is a greatest common divisor of a0 ; a1 ; . . .; an , then we say that a0 þ a1 x þ þ an xn is a primitive polynomial in R½ x. 1.3.11 Problem Let R be a unique factorization domain with unit element 1. Let a0 þ a1 x þ þ an xn and b0 þ b1 x þ þ bm xm be two primitive polynomials in R½ x. Their product is c0 þ c1 x þ þ cn þ m xn þ m ; where 9 c 0 a0 b0 > > = c 1 a1 b0 þ a0 b1 c 2 a2 b0 þ a1 b1 þ a0 b2 > : > ; .. . Then c0 þ c1 x þ þ cn þ m xn þ m is primitive. Proof Suppose to the contrary that c0 þ c1 x þ þ cn þ m xn þ m is imprimitive. We seek a contradiction. Since c0 þ c1 x þ þ cn þ m xn þ m is not primitive, 1 is not a greatest common divisor of c0 ; c1 ; . . .; cn þ m ; and hence there exists an irreducible p such that pjci ði ¼ 0; 1; . . .; n þ mÞ. Since p is irreducible, p is not a unit. Since p is not a unit and 1 is a greatest common divisor of a0 ; a1 ; . . .; an , there exists j 2 f0; 1; . . .; ng such that p-aj and pjal ðl ¼ 0; 1; . . .; j 1Þ. Similarly, there exists k 2 f0; 1; . . .; mg such that p-bk and pjbl ðl ¼ 0; 1; . . .; k 1Þ. Hence a0 þ a1 x þ þ an xn is of the form pð Þ þ pð Þx þ pð Þx2 þ þ pð Þxj1 þ aj x j þ aj þ 1 xj þ 1 þ ; and b0 þ b1 x þ þ bm xm is of the form


39

pð Þ þ pð Þx þ pð Þx2 þ þ pð Þxk1 þ bk xk þ bk þ 1 xk þ 1 þ : Since p is irreducible, p-aj , p-bk , and R is a unique factorization domain, we have p-aj bk . Further, cj þ k ¼ aj þ k b0 þ aj þ k1 b1 þ þ a0 bj þ k ¼ aj þ k b0 þ aj þ k1 b1 þ þ aj þ 1 bk1 þ aj bk þ aj1 bk þ 1 þ þ a0 bj þ k ; so aj bk ¼ cj þ k aj þ k b0 þ aj þ k1 b1 þ þ aj þ 1 bk1 aj1 bk þ 1 þ þ a0 bj þ k ¼ cj þ k aj þ k pð Þ þ aj þ k1 pð Þ þ þ aj þ 1 pð Þ pð Þbk þ 1 þ þ pð Þbj þ k ¼ cj þ k pð Þ pð Þ ¼ cj þ k pð Þ;

and hence aj bk ¼ cj þ k pð Þ: Since pjcj þ k , we have pjaj bk . This is a contradiction.

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1.3.12 Note Let R be a unique factorization domain with unit element 1. It follows that R is an integral domain with unit element 1, and hence it has a field F of quotients. Since F is a field, by 1.3.7, F ½ x is an integral domain with unit element 1. Clearly, R½ x can be considered a subring of F ½ x. Let a0 a1 a2 an þ x þ x2 þ þ xn b0 b1 b2 bn be any member of F ½ x, where each ai is a member of R and each bi is a nonzero member of R. Now we can write a0 a1 a2 an 1 þ x þ x2 þ þ xn ¼ c0 þ c1 x þ c 2 x2 þ þ cn xn ; b b0 b1 b2 bn where b b0 b1 b2 . . .bn 2 R, c0 a0 b1 b2 . . .bn 2 R, c1 b0 a1 b2 . . .bn 2 R; . . ., cn b0 b1 b2 . . .bn1 an 2 R. Let d be a greatest common divisor of c0 ; c1 ; c2 ; . . .; cn . Hence c0 þ c1 x þ c2 x2 þ þ cn xn can be expressed as

40

1 Galois Theory I

d d0 þ d1 x þ d2 x 2 þ þ dn x n ; where each di is a member of R. Clearly, 1 is a greatest common divisor of d0 ; d1 ; d2 ; . . .; dn , and hence d0 þ d1 x þ d2 x2 þ þ dn xn is a primitive polynomial in R½ x. 1.3.13 Conclusion Let R be a unique factorization domain with unit element 1 and let F be its field of quotients. Then every member of F ½ x can be expressed as db f ð xÞ, where f ð xÞ 2 R½ x, b; d 2 R, and f ð xÞ is primitive in R½ x. 1.3.14 Problem Let R be a unique factorization domain with unit element 1, and let F be its field of quotients. Let f ð xÞ 2 R½ x. Suppose that a. f ð xÞ is primitive as an element of R½ x, b. f ð xÞ is irreducible as an element of R½ x. Then f ð xÞ is irreducible as an element of F ½ x. Proof If not, then by 1.3.13, we can suppose to the contrary that f ð xÞ ¼

d1 d2 f1 ð xÞ f2 ð xÞ; b1 b2

where f1 ð xÞ; f2 ð xÞ 2 R½ x, b1 ; b2 ; d1 ; d2 2 R, f1 ð xÞ; f2 ð xÞ are primitive as elements of R½ x, degðf1 ð xÞÞ 1, and degðf2 ð xÞÞ 1. We seek a contradiction. Here b1 b2 f ð xÞ ¼ d1 d2 gð xÞ; where gð xÞ f1 ð xÞf2 ð xÞ. Now, since f1 ð xÞ; f2 ð xÞ are primitive as elements of R½ x, by 1.3.11, gð xÞ is primitive as an element of R½ x. It follows that d1 d2 is a greatest common divisor of the coefficients of the various powers of x in d1 d2 gð xÞð¼b1 b2 f ð xÞÞ. Thus d1 d2 is a greatest common divisor of the coefficients of the various powers of x in b1 b2 f ð xÞ. Since f ð xÞ is primitive as an element of R½ x, b1 b2 is a greatest common divisor of the coefficients of the various powers of x in b1 b2 f ð xÞ. Hence we can suppose that d1 d2 ¼ b1 b2 ð6¼ 0Þ. Now, since b1 b2 f ð xÞ ¼ d1 d2 f1 ð xÞf2 ð xÞ and R½ x is an integral domain, we have f ð xÞ ¼ f1 ð xÞf2 ð xÞ. Next, since degðf1 ð xÞÞ 1, degðf2 ð xÞÞ 1, and f1 ð xÞ; f2 ð xÞ 2 R½ x, it follows that f ð xÞ is not irreducible as an element of R½ x. This is a contradiction. ■ 1.3.15 Problem Let R be a unique factorization domain with unit element 1, and let F be its field of quotients. Let f ð xÞ 2 R½ x. Suppose that a. f ð xÞ is primitive as an element of R½ x, b. f ð xÞ is irreducible as an element of F ½ x. Then f ð xÞ is irreducible as an element of R½ x.


41

Proof Suppose to the contrary that f ð xÞ ¼ f1 ð xÞ f2 ð xÞ; where f1 ð xÞ; f2 ð xÞ 2 R½ xð F ½ xÞ, degðf1 ð xÞÞ 1, and degðf2 ð xÞÞ 1. We seek a contradiction. It follows that f1 ð xÞ; f2 ð xÞ 2 F ½ x. Since f ð xÞ 2 R½ x F ½ x, we have f ð xÞ 2 F ½ x. Since f ð xÞ is irreducible as an element of F ½ x, f ð xÞ ¼ f1 ð xÞ f2 ð xÞ, and f1 ð xÞ; f2 ð xÞ; f ð xÞ 2 F ½ x, we have ðf1 ð xÞ is a unit as an element of F ½ x or f2 ð xÞ is a unit as an element of F ½ xÞ; that is, either degðf1 ð xÞÞ 1 or degðf2 ð xÞÞ 1. This is a contradiction.

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1.3.16 Problem Let R be an integral domain with unit element 1. We know, by 1.3.7, that R½ x is also an integral domain with unit element 1. It is clear that ðÞ for every nonzero f ð xÞ; gð xÞ 2 R½ x, degðf ð xÞgð xÞÞ ¼ degðf ð xÞÞ þ degðgð xÞÞ. Also, if uð xÞ is a unit in R½ x, then uð xÞ is also a unit in R. Proof Let uð xÞ be a unit in R½ x. We have to show that uð xÞ is a unit in R. Since uð xÞ is a unit in the integral domain R½ x, there exists a nonzero vð xÞ in R½ x such that uð xÞvð xÞ ¼ 1; and hence 0 degðuð xÞÞ þ degðvð xÞÞ ¼ degðuð xÞvð xÞÞ ¼ degð1Þ ¼ 0: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} This shows that degðuð xÞÞ ¼ 0, and degðvð xÞÞ ¼ 0. So there exist nonzero a; b in R such that uð xÞ ¼ a and vð xÞ ¼ b. Now, since uð xÞvð xÞ ¼ 1, we have ab ¼ 1. It follows that ðuð xÞ ¼Þa is a unit in R, and hence uð xÞ is a unit in R. ■ 1.3.17 Problem Let R be a unique factorization domain with unit element 1. Let f ð xÞ be a nonzero member of R½ x having degree 1. Suppose that f ð xÞ is primitive as an element of R½ x. Then f ð xÞ can be expressed as a product of finitely many irreducible polynomials in R½ x. Proof Since R½ x can be considered a subring of F ½ x, where F is the field of quotients of members of R, and f ð xÞ 2 R½ x, we can think of f ð xÞ as a nonzero member of F ½ x. Since f ð xÞ is a nonzero member of R½ x having degree 1, f ð xÞ is not a unit in F ½ x. By 1.2.20, f ð xÞ can be expressed as a product of finitely many irreducible

42

1 Galois Theory I

polynomials p1 ð xÞ; p2 ð xÞ; . . .; pn ð xÞ of degree 1 in F ½ x. By 1.3.13, we can suppose that pk ð xÞ ¼

dk f k ð xÞ bk

ðk ¼ 1; 2; . . .; nÞ;

where fk ð xÞ 2 R½ x, bk ; dk 2 R, and fk ð xÞ is primitive in R½ x. Hence b1 b2 . . .bn f ð xÞ ¼ d1 d2 . . .dn gð xÞ; where gð xÞ f1 ð xÞf2 ð xÞ. . .fn ð xÞ. Now, since f1 ð xÞ; f2 ð xÞ; . . .; fn ð xÞ are primitive as elements of R½ x, by 1.3.11, gð xÞ is primitive as an element of R½ x. It follows that d1 d2 . . .dn is a greatest common divisor of the coefficients of the various powers of x in d1 d2 . . .dn gð xÞð¼ b1 b2 . . .bn f ð xÞÞ. Thus d1 d2 . . .dn is a greatest common divisor of the coefficients of the various powers of x in b1 b2 . . .bn f ð xÞ. Since f ð xÞ is primitive as an element of R½ x, b1 b2 . . .bn is a greatest common divisor of the coefficients of the various powers of x in b1 b2 . . .bn f ð xÞ. Hence we can suppose that d1 d2 . . .dn ¼ b1 b2 . . .bn ð6¼ 0Þ: Now, since b1 b2 . . .bn f ð xÞ ¼ d1 d2 . . .dn gð xÞ and R½ x is an integral domain, we have f ð xÞ ¼ gð xÞ, and hence f ð xÞ ¼ f1 ð xÞf2 ð xÞ. . .fn ð xÞ. Since each dbkk fk ð xÞ ¼ pk ð xÞ is irreducible in F ½ x, fk ð xÞ is irreducible in F ½ x. And since fk ð xÞ is primitive in R½ x, by 1.3.15, fk ð xÞ is irreducible as an element of R½ x. Thus f ð xÞ is expressed as a product of finitely many irreducible polynomials in R½ x. ■ 1.3.18 Problem Let R be a unique factorization domain with unit element 1. Let f ð xÞ be a nonzero member of R½ x having degree 1. Suppose that f ð xÞ is primitive as an element of R½ x. Then f ð xÞ can be expressed uniquely as a product of finitely many irreducible polynomials in R½ x. Proof By 1.3.17, it suffices to prove only the uniqueness part of this theorem. To this end, let f ð xÞ ¼ p1 ð xÞp2 ð xÞ. . .pm ð xÞ; where each pi ð xÞði ¼ 1; 2; . . .; mÞ is an irreducible polynomial of degree 1 in R½ x. Let f ð xÞ ¼ p01 ð xÞp02 ð xÞ p0n ð xÞ; where each p0j ð xÞðj ¼ 1; 2; . . .; nÞ is an irreducible polynomial of degree 1 in R½ x. We have to show that


43

1. each pi ð xÞ is an associate of some p0j ð xÞ in R½ x, 2. each p0j ð xÞ is an associate of some pi ð xÞ in R½ x, 3. n ¼ m. Since f ð xÞ is primitive as an element of R½ x and f ð xÞ ¼ p1 ð xÞp2 ð xÞ. . .pm ð xÞ, each pi ð xÞ is primitive as an element of R½ x. Similarly, each p0j ð xÞ is primitive as an element of R½ x. Now, since p1 ð xÞ is an irreducible polynomial in R½ x, by 1.3.14, p1 ð xÞ is irreducible as an element of F ½ x, where F is the field of quotients of members of R. Similarly, each pi ð xÞ is irreducible as an element of F ½ x, and each p0j ð xÞ is irreducible as an element of F ½ x. Since f ð xÞ is a nonzero member of R½ xð F ½ xÞ having degree 1, by 1.2.16, f ð xÞ is not a unit in F ½ x. By 1.2.21, 1. each pi ð xÞ is an associate of some p0j ð xÞ in F ½ x, 2. each p0j ð xÞ is an associate of some pi ð xÞ in F ½ x, 3. n ¼ m. Suppose that pi ð xÞ is an associate of some p0j ð xÞ in F ½ x. It follows that there exist a; b 2 R such that a pi ð xÞ ¼ p0j ð xÞ; b and hence

bpi ð xÞ ¼ ap0j ð xÞ:

Since pi ð xÞ is primitive as an element of R½ x, b is a greatest common divisor of

coefficients of the various powers of x in bpi ð xÞ ¼ ap0j ð xÞ . Thus b is a greatest

common divisor of the coefficients of the various powers of x in ap0j ð xÞ. Since p0j ð xÞ is primitive as an element of R½ x, a is a greatest common divisor of the coefficients of the various powers of x in ap0j ð xÞ. Hence we can suppose that a ¼ bð6¼ 0Þ: Now, since bpi ð xÞ ¼ ap0j ð xÞ and R½ x is an integral domain, pi ð xÞ ¼ p0j ð xÞ, and hence pi ð xÞ is an associate of some p0j ð xÞ in R½ x. Thus each pi ð xÞ is an associate of some p0j ð xÞ in R½ x. Similarly, each p0j ð xÞ is an associate of some pi ð xÞ in R½ x. ■ 1.3.19 Problem Let R be a unique factorization domain with unit element 1. Then R½ x is also a unique factorization domain with unit element 1. Proof Since R is a unique factorization domain with unit element 1, R is an integral domain with unit element 1, and hence by 1.3.7, R½ x is an integral domain with unit element 1. It remains to show that

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1 Galois Theory I

a. every nonzero member of R½ x that is not a unit can be written as a product of finitely many irreducible elements of R½ x, b. the decomposition in part (a) is unique up to the order and associates of the irreducible elements of R½ x. To this end, let us take a nonzero member f ð xÞ of R½ x that is not a unit in R½ x. We have to show that f ð xÞ can be expressed uniquely as a product of finitely many irreducible polynomials in R½ x. Let d be a greatest common divisor of the coefficients of the various powers of x in f ð xÞ. It follows that f ð xÞ takes the form dgð xÞ, where gð xÞ is primitive as an element of R½ x. Case I: degðf ð xÞÞ 1. Since f ð xÞ ¼ dgð xÞ and d 2 R, by 1.3.16, we have degðgð xÞÞ 1. It follows, by 1.3.18, that gð xÞ can be expressed uniquely as a product of finitely many irreducible polynomials in R½ x. Since d 2 R, d is irreducible in R½ x. Now, since f ð xÞ ¼ dgð xÞ, f ð xÞ can be expressed uniquely as a product of finitely many irreducible polynomials in R½ x. Case II: degðf ð xÞÞ ¼ 0, that is, f ð xÞ is a nonzero member of R. It follows that f ð xÞ is irreducible in R½ x. So in all cases,f ð xÞ can be expressed uniquely as a product of finitely many irreducible polynomials in R½ x. ■ 1.3.20 Problem Let R be a unique factorization domain with unit element 1. Then R½x1 ; x2 ; ; xn is also a unique factorization domain with unit element 1. Proof Since R is a unique factorization domain with unit element 1, by 1.3.19, R½x1 is also a unique factorization domain with unit element 1. Now, again by 1.3.19, ðR½x1 ; x2 ¼ÞðR½x1 Þ½x2 is a unique factorization domain with unit element 1. and hence R½x1 ; x2 is a unique factorization domain with unit element 1. Similarly, R½x1 ; x2 ; x3 is a unique factorization domain with unit element 1. Finally, R½x1 ; x2 ; . . .; xn is a unique factorization domain with unit element 1. ■ 1.3.21 Problem Let F be a field. Then F ½x1 ; x2 ; ; xn is a unique factorization domain. Proof From 1.3.7, F is a unique factorization domain, and hence by 1.3.20, F ½x1 ; x2 ; . . .; xn is a unique factorization domain. ■

1.4

Roots of Polynomials

1.4.1 Definition Let F be a field. Let K be a field such that F K. If F is a subfield of K, then we say that K is an extension of F.

1.4 Roots of Polynomials

45

Examples 1. The field R of all real numbers is an extension of the field Q of all rational numbers. 2. The field C of all complex numbers is an extension of the field R of all real numbers. pffiffiffi

3. The field a þ 2b : a; b 2 Q is an extension of Q. 1.4.2 Problem Let F and K be any fields such that K is an extension of F. Let us treat every member of K as a vector and every member of F as a scalar. We define the operation of scalar multiplication as follows: for every f 2 F ð K Þ and every v 2 K, we say that the product fv in the field K is the result of scalar multiplication of the scalar f and vector v. Then K is a vector space over the field F. Proof It suffices to show that 1. for every f1 ; f2 2 F and for every v 2 K, ðf1 þ f2 Þv ¼ f1 v þ f2 v and ðf1 f2 Þv ¼ f1 ðf2 vÞ, 2. for every f 2 F and for every v; w 2 K, f ðv þ wÞ ¼ fv þ fw, 3. for every v 2 K, 1v ¼ v. For 1: Let us take arbitrary f1 ; f2 2 F and v 2 K. We have to show that ðf1 þ f2 Þv ¼ f1 v þ f2 v and ðf1 f2 Þv ¼ f1 ðf2 vÞ. We see that ðf1 þ f2 Þv ¼ f1 v þ f2 v is trivially true, in view of the facts that F K and the right distributive law holds in the field K. Similarly, ðf1 f2 Þv ¼ f1 ðf2 vÞ is trivially true, in view of the facts that F K and the associative law of multiplication holds in the field K. For 2: Let us take arbitrary f 2 F and v; w 2 K. We have to show that f ðv þ wÞ ¼ fv þ fw. This is trivially true, in view of the facts that F K and the left distributive law holds in the field K. For 3: Let us take an arbitrary v 2 K. We have to show that 1v ¼ v. Since F is a subfield of K, the unit element 1 of F is also the unit element of K. Now, 1v ¼ v is trivially true, in view of the fact that F K and the existence of the unit element 1 in the field K. ■ Definition Let F and K be any fields such that K is an extension of F. By 1.4.2, K is a vector space over the field F. If the dimension of this vector space is finite, then we say that K is a finite extension of F. In this case, the dimension of the vector space K over F is denoted by ½K : F and is called the degree of K over F. Example: We have seen that the field C of all complex numbers is an extension of the field R of all real numbers. Hence C is a vector space over the field R. Here

46

1 Galois Theory I

pffiffiffiffiffiffiffi 1; 1 C, and every member of C can be expressed as a linear combination pffiffiffiffiffiffiffi

pffiffiffiffiffiffiffi of vectors 1; 1. Further 1; 1 is a linearly independent set of vectors, in the sense that

pffiffiffiffiffiffiffi a1 þ b 1 ¼ 0; and a; b 2 R ) ða ¼ 0 and b ¼ 0Þ:

pffiffiffiffiffiffiffi Thus 1; 1 is a basis of C. Since the number of elements in the basis

pffiffiffiffiffiffiffi 1; 1 is 2, the dimension of the vector space C over R is 2, which is of course finite. Hence C is a finite extension of R. Also, ½C : R ¼ 2. 1.4.3 Problem Let F; K, and L be any fields such that F K L. Suppose that K is a finite extension of F and L is a finite extension of K. Then a. L is a finite extension of F, b. ½L : F ¼ ½L : K ½K : F . Proof Let ½L : K ¼ m and ½K : F ¼ n. It suffices to construct a basis of the vector space L over F that has mn elements. For the sake of simplicity, let us take m ¼ 2, and n ¼ 3. Since ½L : K ¼ 2, there exists a basis fv1 ; v2 gð LÞ of the vector space L over K. Similarly, there exists a basis fw1 ; w2 ; w3 gð K LÞ of the vector space K over F. It follows that fw1 ; w2 ; w3 g L and fv1 ; v2 g L. Since L is a field, It suffices to show that fv1 w1 ; v1 w2 ; v1 w3 ; v2 w1 ; v2 w2 ; v2 w3 g L. fv1 w1 ; v1 w2 ; v1 w3 ; v2 w1 ; v2 w2 ; v2 w3 g is a basis of the vector space L over F. To this end, we must prove the following: 1. every element of L can be expressed as a linear combination of v1 w1 ; v1 w2 ; v1 w3 ; v2 w1 ; v2 w2 ; v2 w3 with coefficients in F, 2. fv1 w1 ; v1 w2 ; v1 w3 ; v2 w1 ; v2 w2 ; v2 w3 g is a linearly independent set of vectors in the vector space L over F. For 1: Let us take an arbitrary v 2 L. Now, since fv1 ; v2 g is a basis of the vector space L over K, there exist k1 ; k2 2 K such that v ¼ k1 v1 þ k2 v2 . Since k1 2 K, and fw1 ; w2 ; w3 g is a basis of the vector space K over F, there exist f11 ; f12 ; f13 2 F such that k1 ¼ f11 w1 þ f12 w2 þ f13 w3 . Similarly, there exist f21 ; f22 ; f23 2 F such that k2 ¼ f21 w1 þ f22 w2 þ f23 w3 . Hence v ¼ ðf11 w1 þ f12 w2 þ f13 w3 Þv1 þ ðf21 w1 þ f22 w2 þ f23 w3 Þv2 ; that is, v ¼ ðf11 w1 v1 þ f12 w2 v1 þ f13 w3 v1 Þ þ ðf21 w1 v2 þ f22 w2 v2 þ f23 w3 v2 Þ;


47

that is, v ¼ f11 ðv1 w1 Þ þ f12 ðv1 w2 Þ þ f13 ðv1 w3 Þ þ f21 ðv2 w1 Þ þ f22 ðv2 w2 Þ þ f23 ðv2 w3 Þ: Thus v is expressed as a linear combination of v1 w1 ; v1 w2 ; v1 w3 ; v2 w1 ; v2 w2 ; v2 w3 having coefficients in F. For 2: Suppose that f11 ðv1 w1 Þ þ f12 ðv1 w2 Þ þ f13 ðv1 w3 Þ þ f21 ðv2 w1 Þ þ f22 ðv2 w2 Þ þ f23 ðv2 w3 Þ ¼ 0; where each fij 2 F. We have to show that each fij is zero. We have ðf11 ðv1 w1 Þ þ f12 ðv1 w2 Þ þ f13 ðv1 w3 ÞÞ þ ðf21 ðv2 w1 Þ þ f22 ðv2 w2 Þ þ f23 ðv2 w3 ÞÞ ¼ 0; that is, ðf11 w1 þ f12 w2 þ f13 w3 Þv1 þ ðf21 w1 þ f22 w2 þ f23 w3 Þv2 ¼ 0 ðÞ: Since each fij 2 F ð K Þ, each wk 2 K, and K is a field, it follows that ðf11 w1 þ f12 w2 þ f13 w3 Þ 2 K. Similarly, ðf21 w1 þ f22 w2 þ f23 w3 Þ 2 K. Since fv1 ; v2 g is a basis of the vector space L over K, fv1 ; v2 g is linearly independent. Now from ðÞ, f11 w1 þ f12 w2 þ f13 w3 ¼ 0 : f21 w1 þ f22 w2 þ f23 w3 ¼ 0 Since fw1 ; w2 ; w3 g is a basis of the vector space K over F, fw1 ; w2 ; w3 g is linearly independent. Since f11 w1 þ f12 w2 þ f13 w3 ¼ 0; we have f11 ¼ f12 ¼ f13 ¼ 0. Similarly, f21 ¼ f22 ¼ f23 ¼ 0.

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1.4.4 Problem Let F; K, and L be any fields such that L is a finite extension of F, K is an extension of F, and L is an extension of K. Then a. K is a finite extension of F, and L is a finite extension of K, b. ½K : F j½L : F . Proof Since L is a finite extension of F, the dimension of the vector space L over F is finite. So let fv1 ; v2 ; . . .; vn gð LÞ be a basis of the vector space L over K. Since K is an extension of F, it follows by 1.4.2 that K is a vector space over the field F. Since K L, K is a vector space over the field F, and L is a vector space over the field F, we have that K is a linear subspace of L. Since fv1 ; v2 ; . . .; vn gð LÞ is a basis of the vector space L over F, the dimension of the vector space K over F is n, and hence K is a finite extension of F. Since fv1 ; v2 ; . . .; vn gð LÞ is a basis of the vector space L over F, each element of L is a linear combination of v1 ; v2 ; . . .; vn with coefficients in F ð K Þ, and hence

48

1 Galois Theory I

each element of L is a linear combination of v1 ; v2 ; . . .; vn with coefficients in K. This shows that the dimension of the vector space L over K is n, and hence L is a finite extension of K. This proves (a). Now, since K is a finite extension of F, by 1.4.3, ½L : F ¼ ½L : K ½K : F , and hence ½K : F j½L : F . This proves (b). ■ Definition Let F and K be any fields such that K is an extension of F. Let a 2 K. If there exists a nonzero polynomial qð xÞ 2 F ½ x such that ðK3ÞqðaÞ ¼ 0, then we say that a is algebraic over F (Caution: Here the polynomial qð xÞ is a symbol, while qðaÞ is a member of the field K.). 1.4.5 Problem Let F and K be any fields such that K is an extension of F. Then every element of F is algebraic over F. Proof Let us take an arbitrary a 2 F. We have to show that a is algebraic over F. Let us take a þ ð1Þx for qð xÞð2 F ½ xÞ. Clearly, qðaÞ ¼ 0. Thus, a is algebraic over F. ■ 1.4.6 Problem Let F and K be any fields such that K is an extension of F. Let a 2 K. Let M be the collection of all fields L satisfying a. F [ fag L K, b. L is a subfield of K. Clearly, K 2 M, and hence M is a nonempty collection. Also, \ M is a member of M. Thus, \ M is the smallest member of M. Proof Since each member of M is a field, \ M is also a field. Since each member of M contains F [ fag, \ M also contains F [ fag. Since each member of M is contained in K, \ M is also contained in the field K. Now, since \ M is a field, \ M is a subfield of K. Thus by the definition of M, \ M is a member of M. ■ 1.4.7 Problem Let F and K be any fields such that K is an extension of F. Let a 2 K. Let N be the set of all elements of K of the form ðgðaÞÞ1 f ðaÞ, where f ð xÞ; gð xÞ are members of F ½ x and gðaÞ is a nonzero member of K. Then N is a field. Proof Let ðgðaÞÞ1 f ðaÞ 2 N, where f ð xÞ; gð xÞ are members of F ½ x and gðaÞ is a nonzero member of K. Let ðg1 ðaÞÞ1 f1 ðaÞ 2 N, where f1 ð xÞ; g1 ð xÞ are members of F ½ x and g1 ðaÞ is a nonzero member of K. It suffices to show the following: 1. ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ 2 N, 2. ðgðaÞÞ1 f ðaÞ ðg1 ðaÞÞ1 f1 ðaÞ 2 N, 3. if ðgðaÞÞ1 f ðaÞ, ðg1 ðaÞÞ1 f1 ðaÞ are nonzero elements of N, then their product is nonzero, 4. 1 2 N, 5. if ðgðaÞÞ1 f ðaÞ is a nonzero element of N, then there exists b in N such that ðgðaÞÞ1 f ðaÞ b ¼ 1.


49

For 1: Observe that ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðgðaÞÞ1 ðg1 ðaÞÞ1 ðf ðaÞg1 ðaÞ þ gðaÞf1 ðaÞÞ ¼ ðgðaÞg1 ðaÞÞ1 ðf ðaÞg1 ðaÞ þ gðaÞf1 ðaÞÞ ¼ ðgðaÞg1 ðaÞÞ1 ðk1 ðaÞ þ k2 ðaÞÞ; where k1 ð xÞ f ð xÞg1 ð xÞð2 F ½ xÞ and k2 ð xÞ gð xÞf1 ð xÞð2 F ½ xÞ, and hence ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðgðaÞg1 ðaÞÞ1 kðaÞ; where k ð xÞ ðk1 ð xÞ þ k2 ð xÞÞð2 F ½ xÞ. Since gð xÞ; g1 ð xÞ are members of F ½ x, hð xÞ is a member of F ½ x, where hð xÞ gð xÞg1 ð xÞ. It follows that hðaÞ ¼ gðaÞg1 ðaÞ. Since gðaÞ; g1 ðaÞ are nonzero members of K and K. is a field, ðhðaÞ ¼ÞgðaÞg1 ðaÞ is a nonzero member of K, and hence hðaÞ is a nonzero member of K. Thus ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðhðaÞÞ1 kðaÞ; where hð xÞ; kð xÞ are members of F ½ x and hðaÞ is a nonzero member of K. It follows that ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðhðaÞÞ1 k ðaÞ 2 N; and hence

ðgðaÞÞ1 f ðaÞ þ ðg1 ðaÞÞ1 f1 ðaÞ 2 N:

For 2: Observe that ðgðaÞÞ1 f ðaÞ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðgðaÞÞ1 ðg1 ðaÞÞ1 ðf ðaÞf1 ðaÞÞ ¼ ðgðaÞg1 ðaÞÞ1 ðf ðaÞf1 ðaÞÞ ¼ ðgðaÞg1 ðaÞÞ1 ðk ðaÞÞ; where k ð xÞ f ð xÞf1 ð xÞð2 F ½ xÞ. Since gð xÞ; g1 ð xÞ are members of F ½ x, hð xÞ is a member of F ½ x, where hð xÞ gð xÞg1 ð xÞ. It follows that hðaÞ ¼ gðaÞg1 ðaÞ. Since gðaÞ; g1 ðaÞ are nonzero members of K, and K is a field, ðhðaÞ ¼ÞgðaÞg1 ðaÞ is a nonzero member of K, and hence hðaÞ is a nonzero member of K. Thus

ðgðaÞÞ1 f ðaÞ

ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðhðaÞÞ1 kðaÞ;

where hð xÞ; kð xÞ are members of F ½ x, and hðaÞ is a nonzero member of K. It follows that

ðgðaÞÞ1 f ðaÞ ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðhðaÞÞ1 k ðaÞ 2 N;

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1 Galois Theory I

and hence

ðgðaÞÞ1 f ðaÞ

ðg1 ðaÞÞ1 f1 ðaÞ 2 N:

For 3: Let ðgðaÞÞ1 f ðaÞ, ðg1 ðaÞÞ1 f1 ðaÞ be nonzero elements of N. We have to show that ðgðaÞÞ1 f ðaÞ ðg1 ðaÞÞ1 f1 ðaÞ is a nonzero element of N. Suppose to the contrary that

ðgðaÞÞ1 f ðaÞ

ðg1 ðaÞÞ1 f1 ðaÞ ¼ 0: ðÞ

We seek a contradiction. We have seen above that

ðgðaÞÞ1 f ðaÞ

ðg1 ðaÞÞ1 f1 ðaÞ ¼ ðhðaÞÞ1 kðaÞ;

where hð xÞ gð xÞg1 ð xÞð2 F ½ xÞ, kð xÞ f ð xÞf1 ð xÞð2 F ½ xÞ, and hðaÞ is a nonzero member of K. Now from ðÞ, ðhðaÞÞ1 k ðaÞ ¼ 0. Since hðaÞ is a nonzero member of the field K, we have ðf ðaÞf1 ðaÞ ¼ÞkðaÞ ¼ 0, and hence either f ðaÞ ¼ 0 or f1 ðaÞ ¼ 0. It follows that either ðgðaÞÞ1 f ðaÞ ¼ 0 or ðg1 ðaÞÞ1 f1 ðaÞ ¼ 0. This is a contradiction. For 4: Let us take the constant polynomial 1 for f ð xÞ, and again the constant polynomial 1 for gð xÞ. Clearly, ð1 ¼ 11 1 ¼ÞðgðaÞÞ1 f ðaÞ 2 N, and hence 1 2 N. For 5: Let us take an arbitrary nonzero element ðgðaÞÞ1 f ðaÞ of N ð K Þ, where f ð xÞ; gð xÞ are members of F ½ x, and gðaÞ is a nonzero member of K. Since gðaÞ is a nonzero member of the field K, ðgðaÞÞ1 is a nonzero member of K. Next, since ðgðaÞÞ1 f ðaÞ is a nonzero member of the field K, f ðaÞ is a nonzeromember of the field K. This shows that ðf ðaÞÞ1 gðaÞ 2 N. Further, it is clear that ðgðaÞÞ1 f ðaÞ ðf ðaÞÞ1 gðaÞ ¼ 1. Hence ðf ðaÞÞ1 gðaÞ serves the purpose of b. ■ 1.4.8 Problem Let F and K be any fields such that K is an extension of F. Let a 2 K. Let N be the symbol as described in 1.4.7, and M the symbol as described in 1.4.6. Then N ¼ \ M. Proof We must prove: 1. N \ M, 2. \ M N.


51

For 1: By 1.4.6, \ M is a member of M, so it suffices to show that every member of M contains N. To this end, let us take an arbitrary L 2 M. We have to show that N L. Next let us take an arbitrary ðgðaÞÞ1 f ðaÞ, where f ð xÞ; gð xÞ are members of F ½ x, and gðaÞ is a nonzero member of K. We have to show that ðgðaÞÞ1 f ðaÞ 2 L. Since L 2 M, by the definition of M, L is a field satisfying a. F [ fag L K, b. L is a subfield of K. Since f ð xÞ is a member of F ½ x and L is a field containing F [ fag, we have f ðaÞ 2 L. Similarly, gðaÞ 2 L. Now, since gðaÞ is nonzero, ðgðaÞÞ1 2 L. Next, since f ðaÞ 2 L and L is a field, we have ðgðaÞÞ1 f ðaÞ 2 L. For 2: By 1.4.6, \ M is the smallest member of M, so it suffices to show that N is a member of M. By the definition of M, we must prove: a. N is a field, b. F [ fag N K, c. N is a subfield of K. For a: By 1.4.7, N is a field. For b: Let us take an arbitrary a 2 F. We want to show that a 2 N. To this end, let us take the constant polynomial aas f ð xÞð2 F ½ x Þ, and the constant polynomial 1

as gð xÞð2 F ½ xÞ. It is clear that

a ¼ ð1Þ1 a ¼ ðgðaÞÞ1 f ðaÞ 2 N, and hence

a 2 N. Thus we have shown that F N. Now we want to show that a 2 N. To this end, let us take the polynomial 0 þ 1x as f ð xÞð2 F ½ xÞ, and the constant polynomial 1 as gð xÞð2 F ½ xÞ. It is clear that a ¼ ð1Þ1 ð0 þ 1aÞ ¼ ðgðaÞÞ1 f ðaÞ 2 N, and hence a 2 N. Thus we have shown

that F [ fag N. By the definition of N, N K. For c: Since K; N are fields and N K, it follows that N is a subfield of K. ■ Definition Let F and K be any fields such that F K. Suppose that K is an extension of F. Let a 2 K. The smallest subfield of K that contains both F and a is denoted by F ðaÞ, and we say that F ðaÞ is the subfield obtained by adjoining a to F. Thus F F ðaÞ, a 2 F ðaÞ, and F is a field. It follows that ff ðaÞ : f ð xÞ 2 F ½ xg F ðaÞ. ðÞ By 1.4.8, F ðaÞ is equal to the set of all elements of K of the form ðgðaÞÞ1 f ðaÞ, where f ð xÞ; gð xÞ are members of F ½ x, and gðaÞ is a nonzero member of K Further, since F ðaÞ is a field containing the field F as a subfield, F ðaÞ is an extension of F. It follows, by 1.4.2, that F ðaÞ is a vector space over the field F. 1.4.9 Problem Let F and K be any fields such that K is an extension of F. Let a 2 K. Let F ðaÞ be a finite extension of F. Then a is algebraic over F.

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1 Galois Theory I

Proof Case I: a 2 F. Clearly f ðaÞ ¼ 0, where f ð xÞ is the polynomial a þ ð1Þxð2 F ½ xÞ, and hence a is algebraic over F. Case II: a 62 F. We have to show that a is algebraic over F. Suppose to the contrary that a is not algebraic over F. We seek a contradiction. Since a be not algebraic over F, 1; a; a2 ; . . . is a collection of distinct elements

of F ðaÞ. Thus 1; a; a2 ; . . . is an infinite subset of F ðaÞ. Since F ðaÞ is a finite extension ofF, the dimension of the vector space F ðaÞ over F is finite. Now, since

1; a; a2 ; . . . is an infinite subset of F ðaÞ, 1; a; a2 ; . . . is linearly dependent over

F. It follows that there exists a positive integer n such that 1; a; a2 ; . . .; an is linearly dependent over F. Hence, there exist a0 ; a1 ; a2 ; . . .; an 2 F such that not all ai ði ¼ 0; 1; . . .; nÞ are 0, and a0 1 þ a1 a þ a2 a2 þ þ an an ¼ 0: Thus f ðaÞ ¼ 0, where f ð xÞ a0 þ a1 x þ a2 x2 þ þ an xn ð2 F ½ xÞ is such that not all ai ði ¼ 0; 1; ; nÞ are 0. Hence f ð xÞ is nonzero. Thus a is algebraic over F. ■ 1.4.10 Problem Let F be a field. Let gð xÞ 2 F ½ x and gð xÞ 6¼ 0. Let n be the degree of the polynomial gð xÞ. Let us denote the ideal ðgð xÞÞð¼ ff ð xÞgð xÞ : f ð xÞ 2 F ½ xgÞ by V. Then the quotient ring FV½x is a vector space over the field F under the

usual vector addition and scalar multiplication. Further, 1 þ V; x þ V; x2 þ V; ; xn1 þ Vg is a basis of FV½x. And hence n is the dimension of the vector space

F ½ x V .

Proof It suffices to show that 1. for every a; b 2 F, and for every vð xÞ 2 F ½ x, we have ða þ bÞðvð xÞ þ V Þ ¼ aðvð xÞ þ V Þ þ bðvð xÞ þ V Þ and ðabÞðvð xÞ þ V Þ ¼ aðbðvð xÞ þ V ÞÞ, 2. for every a 2 F and for every vð xÞ; wð xÞ 2 F ½ x, we have aððvð xÞ þ V Þ þ ðwð xÞ þ V ÞÞ ¼ aðvð xÞ þ V Þ þ aðwð xÞ þ V Þ, 3. for every vð xÞ 2 F ½ x, 1ðvð xÞ þ V Þ ¼ ðvð xÞ þ V Þ. For 1: Let us take arbitrary a; b 2 F and vð xÞ 2 F ½ x. We have to show that ða þ bÞðvð xÞ þ V Þ ¼ aðvð xÞ þ V Þ þ bðvð xÞ þ V Þ and ðabÞðvðxÞ þ V Þ ¼ aðbðvðxÞ þ V ÞÞ: Here LHS ¼ ða þ bÞðvð xÞ þ V Þ ¼ ða þ bÞvð xÞ þ V ¼ ðavð xÞ þ bvð xÞÞ þ V ¼ ðavð xÞ þ V Þ þ ðbvð xÞ þ V Þ ¼ aðvð xÞ þ V Þ þ bðvð xÞ þ V Þ ¼ RHS


53

and LHS ¼ ðabÞðvð xÞ þ V Þ ¼ ðabÞvð xÞ þ V ¼ aðbvð xÞÞ þ V ¼ aðbvð xÞ þ V Þ ¼ aðbðvð xÞ þ V ÞÞ ¼ RHS: For 2: Let us take arbitrary a 2 F and vð xÞ; wð xÞ 2 F ½ x. We have to show that aððvð xÞ þ V Þ þ ðwð xÞ þ V ÞÞ ¼ aðvð xÞ þ V Þ þ aðwð xÞ þ V Þ. Here LHS ¼ aððvð xÞ þ V Þ þ ðwð xÞ þ V ÞÞ ¼ aððvð xÞ þ wð xÞÞ þ V Þ ¼ aðvð xÞ þ wð xÞÞ þ V ¼ ðavð xÞ þ awð xÞÞ þ V ¼ ðavð xÞ þ V Þ þ ðawð xÞ þ V Þ ¼ aðvð xÞ þ V Þ þ aðwð xÞ þ V Þ ¼ RHS: For 3: Let us take an arbitrary vð xÞ 2 F ½ x. We have to show that 1ðvð xÞ þ V Þ ¼ vð xÞ þ V. Here LHS ¼ 1ðvð xÞ þ V Þ ¼ 1vð xÞ þ V ¼ vð xÞ þ V ¼ RHS: Thus we have shown that FV½x is a vector space over the field F ð F ½ xÞ. It is

clear that 1 þ V; x þ V; x2 þ V; ; xn1 þ V is a subset of FV½x. We shall try to

show that 1 þ V; x þ V; x2 þ V; ; xn1 þ V is a basis of FV½x. To this end, we must show that

1. 1 þ V; x þ V; x2 þ V; ; xn1 þ V is linearly independent,

2. 1 þ V; x þ V; x2 þ V; ; xn1 þ V generates every element of FV½x. For 1: Suppose that a0 ð1 þ V Þ þ a1 ðx þ V Þ þ an1 xn1 þ V ¼ 0 þ V: We have to show that each ak is 0. Since n1 ða0 þ a1 x þ þ an1 xn1 Þ þ V ¼ ða0 1 þ V Þ þ ða1 x þ V Þ þ þ ðan1 x þ V Þ n1 ¼ a0 ð1 þ V Þ þ a1 ðx þ V Þ þ an1 x þ V ¼ 0 þ V ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

we have a0 þ a1 x þ þ an1 xn1 ¼ a0 þ a1 x þ þ an1 xn1 0 2 V |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ff ð xÞgð xÞ : f ð xÞ 2 F ½ xg;

54

1 Galois Theory I

and hence a0 þ a1 x þ þ an1 xn1 is a member of ff ð xÞgð xÞ : f ð xÞ 2 F ½ xg. By 1.2.14, every nonzero member of ff ð xÞgð xÞ : f ð xÞ 2 F ½ xg is of degree

degðgð xÞÞ ¼ n. It follows that either a0 þ a1 x þ þ an1 xn1 ¼ 0 or deg a0 þ a1 x þ þ an1 xn1

nð [ ðn 1ÞÞ: It follows that a0 þ a1 x þ þ an1 xn1 ¼ 0, that is, each ak is 0. For 2: Let us take an arbitrary nonzero member hð xÞ of F ½ x, where hð xÞ b0 þ b1 x þ b2 x2 þ . We have to show that hð xÞ þ V can be expressed as a linear combination of 1 þ V; x þ V; x2 þ V; ; xn1 þ V. By 1.2.14, there exist qð xÞ; r ð xÞ 2 F ½ x such that hð xÞ ¼ qð xÞgð xÞ þ r ð xÞ and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðgð xÞÞÞ. Case I: r ð xÞ ¼ 0. It follows that hð xÞ ¼ qð xÞgð xÞð2 ff ð xÞgð xÞ : f ð xÞ 2 F ½ xg ¼ V Þ, and hence hð xÞ 2 V. We have hð xÞ þ V ¼ 0ð1 þ V Þ þ 0ðx þ V Þ þ 0 x2 þ V þ þ 0 xn1 þ V : Case II: degðr ð xÞÞ\degðgð xÞÞð¼ nÞ. We suppose that r ð xÞ c0 þ c1 x þ þ cn1 xn1 , where not all ci ði ¼ 0; 1; . . .; n 1Þ are 0. It suffices to show that ðqð xÞgð xÞ ¼ hð xÞ r ð xÞ ¼Þhð xÞ c0 þ c1 x þ þ cn1 xn1 is a member of V ð¼ ff ð xÞgð xÞ : f ð xÞ 2 F ½ xgÞ, that is, qð xÞgð xÞ is a member of ff ð xÞgð xÞ : f ð xÞ 2 F ½ xg. This is clearly true.

Thus we have shown that 1 þ V; x þ V; x2 þ V; ; xn1 þ V is a basis of FV½x.

is linearly independent, Since 1 þ V; x þ V; x2 þ V; ; xn1 þ V

2 n1 1 þ V; x þ V; x þ V; ; x þ V is a set of distinct elements, and hence the

number of elements in the basis 1 þ V; x þ V; x2 þ V; ; xn1 þ V of FV½x is n. Thus n is the dimension of the vector space

F ½ x V .

■

Definition Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic over F. It follows that there exists a nonzero polynomial qð xÞ 2 F ½ xð F Þ such that 1. ðK3ÞqðaÞ ¼ 0, 2. degðqð xÞÞ 1, 3. the leading coefficient of qð xÞ is 1. If nð 1Þ is the smallest degree of all such qð xÞ, then we say that a is algebraic of degree n over F. Clearly, every member of F is algebraic of degree 1 over F.


55

1.4.11 Problem Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic of degree n over F. Then there exists a unique polynomial qð xÞ 2 F ½ xð F Þ such that 1. ðK3ÞqðaÞ ¼ 0, 2. n ¼ degðqð xÞÞ 1, 3. the leading coefficient of qð xÞ is 1. The unique polynomial qð xÞ is called the minimal polynomial of a over F. Proof Existence of qð xÞ is clear from the definition of “algebraic of degree n over F.” Uniqueness: Suppose that there exist q1 ð xÞ; q2 ð xÞ 2 F ½ x such that 1. q1 ðaÞ ¼ 0; q2 ðaÞ ¼ 0, 2. n ¼ degðq1 ð xÞÞ ¼ degðq2 ð xÞÞ 1, 3. the leading coefficient of q1 ð xÞ is 1, and the leading coefficient of q2 ð xÞ is 1. We have to show that q1 ð xÞ ¼ q2 ð xÞ. Suppose to the contrary that q1 ð xÞ 6¼ q2 ð xÞ, that is, q1 ð xÞ q2 ð xÞ 6¼ 0. We seek a contradiction. Let us put hð xÞ q1 ð xÞ q2 ð xÞ. Clearly, hð xÞ 6¼ 0. Since q1 ð xÞ; q2 ð xÞ 2 F ½ x and F ½ x is a ring, we have q1 ð xÞ q2 ð xÞ 2 F ½ x, and hence hð xÞ 2 F ½ x. Since n ¼ degðq1 ð xÞÞ ¼ degðq2 ð xÞÞ, the leading coefficient of q1 ð xÞ is 1, and the leading coefficient of q2 ð xÞ is 1, we have degðq1 ð xÞ q2 ð xÞÞ\n, and hence degðhð xÞÞ\n. Since q1 ðaÞ ¼ 0; q2 ðaÞ ¼ 0, and hð xÞ ¼ q1 ð xÞ q2 ð xÞ, we have hðaÞ ¼ q ðaÞ q2 ðaÞ ¼ 0 0 ¼ 0, and hence hðaÞ ¼ 0. |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl1ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Since a is algebraic of degree n over F, hð xÞ 2 F ½ x; hð xÞ 6¼ 0, hðaÞ ¼ 0, degðhð xÞÞ\n, we find that either degðhð xÞÞ\1 or the leading coefficient of hð xÞ is different from 1. Case I: degðhð xÞÞ\1. It follows that hð xÞ is a constant polynomial, and since hðaÞ ¼ 0, we have hð xÞ ¼ 0. This is a contradiction. Case II: the leading coefficient of hð xÞ is different from 1. Here, we can suppose that hð x Þ b0 þ b1 x þ þ b k x k ; where k is a positive integer strictly smaller than n, and bk is a nonzero member of F. Put qð xÞ b1k hð xÞ. Clearly, qð xÞ 2 F ½ x; qð xÞ 6¼ 0, qðaÞ ¼ 0, the leading coefficient of qð xÞ is 1, and 1 k ¼ degðqð xÞÞ. Now, since a is algebraic of degree n over F, we have n k. This is a contradiction. ■ 1.4.12 Problem Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic of degree n over F. Let qð xÞ 2 F ½ xð F Þ. Let qð xÞ be the minimal polynomial of a over F. Then qð xÞ is irreducible over F. Proof Suppose to the contrary that qð xÞ is not irreducible over F. We seek a contradiction.

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1 Galois Theory I

Since qð xÞ is not irreducible over F, there exist r ð xÞ; sð xÞ 2 F ½ x such that 1. qð xÞ ¼ r ð xÞsð xÞ, 2. 1 degðr ð xÞÞ\degðqð xÞÞð¼ nÞ, and 1 degðsð xÞÞ\degðqð xÞÞð¼ nÞ. Since a is algebraic of degree n over F, qð xÞ 2 F ½ xð F Þ, and qð xÞ is the minimal polynomial of a over F, we have qðaÞ ¼ 0, n ¼ degðqð xÞÞ 1, and the leading coefficient of qð xÞ is 1. Since qð xÞ ¼ r ð xÞsð xÞ, we have 0 ¼ qðaÞ ¼ r ðaÞsðaÞ , and hence r ðaÞsðaÞ ¼ 0. Now, since r ðaÞ; sðaÞ 2 K and K is a |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} field, either r ðaÞ ¼ 0 or sðaÞ ¼ 0. Case I: r ðaÞ ¼ 0. Here 1 degðr ð xÞÞ, so degðr1 ð xÞÞ ¼ degðr ð xÞÞð\nÞ, where 1 r1 ð xÞ leading coefficient of rðxÞ r ð xÞ. Thus r1 ð xÞ 2 F ½ x, r1 ðaÞ ¼ 0, and the leading

coefficient of r1 ð xÞ is 1. Further, 1 degðr ð xÞÞ ¼ degðr1 ð xÞÞ, so 1 degðr1 ð xÞÞ. Now, since a is algebraic of degree n over F, we have n degðr1 ð xÞÞ. This is a contradiction. Case II: sðaÞ ¼ 0. This case is similar to Case I. Thus in all cases, we get a contradiction.

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1.4.13 Problem Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic of degree n over F. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be the minimal polynomial of a over F. By 1.4.12, pð xÞ is irreducible over F, and hence by 1.2.22, the ideal ðpð xÞÞð fpð xÞf ð xÞ : f ð xÞ 2 F ½ xgÞ is a maximal ideal of the ring F ½ x. Further, by 1.2.24, the quotient ring ðpFð½xxÞÞ is a field. Also, ff ðaÞ : f ð xÞ 2 F ½ xg F ðaÞ. Let w : f ð xÞ 7! f ðaÞ be a mapping from the ring F ½ x to the field F ðaÞ. Then 1. w : F ½ x ! F ðaÞ is a ring homomorphism, 2. ker ðwÞ ¼ ðpð xÞÞ, where ker ðwÞð ff ð xÞ : f ð xÞ 2 F ½ xand wðf ð xÞÞ ¼ 0g ¼ ff ð xÞ : f ð xÞ 2 F ½ xand f ðaÞ ¼ 0gÞ denotes the kernel of the homomorphism w, and ðpð xÞÞð¼ fpð xÞf ð xÞ : f ð xÞ 2 F ½ xgÞ is the ideal of the ring F ½ x generated by pð xÞ. Proof 1. Let us take arbitrary f ð xÞ; gð xÞ 2 F ½ x, where f ð x Þ a0 þ a1 x þ a2 x 2 þ ; gð x Þ b0 þ b 1 x þ b2 x 2 þ ; each ai 2 F, and each bi 2 F. We have to show that a. wðf ð xÞ þ gð xÞÞ ¼ wðf ð xÞÞ þ wðgð xÞÞ, b. wðf ð xÞgð xÞÞ ¼ wðf ð xÞÞwðgð xÞÞ.


57

For (a): Here LHS ¼ wðf ð xÞ þ gð xÞÞ ¼ w a0 þ a1 x þ a2 x2 þ þ b0 þ b1 x þ b2 x2 þ ¼ w ða0 þ b0 Þ þ ða1 þ b1 Þx þ ða2 þ b2 Þx2 þ ¼ ða0 þ b0 Þ þ ða1 þ b1 Þa þ ða2 þ b2 Þa2 þ ¼ ða0 þ b0 Þ þ ða1 a þ b1 aÞ þ a2 a2 þ b2 a2 þ ¼ a 0 þ a1 a þ a2 a2 þ þ b0 þ b1 a þ b2 a2 þ ¼ f ðaÞ þ gðaÞ ¼ wðf ð xÞÞ þ wðgð xÞÞ ¼ RHS: For (b): Here LHS ¼ wðf ð xÞgð xÞÞ ¼ w a0 þ a1 x þ a2 x2 þ b0 þ b1 x þ b2 x2 þ ¼ w a0 b0 þ ða0 b1 þ a1 b0 Þx þ ða0 b2 þ a1 b1 þ a2 b0 Þx2 þ ¼ a0 b0 þ ða0 b1 þ a1 b0 Þa þ ða0 b2 þ a1 b1 þ a2 b0 Þa2 þ ¼ a0 b0 þ ð a0 b1 a þ a 1 b0 a Þ þ a 0 b2 a2 þ a 1 b1 a2 þ a 2 b0 a2 þ ¼ a0 b0 þ a0 b1 a þ a0 b2 a2 þ þ a1 b0 a þ a1 b1 a2 þ a1 b2 a3 þ þ a2 b0 a2 þ a2 b1 a3 þ a2 b2 a4 þ þ ¼ a0 b0 þ b1 a þ b2 a2 þ þ a1 a b0 þ b1 a þ b2 a 2 þ þ a2 a2 b0 þ b1 a þ b2 a2 þ þ ¼ a0 þ a1 a þ a2 a2 þ b0 þ b1 a þ b2 a2 þ ¼ f ðaÞgðaÞ ¼ wðf ð xÞÞwðgð xÞÞ ¼ RHS: 2. We have to show that a. fpð xÞgð xÞ : gð xÞ 2 F ½ xg ff ð xÞ : f ð xÞ 2 F ½ xand f ðaÞ ¼ 0g, b. ff ð xÞ : f ð xÞ 2 F ½ x and f ðaÞ ¼ 0g fpð xÞgð xÞ : gð xÞ 2 F ½ xg. For (a): Let us take an arbitrary gð xÞ 2 F ½ x. We have to show that pð xÞgð xÞ 2 ff ð xÞ : f ð xÞ 2 F ½ xand f ðaÞ ¼ 0g, that is, pðaÞgðaÞ ¼ wðpð xÞÞwðgð xÞÞ ¼ wðpð xÞgð xÞÞ ¼ 0 ; |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} that is, pðaÞgðaÞ ¼ 0. It suffices to show that pðaÞ ¼ 0. Since a is algebraic of degree n over F and pð xÞ is the minimal polynomial of a over F, we have pðaÞ ¼ 0. For (b): Let us take an arbitrary nonzero f ð xÞ 2 F ½ x such that wðf ð xÞÞ ¼ 0, that is, f ðaÞ ¼ 0. We have to show that f ð xÞ 2 fpð xÞgð xÞ : gð xÞ 2 F ½ xg. Since pð xÞ is the minimal polynomial of a over F, we have pðaÞ ¼ 0, n ¼ degðpð xÞÞ 1, and the leading coefficient of pð xÞ is 1. Since degðpð xÞÞ 1,

58

1 Galois Theory I

pð xÞ is a nonzero member of F ½ x. Now by 1.2.14, there exist qð xÞ; r ð xÞ 2 F ½ x such that f ð xÞ ¼ qð xÞpð xÞ þ r ð xÞ, and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðpð xÞÞÞ. Since f ð xÞ ¼ qð xÞpð xÞ þ r ð xÞ, we have 0 ¼ f ðaÞ ¼ wðf ð xÞÞ ¼ wðqð xÞpð xÞ þ r ð xÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ wðqð xÞpð xÞÞ þ wðr ð xÞÞ ¼ wðqð xÞÞwðpð xÞÞ þ wðr ð xÞÞ ¼ qðaÞpðaÞ þ r ðaÞ ¼ qðaÞ0 þ r ðaÞ ¼ r ðaÞ; and hence r ðaÞ ¼ 0. We claim that r ð xÞ ¼ 0. Suppose to the contrary that r ð xÞ 6¼ 0. We seek a contradiction. Since r ð xÞ 6¼ 0 and r ðaÞ ¼ 0, r ð xÞ is not a constant polynomial, and hence degðr ð xÞÞ 1. Since r ð xÞ 6¼ 0 and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðpð xÞÞÞ, we have 1 degðr ð xÞÞ\degðpð xÞÞ ¼ n, and hence 1 degðr ð xÞÞ\n. |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Put r1 ð xÞ

1 r ð xÞ: leading coefficient of r ð xÞ

Thus r1 ð xÞ 2 F ½ x, r1 ðaÞ ¼ 0, and the leading coefficient of r1 ð xÞ is 1. Further, 1 degðr ð xÞÞ ¼ degðr1 ð xÞÞ, so 1 degðr1 ð xÞÞ. Now, since a is algebraic of degree n over F, we have n degðr1 ð xÞÞ, and hence n degðr ð xÞÞ. This is a contradiction. Thus our claim is true, that is, ðf ð xÞ qð xÞpð xÞ ¼Þr ð xÞ ¼ 0. Hence f ð xÞ ¼ qð xÞpð xÞ 2 fpð xÞgð xÞ : gð xÞ 2 F ½ xg. ■ 1.4.14 Problem Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic of degree n over F. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be the minimal polynomial of a over F. Let w : f ð xÞ 7! f ðaÞ be a mapping from the ring F ½ x to the field F ðaÞ. By 1.4.13, w : F ½ x ! F ðaÞ is a ring homomorphism, and hence by the fundamental theorem of ring homomorphisms, the mapping g : ðf ð xÞ þ ker ðwÞÞ 7! wðf ð xÞÞð¼ f ðaÞÞ is a ring isomorphism from the quotient ring F ½ x F ½ x ker ðwÞ to F ðaÞ. Thus g maps ker ðwÞ onto F ðaÞ. F ½ x In short, the field ker is isomorphic to the field F ðaÞ. ð wÞ

Proof Recall that F ðaÞ is equal to the set of all elements of K of the form ðgðaÞÞ1 f ðaÞ, where f ð xÞ; gð xÞ are members of F ½ x, and gðaÞ is a nonzero member of K. Next, let us take an arbitrary ðgðaÞÞ1 f ðaÞ 2 F ðaÞ, where f ð xÞ; gð xÞ are members of F ½ x, and gðaÞ is a nonzero member of K. From 1.4.13, ðpFð½xxÞÞ is a field,

F ½ x and ker ðwÞ ¼ ðpð xÞÞ, so ker is a field. Since f ð xÞ; gð xÞ are members of F ½ x, ð wÞ F ½ x . Since ðwðgð xÞÞ ¼ÞgðaÞ is nonzero, ðf ð xÞ þ ker ðwÞÞ; ðgð xÞ þ ker ðwÞÞ 2 ker ð wÞ


59

F ½ x gð xÞ 62 ker ðwÞ, and hence gð xÞ þ ker ðwÞ is a nonzero member of the field ker . It ðwÞ

follows that ðf ð xÞhð xÞ þ ker ðwÞÞ ¼ ðf ð xÞ þ ker ðwÞÞðhð xÞ þ ker ðwÞÞ F ½ x ; ¼ ðf ð xÞ þ ker ðwÞÞðgð xÞ þ ker ðwÞÞ1 2 ker ðwÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} where ðhð xÞ þ ker ðwÞÞðgð xÞ þ ker ðwÞÞ ¼ ð1 þ ker ðwÞÞ. Thus ðf ð xÞhð xÞ þ ker ðwÞÞ F ½ x 2 ker , and ðhð xÞgð xÞ þ ker ðwÞÞ ¼ ð1 þ ker ðwÞÞ. It suffices to show that ð wÞ gðf ð xÞhð xÞ þ ker ðwÞÞ ¼ ðgðaÞÞ1 f ðaÞ. Since gðf ð xÞhð xÞ þ ker ðwÞÞ ¼ wðf ð xÞhð xÞÞ ¼ wðf ð xÞÞwðhð xÞÞ ¼ f ðaÞhðaÞ; it suffices to show that f ðaÞhðaÞ ¼ ðgðaÞÞ1 f ðaÞ, that is, gðaÞf ðaÞhðaÞ ¼ f ðaÞ, that is, f ðaÞgðaÞhðaÞ ¼ f ðaÞ. Again, it suffices to show that hðaÞgðaÞ ¼ 1. Since ðhð xÞgð xÞ þ ker ðwÞÞ ¼ ð1 þ ker ðwÞÞ, we have hðaÞgðaÞ ¼ wðhð xÞÞwðgð xÞÞ ¼ wðhð xÞgð xÞÞ ¼ gðhð xÞgð xÞ þ ker ðwÞÞ ¼ gð1 þ ker ðwÞÞ ¼ wð1Þ ¼ 1; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence hðaÞgðaÞ ¼ 1.

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1.4.15 Note Let F and K be any fields such that F K. Suppose that K is an extension of F. Let a be a member of K. Let a be algebraic of degree n over F. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be the minimal of a over F. It follows that polynomial n ¼ degðpð xÞÞ 1. Now by 1.4.10, dim

F ½ x ðpðxÞÞ

¼ n.

Let w : f ð xÞ 7! f ðaÞ be a mapping from ring F ½ x to the field F ðaÞ. By 1.4.14, the mapping g : ðf ð xÞ þ ker ðwÞÞ 7! wðf ð xÞÞð¼ f ðaÞÞ is a ring isomorphism from the F ½ x onto F ðaÞ. Further, by 1.4.13, ker ðwÞ ¼ ðpð xÞÞ. Thus quotient ring ker ð wÞ F ½ x dim ker ¼ n. ð wÞ F ½ x as a vector space over the field F under the usual We can think of ker ðwÞ operations of vector addition and scalar multiplication: For every f ð xÞ; gð xÞ 2 F ½ x and for every a 2 F ð F ½ xÞ,

ðf ð xÞ þ ker ðwÞÞ þ ðgð xÞ þ ker ðwÞÞ ðf ð xÞ þ gð xÞÞ þ ker ðwÞ

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1 Galois Theory I

and aðf ð xÞ þ ker ðwÞÞ ða þ ker ðwÞÞðf ð xÞ þ ker ðwÞÞð¼ af ð xÞ þ ker ðwÞÞ: It suffices to show the following: 1. For every f ð xÞ 2 F ½ x and for every a; b 2 F, ða þ bÞðf ð xÞ þ ker ðwÞÞ ¼ aðf ð xÞ þ ker ðwÞÞ þ bðf ð xÞ þ ker ðwÞÞ and ðabÞðf ð xÞ þ ker ðwÞÞ ¼ aðbðf ð xÞ þ ker ðwÞÞÞ: 2. For every f ð xÞ; gð xÞ 2 F ½ x and for every a 2 F, aððf ð xÞ þ ker ðwÞÞ þ ðgð xÞ þ ker ðwÞÞÞ ¼ aðf ð xÞ þ ker ðwÞÞ þ aðgð xÞ þ ker ðwÞÞ: 3. For every f ð xÞ 2 F ½ x, 1ðf ð xÞ þ ker ðwÞÞ ¼ ðf ð xÞ þ ker ðwÞÞ: For 1: LHS ¼ ða þ bÞðf ð xÞ þ ker ðwÞÞ ¼ ða þ bÞf ð xÞ þ ker ðwÞ ¼ ðaf ð xÞ þ bf ð xÞÞ þ ker ðwÞ ¼ ðaf ð xÞ þ ker ðwÞÞ þ ðbf ð xÞ þ ker ðwÞÞ ¼ aðf ð xÞ þ ker ðwÞÞ þ bðf ð xÞ þ ker ðwÞÞ ¼ RHS: Next, LHS ¼ ðabÞðf ð xÞ þ ker ðwÞÞ ¼ ðabÞf ð xÞ þ ker ðwÞ ¼ aðbf ð xÞÞ þ ker ðwÞ ¼ aðbf ð xÞ þ ker ðwÞÞ ¼ aðbðf ð xÞ þ ker ðwÞÞÞ ¼ RHS: For 2: LHS ¼ aððf ð xÞ þ ker ðwÞÞ þ ðgð xÞ þ ker ðwÞÞÞ ¼ aððf ð xÞ þ gð xÞÞ þ ker ðwÞÞ ¼ aðf ð xÞ þ gð xÞÞ þ ker ðwÞ ¼ ðaf ð xÞ þ agð xÞÞ þ ker ðwÞ ¼ ðaf ð xÞ þ ker ðwÞÞ þ ðagð xÞ þ ker ðwÞÞ ¼ aðf ð xÞ þ ker ðwÞÞ þ aðgð xÞ þ ker ðwÞÞ ¼ RHS:


61

For 3: LHS ¼ 1ðf ð xÞ þ ker ðwÞÞ ¼ 1f ð xÞ þ ker ðwÞ ¼ f ð xÞ þ ker ðwÞ ¼ RHS: F ½ x is a vector space over the field F. Thus ker ðwÞ Since F F ðaÞ and F; F ðaÞ are fields, by 1.4.2, F ðaÞ can be thought of as a vector space over the field F. We shall show that the mapping g : ðf ð xÞ þ ker ðwÞÞ 7! wðf ð xÞÞð¼ f ðaÞÞ is an F ½ x isomorphism from the vector space ker onto the vector space F ðaÞ. ð wÞ F ½ x F ½ x to F ðaÞ, the map g from ker to F ðaÞ is Since g is an isomorphism from ker ð wÞ ðwÞ

one-to-one and onto. Hence it suffices to show that for every f ð xÞ; gð xÞ 2 F ½ x and for every a; b 2 F, gðaðf ðxÞ þ ker ðwÞÞ þ bðgð xÞ þ ker ðwÞÞÞ ¼ agðf ð xÞ þ ker ðwÞÞ þ bgðgð xÞ þ ker ðwÞÞ;

LHS ¼ gðaðf ð xÞ þ ker ðwÞÞ þ bðgð xÞ þ ker ðwÞÞÞ ¼ gððaf ð xÞ þ bgð xÞÞ þ ker ðwÞÞ ¼ wðaf ð xÞ þ bgð xÞÞ ¼ wðaf ð xÞÞ þ wðbgð xÞÞ ¼ wðaÞwðf ð xÞÞ þ wðbÞwðgð xÞÞ ¼ awðf ð xÞÞ þ bwðgð xÞÞ ¼ agðf ð xÞ þ ker ðwÞÞ þ bgðgð xÞ þ ker ðwÞÞ ¼ RHS: Thus the vector space F. It follows that

F ½ x ker ðwÞ

over F is isomorphic to the vector space F ðaÞ over

F ½ x n ¼ dim ¼ dim ðF ðaÞÞ ¼ ½F ðaÞ : F ; ker ðwÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½F ðaÞ : F ¼ n. 1.4.16 Conclusion Let F and K be any fields such that K is an extension of F Let a be a member of K. Let a be algebraic of degree n over F. Then ½F ðaÞ : F ¼ n. In short, a is algebraic of degree ½F ðaÞ : F over F. 1.4.17 Problem Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let a be algebraic over F. Then F ðaÞ is a finite extension of F. Proof Since a is algebraic over F, there exists a nonzero polynomial qð xÞ 2 F ½ x such that ðK3Þ qðaÞ ¼ 0. It follows that degðqð xÞÞ 1. Let n be the smallest degree of all such polynomials qð xÞ. Hence a is algebraic of degree n over F. Now by 1.4.16, ½F ðaÞ : F ¼ n\1. Hence F ðaÞ is a finite extension of F. ■

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1.4.18 Note Let F and K be any fields such that K is an extension of F. Let A be the collection of all elements of K that are algebraic over F. By 1.4.5, F A. Thus F A K. We shall show that A is a field. To this end, let us take arbitrary a; b 2 A. It suffices to show the following: 1. ða bÞ 2 A, 2. ab 2 A, 3. ab1 2 A, provided a; b are nonzero. Since a 2 A, a is algebraic over F, and hence there exists a positive integer m such that a is algebraic of degree m over F. Similarly, there exists a positive integer n such that b is algebraic of degree n over F. Since b is algebraic of degree n over F, by 1.4.11, there exists a unique polynomial qð xÞ 2 F ½ xð F Þ such that 1. ðK3Þ qðbÞ ¼ 0, 2. n ¼ degðqð xÞÞ 1, 3. the leading coefficient of qð xÞ is 1. Suppose that b is algebraic of degree k over the field F ðaÞð K Þ. Since F F ðaÞ, and F ðaÞ is a field, we have F ½ x ðF ðaÞÞ½ x. Now, since qð xÞ 2 F ½ x, we have qð xÞ 2 ðF ðaÞÞ½ x. Since qð xÞ 2 ðF ðaÞÞ½ x, qðbÞ ¼ 0, the leading coefficient of qð xÞ is 1, and b is algebraic of degree k over the field F ðaÞ, we have k degðqð xÞÞ ¼ n: |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Thus k n. Since b is algebraic of degree k over the field F ðaÞð K Þ, by 1.4.16, ½ðF ðaÞÞðbÞ : F ðaÞ ¼ k. Since a is algebraic of degree m over the field F ð K Þ, by 1.4.16, ½F ðaÞ : F ¼ m. Since F F ðaÞ ðF ðaÞÞðbÞ K ðbÞ ¼ K, by 1.4.3, we have ½ðF ðaÞÞðbÞ : F ¼ ½ðF ðaÞÞðbÞ : F ðaÞ½F ðaÞ : F ¼ k½F ðaÞ : F n½F ðaÞ : F ¼ nm: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus ½ðF ðaÞÞðbÞ : F nmð\1Þ. Since ðF ðaÞÞðbÞ is a field containing b and all the elements of F ðaÞð3aÞ, ðF ðaÞÞðbÞ is a field containing a; b. 1. Since ðF ðaÞÞðbÞ is a field containing a; b, ðF ðaÞÞðbÞ is a field containing a b, and hence F ða bÞ ðF ðaÞÞðbÞ. Thus F ða bÞ is a linear subspace of the vector space ðF ðaÞÞðbÞ. It follows that ½F ða bÞ : F ¼ dim ðF ða bÞÞ dim ððF ðaÞÞðbÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ½ðF ðaÞÞðbÞ : F ð nm\1Þ;


63

and hence ½F ða bÞ : F nm. Now, by 1.4.9, a b is algebraic over F, and hence ða bÞ 2 A. Next, by 1.4.16, a b is algebraic of degree ½F ða bÞ : F ð nmÞ over F, and hence a b is algebraic of degree nm over F. 2. Since ðF ðaÞÞðbÞ is a field containing a; b, ðF ðaÞÞðbÞ is a field containing ab. It follows, as above, that ½F ðabÞ : F nm. Now by 1.4.9, ab is algebraic over F, and hence ab 2 A. Next, by 1.4.16, ab is algebraic of degree ½F ðabÞ : F ð nmÞ over F, and hence ab is algebraic of degree nm over F. 3. Suppose that a; b are nonzero. Since ðF ðaÞÞðbÞ is a field containing a; b, ðF ðaÞÞðbÞ is a field containing ab1 . It follows, as above, that ½F ðab1 Þ : F nm. Now by 1.4.9, ab1 is algebraic over F, and hence ab1 2 A. Next, by 1.4.16, ab1 is algebraic of degree ½F ðab1 Þ : F ð nmÞ over F, and hence ab1 is algebraic of degree nm over F. 1.4.19 Conclusion Let F and K be any fields such that K is an extension of F. Let A be the collection of all elements of K that are algebraic over F. Then 1. F A K, 2. A is a subfield of K, and F is a subfield of A, 3. if a is algebraic of degree m over F, and b is algebraic of degree n over F, then all of a b; ab; ab1 ðprovided b is nonzeroÞ are algebraic of degree mn over F. Definition Let F and K be any fields such that F K. Suppose that K is an extension of F. Let a; b 2 K. Since F [ fag F ðaÞ ðF ðaÞÞðbÞ and b 2 ðF ðaÞÞðbÞ, we have F [ fa; bg ðF ðaÞÞðbÞ. So ðF ðaÞÞðbÞ is a field containing F [ fa; bg. The smallest field containing F [ fa; bg is denoted by F ða; bÞ. We have seen that F ða; bÞ ðF ðaÞÞðbÞ. It is clear that F ðaÞ F ða; bÞ, and hence ðF ðaÞÞ [ fbg F ða; bÞ. It follows that ðF ðaÞÞðbÞ F ða; bÞ. Thus we have shown that ðF ðaÞÞðbÞ ¼ F ða; bÞ. Similarly, ðF ðbÞÞðaÞ ¼ F ða; bÞ. Thus ðF ðaÞÞðbÞ ¼ ðF ðbÞÞðaÞ ¼ F ða; bÞ: A similar definition can be supplied for F ða; b; cÞ, etc. Definition Let F and K be any fields such that K is an extension of F. If every element of K is algebraic over F, then we say that K is an algebraic extension of F. 1.4.20 Problem Let F; K, and L be any fields such that F K L. Suppose that K is an algebraic extension of F, and L is an algebraic extension of K. Then L is an algebraic extension of F. Proof Let us take an arbitrary l 2 L. We have to show that l is algebraic over F.

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1 Galois Theory I

Since L is an algebraic extension of K, and l 2 L, there exists a nonzero polynomial k0 þ k1 x þ k2 x2 þ þ kn xn such that each ki 2 K, n is a positive integer, and k0 þ k1 l þ k2 l2 þ þ kn ln ¼ 0: Since k0 2 K and K is an algebraic extension of F, k0 is algebraic over F, and hence by 1.4.17, F ðk0 Þð K Þ is a finite extension of F. Since k1 2 K and K is an algebraic extension of F, k1 is algebraic over F. Now, since F ðk0 Þ is an extension of F, by 1.4.17, ðF ðk0 ; k1 Þ ¼ÞðF ðk0 ÞÞðk1 Þð K Þ is a finite extension of F. Thus F ðk0 ; k1 Þ is an extension of F. Since k2 2 K and K is an algebraic extension of F, k2 is algebraic over F. Now, since F ðk0 ; k1 Þ is an extension of F; by 1.4.17, ðF ðk0 ; k1 ; k2 Þ ¼ÞðF ðk0 ; k1 ÞÞðk2 Þð K Þ is a finite extension of F. Thus F ðk0 ; k1 ; k2 Þ is a finite extension of F, etc. It follows that F ðk0 ; k1 ; ; kn Þð K Þ is a finite extension of F. Since each ki 2 F ðk0 ; k1 ; . . .; kn Þ, the nonzero polynomial k0 þ k1 x þ k2 x2 þ þ kn xn is a member of ðF ðk0 ; k1 ; . . .; kn ÞÞ½ x. Next, k0 þ k1 l þ k2 l2 þ . . . þ kn ln ¼ 0; so l is algebraic over F ðk0 ; k1 ; . . .; kn Þð K Þ. It follows, by 1.4.17, that ðF ðk0 ; k1 ; . . .; kn ÞÞðlÞ is a finite extension of F ðk0 ; k1 ; . . .; kn Þ. Further, F ðk0 ; k1 ; . . .; kn Þ is a finite extension of F, so by 1.4.4, ðF ðk0 ; k1 ; ; kn ÞÞðlÞ is a finite extension of F. Now by 1.4.9, l is algebraic over F. ■ Definition Let a 2 C. Recall that Q C, and the field C is an extension of the field Q. If a is algebraic over Q, then we say that a is an algebraic number. A complex number that is not an algebraic number is called a transcendental number. ðÞ By 1.4.5, every rational number is an algebraic number. By 1.4.19, the collection of all algebraic numbers is a subfield of C. Thus if a is an algebraic number and b is an algebraic number, then all of a b; ab; ab1 ðprovided b is nonzeroÞ are algebraic numbers. 1.4.21 Problem Recall that Q C, and the field C is an extension of the field Q. Let A be the collection of all algebraic numbers. We know from 1.4.19 that Q A C, C is an extension of A, and A is an algebraic extension of Q. Let f ð xÞ be a nonzero member of A½ x, and let að2 CÞ be a root of the polynomial f ð xÞ, in the sense that f ðaÞ ¼ 0. Then a 2 A. In short, the roots of a polynomial whose coefficients are algebraic numbers.


65

Proof Suppose to the contrary that a 62 A. We seek a contradiction. Since C is an extension of A, a 2 C, f ð xÞ is a nonzero member of A½ x, and f ðaÞ ¼ 0, a is algebraic over A, and hence each member of A [ fag is algebraic over A. Observe that the field AðaÞ is an extension of A. Let B be the collection of all elements of AðaÞ that are algebraic over A. By 1.4.19, A B AðaÞ, B is a subfield of AðaÞ, and A is a subfield of B. Since each member of A [ fag is algebraic over A, we have A [ fag B, and hence AðaÞ B. Now, since B AðaÞ, we have B ¼ AðaÞ. Thus every element of AðaÞ is algebraic over A, and hence AðaÞ is an algebraic extension of A. Now, since A is an algebraic extension of Q, by 1.4.20, AðaÞ is an algebraic extension of Q. And since a 2 AðaÞ, a is algebraic over Q, and hence a is an algebraic number. Thus a 2 A. This is a contradiction. ■

1.5

Splitting Fields

1.5.1 Theorem The number e 1 þ number.

1 1!

þ

1 2!

þ

1 3!

þ is a transcendental

Proof (due to Hermite) Suppose to the contrary that e is not a transcendental number. We seek a contradiction. Since e is not a transcendental number, e is an algebraic number, and hence there exists a nonzero polynomial c0 þ c1 x þ c2 x2 þ þ cn xn such that 1. 2. 3. 4. 5.

each ci is an integer, c0 is a positive integer, n is a positive integer, c0 þ c1 e þ c2 e2 þ þ cn en ¼ 0, cn is a nonzero integer.

Let us take a polynomial f ð xÞ 2 R½ x, and let deg ðf ð xÞÞ ¼ r [ 1. It follows that the ðr þ 1Þth derivative f ðr þ 1Þ ð xÞ of f ð xÞ is the zero polynomial. Similarly, f ðr þ 2Þ ð xÞ is the zero polynomial, etc. Put F ð xÞ f ð xÞ þ f 0 ð xÞ þ f 00 ð xÞ þ þ f ðrÞ ð xÞ: By the mean value theorem, there exists a real number h1 2 ð0; 1Þ such that

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1 Galois Theory I

d ðex F ð xÞÞ e F ð 1Þ e F ð 0Þ ¼ ð 1 0Þ dx 1

0

; x¼h1

that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þðex F ð xÞ þ ex F 0 ð xÞÞjx¼h1 ; that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þeh1 ðF 0 ðh1 Þ F ðh1 ÞÞ; that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þeh1 f 0 ðh1 Þ þ f 00 ðh1 Þ þ þ f ðrÞ ðh1 Þ þ f ðr þ 1Þ ðh1 Þ f ðh1 Þ þ f 0 ðh1 Þ þ f 00 ðh1 Þ þ þ f ðrÞ ðh1 Þ ; that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þeh1 f ðr þ 1Þ ðh1 Þ f ðh1 Þ ; that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þeh1 ð0 f ðh1 ÞÞ; that is, e1 F ð1Þ F ð0Þ ¼ ð1 0Þe1h1 f ð1h1 Þ: Similarly, there exists a real number h2 2 ð0; 1Þ such that e2 F ð2Þ F ð0Þ ¼ ð2 0Þe2h2 f ð2h2 Þ: Also, there exists a real number h3 2 ð0; 1Þ such that e3 F ð3Þ F ð0Þ ¼ ð3 0Þe3h3 f ð3h3 Þ: .. . There exists a real number hn 2 ð0; 1Þ such that en F ðnÞ F ð0Þ ¼ ðn 0Þenhn f ðnhn Þ:

1.5 Splitting Fields

67

Thus F ð1Þ eF ð0Þ ¼ e1h1 f ð1h1 Þ; F ð2Þ e2 F ð0Þ ¼ 2e22h2 f ð2h2 Þ; F ð3Þ e3 F ð0Þ ¼ 3e33h3 f ð3h3 Þ; .. .

F ðnÞ en F ð0Þ ¼ nennhn f ðnhn Þ: It follows that c0 F ð0Þ þ c1 F ð1Þ þ c2 F ð2Þ þ þ cn F ðnÞ ¼ c0 F ð0Þ þ c1 eF ð0Þ e1h1 f ð1h1 Þ þ c2 e2 F ð0Þ 2e22h2 f ð2h2 Þ þ þ cn en F ð0Þ nennhn f ðnhn Þ ¼ c0 þ c1 e þ c2 e2 þ þ cn en F ð0Þ c1 e1h1 1f ðh1 Þ þ c2 2e22h2 f ð2h2 Þ þ þ cn nennhn f ðnhn Þ ¼ 0 F ð0Þ c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ ;

and hence c0 F ð0Þ þ c1 F ð1Þ þ c2 F ð2Þ þ þ cn F ðnÞ ¼ c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ :

ðÞ:

Let us take an arbitrary prime p such that 1 n\p, and 1 c0 \p. It follows that p divides neither the integer c0 nor n!, and hence p does not divide the integer c0 ðn!Þp . Next let us take f ð xÞ

1 xp1 ð1 xÞp ð2 xÞp ðn xÞp ð2 R½ xÞ: ðp 1Þ!

Here r ¼ degðf ð xÞÞ ¼ ðp 1Þ þ p þ p þ þ p ¼ ðn þ 1Þp 1: |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} n terms

Thus r ¼ ðn þ 1Þp 1. Observe that

68

1 Galois Theory I 1 f ð xÞ ¼ ðp1 xp1 ð1 xÞp ð2 xÞp ðn xÞp

Þ!

p p1 p p2 2 1 p1 p ¼ ðp1 x 1 1 x þ 1 x Þ!

1

2 p p p2 2 p p1

2 2 xþ 2 x

1

2 p p1 p p2 2 p

3 3 xþ 3 x 1 2 .. .

p p1 p p2 2 p

n n xþ n x 1 2 1 p1 p ¼ ðp1 ð1 2p np þ ðintegerÞx þ ðintegerÞx2 þ Þ Þ! x p integer integer integer ðn!Þ p1 ¼ ðp1 þ ðp1Þ! xp þ ðp1Þ! xp þ 1 þ þ ðp1Þ! xðn þ 1Þp1 ; Þ! x

so f ð xÞ ¼

anp1 ðn þ 1Þp1 ðn!Þp p1 a0 a1 x þ xp þ xp þ 1 þ þ x ; ðp 1Þ! ðp 1Þ! ðp 1Þ! ðp 1Þ!

where each ai is an integer. Now, ðn!Þp a0 a1 Dp xp1 þ D p ð xp Þ þ Dp xp þ 1 þ ðp 1Þ! ðp 1Þ! ðp 1Þ! ðn!Þp a0 a1 ðp þ 1Þ! xðp þ 1Þp 0þ p! þ ¼ ðp 1Þ! ðp 1Þ! ðp 1Þ! ððp þ 1Þ pÞ! a2 ðp þ 2Þ! xðp þ 2Þp þ þ ðp 1Þ! ððp þ 2Þ pÞ! a0 a1 ðp þ 1Þ! a2 ðp þ 2Þ! 2 xþ x þ p! þ ¼ ðp 1Þ! ðp 1Þ! 1! ðp 1Þ! 2! ðp þ 2Þðp þ 1Þp 2 x þ ¼ a0 p þ a1 ðp þ 1Þpx þ a2

2! pþ1 pþ2 ¼ a0 p þ a1 px þ a2 px2 þ ; 1 2

f ð p Þ ð xÞ ¼

so f

ð pÞ

ð x Þ ¼ a0 p þ a1

pþ1 pþ2 px þ a2 px2 þ : 1 2

Here we observe that each coefficient of f ð pÞ ð xÞ is an integer that is divisible by p.


69

Further, f ð p þ 1 Þ ð x Þ ¼ a1

pþ1 pþ2 pþ3 p þ a2 p2x þ a3 p3x2 þ : 1 2 3

Again, we observe that each coefficient of f ðp þ 1Þ ð xÞ is an integer that is divisible by p. Thus for every integer j, and for every integer i p, f ðiÞ ð jÞ is an integer that is divisible by p. [Before going ahead, let us recall the Leibniz rule of differentiation: ðuvÞ0 ¼ u0 v þ uv0 ; ðuvwÞ0 ¼ u0 vw þ uv0 w þ uvw0 ; ðuvwÞ00 ¼ ðu00 vw þ u0 v0 w þ u0 vw0 Þ þ ðu0 v0 w þ uv00 w þ uv0 w0 Þ þ ðu0 vw0 þ uv0 w0 þ uvw00 Þ;

ðuvwÞ00 ¼ u00 vw þ þ uv00 w þ uvw00 þ 2uv0 w0 þ 2u0 vw0 þ 2u0 v0 w ¼ u00 vw þ u0 ð2vw0 þ 2v0 wÞ þ uðv00 w þ vw00 þ 2v0 w0 Þ: Similarly, ðuvwÞðnÞ ¼

P

ðpositive integerÞuðiÞ vð jÞ wðkÞ ði; j; kÞ i; j; k are nonnegative integers iþjþk ¼ n ðnÞ ¼ u ð Þ þ uðn1Þ ð Þ þ uðn2Þ ð Þ þ ¼ vðnÞ ð Þ þ vðn1Þ ð Þ þ vðn2Þ þ ;

etc. Also, for every integer i 2 f0; 1; 2; . . .; p 2g, we have Di ðxp1 Þ x¼0 ¼ 0. Also Dp1 ðxp1 Þ ¼ ðp 1Þ!, and Dp ðxp1 Þ ¼ 0. Next, for every integer i 2 f0; 1; 2; . . .; p 1g; Di ðð1 xÞp Þjx¼1 ¼ 0. Also Dp ðð1 xÞp Þ ¼ ð1Þp ðp!Þ. Similarly, for every integer i 2 f0; 1; 2; ; p 1g, Di ðð2 xÞp Þjx¼2 ¼ 0. Also, p D ðð2 xÞp Þ ¼ ð1Þp ðp!Þ, etc.] Now, since f ð xÞ ¼

1 xp1 ð1 xÞp ð2 xÞp ðn xÞp ; ðp 1Þ!

70

1 Galois Theory I

we have f ðp1Þ ð1Þ ¼

1 ð0 þ 0 þ Þ ¼ 0:: ðp 1Þ!

Similarly, f ðp1Þ ð2Þ ¼ 0, f ðp1Þ ð3Þ ¼ 0, etc. Also, f ðp2Þ ð1Þ ¼ 0, f ðp2Þ ð2Þ ¼ 0, etc. In short, for every i 2 f0; 1; 2; . . .; p 1g, and for every j 2 f1; 2; ; ng, f i ð jÞ ¼ 0. Since f ð xÞ ¼

1 xp1 ð1 xÞp ð2 xÞp ðn xÞp ; ðp 1Þ!

we have f ðp1Þ ð0Þ ¼

1 ððp 1Þ!ð1 0Þp ð2 0Þp ðn 0Þp þ 0 þ 0 þ Þ ¼ ðn!Þp : ðp 1Þ!

Similarly, f ðp2Þ ð0Þ ¼ 0, f ðp3Þ ð0Þ ¼ 0, etc. In short, for every i 2 f0; 1; 2; . . .; p 2g, we have f ðiÞ ð0Þ ¼ 0, and f ðp1Þ ð0Þ ¼ ðn!Þp . Since F ð xÞ ¼ f ð xÞ þ f 0 ð xÞ þ f 00 ð xÞ þ þ f ðrÞ ð xÞ; we have, for every j 2 f1; 2; . . .; ng, F ð jÞ ¼ f ð jÞ þ f 0 ð jÞ þ f 00 ð jÞ þ þ f ðrÞ ð jÞ ¼ f ð jÞ þ f 0 ð jÞ þ f 00 ð jÞ þ þ f ððn þ 1Þp1Þ ð jÞ 0 ¼ f ð jÞ þ f ð jÞ þ f 00 ð jÞ þ þ f ðp1Þ ð jÞ þ f p ð jÞ þ þ f ððn þ 1Þp1Þ ð jÞ ¼ f ð jÞ þ f 0 ð jÞ þ f 00 ð jÞ þ þ f ðp1Þ ð jÞ þ p ðintegerÞ ¼ 0 þ 0 þ 0 þ þ 0 þ p ðintegerÞ ¼ p ðintegerÞ; and hence for every j 2 f1; 2; . . .; ng; F ð jÞ is an integer that is a multiple of p. Since F ð xÞ ¼ f ð xÞ þ f 0 ð xÞ þ f 00 ð xÞ þ þ f ðrÞ ð xÞ;


71

we have F ð0Þ ¼ f ð0Þ þ f 0 ð0Þ þ f 00 ð0Þ þ þ f ðrÞ ð0Þ ¼ f ð0Þ þ f 0 ð0Þ þ f 00 ð0Þ þ þ f ððn þ 1Þp1Þ ð0Þ ¼ f ð0Þ þ f 0 ð0Þ þ f 00 ð0Þ þ þ f ðp2Þ ð0Þ þ f p1 ð0Þ þ f p ð0Þ þ þ f ððn þ 1Þp1Þ ð0Þ ¼ f ð0Þ þ f 0 ð0Þ þ f 00 ð0Þ þ þ f ðp2Þ ð0Þ þ f p1 ð0Þ þ p ðintegerÞ ¼ 0 þ 0 þ 0 þ þ 0 þ ðn!Þp þ p ðintegerÞ ¼ ðn!Þp þ p ðintegerÞ; and hence F ð0Þ is an integer of the form ðn!Þp þ p ðintegerÞ. Since for every j 2 f1; 2; ; ng; F ð jÞ is a multiple of p, F ð0Þ is of the form ðn!Þp þ p ðintegerÞ, and each ci is an integer, it follows that c0 F ð0Þ þ c1 F ð1Þ þ c2 F ð2Þ þ þ cn F ðnÞ ¼ c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ is an integer of the form c0 ðn!Þp þ p ðintegerÞ. Thus c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ is an integer of the form c0 ðn!Þp þ p ðintegerÞ. Observe that 1e1ð1h1 Þ f ð1h1 Þ ¼ 1e1ð1h1 Þ

1 ð1h1 Þp1 ð1 1h1 Þp ð2 1h1 Þp ðn 1h1 Þp ; ðp 1Þ!

so 1ð1h Þ 1e 1 f ð1h1 Þ ¼ 1e1ð1h1 Þ ¼

p1 1 ð1 1h1 Þp ð2 1h1 Þp ðn ðp1Þ! ð1h1 Þ p1 1 1e1ð1h1 Þ ðp1 ðj1 1h1 jj2 1h1 j jn 1h1 jÞp : Þ! ð1h1 Þ

Now, since h1 2 ð0; 1Þ, we have j1 1h1 jj2 1h1 j jn 1h1 j 1 2 n ¼ n!;

1h1 Þp

72

1 Galois Theory I

and hence 1ð1h Þ 1e 1 f ð1h1 Þ 1e1ð1h1 Þ 1 p 1e1ð1h1 Þ ðp1 Þ! n

p1 1 1 p1 ðn!Þp 1e1ð1h1 Þ ðp1 ðn!Þp ðp1Þ! ð1h1 Þ Þ! n n!ÞÞp1 1ð1h1 Þ ðn!Þp ¼ 1e1ð1h1 Þ ðnðn!ÞÞ ðnððp1 ðnðn!ÞÞ 0 Þ! ! 1e

as p ! 1. Thus 1e1ð1h1 Þ f ð1h1 Þ ! 0 as p ! 1. Since 2e2ð1h2 Þ f ð2h2 Þ ¼ 2e2ð1h2 Þ

1 ð2h2 Þp1 ð1 2h2 Þp ð2 2h2 Þp ðn 2h2 Þp ; ðp 1Þ!

we have 2ð1h Þ 2e 2 f ð2h2 Þ ¼ 2e2ð1h2 Þ ¼

p1 1 ð1 2h2 Þp ð2 2h2 Þp ðn ðp1Þ! ð2h2 Þ p1 1 ðj1 2h2 jj2 2h2 j jn 2h2 jÞp : 2e2ð1h2 Þ ðp1 Þ! ð2h2 Þ

2h2 Þp

Now, since h2 2 ð0; 1Þ, we have j1 2h2 jj2 2h2 j jn 2h2 j 1 2 n ¼ n!; and hence 2ð1h Þ 2 2e f ð2h2 Þ 2e2ð1h2 Þ 2e2ð1h2 Þ

1 1 np1 ðn!Þp ð2h2 Þp1 ðn!Þp 2e2ð1h2 Þ ðp 1Þ! ðp 1Þ!

1 ðnðn!ÞÞp1 np ðn!Þp ¼ 2e2ð1h2 Þ ðnðn!ÞÞ ! 2e2ð1h2 Þ ðnðn!ÞÞ 0 ðp 1Þ! ðp 1Þ!

as p ! 1. Thus 2e2ð1h2 Þ f ð2h2 Þ ! 0 as p ! 1. Similarly, 3e3ð1h3 Þ f ð3h3 Þ ! 0 as p ! 1, etc. It follows that c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ ! 0 as p ! 1: Since c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ is an integer of the form c0 ðn!Þp þ p ðintegerÞ and p does not divide the integer c0 ðn!Þp , it follows that


73

c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ is a nonzero integer, and hence c1 1e1ð1h1 Þ f ð1h1 Þ þ c2 2e2ð1h2 Þ f ð2h2 Þ þ þ cn nenð1hn Þ f ðnhn Þ 90 as p ! 1: ■

This is a contradiction.

Definition Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let a 2 K. If ðK3Þf ðaÞ ¼ 0, then we say that a is a root of f ð xÞ. 1.5.2 Theorem Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let a 2 K. Then there exists a nonzero qð xÞ 2 K ½ x such that 1. f ð xÞ ¼ ðx aÞqð xÞ þ f ðaÞ, 2. degðqð xÞÞ ¼ degðf ð xÞÞ 1. This theorem is known as the remainder theorem. Proof Since F K, we have ðf ð xÞ 2ÞF ½ x K ½ x, and hence f ð xÞ 2 K ½ x. It is given that f ð xÞ is nonzero. Since 1; a 2 K, the polynomial x a is a nonzero member of K ½ x. Now, by 1.2.14, there exist qð xÞ; r ð xÞ 2 K ½ x such that f ð xÞ ¼ qð xÞðx aÞ þ r ð xÞ; and ðeither r ð xÞ ¼ 0 or degðr ð xÞÞ\degðx aÞð¼ 1ÞÞ. It follows that either r ð xÞ ¼ 0 or degðr ð xÞÞ ¼ 0. Since r ð xÞ 2 K ½ x, r ð xÞ is a member of K, and hence r ð xÞ ¼ r ðaÞ 2 K. Since f ð xÞ ¼ qð xÞðx aÞ þ r ð xÞ; we have f ðaÞ ¼ qðaÞða aÞ þ r ðaÞ ¼ qðaÞ 0 þ r ðaÞ ¼ r ðaÞ ¼ r ð xÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence f ðaÞ ¼ r ð xÞ. Thus f ð xÞ ¼ qð xÞðx aÞ þ f ðaÞ. This proves (1). Since f ðaÞ 2 K, we have degðf ð xÞÞ ¼ degðqð xÞðx aÞ þ f ðaÞÞ ¼ degðqð xÞðx aÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ degðqð xÞÞ þ degðx aÞ ¼ degðqð xÞÞ þ 1; and hence degðf ð xÞÞ ¼ degðqð xÞÞ þ 1.

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1 Galois Theory I

1.5.3 Theorem Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let a 2 K. Let a be a root of f ð xÞ. Then ðx aÞjf ð xÞ in K ½ x. Proof By 1.5.2, there exists a nonzero qð xÞ 2 K ½ x such that 1. f ð xÞ ¼ ðx aÞqð xÞ þ f ðaÞ, 2. degðqð xÞÞ ¼ degðf ð xÞÞ 1. Since a is a root of f ð xÞ, we have f ð xÞ ðx aÞqð xÞ ¼ f ðaÞ ¼ 0 ; |fflfflfflfflffl{zfflfflfflfflffl} and hence f ð xÞ ðx aÞqð xÞ ¼ 0, that is, f ð xÞ ¼ ðx aÞqð xÞ. Since F K, we have ðf ð xÞ 2ÞF ½ x K ½ x, and hence f ð xÞ 2 K ½ x. Since a 2 K, we have ðx aÞ 2 K ½ x. Also, qð xÞ 2 K ½ x. Next, since f ð xÞ ¼ ðx aÞqð xÞ, it follows that ðx aÞjf ð xÞ in K ½ x. ■ Definition Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let a 2 K. Let m be a positive integer. If ðx aÞm jf ð xÞ in K ½ x, then clearly, f ðaÞ ¼ 0, and hence a is a root of f ð xÞ. If ðx aÞm jf ð xÞ in K ½ x and ðx aÞm þ 1 -f ð xÞ in K ½ x, then we say that a is a root of f ð xÞ of multiplicity m. Caution We count a as m roots. 1.5.4 Theorem Let F and K be any fields such that F K. Suppose that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Suppose that degðf ð xÞÞ ¼ n. Then the number of roots of f ð xÞ in K is n. Proof (Induction on nÞ If f ð xÞ has no root in K, then the number of roots of f ð xÞ in K is 0, and hence the result is trivially true. So we consider the case that there exists a root of f ð xÞ in K. Suppose that degðf ð xÞÞ ¼ 1. We can suppose that f ð xÞ a þ bx, where a; b 2 F and b 6¼ 0. Next, let a; b 2 K such that a þ ba ¼ 0 : a þ bb ¼ 0 We shall show that a ¼ b. Since a þ ba ¼ 0 and b 6¼ 0, we have a ¼ b1 a. Similarly, b ¼ b1 a. It follows that a ¼ b. Thus the result is true for n ¼ 1. Now let us suppose that the result is true for all positive integer values \n. It suffices to show that the result is true for n. Let a be a root of f ð xÞ in K, and let mð 1Þ be its multiplicity. Hence ðx aÞm jf ð xÞ in K ½ x, and ðx aÞm þ 1 -f ð xÞ in K ½ x. It follows that there exists


75

gð xÞ 2 K ½ x such that f ð xÞ ¼ ðx aÞm gð xÞ f ð xÞ ¼ ðx aÞm gð xÞ, we have

and

ðx aÞ-gð xÞ.

Since

n ¼ degðf ð xÞÞ ¼ degððx aÞm gð xÞÞ ¼ degððx aÞm Þ þ degðgð xÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ m þ degðgð xÞÞ 1 þ degðgð xÞÞ [ degðgð xÞÞ; and hence degðgð xÞÞ\n. By the induction hypothesis, the number of roots of gð xÞ in K is degðgð xÞÞ. Since f ð xÞ ¼ ðx aÞm gð xÞ; the number of roots of f ð xÞ in K is equal to m þ ðthe number of roots of gð xÞin K Þð m þ degðgð xÞÞ ¼ nÞ; and hence the number of roots of f ð xÞ in K is n.

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1.5.5 Note Let F and K be any fields such that K is an extension of F. Let pð xÞ be a nonzero member of F ½ x with degðpð xÞÞ 1. Suppose that degðpð xÞÞ ¼ n. Let pð xÞ be irreducible over F. By 1.2.24, the quotient ring FV½x is a field, where V denotes the ideal ðpð xÞÞð¼ ff ð xÞpð xÞ : f ð xÞ 2 F ½ xgÞ. Let w : a 7! ða þ V Þ be a mapping from the field F to the field FV½x. It is clear that w is a ring isomorphism: 1. w : F ! FV½x is one-to-one: To prove this, let wðaÞ ¼ wðbÞ. We have to show that a ¼ b. Since a þ V ¼ wðaÞ ¼ wðbÞ ¼ b þ V, we have a þ V ¼ b þ V, and hence |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ða bÞ 2 V. Since each nonzero element of ð ð a bÞ 2 V ¼ Þ ff ð xÞpð xÞ : f ð xÞ 2 F ½ xg is of degree degðpð xÞÞð 1Þ, we have a b ¼ 0 or degða bÞ 1. Since ða bÞ 2 V, either a b ¼ 0 or deg ða bÞ ¼ 0. It follows that a b ¼ 0, that is, a ¼ b. 2. It is clear that w is a ring homomorphism. Thus we have shown that w : F ! FV½x is a ring isomorphism from the field F to the

field FV½x. It follows that we can identify each element a of F with wðaÞ ð¼ ða þ V ÞÞ

of the field FV½x. It is in this sense that we write F FV½x and treat FV½x as an extension of F. Since degðpð xÞÞ ¼ n, by 1.4.10, n is the dimension of the vector space FV½x, and h i

hence FV½x : F ¼ n. Also, by 1.4.10, 1 þ V; x þ V; x2 þ V; ; xn1 þ V is a

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basis of

F ½ x V .

quotient ring

From the definition of addition and scalar multiplication over the F ½ x V ,

it is clear that

F ½ x ; pðx þ V Þ ¼ pð xÞ þ V ¼ pð xÞ þ ðpð xÞÞ ¼ ðpð xÞÞ ¼ V ¼ 0 þ V 2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} V and hence pðx þ V Þ ¼ 0 þ V. Here pðx þ V Þ ¼ 0 þ V, x þ V is a member of and 0 þ V is the zero element of the field polynomial. Thus pð xÞ has a root in

F ½ x V .

F ½ x V ,

F ½ x V ,

so x þ V is a root of the given

1.5.6 Conclusion Let F and K be any fields such that K is an extension of F. Let pð xÞ be a nonzero member of F ½ x with degðpð xÞÞ 1. Suppose that degðpð xÞÞ ¼ n. Let pð xÞ be irreducible over F. Then there exists a field E such that 1. E is an extension of F, 2. ½E : F ¼ n, 3. pð xÞ has a root in E. 1.5.7 Problem Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Then there exists a field E such that 1. E is a finite extension of F, 2. ½E : F degðf ð xÞÞ, 3. f ð xÞ has a root in E. Proof Since degðf ð xÞÞ 1, f ð xÞ is not a unit in F ½ x, and hence by 1.2.20, there exists an irreducible pð xÞ 2 F ½ x such that 1 degðpð xÞÞ degðf ð xÞÞ and pð xÞjf ð xÞ. It follows, by 1.5.6, that there exists a field E such that 1. E is an extension of F, 2. ½E : F ¼ degðpð xÞÞð degðf ð xÞÞ\1Þ, 3. pð xÞ has a root in E. Since ½E : F \1, E is a finite extension of F. Also ½E : F degðf ð xÞÞ. Since pð xÞ has a root, say a, in E, and pð xÞjf ð xÞ, a is also a root of f ð xÞ. ■ 1.5.8 Note Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x and degðf ð xÞÞ 1. Let degðf ð xÞÞ ¼ n. By 1.5.8, there exists a field E1 such that 1. E1 is a finite extension of F, 2. ½E1 : F n, 3. f ð xÞ has a root, say a1 , in E1 . It follows, by 1.5.3, that ðx a1 Þjf ð xÞ in E1 ½ x, and hence there exists f1 ð xÞ 2 E1 ½ x such that f ð xÞ ¼ ðx a1 Þf1 ð xÞ and degðf1 ð xÞÞ ¼ n 1.


77

By 1.5.9, there exists a field E1 such that 1. E2 is a finite extension of E1 , 2. ½E2 : E1 n 1, 3. f1 ð xÞ has a root, say a2 , in E2 . It follows, by 1.5.3, that ðx a2 Þjf1 ð xÞ in E2 ½ x, and hence there exists f2 ð xÞ 2 E2 ½ x such that f1 ð xÞ ¼ ðx a2 Þf2 ð xÞ and degðf2 ð xÞÞ ¼ ðn 1Þ 1ð¼ n 2Þ. It follows, by 1.4.3, that E2 is a finite extension of F, and ½E2 : F ¼ ½E2 : E1 ½E1 : F ½E2 : E1 n ðn 1Þn ¼ nðn 1Þ: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus ½E2 : F nðn 1Þ. Also, f ð xÞ ¼ ðx a1 Þf1 ð xÞ ¼ f ð xÞ ¼ ðx a1 Þðx a2 Þf2 ð xÞ; so the field E2 contains two roots, a1 ; a2 of f ð xÞ. Similarly, the field E3 contains three roots of f ð xÞ, ½E3 : F nðn 1Þðn 2Þ, and E3 is a finite extension of F. Finally, there exists a field E such that 1. E is a finite extension of F, 2. E contains all the roots of f ð xÞ in K, 3. ½E : F nðn 1Þðn 2Þ 2 1ð¼n!Þ. 1.5.9 Conclusion Let F and K be any fields such that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let degðf ð xÞÞ ¼ n. Suppose that K contains n roots of f ð xÞ. Then there exists a field E such that 1. 2. 3. 4.

E is a finite extension of F, E contains all the roots of f ð xÞ in K, if G is a proper subfield of E, then G does not contain all the roots of f ð xÞ in K, ½E : F n!.

Definition Let F and K be any fields such that F K. Suppose that K is an extension of F. Let f ð xÞ be a nonzero member of F ½ x with degðf ð xÞÞ 1. Let degðf ð xÞÞ ¼ n. Suppose that K contains n roots of f ð xÞ. Let E be a field such that 1. E is a finite extension of F, 2. E contains all the roots of f ð xÞ in K, 3. if G is a proper subfield of E that contains F, then G does not contain all the roots of f ð xÞ in K. Then we say that E is a splitting field over F for f ð xÞ.

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Thus a field E is a splitting field over F for f ð xÞ if and only if E is a minimal finite extension of F in which f ð xÞ can be factored as a product of linear factors in E ½ x. From 1.5.9, ½ðsplitting field over F for f ð xÞÞ : F ðdegðf ð xÞÞÞ!: 1.5.10 Problem Let F; F 0 be any fields. Let s : a 7! a0 be a ring isomorphism from F onto F 0 . Then the map s : a0 þ a1 x þ a2 x2 þ þ an xn 7! a00 þ a01 t þ a02 t2 þ þ a0n tn from the polynomial ring F ½ x to the polynomial ring F 0 ½t is a ring isomorphism from F ½ x onto F 0 ½t such that for every a 2 F, we have s ðaÞ ¼ a0 . Proof s : F ½ x ! F 0 ½t is one-to-one: To show this, suppose that a00 þ a01 t þ a02 t2 þ þ a0n tn ¼ b00 þ b01 t þ b02 t2 þ þ b0n tn : We have to show that ai ¼ bi ði ¼ 0; 1; . . .; nÞ. Since a00 þ a01 t þ a02 t2 þ þ a0n tn ¼ b00 þ b01 t þ b02 t2 þ þ b0n tn ; we have a0i ¼ b0i ði ¼ 0; 1; . . .; nÞ, and hence sðai Þ ¼ sðbi Þði ¼ 0; 1; . . .; nÞ. Since s : F ! F 0 is a ring isomorphism, s : F ! F 0 is one-to-one, and since sðai Þ ¼ sðbi Þði ¼ 0; 1; . . .; nÞ, we have ai ¼ bi ði ¼ 0; 1; . . .; nÞ. s : F ½ x ! F 0 ½t is onto: This is clear. s : F ½ x ! F 0 ½t is a ring homomorphism: This is clear. Thus, s is a ring isomorphism from F ½ x onto F 0 ½t. Also, it is clear that for every a 2 F, s ðaÞ ¼ a0 . ■ 1.5.11 Problem Let F; F 0 be any fields. Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Thus s : f ð xÞ 7! f 0 ðtÞ from the ring F ½ x onto the ring F 0 ½t is an isomorphism. Let pð xÞ 2 F ½ x. It follows that p0 ðtÞ 2 F 0 ½t. Put V ðpð xÞÞ, where ðpð xÞÞ denotes the ideal generated by pð xÞ in F ½ x. Put V 0 ðp0 ðtÞÞ, where ðp0 ðtÞÞ denotes the ideal generated by p0 ðtÞ in F 0 ½t. Let s : f ð xÞ þ V 7! f 0 ðtÞ þ V 0 be the mapping from the quotient ring 0

F ½ x V

to the quotient ring

F 0 ½t V0 .

Then s is an

isomorphism from FV½x onto FV½0t. Also, for every a 2 F, we have s ða þ V Þ ¼ a0 and s ðx þ V Þ ¼ t þ V 0 : 0

Proof s : FV½x ! FV½0t is well defined. To show this, let f ð xÞ; gð xÞ 2 F ½ x be such that f ð xÞ gð xÞ 2 V ð¼ ðpð xÞÞÞ. We have to show that f 0 ðtÞ g0 ðtÞ 2 V 0 . Since


79

f ð xÞ gð xÞ 2 ðpð xÞÞ, there exists hð xÞ 2 F ½ x such that f ð xÞ gð xÞ ¼ pð xÞhð xÞ, and hence f 0 ðtÞ g0 ðtÞ ¼ s ðf ð xÞÞ s ðgð xÞÞ ¼ s ðf ðxÞ gð xÞÞ ¼ s ðpð xÞhð xÞÞ ¼ s ðpð xÞÞs ðhð xÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ p0 ðtÞs ðhð xÞÞ ¼ p0 ðtÞh0 ðtÞ 2 ðp0 ðtÞÞ ¼ V 0 :

Thus f 0 ðtÞ g0 ðtÞ 2 V 0 . 0 s : FV½x ! FV½0t is one-to-one. To show this, let f ð xÞ; gð xÞ 2 F ½ x be such that f 0 ðtÞ g0 ðtÞ 2 V 0 ð¼ ðp0 ðtÞÞÞ. We have to show that f ð xÞ gð xÞ 2 V. Since f 0 ðtÞ g0 ðtÞ 2 ðp0 ðtÞÞ, there exists hð xÞ 2 F ½ x such that s ð f ð x Þ gð x Þ Þ ¼ s ð f ð x Þ Þ s ð gð x Þ Þ ¼ f 0 ðtÞ g0 ðtÞ ¼ p0 ðtÞh0 ðtÞ ¼ s ðpð xÞÞs ðhð xÞÞ ¼ s ðpð xÞhð xÞÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence s ðf ð xÞ gð xÞÞ ¼ s ðpð xÞhð xÞÞ: Since s is one-to-one, we have f ð xÞ gð xÞ ¼ pð xÞhð xÞ 2 ðpð xÞÞ ¼ V, and hence f ð xÞ gð xÞ 2 V. 0 s : FV½x ! FV½0t is onto. This is clear. 0

s : FV½x ! FV½0t is a ring homomorphism. This is clear. 0

Thus s is an isomorphism from FV½x onto FV½0t. By the definition of s , for every a 2 F, s ða þ V Þ ¼ s ðaÞ þ V 0 ¼ a0 þ V 0 , so |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} for every a 2 F, s ða þ V Þ ¼ a0 þ V 0 . Since for every a 2 F, we have a0 2 F 0 , and we identify ðs ða þ V Þ ¼Þa0 þ V 0 with a0 , we can write s ða þ V Þ ¼ a0 . Since the polynomial x is a member of F ½ x, we have s ðx þ V Þ ¼ t þ V 0 :

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1.5.12 Problem Let F and K be any fields such that F K. Suppose that K is an extension of F. Let a be a member of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let n be a positive integer. Let n be the degree of pð xÞ. Let a be a root of pð xÞ in K. Then 1. a is algebraic of degree n over F. 1 2. leading coefficient of pðxÞ pð xÞ is the minimal polynomial of a over F.

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Proof Put p1 ð x Þ

1 pð xÞ: leading coefficient of pð xÞ

Clearly, ðK3Þp1 ðaÞ ¼ 0, 1 n ¼ degðp1 ð xÞÞ, and the leading coefficient of p1 ð xÞ is 1. Let a be algebraic of degree m over F. It follows that m n. We have to show that m ¼ n. Suppose to the contrary that m\n. We seek a contradiction. Since a is algebraic of degree m over F, there exists f ð xÞ 2 F ½ x such that ðK3Þ f ðaÞ ¼ 0, 1 degðf ð xÞÞ ¼ m \n ¼ degðp1 ð xÞÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and the leading coefficient of f ð xÞ is 1. It follows that there exist qð xÞ; r ð xÞ 2 F ½ x such that p1 ð xÞ ¼ f ð xÞqð xÞ þ r ð xÞ and ðr ð xÞ ¼ 0 or degðr ð xÞÞ\degðf ð xÞÞÞ: Since degðf ð xÞÞ\ degðp1 ð xÞÞ, p1 ð xÞ ¼ f ð xÞqð xÞ þ r ð xÞ, and p1 ð xÞ is irreducible over F, we have r ð xÞ 6¼ 0. Also 0 ¼ p1 ðaÞ ¼ f ðaÞqðaÞ þ r ðaÞ ¼ 0qðaÞ þ r ðaÞ ¼ r ðaÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so r ðaÞ ¼ 0. Since a is algebraic of degree m over F, r ð xÞ 6¼ 0, and r ðaÞ ¼ 0, we have ðdegðf ð xÞÞ ¼Þm degðr ð xÞÞ. Since ðr ð xÞ ¼ 0 or degðr ð xÞÞ\degðf ð xÞÞÞ, we have r ð xÞ ¼ 0. This is a contradiction. Thus a is algebraic of degree n over F. Also 1 pð x Þ leading coefficient of pð xÞ is the minimal polynomial of a over F.

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1.5.13 Note Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let n be a positive integer. Let n be the degree of pð xÞ. Let a be a root of pð xÞ in K. By 1.5.12,


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1. a is algebraic of degree n over F. 1 2. leading coefficient of pðxÞ pð xÞ is the minimal polynomial of a over F. Let w : f ð xÞ 7! f ðaÞ be the mapping from the ring F ½ x to the field F ðaÞ. By 1.4.13, w : F ½ x ! F ðaÞ is a ring homomorphism, and hence by the fundamental theorem of ring homomorphism, the mapping w : ðf ð xÞ þ ker ðwÞÞ 7! F ½ x to F ðaÞ. Also, wðf ð xÞÞð¼ f ðaÞÞ is a ring isomorphism from the quotient ring ker ðwÞ F ½ x w maps ker onto F ðaÞ. Put ðwÞ

p1 ð x Þ

1 pð xÞ: leading coefficient of pð xÞ

By 1.4.12, p1 ð xÞ is irreducible over F, and hence by 1.2.22, the ideal fp1 ð xÞf ð xÞ : f ð xÞ 2 F ½ xg n o 1 ¼ leadingcoefficientof p ð x Þf ð x Þ : f ð x Þ 2 F ½ x pð xÞ : 1 ¼ pð x Þ f ð xÞ : f ð xÞ 2 F ½ x leadingcoefficientof pð xÞ

ð p1 ð x Þ Þ

¼ fpð xÞ f ð xÞ : f ð xÞ 2 F ½ xg ¼ ðpð xÞÞÞ is a maximal ideal of the ring F ½ x, and hence ðpð xÞÞ is a maximal ideal of the ring F ½ x. Further, by Note 1.2.24, the quotient ring

F ½ x ðp1 ðxÞÞ

¼ ðpFð½xxÞÞ

is a field. Also,

ff ðaÞ : f ð xÞ 2 F ½ xg F ðaÞ. By 1.4.13, ker ðwÞ ¼ ðp1 ð xÞÞ ¼ ðpð xÞÞ: |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Now, since ker ðwÞ ¼ ðpð xÞÞ, it follows that w : ðf ð xÞ þ ker ðwÞÞ 7! f ðaÞ is a ring isomorphism from the field ðpFð½xxÞÞ onto the field F ðaÞ. Further, for every b 2 F, w ðb þ ðpð xÞÞÞ ¼ w ðb þ ker ðwÞÞ ¼ wðbÞ ¼ b |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and w ðx þ ðpð xÞÞÞ ¼ wð xÞ ¼ a: 1.5.14 Conclusion Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let a be a

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root of pð xÞ in K. Let w : f ð xÞ 7! f ðaÞ be the mapping from the ring F ½ x to the field F ðaÞ. Then 1. w : ðf ð xÞ þ ker ðwÞÞ 7! f ðaÞ is a ring isomorphism from the field field F ðaÞ, 2. for every b 2 F, w ðb þ ðpð xÞÞÞ ¼ b, 3. w ðx þ ðpð xÞÞÞ ¼ a.

F ½ x ðpðxÞÞ

onto the

In short, w is an isomorphism from ðpFð½xxÞÞ onto F ðaÞ such that every element of F is fixed, and the w -image of x þ ðpð xÞÞ is a. 1.5.15 Note Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let a be a root of pð xÞ in K. Let F 0 be any field. Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. We know that s : f ð xÞ 7! f 0 ðtÞ is an isomorphism from the ring F ½ x onto the ring F 0 ½t. Since pð xÞ 2 F ½ x, we have p0 ðtÞ 2 F 0 ½t. Let b be a root of p0 ðtÞ in some extension K 0 of F 0 . Suppose that ðpð xÞÞ denotes the ideal generated by pð xÞ in F ½ x. Suppose that ðp0 ðtÞÞ denotes the ideal generated by p0 ðtÞ in F 0 ½t. Let s : f ð xÞ þ ðpð xÞÞ 7! f 0 ðtÞ þ ðp0 ðtÞÞ 0

be a mapping from the quotient ring ðpFð½xxÞÞ to the quotient ring ðpF0 ð½ttÞÞ. By 1.5.11, s is an isomorphism from ðpFð½xxÞÞ onto s ða þ ðpð xÞÞÞ ¼ a0 þ ðp0 ðtÞÞ and

F 0 ½t ðp0 ðtÞÞ.

Also, for every a 2 F, we have

s ðx þ ðpð xÞÞÞ ¼ t þ ðp0 ðtÞÞ: Let w : f ð xÞ 7! f ðaÞ be a mapping from the ring F ½ x to the field F ðaÞ. Then by 1.5.14, 1. w : ðf ð xÞ þ ðpð xÞÞÞ 7! f ðaÞ is a ring isomorphism from the field field F ðaÞ, 2. for every a 2 F, w ða þ ðpð xÞÞÞ ¼ a, 3. w ðx þ ðpð xÞÞÞ ¼ a.

F ½ x ðpðxÞÞ

onto the

Let h : f 0 ðtÞ 7! f 0 ðbÞ be a mapping from the ring F 0 ½t to the field F 0 ðbÞ. Then by 1.5.14, 1. h : ðf 0 ðtÞ þ ðp0 ðtÞÞÞ 7! f 0 ðbÞ is a ring isomorphism from the field field F 0 ðbÞ,

F 0 ½t ðp0 ðtÞÞ

onto the


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2. for every b 2 F 0 , h ðb þ ðp0 ðtÞÞÞ ¼ b, 3. h ðt þ ðp0 ðtÞÞÞ ¼ b. Since w is a ring isomorphism from phism from F ðaÞ onto ðpFð½xxÞÞ. Now, since s

F ½ x ðpð xÞÞ

onto F ðaÞ, ðw Þ1 is a ring isomor0

is an isomorphism from ðpFð½xxÞÞ onto ðpF0 ð½ttÞÞ, 0

and h is a ring isomorphism from ðpF0 ð½ttÞÞ onto F 0 ðbÞ, the composite h degs deg ðw Þ1 is an isomorphism from F ðaÞ onto F 0 ðbÞ. For every a 2 F,

h s ðw Þ1 ðaÞ ¼ h s ðw Þ1 ðaÞ ¼ h ðs ða þ ðpð xÞÞÞÞ ¼ h ða0 þ ðp0 ðtÞÞÞ ¼ a0 ;

so for every a 2 F, we have rðaÞ ¼ a0 , where r h s ðw Þ1 . Next, rðaÞ ¼ h s ðw Þ1 ðaÞ ¼ h s ðw Þ1 ðaÞ ¼ h ðs ðx þ ðpð xÞÞÞÞ ¼ h ðt þ ðp0 ðtÞÞÞ ¼ b; so rðaÞ ¼ b. 1.5.16 Conclusion Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let a be a root of pð xÞ in K. Let F 0 be any field. Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let b be a root of p0 ðtÞ in some extension K 0 of F 0 . Then there exists an isomorphism r from the field F ðaÞ onto the field F 0 ðbÞ such that 1. rðaÞ ¼ b, 2. for every a 2 F, rðaÞ ¼ a0 . 1.5.17 Note In 1.5.16, let us take F for F 0 and the identity map i : F ! F for s. Thus for every a 2 F, a0 means a, and for every f ð xÞ 2 F ½ x, f 0 ðtÞ means f ðtÞ. Also, p0 ðtÞ means pðtÞ. Thus a; b are the roots of the same polynomial pð xÞ. By 1.5.16, there exists an isomorphism r from the field F ðaÞ onto the field F ðbÞ such that 1. rðaÞ ¼ b, 2. for every a 2 F, rðaÞ ¼ a. 1.5.18 Conclusion Let F and K be any fields such that K is an extension of F. Let a; b be members of K. Let pð xÞ 2 F ½ xð F Þ. Let pð xÞ be irreducible over F. Let

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a; b be any roots of pð xÞ in K. Then there exists an isomorphism r from the field F ðaÞ onto the field F ðbÞ such that 1. rðaÞ ¼ b, 2. for every a 2 F, rðaÞ ¼ a. 1.5.19 Example Let F be the field of all rational numbers, and let K be the field of all complex numbers. Let us take the polynomial x4 þ x2 þ 1 for f ð xÞ in F ½ x. According to 1.5.9, ½ðsplitting field over F for f ð xÞÞ : F ðdegðf ð xÞÞÞ!; so

splitting field over F for x4 þ x2 þ 1 : F deg x4 þ x2 þ 1 ! ¼ 4! ¼ 24;

and hence 1

splitting field over F for x4 þ x2 þ 1 : F 24:

Since 2 x4 þ x2 þ 1 ¼ x2 þ 1 x2 ¼ x2 þ 1 þ x x2 þ 1 x ¼ ðx xÞ x x2 ðx þ xÞ x þ x2 ; pffiffi 3 4 2 þ i where x 1 2 2 , F ðxÞ is a splitting field over F for x þ x þ 1. Since the 2 2 polynomial 1 þ x þ x is a member of F ½ x, 1 þ x þ x is irreducible over F, degð1 þ x þ x2 Þ ¼ 2, and x is a root of 1 þ x þ x2 in K, by 1.5.12, x is algebraic of degree 2 over F, and hence by 1.4.16, ½F ðxÞ : F ¼ 2. Thus

splitting field over F for x4 þ x2 þ 1 : F ¼ 2:

1.5.20 Example Let F be the field of all rational numbers, and let K be the field of all complex numbers. Let us take the polynomial x3 2 for f ð xÞ in F ½ x. According to 1.5.9, ½ðsplitting field over F for f ð xÞÞ : F ðdeg ðf ð xÞÞÞ!; so

splitting field over F for x3 2 : F deg x3 2 ! ¼ 3! ¼ 6;


85

and hence 1

splitting field over F for x3 2 : F 6:

Observe that p p p ffiffiffi ffiffiffi ffiffiffi 3 3 3 x3 2 ¼ x 2 x 2x x 2x2 ; pffiffi 3 þ i x3 2 is a member of F ½ x, x3 2 is where x 1 2 2 . Since the polynomial p pffiffiffi ffiffi ffi 3 irreducible over F, degðx3 2Þ ¼ 3, and 2 is a root of x3 2 in K, by 1.5.12, 3 2 pffiffiffi is algebraic of degree 3 over F, and hence by 1.4.16, F 3 2 : F ¼ 3. By 1.4.4,

h pffiffiffi i 3 F 2 : F j splitting field over F for x3 2 : F : Now, since 6, we have

ffiffiffi p F 3 2 : F ¼ 3, and 1 ½ðsplitting field over F for x3 2Þ : F

splitting field over F for x3 2 : F ¼ 3 or 6: ðÞ

Since members of F

ffiffiffi p 3 2 are real numbers,

p p p ffiffiffi ffiffiffi ffiffiffi 3 3 3 x3 2 ¼ x 2 x 2x x 2x2 ; pffiffiffi pffiffiffi pffiffiffi is not a splitting and 3 2x; 3 2x2 are not real numbers, F 3 2 3 field over F for x 2, and hence 3\½ðsplitting field over F for x3 2Þ : F . It follows from ðÞ that splitting field over F for x3 2 : F ¼ 6: 1.5.21 Example Let F be the field of all rational numbers, and let K be the field of all complex numbers. Let a; b 2 F. Let us take the polynomial x2 þ ax þ b for f ð xÞ in F ½ x. Suppose that a 2 ðK F Þ such that a is a root of x2 þ ax þ b, that is, a2 þ aa þ b ¼ 0. According to 1.5.9, ½ðsplitting field over F for f ð xÞÞ : F ðdegðf ð xÞÞÞ!; so

splitting field over F for x2 þ ax þ b : F deg x2 þ ax þ b ! ¼ 2! ¼ 2;

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1 Galois Theory I

and hence splitting field over F for x2 þ ax þ b : F ¼ 1 or 2: Since a 2 ðK F Þ such that a is a root of x2 þ ax þ b, we have ½ðsplitting field over F for x2 þ ax þ bÞ : F [ 1, and hence

splitting field over F for x2 þ ax þ b : F ¼ 2:

1.5.22 Theorem Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Suppose that ½E : F ¼ 1. Then E ¼ F. Proof Suppose to the contrary that E 6¼ F. We seek a contradiction. Since E is a splitting field over F for gð xÞ, E is a finite extension of F, and hence F E. Now, since E 6¼ F, there exists a nonzero a in E such that a 62 F. Snce 1 2 F, we have a 6¼ 1, and hence f1; agð EÞ is a set of two elements. Clearly, f1; ag is a linearly independent subset of E. Proof Suppose that k1 þ la ¼ 0, where k; l 2 F. We have to show that k ¼ 0 and l ¼ 0. If l 6¼ 0, then a ¼ l1 k 2 F, and hence a 2 F. This is a contradiction. Hence l ¼ 0. Since k1 þ la ¼ 0, we have k ¼ 0. ■ Thus we have shown that f1; ag is a linearly independent subset of E, and the number of elements in f1; ag is 2. It follows that 1 ¼ ½E : F ¼ dim ðE Þ 2. Thus |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} we get a contradiction.

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1.5.23 Note Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Suppose that ½E : F ¼ 1. By 1.5.22, we have E ¼ F. Since E is a splitting field over F for gð xÞ, F is a splitting field over F for gð xÞ, and hence gð xÞ splits into a product of linear factors over F. By 1.5.10, g0 ðtÞ splits into a product of linear factors over F 0 . Next, since E 0 is a splitting field over F 0 for g0 ðtÞ, we have E0 ¼ F 0 . Since s is a ring isomorphism from F onto F 0 , E ¼ F, and E 0 ¼ F 0 , s is a ring isomorphism from E onto E 0 . Let us take an arbitrary a 2 F. By the definition of s, we have sðaÞ ¼ a0 .


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1.5.24 Conclusion Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Suppose that ½E : F ¼ 1. Then there exists a ring isomorphism u from E onto E 0 such that for every a 2 F, uðaÞ ¼ a0 . 1.5.25 Problem Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Suppose that ½E : F ¼ 2. Then there exists a ring isomorphism u from E onto E 0 such that for every a 2 F, uðaÞ ¼ a0 . Proof Since ½E : F ¼ 2, we have ½E : F 6¼ 1, and hence E 6¼ F. It follows that F is a proper subset of E. By 1.3.17, gð xÞ can be expressed as a product of finitely many irreducible polynomials in F ½ x. Since E is a splitting field over F for gð xÞ, and F is a proper subset of E, there exists pð xÞ 2 F ½ x such that 1. degðpð xÞÞ [ 1, 2. pð xÞjgð xÞ, 3. pð xÞ is irreducible over F. Since E is a splitting field over F for gð xÞ, pð xÞjgð xÞ, and pð xÞ is irreducible over F, all roots of pð xÞ are members of E. Since degðpð xÞÞ [ 1, there exists a 2 E such that 1. a 62 F, 2. a is a root of pð xÞ in E. It follows, by 1.5.12, that a is algebraic of degree r over F, where r degðpð xÞÞ ð [ 1Þ. Now by 1.4.16, ½F ðaÞ : F ¼ r. By 1.4.3, 2 ¼ ½E : F ¼ ½E : F ðaÞ½F ðaÞ : F ¼ ½E : F ðaÞr; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½E : F ðaÞ ¼ 2r 1. Now, since ½E : F ðaÞ is a positive integer, we have ½E : F ðaÞ ¼ 1. Since pð xÞ 2 F ½ x, we have p0 ðtÞ 2 F 0 ½t. Since degðpð xÞÞ [ 1, we have degðp0 ðtÞÞ [ 1. Since pð xÞjgð xÞ, we have p0 ðtÞjg0 ðtÞ. Since pð xÞ is irreducible over F, p0 ðtÞ is irreducible over F 0 . It follows that there exists b 2 E0 such that 1. b 62 F 0 , 2. b is a root of p0 ðtÞ in E 0 .

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1 Galois Theory I

Since F F ðaÞ, we have gð xÞ 2 F ½ x ðF ðaÞÞ½ x , and hence gð xÞ 2 ðF ðaÞÞ½ x. |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} Since Eð F ðaÞ F Þ is a splitting field over F for gð xÞ, E is a splitting field over F ðaÞ for gð xÞ. Since E 0 ð F 0 ðbÞ F 0 Þ is a splitting field over F 0 for g0 ðtÞ, E 0 is a splitting field over F 0 ðbÞ for g0 ðtÞ. Now, since ½E : F ðaÞ ¼ 1, by 1.5.24, there exists a ring isomorphism u from E onto E0 such that for every a 2 F ðaÞð F Þ, uðaÞ ¼ a0 . It follows that for every a 2 F, uðaÞ ¼ a0 . ■ 1.5.26 Problem Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Suppose that ½E : F ¼ 3. Then there exists a ring isomorphism u from E onto E 0 such that for every a 2 F, uðaÞ ¼ a0 . Proof Since ½E : F ¼ 3, we have ½E : F 6¼ 1, and hence E 6¼ F. It follows that F is a proper subset of E. By 1.3.17, gð xÞ can be expressed as a product of finitely many irreducible polynomials in F ½ x. Now, since E is a splitting field over F for gð xÞ and F is a proper subset of E, there exists pð xÞ 2 F ½ x such that 1. degðpð xÞÞ [ 1, 2. pð xÞjgð xÞ, 3. pð xÞ is irreducible over F. Since E is a splitting field over F for gð xÞ, pð xÞjgð xÞ, and pð xÞ is irreducible over F, all roots of pð xÞ are members of E. Since degðpð xÞÞ [ 1, there exists a 2 E such that 1. a 62 F, 2. a is a root of pð xÞ in E. It follows, by 1.5.12, that a is algebraic of degree r over F, where r degðpð xÞÞð [ 1Þ. Now by 1.4.16, ½F ðaÞ : F ¼ r. By 1.4.3, 3 ¼ ½E : F ¼ ½E : F ðaÞ½F ðaÞ : F ¼ ½E : F ðaÞr; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½E : F ðaÞ ¼ 3r 2. Now, since ½E : F ðaÞ is a positive integer, we have ½E : F ðaÞ ¼ 1 or 2. Since pð xÞ 2 F ½ x, we have p0 ðtÞ 2 F 0 ½t. Since degðpð xÞÞ [ 1, we have degðp0 ðtÞÞ [ 1. Since pð xÞjgð xÞ, we have p0 ðtÞjg0 ðtÞ. Since pð xÞ is irreducible over F, p0 ðtÞ is irreducible over F 0 . It follows that there exists b 2 E0 such that 1. b 62 F 0 , 2. b is a root of p0 ðtÞ in E 0 .


89

Since F F ðaÞ, we have gð xÞ 2 F ½ x ðF ðaÞÞ½ x , and hence gð xÞ 2 ðF ðaÞÞ½ x. |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} Since E ð F ðaÞ F Þ is a splitting field over F for gð xÞ, E is a splitting field over F ðaÞ for gð xÞ. Since E 0 ð F 0 ðbÞ F 0 Þ is a splitting field over F 0 for g0 ðtÞ, E 0 is a splitting field over F 0 ðbÞ for g0 ðtÞ. Now, since ½E : F ðaÞ ¼ 1 or 2, by 1.5.24 and 1.5.25, there exists a ring isomorphism u from E onto E0 such that for every a 2 F ðaÞð F Þ, uðaÞ ¼ a0 . It follows that for every a 2 F, uðaÞ ¼ a0 . ■ Similarly, we get the following. 1.5.27 Conclusion Let F and K be any fields such that K is an extension of F. Let F 0 and K 0 be any fields such that K 0 is an extension of F 0 . Let s : a 7! a0 be a ring isomorphism from F onto F 0 . For every f ð xÞ 2 F ½ x, we shall denote s ðf ð xÞÞ by f 0 ðtÞ, where s is the same as discussed in 1.5.10. Let gð xÞ 2 F ½ x. It follows that g0 ðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E 0 be a splitting field over F 0 for g0 ðtÞ. Then there exists a ring isomorphism u from E onto E 0 such that for every a 2 F, uðaÞ ¼ a0 . 1.5.28 Note In 1.5.27, let us take F for F 0 , and the identity map i : F ! F for s. Thus for every a 2 F, a0 means a, and for every f ð xÞ 2 F ½ x, f 0 ðtÞ means f ðtÞ. Let gð xÞ 2 F ½ x. It follows that gðtÞ 2 F 0 ½t. Let E be a splitting field over F for gð xÞ. Let E0 be a splitting field over F 0 for g0 ðtÞð¼ gðtÞÞ. Then by 1.5.27, there exists a ring isomorphism u from E onto E0 such that for every a 2 F, uðaÞ ¼ a0 . 1.5.29 Conclusion Let F and K be any fields such that F K. Suppose that K is an extension of F. Let gð xÞ 2 F ½ x. Let E be a splitting field over F for gð xÞ. Let E0 be a splitting field over F for gð xÞ. Then there exists a ring isomorphism u from E onto E 0 such that for every a 2 F, uðaÞ ¼ a0 . Thus the splitting field over F for a polynomial is essentially unique, and hence it is justified in speaking about “the” splitting field.

Exercises 1. Find the greatest common divisor of pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi 5 þ 3 1 and 3 4 1 pffiffiffiffiffiffiffi in J 1 . (Hint: Observe that 5 þ i3 ¼ ð3 i4Þi þ 1 : 3 i4 ¼ 1ð3 i4Þ þ 0

90

1 Galois Theory I

So the required gcd is 1.) 2. Suppose that p is a prime number, and a; b are integers such that pjða2 þ b2 Þ, and p2 -ða2 þ b2 Þ. Show that p can be expressed as a sum of two perfect squares. 3. Show that x3 2 is irreducible in the integral domain Q½ x. 4. Show that if 4n 3 is a prime number, then 4n þ 1 can be expressed as a sum of two perfect squares. 5. Prove that ð72! þ 1Þ is divisible by 73. 6. Suppose that R is a unique factorization domain with unit element 1. Show that R½x; y is also a unique factorization domain. 7. Let F and K be any fields such that K is an extension of F. Suppose that a 2 K. Show that a is algebraic of degree ½F ðaÞ : F over F. pffiffiffi pffiffiffi 8. Prove that 2 þ 3 is algebraic of degree 4 over Q. 9. Show that 10. Show that

splitting field over Q for x2 þ x þ 1 : Q ¼ 2:

pffiffiffi e is a transcendental number.

Chapter 2

Galois Theory II

Roughly, a real number a is called a constructible number if by the application of straightedge and compass we can construct, given a line segment of unit length, a line segment of length a. In some familiar geometric situations, we shall apply the results of Galois theory. In our general development, we shall show that the general polynomial equation of degree five has no solution in radicals.

2.1

Simple Extensions

2.1.1 Definition Let D be an integral domain, that is, D is a commutative ring such that all products of nonzero members of 0 D are nonzero. If for 1 every positive integer m and every nonzero member a of D, ma@ |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} a þ a þ þ aA is nonzero, then we m terms

say that D is of characteristic 0. Definition Let D be an integral domain. If there exists a positive integer m such that for every member a of D, ma ¼ 0, then we say that D is of finite characteristic. 2.1.2 Problem Let D be an integral domain. Let a and b be any nonzero members of D. Then fm : m is a positive integer and ma ¼ 0g ¼ fm : m is a positive integer and mb ¼ 0g:

© Springer Nature Singapore Pte Ltd. 2020 R. Sinha, Galois Theory and Advanced Linear Algebra, https://doi.org/10.1007/978-981-13-9849-0_2

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Proof Let us take an arbitrary positive integer m satisfying ma ¼ 0: We shall show that mb ¼ 0: Since ma ¼ 0, we have aðmbÞ ¼ ðmaÞb ¼ 0b ¼ 0; |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} and hence aðmbÞ ¼ 0: Now, since a and mb are members of the integral domain D and a is nonzero, we have mb ¼ 0: Thus fm : m is a positive integer and ma ¼ 0g fm : m is a positive integer and mb ¼ 0g: Similarly, fm : m is a positive integer and mb ¼ 0g fm : m is a positive integer and ma ¼ 0g: Hence fm : m is a positive integer and ma ¼ 0g ¼ fm : m is a positive integer and mb ¼ 0g: ■ 2.1.3 Note Let D be an integral domain such that D is of finite characteristic. Let b be a nonzero member of D. It follows that fm : m is a positive integer such that for every member a of D; ma ¼ 0g is a nonempty set of positive integers. Also, by 2.1.2, fm : m is a positive integer such that for every member a of D; ma ¼ 0g ¼ fm : m is a positive integer and mb ¼ 0g: Since every set of positive integers has a least member, the smallest member n of fm : m is a positive integer such that for every member a of D; ma ¼ 0g exists. Clearly, n is a prime number. Also, for every nonzero member b of D, fm : m is a positive integer and mb ¼ 0g ¼ fn; 2n; 3n; . . .g:

2.1 Simple Extensions

93

(The number n is called the characteristic of D.) Proof Suppose to the contrary that n is not a prime number. We seek a contradiction. Since n is not a prime number, there exist positive integers n1 ; n2 such that n ¼ n1 n2 and 1\n1 n2 \n: Here, n is the smallest member of fm : m is a positive integer and mb ¼ 0g; so n1 b 6¼ 0; n2 b 6¼ 0; and nb ¼ 0: Since n1 b; n2 b are nonzero members of the integral domain D, we have 0 ¼ 0b ¼ ðnbÞb ¼ ððn1 n2 ÞbÞb ¼ ðn1 bÞðn2 bÞ 6¼ 0; |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence we get a contradiction.

■

Definition Let F be a field. Let f ðxÞ 2 F½x; where f ðxÞ a0 xn þ a1 xn1 þ þ an1 x þ an and ai 2 F ði ¼ 0; 1; . . .; nÞ: It follows that na0 ; ðn 1Þa1 ; . . .; 2an2 ð¼ an2 þ an2 Þ; 1an1 ð¼ an1 Þ are members of F, and hence ðna0 Þxn1 þ ððn 1Þa1 Þxn2 þ þ 1an1 is a member of F½x: The polynomial ðna0 Þxn1 þ ððn 1Þa1 Þxn2 þ þ 1an1 is denoted by f 0 ðxÞ and is called the derivative of f ðxÞ. 2.1.4 Problem Let F be a field. Let f ðxÞ; gðxÞ 2 F½x: Let a 2 F: Suppose that hðxÞ ¼ f ðxÞ þ agðxÞð2 F½xÞ: Then h0 ðxÞ ¼ f 0 ðxÞ þ ag0 ðxÞ: Proof Let f ðxÞ a0 xn þ a1 xn1 þ þ an1 x þ an ; where ai 2 F ði ¼ 0; 1; . . .; nÞ: Next, let

94

2 Galois Theory II

gðxÞ b0 xn þ b1 xn1 þ þ bn1 x þ bn ; where bi 2 F ði ¼ 0; 1; . . .; nÞ: It follows that hðxÞ ¼ c0 xn þ c1 xn1 þ þ cn1 x þ cn ; where ci ai þ abi ði ¼ 0; 1; . . .; nÞ: It follows that f 0 ðxÞ ¼ ðna0 Þxn1 þ ððn 1Þa1 Þxn2 þ þ 1an1 ; g0 ðxÞ ¼ ðnb0 Þxn1 þ ððn 1Þb1 Þxn2 þ þ 1bn1 ; and h0 ðxÞ ¼ ðnc0 Þxn1 þ ððn 1Þc1 Þxn2 þ þ 1cn1 : Here LHS ¼ f 0 ð xÞ þ ag0 ð xÞ ¼ ðna0 þ aðnb0 ÞÞxn1 þ ððn 1Þa1 þ aððn 1Þb1 ÞÞxn2 þ þ ð1an1 þ að1bn1 ÞÞ ¼ ðna0 þ nðab0 ÞÞxn1 þ ððn 1Þa1 þ ðn 1Þðab1 ÞÞxn2 þ þ ð1an1 þ 1ðabn1 ÞÞ ¼ ðnða0 þ ðab0 ÞÞÞxn1 þ ððn 1Þða1 þ ab1 ÞÞxn2 þ þ 1ðan1 þ abn1 Þ ¼ ðnðc0 ÞÞxn1 þ ððn 1Þðc1 ÞÞxn2 þ þ 1ðcn1 Þ ¼ ðnc0 Þxn1 þ ððn 1Þc1 Þxn2 þ þ 1cn1 ¼ h0 ð xÞ ¼ RHS ■ 2.1.5 Problem Let F be a field. Let f ðxÞ; gðxÞ 2 F½x: Suppose that hðxÞ ¼ f ðxÞgðxÞð2 F½xÞ: Then h0 ðxÞ ¼ f 0 ðxÞgðxÞ þ f ðxÞg0 ðxÞ: Proof Let f ðxÞ a0 xn þ a1 xn1 þ þ an1 x þ an ;


95

where ai 2 F ði ¼ 0; 1; . . .; nÞ: Next, let gðxÞ b0 xn þ b1 xn1 þ þ bn1 x þ bn ; where bi 2 F ði ¼ 0; 1; . . .; nÞ: It follows that hðxÞ ¼ c0 x2n þ c1 x2n1 þ þ c2n1 x þ c2n ; where c0 a0 b0 , c1 a0 b1 þ a1 b0 , etc. Here LHS ¼ f 0 ðxÞgðxÞ þ f ðxÞg0 ðxÞ ¼ ðna0 Þxn1 þ ððn 1Þa1 Þxn2 þ b0 xn þ b1 xn1 þ þ bn1 x þ bn þ a0 xn þ a1 xn1 þ þ an1 x þ an ðnb0 Þxn1 þ ððn 1Þb1 Þxn2 þ þ 1bn1 ¼ ððna0 Þb0 Þx2n1 þ ððna0 Þb1 þ ððn 1Þa1 Þb0 Þx2n2 þ þ ða0 ðnb0 ÞÞx2n1 þ ða0 ððn 1Þb1 Þ þ a1 ðnb0 ÞÞx2n2 þ ¼ ððna0 Þb0 þ a0 ðnb0 ÞÞx2n1 þ ððna0 Þb1 þ ððn 1Þa1 Þb0 þ a0 ððn 1Þb1 Þ þ a1 ðnb0 ÞÞx2n2 þ ¼ ðnða0 b0 Þ þ nða0 b0 ÞÞx2n1 þ ðnða0 b1 Þ þ ðn 1Þða1 b0 Þ þ ðn 1Þða0 b1 Þ þ nða1 b0 ÞÞx2n2 þ ¼ ðð2nÞða0 b0 ÞÞx2n1 þ ðnða0 b1 þ a1 b0 Þ þ ðn 1Þða0 b1 þ a1 b0 ÞÞx2n2 þ ¼ ðð2nÞða0 b0 ÞÞx2n1 þ ðð2n 1Þða0 b1 þ a1 b0 ÞÞx2n2 þ ¼ ðð2nÞc0 Þx2n1 þ ðð2n 1Þc1 Þx2n2 þ ¼ h0 ðxÞ ¼ RHS: ■ 2.1.6 Note Let F and K be any fields such that K is an extension of F. Let a be a member of K. Suppose that f ðxÞ ¼ ðx aÞm gðxÞ, where m 2 f2; 3; . . .g, f ðxÞ 2 F½x, and gðxÞ 2 K½x:

96

2 Galois Theory II

It follows that ðx aÞm ; f ðxÞ; gðxÞ 2 K½x: Now by 2.1.5, f 0 ðxÞ ðx aÞ0 ðx aÞðx aÞ ðx aÞ gðxÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼

ðm1Þ factors

þ ðx aÞðx aÞ0 ðx aÞ ðx aÞ gðxÞ þ |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðm2Þ factors

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m terms

þ ðx aÞ ðx aÞ g0 ðxÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m factors

¼ 1 ðx aÞðx aÞ ðx aÞ gðxÞ þ ðx aÞ1 ðx aÞ ðx aÞ gðxÞ þ |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðm1Þ factors

ðm2Þ factors

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m terms


¼ ðx aÞðx aÞ ðx aÞ gðxÞ þ ðx aÞðx aÞ ðx aÞ gðxÞ þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðm1Þ factors

ðm1Þ factors

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m terms


¼ ðx aÞm1 gðxÞ þ ðx aÞm1 gðxÞ þ þ ðx aÞm g0 ðxÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m terms

¼ mðx aÞ gðxÞ þ ðx aÞm g0 ðxÞ ¼ ðx aÞ mðx aÞm2 gðxÞ þ ðx aÞm1 g0 ðxÞ ¼ ðx aÞrðxÞ; m1

rðxÞ mðx aÞm2 gðxÞ þ ðx aÞm1 g0 ðxÞ 2 K½x: Since f ðxÞ ¼ ðx aÞ ðx aÞm1 gðxÞ and f 0 ðxÞ ¼ ðx aÞrðxÞ; ðx aÞ is a common factor of where

f ðxÞ and f 0 ðxÞ: 2.1.7 Conclusion Let F and K be any fields such that K is an extension of F. Let a be a member of K. Let f ðxÞ 2 F½x: Suppose that a is a multiple root of f ðxÞ . Then f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x: 2.1.8 Note Let F and K be any fields such that K is an extension of F. Suppose that f ðxÞ 2 F½x: It follows that f 0 ðxÞ 2 F½x: Suppose that f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x, that is, f ðxÞ and f 0 ðxÞ have a common factor of degree 1 in K½x:


97

It follows that there exists a such that ðx aÞjf ðxÞ and ðx aÞjf 0 ðxÞ: We shall show that ðx aÞ2 jf ðxÞ. Since ðx aÞjf ðxÞ, there exist a positive integer m and a polynomial rðxÞ such that 1. f ðxÞ ¼ ðx aÞm rðxÞ; 2. ðx aÞ-rðxÞ: It suffices to show that m 2: Suppose to the contrary that m ¼ 1. We seek a contradiction. Here f ðxÞ ¼ ðx aÞrðxÞ, so by 2.1.5, f 0 ðxÞ ¼ ðx aÞ0 rðxÞ þ ðx aÞr 0 ðxÞ ¼ 1rðxÞ þ ðx aÞr 0 ðxÞ ¼ rðxÞ þ ðx aÞr 0 ðxÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence f 0 ðxÞ ¼ rðxÞ þ ðx aÞr 0 ðxÞ: It follows that f 0 ðaÞ ¼ r ðaÞ þ ða aÞr 0 ðaÞ ¼ r ðaÞ þ 0r 0 ðaÞ ¼ r ðaÞ: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus f 0 ðaÞ ¼ r ðaÞ. Since ðx aÞ-rðxÞ, we have r ðaÞ 6¼ 0. Since ðx aÞjf 0 ðxÞ, we have f 0 ðaÞ ¼ 0. Since f 0 ðaÞ ¼ 0 and r ðaÞ 6¼ 0, we have f 0 ðaÞ 6¼ r ðaÞ. This is a contradiction. 2.1.9 Conclusion Let F and K be any fields such that K is an extension of F. Suppose that f ðxÞ 2 F½x. Suppose that f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x. Then f ðxÞ has a multiple root. 2.1.10 Problem Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Suppose that f ðxÞ 2 F½x. Let f ðxÞ be irreducible. Then f ðxÞ has no multiple root. Proof Suppose to the contrary that f ðxÞ has a multiple root. We seek a contradiction. Since f ðxÞ has a multiple root, by 2.1.6, f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x. Since f ðxÞ is irreducible, f ðxÞ is the only nontrivial factor of f ðxÞ. Now, since f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x, f ðxÞ is a factor of f 0 ðxÞ, and hence degðf ðxÞÞ degðf 0 ðxÞÞ. Suppose that f ðxÞ a0 xn þ a1 xn1 þ þ an1 x þ an ; where ai 2 F ði ¼ 0; 1; . . .; nÞ, n is a positive integer, and a0 6¼ 0. Since F is of characteristic 0, we have na0 is a nonzero member of F, and hence

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degðf 0 ð xÞÞ ¼ deg ðna0 Þxn1 þ ððn 1Þa1 Þxn2 þ ¼ n 1 \n |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ degða0 xn þ a1 xn1 þ þ an1 x þ an Þ ¼ degðf ð xÞÞ: Thus degðf 0 ðxÞÞ\degðf ðxÞÞ. This is a contradiction.

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2.1.11 Problem Let F and K be any fields such that K is an extension of F. Let F be of characteristic p. By 2.1.3, p is a prime number. Suppose that f ðxÞ 2 F½x. Let f ðxÞ be irreducible. Suppose that f ðxÞ has a multiple root. Then f ðxÞ is of the form gðxp Þ, where gðxÞ 2 F½x. Proof Since f ðxÞ has a multiple root, by 2.1.6, f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x. Since f ðxÞ is irreducible, f ðxÞ is the only nontrivial factor of f ðxÞ. Now, since f ðxÞ and f 0 ðxÞ have a nontrivial common factor in K½x, f ðxÞ is a factor of f 0 ðxÞ, and hence degðf ðxÞÞ degðf 0 ðxÞÞ. But we know that if f 0 ðxÞ is nonzero, then degðf 0 ðxÞÞ\degðf ðxÞÞ, hence f 0 ðxÞ ¼ 0. Suppose that f ðxÞ a0 þ a1 x þ þ ap1 xp1 þ ap xp þ ap þ 1 xp þ 1 þ þ a2p1 x2p1 þ a2p x2p þ a2p þ 1 x2p þ 1 þ þ an xn ; where ai 2 F ði ¼ 0; 1; . . .; nÞ, and n is a positive integer. It suffices to show that a1 ¼ 0; . . .; ap1 ¼ 0; ap þ 1 ¼ 0; a2p1 ¼ 0; etc. Since f 0 ðxÞ ¼ 0, we have a1 þ 2a2 x þ þ ðp 1Þap1 xp2 þ pap xp1 þ ðp þ 1Þap þ 1 xp þ þ ð2p 1Þa2p1 x2p2 þ 2pa2p x2p1 þ ð2p þ 1Þa2p þ 1 x2p þ þ nan xn1 ¼ 0; and hence 0 ¼ a1 , 0 ¼ 2a2 , 0 ¼ ðp 1Þap1 , 0 ¼ ðp þ 1Þap þ 1 , 0 ¼ ð2p 1Þa2p1 , 0 ¼ ð2p þ 1Þa2p þ 1 , etc. Since F is of characteristic p, p-ðp 1Þ, 0 ¼ ðp 1Þap1 , and ap1 2 F, we have ap1 ¼ 0. Similarly, ap þ 1 ¼ 0, a2p1 ¼ 0, etc. ■ 2.1.12 Note Let F and K be any fields such that K is an extension of F. Let F be of characteristic p. Observe that 1xp þ ð1Þx is a member of F½x. Also ð1xp þ ð1ÞxÞ0 ¼ ðp1Þxp1 þ ð1Þ: Since 1 2 F; and F is of characteristic p, we have p1 ¼ 0, and hence ð1xp þ ð1ÞxÞ0 ¼ ðp1Þxp1 þ ð1Þ ¼ 0xp1 þ ð1Þ ¼ 1: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}


99

Thus ð1xp þ ð1ÞxÞ0 ¼ 1. Now, since 1xp þ ð1Þx and 1 have no nontrivial common factor in K½x, 1xp þ ð1Þx and ð1xp þ ð1ÞxÞ0 have no nontrivial common factor in K½x, and hence by 2.1.6, 1xp þ ð1Þx has no multiple root. 2.1.13 Conclusion Let F and K be any fields such that K is an extension of F. Let 2 F be of characteristic p. Then xp x has no multiple root. Similarly, xp x has no 3 multiple root, xp x has no multiple root, etc. 2.1.14 Note Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Let a; b 2 K. Suppose that a, b are algebraic over F. Since a is algebraic over F, there exists a nonzero polynomial qðxÞ 2 F½x such that ðK3ÞqðaÞ ¼ 0. By 1.3.21, there exists an irreducible polynomial f ðxÞ 2 F½x such that f ðxÞjqðxÞ in F½x, ðK3Þ f ðaÞ ¼ 0, and the leading coefficient of f ðxÞ is 1. Similarly, there exists an irreducible polynomial gðxÞ 2 F½x such that ðK3Þ gðbÞ ¼ 0, and the leading coefficient of gðxÞ is 1. Suppose that degðf ðxÞÞ ¼ mð 1Þ and degðgðxÞÞ ¼ nð 1Þ. Let L be a field such that 1. 2. 3. 4.

K L; L is an extension of K, f ðxÞ splits completely in L, gðxÞ splits completely in L.

Suppose that all the m roots of f ðxÞ in L are a; a2 ; . . .; am . Next, suppose that all the n roots of gðxÞ in L are b; b2 ; . . .; bn . Since a; b 2 L and F L, we have Fða; bÞ L. Also f ðxÞ ¼ ðx aÞðx a2 Þ. . .ðx am Þ and gðxÞ ¼ ðx bÞðx b2 Þ. . .ðx bn Þ in L½x. Since F is of characteristic 0, f ðxÞ 2 F½x, f ðxÞ is irreducible, and a; a2 ; . . .; am are the roots of f ðxÞ, by 2.1.10, a; a2 ; . . .; am are distinct. Similarly, b;b2; . . .; bn are distinct. Also, f ðaÞ ¼ 0, gðbÞ ¼ 0, f ðai Þ ¼ 0 ði ¼ 2; 3; . . .; mÞ, and g bj ¼ 0 ðj ¼ 2; 3; . . .; nÞ. Since F is of characteristic 0 and 1 2 F, 1; 1 þ 1; 1 þ 1 þ 1; . . . are distinct members of F, and hence F is an infinite set. Let us take arbitrary i 2 f2; . . .; mg and j 2 f2; . . .; ng. Since b; b2 ; . . .; bn are 1 distinct, b bj 6¼ 0, and hence b bj is a nonzero element of the field L. Observe that there exists a unique k 2 L such that

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ai þ kbj ¼ a þ kb: Proof Existence: Since 1 1 ai þ ð ai aÞ b bj bj ¼ ai b bj þ ðai aÞbj b bj 1 ¼ ai b abj b bj and 1 1 a þ ð ai aÞ b bj b ¼ a b bj þ ðai aÞb b bj 1 ¼ ai b abj b bj ; 1 ð ai aÞ b bj ð2 LÞ is a solution of the k-equation ai þ kbj ¼ a þ kb in L. Uniqueness: Suppose that ai þ k 1 bj ¼ a þ k1 b ; ai þ k 2 bj ¼ a þ k2 b where k1 ; k2 2 L. We have to show that k1 ¼ k2 . Here, ðk1 k2 Þbj ¼ ai þ k1 bj ai þ k2 bj ¼ ða þ k1 bÞ ða þ k2 bÞ ¼ ðk1 k2 Þb; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so ðk1 k2 Þbj ¼ ðk1 k2 Þb, and hence ðk1 k2 Þ b bj ¼ 0. b bj 6¼ 0, we have k1 k2 ¼ 0, and hence k1 ¼ k2 :

Since ■

Thus we have shown that for every i 2 f2; . . .; mg and j 2 f2; . . .; ng, there exists a unique k 2 L ð FÞ such that ai þ kbj ¼ a þ kb. It follows that the collection of all such k is a finite set. Now, since F is an infinite set, there exists a nonzero c 2 F ð Fða; bÞÞ such that for every i 2 f2;. . .; mg and j 2 f2; . . .; ng, ai þ cbj 6¼ a þ cb, and hence for every j 2 f2; . . .; mg, ða þ cbÞ cbj is different from a; a2 ; . . .; am . Since all the m distinct roots of f ðxÞ are a; a2 ; . . .; am , for every j 2 f2; . . .; mg, ða þ cbÞ cbj is not a root of f ðxÞ, and hence for every j 2 f2; . . .; mg, we have f ða þ cbÞ cbj 6¼ 0. Since a, b, c are elements of the field Fða; bÞ, we have ða þ cbÞ 2 Fða; bÞ, and hence Fða þ cbÞ Fða; bÞ L.


101

Now we shall show that Fða; bÞ Fða þ cbÞ. To this end, put hðxÞ f ðða þ cbÞ cxÞð2 ðFða þ cbÞÞ½xÞ: Thus hðxÞ 2 ðFða þ cbÞÞ½x. Also hðbÞ ¼ f ðða þ cbÞ cbÞ ¼ f ðaÞ ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so hðbÞ ¼ 0, and hence ðx bÞ is a factor of the polynomial hðxÞ in L½x. Since gðbÞ ¼ 0, ðx bÞ is a factor of the polynomial gðxÞ in L½x. Thus ðx bÞ is a common factor of the polynomials gðxÞ and hðxÞ in L½x. By 2.1.10, we have ðx bÞ2 -gðxÞ, and hence ðx bÞ2 is not a common factor of the polynomials gðxÞ and hðxÞ. Clearly, for every j 2 f2; . . .; mg, x bj is not a common divisor of the polynomials gðxÞ and hðxÞ. Proof Let us fix an arbitrary j 2 f2; . . .; mg. It follows that h bj ¼ f ða þ cbÞ cbj 6¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence h bj 6¼ 0. Thus x bj -hðxÞ in L½x. It follows that for every j 2 f2; . . .; mg, x bj is not a common divisor of the polynomials gðxÞ and hðxÞ in L½x. ■ Thus ðx bÞ is a greatest common divisor of the polynomials gðxÞ and hðxÞ in L½xð ðFða þ cbÞÞ½xÞ. It follows that ðx bÞ ð2 ðFða; bÞÞ½x L½xÞ divides each member of the set fgðxÞuðxÞ þ hðxÞvðxÞ : uðxÞ; vðxÞ 2 L½xg ð fgðxÞuðxÞ þ hðxÞvðxÞ : uðxÞ; vðxÞ 2 ðFða þ cbÞÞ½xgÞ; and hence ðx bÞ divides each member of fgðxÞuðxÞ þ hðxÞvðxÞ : uðxÞ; vðxÞ 2 ðFða þ cbÞÞ½xg: Since gðxÞ 2 F½xð ðFða þ cbÞÞ½xÞ and hðxÞ 2 ðFða þ cbÞÞ½x, a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x is a member of fgðxÞuðxÞ þ hðxÞvðxÞ : uðxÞ; vðxÞ 2 ðFða þ cbÞÞ½xg:

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Since ðx bÞ divides each member of fgðxÞuðxÞ þ hðxÞvðxÞ : uðxÞ; vðxÞ 2 ðFða þ cbÞÞ½xg; ðx bÞ divides a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x, and hence a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x is nontrivial. Since ðFða þ cbÞÞ½x L½x, a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x divides a greatest common divisor of the polynomials gðxÞ and hðxÞ in L½x, and hence a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x divides ðx bÞ. Now, since ðx bÞ divides a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x, ðx bÞ is a greatest common divisor of the polynomials gðxÞ and hðxÞ in ðFða þ cbÞÞ½x, and hence ðx bÞ 2 ðFða þ cbÞÞ½x. It follows that ðbÞ 2 Fða þ cbÞ, and hence b 2 Fða þ cbÞ. Since ða þ cbÞ 2 Fða þ cbÞ and c 2 F ð Fða þ cbÞÞ, we have a ¼ ðða þ cbÞ cbÞ 2 Fða þ cbÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence a 2 Fða þ cbÞ. Thus F [ fa; bg Fða þ cbÞ, and hence Fða; bÞ Fða þ cbÞ. Next, since Fða þ cbÞ Fða; bÞ, we have FðcÞ ¼ Fða; bÞ, where c ða þ cbÞ 2 Fða; bÞ. 2.1.15 Conclusion I Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Let a; b 2 K. Suppose that a, b are algebraic over F. Then there exists c 2 Fða; bÞ such that Fða; bÞ ¼ FðcÞ. Similarly, we get the following. 2.1.16 Conclusion II Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Let a1 ; a2 ; . . .; an 2 K. Suppose that a1 ; a2 ; . . .; an are algebraic over F. Then there exists c 2 F ða1 ; a2 ; . . .; an Þ such that F ða1 ; a2 ; . . .; an Þ ¼ FðcÞ. Definition Let F and K be any fields such that K is an extension of F. If there exists c 2 K such that K ¼ FðcÞ, then we say that K is a simple extension of F. Now Conclusion II can be stated as follows: 2.1.17 Conclusion III Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Let a1 ; a2 ; . . .; an 2 K. Suppose that a1 ; a2 ; . . .; an are algebraic over F. Then F ða1 ; a2 ; . . .; an Þ is a simple extension of F. Using 1.4.9, we get the following. 2.1.18 Conclusion IV Let F and K be any fields such that K is an extension of F. Let F be of characteristic 0. Let a 2 K. Suppose that FðaÞ is a finite extension of F. Then FðaÞ is a simple extension of F. In short, every finite extension of a field of characteristic 0 is a simple extension.

2.2 Galois Groups

2.2

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Caution: From henceforth, all our fields are of characteristic 0. 2.2.1 Definition Let K be any field. Let r : K ! K be a function. If 1. 2. 3. 4.

for every a; b 2 K, rða þ bÞ ¼ rðaÞ þ rðbÞ, for every a; b 2 K, rðabÞ ¼ rðaÞrðbÞ, r : K ! K is onto, r : K ! K is 1-1, then we say that r is an automorphism of K. Here condition (4) is superfluous. Proof Suppose to the contrary that there exist a; b 2 K such that rðaÞ ¼ rðbÞ, and a ¼ 6 b. We seek a contradiction. Since a 6¼ b, ða bÞ is a nonzero member of K, and hence ða bÞ1 2 K. It follows that ¼ rða bÞr ða bÞ1 1 ¼ rð1Þ ¼ r ða bÞ ða bÞ1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ rða þ ðbÞÞr ða bÞ1 ¼ ðrðaÞ þ rðbÞÞr ða bÞ1 ¼ ðrðaÞ þ ðrðbÞÞÞr ða bÞ1 ¼ ðrðaÞ rðbÞÞr ða bÞ1 ¼ ðrðaÞ rðaÞÞr ða bÞ1 ¼ 0 r ða bÞ1 ¼ 0; and hence 1 ¼ 0. This contradicts the fact that K is a field.

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2.2.2 Note Let K be any field. Let r1 ; . . .; rn be n distinct automorphisms of K. Let a1 ; . . .; an 2 K. Suppose that 1. for every u 2 K, a1 r1 ðuÞ þ þ an rn ðuÞ ¼ 0, 2. not all ai ði ¼ 1; . . .; nÞ are 0. We claim that this is impossible. We seek a contradiction. In the case of n = 1, condition (1) becomes for every u 2 K, a1 r1 ðuÞ ¼ 0, and hence a1 ¼ a1 1 ¼ a1 r1 ð1Þ ¼ 0 : |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} Thus a1 ¼ 0. In this case of n ¼ 1, condition (2) becomes a1 6¼ 0. This is a contradiction. Now we consider the case n ¼ 2. Here condition (1) becomes for every u 2 K, a1 r1 ðuÞ þ a2 r2 ðuÞ ¼ 0. Next, condition (2) becomes (either a1 6¼ 0 or a2 6¼ 0). For definiteness, suppose that a1 6¼ 0. We seek a contradiction.

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Since r1 ; r2 are distinct, there exists a nonzero c 2 K such that r1 ðcÞ 6¼ r2 ðcÞ. Thus r1 ðcÞ; r2 ðcÞ are nonzero, and a1 r1 ðcÞ þ a2 r2 ðcÞ ¼ 0: Since a1 6¼ 0 and r1 ðcÞ 6¼ 0, we have a2 r2 ðcÞ ¼ a1 r1 ðcÞ 6¼ 0; |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} and hence a2 r2 ðcÞ 6¼ 0. This shows that a2 6¼ 0. Observe that for every u 2 K, we have cu 2 K, and hence a2 r2 ðcÞðr2 ðuÞ r1 ðuÞÞ ¼ ða2 r2 ðcÞÞr1 ðuÞ þ a2 r2 ðcÞr2 ðuÞ ¼ a1 r1 ðcÞr1 ðuÞ þ a2 r2 ðcÞr2 ðuÞ ¼ a1 r1 ðcuÞ þ a2 r2 ðcuÞ ¼ 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus for every u 2 K, a2 r2 ðcÞðr2 ðuÞ r1 ðuÞÞ ¼ 0: Now, since a2 ; r2 ðcÞ are nonzero members of K and ðr2 ðuÞ r1 ðuÞÞ is a member of the field K, we have, for every u 2 K, r2 ðuÞ r1 ðuÞ ¼ 0. Thus for every u 2 K, r1 ðuÞ ¼ r2 ðuÞ. Since c 2 K, we have r1 ðcÞ ¼ r2 ðcÞ. This is a contradiction. Next we consider the case n = 3. If a3 ¼ 0, then from the cases discussed above, we get a contradiction. Hence we have to deal with only the case a3 6¼ 0. Since r1 ; r3 are distinct, there exists a nonzero c 2 K such that r1 ðcÞ 6¼ r3 ðcÞ. Thus r1 ðcÞ; r3 ðcÞ are nonzero, and a1 r1 ðcÞ þ a2 r2 ðcÞ þ a3 r3 ðcÞ ¼ 0: Here condition (1) becomes, for every u 2 K, a1 r1 ðuÞ þ a2 r2 ðuÞ þ a3 r3 ðuÞ ¼ 0. Observe that for every u 2 K, we have cu 2 K, and hence a2 ðr2 ðcÞ r1 ðcÞÞ r2 ðuÞ þ a3 ðr3 ðcÞ r1 ðcÞÞ r3 ðuÞ ¼ ða2 r2 ðuÞ a3 r3 ðuÞÞr1 ðcÞ þ a2 r2 ðcÞr2 ðuÞ þ a3 r3 ðcÞr3 ðuÞ ¼ ða1 r1 ðuÞÞr1 ðcÞ þ a2 r2 ðcÞr2 ðuÞ þ a3 r3 ðcÞr3 ðuÞ ¼ a1 r1 ðcÞr1 ðuÞ þ a2 r2 ðcÞr2 ðuÞ þ a3 r3 ðcÞr3 ðuÞ ¼ a1 r1 ðcuÞ þ a2 r2 ðcuÞ þ a3 r3 ðcuÞ ¼ 0 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

2.2 Galois Groups

105

Thus for every u 2 K, b2 r2 ðuÞ þ b3 r3 ðuÞ ¼ 0; where b2 a2 ðr2 ðcÞ r1 ðcÞÞ and b3 a3 ðr3 ðcÞ r1 ðcÞÞ. Since a3 ; ðr3 ðcÞ r1 ðcÞÞ are nonzero members of the field K, ðb3 ¼Þ a3 ðr3 ðcÞ r1 ðcÞÞ is a nonzero member of the field K, and hence not all bi ði ¼ 2; 3Þ are 0. By our earlier case n = 2, we get a contradiction, etc. 2.2.3 Conclusion Let K be any field. Let r1 ; . . .; rn be n distinct automorphisms of K. There do not exist a1 ; . . .; an 2 K such that 1. for every u 2 K, a1 r1 ðuÞ þ þ an rn ðuÞ ¼ 0, 2. not all ai ði ¼ 1; . . .; nÞ are 0. 2.2.4 Problem Let K be any field. Let G be a nonempty collection automorphisms of K. Put KG fa : a 2 K; and for every r 2 G; rðaÞ ¼ ag: Then KG is a subfield of K. Here we say that KG is the fixed field of G. Proof Let us take an arbitrary r 2 G. It follows that r : K ! K is an automorphism of K, and hence rð0Þ ¼ 0 and rð1Þ ¼ 1. This shows that 0; 1 2 KG . Thus KG is a subset of K, and KG contains at least two elements. Let a; b 2 KG . Let us take an arbitrary r 2 G. It follows by the definition of KG that rðaÞ ¼ a and rðbÞ ¼ b. Hence rða þ bÞ ¼ rðaÞ þ rðbÞ ¼ a þ b and rðabÞ ¼ rðaÞrðbÞ ¼ ab. Thus rða þ bÞ ¼ a þ b and rðabÞ ¼ ab. It follows that ða þ bÞ; ab 2 KG . Next, since rðaÞ ¼ ðrðaÞÞ ¼ a, we have rðaÞ ¼ a. This shows that ðaÞ 2 KG . If a 6¼ 0, then rða1 Þ ¼ ðrðaÞÞ1 ¼ a1 . Thus if a is a ■ nonzero element of KG , then a1 2 KG . Hence KG is a subfield of K. 2.2.5 Problem Let K be any field. The collection of all automorphisms of K is denoted by AutðKÞ. Clearly, AutðKÞ is a group. Proof The identity map Id : a 7! a from K onto K is an automorphism of K, and hence Id 2 AutðKÞ. a. Let r; l 2 AutðKÞ. We have to show that ðrlÞ 2 AutðKÞ. Since r 2 AutðKÞ, r is a one-to-one map from K onto K. Similarly, l is a one-to-one map from K onto K. It follows that the composite map ðrlÞ is a one-to-one map from K onto K. Next, let us take arbitrary a, b in K. We have ðrlÞða þ bÞ ¼ rðlða þ bÞÞ ¼ rðlðaÞ þ lðbÞÞ ¼ rðlðaÞÞ þ rðlðbÞÞ ¼ ðrlÞðaÞ þ ðrlÞðbÞ;

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2 Galois Theory II

and hence ðrlÞða þ bÞ ¼ ðrlÞðaÞ þ ðrlÞðbÞ: Similarly, ðrlÞðabÞ ¼ ðrlÞðaÞ ðrlÞðbÞ. Thus ðrlÞ 2 AutðKÞ. b. Let r 2 AutðKÞ. We have to show that r1 2 AutðKÞ. Since r 2 AutðKÞ, r is a one-to-one map from K onto K, and hence r1 is a one-to-one map from K onto K. Next, let us take arbitrary a, b in K. We have to show that 1. r1 ða þ bÞ ¼ r1 ðaÞ þ r1 ðbÞ, that is, a þ b ¼ rðr1 ðaÞ þ r1 ðbÞÞ, 2. r1 ðabÞ ¼ r1 ðaÞ r1 ðbÞ, that is, ab ¼ rðr1 ðaÞ r1 ðbÞÞ. For 1: RHS ¼ r r1 ðaÞ þ r1 ðbÞ ¼ r r1 ðaÞ þ r r1 ðbÞ ¼ a þ b ¼ LHS: For 2: RHS ¼ r r1 ðaÞ r1 ðbÞ ¼ r r1 ðaÞ r r1 ðbÞ ¼ ab ¼ LHS: ■ 2.2.6 Problem Let K be any field. Let F be a subfield of K. Put GðK; FÞ fr : r 2 AutðKÞ; and for every a 2 F; rðaÞ ¼ ag: Then GðK; FÞ is a subgroup of AutðKÞ. Here GðK; FÞ is called the group of automorphisms of K relative to F. Proof The identity map Id : a 7! a from K onto K is an automorphism of K, and hence Id 2 AutðKÞ. Also, for every a 2 F; IdðaÞ ¼ a. Thus Id 2 GðK; FÞ. Let r; l 2 GðK; FÞ. It suffices to show that ðrl1 Þ 2 GðK; FÞ. To this end, let us take an arbitrary a 2 F. It suffices to show that ðrl1 ÞðaÞ ¼ a. Since l 2 GðK; FÞ and a 2 F, we have lðaÞ ¼ a, and hence l1 ðaÞ ¼ a. LHS ¼ rl1 ðaÞ ¼ r l1 ðaÞ ¼ rðaÞ ¼ a ¼ RHS: ■ 2.2.7 Note Let F and K be any fields such that K is a finite extension of F. It follows that ½K : F \1. Hence there exists a basis fu1 ; u2 ; . . .; un g of the vector space K over F, where n ¼ ½K : F . By 2.2.6, GðK; FÞ is a group of automorphisms of K. We claim that the number of elements of GðK; FÞ is n. Suppose to the contrary that the number of elements of GðK; FÞ is [ n. We seek a contradiction. Since the number of elements of GðK; FÞ is [ n, there exist ðn þ 1Þ distinct automorphisms r1 ; r2 ; . . .; rn þ 1 of K. It follows that for every i 2 f1; 2; . . .; ng and

2.2 Galois Groups

107

for every j 2 f1; 2; . . .; n þ 1g, we have rj ðui Þ 2 K. It follows that the following system of n linear equations in ðn þ 1Þ variables x1 ; x2 ; . . .; xn ; xn þ 1 , 9 r1 ðu1 Þx1 þ r2 ðu1 Þx2 þ þ rn ðu1 Þxn þ rn þ 1 ðu1 Þxn þ 1 ¼ 0 > > > r1 ðu2 Þx1 þ r2 ðu2 Þx2 þ þ rn ðu2 Þxn þ rn þ 1 ðu2 Þxn þ 1 ¼ 0 = ; .. > . > > ; r1 ðun Þx1 þ r2 ðun Þx2 þ þ rn ðun Þxn þ rn þ 1 ðun Þxn þ 1 ¼ 0 has a nontrivial solution ðx1 ; x2 ; . . .; xn ; xn þ 1 Þ ¼ ða1 ; a2 ; . . .; an ; an þ 1 Þð6¼ ð0; 0; . . .; 0; 0ÞÞ in K. It follows that 9 r1 ðu1 Þa1 þ r2 ðu1 Þa2 þ þ rn þ 1 ðu1 Þan þ 1 ¼ 0 > > > r1 ðu2 Þa1 þ r2 ðu2 Þa2 þ þ rn þ 1 ðu2 Þan þ 1 ¼ 0 = ; .. > . > > ; r1 ðun Þa1 þ r2 ðun Þa2 þ þ rn þ 1 ðun Þan þ 1 ¼ 0 that is, 9 a1 r1 ðu1 Þ þ a2 r2 ðu1 Þ þ þ an þ 1 rn þ 1 ðu1 Þ ¼ 0 > > > a1 r1 ðu2 Þ þ a2 r2 ðu2 Þ þ þ an þ 1 rn þ 1 ðu2 Þ ¼ 0 = ; .. > . > > ; a1 r1 ðun Þ þ a2 r2 ðun Þ þ þ an þ 1 rn þ 1 ðun Þ ¼ 0 that is, nX þ1

aj rj ðui Þ ¼ 0

ði ¼ 1; . . .; nÞ:

j¼1

Since ða1 ; a2 ; . . .; an ; an þ 1 Þ 6¼ ð0; 0; . . .; 0; 0Þ, not all ai ði ¼ 1; . . .; n þ 1Þ are 0, and hence by 2.2.3, there exists u 2 K such that a1 r1 ðuÞ þ þ an þ 1 rn þ 1 ðuÞ 6¼ 0: Since u 2 K and fu1 ; u2 ; . . .; un g is P a basis of the vector space K over F, there exist b1 ; b2 ; . . .; bn in F such that u ¼ ni¼1 bi ui . It follows that

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2 Galois Theory II

a1 r1 ðuÞ þ þ an þ 1 rn þ 1 ðuÞ ¼ a1 r1

n X

! bi ui

þ þ an þ 1 rn þ 1

i¼1

n X

! bi ui

i¼1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} n n P P ¼ a1 bi r 1 ð ui Þ þ þ an þ 1 bi rn þ 1 ðui Þ i¼1

¼

n P

a1 bi r1 ðui Þ þ i¼1 nP þ1 P n

n P

i¼1

þ an þ 1 bi rn þ 1 ðui Þ i¼1 ! n nP þ1 P ¼ aj bi rj ðui Þ ¼ aj bi rj ðui Þ j¼1 i¼1 ! i¼1 j¼1 n nP þ1 n P P ¼ bi aj rj ðui Þ ¼ ðbi 0Þ ¼ 0; i¼1

j¼1

i¼1

and hence a1 r1 ðuÞ þ þ an þ 1 rn þ 1 ðuÞ ¼ 0: This is a contradiction. 2.2.8 Conclusion Let F and K be any fields such that K is a finite extension of F. Then oðGðK; FÞÞ ½K : F . Definition Let F be a field. By 1.3.7, F ½x1 ; . . .; xn is an integral domain. Its field of quotients is denoted by F ðx1 ; . . .; xn Þ. The members of F ðx1 ; . . .; xn Þ are called rational functions in x1 ; . . .; xn over F. By Sn we shall mean the permutation group fr : r : f1; 2; . . .; ng ! f1; 2; . . .; ng is one-to-one and ontog; which is called the symmetric group of degree n. 2.2.9 Problem Observe that for every r 2 Sn , the mapping r : r ðx1 ; . . .; xn Þ 7! r xrð1Þ ; . . .; xrðnÞ from F ðx1 ; . . .; xn Þ to F ðx1 ; . . .; xn Þ is an automorphism of the field F ðx1 ; . . .; xn Þ: For simplicity, r is also denoted by r. Thus we can treat Sn as a group of automorphisms of the field F ðx1 ; . . .; xn Þ. n Þ r ðx1 ;...;xn Þ where Proof Suppose that pqððxx11 ;...;x ;...;xn Þ ; sðx1 ;...;xn Þ 2 F ðx1 ; . . .; xn Þ, qðx1 ; . . .; xn Þ; r ðx1 ; . . .; xn Þ; sðx1 ; . . .; xn Þ 2 F ½x1 ; . . .; xn . Here

pðx1 ; . . .; xn Þ;

2.2 Galois Groups

109

pð x 1 ; ; x n Þ r ð x 1 ; ; x n Þ þ r qðx1 ; ; xn Þ sðx1 ; ; xn Þ

pðx1 ; ; xn Þsðx1 ; ; xn Þ þ r ðx1 ; ; xn Þqðx1 ; ; xn Þ ¼r qðx1 ; ; xn Þsðx1 ; ; xn Þ p xrð1Þ ; ; xrðnÞ s xrð1Þ ; ; xrðnÞ þ r xrð1Þ ; ; xrðnÞ q xrð1Þ ; ; xrðnÞ ¼ q xrð1Þ ; ; xrðnÞ s xrð1Þ ; ; xrðnÞ p xrð1Þ ; ; xrðnÞ r xrð1Þ ; ; xrðnÞ ¼ þ q xrð1Þ ; ; xrðnÞ s xrð1Þ ; ; xrðnÞ

pð x 1 ; ; x n Þ

r ð x1 ; ; xn Þ ¼r þr ; qð x 1 ; ; x n Þ sðx1 ; ; xn Þ

so

r

pðx1 ; . . .; xn Þ r ðx1 ; . . .; xn Þ þ qðx1 ; . . .; xn Þ sðx1 ; . . .; xn Þ

pðx1 ; . . .; xn Þ

r ðx1 ; . . .; xn Þ ¼r þr : qðx1 ; . . .; xn Þ sðx1 ; . . .; xn Þ

Next, pð x 1 ; ; x n Þ r ð x 1 ; ; x n Þ pðx1 ; ; xn Þr ðx1 ; ; xn Þ ¼ r

qð x 1 ; ; x n Þ s ð x 1 ; ; x n Þ qðx1 ; ; xn Þsðx1 ; ; xn Þ p xrð1Þ ; ; xrðnÞ r xrð1Þ ; ; xrðnÞ p xrð1Þ ; ; xrðnÞ r xrð1Þ ; ; xrðnÞ ¼ ¼ q xrð1Þ ; ; xrðnÞ s xrð1Þ ; ; xrðnÞ q xrð1Þ ; ; xrðnÞ s xrð1Þ ; ; xrðnÞ p ð x ; ; x Þ r ð x ; ; x Þ 1 n 1 n ¼ r

r

; qð x 1 ; ; x n Þ s ð x1 ; ; xn Þ

r

so

r

pðx1 ; . . .; xn Þ r ðx1 ; . . .; xn Þ qðx1 ; . . .; xn Þ sðx1 ; . . .; xn Þ

pðx1 ; . . .; xn Þ

r ðx1 ; . . .; xn Þ ¼r r : qðx1 ; . . .; xn Þ sðx1 ; . . .; xn Þ

Thus r preserves addition and multiplication. r : F ðx1 ; . . .; xn Þ ! F ðx1 ; . . .; xn Þ is one-to-one. To show this, let

r ðr ðx1 ; . . .; xn ÞÞ ¼ r ðsðx1 ; . . .; xn ÞÞ, where r ðx1 ; . . .; xn Þ; sðx1 ; . . .; xn Þ 2 F ðx1 ; . . .; xn Þ. We have to show that r ðx1 ; . . .; xn Þ ¼ sðx1 ; . . .; xn Þ. Since r xrð1Þ ; . . .; xrðnÞ ¼ r ðr ðx1 ; . . .; xn ÞÞ ¼ r ðsðx1 ; . . .; xn ÞÞ ¼ s xrð1Þ ; . . .; xrðnÞ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

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2 Galois Theory II

we have r xrð1Þ ; . . .; xrðnÞ ¼ s xrð1Þ ; . . .; xrðnÞ , and hence r ðx1 ; ; xn Þ ¼ r xr1 ðrð1ÞÞ ; ; xr1 ðrðnÞÞ ¼ r1 r xrð1Þ ; ; xrðnÞ ¼ r1 s xrð1Þ ; ; xrðnÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ s xr1 ðrð1ÞÞ ; ; xr1 ðrðnÞÞ ¼ sðx1 ; ; xn Þ: Thus r ðx1 ; . . .; xn Þ ¼ sðx1 ; . . .; xn Þ. r : F ðx1 ; . . .; xn Þ ! F ðx1 ; . . .; xn Þ is onto. To show this, let us take an arbitrary r ðx1 ; . . .; xn Þ 2 F ðx1 ; . . .; xn Þ. Put sðx1 ; . . .; xn Þ r xr1 ð1Þ ; . . .; xr1 ðnÞ . Here r ðsðx1 ; . . .; xn ÞÞ ¼ r r xr1 ð1Þ ; . . .; xr1 ðnÞ ¼ r xrðr1 ð1ÞÞ ; . . .; xrðr1 ðnÞÞ ¼ r ðx1 ; . . .; xn Þ; so r ðsðx1 ; . . .; xn ÞÞ ¼ r ðx1 ; . . .; xn Þ: Thus r : F ðx1 ; . . .; xn Þ ! F ðx1 ; . . .; xn Þ is an automorphism of F ðx1 ; . . .; xn Þ: ■ 2.2.10 Definition Let F be a field. We know that F ðx1 ; . . .; xn Þ is a field extension of F, and Sn is a group of automorphisms of F ðx1 ; . . .; xn Þ. Here the fixed field of Sn is denoted by S. Thus S ¼ fr ðx1 ; . . .; xn Þ : r ðx1 ; . . .; xn Þ 2 F ðx1 ; . . .; xn Þ; and for every r 2 Sn ; rðr ðx1 ; . . .; xn ÞÞ ¼ r ðx1 ; . . .; xn Þg; that is, S ¼ fr ðx1 ; . . .; xn Þ : r ðx1 ; . . .; xn Þ 2 F ðx1 ; . . .; xn Þ; and for every

r 2 Sn ; r xrð1Þ ; . . .; xrðnÞ ¼ r ðx1 ; . . .; xn Þ : By 2.2.4, S is a subfield of F ðx1 ; . . .; xn Þ. Also F S. Thus F S F ðx1 ; . . .; xn Þ. The members of S are called symmetric rational functions. Thus S is the field of symmetric rational functions. 2.2.11 Example Suppose that n = 3. Here S3 ¼ fr1 ; r2 ; r3 ; r4 ; r5 ; r6 g;

2.2 Galois Groups

111

where

1 1 1 3

r1

2 2 2 2

3 12 ; r2 3 2 3 3 12 ; r6 1 21

3 123 123 ; r3 ; r4 ; r5 1 312 132 3 : 3

Observe that x2 x3 þ x3 x1 þ x1 x2 is a symmetric rational function. Verification: We must show that xri ð2Þ xri ð3Þ þ xri ð3Þ xri ð1Þ þ xri ð1Þ xri ð2Þ ¼ x2 x3 þ x 3 x1 þ x1 x2 For

ði ¼ 1; 2; 3; 4; 5; 6Þ:

i ¼ 1 : LHS ¼ xr1 ð2Þ xr1 ð3Þ þ xr1 ð3Þ xr1 ð1Þ þ xr1 ð1Þ xr1 ð2Þ ¼ x2 x3 þ x3 x1 þ x1 x2 ¼ RHS:

For

i ¼ 2 : LHS ¼ xr2 ð2Þ xr2 ð3Þ þ xr2 ð3Þ xr2 ð1Þ þ xr2 ð1Þ xr2 ð2Þ ¼ x3 x1 þ x1 x2 þ x2 x3 ¼ x2 x3 þ x3 x1 þ x1 x2 ¼ RHS:

For


For


For


For

i ¼ 6 : LHS ¼ xr6 ð2Þ xr6 ð3Þ þ xr6 ð3Þ xr6 ð1Þ þ xr6 ð1Þ xr6 ð2Þ ¼ x1 x3 þ x3 x2 þ x2 x1 ¼ x2 x3 þ x3 x1 þ x1 x2 ¼ RHS: Verified.

Definition Similarly, x1 x2 x3 is a symmetric rational function, and x1 þ x2 þ x3 is a symmetric rational function. The symmetric rational functions x1 þ x2 þ x3 , x2 x3 þ x3 x1 þ x1 x2 , and x1 x2 x3 are called the elementary symmetric functions. The elementary symmetric function x1 þ x2 þ x3 is denoted by a1 , the elementary symmetric function x2 x3 þ x3 x1 þ x1 x2 is denoted by a2 , and the elementary symmetric function x1 x2 x3 is denoted by a3 . Thus ðfa1 ; a2 ; a3 g [ F Þ S, and hence the smallest field F ða1 ; a2 ; a3 Þ containing fa1 ; a2 ; a3 g [ F is contained in S. In short,

112

2 Galois Theory II

F F ða1 ; a2 ; a3 Þ S F ðx1 ; x2 ; x3 Þ: Since F ðx1 ; x2 ; x3 Þ is an extension of the field S, by 2.2.6, GðF ðx1 ; x2 ; x3 Þ; SÞ ( ) r : r 2 AutðF ðx1 ; x2 ; x3 ÞÞ; and for every r ðx1 ; x2 ; x3 Þ ¼ 2 S; rðr ðx1 ; x2 ; x3 ÞÞ ¼ r ðx1 ; x2 ; x3 Þ ( ) r : r 2 AutðF ðx1 ; x2 ; x3 ÞÞ; and for every r ðx1 ; x2 ; x3 Þ ¼ 2 S; r xrð1Þ ; xrð2Þ ; xrð3Þ ¼ r ðx1 ; x2 ; x3 Þ is a group of automorphisms of F ðx1 ; x2 ; x3 Þ. 2.2.12 Problem Clearly, S3 GðF ðx1 ; x2 ; x3 Þ; SÞ. Proof To show this, let us take an arbitrary ri 2 S3 , where i 2 f1; 2; 3; 4; 5; 6g. We have to show that 1. ri 2 AutðF ðx1 ; x2 ; x3 ÞÞ, 2. for every r ðx1 ; x2 ; x3 Þ 2 S; r xri ð1Þ ; xri ð2Þ ; xri ð3Þ ¼ r ðx1 ; x2 ; x3 Þ. For 1: Since we can treat S3 as a group of automorphisms of the field F ðx1 ; x2 ; x3 Þ, and ri 2 S3 , we have ri 2 AutðF ðx1 ; x2 ; x3 ÞÞ. For 2: Let us take an arbitrary r ðx1 ; x2 ; x3 Þ 2 S. Now, since ri 2 S3 , by the ■ definition of S, r xri ð1Þ ; xri ð2Þ ; xri ð3Þ ¼ r ðx1 ; x2 ; x3 Þ. 2.2.13 Note Since S3 GðF ðx1 ; x2 ; x3 Þ; SÞ, we have 3! ¼ oðS3 Þ oðGðF ðx1 ; x2 ; x3 Þ; SÞÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence 3! oðGðF ðx1 ; x2 ; x3 Þ; SÞÞ: Since F ða1 ; a2 ; a3 Þ F ðx1 ; x2 ; x3 Þ, the field F ðx1 ; x2 ; x3 Þ is an extension of F ða1 ; a2 ; a3 Þ. Observe that 3 t þ ða1 Þt2 þ a2 t þ ða3 Þ 2 ðF ða1 ; a2 ; a3 ÞÞ½t and t3 þ ða1 Þt2 þ a2 t þ ða3 Þ ¼ t3 ðx1 þ x2 þ x3 Þt2 þ ðx2 x3 þ x3 x1 þ x1 x2 Þt x1 x2 x3 ¼ ðt x1 Þðt x2 Þðt x3 Þ:

2.2 Galois Groups

113

So t3 þ ða1 Þt2 þ a2 t þ ða3 Þ ¼ ðt x1 Þðt x2 Þðt x3 Þ: It follows that F ðx1 ; x2 ; x3 Þ contains all the roots x1 ; x2 ; x3 of t3 þ ða1 Þt2 þ a2 t þ ða3 Þ in F ðx1 ; x2 ; x3 Þ. We claim that F ðx1 ; x2 ; x3 Þ is a splitting field over F ða1 ; a2 ; a3 Þ for t3 þ ða1 Þt2 þ a2 t þ ða3 Þ. Suppose to the contrary that G is a proper subfield of F ðx1 ; x2 ; x3 Þ that contains all the roots x1 ; x2 ; x3 of t3 þ ða1 Þt2 þ a2 t þ ða3 Þ in F ðx1 ; x2 ; x3 Þ. We seek a contradiction. Since G contains F [ fx1 ; x2 ; x3 g, and G is a field, G contains F ðx1 ; x2 ; x3 Þ. This contradicts the fact that G is a proper subset of F ðx1 ; x2 ; x3 Þ. Hence our claim is substantiated, that is, F ðx1 ; x2 ; x3 Þ is a splitting field over F ða1 ; a2 ; a3 Þ for t3 þ ða1 Þt2 þ a2 t þ ða3 Þ. By 1.5.9, ½ðsplitting field over F ða1 ; a2 ; a3 Þ for t3 þ ða1 Þt2 þ a2 t þ ða3 ÞÞ : F ð1 ; a2 ; a3 Þ ðdegðt3 þ ða1 Þt2 þ a2 t þ ða3 ÞÞÞ!; so ½F ðx1 ; x2 ; x3 Þ : F ða1 ; a2 ; a3 Þ deg t3 þ ða1 Þt2 þ a2 t þ ða3 Þ !; and hence ½F ðx1 ; x2 ; x3 Þ : F ða1 ; a2 ; a3 Þ 3!: Thus F ðx1 ; x2 ; x3 Þ is a finite extension of F ða1 ; a2 ; a3 Þ. Now, since F ða1 ; a2 ; a3 Þ S F ðx1 ; x2 ; x3 Þ, by 1.4.4, S is a finite extension of F ða1 ; a2 ; a3 Þ, and F ðx1 ; x2 ; x3 Þ is a finite extension of S. Also, by 1.4.3, ½F ðx1 ; x2 ; x3 Þ : F ða1 ; a2 ; a3 Þ ¼ ½F ðx1 ; x2 ; x3 Þ : S½S : F ða1 ; a2 ; a3 Þ: Since F ðx1 ; x2 ; x3 Þ is a finite extension of S, by 2.2.8, we have 3! oðGðF ðx1 ; x2 ; x3 Þ; SÞÞ ½F ðx1 ; x2 ; x3 Þ : S ½F ðx1 ; x2 ; x3 Þ : S½S : F ða1 ; a2 ; a3 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ½F ðx1 ; x2 ; x3 Þ : F ða1 ; a2 ; a3 Þ 3!; and hence ½F ðx1 ; x2 ; x3 Þ : S ¼ 3!:

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Also ½F ðx1 ; x2 ; x3 Þ : S ¼ 3!: oðGðF ðx1 ; x2 ; x3 Þ; SÞÞ ¼ 3!ð¼ oðS3 ÞÞ and ½F ðx1 ; x2 ; x3 Þ : S ¼ ½F ðx1 ; x2 ; x3 Þ : S½S : F ða1 ; a2 ; a3 Þ: Since ½F ðx1 ; x2 ; x3 Þ : S ¼ 3!, we have ½S : F ða1 ; a2 ; a3 Þ ¼ 1, and hence S ¼ F ða1 ; a2 ; a3 Þ. Since S3 GðF ðx1 ; x2 ; x3 Þ; SÞ and oðGðF ðx1 ; x2 ; x3 Þ; SÞÞ ¼ 3!ð¼ oðS3 ÞÞ, we have S3 ¼ GðF ðx1 ; x2 ; x3 Þ; SÞ. 2.2.14 Conclusion Let F be a field. Let n be a positive integer. Then 1. 2. 3. 4.

½F ðx1 ; . . .; xn Þ : S ¼ n!; GðF ðx1 ; . . .; xn Þ; SÞ ¼ Sn ; S ¼ F ða1 ; . . .; an Þ; F ðx1 ; . . .; xn Þ is a splitting field over S for tn a1 tn1 þ a2 tn2 . . . þ ð1Þn an , where the symbols have their usual meanings.

2.2.15 Note Let F and K be any fields such K is a finite extension of F. It follows that ½K : F \1. Hence there exists a basis fu1 ; u2 ; . . .; un g of the vector space K over F, where n ¼ ½K : F . By 2.2.6, GðK; FÞ is a group of automorphisms of K, where GðK; FÞ fr : r 2 AutðKÞ; and for every a 2 F; rðaÞ ¼ ag: Further, by 2.2.8, oðGðK; FÞÞ ½K : F ¼ n. Here the fixed field of GðK; FÞ is |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} fa : a 2 K; and for every r 2 GðK; FÞ; rðaÞ ¼ ag ð FÞ; so F ðfixed field of GðK; FÞÞ K. It follows that ðfixed field of GðF; FÞÞ ¼ F: Definition Let F and K be any fields such that F K. Let K be a finite extension of F. If F ¼ ðfixed field of GðK; FÞÞ, that is, ðfixed field of GðK; FÞÞ F, then we say that K is a normal extension of F. Since ðfixed field of GðF; FÞÞ ¼ F, and ½F : F ¼ 1\1, F is a normal extension of F.

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115

2.2.16 Note Let F and K be any fields such that K is a normal extension of F. Let H be a subgroup of the group GðK; FÞ ð AutðKÞÞ. Let KH be the fixed field of H, that is, ðK ÞKH ¼ fa : a 2 K; and for every r 2 H; rðaÞ ¼ ag ð FÞ: Since K is a normal extension of F, K is a finite extension of F, and hence ½K : F \1. Next, by 2.2.7, oðGðK; FÞÞ ½K : F. Since H is a subgroup of the group GðK; FÞ, we have oðHÞ oðGðK; FÞÞ ½K : F\1, and hence oðHÞ\1. |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} This shows that H is a finite subgroup of the finite group GðK; FÞ. Further, F KH K. Observe that GðK; KH Þ ¼ fr : r 2 AutðKÞ; and for every a 2 KH ; rðaÞ ¼ ag ð HÞ: Since F KH K, and K is a finite extension of F, by 1.4.4, K is a finite extension of KH , and hence by 2.2.7, oðGðK; KH ÞÞ ½K : KH \1. Now, since H GðK; KH Þ, we have oðHÞ oðGðK; KH ÞÞ ½K : KH \1; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

ð Þ

and hence oðHÞ ½K : KH . Since 1 ½K : KH \1, there exist a1 ; . . .; am 2 K such that fa1 ; . . .; am g is a basis of the vector space K over the field KH . It follows that for every x 2 K, there exist a1 ; . . .; am 2 KH such that x ¼ ða1 a1 þ þ am am Þ 2 KH ða1 ; . . .; am Þ: Thus K KH ða1 ; . . .; am Þ. Since KH [ fa1 ; . . .; am g K, KH ða1 ; . . .; am Þ K KH ða1 ; . . .; am Þ, and hence |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

we

have

K ¼ KH ða1 ; . . .; am Þ: Since F KH KH ða1 Þ KH ða1 ; . . .; am Þ ¼ K, we have F KH ða1 Þ K. Since K is a normal extension of F, K is a finite extension of F. Since F KH K, by 1.4.4, K is a finite extension of KH . Next, since KH KH ða1 Þ K, by 1.4.4, KH ða1 Þ is a finite extension of KH , and hence by 1.4.9, a1 is algebraic over KH . Similarly, a2 is algebraic over KH , etc. By 2.1.16, there exists a 2 KH ða1 ; ; am Þ such that KH ða1 ; ; am Þ ¼ KH ðaÞ:

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Since K ¼ KH ða1 ; . . .; am Þ, we have a 2 K and K ¼ KH ðaÞ. Since K is a normal extension of F, ðKH ðaÞ ¼ÞK is a finite extension of F, and hence KH ðaÞ is a finite extension of F. Since F KH K ¼ KH ðaÞ, we have F KH KH ðaÞ. Now, since KH ðaÞ is a finite extension of F, by 1.4.4, KH ðaÞ is a finite extension of KH , and hence by 1.4.9, a is algebraic over KH . Let a be algebraic of degree n over KH . By 1.4.11, there exists qðxÞ 2 KH ½x such that qðxÞ is the minimal polynomial of a over KH , that is, 1. ðK3ÞqðaÞ ¼ 0; 2. n ¼ degðqðxÞÞ 1; 3. the leading coefficient of qðxÞ is 1. Again by 1.4.12, qðxÞ is irreducible over KH . Also, since a is algebraic of degree n over KH , by 1.4.16, we have ½K : KH ¼ ½KH ðaÞ : KH ¼ n ¼ degðqðxÞÞ; |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½K : KH ¼ degðqðxÞÞ. Since H is a finite subgroup of the finite group GðK; FÞ, we can suppose that H ¼ fr1 ; r2 ; . . .; rh gð GðK; FÞÞ; oðH Þ ¼ h, and r1 is the identity element of the group GðK; FÞ. It follows that each ri ðaÞ 2 K, and r1 ðaÞ ¼ a. Put a1 r1 ðaÞ þ rh ðaÞ ð2 K Þ; Pr2 ðaÞ þ þ a2 ri ðaÞrj ðaÞ ð2 K Þ; P i\j a3 ri ðaÞrj ðaÞrk ðaÞ ð2 K Þ; i\j\k

.. .

We want to show that a1 2 KH ð¼ fa : a 2 K and for every r 2 H; rðaÞ ¼ agÞ: To this end, let us take an arbitrary ri 2 H, where i 2 f1; 2; . . .; hg. It suffices to show that ri ða1 Þ ¼ a1 , that is, ri ðr1 ðaÞ þ r2 ðaÞ þ þ rh ðaÞÞ ¼ r1 ðaÞ þ r2 ðaÞ þ þ rh ðaÞ: Observe that ri ðr1 ðaÞ þ r2 ðaÞ þ þ rh ðaÞÞ ¼ ri ðr1 ðaÞÞ þ þ ri ðrh ðaÞÞ ¼ ðri r1 ÞðaÞ þ þ ðri rh ÞðaÞ:

2.2 Galois Groups

117

Also, rj 7! ri rj is a one-to-one mapping from fr1 ; r2 ; . . .; rh g onto fr1 ; r2 ; . . .; rh g, so LHS ¼ ri ðr1 ðaÞ þ r2 ðaÞ þ þ rh ðaÞÞ ¼ ðri r1 ÞðaÞ þ þ ðri rh ÞðaÞ ¼ r1 ðaÞ þ r2 ðaÞ þ þ rh ðaÞ ¼ RHS: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus a1 2 KH . Next, we want to show that a2 2 KH ð¼ fa : a 2 K and for every r 2 H; rðaÞ ¼ agÞ. To this end, let us take an arbitrary ri 2 H, where i 2 f1; 2; . . .; hg. It suffices to show that ri ða2 Þ ¼ a2 , that is, ri ððr1 ðaÞr2 ðaÞ þ r1 ðaÞr3 ðaÞ þ Þ þ ðr2 ðaÞr3 ðaÞ þ r2 ðaÞr4 ðaÞ þ Þ þ Þ ¼ ðr1 ðaÞr2 ðaÞ þ r1 ðaÞr3 ðaÞ þ Þ þ ðr2 ðaÞr3 ðaÞ þ r2 ðaÞr4 ðaÞ þ Þ þ : Observe that ri ððr1 ðaÞr2 ðaÞ þ r1 ðaÞr3 ðaÞ þ Þ þ ðr2 ðaÞr3 ðaÞ þ r2 ðaÞr4 ðaÞ þ Þ þ Þ ¼ ðri ðr1 ðaÞr2 ðaÞÞ þ ri ðr1 ðaÞr3 ðaÞÞ þ Þ þ ðri ðr2 ðaÞr3 ðaÞÞ þ ri ðr2 ðaÞr4 ðaÞÞ Þ þ ¼ ðri ðr1 ðaÞÞri ðr2 ðaÞÞ þ ri ðr1 ðaÞÞri ðr3 ðaÞÞ þ Þ þ ðri ðr2 ðaÞÞri ðr3 ðaÞÞ þ ri ðr2 ðaÞÞri ðr4 ðaÞÞ þ Þ þ ¼ ððri r1 ÞðaÞðri r2 ÞðaÞ þ ðri r1 ÞðaÞðri r3 ÞðaÞ þ Þ þ ððri r2 ÞðaÞðri r3 ÞðaÞ þ ðri r2 ÞðaÞðri r4 ÞðaÞ þ Þ þ ¼ ðri r1 ÞðaÞððri r2 ÞðaÞ þ ðri r3 ÞðaÞ þ Þ þ ðri r2 ÞðaÞððri r3 ÞðaÞ þ ðri r4 ÞðaÞ þ Þ þ : Also, rj 7! ri rj is a one-to-one mapping from fr1 ; r2 ; . . .; rh g onto fr1 ; r2 ; . . .; rh g, so ðri r1 ÞðaÞððri r2 ÞðaÞ þ ðri r3 ÞðaÞ þ Þ þ ðri r2 ÞðaÞððri r3 ÞðaÞ þ ðri r4 ÞðaÞ þ Þ þ ¼ ðr1 ðaÞr2 ðaÞ þ r1 ðaÞr3 ðaÞ þ Þ þ ðr2 ðaÞr3 ðaÞ þ r2 ðaÞr4 ðaÞ þ Þ þ ; and hence LHS = RHS. Thus a2 2 KH . Similarly, a3 2 KH , etc. It follows that

xh a1 xh1 þ a2 xh2 þ ð1Þh ah 2 KH ½x:

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Since xh a1 xh1 þ a2 xh2 þ ð1Þh ah ¼ ðx r1 ðaÞÞðx r2 ðaÞÞ ðx rh ðaÞÞ ¼ ðx aÞðx r2 ðaÞÞ ðx rh ðaÞÞ; it follows that a is a root of the polynomial pðxÞ in KH ½x, where pðxÞ xh a1 xh1 þ a2 xh2 þ ð1Þh ah : Hence pðxÞ 2 KH ½x, pðaÞ ¼ 0, h ¼ degðpðxÞÞ 1, and the leading coefficient of pðxÞ is 1. Now, since qðxÞ is the minimal polynomial of a over KH , we have ½K : KH ¼ degðqðxÞÞ degðpðxÞÞ ¼ h ¼ oðH Þ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½K : KH oðH Þ. Since oðH Þ ½K : KH , we have oð H Þ ¼ ½ K : K H Next, from (*), oðGðK; KH ÞÞ ¼ oðH Þ. Now, since H GðK; KH Þ, we have H ¼ GðK; KH Þ . We can substitute GðK; FÞ for H in oðH Þ ¼ ½K : KH . We get oðGðK; FÞÞ ¼ K : KGðK;FÞ . Now observe that KGðK;FÞ ¼ fb : b 2 K; and for every r 2 GðK; FÞ; rðbÞ ¼ bg: Since K is a normal extension of F, we have F ¼ ðfixed field of GðK; FÞÞ ¼ fb : b 2 K and for every r 2 GðK; FÞ; rðbÞ ¼ bg |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ KGðK;FÞ ; and hence KGðK;FÞ ¼ F. Since oðGðK; FÞÞ ¼ K : KGðK;FÞ , we have oðGðK; FÞÞ ¼ ½K : F : We can substitute GðK; FÞ for H in K ¼ KH ðaÞ. We get K ¼ KGðK;FÞ ðaÞ ¼ FðaÞ, |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence K ¼ FðaÞ;

2.2 Galois Groups

119

where a 2 K. Further, h ¼ oðH Þ ¼ oðGðK; FÞÞ ¼ ½K : F ¼ K : KGðK;FÞ ¼ ½K : KH ¼ degðqðxÞÞ ¼ n; so h ¼ n. Now pðxÞ ¼ xh a1 xh1 þ a2 xh2 þ ð1Þh ah becomes pðxÞ ¼ xn a1 xn1 þ a2 xn2 þ ð1Þn an : Next, pðxÞ 2 KH ½x becomes ðxn a1 xn1 þ a2 xn2 þ ð1Þn an ¼Þ pðxÞ 2 F½x, and hence each ai is in F. Also pðaÞ ¼ 0, h ¼ degðpðxÞÞ 1, and the leading coefficient of pðxÞ is 1. Further, xh a1 xh1 þ a2 xh2 þ ð1Þh ah ¼ ðx r1 ðaÞÞðx r2 ðaÞÞ ðx rh ðaÞÞ ¼ ðx aÞðx r2 ðaÞÞ ðx rh ðaÞÞ becomes xn a1 xn1 þ a2 xn2 þ ð1Þn an ¼ ðx r1 ðaÞÞðx r2 ðaÞÞ ðx rn ðaÞÞ ¼ ðx aÞðx r2 ðaÞÞ ðx rn ðaÞÞ; and H ¼ fr1 ; r2 ; ; rh g becomes GðK; FÞ ¼ fr1 ; r2 ; ; rn g: Since each ri ðaÞ is in K and ðF½x3ÞpðxÞ ¼ ðx r1 ðaÞÞðx r2 ðaÞÞ. . . ðx rn ðaÞÞ; K splits the polynomial pðxÞ in F½x into a product of linear factors in K½x. We shall show that K is a splitting field over F for pðxÞ. Assume to the contrary that G is a proper subfield of K ð¼ FðaÞÞ that contains F as well as all the roots of pðxÞ in K. We seek a contradiction. Since r1 ðaÞ is a root of pðxÞ in K, and G contains all the roots of pðxÞ in K, we have r1 ðaÞ 2 G. Now, since r1 ðaÞ ¼ a, we have a 2 G. Thus F [ fag G, and hence K ¼ FðaÞ G. Thus K G. This contradicts the fact that G is a proper |fflfflfflfflfflffl{zfflfflfflfflfflffl} subset of K. Thus we have shown that K is a splitting field over F for pðxÞ. 2.2.17 Conclusion Let F and K be any fields such that K is a normal extension of F. Let H be a subgroup of the group GðK; FÞ ð AutðKÞÞ. Let KH be the fixed field of H. Then

120

1. 2. 3. 4.

2 Galois Theory II

oðH Þ ¼ ½K : KH , H ¼ GðK; KH Þ, oðGðK; FÞÞ ¼ ½K : F , there exists a 2 K such that K ¼ FðaÞ, and K is a splitting field over F for ðx r1 ðaÞÞðx r2 ðaÞÞ ðx rn ðaÞÞ in F½x, where GðK; FÞ ¼ fr1 ; r2 ; . . .; rn g and r1 ðaÞ ¼ a.

2.2.18 Note Let F and K be any fields such that F K. Let f ðxÞ 2 F½x. Let K be a splitting field over F for f ðxÞ. Suppose that degðf ðxÞÞ 1. Let pðxÞ be an irreducible factor of f ðxÞ in F½x. Suppose that all the roots of pðxÞ are a1 ; a2 ; . . .; ar . Since pðxÞ is a factor of f ðxÞ in F½x, all the roots of pðxÞ are the roots of f ðxÞ. Now, since a1 ; a2 ; . . .; ar are the roots of pðxÞ, a1 ; a2 ; . . .; ar are the roots of f ðxÞ. Since K is a splitting field over F for f ðxÞ, K contains all the roots of f ðxÞ. Since a1 ; a2 ; . . .; ar are the roots of f ðxÞ, K contains a1 ; a2 ; . . .; ar . Let us fix an arbitrary i 2 f2; 3; . . .; r g. Since a1 ; ai are members of K, pðxÞ 2 F½xð F Þ, pðxÞ is irreducible over F, and a1 ; ai are the roots of pðxÞ in K, by 1.5.18, there exists an isomorphism si from the field F ða1 Þð K Þ onto the field F ðai Þ such that 1. si ða1 Þ ¼ ai , 2. for every a 2 F, si ðaÞ ¼ a. Since K is a splitting field over F for f ðxÞ, K is a finite extension of F. Since F [ fa1 g K, we have F F ða1 Þ K. Since K is a finite extension of F, by 1.4.4, K is a finite extension of F ða1 Þ. Since f ðxÞ 2 F½xð ðF ða1 ÞÞ½xÞ, we have f ðxÞ 2 ðF ða1 ÞÞ½x. We want to show that K is a splitting field over F ða1 Þ for f ðxÞ. To this end, let us take a proper subfield G of K that contains F ða1 Þð F Þ. It suffices to show that G does not contain all the roots of f ðxÞ. Since G is a proper subfield of K that contains F, and K is a splitting field over F for f ðxÞ, G does not contain all the roots of f ðxÞ. Thus K is a splitting field for f ðxÞ considered a polynomial over F1 , where F1 F ða1 Þ. Similarly, K is a splitting field for f ðxÞ considered a polynomial over ðF1 Þ0 , where ðF1 Þ0 F ðai Þ. Now, by 1.5.27, there exists a ring isomorphism ri from K onto K such that for every a 2 F1 ð¼ F ða1 Þ F [ fa1 gÞ, ri ðaÞ ¼ si ðaÞ. Hence for every a 2 F, ri ðaÞ ¼ si ðaÞ ¼ a, and hence for every a 2 F, |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ri ðaÞ ¼ a. Also, ri ða1 Þ ¼ si ða1 Þ ¼ ai . Thus ri ða1 Þ ¼ ai . Since ri is a ring iso|fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} morphism from the field K onto K, we have ri 2 AutðKÞ. Next, since for every a 2 F, ri ðaÞ ¼ a, we have ri 2 GðK; FÞ. 2.2.19 Conclusion Let F and K be any fields such that F K. Let f ðxÞ 2 F½x. Let K be a splitting field over F for f ðxÞ. Suppose that degðf ðxÞÞ 1. Let pðxÞ be an irreducible factor of f ðxÞ in F½x. Suppose that all the roots of pðxÞ are a1 ; a2 ; . . .; ar . Then for every i 2 f1; 2; . . .; r g, there exists ri 2 GðK; FÞ such that ri ða1 Þ ¼ ai .

2.2 Galois Groups

121

2.2.20 Problem Let F and K be any fields such that K is an extension of F. Let f ðxÞ 2 F½x. Let K be a splitting field over F for f ðxÞ. Suppose that degðf ðxÞÞ 1. Then K is a normal extension of F. Proof Case I: f ðxÞ splits into linear factors over F. Since f ðxÞ splits into linear factors over F, F is a splitting field over F for f ðxÞ. Now, since K is a splitting field over F for f ðxÞ, by 1.5.29, K ¼ F. Since F is a normal extension of F, K is a normal extension of F. Case II: f ðxÞ does not split into linear factors over F. For induction on ½K : F , let us assume that for every pair of fields K1 ; F1 of degree \½K : F , ðK1 is a splitting field over F1 of some polynomial in F1 ½xÞ ) ðK1 is a normal extension of F1 Þ: ð Þ Since f ðxÞ does not split into linear factors over F, by 1.2.21, there exists an irreducible factor pðxÞ of f ðxÞ in F½x such that degðpðxÞÞ 2. Suppose that all the roots of pðxÞ are a1 ; a2 ; . . .; ar , where r degðpðxÞÞð 2Þ. Since K is a splitting field over F for f ðxÞ, K is a finite extension of F. Since a1 is a root of pðxÞ, and pðxÞ is irreducible over F, a1 is a nonzero. Since pðxÞ is a factor of f ðxÞ in F½x, all the roots of pðxÞ are roots of f ðxÞ. Since a1 ; a2 ; . . .; ar are the roots of pðxÞ, a1 ; a2 ; . . .; ar are the roots of f ðxÞ. Since K is a splitting field over F for f ðxÞ, K contains all the roots of f ðxÞ. Since a1 ; a2 ; . . .; ar are the roots of f ðxÞ, K contains a1 ; a2 ; . . .; ar . It follows that F [ fa1 g K, and hence F F ða1 Þ K. Now, since K is a finite extension of F, by 1.4.4, F ða1 Þ is a finite extension of F, and hence by 1.4.9, a1 is algebraic over F. By 1.4.3, ½K : F ¼ ½K : F ða1 Þ½F ða1 Þ : F : Since a1 is a root of pðxÞ, we have pða1 Þ ¼ 0. Further, since pðxÞ is irreducible in F½x and r ¼ degðpðxÞÞ 2, a1 is algebraic of degree r ð 2Þ over F. Now, by 1.4.16, ½F ða1 Þ : F ¼ r, and hence ½K : F ¼ ½K : F ða1 Þ½F ða1 Þ : F ¼ ½K : F ða1 Þr [ ½K : F ða1 Þ1 ¼ ½K : F ða1 Þ: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus ½K : F ða1 Þ\½K : F . Since F F ða1 Þ K and K is a finite extension of F, by 1.4.4, K is a finite extension of F ða1 Þ. Since f ðxÞ 2 F½xð ðF ða1 ÞÞ½xÞ, we have f ðxÞ 2 ðF ða1 ÞÞ½x. We want to show that K is a splitting field over F ða1 Þ for f ðxÞ. To this end, let us take a proper subfield G of K that contains F ða1 Þð F Þ. It suffices to show that G does not contain all the roots of f ðxÞ. Since G is a proper subfield of K that contains F, and K is a splitting field over F for f ðxÞ, G does not contain all the roots of f ðxÞ.

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Thus K is a splitting field over F ða1 Þ for f ðxÞ. Next, since ½K : F ða1 Þ\½K : F , by the induction hypothesis (*), K is a normal extension of F ða1 Þ. We claim that K is a normal extension of F. Suppose to the contrary that K is not a normal extension of F. We seek a contradiction. Since K is a finite extension of F, and K is not a normal extension of F, we have ðfixed field of GðK; FÞÞ 6 F. It follows that there exists h 2 ðfixed field ofGðK; FÞÞ such that h 62 F. Since K is a normal extension of F ða1 Þ, we have ðfixed field of GðK; F ða1 ÞÞÞ ¼ F ða1 Þ: Since F F ða1 Þ, we have GðK; F ða1 ÞÞ ¼

fr : r 2 AutðK Þ; for every a 2 F ða1 Þ; rðaÞ ¼ ag fr : r 2 AutðK Þ; for every a 2 F; rðaÞ ¼ ag |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ GðK; F Þ; and hence GðK; F ða1 ÞÞ GðK; FÞ: It follows that h 2 ðfixed field of GðK; F ÞÞ ¼

fa : a 2 K; for every r 2 GðK; F Þ; rðaÞ ¼ ag fa : a 2 K; for ever r 2 GðK; F ða1 ÞÞ; rðaÞ ¼ ag |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

¼ ðfixed field of GðK; F ða1 ÞÞÞ ¼ F ða1 Þ; and hence h 2 F ða1 Þ. Also h 62 F.

n o Since a1 is algebraic of degree r ð 2Þ over F, 1; a1 ; ða1 Þ2 ; . . .; ða1 Þr1 is a

linearly independent set of vectors in the vector space F ða1 Þ over the field F. Proof Suppose to the contrary that there exist c0 ; c1 ; . . .; cr1 in F such that not all the ci are zero and c0 1 þ c1 a1 þ þ cr1 ða1 Þr1 ¼ 0: We seek a contradiction. Here, it follows that qðxÞ c0 þ c1 x þ þ cr1 xr1 is a nonzero polynomial in F½x such that qða1 Þ ¼ 0; and degðqðxÞÞ r 1\r. This contradicts the fact that a1 is algebraic of degree r over F. ■

2.2 Galois Groups

123

Thus, we have shown that

n o 1; a1 ; ða1 Þ2 ; . . .; ða1 Þr1 is a linearly independent

set of vectors in the vector space F ða1 Þ over the field F. Since ½F ða1 Þ : F ¼ r, the dimension of the vector n o space F ða1 Þ over the field F is r. Now, since 2 r1 1; a1 ; ða1 Þ ; . . .; ða1 Þ is a linearly independent set of vectors in the vector n o space F ða1 Þ over the field F, 1; a1 ; ða1 Þ2 ; . . .; ða1 Þr1 constitutes a basis for the vector space F ða1 Þ over the field F. Now, since h 2 F ða1 Þ, there exist k0 ; k1 ; . . .; kr1 in F such that h ¼ k0 1 þ k1 a1 þ þ kr1 ða1 Þr1 : Since K is a splitting field over F for f ðxÞ, degðf ðxÞÞ 1, pðxÞ is an irreducible factor of f ðxÞ in F½x, and all the roots of pðxÞ are a1 ; a2 ; . . .; ar , by 2.2.19, for every i 2 f1; 2; . . .; r g, there exists ri 2 GðK; FÞ such that ri ða1 Þ ¼ ai . It follows that for every i 2 f1; 2; . . .; r g, ri ðhÞ ¼ ri k0 1 þ k1 a1 þ þ kr1 ða1 Þr1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ri ðk0 1Þ þ ri ðk1 a1 Þ þ þ ri kr1 ða1 Þr1 ¼ ri ðk0 Þri ð1Þ þ ri ðk1 Þri ða1 Þ þ þ ri ðkr1 Þðri ða1 ÞÞr1 ¼ ri ðk0 Þri ð1Þ þ ri ðk1 Þai þ þ ri ðkr1 Þðai Þr1 ¼ ri ðk0 Þ1 þ ri ðk1 Þai þ þ ri ðkr1 Þðai Þr1 ¼ k0 1 þ k1 ai þ þ kr1 ðai Þr1 ; and hence ri ðhÞ ¼ k0 1 þ k1 ai þ þ kr1 ðai Þr1

ði ¼ 1; 2; . . .; r Þ:

Since h 2 ðfixed field of GðK; FÞÞ ¼ fa : a 2 K; and for every r 2 GðK; FÞ; rðaÞ ¼ ag; and each ri is in GðK; FÞ, we have ri ðhÞ ¼ h ði ¼ 1; 2; . . .; r Þ. It follows, from (*) that h ¼ k0 1 þ k1 ai þ þ kr1 ðai Þr1 This

shows

that

a1 ; a2 ; . . .; ar |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

are

ði ¼ 1; 2; . . .; r Þ: roots

of

the

polynomial

r in population

ðk0 hÞ þ k1 x þ þ kr1 xr1 . Here ðk0 hÞ þ k1 x þ þ kr1 xr1 is a polynomial of degree at most ðr 1Þð\r Þ, so ðk0 hÞ þ k1 x þ þ kr1 xr1 is the

124

2 Galois Theory II

zero polynomial of K½x. It follows that ðk0 hÞ ¼ 0, and hence h ¼ k0 . Now, since k0 2 F, we have h 2 F. This is a contradiction. Thus our claim is substantiated, and hence K is a normal extension of F. So in all cases, K is a normal extension of F. ■ Definition Let F and K be any fields such that K is an extension of F. Let f ðxÞ 2 F½x. Let K be a splitting field over F for f ðxÞ. The group GðK; FÞð¼ fr : r 2 AutðKÞ; and for every a 2 F; rðaÞ ¼ agÞ is called the Galois group of f ðxÞ. 2.2.21 Note Let F and K be any fields such that K is an extension of F. Let f ðxÞ be a nonzero member of F½x. Let K be a splitting field over F for f ðxÞ. Suppose that degðf ðxÞÞ 1. Let T be a subfield of K that contains F. Put GðK; T Þ fr : r 2 GðK; FÞ; and for every t 2 T; rðtÞ ¼ tg: Clearly, GðK; T Þ is a subgroup of the group GðK; FÞ. For every subgroup H of the group GðK; FÞ, put KH fa : a 2 K; and for every r 2 H; rðaÞ ¼ ag: Clearly, K is a splitting field over T for f ðxÞ. Proof Since K is a splitting field over F for f ðxÞ, K is a finite extension of F. Now since F T K, by 1.4.4, K is a finite extension of T. Let G be a proper subfield of K which contains T ð F Þ. We have to show that G does not contain all the roots of f ðxÞ. Since G is a proper subfield of K which contains F, and K is a splitting field over F for f ðxÞ, G does not contain all the roots of f ðxÞ. ■ Thus we have shown that K is a splitting field over T for f ðxÞ. Now, by 2.2.20, K is a normal extension of T, and hence, by definition of normal extension, T ¼ ðfixed field of GðK; T ÞÞ ¼ fa : a 2 K; and for every r 2 GðK; T Þ; rðaÞ ¼ ag |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ KGðK;T Þ ; and hence T ¼ KGðK;T Þ . Since K is a normal extension of T, by 2.2.17, oðGðK; T ÞÞ ¼ ½K : T , and H ¼ GðK; KH Þ . Since K is a splitting field over F for f ðxÞ, by 2.2.20, K is a normal extension of F, and hence, by 2.2.17, oðGðK; FÞÞ ¼ ½K : F . Since K is a finite extension of F, and F T K, by 1.4.4, and 1.4.3, we have oðGðK; FÞÞ ¼ ½K : F ¼ ½K : T ½T : F ¼ oðGðK; T ÞÞ½T : F ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

2.2 Galois Groups

125

and hence, ½T : F ¼

oðGðK; FÞÞ ¼ ðindex of subgroup GðK; T Þof the group GðK; FÞÞ: oðGðK; T ÞÞ

2.2.22 Conclusion Let F and K be any fields such that F K. Let K be an extension of F. Let f ðxÞ be a nonzero member of F½x. Let K be a splitting field over F for f ðxÞ. Suppose that degðf ðxÞÞ 1. Let T be a subfield of K which contains F. Put GðK; T Þ fr : r 2 GðK; FÞ and for every t 2 T; rðtÞ ¼ tg: Clearly, GðK; T Þ is a subgroup of the group GðK; FÞ. For every subgroup H of the group GðK; FÞ, put KH fa : a 2 K; and for every r 2 H; rðaÞ ¼ ag: Then: 1. T ¼ KGðK;T Þ , that is, the mapping U : H 7! KH from the collection of all subgroups of the group GðK; FÞ to the collection of all subfields of K that contain F is onto. 2. H ¼ GðK; KH Þ, that is, the mapping W : T 7! GðK; T Þ from the collection of all subfields of K that contain F to the collection of all subgroups of the group GðK; FÞ is onto. Also ðW UÞðH Þ ¼ WðUðH ÞÞ ¼ WðKH Þ ¼ GðK; KH Þ ¼ H, so ðW UÞðH Þ ¼ H. Next, ðU WÞðtÞ ¼ UðWðtÞÞ ¼ UðGðK; T ÞÞ ¼ KGðK;T Þ ¼ T, so ðU WÞðtÞ ¼ T. Thus W1 ¼ U. 3. It follows that W : T 7! GðK; T Þ is a one-to-one correspondence from the collection of all subfields of K that contain F onto the collection of all subgroups of the group GðK; FÞ. 4. oðGðK; T ÞÞ ¼ ½K : T . 5. ½T : F is equal to the index of the subgroup GðK; T Þ in the group GðK; FÞ that Þ is, ½T : F ¼ ooððGðK;FÞ GðK;T ÞÞ. 6. oðGðK; FÞÞ ¼ ½K : F ¼ ½K : T ½T : F ¼ oðGðK; T ÞÞ½T : F : 2.2.23 Problem Let F and K be any fields such that K is an extension of F. Let T be a subfield of K that contains F. Suppose that T is a normal extension of F. Let r 2 GðK; FÞ. Then rðtÞ T. Proof Suppose to the contrary that there exists h 2 T such that rðhÞ 62 T. We seek a contradiction.

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2 Galois Theory II

Since T is a normal extension of F, by 2.2.17, there exists a 2 T such that T ¼ FðaÞ and T is a splitting field over F for ðx r1 ðaÞÞðx r2 ðaÞÞ. . .ðx rn ðaÞÞ in F½x, where GðT; F Þ ¼ fr1 ; r2 ; . . .; rn g and r1 ðaÞ ¼ a. Thus T is a splitting field over F for pðxÞ 2 F½x, where pðxÞ ðx r1 ðaÞÞðx r2 ðaÞÞ . . . ðx rn ðaÞÞð¼ ðx aÞðx r2 ðaÞÞ . . . ðx rn ðaÞÞÞ: Suppose that pðxÞ xn þ b1 xn1 þ þ bn ; where each bi is in F. Clearly, pðaÞ ¼ 0. Now, pðrðaÞÞ ðrðaÞÞn þ b1 ðrðaÞÞn1 þ þ bn ¼ rða Þ þ b1 rðan1 Þ þ þ bn rðan Þ þ rðb1 Þrðan1 Þ þ þ rðbn Þ ¼ rðan Þ þ rðb1 an1 Þ þ þ rðbn Þ n ¼ rða þ b1 an1 þ þ bn Þ ¼ rðpðaÞÞ ¼ rð0Þ ¼ 0; n

so pðrðaÞÞ ¼ 0, and hence rðaÞ is a root of pðxÞ. Now, since T is a splitting field over F for pðxÞ 2 F½x, we have rðaÞ 2 T ð¼ FðaÞÞ, and hence rðaÞ 2 FðaÞ. Since T is a normal extension of F, ðFðaÞ ¼ÞT is a finite extension of F, and hence FðaÞ is a finite extension of F. Since rðhÞ 62 T ¼ FðaÞ, we have rðhÞ 62 FðaÞ. Since h 2 T ¼ FðaÞ, we have h 2 FðaÞ. Since FðaÞ is a finite extension of F, by 1.4.9, a is algebraic over F. Let að2 FðaÞÞ be algebraic of degree n over F. It follows, by 1.4.2, that ½FðaÞ : F ¼ n, and hence n is the dimension of the vector space

FðaÞ over the field

F. Clearly, 1; a; a2 ; . . .; an1 is a linearly independent set of vectors for the vector space FðaÞ over the field F. Proof Suppose to the contrary that there exist k0 ; k1 ; . . .; kn1 2 F such that not all the ki are zero, and k0 1 þ k1 a þ þ kn1 an1 ¼ 0: We see, a contradiction. It follows that a is a root of the nonzero polynomial k0 þ k1 x þ þ kn1 xn1 in F½x. Further, the degree of this polynomial is strictly smaller than n. This contradicts the fact that a is algebraic of degree n over F. ■

2 n1 is a linearly independent set of Thus we have shown that 1; a; a ; . . .; a vectors for the vector space FðaÞ over the field F. Now, since

n is the dimension of the vector space FðaÞ over the field F, 1; a; a2 ; . . .; an1 is a basis of the vector space FðaÞ over the field F. Next, since h 2 FðaÞ, there exist c0 ; c1 ; . . .; cn1 2 F such that

2.2 Galois Groups

127

h ¼ c0 1 þ c1 a þ þ cn1 an1 : Hence rðhÞ ¼ r c0 1 þ c1 a þ þ cn1 an1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ rðc0 1Þ þ rðc1 aÞ þ þ rðcn1 an1 Þ ¼ rðc0 Þrð1Þ þ rðc1 ÞrðaÞ þ þ rðcn1 Þrðan1 Þ ¼ rðc0 Þrð1Þ þ rðc1 ÞrðaÞ þ þ rðcn1 ÞðrðaÞÞn1 ¼ c0 rð1Þ þ c1 rðaÞ þ þ cn1 ðrðaÞÞn1 ¼ c0 1 þ c1 rðaÞ þ þ cn1 ðrðaÞÞn1 ; so FðaÞ63 rðhÞ ¼ c0 1 þ c1 rðaÞ þ þ cn1 ðrðaÞÞn1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and

hence

6 FðaÞ. c0 1 þ c1 rðaÞ þ þ cn1 ðrðaÞÞn1 2

rðaÞ 62 FðaÞ. This is a contradiction.

It

follows

that ■

2.2.24 Problem Let F and K be any fields such that K is an extension of F. Let T be a subfield of K that contains F. Suppose that T is a normal extension of F. Clearly, GðK; T Þ is a subgroup of the group GðK; FÞ. Also, GðK; T Þ is a normal subgroup of the group GðK; FÞ. Proof Let us take any s 2 GðK; T Þ and r 2 GðK; FÞ. We have to show that r1 sr 2 GðK; T Þ. To this end, let us take any t 2 T. It suffices to show that ðr1 srÞðtÞ ¼ t, that is, r1 ðsðrðtÞÞÞ ¼ t, that is, sðrðtÞÞ ¼ rðtÞ. Since s 2 GðK; T Þ, it is enough to show that rðtÞ 2 T. By 2.2.23, rðtÞ T. Now, since rðtÞ 2 rðtÞ, we have rðtÞ 2 T. ■ 2.2.25 Note Let F and K be any fields such that K is an extension of F. Let T be a subfield of K that contains F. Suppose that T is a normal extension of F. By 2.2.24, GðK; T Þ is a normal subgroup of the group GðK; FÞ. Hence GðK;FÞ GðK;T Þ is a quotient group. Also, GðT; F Þ is a group. Take an arbitrary r 2 GðK; FÞ. By 2.2.23, rðtÞ T. Since r 2 GðK; FÞ and GðK; FÞ is a group, the inverse function r1 is in GðK; FÞ, and hence by 2.2.23, r1 ðtÞ T. It follows that T rðtÞ. Thus rðtÞ ¼ T. Now, since r 2 GðK; FÞ and F T K, we have ðrjT Þ 2 GðT; F Þ. Thus g : r 7! ðrjT Þ is a mapping from group GðK; FÞ to the group GðT; F Þ. η preserves the binary operation: To show this, let us take arbitrary r; l 2 GðK; FÞ. We have to show that ðrlÞjT ¼ ðrjT ÞðljT Þ.

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2 Galois Theory II

Let us take an arbitrary a 2 T. We have to show that ððrlÞjT ÞðaÞ that is, ðrlÞðaÞ ¼ ðrjT ÞððljT ÞðaÞÞ, that is, ¼ ððrjT ÞðljT ÞÞðaÞ, ðrlÞðaÞ ¼ ðrjT ÞðlðaÞÞ. Since l 2 GðK; FÞ, as above, we have lðtÞ ¼ T. Since a 2 T, we have lðaÞ 2 lðtÞ ¼ T, and hence lðaÞ 2 T. It follows that ðrjT ÞðlðaÞÞ ¼ rðlðaÞÞ ¼ ðrlÞðaÞ. Thus g preserves the binary operation. kerðgÞ ¼ GðK; TÞ: Let us take an arbitrary r 2 kerðgÞ, that is, r 2 GðK; FÞ and ðrjT Þ ¼ IdT . Since r 2 GðK; FÞ, we have r 2 AutðKÞ. Next, since ðrjT Þ ¼ IdT , for every t 2 T, rðtÞ ¼ t. This shows that r 2 GðK; T Þ. Thus kerðgÞ GðK; T Þ. Let us take an arbitrary r 2 GðK; T Þ, that is, r 2 AutðKÞ and ðrjT Þ ¼ IdT . We have to show that r 2 kerðgÞ, that is, r 2 GðK; FÞ and ðrjT Þ ¼ IdT . It remains to show that ðrjF Þ ¼ IdF . Since ðrjT Þ ¼ IdT and F T, we have ðrjF Þ ¼ IdF . Thus GðK; T Þ kerðÞ. Hence kerðgÞ ¼ GðK; T Þ. Since g : GðK; FÞ ! GðT; F Þ preserves the group binary operations, g : GðK; FÞ ! GðT; F Þ is a homomorphism from GðK; FÞ onto gðGðK; FÞÞ, and hence by the theorem of group homomorphisms, the quotient group fundamental GðK;FÞ kerðgÞ

GðK;FÞ ¼ GðK;FÞ GðK;T Þ is isomorphic to gðGðK; FÞÞ. It follows that GðK;T Þ is isomorphic to Þ GðK;FÞ oðGðK;FÞÞ gðGðK; FÞÞ, and hence ooððGðK;FÞ GðK;T ÞÞ ¼ o GðK;T Þ ¼ oðgðGðK; FÞÞÞ. Thus oðGðK;T ÞÞ ¼

oðgðGðK; FÞÞÞ. By 2.2.21, oðGðK; FÞÞ ¼ oðgðGðK; FÞÞÞ oðGðT; F ÞÞ: ½T : F ¼ oðGðK; T ÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Since T is a normal extension of F, by 2.2.17, oðGðT; F ÞÞ ¼ ½T : F , and hence ½T : F ¼

oðGðK; FÞÞ ¼ oðgðGðK; FÞÞÞ oðGðT; F ÞÞ ¼ ½T : F : oðGðK; T ÞÞ

This shows that oðgðGðK; FÞÞÞ ¼ oðGðT; F ÞÞ: Since g : GðK; FÞ ! GðT; F Þ, we have gðGðK; FÞÞ GðT; F Þ. Since oðgðGðK; FÞÞÞ ¼ oðGðT; F ÞÞ, we have gðGðK; FÞÞ ¼ GðT; F Þ, and since GðK;FÞ GðK;T Þ is isomorphic to gðGðK; FÞÞ,

GðK;FÞ GðK;T Þ

is isomorphic to GðT; F Þ.

2.2.26 Conclusion Let F and K be any fields such that K is an extension of F. Let T be a subfield of K that contains F. Suppose that T is a normal extension of F. Then the quotient group GðK;FÞ GðK;T Þ is isomorphic to the group GðT; F Þ. This result is known as the fundamental theorem of Galois theory.

2.3 Applications of Galois Theory

2.3

129

Applications of Galois Theory

2.3.1 Definition Let G be a group. If there exists a finite collection fN0 ; N1 ; . . .; Nk g of subgroups of G such that 1. G ¼ N0 N1 Nk ¼ feg, where e denotes the identity element of G, 2. for every i ¼ 1; . . .; k; Ni is a normal subgroup of Ni1 , 3. for every i ¼ 1; . . .; k, the quotient group NNi1i is abelian, then we say that G is solvable. Definition Let G be a group. Let a; b 2 G. By the commutator of a and b we mean a1 b1 ab. Let C be the collection of all commutators in G. Then the subgroup G0 of G generated by all the commutators in G is the smallest subgroup of G containing C. Clearly, G0 is equal to the collection of all finite products of the members in C or 1 their inverses. Observe that for every a; b 2 G, ða1 b1 abÞ ¼ b1 a1 ba, so the 0 inverse of a member of C is also a member of C. Hence G is equal to the collection of all finite products of the members in C. Here G0 is called the commutator subgroup of G. 2.3.2 Problem Let G be a group. Then the commutator subgroup G0 of G is a normal subgroup of G. Proof Let C be the collection of all commutators in G. Let u 2 G0 and g 2 G. We have to show that g1 ug 2 G0 . Observe that g1 ug ¼ u u1 g1 ug : Since u; g 2 G, we have u1 g1 ug 2 C. By the definition of G0 , we have C G0 . Now, since u1 g1 ug 2 C, we have u1 g1 ug 2 G0 . Since u; u1 g1 ug 2 G0 and G0 is a group, we have ðg1 ug ¼Þuðu1 g1 ugÞ 2 G0 , and hence ■ g1 ug 2 G0 . 2.3.3 Problem Let G be a group. By 2.3.2, G0 is a normal subgroup of G. Then the quotient group GG0 is an abelian group. Proof Let us take any a; b 2 G. We have to show that ðaG0 ÞðbG0 Þ ¼ ðbG0 ÞðaG0 Þ, that is, ðabÞG0 ¼ ðbaÞG0 , that is, ðabÞ1 ðbaÞ 2 G0 , that is, b1 a1 ba 2 G0 . Since b1 a1 ba 2 C, where C is the collection of all commutators in G and ■ C G0 , we have b1 a1 ba 2 G0 . 2.3.4 Problem Let G be a group. Let M be a normal subgroup of G. Suppose that G the quotient group M is an abelian group. Then G0 M.

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2 Galois Theory II

Proof It suffices to show that C M. To this end, let us take any a; b 2 G. We have to show that b1 a1 ba 2 M, that is, ðabÞ1 ba 2 M, that is, ðabÞM ¼ ðbaÞM, G is an abelian that is, ðaM ÞðbM Þ ¼ ðbM ÞðaM Þ. This is known to be true, because, M group. ■ Definition Let G be a group. Let C be a subgroup of G. If for every automorphism T of G, T ðCÞ C, then we say that C is a characteristic subgroup of G. 2.3.5 Problem Let G be a group. Then the commutator subgroup G0 of G is a characteristic subgroup of G. Proof To show this, let us take an arbitrary automorphism T of G. We have to show that T ðG0 Þ G0 . To this end, let us take arbitrary ðc1 . . .cn Þ 2 G0 , where each ci is a commutator in G. We have to show that ðT ðc1 Þ. . .T ðcn Þ ¼ÞT ðc1 . . .cn Þ 2 G0 . It suffices to show that each T ðci Þ is a commutator in G. Since ci is a commutator in G, there exist ai ; bi 2 G such that ci ¼ ðai Þ1 ðbi Þ1 ai bi . Now, since T is an automorphism of G, we have T ðci Þ ¼ T ðai Þ1 ðbi Þ1 ai bi ¼ T ðai Þ1 T ðbi Þ1 T ðai ÞT ðbi Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðT ðai ÞÞ1 ðT ðbi ÞÞ1 T ðai ÞT ðbi Þ; and hence T ðci Þ ¼ ðT ðai ÞÞ1 ðT ðbi ÞÞ1 T ðai ÞT ðbi Þ. Since ðT ðai ÞÞ1 ðT ðbi ÞÞ1 T ðai ÞT ðbi Þ is a commutator in G, T ðci Þ is a commutator in G. ■ 0 2.3.6 Problem Let G be a group. Then the subgroup ðG0 Þ Gð2Þ of G is normal. Similarly, for every positive integer n, GðnÞ is a normal subgroup of G. Proof To show this, let us take an arbitrary g 2 G. We have to show that g1 ðG0 Þ0 g ðG0 Þ0 , that is, T ðG0 Þ0 ðG0 Þ0 , where T is the automorphism x 7! g1 xg of G. By 2.3.5, G0 is a characteristic subgroup of G. Again by 2.3.5, ðG0 Þ0 is a characteristic subgroup of G0 . It follows that ðG0 Þ0 is a subgroup of G. Since G0 is a characteristic subgroup of G, and T is an automorphism of G, we have T ðG0 Þ G0 . Now, since T : G ! G is an automorphism, its restriction TjG0 is an automorphism of G0 . Next, since ðG0 Þ0 is a characteristic subgroup of G0 , we have 0 0 0 T ðG0 Þ ¼ TjG0 ðG0 Þ ðG0 Þ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence T ðG0 Þ0 ðG0 Þ0 .

■

2.3.7 Note Let G be a group. Let G be solvable. It follows that there exists a finite collection fN0 ; N1 ; . . .; Nk g of subgroups of G such that


131

1. G ¼ N0 N1 . . . Nk ¼ feg, where e denotes the identity element of G, 2. for every i ¼ 1; . . .; k; Ni is a normal subgroup of Ni1 , 3. for every i ¼ 1; . . .; k; the quotient group NNi1i is abelian. Since N1 is a normal subgroup of N0 , and the quotient group 0

N0 N1

is abelian, by

N2 N3

is abelian, by

2.3.4, ðN0 Þ N1 . Since N2 is a normal subgroup of N1 and the quotient group NN12 is abelian, by 2.3.4, ðN1 Þ0 N2 . Since ðN0 Þ0 N1 , we have

0 Gð2Þ ¼ ðN0 Þð2Þ ¼ ðN0 Þ0 ðN1 Þ0 N2 ; |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence Gð2Þ N2 . Since N3 is a normal subgroup of N2 , and the quotient group 0

0

2.3.4, ðN2 Þ N3 . Since ðN1 Þ N2 , we have

ð2Þ 0 Gð3Þ ¼ ðN0 Þð3Þ ¼ ðN0 Þ0 ðN1 Þð2Þ ¼ ðN1 Þ0 ðN2 Þ0 N3 ; |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence Gð3Þ N3 , etc. Hence feg ¼ GðkÞ Nk ¼ feg. Thus GðkÞ ¼ feg. |fflfflfflfflfflffl{zfflfflfflfflfflffl} 2.3.8 Conclusion Let G be solvable. Then there exists a positive integer k such that GðkÞ ¼ feg. 2.3.9 Problem Let G be a group. Let k be a positive integer. Suppose that GðkÞ ¼ feg, where e denotes the identity element of G. Then G is solvable. It also follows that G0 is solvable, etc. Proof By 2.3.6, for every i 2 f1; 2; . . .; kg, GðiÞ is a normal subgroup of G. Now,

since G is a normal subgroup of G, Gð0Þ ; Gð1Þ ; . . .; GðkÞ is a collection of normal subgroups of G, where Gð0Þ G. Further, G ¼ Gð0Þ Gð1Þ . . . GðkÞ ¼ feg: By 2.3.2, for every i ¼ 1; . . .; k; GðiÞ is a normal subgroup of Gði1Þ . By 2.3.3, for ði1Þ every i ¼ 1; . . .; k, the quotient group GGðiÞ is abelian. Thus G is solvable. ■ 2.3.10 Problem Let G, G be any groups. Let f : G ! G be a homomorphism from G onto G. Thus G is the homomorphic f-image of G. Let G be solvable. Then G is solvable. Proof Since G is solvable, by 2.3.8, there exists a positive integer k such that GðkÞ ¼ feg, where e denotes the identity element of G. By 2.3.9, it suffices to show that G

ðk Þ

¼ feg, where e denotes the identity element of G.

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2 Galois Theory II

Now, since f : G ! G is a homomorphism from G onto G, we have f ðGÞ ¼ G and f ðeÞ ¼ e. Thus it is enough to show that ðf ðGÞÞðkÞ ff ðeÞg. Since GðkÞ ¼ feg, we have ff ðeÞg ¼ f GðkÞ , and hence it suffices to show that ðf ðGÞÞðkÞ f GðkÞ . Clearly, ðf ðGÞÞ0 f ðG0 Þ. Proof By 2.3.1, G0 is a normal subgroup of G. We first show that f ðG0 Þ is a normal subgroup of G. To this end, let us take arbitrary g 2 G and f ðaÞ 2 f ðG0 Þ, where a 2 G0 . We have to show that ðgÞ1 ðf ðaÞÞg 2 f ðG0 Þ. Since g 2 G ¼ f ðGÞ, there exists 2 G such that g ¼ f ðgÞ. Now, ðgÞ1 ðf ðaÞÞg ¼ ðf ðgÞÞ1 f ðaÞðf ðgÞÞ ¼ f g1 f ðaÞf ðgÞ ¼ f g1 ag ; so ðgÞ1 ðf ðaÞÞg ¼ f ðg1 agÞ. Since G0 is a normal subgroup of G, a 2 G0 , and g 2 G, we have g1 ag 2 G0 , and hence ðgÞ1 ðf ðaÞÞg ¼ f g1 ag 2 f ðG0 Þ : |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus ðgÞ1 ðf ðaÞÞg 2 f ðG0 Þ.

Next we shall show that the quotient group f ðGG0 Þ ¼ ffððGG0ÞÞ is an abelian group.

To this end, let us take any a; b 2 G. We have to show that ðf ðaÞf ðG0 ÞÞ ðf ðbÞf ðG0 ÞÞ ¼ ðf ðbÞf ðG0 ÞÞðf ðaÞf ðG0 ÞÞ, that is, ðf ðaÞf ðbÞÞf ðG0 Þ ¼ ðf ðbÞf ðaÞÞ 1 0 0 0 f ðG0 Þ, thatis, ðf ðabÞ Þf ðG Þ ¼ ðf ðbaÞÞf ðG Þ, that is, ðf ðbaÞÞ ðf ðabÞÞ 2 f ðG Þ, that is, f ðbaÞ1 f ðabÞ 2 f ðG0 Þ, that is, f ða1 b1 Þf ðabÞ 2 f ðG0 Þ, that is,

f ða1 b1 abÞ 2 f ðG0 Þ. It suffices to show that a1 b1 ab 2 G0 . Since a1 b1 ab is a commutator of a and b, we have a1 b1 ab 2 G0 . Since ffððGG0ÞÞ is an abelian group, by 2.3.4, ðf ðGÞÞ0 f ðG0 Þ . By 2.3.9, G0 is solvable. Also f jG0 is a homomorphism. So as above, 0 0 0 0 ðf ðGÞÞ00 ¼ ðf ðGÞÞ0 f jG0 ðG0 Þ f jG0 ðG0 Þ ¼ f ðG0 Þ ¼ f ðG00 Þ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so ðf ðGÞÞ0 f ðG00 Þ . Finally, we get ðf ðGÞÞðkÞ f GðkÞ .

■

2.3.11 Problem Let G be a group. Let N be a normal subgroup of G. Then N 0 is also a normal subgroup of G. Proof Since N 0 is a subgroup of N, and N is a subgroup of G, N 0 is a subgroup of G. Next, let us take arbitrary g 2 G and ðc1 . . .cn Þ 2 N 0 , where each ci is a commutator in N. We have to show that


133

1 1 1 g c1 g g c2 g . . . g cn g ¼ g1 ðc1 . . .cn Þg 2 N 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} It suffices to show that each g1 ci g is in N 0 , that is, T ðci Þ 2 N 0 , where T is the automorphism x 7! g1 xg of N. By 2.3.5, N 0 is a characteristic subgroup of N, and T is the automorphism x 7! g1 xg of N, so T ðN 0 Þ N 0 . Now, since ci is a commutator in N, and N 0 contains all commutators in N, we have ci 2 N 0 , and hence T ðci Þ 2 T ðN 0 Þ N 0 . Thus, T ðci Þ 2 N 0 . ■ 2.3.12 Problem Suppose that n 2 f5; 6; 7; 8; . . .g. Let Sn be the symmetric group of all permutations of n symbols 1; 2; . . .; n. Then 1. ðSn Þ0 contains all 3-cycles, 2. ðSn Þ00 contains all 3-cycles, etc. In short, for every n 5 and for every k 1, ðSn ÞðkÞ contains all 3-cycles. It follows that for every k 1, ðSn ÞðkÞ 6¼ feg, and hence by 2.3.9, Sn is not solvable when n 5. Proof 1 Let us take an arbitrary 3-cycle ði1 i2 i3 Þ in Sn , where i1 ; i2 ; i3 are three 0 distinct members of f1; 2; . . .; ng. We have to show that ði1 i2 i3 Þ 2 ð Sn Þ . 1 2 3 4 5. . . Since n 2 f5; 6; 7; 8; . . .g, the 3-cycle ð1 4 5Þ is in Sn . 4 2 3 5 1. . . Observe that the 3-cycle (135) is a commutator in Sn . Proof Since ð123Þ1 ð145Þ1 ð123Þð145Þ 123456 1 123456 1 123456 123456 ¼ 231456 423516 231456 423516 1 123456 123456 123456 123456 ¼ 312456 423516 231456 423516 123456 123456 123456 123456 ¼ 312456 523146 231456 423516 123456 123456 123456 ¼ 523146 312456 431526 123456 123456 123456 ¼ ¼ ¼ ð135Þ 312456 135426 325416

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2 Galois Theory II

we have ð1 3 5Þ ¼ ð1 2 3Þ1 ð1 4 5Þ1 ð1 2 3Þð1 4 5Þ. Since ð1 2 3Þ; ð1 4 5Þ 2 Sn , ðð1 3 5Þ ¼Þð1 2 3Þ1 ð1 4 5Þ1 ð1 2 3Þð1 4 5Þ is a commutator in Sn , and hence ð1 3 5Þ is a commutator in Sn .

■

It follows that ð1 3 5Þ 2 ðSn Þ0 . By 2.3.11, ðSn Þ0 is a normal subgroup of Sn . There exists a permutation j of 1; 2; . . .; n such that jð1Þ ¼ i1 ; jð3Þ ¼ i2 , and jð5Þ ¼ i3 . Thus j 2 Sn . Since ð1 3 5Þ 2 ðSn Þ0 , j 2 Sn , and ðSn Þ0 is a normal subgroup of Sn , we have jð1 3 5Þj1 2 ðSn Þ0 . It suffices to show that jð1 3 5Þj1 ¼ ði1 i2 i3 Þ: For this we must prove that 8 > >
> : 1 ðjð1 3 5Þj ÞðlÞ ¼ l when l 2 f1; 2; . . .; ng fi1 ; i2 ; i3 g: Here jð1 3 5Þj1 ði1 Þ ¼ jð1 3 5Þ j1 ði1 Þ ¼ jðð1 3 5Þð1ÞÞ ¼ jð3Þ ¼ i2 ; jð1 3 5Þj1 ði2 Þ ¼ jð1 3 5Þ j1 ði2 Þ ¼ jðð1 3 5Þð3ÞÞ ¼ jð5Þ ¼ i3 ;

and

jð1 3 5Þj1 ði3 Þ ¼ jð1 3 5Þ j1 ði3 Þ ¼ jðð1 3 5Þð5ÞÞ ¼ jð1Þ ¼ i1 :

Suppose that l 2 f1; 2; . . .; ng fi1 ; i2 ; i3 g. ðjð1 3 5Þj1 ÞðlÞ ¼ l. Here

It

suffices

to

show

that

jð1 3 5Þj1 ðlÞ ¼ ðjð1 3 5ÞÞ j1 ðlÞ ¼ ðjð1 3 5ÞÞðmÞ;

where m 2 f1; 2; . . .; ng f1; 3; 5g, and jðmÞ ¼ l. It follows that ð1 3 5ÞðmÞ ¼ m, and hence LHS ¼ jð1 3 5Þj1 ðlÞ ¼ ðjð1 3 5ÞÞðmÞ ¼ jðð1 3 5ÞðmÞÞ ¼ jðmÞ ¼ l ¼ RHS: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2: Let us take an arbitrary 3-cycle ði1 i2 i3 Þ in ðSn Þ0 , where i1 ; i2 ; i3 are three distinct 0 members of f1; 2; . . .; ng. We have to show that ði1 i2 i3 Þ 2 ðSn Þ0 . Since 7; 8; . . .g, by assumption the 3-cycle n 2 f5; 6; 1 2 3 4 5. . . ð 1 4 5Þ is inðSn Þ0 . 4 2 3 5 1. . .


135

Observe that the 3-cycle ð1 3 5Þ is a commutator in ðSn Þ0 . Proof Since ð123Þ1 ð145Þ1 ð123Þð145Þ 123456 1 123456 1 123456 123456 ¼ 423516 231456 423516 231456 1 123456 123456 123456 123456 ¼ 423516 312456 423516 231456 123456 123456 123456 123456 ¼ 312456 523146 231456 423516 123456 123456 123456 ¼ 312456 523146 431526 123456 123456 123456 ¼ ¼ ¼ ð135Þ; 312456 135426 325416 we have ð1 3 5Þ ¼ ð1 2 3Þ1 ð1 4 5Þ1 ð1 2 3Þð1 4 5Þ. Since ð1 2 3Þ; ð1 4 5Þ 2 ðSn Þ0 , ðð1 3 5Þ ¼Þð1 2 3Þ1 ð1 4 5Þ1 ð1 2 3Þð1 4 5Þ is a commutator in ðSn Þ0 , and hence ð1 3 5Þ is a commutator in ðSn Þ0 . 0 0 It follows that ð1 3 5Þ 2 ðSn Þ0 . By two applications of 2.3.11, ðSn Þ0 is a normal subgroup of Sn . There exists a permutation j of 1; 2; . . .; n such that 0 jð1Þ ¼ i1 ; jð3Þ ¼ i2 , and jð5Þ ¼ i3 . Thus j 2 Sn . Since ð1 3 5Þ 2 ðSn Þ0 , j 2 Sn , 0 0 and ðSn Þ0 is a normal subgroup of Sn , we have jð1 3 5Þj1 2 ðSn Þ0 . It suffices to show that jð1 3 5Þj1 ¼ ði1 i2 i3 Þ: For this we must prove 8 > >
> : 1 ðjð1 3 5Þj ÞðlÞ ¼ l when l 2 f1; 2; . . .; ng fi1 ; i2 ; i3 g: Here

jð1 3 5Þj1 ði1 Þ ¼ jð1 3 5Þ j1 ði1 Þ ¼ jðð1 3 5Þð1ÞÞ ¼ jð3Þ ¼ i2 ; jð1 3 5Þj1 ði2 Þ ¼ jð1 3 5Þ j1 ði2 Þ ¼ jðð1 3 5Þð3ÞÞ ¼ jð5Þ ¼ i3 ;

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2 Galois Theory II

and

jð1 3 5Þj1 ði3 Þ ¼ jð1 3 5Þ j1 ði3 Þ ¼ jðð1 3 5Þð5ÞÞ ¼ jð1Þ ¼ i1 :

Suppose that l 2 f1; 2; . . .; ng fi1 ; i2 ; i3 g. ðjð1 3 5Þj1 ÞðlÞ ¼ l. Here

It

suffices

to

show

that

jð1 3 5Þj1 ðlÞ ¼ ðjð1 3 5ÞÞ j1 ðlÞ ¼ ðjð1 3 5ÞÞðmÞ;

where m 2 f1; 2; . . .; ng f1; 3; 5g and jðmÞ ¼ l. It follows that ð1 3 5ÞðmÞ ¼ m, and hence LHS ¼ jð1 3 5Þj1 ðlÞ ¼ ðjð1 3 5ÞÞðmÞ ¼ jðð1 3 5ÞðmÞÞ ¼ jðmÞ ¼ l ¼ RHS: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ■ 2.3.13 Example Let F be the field of all real numbers, and let K be the field of all complex numbers. Clearly, F is a subfield of K. Next, GðK; FÞ ¼ fr : r 2 AutðKÞ; and for every a 2 F; rðaÞ ¼ ag: Suppose that r 2 AutðKÞ. Here 1; i 2 K. Since r 2 AutðKÞ, r : K ! K is an automorphism, and hence 1 ¼ rð1Þ ¼ rð1Þ ¼ r i2 ¼ rðiiÞ ¼ rðiÞrðiÞ: Thus rðiÞrðiÞ ¼ 1, where rðiÞ is a complex number. It follows that rðiÞ ¼ i or rðiÞ ¼ i. Case I: rðiÞ ¼ i. For every real a; b, we have rðaÞ ¼ a and rðbÞ ¼ b. It follows that rða þ ibÞ ¼ rðaÞ þ rðibÞ ¼ a þ rðibÞ ¼ a þ rðiÞrðbÞ ¼ a þ rðiÞb ¼ a þ ib; and hence rða þ ibÞ ¼ a þ ib. This shows that in this case, r is equal to the identity mapping of K. Let us denote this r by r1 . Case II: rðiÞ ¼ i. For every real a; b, we have rðaÞ ¼ a and rðbÞ ¼ b. It follows that rða þ ibÞ ¼ rðaÞ þ rðibÞ ¼ a þ rðibÞ ¼ a þ rðiÞrðbÞ ¼ a þ rðiÞb ¼ a þ ðiÞb ¼ a ib;


137

and hence rða þ ibÞ ¼ a ib. This shows that in this case, r is equal to the complex-conjugation mapping of K. Let us denote this r by r2 . Thus GðK; FÞ ¼ fr1 ; r2 g. Hence oðGðK; FÞÞ ¼ oðfr1 ; r2 gÞ ¼ 2. Further, ðfixed field of GðK; F ÞÞ ¼ fa : a 2 K; and for every r 2 GðK; F Þ; rðaÞ ¼ ag ¼ fa : a 2 K; and for every r 2 fr1 ; r2 g; rðaÞ ¼ ag ¼ fa : a 2 K; r1 ðaÞ ¼ aandr2 ðaÞ ¼ ag ¼ fa : a 2 K; r2 ðaÞ ¼ ag ¼ fa : a 2 K; a ¼ ag ¼ ðthe set of all real numbersÞ ¼ F; so the fixed field of GðK; FÞ is F.

pffiffiffi 2.3.14 Example Let F0 be the field of all rational numbers and K ¼ F0 3 2 , pffiffiffi pffiffiffi where 3 2 is the real cube root of 2. By 1.5.20, we have F0 3 2 : F0 ¼ 3. Now, n pffiffiffi pffiffiffi pffiffiffi2 o by 1.4.5, 3 2 is algebraic of degree 3 over F0 . It follows that 1; 3 2; 3 2 is a ffiffiffi p 3 linearly independent set of vectors in the vector space F0 2 over F0 . Since ffiffiffi p pffiffiffi F0 3 2 : F0 ¼ 3, the dimension of the vector space F0 3 2 over F0 is 3. It n pffiffiffi pffiffiffi2 o pffiffiffi follows that 1; 3 2; 3 2 is a basis of the vector space F0 3 2 over F0 . Hence pffiffiffi pffiffiffi2 p ffiffiffi 3 3 3 F0 2 ¼ a0 þ a1 2 þ a2 2 : a0 ; a1 ; a2 2 F0 : Next, GðK; F0 Þ ¼ fr : r 2 AutðKÞ; and for every a 2 F0 ; rðaÞ ¼ ag: Suppose that r 2 GðK; F0 Þ. pffiffiffi pffiffiffi2 Here 1; 3 2; 3 2 2 K. Since r 2 GðK; F0 Þ, we have r 2 AutðKÞ, that is, r : K ! K is an automorphism, and hence 2 ¼ rð2Þ ¼ r

pffiffiffipffiffiffipffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 3 3 3 3 2 2 2 ¼r 2 r 2 r 2 :

pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi Thus r 3 2 r 3 2 r 3 2 ¼ 2. Now, since r : F0 3 2 ! F0 3 2 and pffiffiffi pffiffiffi pffiffiffi F0 3 2 R, we have r 3 2 ¼ 3 2. Now, for every a0 ; a1 ; a2 2 F0 , pffiffiffi2 pffiffiffi pffiffiffi2 pffiffiffi r a0 þ a 1 3 2 þ a 2 3 2 ¼ rð a 0 Þ þ r a 1 3 2 þ r a 2 3 2 pffiffiffi pffiffiffi pffiffiffi ¼ rða0 Þ þ rða1 Þr 3 2 þ rða2 Þr 3 2 r 3 2 p ffiffi ffi ffiffi ffi p pffiffiffi ¼ a0 þ a1 r 3 2 þ a 2 r 3 2 r 3 2 pffiffiffi pffiffiffipffiffiffi pffiffiffi pffiffiffi2 ¼ a 0 þ a1 3 2 þ a2 3 2 3 2 ¼ a 0 þ a1 3 2 þ a2 3 2 ;

138

2 Galois Theory II

that is, for every a0 ; a1 ; a2 2 F0 , we have pffiffiffi2 pffiffiffi2 p ffiffiffi p ffiffiffi 3 3 3 3 r a0 þ a1 2 þ a2 2 ¼ a0 þ a1 2 þ a2 2 : This shows that r is equal to the identity mapping Id of K. Thus GðK; F0 Þ ¼ fIdg. Hence oðGðK; F0 ÞÞ ¼ oðfIdgÞ ¼ 1. Further, ðfixed field of GðK; F0 ÞÞ ¼ fa : a 2 K; and for every r 2 GðK; F0 Þ; rðaÞ ¼ ag ¼ fa : a 2 K; and for every r 2 fIdg; rðpaffiffiÞffi¼ ag ¼ fa : a 2 K; IdðaÞ ¼ ag ¼ K ¼ F0 3 2 ; pffiffiffi pffiffiffi so the fixed field of G F0 3 2 ; F0 is F0 3 2 . 2pi

2.3.15 Example Let F0 be the field of all rational numbers. Let us denote e 5 by a. 5 1 ¼ 11 It follows that a5 ¼ 1. Next, a4 þ a3 þ a2 þ a þ 1 ¼ aa1 a1 ¼ 0, so a is a 4 3 2 root of the polynomial x þ x þ x þ x þ 1 2 F0 ½x. Here x4 þ x3 þ x2 þ x þ 1 is an irreducible polynomial over the field of rational numbers. Proof Put x y þ 1. It suffices to show that ðy þ 1Þ4 þ ðy þ 1Þ3 þ ðy þ 1Þ2 þ ðy þ 1Þ þ 1 is an irreducible polynomial over the field of rational numbers. Observe that ðy þ 1Þ4 þ ðy þ 1Þ3 þ ðy þ 1Þ2 þ ðy þ 1Þ þ 1 ¼ ðy þ 4y þ 6y þ 4y þ 1Þ þ ðy3 þ 3y2 þ 3y þ 1Þ þ ðy2 þ 2y þ 1Þ þ ðy þ 1Þ þ 1 ¼ 5 þ 10y þ 10y2 þ 5y3 þ y4 ¼ 5ð1 þ 2y þ 2y2 þ y3 Þ þ y4 : 4

3

2

By 1.3.5, 5 þ 10y þ 10y2 þ 5y3 þ y4 is irreducible over the field of rational numbers, and ðy þ 1Þ4 þ ðy þ 1Þ3 þ ðy þ 1Þ2 þ ðy þ 1Þ þ 1 ¼ 5 þ 10y þ 10y2 þ 5y3 þ y4 ; so ðy þ 1Þ4 þ ðy þ 1Þ3 þ ðy þ 1Þ2 þ ðy þ 1Þ þ 1 is irreducible over the field of rational numbers.

■


139

Thus we have shown that x4 þ x3 þ x2 þ x þ 1 is an irreducible polynomial over the field of rational numbers. Now, by 1.5.12, a is algebraic of degree 4 over F0 , and hence by

1.4.16, ½F0 ðaÞ : F0 ¼ 4. Since a is algebraic of degree 4 over F0 , 2 3 1; a; a ; a is a linearly independent set of vectors in the vector space F0 ðaÞ over F0 . Since ½F0 ðaÞ : F0 ¼ 4, the dimension of the vector space F0 ðaÞ over F0 is 4. It follows that 1; a; a2 ; a3 is a basis of the vector space F0 ðaÞ over F0 . Hence

F 0 ð aÞ ¼ a0 þ a1 a þ a2 a2 þ a3 a3 : a0 ; a1 ; a2 2 F 0 : Next, GðK; F0 Þ ¼ fr : r 2 AutðKÞ; and for every a 2 F0 ; rðaÞ ¼ ag: Suppose that r 2 GðK; F0 Þ. Here 1; a; a2 ; a3 2 K. Also a5 ¼ 1 . Since r 2 GðK; F0 Þ, we have r 2 AutðKÞ, that is, r : K ! K is an automorphism, and hence 1 ¼ rð1Þ ¼ r a5 ¼ rðaaaaaÞ ¼ rðaÞrðaÞrðaÞrðaÞrðaÞ ¼ ðrðaÞÞ5 : Thus ðrðaÞÞ5 ¼ 1. Now, since r : F0 ðaÞ ! F0 ðaÞ and F0 ðaÞ C, we have rðaÞ ¼ 1 or a or a2 or a3 or a4 . Since r : K ! K is an automorphism, r is one-to-one. Since a 6¼ 1, we have rðaÞ 6¼ rð1Þ ¼ 1, and hence rðaÞ 6¼ 1. Since |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} rðaÞ ¼ 1 or a or a2 or a3 or a4 , we have rðaÞ ¼ a or a2 or a3 or a4 : Case I: rðaÞ ¼ a. For every a0 ; a1 ; a2 ; a3 2 F0 , rða0 þ a1 a þ a2 a2 þ a3 a3 Þ ¼ rða0 Þ þ rða1 aÞ þ rða2 a2 Þ þ rða2 a3 Þ ¼ rða0 Þ þ rða1 ÞrðaÞ þ rða2 ÞðrðaÞÞ2 þ rða3 ÞðrðaÞÞ3 ¼ a0 þ a1 a þ a2 ðaÞ2 þ a3 ðaÞ3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 ; that is, for every a0 ; a1 ; a2 ; a3 2 F0 , we have r a0 þ a1 a þ a2 a 2 þ a3 a3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 : This shows that r is equal to the identity mapping Id of K. Let us denote this r by r1 . Thus r1 ðaÞ ¼ a and r1 2 GðK; F0 Þ. Case II: rðaÞ ¼ a2 . For every a0 ; a1 ; a2 ; a3 2 F0 ,

140

2 Galois Theory II

r a0 þ a1 a þ a2 a2 þ a3 a 3 ¼ rða0 Þ þ rða1 aÞ þ r a2 a2 þ r a2 a3 ¼ rða0 Þ þ rða1 ÞrðaÞ þ rða2 ÞðrðaÞÞ2 þ rða3 ÞðrðaÞÞ3 2 3 ¼ a0 þ a1 a2 þ a2 a2 þ a3 a 2 1 ¼ a0 þ a1 a2 þ a2 þ a 3 a a ¼ a0 þ a1 a2 þ a2 1 a a2 a3 þ a3 a ¼ ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 ; that is, for every a0 ; a1 ; a2 ; a3 2 F0 , we have r a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 : Let us denote this r by r2 . Thus r2 ðaÞ ¼ a2 , and for every a0 ; a1 ; a2 ; a3 2 F0 , we have r2 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 ð2 F0 ðaÞÞ 1 1 ¼ a0 þ a 1 a2 þ a 2 þ a 3 a ¼ a 2 þ a 0 a þ a 3 a2 þ a 1 a3 a a ¼ a0 þ a3 a þ a1 a2 þ a2 a4 : Also r2 2 GðK; F0 Þ. Proof r2 : K ! K is one-to-one: Let r 2 a 0 þ a 1 a þ a 2 a 2 þ a 3 a 3 ¼ r2 b 0 þ b 1 a þ b 2 a 2 þ b 3 a 3 ; where each ai ; bi is in F0 . We have to show that for every i 2 f0; 1; 2; 3g, ai ¼ bi . Here ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 ¼ ðb0 b2 Þ þ ðb3 b2 Þa þ ðb1 b2 Þa2 b2 a3 :

Now, since 1; a; a2 ; a3 is a basis of the vector space F0 ðaÞ over F0 , we have 9 a0 a2 ¼ b0 b2 > > = a3 a2 ¼ b3 b2 ; a1 a2 ¼ b1 b2 > > ; a2 ¼ b2


that is, for every i 2 f0; 1; 2; 3g, ai ¼ bi .

141

■

r2 : K ! K is onto: Let us take an arbitrary sum b0 þ b1 a þ b2 a2 þ b3 a3 2 K, where each bi 2 F0 . Since r2 ðb0 b3 Þ þ ðb2 b3 Þa b3 a2 þ ðb1 b3 Þa3 ¼ ððb0 b3 Þ ðb3 ÞÞ þ ððb1 b3 Þ ðb3 ÞÞa þ ððb2 b3 Þ ðb3 ÞÞa2 ðb3 Þa3 ¼ b0 þ b1 a þ b2 a 2 þ b3 a3 ; it follows that r2 ðb0 b3 Þ þ ðb2 b3 Þa b3 a2 þ ðb1 b3 Þa3 ¼ b0 þ b1 a þ b2 a2 þ b3 a3 ; where ðb0 b3 Þ þ ðb2 b3 Þa b3 a2 þ ðb1 b3 Þa3 2 K. It is clear that r2 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a3 a þ a1 a2 þ a2 a4 Þ preserves addition. We claim that r2 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a3 a þ a1 a2 þ a2 a4 Þ preserves multiplication. We have to show that r2 a0 þ a1 a þ a2 a2 þ a3 a3 b0 þ b1 a þ b2 a2 þ b3 a3 1 1 ¼ a0 þ a1 a2 þ a2 þ a3 a b0 þ b1 a2 þ b2 þ b3 a : a a

Since ða0 þ a1 a þ a2 a2 þ a3 a3 Þðb0 þ b1 a þ b2 a2 þ b3 a3 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b1 þ a1 b0 þ a3 b3 Þa þ ða0 b2 þ a1 b1 þ a2 b0 Þa2 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa3 þ ða1 b3 þ a2 b2 þ a3 b1 Þð1 a a2 a3 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 ða1 b3 þ a2 b2 þ a3 b1 ÞÞ þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa2 þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa3 ;

we have

142

2 Galois Theory II

LHS ¼ ða0 b0 þ a2 b3 þ a3 b2 ða1 b3 þ a2 b2 þ a3 b1 ÞÞ þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa2 þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa4 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa ða0 b1 þ a1 b0 þ a3 b3 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þð1 a a2 a4 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa þ ða0 b1 þ a1 b0 þ a3 b3 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þa3 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa þ ða0 b1 þ a1 b0 þ a3 b3 Þa2 þ ða1 b3 þ a2 b2 þ a3 b1 Þa3 þ ða0 b2 þ a1 b1 þ a2 b0 Þa4 ;

and RHS ¼ a0 þ a1 a2 þ a2 1a þ a3 a b0 þ b1 a2 þ b2 1a þ b3 a ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa þ ða0 b1 þ a1 b0 þ a3 b3 Þa2 þ ða1 b3 þ a2 b2 þ a3 b1 Þa3 þ ða0 b2 þ a1 b1 þ a2 b0 Þa4 ;

so LHS = RHS. Finally, let a0 2 F0 . We have to show that r2 ða0 Þ ¼ a0 . Here, LHS ¼ r2 ða0 Þ ¼ r2 a0 þ 0a þ 0a2 þ 0a3 ¼ a0 þ 0a þ 0a2 þ 0a4 ¼ a0 ¼ RHS: Thus we have shown that r2 2 GðK; F0 Þ. Case III: rðaÞ ¼ a3 . For every a0 ; a1 ; a2 ; a3 2 F0 , r a0 þ a1 a þ a2 a2 þ a3 a3 ¼ rða0 Þ þ rða1 aÞ þ r a2 a2 þ r a2 a3 ¼ rða0 Þ þ rða1 ÞrðaÞ þ rða2 ÞðrðaÞÞ2 þ rða3 ÞðrðaÞÞ3 2 3 1 ¼ a0 þ a1 a3 þ a2 a3 þ a3 a3 ¼ a0 þ a1 a3 þ a2 a þ a3 a 3 2 3 ¼ a0 þ a1 a þ a2 a þ a3 1 a a a ¼ ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ; that is, for every a0 ; a1 ; a2 ; a3 2 F0 , we have r a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 :


143

Let us denote this r by r3 . Thus r3 ðaÞ ¼ a3 , and for every a0 ; a1 ; a2 ; a3 2 F0 , we have r3 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ð2 F0 ðaÞÞ 1 1 ¼ a0 þ a 1 a3 þ a 2 a þ a 3 ¼ 2 a1 þ a 3 a þ a 0 a2 þ a 2 a3 a a ¼ a0 þ a2 a þ a1 a3 þ a3 a4 : Also r3 2 GðK; F0 Þ. Proof r3 : K ! K is one-to-one: Let r 3 a 0 þ a 1 a þ a 2 a 2 þ a 3 a 3 ¼ r3 b 0 þ b 1 a þ b 2 a 2 þ b 3 a 3 ; where each ai ; bi is in F0 . We have to show that for every i 2 f0; 1; 2; 3g, ai ¼ bi . Here ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ¼ ðb0 b3 Þ þ ðb2 b3 Þa b3 a2 þ ðb1 b3 Þa3 :

Now, since 1; a; a2 ; a3 is a basis of the vector space F0 ðaÞ over F0 , we have a0 a3 a2 a3 a3 a1 a3

9 ¼ b0 b3 > > = ¼ b2 b3 ; ¼ b3 > > ; ¼ b1 b3

that is, for every i 2 f0; 1; 2; 3g, ai ¼ bi . r3 : K ! K is onto: Let us take an arbitrary sum b0 þ b1 a þ b2 a2 þ b3 a3 2 K, where each bi is in F0 . Since r3 ðb0 b2 Þ þ ðb3 b2 Þa þ ðb1 b2 Þa2 b2 a3 ¼ ððb0 b2 Þ ðb2 ÞÞ þ ððb1 b2 Þ ðb2 ÞÞa ðb2 Þa2 þ ððb3 b2 Þ ðb2 ÞÞa3 ¼ b0 þ b1 a þ b2 a2 þ b3 a3 ; it follows that

144

2 Galois Theory II

r3 ðb0 b2 Þ þ ðb3 b2 Þa þ ðb1 b2 Þa2 b2 a3 ¼ b0 þ b1 a þ b2 a2 þ b3 a3 ; where ðb0 b2 Þ þ ðb3 b2 Þa þ ðb1 b2 Þa2 b2 a3 2 K. It is clear that r3 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a2 a þ a1 a3 þ a3 a4 Þ preserves addition. We claim that r3 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a2 a þ a1 a3 þ a3 a4 Þ multiplication: We have to show that

preserves

þ a1 a þ a2 a2 þ a3 a3 b0 þ b1 a þ b2 a2 þ b3 a3 1 1 ¼ a0 þ a1 a3 þ a2 a þ a3 b0 þ b1 a3 þ b2 a þ b3 : a a

r3

0

Since ð a0 þ a1 a þ a2 a2 þ a3 a3 Þ ð b0 þ b1 a þ b2 a2 þ b3 a3 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b1 þ a1 b0 þ a3 b3 Þa þ ða0 b2 þ a1 b1 þ a2 b0 Þa2 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa3 þ ða1 b3 þ a2 b2 þ a3 b1 Þð1 a a2 a3 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 ða1 b3 þ a2 b2 þ a3 b1 ÞÞ þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa2 þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa3 ; we have LHS ¼ ða0 b0 þ a2 b3 þ a3 b2 ða1 b3 þ a2 b2 þ a3 b1 ÞÞ þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa3 þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa4 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þa þ ða0 b1 þ a1 b0 þ a3 b3 Þa3 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þ 1 a a3 a4 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þa þ ða0 b1 þ a1 b0 þ a3 b3 Þa3 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þa2 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þa þ ða1 b3 þ a2 b2 þ a3 b1 Þa2 þ ða0 b1 þ a1 b0 þ a3 b3 Þa3 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa4 ;

and


145

RHS ¼ a0 þ a1 a3 þ a2 a þ a3 1a b0 þ b1 a3 þ b2 a þ b3 1a ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b2 þ a1 b1 þ a2 b0 Þa þ ða1 b3 þ a2 b2 þ a3 b1 Þa2 þ ða0 b1 þ a1 b0 þ a3 b3 Þa3 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa4 ; so LHS = RHS. Finally, let a0 2 F0 . We have to show that r3 ða0 Þ ¼ a0 . Here LHS ¼ r3 ða0 Þ ¼ r3 a0 þ 0a þ 0a2 þ 0a3 ¼ a0 þ 0a þ 0a3 þ 0a4 ¼ a0 ¼ RHS: Thus we have shown that r3 2 GðK; F0 Þ.

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Case IV: rðaÞ ¼ a4 . For every a0 ; a1 ; a2 ; a3 2 F0 , r a0 þ a1 a þ a2 a2 þ a3 a3 ¼ rða0 Þ þ rða1 aÞ þ r a2 a2 þ r a2 a3 ¼ rða0 Þ þ rða1 ÞrðaÞ þ rða2 ÞðrðaÞÞ2 þ rða3 ÞðrðaÞÞ3 2 3 1 ¼ a0 þ a1 a4 þ a2 a4 þ a3 a4 ¼ a0 þ a 1 þ a2 a3 þ a3 a2 a ¼ a0 þ a1 1 a a2 a3 þ a2 a3 þ a3 a2 ¼ ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ; that is, for every a0 ; a1 ; a2 ; a3 2 F0 , we have r a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 : Let us denote this r by r4 . Thus r4 ðaÞ ¼ a4 , and for every a0 ; a1 ; a2 ; a3 2 F0 , we have r4 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ð2 F0 ðaÞÞ 1 1 ¼ a0 þ a1 þ a 2 a3 þ a3 a2 ¼ 3 a3 þ a2 a þ a1 a2 þ a0 a 3 a a ¼ a0 þ a3 a2 þ a2 a3 þ a1 a4 : Also r4 2 GðK; F0 Þ. Proof r4 : K ! K is one-to-one: Let r4 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ r4 b0 þ b1 a þ b2 a2 þ b3 a3 ; where each ai ; bi is in F0 . We have to show that for every i 2 f0; 1; 2; 3g, ai ¼ bi . Here

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2 Galois Theory II

ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ ðb0 b1 Þ b1 a þ ðb3 b1 Þa2 þ ðb2 b1 Þa3 :

Now, since 1; a; a2 ; a3 is a basis of the vector space F0 ðaÞ over F0 , we have a0 a1 a1 a3 a1 a2 a1

9 ¼ b0 b1 > > = ¼ b1 ; ¼ b3 b1 > > ; ¼ b2 b1

that is, for every i 2 f0; 1; 2; 3g, ai ¼ bi . r4 : K ! K is onto: Let us take an arbitrary sum b0 þ b1 a þ b2 a2 þ b3 a3 2 K, where each bi is in F0 . Since r4 ðb0 b1 Þ b1 a þ ðb3 b1 Þa2 þ ðb2 b1 Þa3 ¼ ððb0 b1 Þ ðb1 ÞÞ ðb1 Þa þ ððb2 b1 Þ ðb1 ÞÞa2 þ ððb3 b1 Þ ðb1 ÞÞa3 ¼ b0 þ b1 a þ b2 a2 þ b3 a3 ; it follows that r4 ðb0 b1 Þ b1 a þ ðb3 b1 Þa2 þ ðb2 b1 Þa3 ¼ b0 þ b1 a þ b2 a2 þ b3 a3 ; where ðb0 b1 Þ b1 a þ ðb3 b1 Þa2 þ ðb2 b1 Þa3 2 K. It is clear that r4 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a3 a2 þ a2 a3 þ a1 a4 Þ preserves addition. r4 : ða0 þ a1 a þ a2 a2 þ a3 a3 Þ 7! ða0 þ a3 a2 þ a2 a3 þ a1 a4 Þ preserves multiplication: We have to show that r4 a0 þ a1 a þ a2 a2 þ a3 a3 b0 þ b1 a þ b2 a2 þ b3 a3 1 1 3 2 3 2 ¼ a0 þ a1 þ a2 a þ a3 a b0 þ b1 þ b2 a þ b3 a : a a

Since ða0 þ a1 a þ a2 a2 þ a3 a3 Þðb0 þ b1 a þ b2 a2 þ b3 a3 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b1 þ a1 b0 þ a3 b3 Þa þ ða0 b2 þ a1 b1 þ a2 b0 Þa2 þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa3 þ ða1 b3 þ a2 b2 þ a3 b1 Þð1 a a2 a3 Þ ¼ ð a0 b0 þ a2 b3 þ a3 b2 ð a1 b3 þ a2 b2 þ a3 b1 Þ Þ þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa2 þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa3 ; we have


147

LHS ¼ ða0 b0 þ a2 b3 þ a3 b2 ða1 b3 þ a2 b2 þ a3 b1 ÞÞ þ ðða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa2 þ ðða0 b2 þ a1 b1 þ a2 b0 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa3 þ ðða0 b1 þ a1 b0 þ a3 b3 Þ ða1 b3 þ a2 b2 þ a3 b1 ÞÞa4 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa3 þ ða0 b1 þ a1 b0 þ a3 b3 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þ ð1 a2 a3 a4 Þ ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa3 þ ða0 b1 þ a1 b0 þ a3 b3 Þa4 þ ða1 b3 þ a2 b2 þ a3 b1 Þa ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða1 b3 þ a2 b2 þ a3 b1 Þa þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa3 þ ða0 b1 þ a1 b0 þ a3 b3 Þa4 ; and RHS ¼ a0 þ a1 1a þ a2 a3 þ a3 a2 b0 þ b1 1a þ b2 a3 þ b3 a2 ¼ ða0 b0 þ a2 b3 þ a3 b2 Þ þ ða1 b3 þ a3 b1 þ a2 b2 Þa þ ða0 b3 þ a1 b2 þ a2 b1 þ a3 b0 Þa2 þ ða0 b2 þ a1 b1 þ a2 b0 Þa3 þ ða0 b1 þ a1 b0 þ a3 b3 Þa4 ; so LHS = RHS. Finally, let a0 2 F0 . We have to show that r4 ða0 Þ ¼ a0 . Here LHS ¼ r4 ða0 Þ ¼ r4 a0 þ 0a þ 0a2 þ 0a3 ¼ a0 þ 0a2 þ 0a3 þ 0a4 ¼ a0 ¼ RHS: Thus we have shown that r4 2 GðK; F0 Þ. Hence GðK; F0 Þ ¼ fr1 ; r2 ; r3 ; r4 g: Next, oðGðK; F0 ÞÞ ¼ oðfr1 ; r2 ; r3 ; r4 gÞ ¼ 4. Further, ðfixed field of GðK; F0 ÞÞ ¼ fa : a 2 K; and for every r 2 GðK; F0 Þ; rðaÞ ¼ ag ¼ fa : a 2 K; and for every r 2 fr1 ; r2 ; r3 ; r4 g; rðaÞ ¼ ag ¼ fa : a 2 K; r1 ðaÞ ¼ a; r2 ðaÞ ¼ a; r3 ðaÞ ¼ a; r4 ðaÞ ¼ ag ¼ fa : a 2 K; r2 ðaÞ ¼ a; r3 ðaÞ ¼ a; r4 ðaÞ ¼ ag

¼ a0 þ a1 a þ a2 a2 þ a3 a3 : a0 ; a1 a2 ; a3 2 F0 ; ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 ; ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 ; ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 g

¼ a0 þ a1 a þ a2 a2 þ a3 a3 : a0 2 F0 ; a1 ¼ a2 ¼ a3 ¼ 0 ¼ F0 ;

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2 Galois Theory II

so the fixed field of GðF0 ðaÞ; F0 Þ is F0 . Observe that ðr2 Þ2 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ r2 r2 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ r2 ða0 a2 Þ þ ða3 a2 Þa þ ða1 a2 Þa2 a2 a3 ¼ ðða0 a2 Þ ða1 a2 ÞÞ þ ðða2 Þ ða1 a2 ÞÞa þ ðða3 a2 Þ ða1 a2 ÞÞa2 ða1 a2 Þa3 ¼ ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ r4 a0 þ a1 a þ a2 a2 þ a3 a3 ; so ðr2 Þ2 ¼ r4 . Next, ðr2 Þ3 a0 þ a1 a þ a2 a2 þ a3 a3 ¼ r2 ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ ðða0 a1 Þ ða3 a1 ÞÞ þ ðða2 a1 Þ ða3 a1 ÞÞa þ ðða1 Þ ða3 a1 ÞÞa2 ða3 a1 Þa3 ¼ ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ¼ r3 a0 þ a1 a þ a2 a2 þ a3 a3 ; so ðr2 Þ3 ¼ r3 . Finally, ð r2 Þ 4 a 0 þ a 1 a þ a 2 a 2 þ a 3 a 3 ¼ r2 ða0 a3 Þ þ ða2 a3 Þa a3 a2 þ ða1 a3 Þa3 ¼ ðða0 a3 Þ ða3 ÞÞ þ ðða1 a3 Þ ða3 ÞÞa þ ðða2 a3 Þ ða3 ÞÞa2 ða3 Þa3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3 ¼ r1 a0 þ a1 a þ a2 a2 þ a3 a3 ; so ðr2 Þ4 ¼ r1 . Thus n o GðK; F0 Þ ¼ fr1 ; r2 ; r3 ; r4 g ¼ r2 ; ðr2 Þ2 ; ðr2 Þ3 ; ðr2 Þ4 : It follows that GðK; F0 Þ is a cyclic group generated by r2 : a0 þ a1 a þ a2 a2 þ a3 a3 7! a0 þ a3 a þ a1 a2 þ a2 a4 :


149

Since ð r4 Þ 2 a 0 þ a 1 a þ a 2 a 2 þ a 3 a 3 ¼ r4 ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ ðða0 a1 Þ ða1 ÞÞ ða1 Þa þ ðða2 a1 Þ ða1 ÞÞa2 þ ðða3 a1 Þ ða1 ÞÞa3 ¼ a0 þ a1 a þ a 2 a2 þ a3 a3 ¼ r1 a0 þ a1 a þ a2 a2 þ a3 a3 ; we have ðr4 Þ2 ¼ r1 , and hence fr1 ; r4 g is a subgroup of GðK; F0 Þ. Here ðfixed field of fr1 ; r4 gÞ ¼ fa : a 2 K; and for every r 2 fr1 ; r4 g; rðaÞ ¼ ag ¼ fa : a 2 K; r1 ðaÞ ¼ a; r4 ðaÞ ¼ ag ¼ fa : a 2 K; r4 ðaÞ ¼ ag

¼ a0 þ a1 a þ a2 a2 þ a3 a3 : a0 ; a1 a2 ; a3 2 F0 ;ða0 a1 Þ a1 a þ ða3 a1 Þa2 þ ða2 a1 Þa3 ¼ a0 þ a1 a þ a2 a2 þ a3 a3

¼ a0 þ a1 a þ a2 a2 þ a3 a3 : a0 2 F0 ; a1 ¼ 0; a2 ¼ a3

¼ a0 þ a2 a2 þ a3 : a0 ; a2 2 F 0 ;

so the fixed field of fr1 ; r4 g is a0 þ a2 ða2 þ a3 Þ : a0 ; a2 2 F0 . 2.3.16 Problem Let n be a positive integer 2. Let F be a field such that F C. Suppose that F contains all the nthroots of unity, that is, F contains all the roots of 2 n1 2pi 2pi 2pi F. Let a be a the polynomial xn 1, that is, 1; e n ; e n ; . . .; e n nonzero member of F. Let u be a root of the polynomial xn a in C, that is, un ¼ a. Then FðuÞ is the splitting field over F for xn a. Proof We must prove: 1. FðuÞ is a finite extension of F, 2. FðuÞ contains all the roots of xn a in C, 3. if G is a proper subfield of FðuÞ that contains F, then G does not contain all the roots of xn a in C. For 1: Since uð2 CÞ is a root of the polynomial ðxn aÞ 2 F½x, u is algebraic over F, and hence by 1.4.17, FðuÞ is a finite extension of F. 2pi 2 2pi n1 2pi For 2: Here we have to show that u1; ue n ; u e n ; . . .; u e n FðuÞ. 2pi 2 2pi n1 2pi Since 1; e n ; e n ; . . .; e n F FðuÞ, u 2 FðuÞ, and FðuÞ is a field, we have

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2 Galois Theory II

2pi n

u1; ue ; u e

2pi n

2

2pi n1 ; . . .; u e n FðuÞ:

For 3: Suppose to the contrary that G is a subfield of FðuÞ such that G 6¼ FðuÞ, G contains F, and G contains

2pi 2 2pi n1 2pi u1; ue n ; u e n ; . . .; u e n ð3uÞ:

We seek a contradiction. Since G is a subfield of FðuÞ, we have G FðuÞ. Since G is a field that contains F [ fug, we have FðuÞ G, and hence G ¼ FðuÞ. This is a contradiction. ■ 2.3.17 Problem Let n be a positive integer 2. Let F be a field such that F C. Suppose that F contains all the nthroots of unity, that is, F contains all the roots of 2pi 2 2pi n1 2pi the polynomial xn 1, that is, 1; e n ; e n ; . . .; e n F. Let a be a nonzero member of F. Hence ðxn aÞ 2 F½x. Then the Galois group of xn a over F is abelian. Proof By 2.3.16, FðuÞ is the splitting field over F for xn a, where un ¼ a and u 2 C, and hence FðuÞcontains all the roots of xn a in C. Further, the set of all 2pi 2 2pi n1 2pi ð FðuÞÞ. the roots of xn a is u1; ue n ; u e n ; . . .; u e n is the group The Galois group of xn a GðFðuÞ; F Þð¼ fr : r 2 AutðFðuÞÞ; and for every a 2 F; rðaÞ ¼ agÞ:We have to show that GðFðuÞ; F Þ is abelian. To this end, let us take an automorphism r : FðuÞ ! FðuÞ such that for every a 2 F; rðaÞ ¼ a. Next let us take an automorphism s : FðuÞ ! FðuÞ such that for every a 2 F; sðaÞ ¼ a. We have to show that for every b 2 FðuÞ, sðrðbÞÞ ¼ rðsðbÞÞ. Observe that ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg is a subfield of FðuÞ. Proof Let s; t 2 FðuÞ, where sðrðsÞÞ ¼ rðsðsÞÞ, and sðrðtÞÞ ¼ rðsðtÞÞ. It suffices to show that 1. sðrðs tÞÞ ¼ rðsðs tÞÞ; 2. sðrðstÞÞ ¼ rðsðstÞÞ; 3. if s, t are nonzero, then sðrðst1 ÞÞ ¼ rðsðst1 ÞÞ. For 1: LHS ¼ sðrðs tÞÞ ¼ sðrðsÞ rðtÞÞ ¼ sðrðsÞÞ sðrðtÞÞ ¼ rðsðsÞÞ sðrðtÞÞ ¼ rðsðsÞÞ rðsðtÞÞ ¼ rðsðsÞ sðtÞÞ ¼ rðsðs tÞÞ ¼ RHS:


151

For 2: LHS ¼ sðrðstÞÞ ¼ sðrðsÞrðtÞÞ ¼ sðrðsÞÞsðrðtÞÞ ¼ rðsðsÞÞsðrðtÞÞ ¼ rðsðsÞÞrðsðtÞÞ ¼ rðsðsÞsðtÞÞ ¼ rððstÞÞ ¼ RHS: For 3: Let s, t be nonzero. We have LHS ¼ sðrðst1 ÞÞ ¼ s rðsÞðrðtÞÞ1 ¼ sðrðsÞÞðsðrðtÞÞÞ1 ¼ rðsðsÞÞðsðrðtÞÞÞ1 ¼ rðsðsÞÞðrðsðtÞÞÞ1 ¼ r sðsÞðsðtÞÞ1 ¼ rðsðst1 ÞÞ ¼ RHS: ■ We have shown that ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg is a subfield of FðuÞ. It is clear that F ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg: Since un ¼ a, we have ðrðuÞÞn ¼ rðun Þ ¼ rðaÞ ¼ a; |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} and hence ðrðuÞÞn a ¼ 0. Thus rðuÞ is a root of xn a inC. Now, since the set of 2pi 2 2pi n1 2pi , there exists an integer all the roots of xn a is u1; ue n ; u e n ; . . .; u e n 2pi k k 2 f0; 1; . . .; n 1g such that rðuÞ ¼ u e n . Similarly, there exists an integer 2pi l l 2 f0; 1; . . .; n 1g such that sðuÞ ¼ u e n . Clearly, sðrðuÞÞ ¼ rðsðuÞÞ. Proof 2pi k 2pi l 2pi k 2pi k LHS ¼ sðrðuÞÞ ¼ s u e n ¼ sðuÞs e n ¼ sðuÞ e n ¼ u e n 2pi k 2pi k þ l ; en ¼ u en and 2pi l 2pi l 2pi l RHS ¼ rðsðuÞÞ ¼ r u e n ¼ rðuÞr e n ¼ rðuÞ e n 2pi k 2pi l 2pi k þ l ¼ u en en ¼ u en : Thus LHS = RHS.

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We have shown that sðrðuÞÞ ¼ rðsðuÞÞ, and hence u 2 ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg. Thus ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg is a field containing F [ fug, and hence FðuÞ ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞgð FðuÞÞ: Thus FðuÞ ¼ ft : t 2 FðuÞ; sðrðtÞÞ ¼ rðsðtÞÞg. Thus for every b 2 FðuÞ, sðrðbÞÞ ¼ rðsðbÞÞ. ■

2.4

Solvability By Radicals

2.4.1 Definition Let F, K be any fields such that K is an extension of F. Let pðxÞ 2 F½x. Let degðpðxÞÞ ¼ n. Suppose that K contains all the n roots of pðxÞ in K. If there exists a finite sequence ðx1 ; r1 Þ; ðx2 ; r2 Þ; . . .; ðxk ; rk Þ such that 1. 2. 3. 4.

each xi is a member of K, each ri is an integer 2,

ðx1 Þr1 2 F; ðx2 Þr2 2 F ðx1 Þ; ðx3 Þr3 2 F ðx1 ; x2 Þ; . . .; ðxk Þrk 2 F ðx1 ; x2 ; . . .; xk1 Þ;

F ðx1 ; x2 ; . . .; xk1 ; xk Þ contains all the n roots of pðxÞ in K,

then we say that pðxÞ is solvable by radicals over F. 2.4.2 Example Let us take Q for F, C for K, and x5 2x3 x2 þ 2 for pðxÞ. Since x5 2x3 x2 þ 2 ¼ x3 x2 2 x2 2 pffiffiffi pffiffiffi ¼ x2 2 x3 1 ¼ x 2 x þ 2 x3 1 pffiffiffi pffiffiffi ¼ x 2 x þ 2 ð x 1Þ x2 þ x þ 1 pffiffiffi 2 ! pffiffiffi pffiffiffi 3 1 2 i ¼ x 2 x þ 2 ð x 1Þ xþ 2 2 pffiffiffi pffiffiffi pffiffiffi pffiffiffi 3 3 1 1 þi i ¼ ðx 1Þ x 2 x þ 2 x x ; 2 2 2 2

2.4 Solvability By Radicals

153

we have pffiffiffi pffiffiffi x5 2x3 x2 þ 2 ¼ ðx 1Þ x 2 x þ 2 ðx xÞ x x2 ; pffiffi pffiffiffi pffiffiffi 3 the roots of pðxÞ are 1, 2; 2; x; x2 . where x 1 2 þ i 2 . It follows that all pffiffiffi They all are members of C. Let us take 2 for x1 and x for x2 . Let us take 2 for r1 pffiffiffi and 3 for r2 . Put F1 Q 2 and F2 F1 ðxÞ. Let us take k ¼ 2. Now all six conditions of the above definition are satisfied, so x5 2x3 x2 þ 2 is solvable by radicals over Q.

2.4.3 Example Let us take the general cubic polynomial x 3 þ a1 x 2 þ a2 x þ a3 over the field F0 of all rational numbers. By F0 ða1 ; a2 ; a3 Þ we shall mean the field of rational functions in a1 ; a2 ; a3 over F0 . Since x 3 þ a1 x 2 þ a2 x þ a3 a 2 a 3 a1 1 1 ¼ x3 þ 3x2 þ 3x þ 3 3 3 a 2 a 3 1 1 3x þ a2 x þ a3 3 3 ! a 3 a1 3 ða1 Þ2 1 ¼ xþ þ x a2 þ a3 3 3 3 ! ! a 3 a1 3 ða1 Þ2 a1 ð a1 Þ 2 a1 1 ¼ xþ þ a3 þ a2 xþ a2 3 3 3 3 3 3 ¼ y3 þ py þ q; 2 2 p a2 ða31 Þ ð2 F0 ða1 ; a2 ; a3 ÞÞ, q a2 ða31 Þ where y x þ a31 , a 3 2ða1 Þ3 a a a1 1 ¼ 27 13 2 þ a3 ð2 F0 ða1 ; a2 ; a3 ÞÞ, we have 3 þ a3 3 x3 þ a1 x2 þ a2 x þ a3 ¼ y3 þ py þ q: Put y u þ v, where uv ¼ p 3 . It follows that y3 þ py þ q ¼ u3 þ v3 þ 3uvy þ py þ q ¼ u3 þ v3 þ q:

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Hence x 3 þ a1 x 2 þ a2 x þ a3 ¼ 0 is equivalent to y3 þ py þ q ¼ 0; which, in turn, is equivalent to the simultaneous equations u3 þ v 3 þ q ¼ 0 ; uv ¼ p 3 that is, u3 þ v3 ¼ q 3 ; u3 v3 ¼ p 27 that is, ) u3 þ v3 ¼ q 3 : 2 ðu3 v3 Þ ¼ q2 4 p 27 Now we can take qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 3 > > q þ q2 þ 4p u ¼ = 27 : qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 3 > > ; v3 ¼ 12 q q2 þ 4p 27 3

1 2

It follows that we can take rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 9 > 3 1 1 3 = u ¼ 2 q þ 27 p þ 14 q2 > rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi >: 3 ; v ¼ 1 q 1 p3 þ 1 q2 > 2

27

4

Hence all the roots of x3 þ a1 x2 þ a2 x þ a3 are

a1 a1 a1 þ ðu þ vÞ; þ ux þ vx2 ; þ ux2 þ vx : 3 3 3

This is known as Cardan’s formula.


155

2.4.4 Example Let us take the general biquadratic polynomial x 4 þ a1 x 3 þ a2 x 2 þ a3 x þ a4 over the field F0 of all rational numbers. By F0 ða1 ; a2 ; a3 ; a4 Þ we shall mean the field of rational functions in a1 ; a2 ; a3 ; a3 over F0 . Since x 4 þ a1 x 3 þ a2 x 2 þ a3 x þ a4 a 2 a 3 a 4 a1 1 1 1 ¼ x4 þ 4x3 þ 6x2 þ 4x þ 4 4 4 4 a 2 a 3 a 4 1 1 1 6x2 4x þ a2 x 2 þ a3 x þ a4 4 4 4 a 2 a 3 a 4 a1 4 1 1 1 ¼ xþ þ x 2 a2 6 þ x a3 4 þ a4 4 4 4 4 a1 2 a1 ¼ y4 þ y ð Þ þ y ð Þ þ ð Þ ¼ y4 þ py2 þ qy þ r; 4 4 where y x þ a41 , and p; q; r 2 F0 ða1 ; a2 ; a3 ; a4 Þ. It suffices to solve the equation y4 þ py2 þ qy þ r ¼ 0; that is,

y2 þ

p 2 p2 ¼ qy r; 2 4

or 2 p p2 y2 þ þ m ¼ m2 þ 2y2 þ p m þ qy r; 2 4 where m is to be determined. Here

y2 þ

2 p p2 þ m ¼ 2my2 qy þ m2 þ pm þ r : 2 4

The equation is solved when the quadratic expression on the right-hand side of the above equation is a perfect square, that is, if its discriminant is zero, that is, p2 ðqÞ2 4ð2mÞ m2 þ pm þ r ¼ 0; 4

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2 Galois Theory II

that is,

p2 q2 r m ¼ 0: m þ pm þ 4 8 3

2

This is a cubic equation in m, so all its roots can be found as in Example 2.4.3. This is known as Ferrari’s method. 2.4.5 Note Let F, K be any fields such that K is an extension of F. Suppose that for every positive integer l, F contains all the lth roots of unity. Let pðxÞ 2 F½x. Let degðpðxÞÞ ¼ n. Suppose that K contains all the n roots of pðxÞ. Suppose that pðxÞ is solvable by radicals over F. Hence there exists a finite sequence ðx1 ; r1 Þ; ðx2 ; r2 Þ; . . .; ðxk ; rk Þ such that 1. each xi is a nonzero member of K, 2. each ri is an integer 2, ðx1 Þr1 2 F; ðx2 Þr2 2 F ðx1 Þ; ðx3 Þr3 2 F ðx1 ; x2 Þ; . . .; ðxk Þrk 2 F ðx1 ; x2 ; . . .; xk1 Þ; 4. F ðx1 ; x2 ; . . .; xk1 ; xk Þ contains all the n roots of pðxÞ in K. 3:

Let L be the splitting field over F for pðxÞ. It follows that L is the smallest field containing all the n roots of pðxÞ in K. Now since F ðx1 ; x2 ; . . .; xk1 ; xk Þ is a field containing all the n roots of pðxÞ in K, we have L F ðx1 ; x2 ; . . .; xk1 ; xk Þ. Since ðx1 Þr1 2 F, there exists a nonzero a 2 F such that x1 is a root of the polynomial ðxr1 aÞ 2 F½x. By assumption, F contains all the r1 th roots of unity. So by 2.3.16, F ðx1 Þ is the splitting field over F for xr1 a, and hence by 2.2.20, F ðx1 Þ is a normal extension of F. Thus F ðx1 Þ is a normal extension of F. Since ðx2 Þr2 2 F ðx1 Þ, there exists b 2 F ðx1 Þ such that x2 is a root of the polynomial equation ðxr2 bÞ 2 ðF ðx1 ÞÞ½x. By assumption, F ð F ðx1 ÞÞ contains all the r2 th roots of unity, so F ðx1 Þ contains all the r2 th roots of unity. Now by 2.3.16, ðF ðx1 ÞÞðx2 Þð¼ F ðx1 ; x2 ÞÞ is the splitting field over ðF ðx1 ÞÞ for xr2 b, and hence by 2.2.20, F ðx1 ; x2 Þ is a normal extension of F ðx1 Þ. Similarly, F ðx1 ; x2 ; x3 Þ is a normal extension of F ðx1 ; x2 Þ, etc. Since F ðx1 Þ is a normal extension of F, by 2.2.24, GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ÞÞ is a normal subgroup of the group GðF ðx1 ; x2 ; . . .; xk Þ; F Þ. Similarly, GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ÞÞ is a normal subgroup of the group GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ÞÞ, GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; x3 ÞÞ is a normal subgroup of the group GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ÞÞ, etc. Thus


157

fGðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x1 ; x2 ; . . .; xk1 ÞÞ; . . .; GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ÞÞ; GðF ðx1 ; x2 ; . . .; xk Þ; F Þg is a collection of subgroups of GðF ðx1 ; x2 ; . . .; xk Þ; F Þ such that 1. GðF ðx1 ; x2 ; . . .; xk Þ; F Þ GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ÞÞ GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x1 ; x2 ; . . .; xk1 ÞÞ fIdg; where Id denotes the identity automorphism of F ðx1 ; x2 ; . . .; xk Þ. 2. for every i ¼ 1; . . .; k, GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi ÞÞ is a normal subgroup of GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ. By 2.2.26, for every i ¼ 1; . . .; k, the quotient group GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi ÞÞ is isomorphic onto the group GðF ðx1 ; x2 ; ; xi Þ; F ðx1 ; x2 ; ; xi1 ÞÞ ð¼ GððF ðx1 ; x2 ; ; xi1 ÞÞðxi Þ; F ðx1 ; x2 ; ; xi1 ÞÞÞ: We want to show that, for every i ¼ 1; . . .; k, the quotient group GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ GðF ðx1 ; x2 ; . . .; xk Þ; F ðx1 ; x2 ; . . .; xi ÞÞ is abelian. It suffices to show that for every i ¼ 1; . . .; k, GððF ðx1 ; x2 ; . . .; xi1 ÞÞðxi Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ is abelian, that is, each GðF ðx1 ; x2 ; . . .; xi Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ is abelian. To this end, we shall apply 2.3.17. Since ðxi Þri 2 F ðx1 ; x2 ; . . .; xi1 Þ, there exists a nonzero a2 F ðx1 ; x2 ; . . .; xi1 Þ such that xi is a root of the polynomial ðxri aÞ 2 ðF ðx1 ; x2 ; . . .; xi1 ÞÞ½x. By assumption, F ðx1 ; x2 ; . . .; xi1 Þ contains all the ri th roots of unity. So by 2.3.16, ðF ðx1 ; x2 ; . . .; xi1 ÞÞðxi Þð¼ F ðx1 ; x2 ; . . .; xi ÞÞ is the splitting field over F ðx1 ; x2 ; . . .; xi1 Þ for ðxri aÞ. By 2.3.17, the Galois group of xri a over F ðx1 ; x2 ; . . .; xi1 Þ is abelian.Here the Galois group of xri a is GðF ðx1 ; x2 ; . . .; xi Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ; so GðF ðx1 ; x2 ; . . .; xi Þ; F ðx1 ; x2 ; . . .; xi1 ÞÞ is abelian. Thus we have shown that GðF ðx1 ; x2 ; . . .; xk Þ; F Þ is a solvable group. We want to show that the Galois group over F of pðxÞ is a solvable group, that is, GðL; F Þ is a solvable group.

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2 Galois Theory II

Since L is the splitting field over F for pðxÞð2 F½xÞ, by 2.2.20, L is a normal extension of F. Now, since L F ðx1 ; x2 ; . . .; xk1 ; xk Þ, by 2.2.24, GðF ðx1 ; x2 ; . . .; xk Þ; LÞ is a normal subgroup of the group GðF ðx1 ;x2 ;...;xk Þ;F Þ GðF ðx1 ; x2 ; . . .; xk Þ; F Þ. Further, by 2.2.25, the quotient group GðF ðx1 ;x2 ;...;xk Þ;LÞ is 2 ;...;xk Þ;F Þ isomorphic to the group GðL; F Þ. Since the quotient group GGððFFððxx11;x ;x2 ;...;xk Þ;LÞ is a homomorphic image of GðF ðx1 ; x2 ; . . .; xk Þ; F Þ, the group GðL; F Þ is a homomorphic image of GðF ðx1 ; x2 ; . . .; xk Þ; F Þ. Next, since GðF ðx1 ; x2 ; . . .; xk Þ; F Þ is a solvable group, the group GðL; F Þ is a homomorphic image of a solvable group. It follows, by 2.3.10, that the group GðL; F Þ is solvable.

2.4.6 Conclusion Let F, K be any fields such that K is an extension of F. Suppose that for every positive integer l, F contains all the lth roots of unity. Let pðxÞ 2 F½x. Let degðpðxÞÞ ¼ n. Suppose that K contains all the n roots of pðxÞ. Suppose that pðxÞ is solvable by radicals over F. Then the Galois group over F of pðxÞ is a solvable group. 2.4.7 Note Let F be any field. By the general polynomial xn þ a1 xn1 þ þ an of degree n over F, we mean the following: F ða1 ; . . .; an Þ is the field of all rational functions in n variables a1 ; . . .; an over F, and xn þ a1 xn1 þ þ an is a polynomial in x over the field F ða1 ; . . .; an Þ ð3a1 ; . . .; an Þ. If xn þ a1 xn1 þ þ an is solvable by radicals over F ða1 ; . . .; an Þ, then we say that the general polynomial xn þ a1 xn1 þ þ an of degree n over F is solvable by radicals. By 2.2.14, the splitting field over F ða1 ; . . .; ð1Þn an Þð¼ F ða1 ; . . .; an ÞÞ for n t þ a1 tn1 þ a2 tn2 þ þ an is F ðx1 ; . . .; xn Þ, and Sn ¼ GðF ðx1 ; ; xn Þ; F ða1 ; ; an ÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼G

splitting field over F ða1 ; ; an Þfor tn þ a1 tn1 þ a2 tn2

¼ Galois group of t þ a1 t n

n1

!

þ þ an ; F ð a1 ; ; an Þ þ a2 t þ þ an : n2

It follows that the Galois group of tn þ a1 tn1 þ a2 tn2 þ þ an is Sn . Next, by 2.3.12, Sn is not solvable for n 5; so the Galois group of tn þ a1 tn1 þ a2 tn2 þ þ an is not solvable for n 5, and hence by 2.4.6, tn þ a1 tn1 þ a2 tn2 þ þ an is not solvable by radicals over F for n 5. 2.4.8 Conclusion Let F be any field. Suppose that for every positive integer l, F contains all the lth roots of unity. Let n 5. Let xn þ a1 xn1 þ þ an be the general polynomial of degree n over F. Then tn þ a1 tn1 þ a2 tn2 þ þ an is not solvable by radicals over F.


159

Roughly speaking, for n 5, there exists no formula for the roots of tn þ a1 tn1 þ a2 tn2 þ þ an involving only a combination of mth roots of rational functions of a1 ; ; an , for various values of m. This result is due to Niels Henrik Abel (1802–1829). 2.4.9 Note Here we shall recapitulate something about high-school “construction geometry,” We shall assume that we have a straightedge (that is, an ungraduated scale), a compass (that is, an instrument with two arms, one with a metallic needle end, and another with a pencil’s lead end). We also assume that we are given the measure of a “unit distance.” Some of the well-known constructions are sketched below: 1. In Fig. 2.1, the perpendicular bisector of a given line segment is constructed. 2. In Fig. 2.2, the perpendicular at given point of a line is constructed. 3. In Fig. 2.3, the perpendicular line from a given point on a given line is constructed. 4. In Fig. 2.4, a line parallel to a given line and passing through a given point is constructed. Here, construction (3) and then construction (2) are made. 5. In Fig. 2.5, a given line segment is divided into three equal parts. Similarly, a given line segment can be divided into any number of equal parts. It follows that every rational number is a “constructible number.” In other words, Q W, where W denotes the collection of all constructible numbers. It is easy to

Fig. 2.1 Perpendicular bisector of a given line segment

Fig. 2.2 Perpendicular at a given point of a line

160

Fig. 2.3 Perpendicular line drawn from a given point on a given line

Fig. 2.4 Line parallel to a given line and passing through a given point

Fig. 2.5 A given line segment to be divided into three equal parts

Fig. 2.6 Construction for squaring the size of a line segment

2 Galois Theory II


161

Fig. 2.7 Construction for reciprocal of the size of a line segment

observe that if a and b are constructible numbers, then a þ b and a b are constructible numbers. Now we shall show that if a is a constructible number, then so is a2 (Fig. 2.6). We first construct a right triangle ABC one of whose legs, AB, is of length 1, and the other leg, AC, is of length a. Now we draw a line perpendicular to BC at C. Suppose that this line meets AB at D. Thus we get two similar triangles, ACB and ADC. It follows that a AC AD AD ¼ ¼ ¼ ; 1 |fflfflfflfflffl AB ffl{zfflfflfflfflffl AC a ffl} and hence AD ¼ a2 . Thus a2 is a constructible number. 2

2

Now, since ab ¼ ða þ bÞ 4þ ðabÞ , we get the following result: if a and b are constructible numbers, then so is ab. We want to show that W is a field. For this, it suffices to show that if a is a positive constructible number, then so is 1a (Fig. 2.7). We first construct a right triangle ABC one of whose legs, AB, is of length a, and the other leg, AC, is of length 1. Now we draw a line perpendicular to BC at C. Suppose that this line meets AB at D. Thus we get two similar triangles, ACB and ADC. It follows that 1 AC AD AD ¼ ¼ ¼ ; a |fflfflfflfflffl AB ffl{zfflfflfflfflffl AC 1 ffl} and hence AD ¼ 1a. Thus 1a is a constructible number. Thus W is a subfield of R that contains Q. 2.4.10 Definition Let F be a subfield of R. By the plane of F we mean the Cartesian product F F R2 . Observe that the straight line joining the point ðx1 ; y1 Þð2 F F Þ and the point ðx2 ; y2 Þð2 F F Þ is y y1 ¼

y2 y1 ðx x1 Þ; x2 x1

which is of the form ax þ by þ c ¼ 0, where a; b; c 2 F. Similarly, every equation of the form ax þ by þ c ¼ 0 represents a straight line passing through two points of the plane of F. Such straight lines are called straight lines in F.

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2 Galois Theory II

It is clear that if two straight lines in F intersect in the real plane R2 , then their point of intersection is a point in the plane of F. Every circle having center at a point of the plane of F and radius an element of F is of the form x2 þ y2 þ ax þ by þ c ¼ 0, where a; b; c 2 F. Such circles are called circles in F. It is clear that if a circle in F and a straight line in F intersect in the real plane R2 , then their points of intersection either points in the plane of F or points in the pare ffiffiffi plane of the field extension F c of F, for some positive c 2 F. Similarly, it is clear that if two circles in F intersect in the real plane R2 , then their points of intersection either points in the plane of F or points in the plane pare ffiffiffi of the field extension F c of F, for some positive c 2 F. Thus, if a straight line or a circle in the field F intersects another straight line or a circle in the field F in the real plane R2 , then there exists a positive real number c1 such that their point(s) of intersection are points in the plane of the field extension F ðc1 Þ of F, where ðc1 Þ2 2 F. As above, if a straight line or a circle in the field F ðc1 Þ intersects another straight line or a circle in the field F ðc1 Þ in the real plane R2 , then there exists a positive real number c2 such that their point(s) of intersection are points in the plane of the field extension ðF ðc1 ÞÞðc2 Þð¼ F ðc1 ; c2 ÞÞ of F ðc1 Þ, where ðc2 Þ2 2 F ðc1 Þ, etc. Hence, if a point is constructible from F, then there exists a finite sequence c1 ; . . .; cn of real numbers such that 1. ðc1 Þ2 2 F; ðc2 Þ2 2 F ðc1 Þ; ðc3 Þ2 2 F ðc1 ; c2 Þ; . . .; ðcn Þ2 2 F ðc1 ; . . .; cn1 Þ; 2. the point is in the plane of F ðc1 ; . . .; cn Þ. Since ðc1 Þ2 2 F, there exists a 2 F such that c1 is a root of the polynomial ðx aÞ 2 FðxÞ. Here degðx2 aÞ ¼ 2, so c1 is algebraic of degree 1 or 2, and hence by 1.4.16, ½F ðc1 Þ; F ¼ 1 or 2. Similarly ; ½F ðc1 ; c2 Þ; F ðc1 Þ ¼ 1 or 2, and hence by 1.4.3, ½F ðc1 ; c2 Þ; F ¼ ½F ðc1 ; c2 Þ; F ðc1 Þ½F ðc1 Þ; F ð¼ 1or2or22 Þ. Thus ½F ðc1 ; c2 Þ; F ¼ 1 or 2 or 22 . Similarly, ½F ðc1 ; c2 ; c3 Þ; F ¼ 1 or 2 or 22 or 23 , etc. 2

2.4.11 Conclusion Suppose that a real number a is constructible. Then there exist an extension K of Q and a nonnegative integer k such that a 2 K and a is algebraic of degree 2k . 2.4.12 Theorem It is impossible, by straightedge and a compass alone, to trisect the angle 60 . Proof Suppose to the contrary that the angle 20 is constructible. We seek a contradiction.

Fig. 2.8 Construction for the size of cos 20 , provided 20 angle is constructible


163

Let us draw a circle of radius 1 with center at the vertex of the 20 angle. Now draw the foot of perpendicular as shown in Fig. 2.8: Thus cos 20 is constructible. On using the formula cos 3h ¼ 4 cos3 h 3 cos h, we get 12 ¼ 4 cos3 20 3 cos 20 , and hence 8x3 6x 1 ¼ 0, where x cos 20 . Thus cos 20 is a root of the polynomial 8x3 6x 1ð2 Q½xÞ. We claim that 8x3 6x 1 is an irreducible polynomial over the field of rational numbers. Proof Put x y 1. It suffices to show that 8ðy 1Þ3 6ðy 1Þ 1 is irreducible over the field of rational numbers. Observe that 8ðy 1Þ3 6ðy 1Þ 1 ¼ 8 y3 3y2 þ 3y 1 6y þ 6 1 ¼ 3 þ 18y 24y2 þ 8y3 ¼ 3 1 þ 6y 8y2 þ 8y3 ; so 8ðy 1Þ3 6ðy 1Þ 1 ¼ 3 1 þ 6y 8y2 þ 8y3 : By 1.3.5, 3ð1 þ 6y 8y2 Þ þ 8y3 is irreducible over the field of rational numbers, and hence ■ 8ðy 1Þ3 6ðy 1Þ 1 is irreducible over the field of rational numbers. Thus we have shown that 8x3 6x 1 is an irreducible polynomial over the field of rational numbers. Then by 1.5.12, cos 20 is algebraic of degree 3 over Q. Since cos 20 is constructible, by 2.4.12, cos 20 is algebraic of a degree of the form ■ 2k . This contradicts the fact that cos 20 is algebraic of degree 3 over Q. 2.4.13 Theorem It is impossible by straightedge and a compass alone to duplicate the cube in the sense of constructing an edge of a cube whose volume is twice the volume of a given cube. Proof For simplicity, suppose that the volume of the given cube is 1. We have to construct a length a such that a3 ¼ 2. Suppose that a is constructible. We seek a contradiction. Here a is a root of the polynomial x3 2ð2 Q½xÞ. We claim that x3 2 is an irreducible polynomial over the field of rational numbers. Proof Put x y 1. It suffices to show that ðy 1Þ3 2 is an irreducible polynomial over the field of rational numbers.

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2 Galois Theory II

Observe that ðy 1Þ3 2 ¼ y3 3y2 þ 3y 1 2 ¼ 3 þ 3y 3y2 þ y3 ¼ 3 1 þ y y2 þ y3 ; so ðy 1Þ3 2 ¼ 3 1 þ y y2 þ y3 : By 1.3.5, 3ð1 þ y y2 Þ þ y3 is irreducible over the field of rational numbers, and hence ■ ðy 1Þ3 2 is irreducible over the field of rational numbers. Thus we have shown that x3 2 is an irreducible polynomial over the field of rational numbers. So by 1.5.12, a is algebraic of degree 3 over Q. Since a is constructible, by 2.4.12, a is algebraic of a degree of the form 2k . This contradicts the fact that a is algebraic of degree 3 over Q. ■ 2.4.14 Theorem It is impossible, by straightedge and a compass alone, to construct a regular septagon. Proof Construction of a regular septagon requires the construction of an angle 2p 7. 2p Suppose that the angle 2p is constructible, and hence 2 cos is constructible. We 7 7 seek a contradiction. Put h 2p . 7 It follows that 4h ¼ 2p 3h; and hence 2ð2 sin h cos hÞ cos 2h ¼ 2 sin 2h cos 2h ¼ sin 4h ¼ sinð2p 3hÞ ¼ sin 3h |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 3 sin h þ 4ðsin hÞ3 ¼ sin h 3 þ 4 sin2 h : This shows that cos h cos 2hffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ¼ 3 þ 4 sin2 fflh} 2 cos h ð2 cos hÞ2 2 ¼ 4 cos hð2 cos2 h 1Þ ¼ 4|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ¼ 3 þ 4ð1 cos2 hÞ ¼ 4 cos2 h þ 1;

that is, y y2 2 ¼ y2 þ 1;


165

or y3 þ y2 2y 1 ¼ 0; where y 2 cos h. Thus 2 cos h is a root of the polynomial x3 þ x2 2x 1ð2 Q½xÞ. Next, we claim that x3 þ x2 2x 1 is an irreducible polynomial over the field of rational numbers. Proof Put x y þ 2. It suffices to show that ðy þ 2Þ3 þ ðy þ 2Þ2 2ðy þ 2Þ 1 is an irreducible polynomial over the field of rational numbers. Observe that ðy þ 2Þ3 þ ðy þ 2Þ2 2ðy þ 2Þ 1 ¼ 7 þ 14y þ 7y2 þ y3 ¼ 7 1 þ 2y þ y2 þ y3 ; so ðy þ 2Þ3 þ ðy þ 2Þ2 2ðy þ 2Þ 1 ¼ 7 1 þ 2y þ y2 þ y3 : By 1.3.5, 7ð1 þ 2y þ y2 Þ þ y3 is irreducible over the field of rational numbers, and hence ðy þ 2Þ3 þ ðy þ 2Þ2 2ðy þ 2Þ 1 is irreducible over the field of rational numbers. Thus we have shown that x3 þ x2 2x 1 is an irreducible polynomial over the field of rational numbers. So by 1.5.12, 2 cos h is algebraic of degree 3 over Q. Since 2 cos h is constructible, by 2.4.11, 2 cos h is algebraic of a degree of the form 2k . This contradicts the fact that 2 cos h is algebraic of degree 3 over Q. ■ Exercises 1. Let F be a field. Let f ðxÞ; gðxÞ; kðxÞ 2 F½x. Let a 2 F. Suppose that hðxÞ ¼ f ðxÞ þ k ðaÞgðxÞð2 F½xÞ. Then h0 ðxÞ ¼ f 0 ðxÞ þ kðaÞg0 ðxÞ: 2. Show that every finite extension of a field of characteristic 0 is a simple extension. 3. Let F and K be any fields such that K is an extension of F. Let F be of 2 characteristic p. Suppose that a is a member of K such that aðp 1Þ ¼ 1. Show

2 that p2 aðp 1Þ 6¼ 1. 4. Suppose that K is a finite extension of F. Show that the order of the group of automorphisms of K relative to F cannot be greater than ½K : F . 5. Suppose that K is a finite extension of F. Suppose that T is a subfield of K that contains F. Suppose that T is a normal extension of F. Show that GðK; T Þ is a normal subgroup of GðK; FÞ.

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6. Show that for all positive integers m, n, the group ðSm ÞðnÞ is a normal subgroup of the symmetric group Sm . pffiffiffi pffiffiffi 7. Show that Q 3 2 is the fixed field of G Q 3 2 ; Q . 8. Let F be a field such that F C. Suppose that F contains all the nth roots of unity. Show that the Galois group of x5 5 over F is abelian. 9. Show that it is impossible, by straightedge and a compass alone, to construct the angle 10 . 10. Suppose that a; b are nonzero constructible numbers. Show that a2 b 2 a 2 þ b2 is a constructible number.

Chapter 3

Linear Transformations

The subject matter in this chapter is also known as linear algebra. As we shall see, the theory of matrices is intimately related to linear algebra. Its applications to other branches of knowledge is overwhelming. That is why it is considered an independent subfield of mathematics, exciting on its own.

3.1

Eigenvalues

3.1.1 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Suppose that for every v 2 V, hTðvÞ; vi ¼ 0. Then T ¼ 0. Proof We have to show that T ¼ 0. To this end, let us fix an arbitrary w 2 V. We have to show that TðwÞ ¼ 0, that is, hTðwÞ; TðwÞi ¼ 0. Let us take arbitrary u; v 2 V. By the given condition, we have hTðuÞ; vi þ hTðvÞ; ui ¼ 0 þ hTðuÞ; vi þ hTðvÞ; ui þ 0 ¼ hTðuÞ; ui þ hTðuÞ; vi þ hTðvÞ; ui þ hTðvÞ; vi ¼ hTðuÞ þ TðvÞ; u þ vi ¼ hT ðu þ vÞ; u þ vi ¼ 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus for every u; v 2 V, hTðuÞ; vi þ hTðvÞ; ui ¼ 0: It follows that for every u; v 2 V,


167

168

3 Linear Transformations

2ihTðvÞ; ui ¼ ihTðvÞ; ui þ ihTðvÞ; ui ¼ ihTðuÞ; vi þ ihTðvÞ; ui ¼ ihTðuÞ; vi þ ihTðvÞ; ui ¼ hTðuÞ; ivi þ ihTðvÞ; ui ¼ hTðuÞ; ivi þ hiTðvÞ; ui ¼ hTðuÞ; ivi þ hT ðivÞ; ui ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence for every u; v 2 V, hTðvÞ; ui ¼ 0: It follows that hTðwÞ; TðwÞi ¼ 0: ■ Definition Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. If for every u; v 2 V, hTðuÞ; TðvÞi ¼ hu; vi, then we say that T is unitary. 3.1.2 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a unitary linear transformation. Suppose that for every v 2 V, hTðvÞ; TðvÞi ¼ hv; vi. Then T is unitary. Proof Let us take any u; v 2 V. By the given condition, we have hu; ui þ hTðuÞ; TðvÞi þ hTðvÞ; TðuÞi þ hv; vi ¼ hTðuÞ; TðuÞi þ hTðuÞ; TðvÞi þ hTðvÞ; TðuÞi þ hTðvÞ; TðvÞi ¼ hTðuÞ þ TðvÞ; TðuÞ þ TðvÞi ¼ hT ðu þ vÞ; T ðu þ vÞi ¼ hu þ v; u þ vi |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ hu; ui þ hu; vi þ hv; ui þ hv; vi: Thus for every u; v 2 V, hTðuÞ; TðvÞi þ hTðvÞ; TðuÞi ¼ hu; vi þ hv; ui: It follows that for every u; v 2 V,

3.1 Eigenvalues

169

2ihTðvÞ; TðuÞi ihu; vi ihv; ui ¼ iðhTðvÞ; TðuÞi hu; vi hv; uiÞ þ ihTðvÞ; TðuÞi ¼ ihTðuÞ; TðvÞi þ ihTðvÞ; TðuÞi ¼ ihTðuÞ; TðvÞi þ hiTðvÞ; TðuÞi ¼ hTðuÞ; iTðvÞi þ hiTðvÞ; TðuÞi ¼ hTðuÞ; T ðivÞi þ hT ðivÞ; TðuÞi ¼ hu; ivi þ hiv; ui |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ihu; vi þ hiv; ui ¼ ihu; vi þ ihv; ui; and hence for every u; v 2 V, hTðvÞ; TðuÞi ¼ hv; ui: ■

Thus T is unitary.

3.1.3 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a unitary linear transformation. Let fv1 ; . . .; vn g be any orthonormal basis of V. Then fT ðv1 Þ; . . .; T ðvn Þg is an orthonormal basis of V. Proof Let a1 T ðv1 Þ þ þ an T ðvn Þ ¼ 0: It follows that a1 ¼ a1 1 þ a2 0 þ þ an 0 ¼ a1 h v 1 ; v 1 i þ a2 h v 2 ; v 1 i þ þ an h v n ; v 1 i ¼ a1 hT ðv1 Þ; T ðv1 Þi þ a2 hT ðv2 Þ; T ðv1 Þi þ þ an hT ðvn Þ; T ðv1 Þi ¼ ha1 T ðv1 Þ þ þ an T ðvn Þ; T ðv1 Þi ¼ h0; T ðv1 Þi ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence a1 ¼ 0. Similarly, a2 ¼ 0; . . .; an ¼ 0. Thus we have shown that T ðv1 Þ; . . .; T ðvn Þ are linearly independent. Since T ðv1 Þ; . . .; T ðvn Þ are linearly independent, T ðv1 Þ; . . .; T ðvn Þ are distinct members of V. It follows that f T ðv1 Þ; . ..; Tðvn Þg is a basis of V. Next, for distinct indices i; j 2 f1; . . .; ng, T ðvi Þ; T vj ¼ vi ; vj ¼ 0. Also, for every index i 2 f1; . . .; ng, hT ðvi Þ; T ðvi Þi ¼ ■ hvi ; vi i ¼ 1. Thus fT ðv1 Þ; . . .; T ðvn Þg is an orthonormal basis of V. 3.1.4 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Suppose that T sends every orthonormal basis of V to an orthonormal basis of V. Then T is unitary.

170


Proof Since V is an n-dimensional inner product space, there exists an orthonormal basis fv1 ; . . .; vn g ofPV. By assumption,PfT ðv1 Þ; . . .; T ðvn Þg is also an orthonormal basis of V. Let u ni¼1 ai vi and w nj¼1 bj vj be any members of V. We have to show that hTðuÞ; TðwÞi ¼ hu; wi: Here, * LHS ¼ hTðuÞ; TðwÞi ¼

T *

¼ ¼

X

n X

! ai v i ; T

i¼1 n X i¼1

n X

!+ bj vj

j¼1

!

n X

ai T ð v i Þ ;

bj T vj

!+

j¼1

ai b| T ðvi Þ; T vj

i;j

¼

X

ai b| dij ¼

i;j

n X

ai bi ;

i¼1

and * RHS ¼ hu; wi ¼

n X i¼1

ai v i ;

n X

+ b j vj

¼

X

n X X vi ; vj ¼ dij ¼ ai b ai b ai b | j i

i;j

j¼1

i;j

i¼1

■

so LHS = RHS.

3.1.5 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Let v 2 V. Then there exists a unique w 2 V such that u 2 V ) hu; wi ¼ hTðuÞ; vi: We denote w by T ðvÞ. Thus T : V ! V, and for every u; v 2 V, hu; T ðvÞi ¼ hTðuÞ; vi. Also, T : V ! V is linear. Proof Existence: Since V is an n-dimensional inner product space, there exists an orthonormal basis fu1 ; . . .; un g of V. Put w hT ðu1 Þ; viu1 þ þ hT ðun Þ; viun : Let us fix an arbitrary u

Pn i¼1

ai ui . We have to show that

3.1 Eigenvalues

*

n X i¼1

171

n X ai ui ; T u| ; v uj

+

* ¼

j¼1

n X

T *

LHS ¼

! + ai ui ; v ;

i¼1 n X

n X ai ui ; T u| ; v uj

i¼1

+

j¼1

X X ij ¼ ai T uj ; v ui ; uj ¼ ai T uj ; v d i;j

¼

n X

* ai hT ðui Þ; vi ¼

i¼1

* ¼

T

n X

+

ai T ðui Þ; v

i¼1

! + ai ui ; v

i;j n X

¼ RHS:

i¼1

Uniqueness: Suppose that there exist w1 ; w2 2 V such that u 2 V ) hu; w1 i ¼ hTðuÞ; vi; and hu; w2 i ¼ hTðuÞ; vi: We have to show that w1 ¼ w2 , that is, hw1 w2 ; w1 w2 i ¼ 0. Here u 2 V ) hu; w1 i ¼ hu; w2 i; so for every u 2 V, hu; w1 w2 i ¼ 0. It follows that hw1 w2 ; w1 w2 i ¼ 0. Linearity: Let us take arbitrary v1 ; v2 2 V: Let a; b be arbitrary complex numbers. We have to show that T ðav1 þ bv2 Þ ¼ aT ðv1 Þ þ bT ðv2 Þ: It suffices to show that for every u 2 V, hu; T ðav1 þ bv2 Þi ¼ hu; aT ðv1 Þ þ bT ðv2 Þi: To this end, let us fix an arbitrary u 2 V. We have to show that hu; T ðav1 þ bv2 Þi ¼ hu; aT ðv1 Þ þ bT ðv2 Þi; hTðuÞ; v2 i LHS ¼ hu; T ðav1 þ bv2 Þi ¼ hTðuÞ; av1 þ bv2 i ¼ ahTðuÞ; v1 i þ b ¼ ahu; T ðv1 Þi þ bhu; T ðv2 Þi ¼ hu; aT ðv1 Þ þ bT ðv2 Þi ¼ RHS:

■ Definition Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. By 3.1.5, T : V ! V is a linear transformation such that for every u; v 2 V, hu; T ðvÞi ¼ hTðuÞ; vi. Here T is called the Hermitian adjoint of T.

172


3.1.6 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Then ðT Þ ¼ T. Proof Let us take an arbitrary v 2 V. We have to show that ðT Þ ðvÞ ¼ TðvÞ: To this end, let us take an arbitrary u 2 V. It suffices to show that hu; ðT Þ ðvÞi ¼ hu; TðvÞi; LHS ¼ hu; ðT Þ ðvÞi ¼ hT ðuÞ; vi ¼

hv; T ðuÞi

:

¼ hTðvÞ; ui ¼ hu; TðvÞi ¼ RHS: ■

3.1.7 Problem Let V be an n-dimensional inner product space. Let S : V ! V and T : V ! V be linear transformations. Let k; l be any complex numbers. Then T . ðkS þ lT Þ ¼ kS þ l Proof Let us take an arbitrary v 2 V. We have to show that T ðvÞ; ðkS þ lT Þ ðvÞ ¼ kS þ l that is, T ðvÞ: ðkS þ lT Þ ðvÞ ¼ kS ðvÞ þ l To this end, let us take an arbitrary u 2 V. It suffices to show that T ðvÞ : hu; ðkS þ lT Þ ðvÞi ¼ u; kS ðvÞ þ l LHS ¼ hu; ðkS þ lT Þ ðvÞi ¼ hðkS þ lT ÞðuÞ; vi ¼ hkSðuÞ þ lTðuÞ; vi ¼ khSðuÞ; vi þ lhTðuÞ; vi ¼ khu; S ðvÞi þ lhu; T ðvÞi T ðvÞ ¼ RHS: ¼ u; kS ðvÞ þ l ■ 3.1.8 Problem Let V be an n-dimensional inner product space. Let S : V ! V and T : V ! V be linear transformations. Then ðST Þ ¼ T S : Proof Let us take an arbitrary v 2 V. We have to show that ðST Þ ðvÞ ¼ ðT S ÞðvÞ;

3.1 Eigenvalues

173

that is, ðST Þ ðvÞ ¼ T ðS ðvÞÞ: To this end, let us take an arbitrary u 2 V. It suffices to show that hu; ðST Þ ðvÞi ¼ hu; T ðS ðvÞÞi: LHS ¼ hu; ðST Þ ðvÞi ¼ hðST ÞðuÞ; vi ¼ hSðTðuÞÞ; vi ¼ hTðuÞ; S ðvÞi ¼ hu; T ðS ðvÞÞi ¼ RHS: ■ 3.1.9 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a unitary linear transformation. Then T T ¼ I. Proof Let us take an arbitrary v 2 V. We have to show that T ðTðvÞÞ ¼ v. To this end, let us take an arbitrary u 2 V. It suffices to show that hu; T ðTðvÞÞi ¼ hu; vi: RHS ¼ hu; vi ¼ hTðuÞ; TðvÞi ¼ hu; T ðTðvÞÞi ¼ LHS: ■ 3.1.10 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation such that T T ¼ I. Then T is unitary. Proof Let us take arbitrary u; v 2 V. We have to show that hTðuÞ; TðvÞi ¼ hu; vi: LHS ¼ hTðuÞ; TðvÞi ¼ hu; T ðTðvÞÞi ¼ hu; ðT T ÞðvÞi ¼ hu; IðvÞi ¼ hu; vi ¼ RHS: ■ 3.1.11 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Let fv1 ; . . .; vn g be an orthonormal basis of V. Let aij be the matrix of T relative to the basis fv1 ; . . .; vn g, in the sense that T ðv1 Þ ¼ a11 v1 þ a21 v2 þ þ an1 vn ¼

n P

ai1 vi ;

i¼1

T ðv2 Þ ¼ a12 v1 þ a22 v2 þ þ an2 vn ; .. . T ðvn Þ ¼ a1n v1 þ a2n v2 þ þ ann vn :

P In short, T vj ¼ ni¼1 aij vi . Then the matrix of T relative to the basis fv1 ; . . .; vn g is bij , where bij ¼ a|i . P In short, T vj ¼ ni¼1 bij vi .

174


Proof By the proof of 3.1.5, T ðv1 Þ ¼ hT ðv1 Þ; v1 i v1 þ þ hT ðvn Þ; v1 i vn ; T ðv2 Þ ¼ hT ðv1 Þ; v2 i v1 þ þ hT ðvn Þ; v2 i vn ; .. . T ðvn Þ ¼ hT ðv1 Þ; vn i v1 þ þ hT ðvn Þ; vn i vn :

Since T ð v1 Þ ¼ ¼ ¼ ¼ ¼

n X i¼1 n X i¼1 n X i¼1 n X i¼1 n X

hT ðvi Þ; v1 ivi ha1i v1 þ a2i v2 þ þ ani vn ; v1 ivi a1i hv1 ; v1 i þ a2i hv2 ; v1 i þ þ ani hvn ; v1 ivi a1i 1 þ ai2 0 þ þ ain 0vi a1i vi ¼ a11 v1 þ a12 v2 þ þ a1n vn ;

i¼1

we have

T ðv1 Þ ¼ a11 v1 þ a12 v2 þ þ a1n vn ¼

n X

! a1i vi ;

i¼1

Similarly, T ðv2 Þ ¼ a21 v1 þ a22 v2 þ þ a2n vn ; P etc. In short, T vj ¼ ni¼1 a|i vi . If the matrix of T relative to the basis ■ fv1 ; . . .; vn g is bij , then bij ¼ a|i . 3.1.12 Problem Let V be an n-dimensional inner product space. Let T : V ! V be linear transformation. Let fv1 ; . . .; vn g be an orthonormal basis of V. Let a unitary P aij be the matrix of T relative to the basis fv1 ; . . .; vn g. Then nj¼1 aji a|k ¼ dik . Proof It is given that T ðvi Þ ¼

n X j¼1

aji vj :

3.1 Eigenvalues

175

By 3.1.11, n X T vj ¼ a|k vk : k¼1

Since T : V ! V is unitary, by 3.1.9, T T ¼ I. It follows that n X

dki vk ¼ vi ¼ I ðvi Þ ¼ ðT T Þðvi Þ |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} k¼1 ! n n X X ¼ T ðT ðvi ÞÞ ¼ T aji vj ¼ aji T vj j¼1

¼

n X j¼1

¼

aji

n X

a|k vk

k¼1

n n X X k¼1

!

n n X X j¼1

!

aji a|k vk

j¼1

¼

j¼1

¼

k¼1

n n X X j¼1

!

aji a|k vk ! aji a|k vk ;

j¼1

and hence n X

aji a|k ¼ dki :

j¼1

■ Definition Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. If T ¼ T, then we say that T is Hermitian. 3.1.13 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a Hermitian linear transformation. Then all its eigenvalues are real. Proof Let k be an eigenvalue of T. We have to show that k is real, that is, k ¼ k. Since k is an eigenvalue of T, there exists a nonzero v 2 V such that TðvÞ ¼ kv. Since T : V ! V is Hermitian, we have T ¼ T. Now, khv; vi; khv; vi ¼ hkv; vi ¼ hTðvÞ; vi ¼ hv; T ðvÞi ¼ hv; TðvÞi ¼ hv; kvi ¼ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so

k k hv; vi ¼ 0;

and hence, k ¼ k or hv; vi ¼ 0. Since v is nonzero, hv; vi 6¼ 0, and hence k ¼ k. ■

176


3.1.14 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Then ðT T ÞðvÞ ¼ 0 ) TðvÞ ¼ 0: Proof Let v 2 V be such that ðT T ÞðvÞ ¼ 0. We have to show that TðvÞ ¼ 0, that is, hTðvÞ; TðvÞi ¼ 0. Since ðT T ÞðvÞ ¼ 0, we have hTðvÞ; TðvÞi ¼ hv; T ðTðvÞÞi ¼ hv; ðT T ÞðvÞi ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence hTðvÞ; TðvÞi ¼ 0.

■

3.1.15 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a Hermitian linear transformation. Let k be a positive integer. Then T k ðvÞ ¼ 0 ) TðvÞ ¼ 0: Proof For k ¼ 1, the theorem is trivial. So we consider the case k ¼ 2. Since T : V ! V is Hermitian, we have T ¼ T. Now, 0 ¼ T k ðvÞ ¼ T 2 ðvÞ ¼ ðTT ÞðvÞ ¼ ðT T ÞðvÞ; |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} so ðT T ÞðvÞ ¼ 0. It follows from 3.1.14 that TðvÞ ¼ 0. Next, we consider the case k ¼ 3. Here 0 ¼ T k ðvÞ ¼ T 3 ðvÞ ¼ ðTTT ÞðvÞ ¼ ðT TT ÞðvÞ ¼ ðT T ÞðTðvÞÞ; |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} so ðT T ÞðTðvÞÞ ¼ 0. It follows from 3.1.14 that T ðTðvÞÞ ¼ 0, that is, T 2 ðvÞ ¼ 0. Since the theorem has been proved for k ¼ 2, we have TðvÞ ¼ 0, etc. ■ Definition Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. If T T ¼ TT , then we say that T is normal. 3.1.16 Theorem Let V be an n-dimensional inner product space. Let T : V ! V be a unitary linear transformation. Then T is normal. Proof We have to show that T T ¼ TT . Since T : V ! V is unitary, by 3.1.9, T T ¼ I. It follows that T 1 ¼ T : LHS ¼ T T ¼ T 1 T ¼ I ¼ TT 1 ¼ TT ¼ RHS: ■ 3.1.17 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a Hermitian linear transformation. Then T is normal.

3.1 Eigenvalues

177

Proof We have to show that T T ¼ TT . Since T : V ! V is Hermitian, T ¼ T: LHS ¼ T T ¼ TT ¼ TT ¼ RHS: ■ Definition Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. If T ¼ T, then we say that T is skew-Hermitian. 3.1.18 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a skew-Hermitian linear transformation. Then T is normal. Proof We have to show that T T ¼ TT . Since T : V ! V is skew-Hermitian, T ¼ T: LHS ¼ T T ¼ ðT ÞT ¼ T ðT Þ ¼ TT ¼ RHS: ■ 3.1.19 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Then TðvÞ ¼ 0 ) T ðvÞ ¼ 0: Proof Let v 2 V be such that TðvÞ ¼ 0. We have to show that T ðvÞ ¼ 0, that is, hT ðvÞ; T ðvÞi ¼ 0, that is, hT ðT ðvÞÞ; vi ¼ 0, that is, hðTT ÞðvÞ; vi ¼ 0. Since T : V ! V is normal, we have T T ¼ TT . It suffices to show that hðT T ÞðvÞ; vi ¼ 0: LHS ¼ hðT T ÞðvÞ; vi ¼ hT ðTðvÞÞ; vi ¼ hT ð0Þ; vi ¼ h0; vi ¼ 0 ¼ RHS: ■ 3.1.20 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let k be any complex number. Then ðT kI Þ : V ! V is a normal linear transformation. Proof By 3.1.7, ðT kI Þ ¼ T kI . Since for every u; v 2 V, hu; I ðvÞi ¼ hIðuÞ; vi ¼ hu; vi ¼ hu; IðvÞi, we have, for every u; v 2 V, hu; I ðvÞi ¼ hu; IðvÞi. It follows that I ¼ I. Thus ðT kI Þ ¼ T kI. It suffices to show that

kI ; T kI ðT kI Þ ¼ ðT kI Þ T

that is, T T kT kT þ jkj2 I ¼ TT kT kT þ jkj2 I;

178


that is, T T ¼ TT : This is known to be true, because T : V ! V is normal.

■

3.1.21 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let k be an eigenvalue of T. Let v be an eigenvector belonging to k, in the sense that v is nonzero, and TðvÞ ¼ kv. Then T ðvÞ ¼ kv. Proof It suffices to show that T kI ðvÞ ¼ 0, that is, T kI ðvÞ ¼ 0, that is, ðT kI Þ ðvÞ ¼ 0. Since TðvÞ ¼ kv, we have ðT kI ÞðvÞ ¼ 0. By 3.1.20, ðT kI Þ : V ! V is a normal linear transformation. Now, since ðT kI ÞðvÞ ¼ 0, by 3.1.19, ðT kI Þ ðvÞ ¼ 0. ■ 3.1.22 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a unitary linear transformation. Let k be an eigenvalue of T. Then jkj ¼ 1. Proof Since T is unitary, by 3.1.16, T is normal. Since k is an eigenvalue of T, there exists a nonzero v 2 V such that TðvÞ ¼ kv. Now, by 3.1.21, T ðvÞ ¼ kv. Since T is unitary, by 3.1.9, T T ¼ I, and hence jkj2 v ¼ kk v ¼ k kv ¼ kðT ðvÞÞ ¼ T ðkvÞ ¼ T ðTðvÞÞ ¼ ðT T ÞðvÞ ¼ IðvÞ ¼ v: |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} It follows that jkj2 1 v ¼ 0: Now since v is nonzero, we have jkj2 1 ¼ 0, that is, jkj ¼ 1.

■

3.1.23 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let k be a positive integer. Then T k ðvÞ ¼ 0 ) TðvÞ ¼ 0: Proof For k ¼ 1, the theorem is trivial. So we consider the case k ¼ 2. Since T : V ! V is normal, we have TT ¼ T T. Since ðT T Þ ¼ T ðT Þ ¼ T T; we have ðT T Þ ¼ T T, and hence T T is a Hermitian linear transformation. Now suppose that T k ðvÞ ¼ 0, that is, T 2 ðvÞ ¼ 0. We have to show that TðvÞ ¼ 0. Since

3.1 Eigenvalues

179

ðT T Þ2 ðvÞ ¼ ððT T ÞðT T ÞÞðvÞ ¼ ðT ðTT ÞT ÞðvÞ ¼ ðT ðT T ÞT ÞðvÞ ¼ ðT T ÞððTT ÞðvÞÞ ¼ ðT T Þ T 2 ðvÞ ¼ ðT T Þð0Þ ¼ 0; we have ðT T Þ2 ðvÞ ¼ 0. Now, since T T is Hermitian, by 3.1.15, ðT T ÞðvÞ ¼ 0, and hence by 3.1.14, TðvÞ ¼ 0. Next, we consider the case k ¼ 3. Suppose that T k ðvÞ ¼ 0, that is, T 3 ðvÞ ¼ 0. We have to show that TðvÞ ¼ 0. Since T is normal, we have ðT T Þ3 ðvÞ ¼ ððT T ÞðT T ÞðT T ÞÞðvÞ ¼ ðT Þ3 T 3 ðvÞ ¼ ðT Þ3 T 3 ðvÞ ¼ ðT Þ3 ð0Þ ¼ 0; and hence ðT T Þ3 ðvÞ ¼ 0. Now, since T T is Hermitian, by 3.1.15, ðT T ÞðvÞ ¼ 0, and hence by 3.1.14, TðvÞ ¼ 0, etc. ■ 3.1.24 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let k be any complex number. Let k be a positive integer. Then ðT kI Þk ðvÞ ¼ 0 ) TðvÞ ¼ kv: Proof By 3.1.20, ðT kI Þ : V ! V is a normal linear transformation. Now, by 3.1.23, ðT kI Þk ðvÞ ¼ 0 ) ðT kI ÞðvÞ ¼ 0; and hence ðT kI Þk ðvÞ ¼ 0 ) TðvÞ ¼ kv: ■ 3.1.25 Problem Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let k; l be two distinct eigenvalues of T. Let v; w 2 V be such that TðvÞ ¼ kv and TðwÞ ¼ lw. Then hv; wi ¼ 0. Proof If v ¼ 0 or w ¼ 0, then the theorem is trivial. So we consider the case that v and w both are nonzero. Since l is an eigenvalue of T and TðwÞ ¼ lw, w is an w. Hence eigenvector belonging to l. It follows, from 3.1.21, that T ðwÞ ¼ l wi ¼ hv; T ðwÞi ¼ hTðvÞ; wi ¼ hkv; wi ¼ khv; wi: lhv; wi ¼ hv; l |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

180


Thus ðl kÞhv; wi ¼ 0: Now, since l 6¼ k, we have hv; wi ¼ 0.

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Definition Let A be a ring. Let A be a vector space over the field C of complex numbers. If for every a; b 2 A, and, for every complex number a, aðabÞ ¼ ðaaÞb ¼ aðabÞ; then we say that A is an algebra. Let V be any vector space over the field C of complex numbers. Let AðVÞ be the collection of all linear transformations from V to V. We know that AðVÞ is an algebra with unit element I. If dim V ¼ n, then dim AðVÞ ¼ n2 . Definition Let A be an algebra with unit element e. Let pðxÞ a0 þ a1 x þ þ an xn be any polynomial in x with complex coefficients ai . Let a 2 A. By a satisfies pðxÞ, we mean a0 e þ a1 a þ þ an an ¼ 0: ðIn short; pðaÞ ¼ 0:Þ 3.1.26 Problem Let A be an algebra with unit element e. Let m be the dimension of A. Let a 2 A. Then there exists a nontrivial polynomial pðxÞ such that a satisfies pðxÞ. Also, the degree of pðxÞ is not greater than m. Proof If any two of e; a; a2 ; . . .; am are equal, say a2 ¼ a5 , then the polynomial 1x2 þ ð1Þx5 serves the purpose of pðxÞ. Finally, we consider the case that e; a; a2 ; . . .; am are ðm þ 1Þ distinct members of A. Since m is the dimension of A; e; a; a2 ; . . .; am are linearly dependent, and hence there exist complex numbers a0 ; a1 ; . . .; am , not all zero, such that a0 e þ a1 a þ þ am am ¼ 0: It follows that the polynomial a0 þ a1 x þ þ am xm serves the purpose of pðxÞ. Here, the degree of pðxÞ is not greater than m. ■ 3.1.27 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Then there exists a nontrivial polynomial pðxÞ of degree n2 such that pðTÞ ¼ 0. Proof We know that dim AðVÞ ¼ n2 , so by 3.1.26, there exists a nontrivial poly■ nomial pðxÞ of degree n2 such that pðTÞ ¼ 0.

3.1 Eigenvalues

181

Definition Let V be any n-dimensional vector space. Let T 2 AðVÞ. A nontrivial polynomial pðxÞ of lowest degree such that pðTÞ ¼ 0 is called a minimal polynomial of T. If pðxÞ is a minimal polynomial of T, and T satisfies another polynomial hðxÞ, then pðxÞ divides hðxÞ. 3.1.28 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let T be invertible. Suppose that a0 þ a1 x þ þ am xm is a minimal polynomial of T, where am 6¼ 0. Then a0 6¼ 0. Proof Suppose to the contrary that a0 ¼ 0. We seek a contradiction. Since a1 x þ þ am xm is a minimal polynomial of T, we have a1 T þ þ am T m ¼ 0; and hence a1 I þ a2 T þ am T m1 ¼ T ða1 T þ þ am T m Þ ¼ 0T ¼ 0: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus a1 I þ a2 T þ am T m1 ¼ 0: Hence hðTÞ ¼ 0, where hðxÞ a1 þ a2 x þ am xm1 . Now, since am 6¼ 0, deg hðxÞ ¼ m 1 \m ¼ degða1 x þ þ am xm Þ: |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} Since deg hðxÞ\degða1 x þ þ am xm Þ and a1 x þ þ am xm is a minimal polynomial of T, we have hðTÞ 6¼ 0. This is a contradiction. ■ 3.1.29 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Suppose that a0 þ a1 x þ þ am xm is a minimal polynomial of T, where am 6¼ 0 and a0 6¼ 0. Then T 1 exists. Proof Since a0 þ a1 x þ þ am xm is a minimal polynomial of T, we have a0 I þ a1 T þ þ am T m ¼ 0: It follows that I¼

a1 a2 2 am m Tþ T þ þ T ; a0 a0 a0

182


or T This shows that

a1 a2 am m1 þ T þ þ T ¼ I: a0 a0 a0

a1 a0

þ

a2 a0

T þ þ

am a0

T m1 is the inverse of T.

■

3.1.30 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Suppose that T 1 does not exist. Then there exists a nonzero S 2 AðVÞ such that ST ¼ TS ¼ 0. Proof Suppose that a0 þ a1 x þ þ am xm is a minimal polynomial of T, where am 6¼ 0. By 3.1.29, a0 ¼ 0. Hence a1 T þ þ am T m ¼ 0; or

a1 I þ a2 T þ þ am T m1 T ¼ T a1 I þ a2 T þ þ am T m1 ¼ 0:

Thus ST ¼ TS ¼ 0, where S a1 I þ a2 T þ þ am T m1 ð2 AðVÞÞ. Since a0 þ a1 x þ þ am xm is a minimal polynomial of T and am 6¼ 0, we have ■ a1 I þ a2 T þ þ am T m1 6¼ 0, and hence S 6¼ 0. 3.1.31 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Suppose that T 1 does not exist. Then there exists a nonzero v 2 V such that TðvÞ ¼ 0. Proof By 3.1.30, there exists a nonzero S 2 AðVÞ such that TS ¼ 0. Since S is nonzero, there exists u 2 V such that SðuÞ 6¼ 0. Now, since TS ¼ 0, we have T ðSðuÞÞ ¼ ðTSÞðuÞ ¼ 0ðuÞ ¼ 0; |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence TðvÞ ¼ 0, where v SðuÞ ð6¼ 0Þ.

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3.1.32 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Suppose that there exists a nonzero v 2 V such that TðvÞ ¼ 0. Then T 1 does not exist. Proof Suppose to the contrary that T 1 exists. We seek a contradiction. Suppose that a0 þ a1 x þ þ am xm is a minimal polynomial of T, where am 6¼ 0. Now, by 3.1.28, a0 6¼ 0. Since a0 þ a1 x þ þ am xm is a minimal polynomial of T and am 6¼ 0, we have a0 I þ a1 T þ þ am T m ¼ 0:

3.1 Eigenvalues

183

It follows that a0 v ¼ a0 v þ a1 0 þ þ am 0 ¼ a0 IðvÞ þ a1 TðvÞ þ þ am T m ðvÞ ¼ ða0 I þ a1 T þ þ am T m ÞðvÞ ¼ 0ðvÞ ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence a0 v ¼ 0. Now, since v is nonzero, a0 ¼ 0. This is a contradiction.

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3.1.33 Problem Let V be any n-dimensional vector space. Let S; T 2 AðVÞ. Let v1 ; . . .; vn be any basis of V. Let mðSÞ S relative to the basis be the matrix of P v1 ; . . .; vn , in the sense that mðSÞ ¼ aij nn , where S vj ni¼1 aij vi . Let mðTÞ be the matrix of T relative to the basis v1 ; . . .; vn . Then mðST Þ ¼ mðSÞmðTÞ: P Proof Let mðTÞ ¼ bij nn , where T vj ni¼1 bij vi . It suffices to show that ! n n X X ðST Þ vj ¼ aki bij vk : i¼1

k¼1

n X bij vi LHS ¼ ðST Þ vj ¼ S T vj ¼ S i¼1

¼

n X

bij Sðvi Þ ¼

i¼1

¼ ¼

k¼1

n n X X k¼1

bij

i¼1

n n X X i¼1

n X

! :

aki vk

k¼1

!

bij aki vk

n X

!

¼

!

n n X X k¼1

! bij aki vk

i¼1

¼

n n X X k¼1

! bij aki vk

i¼1

aki bij vk ¼ RHS:

i¼1

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3.1.34 Problem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let v1 ; . . .; vn be any basis of V. Suppose that T 1 exists. Let mðTÞ be the matrix of T relative to the basis v1 ; . . .; vn . Then m T 1 ¼ ðmðTÞÞ1 : Proof Here, it suffices to show that mðT 1 ÞmðTÞ ¼ dij nn . Since T 1 T ¼ I, by Problem 3.1.33, we have dij nn ¼ mðIÞ ¼ m T 1 T ¼ m T 1 mðTÞ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence mðT 1 ÞmðTÞ ¼ dij nn :

■

184


3.1.35 Theorem a. Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let v1 ; . . .; vn and w1 ; . . .; wn be any two bases of V. Let m1 ðTÞ be the matrix of T relative to the basis v1 ; . . .; vn . Let m2 ðTÞ be the matrix of T relative to the basis w1 ; . . .; wn . Let S : V ! V be the linear transformation such that for every i 2 f1; . . .; ng, Sðvi Þ ¼ wi . Then m2 ðTÞ ¼ ðm1 ðSÞÞ1 m1 ðTÞm1 ðSÞ: b. Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let v1 ; . . .; vn be any basis of V. Let A aij nn be the matrix of T relative to the basis v1 ; . . .; vn . Let P pij nn be any invertible matrix. Then there exists a basis w1 ; . . .; wn of V such that P1 AP is the matrix of T relative to the basis w1 ; . . .; wn . P Proof (a) Let m1 ðTÞ ¼ aij nn , where T vj ni¼1 aij vi . Let m2 ðTÞ ¼ bij nn , Pn where T wj i¼1 bij wi . Clearly, S is invertible, that is, S1 exists. Proof Suppose that SðvÞ ¼ 0. It suffices to show that v ¼ 0. There exist scalars a1 ; . . .; an such that v ¼ a1 v1 þ þ an vn . It follows that 0 ¼ SðvÞ ¼ Sða1 v1 þ þ an vn Þ ¼ a1 Sðv1 Þ þ þ an Sðvn Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a1 w 1 þ þ an w n ; so a1 w1 þ þ an wn ¼ 0. Now, since w1 ; . . .; wn is a basis of V, each ai is 0, and hence v ¼ a1 v1 þ þ an vn ¼ 0. Thus v ¼ 0. ■ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Since for every j 2 f1; . . .; ng, n X T wj ¼ bij wi ; i¼1

we have, for every j 2 f1; . . .; ng, ! n n n X X X ðTSÞ vj ¼ T S vj ¼ T wj ¼ bij wi ¼ bij Sðvi Þ ¼ S bij vi ; i¼1 |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl}

i¼1

i¼1

3.1 Eigenvalues

185

and hence for every j 2 f1; . . .; ng, ! n X ðTSÞ vj ¼ S bij vi : i¼1

It follows that for every j 2 f1; . . .; ng,

n X bij vi : S1 TS vj ¼ S1 ðTSÞ vj ¼ i¼1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl}

Thus for every j 2 f1; . . .; ng, n 1 X S TS vj ¼ bij vi : i¼1

It follows that m1 S1 TS ¼ bij ¼ m2 ðTÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} where m1 ðS1 TSÞ is the matrix of S1 TS relative to the basis v1 ; . . .; vn . Thus m2 ðTÞ ¼ m1 S1 TS : By 3.1.33, and 3.1.34, m2 ðTÞ ¼ m1 S1 TS ¼ m1 S1 m1 ðTÞm1 ðSÞ ¼ ðm1 ðSÞÞ1 m1 ðTÞm1 ðSÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence m2 ðTÞ ¼ ðm1 ðSÞÞ1 m1 ðTÞm1 ðSÞ: P Proof (b) Put wj ni¼1 pij vi ðj ¼ 1; . . .; nÞ. Clearly, fw1 ; . . .; wn g is linearly independent. Proof To show this, suppose that a1 w1 þ þ an wn ¼ 0. We have to show that each ai equals 0. Since

186


n n P P

n

n P P pki ai vk ¼ ai pki vk ¼ ai pki vk k¼1 i¼1 k¼1 i¼1 k¼1 i¼1

n n n n n X P P P P ¼ ai pki vk ¼ ai pki vk ¼ ai wi ¼ 0; i¼1

n n P P

i¼1

k¼1

k¼1

i¼1 |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl}

we have n n X X k¼1

! pki ai vk ¼ 0:

i¼1

Since v1 ; . . .; vn is a basis of V, we have, for every k 2 f1; . . .; ng, n X

pki ai ¼ 0;

i¼1

2 3 2 3 a1 0 6 . 7 . and hence pij nn 4 .. 5 ¼ 4 .. 5, that is, P½a1 ; . . .; an T ¼ ½0; . . .; 0T . It follows 0 an that ½a1 ; . . .; an T ¼ I ½a1 ; . . .; an T ¼ P1 P ½a1 ; . . .; an T ¼ P1 P½a1 ; . . .; an T ¼ P1 ½0; . . .; 0T ¼ ½0; . . .; 0T ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ½a1 ; . . .; an T ¼ ½0; . . .; 0T . This shows that each ai is 0.

■

Thus, we have shown that fw1 ; . . .; wn g is linearly independent. Now, since V is an n-dimensional vector space, w1 ; . . .; wn is a basis of V. It remains to show that P1 AP is the matrix of T relative to the basis w1 ; . . .; wn . Let S : V ! V be the linear transformation defined as follows: for every i 2 f1; . . .; ng, Sðvi Þ ¼ wi . Since Sðvi Þ ¼ wi ¼ |fflfflfflfflfflffl{zfflfflfflfflfflffl}

n X

pki vk ;

k¼1

P we have Sðvi Þ ¼ nk¼1 pki vk , and hence the matrix of S relative to the basis v1 ; . . .; vn is ½pki ð¼ PÞ. Now, by 3.1.35(a), P1 AP is the matrix of T relative to the basis w1 ; . . .; wn . ■

3.2 Canonical Forms

3.2

187

Canonical Forms

Definition Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let V1 be any subspace of V. If T ðV1 Þ V1 , then we say that V1 is invariant under T. 3.2.1 Theorem Let V be any n-dimensional vector space. Let T 2 AðVÞ. Let V1 and V2 be any subspaces of V. Suppose that V ¼ V1 V2 , in the sense that every v 2 V can be expressed uniquely as v1 þ v2 , where v1 2 V1 and v2 2 V2 . Suppose that V1 is invariant under T, and V2 is invariant under T, in the sense that the restriction TjV1 isin AðV1 Þ and the restriction TjV2 is in AðV2 Þ. Let p1 ðxÞ be a minimal polynomial of TjV1 , and p2 ðxÞ a minimal polynomial of TjV2 . Then the least common multiple of p1 ðxÞ and p2 ðxÞ is a minimal polynomial of T. Proof Let pðxÞ be a minimal polynomial of T. It suffices to show that 1. p1 ðxÞ divides pðxÞ, that is, p TjV1 ¼ 0, 2. p2 ðxÞ divides pðxÞ, that is, p TjV2 ¼ 0, 3. if p1 ðxÞ divides qðxÞ, and p2 ðxÞ divides qðxÞ, then pðxÞ divides qðxÞ, that is, ðp1 ðxÞjqðxÞ and p2 ðxÞjqðxÞÞ ) qðTÞ ¼ 0: For 1: Since p1 ðxÞ is a minimal polynomial of T1 , we have p1 ðT1 Þ ¼ 0. Clearly, for every polynomial qðxÞ, qðTÞjV1 ¼ q TjV1 . Proof Suppose that qðxÞ a0 þ a1 x þ þ am xm and v 2 V1 . We have to show that

m a2 T ðTðvÞ a0 IðvÞ þ a 1 TðvÞ þ Þ þ am T ðvÞ

m ¼ a0 IjV1 ðvÞ þ a1 TjV1 ðvÞ þ a2 TjV1 TjV1 ðvÞ þ am TjV1 ðvÞ: m RHS ¼ a0 IjV1 ðvÞ þ a1 TjV1 ðvÞ þ a2 TjV1 TjV1 ðvÞ þ am TjV1 ðvÞ m ¼ a0 v þ a1 TðvÞ þ a2 TjV1 ðTðvÞÞ þ am TjV1 ðvÞ m ¼ a0 v þ a1 TðvÞ þ a2 T ðTðvÞÞ þ am TjV1 ðvÞ m ¼ a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ am TjV1 ðvÞ .. . ¼ a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ am T m ðvÞ ¼ LHS: ■

188


Similarly, for every polynomial qðxÞ, we have qðTÞjV2 ¼ q TjV2 . Since pðxÞ is a minimal polynomial of T, we have pðTÞ ¼ 0, and hence p TjV1 ¼ pðTÞjV1 ¼ 0 : |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} Thus p TjV1 ¼ 0. For 2: The proof is similar to (1). For 3: Suppose that p1 ðxÞ divides qðxÞ and p2 ðxÞ divides qðxÞ. We have to show that qðTÞ ¼ 0. To this end, let us take an arbitrary v 2 V. We have to show that ðqðTÞÞðvÞ ¼ 0. Since v 2 V and V ¼ V1 V2 , there exist v1 2 V1 and v2 2 V2 such that v ¼ v1 þ v2 . We have to show that ðqðTÞÞðv1 Þ þ ðqðTÞÞðv2 Þ ¼ ðqðTÞÞðv1 þ v2 Þ ¼ 0 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} that is, ðqðTÞÞðv1 Þ þ ðqðTÞÞðv2 Þ ¼ 0: It suffices to show that

qðTÞjV1 ¼ 0 ; qðTÞjV2 ¼ 0 that is, 9 q TjV1 ¼ 0 = : q TjV2 ¼ 0 ; Since p1 ðxÞ is a minimal polynomial of TjV1 , we have p1 TjV1 ¼ 0. Now, since p1 ðxÞ divides qðxÞ, we have q TjV1 ¼ 0. Similarly, q TjV2 ¼ 0. ■ 3.2.2 Note Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let pðxÞ 2 F ½ x. Let pðxÞ be the minimal polynomial of T over F. By 1.3.19, we can write pðxÞ ¼ ðq1 ðxÞÞl1 ðq2 ðxÞÞl2 . . . ðqk ðxÞÞlk ;

3.2 Canonical Forms

189

where q1 ðxÞ; q2 ðxÞ; . . .; qk ðxÞ are distinct irreducible polynomials (of degree 1) in F ½ x, and l1 ; l2 ; . . .; lk are positive integers. Put n o V1 v : v 2 V and ðq1 ðTÞÞl1 ðvÞ ¼ 0 ; n o V2 v : v 2 V and ðq2 ðTÞÞl2 ðvÞ ¼ 0 ; etc: It is clear that each Vi is a linear subspace of V. It follows that V1 þ V2 þ þ Vk V:

ðÞ

Next suppose that v 2 V1 . It follows that ðq1 ðTÞÞl1 ðvÞ ¼ 0. Now ðq1 ðTÞÞl1 ðTðvÞÞ ¼ ðq1 ðTÞÞl1 T ðvÞ ¼ T ðq1 ðTÞÞl1 ðvÞ ¼ T ðq1 ðTÞÞl1 ðvÞ ¼ T ð0Þ ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so ðq1 ðTÞÞl1 ðTðvÞÞ ¼ 0; and hence TðvÞ 2 V1 . This shows that V1 is invariant under T. Similarly, V2 is invariant under T, etc. We assume that k [ 1. Put h1 ðxÞ ðq2 ðxÞÞl2 ðq3 ðxÞÞl3 ðqk ðxÞÞlk ; h2 ðxÞ ðq1 ðxÞÞl1 ðq3 ðxÞÞl3 ðqk ðxÞÞlk ; .. . hk ðxÞ ðq1 ðxÞÞl1 ðq2 ðxÞÞl2 ðqk1 ðxÞÞlk1 : Observe that the greatest common divisor of h1 ðxÞ; h2 ðxÞ; . . .; hk ðxÞ is 1. So by 1.1.5, there exist a1 ðxÞ; a2 ðxÞ; . . .; ak ðxÞ 2 F ½ x such that 1 ¼ h1 ðxÞa1 ðxÞ þ h2 ðxÞa2 ðxÞ þ þ hk ðxÞak ðxÞ; and hence

190


I ¼ ðh1 ðTÞÞ ða1 ðTÞÞ þ ðh2 ðTÞÞ ða2 ðTÞÞ þ þ ðhk ðTÞÞ ðak ðTÞÞ: It follows that for every v 2 V, we have v ¼ IðvÞ ¼ ððh1 ðTÞÞ ða1 ðTÞÞ þ ðh2 ðTÞÞ ða2 ðTÞÞ þ þ ðhk ðTÞÞ ðak ðTÞÞÞðvÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðh1 ðTÞÞðða1 ðTÞÞðvÞÞ þ ðh2 ðTÞÞðða2 ðTÞÞðvÞÞ þ þ ðhk ðTÞÞððak ðTÞÞðvÞÞ; and hence for every v 2 V, there exist w1 ; w2 ; . . .; wk 2 V such that v ¼ ðh1 ðTÞÞðw1 Þ þ ðh2 ðTÞÞðw2 Þ þ þ ðhk ðTÞÞðwk Þ: ðÞ Since q1 ðxÞ is a polynomial of degree 1 in F ½ x and l1 is a positive integer, we have degðh1 ð xÞÞ ¼ deg ðq2 ð xÞÞl2 ðq3 ð xÞÞl3 ðqk ð xÞÞlk ¼ l2 degðq2 ð xÞÞ þ þ lk degðqk ð xÞÞ\l1 degðq1 ð xÞÞ þ þ lk degðqk ð xÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ deg ðq1 ð xÞÞl1 ðq2 ð xÞÞl2 ðqk ð xÞÞlk ¼ degðpð xÞÞ; and hence degðh1 ðxÞÞ\degðpðxÞÞ. Now since pðxÞ is the minimal polynomial of T over F, we have pðTÞ ¼ 0 and h1 ðTÞ 6¼ 0. Since h1 ðTÞ 6¼ 0 and h1 ðTÞ : V ! V, there exists v1 2 V such that ðh1 ðTÞÞðv1 Þ 6¼ 0. Since pðxÞ ¼ ðq1 ðxÞÞl1 h1 ðxÞ, we have 0 ¼ pðTÞ ¼ ðq1 ðTÞÞl1 h1 ðTÞ , and hence ðq1 ðTÞÞl1 h1 ðTÞ ¼ 0. It follows that |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðq1 ðTÞÞl1 ððh1 ðTÞÞðv1 ÞÞ ¼ ðq1 ðTÞÞl1 h1 ðTÞ ðv1 Þ ¼ 0ðv1 Þ ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ðq1 ðTÞÞl1 ðw1 Þ ¼ 0, where w1 ðh1 ðTÞÞðv1 Þ ð6¼ 0Þ. Hence by the definition of V1 , we have w1 2 V1 . Also w1 6¼ 0. This shows that V1 6¼ f0g. Similarly, V2 6¼ f0g, etc. Since h1 ðTÞ 6¼ 0 and h1 ðTÞ : V ! V, we have ðh1 ðTÞÞðVÞ 6¼ f0g. Similarly, ðh2 ðTÞÞðVÞ 6¼ f0g, etc. Clearly, ðh1 ðTÞÞðVÞ V1 . Proof To show this, let us take an arbitrary u1 2 V. We have to show that n o ðh1 ðTÞÞðu1 Þ 2 V1 ¼ v : v 2 V and ðq1 ðTÞÞl1 ðvÞ ¼ 0 ;

3.2 Canonical Forms

that is,

191

ðq1 ðTÞÞl1 ððh1 ðTÞÞðu1 ÞÞ ¼ 0, that is, ðq1 ðTÞÞl1 ðh1 ðTÞÞ ðu1 Þ ¼ 0.

Since pðxÞ ¼ ðq1 ðxÞÞl1 h1 ðxÞ, we have 0 ¼ pðTÞ ¼ ðq1 ðTÞÞl1 h1 ðTÞ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ðq1 ðTÞÞl1 h1 ðTÞ ¼ 0. It follows that LHS ¼ ðq1 ðTÞÞl1 ðh1 ðTÞÞ ðu1 Þ ¼ ð0Þðu1 Þ ¼ 0 ¼ RHS: ■ Similarly, ðh2 ðTÞÞðVÞ V2 , etc. Now, since w1 ; w2 ; . . .; wk 2 V, we have ðh1 ðTÞÞðw1 Þ 2 V1 , ðh2 ðTÞÞðw2 Þ 2 V2 ; . . .; ðhk ðTÞÞðwk Þ 2 Vk . Hence from (**), for every v 2 V, there exist v1 2 V1 ; v2 2 V2 ; . . .; vk 2 Vk such that v ¼ v1 þ v2 þ þ vk : This proves that V V1 þ V2 þ þ Vk : Now, from (*), we have V ¼ V1 þ V2 þ þ Vk : ð Þ Observe that if v 2 V2 , then ðq2 ðTÞÞl2 ðvÞ ¼ 0, and hence ðh1 ðTÞÞðvÞ ðq2 ðTÞÞl2 ðvÞ ¼ ðq3 ðTÞÞl3 ðqk ðTÞÞlk ð0Þ ¼ ðq3 ðTÞÞl3 ðqk ðTÞÞlk |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0: Thus v2 2 V2 ) ðh1 ðTÞÞðv2 Þ ¼ 0: Similarly, v3 2 V3 ) ðh1 ðTÞÞðv3 Þ ¼ 0; etc. In short, for any distinct i; j 2 f1; 2; . . .; k g, we have ðhi ðTÞÞ Vj ¼ f0g.

192


We claim that V ¼ V1 V2 Vk : In view of (***), it suffices to show that if each vi is in Vi , and v1 þ v2 þ þ vk ¼ 0, then each vi equals 0. Suppose to the contrary that there exist vi 2 Vi ði ¼ 1; 2; . . .; kÞ such that v1 þ v2 þ þ vk ¼ 0 and v1 6¼ 0. We seek a contradiction. Since v1 þ v2 þ þ vk ¼ 0, we have

ðq2 ðTÞÞl2 ðq3 ðTÞÞl3 ðqk ðTÞÞlk ðv1 Þ ¼ ðh1 ðTÞÞðv1 Þ þ 0 þ þ 0 ¼ ðh1 ðTÞÞðv1 Þ þ ðh1 ðTÞÞðv2 Þ þ þ ðh1 ðTÞÞðvk Þ ¼ ðh1 ðTÞÞðv1 þ v2 þ þ vk Þ ¼ ðh1 ðTÞÞð0Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ 0;

and hence ðq2 ðTÞÞl2 ðq3 ðTÞÞl3 ðqk ðTÞÞlk ðv1 Þ ¼ 0: n o Since v1 2 V1 ¼ v : v 2 V and ðq1 ðTÞÞl1 ðvÞ ¼ 0 , we have ðq1 ðTÞÞl1 ðv1 Þ ¼ 0. Observe that the greatest common divisor of ðq1 ðxÞÞl1 ; ðq2 ðxÞÞl2 ðq3 ðxÞÞl3 ðqk ðxÞÞlk is 1. So by 1.1.5, there exist b1 ðxÞ; b2 ðxÞ 2 F ½ x such that 1 ¼ b1 ðxÞ ðq1 ðxÞÞl1 þ b2 ðxÞ ðq2 ðxÞÞl2 ðq3 ðxÞÞl3 ðqk ðxÞÞlk ; and hence I ¼ ðb1 ðTÞÞ ðq1 ðTÞÞl1 þ ðb2 ðTÞÞ ðq2 ðTÞÞl2 ðq3 ðTÞÞl3 ðqk ðTÞÞlk : It follows that

3.2 Canonical Forms

193

v1 ¼ I ð v1 Þ ¼

ðb1 ðTÞÞ ðq1 ðTÞÞl1 þ ðb2 ðTÞÞ

ðq2 ðTÞÞl2 ðq3 ðTÞÞl3 ðqk ðTÞÞlk ðv1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðb1 ðTÞÞ ðq1 ðTÞÞl1 ðv1 Þ þ ðb2 ðTÞÞ ðq2 ðTÞÞl2 ðq3 ðTÞÞl3 ðqk ðTÞÞlk ðv1 Þ ¼ ðb1 ðTÞÞ ðq1 ðTÞÞl1 ðv1 Þ þ ðb2 ðTÞÞð0Þ ¼ ðb1 ðTÞÞð0Þ þ ðb2 ðTÞÞð0Þ ¼ 0 þ 0 ¼ 0;

and hence v1 ¼ 0. This is a contradiction. Thus we have shown that V ¼ V1 V2 Vk . l1 Observe that q1 TjV1 ¼ 0. Proof To show this, let us take an arbitrary v1 2 V1 . We have to show that l1 ðv1 Þ ¼ 0. q1 TjV1 Since n o v1 2 V1 ¼ v : v 2 V and ðq1 ðTÞÞl1 ðvÞ ¼ 0 ; we have ðq1 ðTÞÞl1 ðv1 Þ ¼ 0. Now l1 l1

ðv1 Þ ¼ q1 TjV1 ðv1 Þ ¼ ðq1 ðT ðv1 ÞÞÞl1 q1 TjV1 ¼ ðq1 ðTÞÞl1 ðv1 Þ ¼ 0 ¼ RHS:

LHS ¼

■ It follows that the minimal polynomial of TjV1 is of the form ðq1 ðxÞÞm1 , where m1 is a positive integer l1 . Similarly, the minimal polynomial of TjV2 is of the form ðq2 ðxÞÞm2 , where m2 is a positive integer l2 , etc. Now, since q1 ðxÞ; q2 ðxÞ; . . .; qk ðxÞ are distinct irreducible polynomials, the least common multiple of ðq1 ðxÞÞm1 ; ðq2 ðxÞÞm2 ; . . .; ðqk ðxÞÞmk is ðq1 ðxÞÞm1 ðq2 ðxÞÞm2 . . .ðqk ðxÞÞmk . By 3.2.1, the minimal polynomial of T is the least common multiple of ðq1 ðxÞÞm1 ; ðq2 ðxÞÞm2 ; . . .; ðqk ðxÞÞmk , and hence the minimal polynomial of T is ðq1 ðxÞÞm1 ðq2 ðxÞÞm2 . . . ðqk ðxÞÞmk . Since ðq1 ðxÞÞl1 ðq2 ðxÞÞl2 . . . ðqk ðxÞÞlk

194


is the minimal polynomial of T over F, and each mi li , we have m1 ¼ l1 ; . . .; mk ¼ lk . Since the minimal polynomial of TjV1 is of the form ðq1 ðxÞÞm1 , and m1 ¼ l1 , the minimal polynomial of TjV1 is ðq1 ðxÞÞl1 . Similarly, the minimal polynomial of TjV2 is ðq2 ðxÞÞl2 , etc. 3.2.3 Conclusion Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let pðxÞ 2 F ½ x. Let pðxÞ be the minimal polynomial of T over F. Suppose that pðxÞ ¼ ðq1 ðxÞÞl1 ðq2 ðxÞÞl2 . . .ðqk ðxÞÞlk ; where q1 ðxÞ; q2 ðxÞ; . . .; qk ðxÞ are distinct irreducible polynomials (of degree 1) in F ½ x, and l1 ; l2 ; . . .; lk are positive integers. Put n o V1 v : v 2 V and ðq1 ðTÞÞl1 ðvÞ ¼ 0 ; V2 n o v : v 2 V and ðq2 ðTÞÞl2 ðvÞ ¼ 0 ; etc: Then, 1. 2. 3. 4.

each Vi is a nontrivial linear subspace of V, each Vi is invariant under T, V ¼ V1 V2 Vk , for each i ¼ 1; 2; . . .; k, the minimal polynomial of TjV1 is ðqi ðxÞÞli .

3.2.4 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let k 2 F. Suppose that k is an eigenvalue of T. Then ðkI T Þ : V ! V is not invertible. Proof Since k is an eigenvalue of T, there exists a nonzero v 2 V such that TðvÞ ¼ kv, and hence ðkI T ÞðvÞ ¼ 0 ð¼ ðkI T Þð0ÞÞ. Here v 6¼ 0, and ðkI T ÞðvÞ ¼ ðkI T Þð0Þ, ðkI T Þ : V ! V is not one-to-one, and hence ðkI T Þ is not invertible. ■ 3.2.5 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let k 2 F. Suppose that ðkI T Þ : V ! V is not invertible, that is, ðkI T Þ is singular. Then k is an eigenvalue of T. Proof Here ðkI T Þ : V ! V is not invertible, that is, ðkI T Þ1 does not exist, so by 3.1.31, there exists a nonzero v 2 V such that ðkI T ÞðvÞ ¼ 0. It follows that TðvÞ ¼ kv, where v 6¼ 0. Hence k is an eigenvalue of T. ■ 3.2.6 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let k 2 F. Suppose that k is an eigenvalue of T. Let qðxÞ 2 F ½ x. (It follows that qðkÞ 2 F and qðTÞ 2 AðVÞ.) Then qðkÞ is an eigenvalue of qðTÞ.

3.2 Canonical Forms

195

Proof Suppose that qðxÞ ¼ a0 þ a1 x þ þ an xn ; where each ai is in F. It follows that qð kÞ ¼ a0 þ a1 k þ þ an kn and qðTÞ ¼ a0 I þ a1 T þ þ an T n : Since k is an eigenvalue of T, there exists a nonzero v 2 V such that TðvÞ ¼ kv. It suffices to show that a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðTðvÞÞ þ þ an T n ðvÞ ¼ ða0 I þ a1 T þ þ an T n ÞðvÞ ¼ ðqðTÞÞðvÞ ¼ ðqðkÞÞv |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ða0 þ a1 k þ þ an kn Þv; that is, a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n ðvÞ ¼ ða0 þ a1 k þ þ an kn Þv: LHS ¼ a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n ðvÞ ¼ a0 v þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n ðvÞ ¼ a0 v þ a1 ðkvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n1 ðTðvÞÞ ¼ a0 v þ a1 ðkvÞ þ a2 T ðkvÞ þ a3 T ðT ðkvÞÞ þ þ an T n1 ðkvÞ ¼ a0 v þ a1 ðkvÞ þ a2 kTðvÞ þ a3 kT ðTðvÞÞ þ þ an kT n1 ðvÞ ¼ a0 v þ a1 ðkvÞ þ a2 kðkvÞ þ a3 kT ðkvÞ þ þ an kT n2 ðkvÞ ¼ a0 v þ a1 ðkvÞ þ a2 k2 v þ a3 k2 TðvÞ þ þ an k2 T n2 ðvÞ .. . 2 ¼ a0 v þ a1 ðkvÞ þ a2 k v þ a3 k3 v þ þ an ðkn vÞ ¼ a0 þ a1 k þ a2 k2 þ a3 k3 þ þ an kn v ¼ RHS: ■ 3.2.7 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let pðxÞ be the minimal polynomial of T over F. Let k F. Suppose that k is an eigenvalue of T. Then k is a root of pðxÞ, that is, pðkÞ ¼ 0. (Since the number roots of pðxÞ is finite, the number of eigenvalues of T is finite.)

196


Proof Since pðxÞ is the minimal polynomial of T over F, we have pðxÞ 2 F ½ x and pðTÞ ¼ 0. Suppose that pðxÞ ¼ a0 þ a1 x þ þ an xn ; where each ai is in F. It follows that pð kÞ ¼ a0 þ a1 k þ þ an kn and pðTÞ ¼ a0 I þ a1 T þ þ an T n : Since k is an eigenvalue of T, there exists a nonzero v 2 V such that TðvÞ ¼ kv. We claim that ðpðTÞÞðvÞ ¼ ðpðkÞÞv, that is, a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðTðvÞÞ þ þ an T n ðvÞ ¼ ða0 þ a1 k þ þ an kn Þv; LHS ¼ a0 IðvÞ þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n ðvÞ ¼ a0 v þ a1 TðvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n ðvÞ ¼ a0 v þ a1 ðkvÞ þ a2 T ðTðvÞÞ þ a3 T ðT ðTðvÞÞÞ þ þ an T n1 ðTðvÞÞ ¼ a0 v þ a1 ðkvÞ þ a2 T ðkvÞ þ a3 T ðT ðkvÞÞ þ þ an T n1 ðkvÞ ¼ a0 v þ a1 ðkvÞ þ a2 kTðvÞ þ a3 kT ðTðvÞÞ þ þ an kT n1 ðvÞ ¼ a0 v þ a1 ðkvÞ þ a2 kðkvÞ þ a3 kT ðkvÞ þ þ an kT n2 ðkvÞ ¼ a0 v þ a1 ðkvÞ þ a2 k2 v þ a3 k2 TðvÞ þ þ an k2 T n2 ðvÞ .. . 2 ¼ a0 v þ a1 ðkvÞ þ a2 k v þ a3 k3 v þ þ an ðkn vÞ ¼ a0 þ a1 k þ a2 k2 þ a3 k3 þ þ an kn v ¼ RHS: Thus, 0 ¼ 0ðvÞ ¼ ðpðTÞÞðvÞ ¼ ðpðkÞÞv . Now, since ðpðkÞÞv ¼ 0 and v is non|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} zero, we have pðkÞ ¼ 0.

■

3.2.8 Problem Let V be any n-dimensional vector space over the field F. Let S; T 2 AðVÞ. Let S : V ! V be invertible. Then T and S1 T S have the same minimal polynomial. Proof Let pðxÞ be the minimal polynomial of T, and qðxÞ the minimal polynomial of S1 T S. It suffices to show that 1. qðxÞjpðxÞ, that is, pðS1 T SÞ ¼ 0, 2. pðxÞjqðxÞ, that is, qðTÞ ¼ 0,

3.2 Canonical Forms

197

For 1: Since pðxÞ is the minimal polynomial of T, we have pðTÞ ¼ 0. Suppose that pðxÞ ¼ a0 þ a1 x þ þ an xn ; where each ai is in F. It follows that 0 ¼ pðTÞ ¼ a0 I þ a1 T þ þ an T n |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and p S1 T S ¼ a0 I þ a1 S1 T S þ a2 S1 T S S1 T S þ a3 S1 T S S1 T S S1 T S n þ þ an S1 T S : Hence p S1 T S ¼ a0 I þ a1 S1 T S þ a2 S1 T 2 S þ a3 S1 T 3 S þ þ an S1 T n S : Now, since a0 I þ a1 S1 T S þ a2 S1 T 2 S þ a3 S1 T 3 S þ þ an S1 T n S ¼ S1 ða0 I Þ S þ S1 ða1 T Þ S þ S1 a2 T 2 S þ S1 a3 T 3 S þ þ S1 ðan T n Þ S ¼ S1 a0 I þ a1 T þ a2 T 2 þ a3 T 3 þ þ an T n S ¼ S1 ðpðTÞÞ S ¼ S1 0 S ¼ 0; we have pðS1 T SÞ ¼ 0. For 2: Since qðxÞ is the minimal polynomial of S1 T S, we have qðS1 T SÞ ¼ 0. Suppose that qðxÞ ¼ b0 þ b1 x þ þ bm xm ; where each bi is in F. It follows that qðTÞ ¼ b0 I þ b1 T þ b2 T 2 þ þ bm T m

198


and m 0 ¼ q S1 T S ¼ b0 I þ b1 S1 T S þ þ bm S1 T S |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ b0 ðS1 I SÞ þ b1 ðS1 T SÞ þ þ bm ðS1 T m SÞ ¼ S1 ðb0 I Þ S þ S1 ðb1 T Þ S þ þ S1 ðbm T m Þ S ¼ S1 ðb0 I þ b1 T þ b2 T 2 þ þ bm T m Þ S and hence 0 ¼ S1 b0 I þ b1 T þ b2 T 2 þ þ bm T m S: It follows that qðTÞ ¼ b0 I þ b1 T þ b2 T 2 þ þ bm T m ¼ S 0 S1 ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence qðTÞ ¼ 0.

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Definition Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let k 2 F. Suppose that k is an eigenvalue of T. Let v be a nonzero vector in V. If TðxÞ ¼ kx, then we say that v is an eigenvector of T belonging to the eigenvalue k. 3.2.9 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Let k1 ; k2 ; . . .; kk be distinct eigenvalues of T. Suppose that for every i 2 f1; 2; . . .; k g, vi is an eigenvector of T belonging to the eigenvalue ki . Then fv1 ; v2 ; . . .; vk g is a linearly independent set of vectors over F. Proof Suppose to the contrary (after suitably rearranging the indices) that v 1 ¼ a2 v 2 þ a3 v 3 þ . . . þ al v l |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðl1Þ terms

is the shortest linear relation, where a2 ; a3 ; . . .; al are nonzero members of F. We seek a contradiction. Since v1 ¼ a2 v2 þ a3 v3 þ þ al vl , we have k1 a2 v 2 þ k1 a3 v 3 þ þ k 1 al v l ¼ k1 ð a2 v 2 þ a3 v 3 þ þ al v l Þ ¼ k1 v 1 ¼ T ð v 1 Þ ¼ T ð a2 v 2 þ a3 v 3 þ þ al v l Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a2 T ð v 2 Þ þ a3 T ð v 3 Þ þ þ al T ð v l Þ ¼ a2 k2 v 2 þ a 3 k3 v 3 þ þ al kl v l ;

3.2 Canonical Forms

199

and hence k1 a2 v 2 þ k1 a3 v 3 þ þ k1 al v l ¼ a2 k2 v 2 þ a3 k3 v 3 þ þ al kl v l : This shows that ðk1 k2 Þa2 v2 þ ðk1 k3 Þa3 v3 þ þ ðk1 kl Þal vl ¼ 0: Since k1 6¼ k2 and a2 6¼ 0, we have v2 ¼

ðk1 k3 Þa3 ðk1 kl Þal v3 þ þ vl : ðk1 k2 Þa2 ðk1 k2 Þa2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðl2Þ terms

This contradicts the fact that v 1 ¼ a2 v 2 þ a3 v 3 þ þ al v l |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ðl1Þ terms

is the shortest linear relation.

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3.2.10 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Then the number of distinct eigenvalues of T is n. Proof Let k1 ; k2 ; . . .; kk be distinct eigenvalues of T. We have to show that k n, that is, k dimðVÞ. Suppose that for every i 2 f1; 2; . . .; kg, vi is an eigenvector of T belonging to the eigenvalue ki . By 3.2.9, fv1 ; v2 ; . . .; vk g is a linearly independent set of vectors over F. It follows that the number of elements in fv1 ; v2 ; . . .; vk g is dimðVÞ. Since fv1 ; v2 ; . . .; vk g is a linearly independent set of vectors, the number of elements in ■ fv1 ; v2 ; . . .; vk g is k, and hence k dimðVÞ. 3.2.11 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T has n distinct eigenvalues in F. Then there exists a basis of V over F such that each member of the basis is an eigenvector of T. Proof Let k1 ; k2 ; . . .; kn be distinct eigenvalues of T. Suppose that for every i 2 f1; 2; . . .; ng, vi is an eigenvector of T belonging to the eigenvalue ki . By 3.2.9, fv1 ; v2 ; . . .; vn g is a linearly independent set of vectors over F. Now, since the number of elements in fv1 ; v2 ; . . .; vn g is equal to dimðVÞ, fv1 ; v2 ; . . .; vn g constitutes a basis of V. ■ Definition Let V be any n-dimensional vector space over the field F. Let S; T 2 AðVÞ. If there exists C 2 AðVÞ such that C 1 exists, and C 1 S C ¼ T, then we say that S is similar to T, and we write S T. 3.2.12 Problem is an equivalence relation over AðVÞ.

200


And hence AðVÞ is partitioned into equivalence classes. Each equivalence class is called a similarity class. Proof a. Let us take an arbitrary T 2 AðVÞ. Since I 1 T I ¼ T, we have T T. Thus

is reflexive over AðVÞ. b. Let us take arbitrary S; T 2 AðVÞ satisfying S T. It follows that there exists C 2 AðVÞ such that C1 exists, and C1 S C ¼ T. It follows that 1 ðC1 Þ ð¼ C 2 AðVÞÞ and C T C 1 ¼ S. Thus S T. Hence is symmetric. c. Let us take arbitrary R; S; T 2 AðVÞ satisfying R S and S T. We have to show that R T. Since R S, there exists C 2 AðVÞ such that C 1 exists, and C 1 R C ¼ S. Again there exists D 2 AðVÞ such that D1 exists, and ðC DÞ1 R ðC DÞ ¼ D1 C 1 R ðC DÞ 1 ¼ D1 C 1 R C D ¼ D S D¼T: |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} Thus E 1 R E ¼ T, where E ðC DÞ 2 AðVÞ. This proves that is an equivalence relation over AðVÞ.

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3.2.13 Problem Let V be any n-dimensional vector space over F. Let T 2 AðVÞ. b: Let W be any subspace of V. Suppose that W is invariant under T. Then T V V ðv þ W Þ 7! ðTðvÞ þ W Þ from the quotient space W to W is a linear transformation. b is a well-defined function from the quotient space V to V : To show this, let Proof T W W us take arbitrary u; v 2 V such that ðu þ W Þ ¼ ðv þ W Þ, that is, ðu vÞ 2 W. We have to show that ðTðuÞ þ W Þ ¼ ðTðvÞ þ W Þ, that is, ðTðuÞ TðvÞÞ 2 W, that is, T ðu vÞ 2 W. Since ðu vÞ 2 W and W is invariant under T, we have T ðu vÞ 2 W. b : V ! V is linear: To show this, let us take arbitrary u; v 2 V and a; b 2 F. T W W We have to show that T ðau þ bvÞ þ W ¼ aðTðuÞ þ W Þ þ bðTðvÞ þ W Þ : LHS ¼ T ðau þ bvÞ þ W ¼ ðaTðuÞ þ bTðvÞÞ þ W ¼ aðTðuÞ þ W Þ þ bðTðvÞ þ W Þ ¼ RHS:

■

3.2.14 Problem Let V be any n-dimensional vector space over F. Let T 2 AðVÞ. Let W be any subspace of V. Suppose that W is invariant under T. Let 2 F ½ x. pðxÞ V V b , as defined in 3.2.13, is a member of A b Here T W , and hence p T 2 A W . Also b is the pðTÞ 2 AðVÞ. Suppose that pðTÞ is the zero element of AðVÞ. Then p T V zero element of A W .

3.2 Canonical Forms

201

Proof Suppose that pðxÞ ¼ a0 þ a1 x þ þ an xn ; where each ai is in F. It follows that AðVÞ 3 0 ¼ pðTÞ ¼ a0 I þ a1 T þ þ an T n |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and

b þ a2 T b T b þ a3 T b T b T b þ þ an T n : b ¼ a0bI þ a1 T p T

n b b b b b b b We have to show V that a0 I þ a1 T þ a2 T T þ a3 T T T þ þ an T is the zero element of A W . To this end, let us take an arbitrary v 2 V. We have to show that

b þ a2 T b T b þ a3 T b T b T b þ þ an T n ðv þ W Þ ¼ ð0 þ W Þ; a0bI þ a1 T

that is, a0Î ðv þ W Þ þ a1 T^ ðv þ W Þ þ a2 T^ T^ ðv þ W Þ þ a3 T^ T^ T^ ðv þ W Þ þ þ an T^ n ðv þ W Þ ¼ ð0 þ W Þ: LHS ¼ a0Î ðv þ W Þ þ a1 T^ ðv þ W Þ þ a2 T^ T^ ðv þ W Þ þ a3 T^ T^ T^ ðv þ W Þ þ þ an T^ n ðv þ W Þ ¼ a0 ðIðvÞ þ W Þ þ a1 ðTðvÞ þ W Þ þ a2 T^ T^ ðv þ W Þ þ a3 T^ T^ T^ ðv þ W Þ þ þ an T^ n1 T^ ðv þ W Þ ¼ a0 ðv þ W Þ þ a1 ðTðvÞ þ W Þ þ a2 T^ ðTðvÞ þ W Þ þ a3 T^ T^ ðTðvÞ þ W Þ þ þ an T^ n1 ðTðvÞ þ W Þ ¼ a0 ðv þ W Þ þ a1 ðTðvÞ þ W Þ þ a2 ðT ðTðvÞÞ þ W Þ þ a3 T^ ðT ðTðvÞÞ þ W Þ þ þ an T^ n2 ðT ðTðvÞÞ þ W Þ ¼ a0 ðv þ W Þ þ a1 ðTðvÞ þ W Þ þ a2 T 2 ðvÞ þ W þ a3 T^ T 2 ðvÞ þ W þ þ an T^ n2 T 2 ðvÞ þ W ¼ a0 ðv þ W Þ þ a1 ðTðvÞ þ W Þ þ a2 T 2 ðvÞ þ W þ a3 T 3 ðvÞ þ W þ þ an ðT n ðvÞ þ W Þ ¼ ða0 v þ W Þ þ ða1 TðvÞ þ W Þ þ a2 T 2 ðvÞ þ W þ a3 T 3 ðvÞ þ W þ þ ðan T n ðvÞ þ W Þ ¼ a0 v þ a1 TðvÞ þ a2 T 2 ðvÞ þ a3 T 3 ðvÞ þ þ an T n ðvÞ þ W ¼ ða0 I þ a1 T þ þ an T n ÞðvÞ þ W ¼ 0ðvÞ þ W ¼ 0 þ W ¼ RHS:

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202


3.2.15 Problem Let V be any n-dimensional vector space over F. Let T 2 AðVÞ. Let W be any subspace of V. Suppose that W is invariant under T. Let b , as defined in 3.2.13, is a member of A V , and hence pðxÞ; qðxÞ 2 F ½ x. Here T W V V b b p T 2A . Also pðTÞ 2 AðVÞ. Similarly, q T 2 A and qðTÞ 2 AðVÞ. W

W

Suppose that pðxÞ is the minimal polynomial of T : V ! V over F, and qðxÞ is the b : V ! V over F. Then qðxÞjpðxÞ. minimal polynomial of T W W Proof Since pðxÞ is the minimal polynomial of T : V ! V over F, pðTÞ is the zero V b element of AðVÞ, and hence by 3.2.14, p T is the zero element of A W . Now, b : V ! V over F, we have qðxÞjpðxÞ. ■ since qðxÞ is the minimal polynomial of T W W 3.2.16 Theorem Let V be any n-dimensional vector space over F. Let T 2 AðVÞ. Suppose that all the roots of the minimal polynomial of T over F are in F. Then there exists a basis of V in which the matrix of T is such that all its entries above the diagonal are zero. In short, there exists a basis of V in which the matrix of T is triangular. Proof (Induction on n): The theorem is trivially true when n ¼ 1. Now let us assume that the theorem is true for all vector spaces of dimension ðn 1Þ. Let k1 2 F. Let k1 be an eigenvalue of T. Since k1 is an eigenvalue of T, there exists a nonzero vector v1 2 V such that T ðv1 Þ ¼ k1 v1 . Put W fav1 : a 2 F g: Clearly, W is a linear subspace of V. Also, since v1 is nonzero, W is a one-dimensional space. Next, since T ðv1 Þ ¼ k1 v1 , W is invariant under T. It follows that

V dim ¼ dimðVÞ dimðWÞ ¼ dimðVÞ 1 ¼ n 1: W V is an ðn 1Þ-dimensional vector space over F. Thus W Suppose that pðxÞ is the minimal polynomial of T : V ! V over F, and qðxÞ is b : V ! V over F, where T b : ðv þ W Þ 7! ðTðvÞ þ W Þ the minimal polynomial of T W W V V from the quotient space W to W is a linear transformation. By 3.2.15, qðxÞjpðxÞ, and hence every root of qðxÞ is a root of pðxÞ. Now, since all the roots of the minimal polynomial pðxÞ of T over F are in F, all the roots of qðxÞ are in F. b 2 A V , and all the roots of Since WV is an ðn 1Þ-dimensional vector space, T W b over F are in F, it follows by our induction the minimal polynomial qðxÞ of T V hypothesis that there exists a basis fv2 þ W; v3 þ W; . . .; vn þ W g of W in which the

3.2 Canonical Forms

203

b is such that all its entries above the diagonal are zero. So there exists a matrix of T matrix 2

a22 6 a32 6 6 4 an2

a23 a33 an3

a24 a34 .. .

an4

3 a2n a3n 7 7 7 5 ann ðn1Þðn1Þ

such that all its entries above the diagonal are zero, and b ðv2 þ W Þ ¼ a22 ðv2 þ W Þ; T b T ðv3 þ W Þ ¼ a32 ðv2 þ W Þ þ a33 ðv3 þ W Þ; .. .

b ðvn þ W Þ ¼ an2 ðv2 þ W Þ þ an3 ðv3 þ W Þ þ þ ann ðvn þ W Þ: T It follows that T ðv2 Þ þ W ¼ a22 v2 þ W; T ðv3 Þ þ W ¼ ða32 v2 þ a33 v3 Þ þ W; .. .

T ðvn Þ þ W ¼ ðan2 v2 þ an3 v3 þ þ ann vn Þ þ W; and hence T ðv2 Þ a22 v2 2 W T ðv3 Þ ða32 v2 þ a33 v3 Þ 2 W .. .

T ðvn Þ ðan2 v2 þ an3 v3 þ þ ann vn Þ 2 W

9 > > > = : > > > ;

Thus T ðv2 Þ a22 v2 2 fav1 : a 2 F g T ðv3 Þ ða32 v2 þ a33 v3 Þ 2 fav1 : a 2 F g .. .

T ðvn Þ ðan2 v2 þ an3 v3 þ þ ann vn Þ 2 fav1 : a 2 F g

9 > > > = > > > ;

:

It follows that there exists a21 2 F such that T ðv2 Þ a22 v2 ¼ a21 v1 , and hence T ðv2 Þ ¼ a21 v1 þ a22 v2 . Similarly, there exists a31 2 F such that T ðv3 Þ ¼ a31 v1 þ a32 v2 þ a33 v3 , etc. Also T ðv1 Þ ¼ k1 v1 . Thus

204


T ð v 1 Þ ¼ k1 v 1 T ðv2 Þ ¼ a21 v1 þ a22 v2 T ðv3 Þ ¼ a31 v1 þ a32 v2 þ a33 v3 .. .

T ðvn Þ ¼ an1 v1 þ an2 v2 þ þ an3 vn

9 > > > > > = : > > > > > ;

ð Þ

Clearly, fv1 ; v2 ; . . .; vn g is a linearly independent set of vectors in V. Proof To show this, suppose that a1 v1 þ a2 v2 þ þ an vn ¼ 0:

ð Þ

We have to show that each ai equals 0. Here a2 v2 þ þ an vn ¼ ða1 Þv1 2 fav1 : a 2 F g ¼ W; so a2 ðv2 þ W Þ þ þ an ðvn þ W Þ ¼ 0 þ W: V Now, since fv2 þ W; v3 þ W; . . .; vn þ W g is a basis of W , fv2 þ W; v3 þ V W; . . .; vn þ Wg is a linearly independent set of vectors in W , and hence a2 ¼ 0; a3 ¼ 0, and an ¼ 0. It remains to show that a1 ¼ 0. From (*),

a1 v1 ¼ a1 v1 þ 0v2 þ þ 0vn ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence a1 v1 ¼ 0. Now, since v1 is nonzero, we have a1 ¼ 0. ■ Thus we have shown that fv1 ; v2 ; . . .; vn g is a linearly independent set of vectors in the n-dimensional vector space V. It follows that fv1 ; v2 ; . . .; vn g is a basis of V. Next, from (*), the matrix of T relative to the basis fv1 ; v2 ; . . .; vn g is triangular. ■ 3.2.17 Problem Let F be a field. Let A be an n n matrix with entries in F. Suppose that all its eigenvalues are in F. Then there exists an invertible n n matrix C with entries in F such that C1 AC is a triangular matrix. Such matrices of a particularly nice form are called canonical forms. In short, we say that A can be brought to triangular form over F by similarity. Proof We know that F n constitutes a vector space over F under pointwise addition and pointwise scalar multiplication. Put

3.2 Canonical Forms

205

0

1

0

1

v1 @1; 0; 0; . . .; 0 A; v2 @0; 1; 0; . . .; 0A; etc: |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} n

n

We know that fv1 ; v2 ; . . .; vn g is a basis of F n . Suppose that A aij nn , where each aij is in F. Suppose that T : F n ! F n is a linear transformation such that T ðv1 Þ ¼ a11 v1 þ a12 v2 þ þ a1n vn ; T ðv2 Þ ¼ a21 v1 þ a22 v2 þ þ a2n vn ; etc: It follows that m1 ðTÞ ¼ aij nn , where m1 ðTÞ is the matrix of T relative to the basis v1 ; . . .; vn . Since all the eigenvalues of the matrix aij nn are in F, all the eigenvalues of T are in F, and hence by 3.2.7, all the roots of the minimal polynomial of T over F are in F. Hence by 3.2.16, there exists a basis w1 ; . . .; wn of F n such that the matrix m2 ðTÞ of T relative to the basis w1 ; . . .; wn is triangular. Let S : F n ! F n be the linear transformation such that for every i 2 f1; . . .; ng, Sðvi Þ ¼ wi . By 3.1.35, m2 ðTÞ ¼ ðm1 ðSÞÞ1 m1 ðTÞm1 ðSÞ: Thus m2 ðTÞ ¼ C 1 AC; where C m1 ðSÞ. Now, since the matrix m2 ðTÞ of T is triangular, C 1 AC is a triangular matrix. ■ 3.2.18 Problem Let A be a triangular matrix with entries in the field F. Suppose that no entry on the diagonal is 0. Then A is invertible. Proof Let A aij nn , where i \ j ) aij ¼ 0. Next, suppose that each aii is nonzero. We have to show that aij nn is invertible. It suffices to show that det aij nn 6¼ 0. Since aij nn is a triangular matrix, we have det aij nn ¼ a11 a22 . . . ann , and since each aii is nonzero, we have det aij nn 6¼ 0. ■ 3.2.19 Problem Let A be a triangular matrix with entries in the field F. Suppose that some entry on the diagonal is 0. Then A1 does not exist. Proof Let A aij nn , where i\j ) aij ¼ 0. Next, there exists i 2 f1; 2; . . .; ng such that aii ¼ 0. We have to show that aij nn is not invertible. It suffices to show that det aij nn ¼ 0. Since aij nn is a triangular matrix, we have det aij nn ¼ a11 a22 . . . ann , and since aii ¼ 0, we have det aij nn ¼ 0. ■

206


3.2.20 Problem Let aij nn be a triangular matrix with entries in the field F. Then the set of all eigenvalues of aij nn is fa11 ; a22 ; . . .; ann g. Proof Let k be an eigenvalue of aij nn . It follows that aij nn kI is singular, that is, det aij nn kI ¼ 0. Since aij nn is a triangular matrix, we have det aij nn kI ¼ ða11 kÞða22 kÞ ðann kÞ. Since det aij nn kI ¼ 0, we have ða11 kÞða22 kÞ ðann kÞ ¼ 0: This shows that k 2 fa11; a22 ; . . .; ann g. Conversely, since det aij nn a11 I ¼ ða11 a11 Þða22 a11 Þ ðann a11 Þ ¼ 0, it follows that aij nn a11 I is singular, and hence a11 is an eigenvalue of aij nn . Similarly, a22 is an eigenvalue of aij nn , etc. ■ 3.2.21 Theorem Let V be any n-dimensional vector space over F. Let T 2 AðVÞ. Suppose that all the roots of the minimal polynomial of T over F are in F. Then there exists a polynomial pðxÞ in F ½ x such that pðxÞ is of degree n, and pðTÞ ¼ 0. Proof By 3.2.16, there exists a basis fv1 ; v2 ; . . .; vn g of V such that the matrix aij nn of T relative to the basis fv1 ; v2 ; . . .; vn g is triangular. It follows that i \ j ) aij ¼ 0; and T ðv1 Þ ¼ a11 v1 ; T ðv2 Þ ¼ a21 v1 þ a22 v2 ; T ðv3 Þ ¼ a31 v1 þ a32 v2 þ a33 v3 ; .. .: For every i 2 f1; 2; . . .; ng, put ki aii . By 3.2.20, the set of all eigenvalues of aij nn is fk1 ; k2 ; . . .; kn g. Further, ðT k1 I Þðv1 Þ ¼ 0; ðT k2 I Þðv2 Þ ¼ a21 v1 ; ðT k3 I Þðv3 Þ ¼ a31 v1 þ a32 v2 ; .. . Observe that

3.2 Canonical Forms

207

ððT k1 I Þ ðT k2 I ÞÞðv2 Þ ¼ ðT k1 I ÞððT k2 I Þðv2 ÞÞ ¼ ðT k1 I Þða21 v1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a21 ðT k1 I Þðv1 Þ ¼ a21 0 ¼ 0; ððT k1 I Þ ðT k2 I Þ ðT k3 I ÞÞðv3 Þ ðððT k1 I Þ ðT k2 I ÞÞÞððT k3 I Þðv3 ÞÞ ¼ ðððT k1 I Þ ðT k2 I ÞÞÞ ¼ ða31 v1 þ a32 v2 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a31 ððT k1 I Þ ðT k2 I ÞÞðv1 Þ þ a32 ððT k1 I Þ ðT k2 I ÞÞðv2 Þ ¼ a31 ððT k1 I Þ ðT k2 I ÞÞðv1 Þ þ a32 0 ¼ a31 ððT k1 I Þ ðT k2 I ÞÞðv1 Þ ¼ a31 ðT T k1 T k2 T þ k1 k2 I Þðv1 Þ ¼ a31 ððT k2 I Þ ðT k1 I ÞÞðv1 Þ ¼ a31 ðT k2 I ÞððT k1 I Þðv1 ÞÞ ¼ a31 ðT k2 I Þð0Þ ¼ 0; etc: Thus ðT k1 I Þðv1 Þ ¼ 0; ððT k1 I Þ ðT k2 I ÞÞðv2 Þ ¼ 0; ððT k1 I Þ ðT k2 I Þ ðT k3 I ÞÞðv3 Þ ¼ 0; .. . Now, since each ðT ki I Þ commutes with each T kj I , we have 9 ððT k1 I Þ ðT k2 I Þ ðT kn I ÞÞðv1 Þ ¼ 0 > > > ððT k1 I Þ ðT k2 I Þ ðT kn I ÞÞðv2 Þ ¼ 0 = : .. > . > > ; ððT k1 I Þ ðT k2 I Þ ðT kn I ÞÞðvn Þ ¼ 0 Next, since fv1 ; v2 ; . . .; vn g is a basis of V, for every v 2 V, we have ððT k1 I Þ ðT k2 I Þ ðT kn I ÞÞðvÞ ¼ 0: This shows that ðT k1 I Þ ðT k2 I Þ ðT kn I Þ ¼ 0, and hence pðTÞ ¼ 0, where pðxÞ ðx k1 Þðx k2 Þ ðx kn Þ: Since each ki is in F, ðpðxÞ Þðx k1 Þðx k2 Þ ðx kn Þ is a polynomial in F ½ x, and it is of degree n. Thus pðxÞ is a polynomial in F ½ x of degree n. ■ Definition Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. If there exists a positive integer m such that T m ¼ 0, then we say that T is nilpotent. 3.2.22 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Let k be an eigenvalue of T. Then k ¼ 0.

208


Proof Suppose to the contrary that k 6¼ 0. We seek a contradiction. Since k is an eigenvalue of T, there exists a nonzero v in V such that TðvÞ ¼ kv. Since T is nilpotent, there exists a positive integer m such that T m ¼ 0. It follows that T m ðvÞ ¼ 0. Hence

k : k is a positive integer; and T k ðvÞ ¼ 0

is a nonempty set of positive integers. It follows that min k : k is a positive integer; and T k ðvÞ ¼ 0 exists. Put n min k : k is a positive integer; and T k ðvÞ ¼ 0 : It follows that T n1 ðvÞ 6¼ 0 and T n ðvÞ ¼ 0. Now, kT n1 ðvÞ ¼ T n1 ðkvÞ ¼ T n1 ðTðvÞÞ ¼ T n ðvÞ ¼ 0 : |fflfflfflfflfflffl{zfflfflfflfflfflffl} Thus kT n1 ðvÞ ¼ 0. Since k 6¼ 0, we have T n1 ðvÞ ¼ 0. This is a contradiction. ■ 3.2.23 Problem Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Let a0 ; a1 ; . . .; an 2 F. Let a0 6¼ 0. Then a0 I þ a1 T þ þ an T n is invertible, and ða0 I þ a1 T þ þ an T n Þ1 is a polynomial in the linear transformation S, where S a1 T þ þ an T n . Proof We have to show that a0 I þ S is invertible. Since T is nilpotent, there exists a positive integer m such that T m ¼ 0. It follows that r m ) T r ¼ 0; and hence Sm ¼ ð a1 T þ þ an T n Þ m ¼ ða1 T þ þ an T n Þ ða1 T þ þ an T n Þ ða1 T þ þ an T n Þ ¼ 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} m factors

3.2 Canonical Forms

209

Thus Sm ¼ 0. Observe that Iþ þ

1 ða0 Þ2

S a10 I ða1Þ2 S þ ða1Þ3 S2 þ ð1Þm1 ða01Þm Sm1 : 0 0 ¼ a10 I ða1Þ2 S þ ða1Þ3 S2 þ ð1Þm1 ða01Þm Sm1 1 a0

0

0

1 S3 ð1Þm1 ða01Þm Sm1 þ ð1Þm1 ða Þ1m þ 1 ða0 Þ4 0 þ ð1Þm1 ða Þ1m þ 1 Sm ¼ a10 I þ ð1Þm1 ða Þ1m þ 1 0 ¼ a10 I; 0 0

S ða1Þ3 S2 þ 0

¼ a10 I

Sm

so

1 Iþ S a0

! 1 1 1 2 1 I Sþ S þ ð1Þm1 Sm1 2 3 a0 ð a0 Þ m ð a0 Þ ð a0 Þ

¼

1 I: a0

This shows that ð a0 I þ S Þ

1

¼

! 1 1 1 2 m1 1 m1 ; I Sþ S þ ð1Þ S a0 ð a0 Þ m ða0 Þ2 ð a0 Þ 3

and hence a0 I þ S is invertible.

■

Definition Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Since T is nilpotent, there exists a positive integer m such that T m ¼ 0. Hence

k : k is a positive integer; and T k ¼ 0

is a nonempty set of positive integers. It follows that min k : k is a positive integer; and T k ¼ 0 exists. Put n min k : k is a positive integer; and T k ¼ 0 : It follows that T n1 6¼ 0 and T n ¼ 0. Here n is called the index of nilpotence of T. 3.2.24 Note Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Let n1 be a positive integer. Let n1 be the index of nilpotence of T. It follows that T n1 1 6¼ 0 and T n1 ¼ 0. Since T n1 1 : V ! V is a nonzero function, there exists a nonzero v 2 V such that T n1 1 ðvÞ 6¼ 0 and T n1 ðvÞ ¼ 0. Clearly, v; TðvÞ; T 2 ðvÞ; . . .; T n1 1 ðvÞ are linearly independent over F.

210


Proof Suppose to the contrary that v; TðvÞ; T 2 ðvÞ; . . .; T n1 1 ðvÞ are linearly dependent over F. We seek a contradiction. It follows that there exist a1 ; a2 ; . . .; an1 2 F such that not all the ai are zero, and a1 v þ a2 TðvÞ þ a3 T 2 ðvÞ þ þ an1 T n1 1 ðvÞ ¼ 0: Suppose that as is the first nonzero ai . It follows that s n1 , as 6¼ 0, and ðr\s ) ar ¼ 0Þ. Hence

as I þ as þ 1 T þ as þ 2 T 2 þ þ an1 T n1 s T s1 ðvÞ ¼ as T s1 ðvÞ þ as þ 1 T s ðvÞ þ þ an1 T n1 1 ðvÞ ¼ 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

This shows that

as I þ as þ 1 T þ as þ 2 T 2 þ þ an1 T n1 s T s1 ðvÞ ¼ 0: ðÞ

Since as 6¼ 0 and T is nilpotent, by 3.2.23,

as I þ as þ 1 T þ as þ 2 T 2 þ þ an1 T n1 s

1

exists. Now, from (*), T s1 ðvÞ ¼ 0. Since s n1 , we have ðs 1Þ ðn1 1Þ. ■ Since T n1 1 ðvÞ 6¼ 0, we have T s1 ðvÞ 6¼ 0. This is a contradiction. Put v1 v; v2 TðvÞ; v3 T 2 ðvÞ; . . .; vn1 T n1 1 ðvÞ: Since v; TðvÞ; T 2 ðvÞ; . . .; T n1 1 ðvÞ are linearly independent over F, it follows that fv1 ; v2 ; . . .; vn1 g is a linearly independent set of vectors in V. Suppose that V1 is the linear span of fv1 ; v2 ; . . .; vn1 g. It follows that V1 is an n1 -dimensional linear subspace of V, and fv1 ; v2 ; . . .; vn1 g is a basis of V1 . Observe that T ðv1 Þ ¼ TðvÞ ¼ v2 2 V1 ; T ðv2 Þ ¼ T ðTðvÞÞ ¼ T 2 ðvÞ ¼ v3 2 V1 ; .. .

T ðvn1 Þ ¼ T ðT n1 1 ðvÞÞ ¼ T n1 ðvÞ ¼ 0 2 V1 ; and hence fT ðv1 Þ; T ðv2 Þ; . . .; T ðvn1 Þg V1 . Now, since fv1 ; v2 ; . . .; vn1 g is a basis of V1 , we have ðw 2 V1 ) TðwÞ 2 V1 Þ. Thus we have shown that V1 is invariant under T.

3.2 Canonical Forms

211

3.2.24.1 There exists a linear subspace W of V, of largest possible dimension, such that 1. V1 \ W ¼ f0g, 2. W is invariant under T. Proof Since V1 is an n1 -dimensional linear subspace of the n-dimensional vector space V, and fv1 ; v2 ; . . .; vn1 g is a basis of V1 , there exists a basis fv1 ; v2 ; . . .; vn1 ; wn1 þ 1 ; wn1 þ 2 ; . . .; wn g of V. Let W1 be the linear span of fwn1 þ 1 ; wn1 þ 2 ; . . .; wn g. It is clear that W1 is a linear subspace of V, dimðW1 Þ ¼ n n1 ( 1), and V ¼ V1 W1 . Thus V1 \ W1 ¼ f0g. Since fv1 ; v2 ; . . .; vn1 ; wn1 þ 1 ; wn1 þ 2 ; . . .; wn g is a basis of V, we have wn1 þ 1 6¼ 0. Now, since T n1 ¼ 0, we have that

k : k is a positive integer; and T k ðwn1 þ 1 Þ ¼ 0

is a nonempty set of positive integers, and it follows that min k : k is a positive integer; and T k ðwn1 þ 1 Þ ¼ 0 exists. Put n2 min k : k is a positive integer; and T k ðwn1 þ 1 Þ ¼ 0 : It follows that T n2 1 ðwn1 þ 1 Þ 6¼ 0 and T n2 ðwn1 þ 1 Þ ¼ 0. Put w wn1 þ 1 . We have w 2 W1 ; T n2 1 ðwÞ 6¼ 0 and T n2 ðwÞ ¼ 0. Clearly, w; TðwÞ; T 2 ðwÞ; . . .; T n2 1 ðwÞ are linearly independent over F. Proof Suppose to the contrary that w; TðwÞ; T 2 ðwÞ; . . .; T n2 1 ðwÞ are linearly dependent over F. We seek a contradiction. It follows that there exist a1 ; a2 ; . . .; an2 2 F such that not all the ai are zero and a1 w þ a2 TðwÞ þ a3 T 2 ðwÞ þ þ an2 T n2 1 ðwÞ ¼ 0: Suppose that as is the first nonzero ai . It follows that s n2 , as 6¼ 0, and ðr\s ) ar ¼ 0Þ. Hence

as I þ as þ 1 T þ as þ 2 T 2 þ þ an2 T n2 s T s1 ðwÞ

¼ as T s1 ðwÞ þ as þ 1 T s ðwÞ þ þ an2 T n2 1 ðwÞ ¼ 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

212


This shows that

as I þ as þ 1 T þ as þ 2 T 2 þ þ an2 T n2 s T s1 ðwÞ ¼ 0: ðÞ

Since as 6¼ 0 and T is nilpotent, by 3.2.23,

as I þ as þ 1 T þ as þ 2 T 2 þ þ an2 T n2 s

1

exists. Now, from (*), T s1 ðwÞ ¼ 0. Since s n2 , we ðs 1Þ ðn2 1Þ. Since T n2 1 ðwÞ 6¼ 0, we have T s1 ðwÞ 6¼ 0. This is a contradiction.

have ■

Put w1 w; w2 TðwÞ; w3 T 2 ðwÞ; . . .; wn2 T n2 1 ðwÞ: Since w; TðwÞ; T 2 ðwÞ; . . .; T n2 1 ðwÞ are linearly independent over F, it follows that fw1 ; w2 ; . . .; wn2 g is a linearly independent set of vectors in V. Suppose that W1 is the linear span of fw1 ; w2 ; . . .; wn2 g. It follows that W1 is an n2 -dimensional linear subspace of V, and fw1 ; w2 ; . . .; wn2 g is a basis of W1 . Observe that T ðw1 Þ ¼ TðwÞ ¼ w2 2 W1 ; T ðw2 Þ ¼ T ðTðwÞÞ ¼ T 2 ðwÞ ¼ w3 2 W1 ; .. .

T ðwn2 Þ ¼ T ðT n2 1 ðwÞÞ ¼ T n2 ðwÞ ¼ 0 2 W1 ; and hence fT ðw1 Þ; T ðw2 Þ; . . .; T ðwn2 Þg W1 . Now, since fw1 ; w2 ; . . .; wn2 g is a basis of W1 , we have ðz 2 W1 ) TðzÞ 2 W1 Þ. Thus we have shown that W1 is invariant under T. Hence there exists a linear subspace W1 of V such that 1. V1 \ W1 ¼ f0g, 2. W1 is invariant under T. It follows that there exists a linear subspace W of V, of largest possible dimension, such that 1. V1 \ W ¼ f0g, 2. W is invariant under T.

■

3.2.24.2 Suppose that u 2 V1 . Let k be a positive integer such that k n1 . Suppose that T ðn1 kÞ ðuÞ ¼ 0. Then there exists u0 2 V1 such that T k ðu0 Þ ¼ u.

3.2 Canonical Forms

213

Proof Since u 2 V1 and V1 is the linear span of v; TðvÞ; T 2 ðvÞ; . . .; T n1 1 ðvÞ , there exist a1 ; a2 ; . . .; an1 2 F such that u ¼ a1 v þ a2 TðvÞ þ a3 T 2 ðvÞ þ þ an1 T n1 1 ðvÞ: Since 0 ¼ T ðn1 kÞ ðuÞ |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl}

¼ T ðn1 kÞ a1 v þ a2 TðvÞ þ a3 T 2 ðvÞ þ þ ak T k1 ðvÞ þ ak þ 1 T k ðvÞ þ þ an1 T n1 1 ðvÞ ¼ a1 T ðn1 kÞ ðvÞ þ a2 T ðn1 kÞ þ 1 ðvÞ þ a3 T ðn1 kÞ þ 2 ðvÞ þ þ ak T n1 1 ðvÞ þ ak þ 1 T n1 ðvÞ þ ak þ 2 T n1 þ 1 ðvÞ þ þ an1 T ðn1 kÞ þ ðn1 1Þ ðvÞ ðn1 kÞ ¼ a1 T ðvÞ þ a2 T ðn1 kÞ þ 1 ðvÞ þ a3 T ðn1 kÞ þ 2 ðvÞ þ þ ak T n1 1 ðvÞ þ ak þ 1 0ðvÞ þ ak þ 2 0ðvÞ þ þ an1 0ðvÞ ¼ a1 T ðn1 kÞ ðvÞ þ a2 T ðn1 kÞ þ 1 ðvÞ þ a3 T ðn1 kÞ þ 2 ðvÞ þ þ ak T n1 1 ðvÞ; we have a1 T ðn1 kÞ ðvÞ þ a2 T ðn1 kÞ þ 1 ðvÞ þ a3 T ðn1 kÞ þ 2 ðvÞ þ þ ak T n1 1 ðvÞ ¼ 0: Next, since v; TðvÞ; T 2 ðvÞ; . . .; T n1 1 ðvÞ is a basis of V1 , we have a1 ¼ 0; a2 ¼ 0; . . .; ak ¼ 0. It follows that u ¼ ak þ 1 T k ðvÞ þ ak þ 2 T k þ 1 ðvÞ þ þ an1 T n1 1 ðvÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ T k ak þ 1 v þ ak þ 2 TðvÞ þ þ an1 T n1 k1 ðvÞ ¼ T k ðak þ 1 v1 þ ak þ 2 v2 þ þ an1 vn1 k Þ ¼ T k ðu0 Þ; where u0 ðak þ 1 v1 þ ak þ 2 v2 þ þ an1 vn1 k Þ 2 ½v1 ; v2 ; . . .; vn1 ¼ V1 . Thus ■ u0 2 V1 and T k ðu0 Þ ¼ u. 3.2.24.3 There exists a linear subspace W of V, of largest possible dimension, such that 1. V ¼ V1 W, 2. W is invariant under T. Proof By 3.2.24.1, there exists a linear subspace W of V such that 1. V1 \ W ¼ f0g, 2. W is invariant under T. It suffices to show that V V1 þ W.

214


Suppose to the contrary that there exists z 2 V such that z 62 ðV1 þ W Þ. We seek a contradiction. Clearly, ðV1 þ W Þ T 1 ðV1 þ W Þ. Proof To show this, let us take arbitrary x 2 V1 and y 2 W. We have to show that ðTðxÞ þ TðyÞ ¼Þ T ðx þ yÞ 2 ðV1 þ W Þ; that is, ðTðxÞ þ TðyÞÞ 2 ðV1 þ W Þ. It suffices to show that TðxÞ 2 V1 and TðyÞ 2 W. Since V1 is invariant under T, and x 2 V1 , we have TðxÞ 2 V1 . Since W is invariant under T, and y 2 W, we have TðyÞ 2 W. ■ 1 2 1 Clearly, T ðV1 þ W Þ ðT Þ ðV1 þ W Þ. Proof To show this, let us take arbitrary x 2 V1 and y 2 W such that T ðx þ yÞ 2 ðV1 þ W Þ. We have to show that ðT 2 ðxÞ þ T 2 ðyÞ ¼ÞT 2 ðx þ yÞ 2 ðV1 þ W Þ, that is, ðT 2 ðxÞ þ T 2 ðyÞÞ 2 ðV1 þ W Þ. It suffices to show that T 2 ðxÞ 2 V1 and T 2 ðyÞ 2 W. Since V1 is invariant under T, and x 2 V1 , we have TðxÞ 2 V1 , and hence ðT 2 ðxÞ ¼ÞT ðTðxÞÞ 2 V1 : Since W is invariant under T, and y 2 W, we have TðyÞ 2 W, and hence ðT 2 ðyÞ ¼ÞT ðTðYÞÞ 2 W. ■ 2 1 3 1 Clearly, ðT Þ ðV1 þ W Þ ðT Þ ðV1 þ W Þ. Proof To show this, let us take arbitrary x 2 V1 and y 2 W such that T 2 ðx þ yÞ 2 ðV1 þ W Þ. We have to show that

T 3 ðxÞ þ T 3 ðyÞ ¼ T 3 ðx þ yÞ 2 ðV1 þ W Þ;

that is, ðT 3 ðxÞ þ T 3 ðyÞÞ 2 ðV1 þ W Þ. It suffices to show that T 3 ðxÞ 2 V1 and T 3 ðyÞ 2 W. Since V1 is invariant under T, and x 2 V1 , we have TðxÞ 2 V1 , and hence ðT 2 ðxÞ ¼ÞT ðTðxÞÞ 2 V1 . Now, ðT 3 ðxÞ ¼ÞT ðT 2 ðxÞÞ 2 V1 , so T 3 ðxÞ 2 V1 . Similarly, T 3 ðyÞ 2 W. 1 1 Thus we have shown that ðT 2 Þ ðV1 þ W Þ ðT 3 Þ ðV1 þ W Þ, etc. ■ Hence 1 1 z 62 ðV1 þ W Þ T 1 ðV1 þ W Þ T 2 ðV1 þ W Þ T 3 ðV1 þ W Þ ðT n1 Þ1 ðV1 þ W Þ ¼ V: Now, since z 2 V, there exists a positive integer k such that 1. k n1 , 1 2. z 2 T k ðV1 þ W Þ, 3. r\k ) z 62 ðT r Þ1 ðV1 þ W Þ. 1 Since z 2 T k ðV1 þ W Þ, we have T k ðzÞ 2 ðV1 þ W Þ, and hence there k exist u 2 V1 and w 2 W such that T ðzÞ ¼ u þ w. It follows that

3.2 Canonical Forms

215

0 ¼ 0ðzÞ ¼ ðT n1 ÞðzÞ ¼ T n1 k T k ðzÞ ¼ T n1 k ðu þ wÞ ¼ T n1 k ðuÞ þ T n1 k ðwÞ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence T n1 k ðuÞ ¼ T n1 k ðwÞ. Since u 2 V1 , and V1 is invariant under T, we is invariant under T, we have have T n1 k ðuÞ 2 V1 . Since w 2 W, and Wn k n1 k n1 k 1 ðwÞ 2 W, and hence T ðuÞ ¼ T ðwÞ 2 W. Thus T T n1 k ðuÞ 2 V1 \ W ¼ f0g: |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} n1 k It follows that Tk ðuÞ ¼ 0. Now by 3.2.24.2, there exists u0 2 V1 such that T ðu0 Þ ¼ u ¼ T ðzÞ w , and hence T k ðz u0 Þ ¼ w ð2 W Þ. Thus T k ðz u0 Þ 2 W. Next, since W is invariant under T, we have k

m k ) T ðmkÞ T k ðz u0 Þ 2 W ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence m k ) T m ðz u0 Þ 2 W: It follows that T k ðz u0 Þ 2 W. Let us take an arbitrary r\k. Clearly, T r ðz u0 Þ 62 ðV1 þ W Þ. (*) Proof Suppose to the contrary that ðT r ðzÞ T r ðu0 Þ ¼ÞT r ðz u0 Þ 2 ðV1 þ W Þ. We seek a contradiction. Since u0 2 V1 , and V1 is invariant under T, we have T r ðu0 Þ 2 V1 . Now, since ðT r ðzÞ T r ðu0 ÞÞ 2 ðV1 þ W Þ, we have T r ðzÞ ¼ T r ðu0 Þ þ ðT r ðzÞ T r ðu0 ÞÞ ¼ V1 þ ðV1 þ W Þ ¼ ðV1 þ V1 Þ þ W |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ V1 þ W; and hence ðT r ÞðzÞ 2 ðV1 þ W Þ. Since r\k, we have by point 3 above that ðT r ÞðzÞ 62 ðV1 þ W Þ. This is a contradiction. ■ Thus we have shown that r\k ) T r ðz u0 Þ 62 ðV1 þ W Þ: ðAÞ Clearly, ðz u0 Þ 62 ðV1 þ W Þ. Proof Suppose to the contrary that ðz u0 Þ 2 ðV1 þ W Þ. We seek a contradiction. Since u0 2 V1 and ðz u0 Þ 2 ðV1 þ W Þ, we have

216


z 2 u0 þ ðV1 þ W Þ ¼ ðu0 þ V1 Þ þ W ðV1 þ V1 Þ þ W ¼ V1 þ W; and hence z 2 ðV1 þ W Þ. This is a contradiction.

■

Thus we have shown that ðz u0 Þ 62 ðV1 þ W Þ ð f0g þ W ¼ W Þ. Hence ðz u0 Þ 62 W. Similarly, from (A), T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ 62 ðV1 þ W Þ ð f0g þ W ¼ W Þ: Thus

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ W c :

Suppose that W2 is the linear span of

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W:

Since ðz u0 Þ 62 W, the dimension of the linear span of

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W

is strictly greater than dimðWÞ, and hence dimðW2 Þ [ dimðWÞ: It follows that either ðV1 \ W2 6¼ f0gÞ or W2 is not invariant under T: ðÞ Observe that T ð z u0 Þ 2

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W W2 ;

so T ðz u0 Þ 2 W2 . Next, T ðT ðz u0 ÞÞ 2

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W W2 ;

so T ðT ðz u0 ÞÞ 2 W2 . Similarly, T ðT 2 ðz u0 ÞÞ 2 W2 ; . . .. Next, T T k1 ðz u0 ÞÞ ¼ T k ðz u0 Þ 2 W W2 , so T T k1 ðz u0 Þ 2 W2 . Thus T

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ W2 :

3.2 Canonical Forms

217

Since W is invariant under T, we have TðWÞ W W2 . Thus T ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W ¼ T ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ TðWÞ W2 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Hence T

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W W2 :

Now, since T is a linear transformation and W2 is a linear space, we have T ðW 2 Þ ¼

T ðlinear span of ðfðz u0 Þ; T ðz u0 Þ; ; T ðz u0 Þ; . . .; T k1 ðz u0 Þ [ W W2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2

and hence T ðW2 Þ W2 . This shows that W2 is invariant under T. It follows from (**) that V1 \ W2 6¼ f0g. Since V1 \ W2 6¼ f0g, there exists a nonzero ( z0 2 W2 ¼ linear span of

ðz u0 Þ; T ðz u0 Þ; T 2 ðz u0 Þ; . . .; T k1 ðz u0 Þ

)

!! [W

such that z0 2 V1 . It follows that there exist a1 ; a2 ; . . .; ak 2 F and w 2 W such that ð0 6¼Þ z0 ¼ a1 ðz u0 Þ þ a2 T ðz u0 Þ þ þ ak T k1 ðz u0 Þ þ w : Clearly, not all of a1 ; a2 ; . . .; ak are zero, Proof Suppose to the contrary that each ai is zero. We seek a contradiction. Since each ai is zero, we have V1 3 z0 ¼ 0ðz u0 Þ þ 0T ðz u0 Þ þ þ 0T k1 ðz u0 Þ þ w ¼ w ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence w 2 V1 . Also w 2 W. It follows that 2 V1 \ W ð¼ f0gÞ, and hence z0 ¼ 0. This is a contradiction.

ðz0 ¼ Þ w ■

Suppose that as is the first nonzero ai , where s k. It follows from (*) that T s1 ðz u0 Þ 62 ðV1 þ W Þ. Also

218


ð0 6¼Þ z0 ¼ as T s1 ðz u0 Þ þ as þ 1 T s ðz u0 Þ þ þ ak T k1 ðz u0 Þ þ w |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ as I þ as þ 1 T þ þ ak T ks ðT s1 ðz u0 ÞÞ þ w ; so

as I þ as þ 1 T þ þ ak T ks T s1 ðz u0 Þ ¼ z0 w :

1 By 3.2.23, as I þ as þ 1 T þ þ ak T ks exists and is a polynomial pðSÞ in S, where S as þ 1 T þ þ ak T ks : Thus ðV1 þ W Þ 63 T s1 ðz u0 Þ ¼ ðpðSÞÞðz0 w Þ ¼ ðpðSÞÞðz0 Þ ðpðSÞÞðw Þ: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Since V1 is invariant under T, V1 is invariant under ðas þ 1 T þ þ ak T ks Þ ð¼ SÞ, and hence V1 is invariant under S. Now, since pðSÞ is a polynomial in S, V1 is invariant under pðSÞ. Next, since z0 2 V1 , we have ðpðSÞÞðz0 Þ 2 V1 . Since ððpðSÞÞðz0 Þ ðpðSÞÞðw ÞÞ 62 ðV1 þ W Þ, we have ðpðSÞÞðw Þ 62 W. Since W is invariant under T, W is invariant under ðas þ 1 T þ þ ak T ks Þ ð¼ SÞ, and hence W is invariant under S. Since pðSÞ is a polynomial in S, W is invariant under pðSÞ. Next, since w 2 W, we have ðpðSÞÞðw Þ 2 W: This is a contradiction.

■

3.2.24.4 Note Let V1 ; V2 ; . . .; Vk be linear subspaces of V such that 1. V ¼ V1 V2 Vk , 2. each Vi is invariant under T.

Suppose that dimðV1 Þ ¼ n1 , and v11 ; v21 ; . . .; vn11 is a basis of V1 . Suppose that dimðV2 Þ ¼ n2 , and v12 ; v22 ; . . .; vn22 is a basis of V2 , etc. Since V ¼ V1 V2 Vk , we have n ¼ dimðVÞ ¼ n1 þ n2 þ þ nk ; and

3.2 Canonical Forms

219

v11 ; v21 ; . . .; vn11 ; v12 ; v22 ; . . .; vn22 ; . . .; v1k ; v2k ; . . .; vnk k

is a basis of V. Since v11 2 V1 , and V1 is invariant under T, we have T v11 2 V1 . Now, since v11 ; v21 ; . . .; vn11 is a basis of V1 , there exist a11 ; a21 ; . . .; an11 2 F such that T v11 ¼ a11 v11 þ a21 v21 þ þ an11 vn11 : Hence T v11 ¼ a11 v11 þ a21 v21 þ þ an11 vn11 þ 0v12 þ 0v22 þ þ 0vn22 þ þ 0v1k þ 0v2k þ þ 0vnk k : Similarly, T v21 ¼ a12 v11 þ þ an21 vn11 þ 0v12 þ þ 0vn22 þ þ 0v1k þ þ 0vnk k ; .. . n1 1 1 n T v1 ¼ an1 v1 þ þ ann11 v11 þ 0v12 þ þ 0vn22 þ þ 0v1k þ þ 0vnk k ; 1 1 T v2 ¼ 0v1 þ þ 0vn11 þ b11 v12 þ þ bn12 vn22 þ þ 0v1k þ þ 0vnk k ; T v22 ¼ 0v11 þ þ 0vn11 þ b12 v12 þ þ bn22 vn22 þ þ 0v1k þ þ 0vnk k ; .. . n2 1 T v2 ¼ 0v1 þ þ 0vn11 þ b1n2 v12 þ þ bnn22 vn22 þ þ 0v1k þ þ 0vnk k ; .. .:

Thus the matrix of T ð2 AðVÞÞ relative to the basis

v11 ; v21 ; . . .; vn11 ; v12 ; v22 ; . . .; vn22 ; . . .; v1k ; v2k ; . . .; vnk k

is the n n matrix in the canonical form 2

A1 6 0 4

0 A2

0

0

3 0 0 7 5; .. .

a21 a22 .. .

3 7 5 .. .

where 2

a11 6 a1 A1 4 2 .. .

n1 n1

;

220


2

b11 6 1 A2 4 b2 .. .

b21 b22 .. .

3 7 5 .. .

; etc:

n2 n2

Since

v11 ¼ T v11 ¼ a11 v11 þ a21 v21 þ þ an11 vn11 ; TjV1 v21 ¼ T v21 ¼ a12 v11 þ a22 v21 þ þ an21 vn11 ; .. . TjV1 vn11 ¼ T vn11 ¼ a1n1 v11 þ a2n1 v21 þ þ ann11 vn11 ; TjV1

A1 is the n1 n1 matrix of the linear transformation TjV1 induced by T on V1 . Similarly, A2 is the n2 n2 matrix of the linear transformation TjV2 induced by T on V2 , etc. 3.2.24.5 Conclusion Let V1 ; V2 ; . . .; Vk be linear subspaces of V such that 1. V ¼ V1 V2 Vk , 2. each Vi is invariant under T. Then there exist a basis v11 ; v21 ; . . .; vn11 of V1 , a basis v12 ; v22 ; . . .; vn22 of 1 2 V2 ; . . ., a basis vk ; vk ; . . .; vnk k of Vk such that the matrix of T ð2 AðVÞÞ relative to the basis v11 ; v21 ; . . .; vn11 ; v12 ; v22 ; . . .; vn22 ; . . .; v1k ; v2k ; . . .; vnk k has the canonical form 2

A1 6 0 4

0 A2

0

0

3 0 0 7 5 .. .

; nn

where A is the matrix of the linear transformation TjV1 relative to 1 2 1 n1 v ; v ; . . .; v1 , A2 is the matrix of the linear transformation TjV2 relative to 11 12 v2 ; v2 ; . . .; vn22 , etc. Also v21 ¼ T v11 , v31 ¼ T 2 v11 ; . . .; v22 ¼ T v12 , v31 ¼ T 2 v12 ; . . .; etc. Hence A1 takes the form 2

0 60 6. 6. 6. 6. 6 .. 6 6 .. 4. 0

1 0

0 1

0 .. . .. . 0

0 0 .. . 0

..

.

3 0 07 7 7 07 7 : 07 7 7 15 0 tt

3.2 Canonical Forms

221

Notation This matrix is denoted by Mt . Thus 2

0 60 6. 6. 6. Mt 6 6 ... 6 6 .. 4. 0

1 0 0 .. . .. . 0

3 0 07 7 .. 7 0 . 07 7 : 0 07 7 7 .. . 15 0 0 tt 0 1

Similarly, A2 takes the form Ms for some positive integer s, etc. 3.2.24.6 Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Let n1 be a positive integer. Let n1 be the index of nilpotence of T. By 3.2.24.3, there exist linear subspaces V1 , W of V such that 1. V ¼ V1 W, 2. V1 , W are invariant under T.

Now, by 3.2.24.4, there exists a basis v11 ; v21 ; . . .; vn11 of V1 such that for every basis B of W, the matrix of T ð2 AðVÞÞ relative to the basis v11 ; v21 ; . . .; vn11 [ B has the canonical form

Mn1 0

0 A2

; nn

where Mn1 is the n1 n1 matrix of the linear transformation TjV1 relative to the basis v11 ; v21 ; . . .; vn11 , and A2 is the ðn n1 Þ ðn n1 Þ matrix of the linear transformation TjW relative to the “arbitrary” basis B. n Since T n1 ¼ 0 and W is invariant under T, we have TjW 1 ¼ 0, and hence n there exists a smallest integer n2 such that n2 n1 and TjW 2 ¼ 0. Thus ðT2 Þn2 ¼ 0, where T2 TjW 2 AðWÞ. Also, n2 is the index of nilpotence of T2 . Again by 3.2.24.3, there exist linear subspaces V2 , X of W such that 1. W ¼ V2 X, 2. V2 , X are invariant under T2 .

Now, by 3.2.24.4, there exists a basis v12 ; v22 ; . . .; vn22 of V2 such that for every basis C of X, the matrix of T2 ð2 AðWÞÞ relative to the basis v12 ; v22 ; . . .; vn22 [ C has the canonical form

Mn2 0

0 A3

ðnn1 Þðnn1 Þ

;

222


where Mn2 is the n2 n2 matrix of the linear transformation TjV2 relative to the basis v12 ; v22 ; . . .; vn22 , and A3 is the ððn n1 Þ n2 Þ ððn n1 Þ n2 Þ matrix of the linear transformation T2 relative to the “arbitrary” basis C. Now, since A2 is the ðn n1 Þ ðn n1 Þ matrix of the linear transformation T2 relative to the “arbitrary” basis B, the matrix of T ð2 AðVÞÞ relative to the basis v11 ; v21 ; . . .; vn11 [ v12 ; v22 ; . . .; vn22 [ C has the canonical form 2

Mn1 4 0 0

3 0 0 5 : A3 nn

0 Mn2 0

We can repeat the above process finitely many times, obtaining finally the following result. 3.2.25 Conclusion Let V be any n-dimensional vector space over the field F. Let T 2 AðVÞ. Suppose that T is nilpotent. Then there exist linear subspaces V1 ; V2 ; . . .; Vk of V such that 1. V ¼ V1 V2 Vk , 2. each Vi is invariant under T. Also there exist a basis v11 ; v21 ; . . .; vn11 of V1 , a basis v12 ; v22 ; . . .; vn22 of V2 ; . . ., 1 2 a basis vk ; vk ; . . .; vnk k of Vk such that the matrix of T ð2 AðVÞÞ relative to the basis v11 ; v21 ; . . .; vn11 ; v12 ; v22 ; . . .; vn22 ; . . .; v1k ; v2k ; . . .; vnk k has the canonical form 2

Mn1

6 4 0 0

0 .. . 0

0

3

7 0 5 ; Mnk nn

is the matrix of the linear transformation TjV1 relative to where M 1 2 n1 n1 v ; v ; . . .; v1 , Mn2 is the matrix of the linear transformation TjV2 relative to 11 12 v2 ; v2 ; . . .; vn22 , etc. Further, n ¼ n1 þ þ nk and n1 n2 nk :

3.3 The Cayley–Hamilton Theorem

3.3

223

The Cayley–Hamilton Theorem

3.3.1 Definition By the transpose of an m n matrix A aij , we mean the n m matrix whose ði; jÞ-entry is aji : The transpose of A is denoted by AT . By the conjugate of an m n matrix A aij having complex numbers as entries, we mean the m n matrix whose ði; jÞ-entry is ai| where ai| denotes the complex conjugate of the complex number aij . The conjugate of A is denoted by A. By A we T mean A , and this matrix is called the conjugate transpose of A. The n n matrix whose ði; jÞ-entry is 1 if i ¼ j 0 if i 6¼ j

is denoted by In , or simply I. By a scalar matrix we mean a scalar multiple of I. By a zero matrix we mean a matrix each entry of which is 0. Definition Let A aij be a square complex matrix. If i 6¼ j ) aij ¼ 0 , then we say that A is diagonal. If i [ j ) aij ¼ 0 , then we say that A is upper triangular. If AT ¼ A; then we say that A is symmetric. If AT A ¼ AAT ¼ I, then we say that A is orthogonal. If A A ¼ AA ¼ I, then we say that A is unitary. If A ¼ A, then we say that A is Hermitian. If A A ¼ AA , then we say that A is normal. Definition Let A, B be square complex matrices of the same size. If there exists an invertible matrix P such that P1 AP ¼ B, then we say that A and B are similar. Clearly, the relation of similarity is an equivalence relation. Definition A submatrix of a matrix is obtained by suppressing some rows and/or suppressing some columns from the given matrix. Some nomenclatures are selfexplanatory. For example, if 2

1 A 44 0 is the given matrix, then 0 form matrices like

3 5

2

3 0 3

1

3

3 5 5

2

is a submatrix of A. Using submatrices, we can

½1 ½ 3 4 4 0 0 3

1 3 5

2

3 5;

224


which is an example of a partitioned form of A into a 2 2 block matrix. The manipulation of matrices in partitioned form is a basic technique in linear algebra. For example: 2

3 2 3 ½1 ½ 3 1 ½1 ½3 ½1 3 3 4 4 5 0 0 5 5 4 4 5 0 0 3 3 2 2 22 23 3 3 2 4 0 5 ½ 1 ½ 1 þ ½ 3 1 ½ 1 ½ 3 þ ½ 3 1 ½ 1 ½ 1 þ ½ 3 1 6 7 3 0 23 7 ¼6 3 3 3 4 4 5 4 4 0 4 0 5 0 5 0 5 5 ½1 þ ½3 þ ½1 þ 0 0 3 0 3 2 0 3 2 3 2 2 3 3 23 2 2 ½12 ½3þ ½ 3 ½1 þ 19 ½13 ½0 14 5 5 ½1 þ 6 5 26 5 9 69 ¼4 4 ¼4 4 : 0 12 4 5 5 5 5 þ þ þ 29 29 0 12 0 12 0 6 6 5 5 23 23

Further, 2

32 3 1 1 3 0 35 54 4 0 3 2 0 3

1 44 0

1

3

2

13 3 5 4 4 ¼ 5 12 2

0 69 5

6

3

14 5 26 5 : 5 29 5

3.3.2 Note By an elementary row operation for matrices, we mean any one of the following: 1. interchange any two rows, 2. multiply a row by a nonzero constant, 3. add a multiple of a row to another row. Definition An n n (or n-square for short) matrix E is called an elementary matrix if there exists an elementary row operation R such that E is obtained by a single application of R on the unit matrix In . For n ¼ 3, examples of elementary matrices are 2

1 40 0

0 0 1

3 2 0 1 0 1 5; 4 0 3 0 0 0

3 2 0 1 0 5; 4 0 1 0

0 1 0

3 0 5 5: 1

3.3.3 Example Observe that

1 2 4 5 1 0

3 R2 !R2 þ ð4ÞR1 1 2 3 ; ! 6 0 3 6 0 R2 !R2 þ ð4ÞR1 1 0 ; ! 4 1 1


225

and

1 4

0 1

2 3 1 ¼ 5 6 0

2 3

1 R2 !3 R2 3 1 ! 6 0

2 1

1 4

3 : 6

Again

1 2 0 3

0 R2 !31 R2 1 ! 0 1

1 0

0 1 3

3 ; 2

;

and

1 0

0

1 3

1 0

2 3

3 1 2 ¼ 6 0 1

3 : 2

Next,

1 2 0 1 1 0

3 R1 !R1 þ ð2ÞR2 1 ! 2 0 0 R1 !R1 þ ð2ÞR2 1 ! 0 1

0 1 ; 1 2 2 ; 1

and

1 0

2 1

1 0

2 3 1 0 ¼ 1 2 0 1

1 : 2

1 0 Now we apply certain elementary column operations on 0 1 things simpler. The following are self-explanatory:

1 0 1 C3 !C3 þ ð2ÞC2 1 0 1 ; ! 20 1 23 2 0 1 03 1 0 0 1 0 0 3 !C3 þ ð2ÞC2 4 0 1 0 5 C ! 4 0 1 2 5; 0 0 12 3 0 0 1 1 0 0 1 0 1 4 1 0 1 0 1 2 5 ¼ : 0 1 2 0 1 0 0 0 1

1 to make 2

226


Next,

1 0 0 2 1 1 0 40 1 0 0

1 0

0 1

1 C3 !C3 þ 1C1 1 0 0 ; ! 03 2 0 1 03 0 1 0 1 C3 !C3 þ 1C1 0 5 !4 0 1 0 5; 1 2 3 0 0 1 1 0 1 1 0 0 1 4 0 1 05 ¼ : 0 0 1 0 0 0 1

If we collect the above information, we get

3 R2 !R2 þ ð4ÞR1 1 ! 6 0

1 2 4 5

R1 !R1 þ ð2ÞR2

!

1 2

1 0 0 1

1 R2 !3 R2 1 2 3 ! 3 6 0

C3 !C3 þ ð2ÞC2

!

1 0

0 1

2 3 1 2

1 C3 !C3 þ 1C1 1 ! 0 0

0 1

0 0

and

1 0

¼

2 1 1 0

1 0

0 1 3

0 : 0

0 1

1 4

0 1

1 4

2 5

02

1 0 3 @4 0 1 6 0 0

32 0 1 0 2 54 0 1 1 0 0

31 1 0 5A 1

3.3.4 Conclusion Let A be a nonzero m n matrix with complex numbers as entries. Then there exist an invertible m m matrix P, an invertible n n matrix Q, and a positive integer r such that Ir 0 1. PAQ ¼ ; 0 0 mn 2. P is a product of elementary matrices of size m m, 3. Q is a product of elementary matrices of size n n. 4. r is the rank of A. 3.3.5 Observe that T Ir 0 I Q A P ¼ ðPAQÞ ¼ ¼ r 0 0 mn 0 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} T

T

T

T

0 0

; nm


227

so

I R A S¼ r 0 T

0 0

; nm

where R QT and S PT . Since Q is invertible, ðR ¼Þ QT is invertible, and hence R is invertible. Similarly, S is invertible. It follows that rankðAT Þ ¼ r ð¼ rankðAÞÞ. Thus rankðAT Þ ¼ rankðAÞ. Similarly, rank A ¼ rankðAÞ and rankðA Þ ¼ rankðAÞ. A B 3.3.6 Note It is easy to see that for a partitioned matrix , we have 0 C det

A 0

B ¼ detðAÞ detðCÞ: C

It is also written as A 0

B ¼ detðAÞ detðCÞ: C

3.3.7 Example 2 3 2 3 a11 a12 a13 b11 b12 4 a21 a22 a23 5 4 b21 b22 5 a31 a32 a33 b31 b32 c11 c12 0 0 0 c21 c22 0 0 0 a11 a12 a13 b12 a11 a12 a13 b11 a a a b a a a b ¼ c21 21 22 23 22 þ c22 21 22 23 21 a31 a32 a33 b32 a31 a32 a33 b31 0 0 0 c12 1 1 0 00 0 0 c11 a11 a12 a13 a11 a12 a13 ¼ c21 @c12 a21 a22 a23 A þ c22 @c11 a21 a22 a23 A a31 a32 a33 a31 a32 a33 a11 a12 a13 a11 a12 a13 c c ¼ a21 a22 a23 ðc11 c22 c12 c21 Þ ¼ a21 a22 a23 11 12 ; a31 a32 a33 a31 a32 a33 c21 c22 and hence

228


2 3 2 a11 a12 a13 b11 4 a21 a22 a23 5 4 b21 a31 a32 a33 b31 0 0 0 c11 0 0 0 c21

3 b12 b22 5 a11 b32 ¼ a21 c12 a31 c22

a12 a22 a32

a13 c a23 11 c a33 21

c12 : c22

3.3.8 Problem Let V be an n-dimensional vector space over C. Let A : V ! V be a linear transformation. Let k1 and k2 be distinct eigenvalues of A. Let v1 be an eigenvector corresponding to the eigenvalue k1 . Let v2 be an eigenvector corresponding to the eigenvalue k2 . Then v1 ; v2 are linearly independent. Proof Suppose to the contrary that v1 ¼ kv2 , where k is a complex number. We seek a contradiction. Since v2 is an eigenvector corresponding to the eigenvalue k2 , we have v2 6¼ 0. Similarly,v1 6¼ 0. Now, since v1 ¼ kv2 , we have k 6¼ 0. Also, Aðv1 Þ ¼ k1 v1 and Aðv2 Þ ¼ k2 v2 . Hence kAðv2 Þ ¼ Aðkv2 Þ ¼ Aðv1 Þ ¼ k1 v1 ¼ k1 kv2 : |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} Since kAðv2 Þ ¼ k1 kv2 and k is nonzero, we have k2 v2 ¼ Aðv2 Þ ¼ k1 v2 , and |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} hence ðk2 k1 Þv2 ¼ 0. Since k1 and k2 are distinct, k2 k1 is nonzero. Now, since ðk2 k1 Þv2 ¼ 0, we have v2 ¼ 0. This is a contradiction. ■ 3.3.9 Conclusion The eigenvectors corresponding to distinct eigenvalues are linearly independent. 3.3.10 Note By Cn we shall mean the collection of all column matrices of size n 1 with complex entries. Such matrices are also called column vectors. We know that Cn is a vector space over C. For every x ½x1 ; . . .; xn T and y ½y1 ; . . .; yn T in Cn , we define hx; yi x1 y1 þ . . . þ xn yn ¼ xT y ¼ y x : Clearly, Cn is an inner product space. For every x ½x1 ; . . .; xn T in Cn , we define the length x of x as follows: k xk

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi jx1 j2 þ þ jxn j2 ¼ hx; xi :

In Cn , an m-tuple ðv1 ; . . .; vm Þ of vectors in Cn can be thought of as a matrix vij nm , where v1 ½v11 ; . . .; vn1 T , v1 ½v12 ; . . .; vn2 T , etc. 3.3.11 Note Let x be a nonzero vector of Cn .


229

There exist distinct nonzero vectors v2 ; . . .; vn in Cn such that fx; v2 ; . . .; vn g is a basis of Cn . Assume that for some complex number a, ax þ v2 is orthogonal to x, that is, ðahx; xi þ hv2 ; xi ¼Þ hax þ v2 ; xi ¼ 0: Since x is nonzero, hx; xi is nonzero. This shows that to x. Let us put y2

hv2 ;xi hx;xi x þ v2

is orthogonal

h v2 ; xi x þ v2 : hx; xi

Since fx; v2 ; . . .; vn g is a basis of Cn ,

n o v2 ;xi x; hhx;x is a basis of i x þ v2 ; v3 ; . . .; vn

v2 ;xi Cn , and hence fx; y2 ; v3 ; . . .; vn g is a basis of Cn . Since, ðy2 ¼Þ hhx;x i x þ v2 is orthogonal to x, it follows that y2 is orthogonal to x. Thus fx; y2 ; v3 ; . . .; vn g is a basis of Cn such that y2 is orthogonal to x. Assume that for some complex numbers a; b, ax þ by2 þ v3 is orthogonal to x and y2 , that is,

ðahx; xi þ hv3 ; xi ¼ ahx; xi þ b0 þ hv3 ; xi ¼ ahx; xi þ bhy2 ; xi þ hv3 ; xi ¼Þ hax þ by2 þ v3 ; xi ¼ 0 and ðbhy2 ; y2 i þ hv3 ; y2 i ¼ a0 þ bhy2 ; y2 i þ hv3 ; y2 i ¼ ahx; y2 i þ bhy2 ; y2 i þ hv3 ; y2 i ¼Þ hax þ by2 þ v3 ; y2 i ¼ 0: hv3 ;y2 i 3 ;xi It follows that hhvx;x i x hy2 ;y2 i y2 þ v3 is orthogonal to x and y2 . Let us put

y3

h v3 ; y2 i h v3 ; xi x y2 þ v3 : hx; xi h y2 ; y2 i n

hv3 ;y2 i 3 ;xi Since fx; y2 ; v3 ; . . .; vn g is a basis of Cn , x; y2 ; hhvx;x i x hy2 ;y2 i y2 þ v3 ; v4 ; . . .; vn

o

is a basis of Cn , and hence fx; y2 ; y3 ; v4 ; . . .; vn g is a basis of Cn . Since ðy3 ¼Þ hv3 ;xi hx;xi x

hhyv32 ;y;y22 ii y2 þ v3 is orthogonal to x and y2 , it follows that y3 is orthogonal to x and y2 . Thus fx; y2 ; y3 ; v4 ; . . .; vn g is a basis of Cn such that the members of fx; y2 ; y3 g are mutually orthogonal, etc. Finally, we get a basis fx; y2 ; y3 ; y4 ; . . .; yn g of Cn such that the members of fx; y2 ; y3 ; y4 ; . . .; yn g are mutually orthogonal.

230


n It follows that

1 1 1 1 1 kxk x; ky2 k y2 ; ky3 k y3 ; ky4 k y4 ; . . .; kyn k yn

o is an orthonormal basis

of C . n

3.3.12 Conclusion Let u1 ; . . .; uk be orthonormal vectors in Cn . Then there exist uk þ 1 ; . . .; un in Cn such that fu1 ; . . .; uk ; uk þ 1 ; . . .; un g is an orthonormal basis of |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} nk

Cn . By the definition of unitary matrix ðA A ¼ AA ¼ I Þ, the n-square matrix ½u1 ; . . .; uk ; uk þ 1 ; . . .; un is unitary. The construction used here is known as the Gram–Schmidt orthogonalization process. 3.3.13 Note Let A aij be an n-square complex matrix. For every x ½x1 ; . . .; xn T in Cn , the product Ax of matrices A and x is a matrix of size n 1, and hence Ax is a member of Cn . Let us define a function A : Cn ! Cn as follows: for every x ½x1 ; . . .; xn T in Cn , AðxÞ Ax: Clearly, A : Cn ! Cn is a linear transformation. 3.3.14 Note Let A : Cn ! Cn be a linear transformation. Put e1 ½1; 0; . . .; 0T ð2 Cn Þ; e2 ½0; 1; 0; . . .; 0T ð2 Cn Þ; etc: n Clearly, fe1 ; . . .; en g is an orthonormal basis of the inner product space C . This n basis is called the standard basis of C : Let A aij be the matrix of A relative to the basis fe1 ; . . .; en g. Hence

0

2

3 1 a11 B 6 . 7 TC 7 C ai1 ei ¼ 6 Aðe1 Þ ¼ a11 e1 þ a21 e2 þ þ an1 en B @¼ 4 .. 5 ¼ ½a11 ; . . .; an1 A; i¼1 an1 0 2 3 1 a12 n B X 6 . 7 C TC 7 Aðe2 Þ ¼ a12 e1 þ a22 e2 þ þ an2 an B ai2 ei ¼ 6 @¼ 4 .. 5 ¼ ½a12 ; . . .; an2 A; i¼1 an2 n X

etc: T In short, A ej ¼ a1j ; . . .; anj . By 3.3.13, b : x 7! Ax A


231

is a linear transformation from Cn to Cn . It follows that 2

3 a11 6 7 b ðe1 Þ ¼ Ae1 ¼ 6 .. 7 ¼ Aðe1 Þ; A 4 . 5 an1 2 3 a12 6 7 b ðe2 Þ ¼ Ae2 ¼ 6 .. 7 ¼ Aðe2 Þ; etc: A 4 . 5 an2 b ¼ A. It follows that A

3.3.15 Conclusion The n-square complex matrix A aij can be represented by the linear transformation x 7! Ax from Cn to Cn . 3.3.16 Note Let a1 ; a2 ; a3 ; a4 be any complex numbers. Observe that 1 a1

1 ¼ ða2 a1 Þ: a2

Next, 1 1 1 a2 a3 a1 ða1 Þ2 ða2 Þ2 ða3 Þ2 1 1 1 ¼ a1 a2 a3 ðR3 ! R3 a1 R2 Þ 0 a2 ð a2 a1 Þ a3 ð a3 a1 Þ 1 1 1 ¼ 0 a2 a1 a3 a1 ðR2 ! R2 a1 R1 Þ 0 a2 ð a2 a1 Þ a3 ð a3 a1 Þ 1 1 1 ¼ ða2 a1 Þða3 a1 Þ 0 1 1 0 a2 a3 1 1 ¼ ða2 a1 Þða3 a1 Þða3 a2 Þ; ¼ ða2 a1 Þða3 a1 Þ a a 2

3

232


so 1 a1 ða1 Þ2

1 a2 ða2 Þ2

1 a3 ¼ ða2 a1 Þða3 a1 Þða3 a2 Þ: ð a3 Þ 2

Again, 1 1 1 1 a a2 a3 a4 1 ða1 Þ2 ða2 Þ2 ða3 Þ2 ða4 Þ2 ða Þ3 ða Þ3 ða Þ3 ða Þ3 1 2 3 4 0 1 1 1 1 1 R4 ! R4 a1 R3 0 ða a Þ ð a3 a1 Þ ð a4 a1 Þ B 2 1 C ¼ @ R3 ! R3 a1 R2 A 0 a2 ð a2 a1 Þ a3 ð a3 a 1 Þ a 4 ð a4 a1 Þ 0 ða Þ2 ða a Þ ða Þ2 ða a Þ ða Þ2 ða a Þ R2 ! R2 a1 R1 2 2 1 3 3 1 4 4 1 1 1 1 1 0 1 1 1 ¼ ða2 a1 Þða3 a1 Þða4 a1 Þ 0 a2 a3 a4 0 ða Þ2 ða Þ2 ða Þ2 2 3 4 1 1 1 ¼ ða2 a1 Þða3 a1 Þða4 a1 Þ a2 a3 a4 ða2 Þ2 ða3 Þ2 ða4 Þ2 ¼ ða2 a1 Þða3 a1 Þða4 a1 Þ ða3 a2 Þða4 a2 Þða4 a3 Þ; so 1 1 1 1 a a2 a3 a4 1 ða1 Þ2 ða2 Þ2 ða3 Þ2 ða4 Þ2 ða Þ3 ða Þ3 ða Þ3 ða Þ3 1 2 3 4 ¼ ða2 a1 Þða3 a1 Þða3 a2 Þða4 a1 Þða4 a2 Þða4 a3 Þ; etc: 3.3.17 Conclusion If each ai is a complex number, then

1 a1 .. .

1 a2 .. .

ða1 Þn1 ða2 Þn1

Y ¼ aj ai : nj[i1 n1

1 an .. .

ð an Þ


233

3.3.18 Note Let A aij be an n-square complex matrix. Let k1 ; . . .; kn be the eigenvalues of A. Suppose that all the eigenvalues of A are distinct. Let B bij be an n-square complex matrix such that AB ¼ BA, that is, B commutes with A. Since k1 is an eigenvalue of A, there exists a nonzero u1 2 Cn such that Au1 ¼ k1 u1 . Similarly, there exists a nonzero u2 2 Cn such that Au2 ¼ k2 u2 , etc. Since all the eigenvalues of A are distinct, by 3.3.9, fu1 ; . . .; un g is linearly independent, and hence fu1 ; . . .; un g is a basis of Cn . It follows that the n n matrix ðu1 ; . . .; un Þ in invertible. Next, ðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 ðAðu1 ; . . .; un ÞÞ ¼ ðu1 ; . . .; un Þ1 ðAu1 ; . . .; Aun Þ ¼ ðu1 ; . . .; un Þ1 ðk1 u1 ; . . .; kn un Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðu1 ; . . .; un Þ1 ðk1 u1 Þ; . . .; ðu1 ; . . .; un Þ1 ðkn un Þ ¼ k1 ðu1 ; . . .; un Þ1 u1 ; . . .; kn ðu1 ; . . .; un Þ1 un ¼ ðk1 e1 ; . . .; kn en Þ ¼ diagðk1 ; . . .; kn Þ; so ðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ¼ diagðk1 ; . . .; kn Þ: Next, ðdiagðk1 ; . . .; kn ÞÞ ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 A ðu1 ; . . .; un Þðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 AIBðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 ABðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 BAðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 BIAðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 B ðu1 ; . . .; un Þðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ¼ ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ðdiagðk1 ; . . .; kn ÞÞ;

:

so ðdiagðk1 ; . . .; kn ÞÞD ¼ Dðdiagðk1 ; . . .; kn ÞÞ; where D ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ: Suppose that D ðv1 ; . . .; vn Þ, where each vi ½v1i ; . . .; vni T is in Cn . It follows that

234


ðdiagðk1 ; . . .; kn ÞÞD ¼ ðdiagðk1 ; . . .; kn ÞÞðv1 ; . . .; vn Þ ¼ ððdiagðk1 ; . . .; kn ÞÞv1 ; . . .; ðdiagðk1 ; . . .; kn ÞÞvn Þ ¼ ðdiagðk1 ; . . .; kn ÞÞ½v11 ; . . .; vn1 T ; . . .; ðdiagðk1 ; . . .; kn ÞÞ½v1n ; . . .; vnn T ¼ ½k1 v11 ; . . .; kn vn1 T ; . . .; ½k1 v1n ; . . .; kn vnn T 2 3 k1 v11 k1 v1n 6 . .. 7 .. 7 ¼6 . 5; . 4 .. kn vn1 kn vnn and Dðdiagðk1 ; . . .; kn ÞÞ ¼ ðv1 ; . . .; vn Þðdiagðk1 ; . . .; kn ÞÞ ¼ ðv1 ; . . .; vn Þðk1 e1 ; . . .; kn en Þ ¼ ððv1 ; . . .; vn Þðk1 e1 Þ; . . .; ðv1 ; . . .; vn Þðkn en ÞÞ ¼ ðk1 ðv1 ; . . .; vn Þe1 ; . . .; kn ðv1 ; . . .; vn Þen Þ ¼ ðk1 v1 ; . . .; kn vn Þ ¼ k1 v1 þ þ kn vn 2 3 k1 v11 kn v1n 6 . .. 7 .. 7 ¼6 . 5: . 4 .. k1 vn1 kn vnn Next, since 2

k1 v11 6 .. 4 .

..

kn vn1

.

3 k1 v1n .. 7 diagðk1 ; . . .; kn ÞÞD ¼ Dðdiagðk1 ; . . .; kn ÞÞ . 5 ¼ ð|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} kn vnn 2 3 k1 v11 kn v1n 6 . .. 7 .. ¼ 4 .. . 5; . k1 vn1 kn vnn

we have 2

k1 v11 6 .. 4 .

kn vn1

3 2 k1 v1n k1 v11 .. 7 6 .. .. . 5¼4 . . k1 vn1 kn vnn

..

.

3 kn v1n .. 7 . 5: kn vnn

Now, since all the eigenvalues k1 ; . . .; kn of A are distinct, i 6¼ j ) vij ¼ 0. Next, since D ¼ ðv1 ; . . .; vn Þ; and each vi equals ½v1i ; . . .; vni T , it follows that


235

ðu1 ; . . .; un Þ1 Bðu1 ; . . .; un Þ ¼ D is a diagonal matrix, and hence P1 BP is a

diagonal matrix, where P ðu1 ; . . .; un Þ. Since P1 BP is a diagonal matrix, we can suppose that P1 BP diagðl1 ; . . .; ln Þ; where each li is a complex number. Let us consider the following system of linear equations in n variables x0 ; x1 ; . . .; xn1 : 9 1x0 þ k1 x1 þ ðk1 Þ2 x2 þ þ ðk1 Þn1 xn1 ¼ l1 > > > 1x0 þ k2 x1 þ ðk2 Þ2 x2 þ þ ðk2 Þn1 xn1 ¼ l2 = : .. > > . > ; 1x0 þ kn x1 þ ðkn Þ2 x2 þ þ ðkn Þn1 xn1 ¼ ln Since k1 ; . . .; kn are distinct, by 3.3.17, we have 1 1 .. . 1

k1 ðk1 Þ2 ðk1 Þn1 Y k2 ðk2 Þ2 ðk2 Þn1 kj ki 6¼ 0; ¼ .. .. .. nj[i1 . . . 2 n1 kn ðkn Þ ð kn Þ

and the above system of linear equations has a unique solution ðx0 ; x1 ; . . .; xn1 Þ ¼ ða0 ; a1 ; . . .; an1 Þ. Hence 9 1a0 þ k1 a1 þ ðk1 Þ2 a2 þ þ ðk1 Þn1 an1 ¼ l1 > > > 1a0 þ k2 a1 þ ðk2 Þ2 a2 þ þ ðk2 Þn1 an1 ¼ l2 = ; .. > > . > ; 1a0 þ kn a1 þ ðkn Þ2 a2 þ þ ðkn Þn1 an1 ¼ ln that is, 9 1a0 þ a1 k1 þ a2 ðk1 Þ2 þ þ an1 ðk1 Þn1 ¼ l1 > > > 1a0 þ a1 k2 þ a2 ðk2 Þ2 þ þ an1 ðk2 Þn1 ¼ l2 = : .. > > . > ; 1a0 þ a1 kn þ a2 ðkn Þ2 þ þ an1 ðkn Þn1 ¼ ln Let us denote the polynomial

236


1a0 þ a1 x þ a2 x2 þ þ an1 xn1 by pðxÞ. Observe that degðpðxÞÞ ðn 1Þ. Also, 9 p ð k 1 Þ ¼ l1 > > > p ð k 2 Þ ¼ l2 = : .. > . > > ; p ð k n Þ ¼ ln Next, pðAÞ ¼ a0 I þ a1 A þ a2 A2 þ þ an1 An1 and pðP1 APÞ ¼ a0 I þ a1 ðP1 APÞ þ a2 ðP1 APÞðP1 APÞ þ þ an1 ðP1 APÞ ¼ a0 ðP1 IPÞ þ a1 ðP1 APÞ þ a2 ðP1 A2 PÞ þ þ an1 ðP1 An1 PÞ ¼ P1 ða0 I þ a1 A þ a2 A2 þ þ an1 An1 ÞP ¼ P1 pðAÞP;

n1

so pðP1 APÞ ¼ P1 pðAÞP. Since 1 P AP ¼ ðu1 ; . . .; un Þ1 Aðu1 ; . . .; un Þ ¼ diagðk1 ; . . .; kn Þ; we have P1 pðAÞP ¼ p P1 AP ¼ pðdiagðk1 ; . . .; kn ÞÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a0 I þ a1 diagðk1 ; . . .; kn Þ þ a2 ðdiagðk1 ; . . .; kn ÞÞ2 þ þ an1 ðdiagðk1 ; . ..; kn ÞÞn1

¼ a0 diagð1; . . .; 1Þ þ a1 diagðk1 ; . . .; kn Þ þ a2 diag ðk1 Þ2 ; . . .; ðkn Þ2 þ þ an1 diag ðk1 Þn1 ; . . .; ðkn Þn1 ¼ diagða0 ; . . .; a0 Þ þ diagða1 k1 ; . . .; a1 kn Þ þ diag a2 ðk1 Þ2 ; . . .; a2 ðkn Þ2 þ þ diag an1 ðk1 Þn1 ; . . .; an1 ðkn Þn1 ¼ diag

a0 þ a1 k1 þ a2 ðk1 Þ2 þ þ an1 ðk1 Þn1 ; . . .; a0 þ a1 kn þ a2 ðkn Þ2 þ þ an1 ðkn Þn1 ¼ diagðpðk1 Þ; . . .; pðkn ÞÞ ¼ diagðl1 ; . . .; ln Þ ¼ P1 BP;

!


237

and hence P1 pðAÞP ¼ P1 BP: It follows that pðAÞ ¼ B.

3.3.19 Conclusion Let A aij be an n-square complex matrix. Suppose that all the eigenvalues of A are distinct. Let B bij be an n-square complex matrix such that B commutes with A. Then there exists a polynomial pðxÞ such that 1. degðpðxÞÞ n 1, 2. pðAÞ ¼ B. 3.3.20 Problem Let A aij and B bij be any n-square complex matrices. Suppose that A commutes with B, that is, AB ¼ BA. Then there exists a unitary matrix U such that U AU and U BU are both upper triangular matrices. Proof (Induction on n) The assertion is trivially true for n ¼ 1. Next, suppose that the assertion is true for n 1. We have to show that the assertion is true for n. Let us take any eigenvalue l of B. Clearly, fv : v 2 Cn and Bv ¼ lvg is a linear subspace of Cn such that its dimension is 1: Let A : v 7! Av be a mapping from Cn to Cn . Clearly, A : Cn ! Cn is a linear transformation. Observe that for every v 2 Cn satisfying Bv ¼ lv, we have BðAðvÞÞ ¼ BðAvÞ ¼ ðBAÞv ¼ ðABÞv ¼ AðBvÞ ¼ AðlvÞ ¼ lðAvÞ ¼ lðAðvÞÞ; and hence BðAðvÞÞ ¼ lðAðvÞÞ. This shows that the subspace fv : v 2 Cn and Bv ¼ lvg is invariant under A : Cn ! Cn . Hence the restriction AjVl : Vl ! Vl ; where Vl fv : v 2 Cn and Bv ¼ lvg, is a linear transformation. Also, dim Vl 1. Let k be an eigenvalue of AjVl . Then there exists a nonzero vector v1 in Vl such that Av1 ¼ Aðv1 Þ ¼ AjVl ðv1 Þ ¼ kv1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence Av1 ¼ kv1 . Next, since v1 is in Vl ð¼ fv : v 2 Cn and Bv ¼ lvgÞ, we have Bv1 ¼ lv1 . Since v1 6¼ 0, w1 v11 v1 is a unit vector. Also, Aw1 ¼ kw1 and Bw1 ¼ lw1 . By 3.3.12, there exist w2 ; . . .; wn in Cn such that fw1 ; w2 ; . . .; wn g is an |fflfflfflfflfflffl{zfflfflfflfflfflffl} n1

238


orthonormal basis of Cn . By the definition of unitary matrix ðA A ¼ AA ¼ I Þ, the n-square matrix ½w1 ; w2 ; . . .; wn is unitary. Observe that ½w1 ; w2 ; . . .; wn A½w1 ; w2 ; . . .; wn ¼ ½w1 ; w2 ; . . .; wn ðA½w1 ; w2 ; . . .; wn Þ ¼ ½w1 ; w2 ; . . .; wn ½Aw1 ; Aw2 ; . . .; Awn ¼ ½w1 ; w2 ; . . .; wn ½kw1 ; Aw2 ; . . .; Awn

¼ ½w1 ; w2 ; . . .; wn T ½kw1 ; Aw2 ; . . .; Awn ¼ ½w1 ; w2 ; . . .; wn T ðkw1 Þ; ½w1 ; w2 ; . . .; wn T ðAw2 Þ; . . .; ½w1 ; w2 ; . . .; wn T ðAwn Þ ¼ k ½w1 ; w2 ; . . .; wn T w1 ; ½w1 ; w2 ; . . .; wn T ðAw2 Þ; . . .; ½w1 ; w2 ; . . .; wn T ðAwn Þ ¼ k½hw1 ; w1 i; hw1 ; w2 i; . . .; hw1 ; wn iT ; ½hAw2 ; w1 i; hAw2 ; w2 i; . . .; hAw2 ; wn iT ; . . .; ½hAwn ; w1 i; hAwn ; w2 i; . . .; hAwn ; wn iT ¼ k½1; 0; . . .; 0T ; ½hAw2 ; w1 i; hAw2 ; w2 i; . . .; hAw2 ; wn iT ; . . .; ½hAwn ; w1 i; hAwn ; w2 i; . . .; hAwn ; wn iT ¼ ½k; 0; . . .; 0T ; ½hAw2 ; w1 i; hAw2 ; w2 i; . . .; hAw2 ; wn iT ; . . .; ½hAwn ; w1 i; hAwn ; w2 i; . . .; hAwn ; wn iT ; and ½k; 0; . . .; 0T ; ½hAw2 ; w1 i; hAw2 ; w2 i; . . .; hAw2 ; wn iT ; . . .; ½hAwn ; w1 i; hAwn ; w2 i; . . .; hAwn ; wn iT is of the form 2

k 60 6. 4 ..

.. .

0

3 7 .. 7 ; .5

so ½w1 ; w2 ; . . .; wn A½w1 ; w2 ; . . .; wn is of the partitioned form

k 0

a ; C

where C is a matrix of size ðn 1Þ ðn 1Þ, and a is a matrix of size 1 ðn 1Þ.


239

Next, ½w1 ; w2 ; . . .; wn B½w1 ; w2 ; . . .; wn ¼ ½w1 ; w2 ; . . .; wn ðB½w1 ; w2 ; . . .; wn Þ

¼ ½w1 ; w2 ; . . .; wn ½Bw1 ; Bw2 ; . . .; Bwn

¼ ½w1 ; w2 ; . . .; wn ½lw1 ; Bw2 ; . . .; Bwn

¼ ½w1 ; w2 ; . . .; wn T ½lw1 ; Bw2 ; . . .; Bwn ¼ ½w1 ; w2 ; . . .; wn T ðlw1 Þ; ½w1 ; w2 ; . . .; wn T ðBw2 Þ; . . .; ½w1 ; w2 ; . . .; wn T ðBwn Þ ¼ l ½w1 ; w2 ; . . .; wn T w1 ; ½w1 ; w2 ; . . .; wn T ðBw2 Þ; . . .; ½w1 ; w2 ; . . .; wn T ðBwn Þ ¼ l½hw1 ; w1 i; hw1 ; w2 i; . . .; hw1 ; wn iT ; ½hBw2 ; w1 i; hBw2 ; w2 i; . . .; hBw2 ; wn iT ; . . .; ½hBwn ; w1 i; hBwn ; w2 i; . . .; hBwn ; wn iT ¼ l½1; 0; . . .; 0T ; ½hBw2 ; w1 i; hBw2 ; w2 i; . . .; hBw2 ; wn iT ; . . .; ½hBwn ; w1 i; hBwn ; w2 i; . . .; hBwn ; wn iT ¼ ½l; 0; . . .; 0T ; ½hBw2 ; w1 i; hBw2 ; w2 i; . . .; hBw2 ; wn iT ; . . .; ½hBwn ; w1 i; hBwn ; w2 i; . . .; hBwn ; wn iT and

½l; 0; . . .; 0T ; ½hBw2 ; w1 i; hBw2 ; w2 i; . . .; hBw2 ; wn iT ; . . .; ½hBwn ; w1 i; hBwn ; w2 i; . . .; hBwn ; wn iT

is of the form 2

l 60 6. 4 ..

.. .

0

3 7 .. 7 ; .5

so ½w1 ; w2 ; . . .; wn B½w1 ; w2 ; . . .; wn is of the partitioned form

l 0

b ; D

240


where D is a matrix of size ðn 1Þ ðn 1Þ, and b is a matrix of size 1 ðn 1Þ. It follows that

lk

la þ bC

l

b

k

a

¼ 0 DC 0 D 0 C ¼ ð½w1 ; w2 ; . . .; wn B½w1 ; w2 ; . . .; wn Þð½w1 ; w2 ; . . .; wn A½w1 ; w2 ; . . .; wn Þ ¼ ½w1 ; w2 ; . . .; wn Bð½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn ÞA½w1 ; w2 ; . . .; wn

¼ ½w1 ; w2 ; . . .; wn BIA½w1 ; w2 ; . . .; wn ¼ w1 ; w2 ; . . .; wn BA½w1 ; w2 ; . . .; wn ¼ ½w1 ; w2 ; . . .; wn AB½w1 ; w2 ; . . .; wn ¼ ½w1 ; w2 ; . . .; wn AIB½w1 ; w2 ; . . .; wn

¼ ½w1 ; w2 ; . . .; wn Að½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn ÞB½w1 ; w2 ; . . .; wn

ð½w1 ; w2 ; . . .; wn A½w1 ; w2 ; . . .; wn Þð½w1 ; w2 ; . . .; wn B½w1 ; w2 ; . . .; wn Þ # " #" ¼ l b k a ¼ 0 D 0 C |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} kl kb þ aD ¼ ; 0 CD and hence

lk 0

la þ bC DC

¼

kl 0

kb þ aD : CD

This shows that CD ¼ DC. Further, C and D are ðn 1Þ-square complex matrices. By the induction hypothesis, there exists a unitary matrix V of size ðn 1Þ ðn 1Þ such that V CV and V DV are both upper triangular matrices of size ðn 1Þ ðn 1Þ. It follows that

1 0

0 V

is a partitioned form of an n n matrix, and hence 1 ½w1 ; w2 ; . . .; wn 0

0 V


241

is an n n matrix. Next,

1 0 1 0 ½w1 ; w2 ; . . .; wn A ½w1 ; w2 ; . . .; wn 0 V 0 V

1 0 1 0 ½w1 ; w2 ; . . .; wn A ½w1 ; w2 ; . . .; wn ¼ 0 V 0 V

1 0 1 0 ¼ ½ ; w ; . . .; w ½ ; w ; . . .; w w A w 1 2 n 1 2 n 0 V 0 V 1 0 1 0 ¼ ð½w1 ; w2 ; . . .; wn A½w1 ; w2 ; . . .; wn Þ 0 V 0 V 1 0 k a 1 0 ¼ 0 V 0 C 0 V

1 0 k a 1 0 ¼ 0 V 0 C 0 V 1 0 k aV k aV ¼ ¼ ; 0 V 0 CV 0 V CV so

½w1 ; w2 ; . . .; wn

1 0

0 V

1 A ½w1 ; w2 ; . . .; wn 0

0 V

¼

k aV : 0 V CV

Now, since V CV is an upper triangular matrix of size ðn 1Þ ðn 1Þ, k aV is an upper triangular matrix of size n n, and hence U AU is an 0 V CV upper triangular matrix of size n n, where

U ¼ ½w1 ; w2 ; . . .; wn

1 0

0 : V

Similarly, U BU is an upper triangular matrix of size n n. It remains to show that U is unitary, that is, U U ¼ UU ¼ I. Observe that

242


1 0 1 0 U U ¼ ½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn 0 V 0 V

1 0 1 0 ¼ ½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn 0 V

0 V

1 0 1 0 ¼ ½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn 0 V 0 V 1 0 1 0 ¼ ð½w1 ; w2 ; . . .; wn ½w1 ; w2 ; . . .; wn Þ 0 V 0 V 1 0 1 0 1 0 1 0 1 0 1 0 ¼ ¼ I ¼ ¼ ¼ I; 0 V 0 In1 0 V 0 V 0 V 0 V V

■ so U U ¼ I. Similarly, UU ¼ I. 3.3.21 Problem Let A aij be an n-square complex matrix. Then there exists a unitary matrix U such that U AU is an upper triangular matrix. Proof Since A commutes with A, by 3.3.20, there exists a unitary matrix U such that U AU is an upper triangular matrix. ■ 3.3.22 Theorem Let A aij be an n-square complex matrix. Then there exists a unitary matrix U such that 1. U AU is an upper triangular matrix, 2. the eigenvalues of A are the diagonal entries of U AU. This result is due to Issai Schur (1875–1941). Proof By 3.3.21, there exists a unitary matrix U such that U AU is an upper triangular matrix, say C. Since U is unitary, U U ¼ UU ¼ I, we have U 1 ¼ U . Thus C ¼ U 1 AU, and hence A ¼ UCU 1 . Now, A kI ¼ UCU 1 kI ¼ UCU 1 kUU 1 ¼ ðUC kU ÞU 1 ¼ ðUC kðUI ÞÞU 1 ¼ ðUC U ðkI ÞÞU 1 ¼ U ðC kI ÞU 1 ; so ðA kI Þ ¼ U ðC kI ÞU 1 , and hence detðA kI Þ ¼ det U ðC kI ÞU 1 ¼ detðUÞ detðC kI Þ det U 1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ detðUÞ detðC kI Þ ¼ detðC kI Þ:

1 detðUÞ


243

Thus detðA kI Þ ¼ detðC kI Þ: Since C is an upper triangular matrix, we have detðA kI Þ ¼ detðC kI Þ ¼ ðc1 kÞðc2 kÞ ðcn kÞ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} where the diagonal entries of ðU AU ¼ÞC are c1 ; c2 ; . . .; cn . Also, detðA kI Þ ¼ ðc1 kÞðc2 kÞ ðcn kÞ: Hence the roots of the polynomial detðA kI Þ in k are c1 ; c2 ; . . .; cn . This shows that c1 ; c2 ; . . .; cn are the eigenvalues of A. ■ 3.3.23 Problem If C is a normal upper triangular matrix of size n n, then C is diagonal. Proof (Induction on n) The assertion is trivially true for n ¼ 1. Next suppose that the assertion is true for n 1. We have to show that the assertion is true for n. Let C be any normal upper triangular matrix of size n n. We have to show that C is diagonal. Since C is an upper triangular matrix, C is of the form

k 0

a ; D

where D is an upper triangular matrix of size ðn 1Þ ðn 1Þ, a is a matrix of size 1 ðn 1Þ, and k is a complex number. It suffices to show that a ¼ 0, and D is a diagonal matrix of size ðn 1Þ ðn 1Þ. Here, C ¼

k 0

a D

¼

k a 0 D

T

¼

k a

0 ; D

so C ¼

k a

0 : D

Now, since C is normal, we have

kk a k

k ka k a k a k 0 ¼ |fflfflfflfflfflfflffl C Cffl{zfflfflfflfflfflfflffl ¼ CCffl} ¼ ¼ 0 D 0 D a a aþD D a D kk þ aa aD ¼ ; Da DD

0 D

244


and hence

kk ka kk þ aa ¼ a k a aþD D Da

aD : DD

It follows that

kk ¼ kk þ aa ; a a þ D D ¼ DD that is,

aa ¼ 0 : a a þ D D ¼ DD Hence

a¼0 ; a a þ D D ¼ DD that is,

a¼0 ; D D ¼ DD that is, a ¼ 0, and D is normal. Since D is a normal upper triangular matrix of size ðn 1Þ ðn 1Þ, it follows by the induction hypothesis that D is diagonal. ■ 3.3.24 Problem Let A aij be an n-square complex matrix. Suppose that A is a normal matrix. Then there exists a unitary matrix U such that 1. U AU is a diagonal matrix, 2. the eigenvalues of A are the diagonal entries of U AU. Proof By 3.3.22, there exists a unitary matrix U such that 1. U AU is an upper triangular matrix, 2. the eigenvalues of A are the diagonal entries of U AU. It suffices to show that U AU is a diagonal matrix. Let us denote the upper triangular matrix U AU by C. Since U is unitary, we have U U ¼ UU ¼ I. It follows that U 1 ¼ U . Now, since C ¼ U AU, we have UCU ¼ U ðU AU ÞU ¼ ðUU ÞAðUU Þ ¼ IAI ¼ A; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}


245

and hence A ¼ UCU . It follows that A ¼ ðUCU Þ ¼ ðU Þ C U ¼ UC U ; |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} and hence A ¼ UC U . Since A is a normal matrix, we have U ðC C ÞU 1 ¼ U ðC C ÞU ¼ UC ICU ¼ UC ðU U ÞCU ¼ ðUC U ÞðUCU Þ ¼ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} A A ¼ AA ¼ AðUC U Þ ¼ ðUCU ÞðUC U Þ ¼ UC ðU U ÞC U ¼ UCIC U ¼ U ðCC ÞU ¼ U ðCC ÞU 1 ; and hence U ðC C ÞU 1 ¼ U ðCC ÞU 1 . It follows that C C ¼ CC , and hence C is normal. Since C is a normal upper triangular matrix, by 3.3.23, ðU AU ¼ÞC is diagonal, and hence U AU is diagonal. ■ Note 3.3.25 Problem Let A aij be an n-square complex matrix. Let U be a unitary matrix such that U AU is a diagonal matrix. Then A is a normal matrix. Proof We have to show that A A ¼ AA . Let us denote the diagonal matrix U AU by diagðk1 ; . . .; kn Þ. Since U is unitary, we have U U ¼ UU ¼ I. Now, since U ðdiagðk1 ; . . .; kn ÞÞU ¼ U ðU AU ÞU ¼ ðUU ÞAðUU Þ ¼ IAI ¼ A; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} we have A ¼ U ðdiagðk1 ; . . .; kn ÞÞU . It follows that A ¼ ðU ðdiagðk1 ; . . .; kn ÞÞU Þ ¼ ðU Þ ðdiagðk1 ; . . .; kn ÞÞ U ¼ U ðdiagðk1 ; . . .; kn ÞÞ U ¼ U diag k1 ; . . .; kn U ;

so A ¼ U diag k1 ; . . .; kn U . Here

246


A A ¼ U diag k1 ; . . .; kn U A ¼ U diag k1 ; . . .; kn U ðU ðdiagðk1 ; . . .; kn ÞÞU Þ ¼ U diag k1 ; . . .; kn ðU U Þðdiagðk1 ; . . .; kn ÞÞU ¼ U diag k1 ; . . .; kn I ðdiagðk1 ; . . .; kn ÞÞU ¼ U diag k1 ; . . .; kn ðdiagðk1 ; . . .; kn ÞÞ U ¼ U diag jk1 j2 ; . . .; jkn j2 U ; and AA ¼ A U diag k1 ; . . .; kn U ¼ ðU ðdiagðk1 ; . . .; kn ÞÞU Þ U diag k1 ; . . .; kn U ¼ U ðdiagðk1 ; . . .; kn ÞÞðU U Þ diag k1 ; . . .; kn U ¼ U ðdiagðk1 ; . . .; kn ÞÞI diag k1 ; . . .; kn U ¼ U ðdiagðk1 ; . . .; kn ÞÞ diag k1 ; . . .; kn U ¼ U diag jk1 j2 ; . . .; jkn j2 U ; so A A ¼ AA .

■ 3.3.26 Theorem Let A aij be an n-square complex matrix. Let A be a Hermitian matrix, that is, A ¼ A. Then the eigenvalues of A are real numbers. Proof Let k1 ; . . .; kn be the eigenvalues of A. It suffices to show that diag k1 ; . . .; kn ¼ diagðk1 ; . . .; kn Þ, that is, ðdiagðk1 ; . . .; kn ÞÞ ¼ diagðk1 ; . . .; kn Þ: Since A ¼ A, we have A A ¼ AA ¼ AA , and hence A A ¼ AA . Thus A is normal. Now, by 3.3.24, there exists a unitary matrix U such that U AU ¼ diagðk1 ; . . .; kn Þ: Hence ðdiagðk1 ; . . .; kn ÞÞ ¼ ðU AU Þ ¼ U A ðU Þ ¼ U A U ¼ U AU |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ diagðk1 ; . . .; kn Þ: Thus ðdiagðk1 ; . . .; kn ÞÞ ¼ diagðk1 ; . . .; kn Þ.

■


247

3.3.27 Theorem Let A aij be an n-square complex matrix. Let A be a normal matrix. Suppose that all the eigenvalues of A are real numbers. Then A is a Hermitian matrix, that is, A ¼ A. Proof By 3.3.22, there exists a unitary matrix U such that 1. U AU is an upper triangular matrix, 2. the eigenvalues of A are the diagonal entries of U AU. Let us denote U AU by C. Thus C ¼ U AU. Since U is unitary, we have U U ¼ UU ¼ I. It follows that U 1 ¼ U and

UCU ¼ U ðU AU ÞU ¼ ðUU ÞAðUU Þ ¼ IAI ¼ A; and hence A ¼ UCU . Next, A ¼ ðUCU Þ ¼ ðU Þ C U ¼ UC U ; so A ¼ UC U . Thus it suffices to show that C ¼ C. Since A is normal, we have U ðC C ÞU 1 ¼ UC CU ¼ UC ICU ¼ UC ðU U ÞCU ¼ ðUC U ÞðUCU Þ ¼ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} A A ¼ AA ¼ ðUCU ÞðUC U Þ ¼ UC ðU U ÞC U ¼ UCIC U ¼ UCC U ¼ U ðCC ÞU 1 ; and hence U ðC C ÞU 1 ¼ U ðCC ÞU 1 . It follows that C C ¼ CC , and hence C is normal. Now, since C is an upper triangular matrix, by 3.3.23, C is diagonal. Since C is diagonal, and the diagonal entries ofC are real numbers, we have C ¼ C. ■ Definition Let A aij be an n-square complex matrix. If for every x 2 Cn , the 1 1 matrix x Ax has a nonnegative real number as its sole entry, then we say that A is a positive semidefinite matrix or A is a nonnegative definite matrix. This is expressed as A 0. In short, A is a positive semidefinite matrix if for every x 2 Cn , x Ax 0. 3.3.28 Problem Let A aij be an n-square complex matrix. Suppose that A is positive semidefinite. Then A is a Hermitian matrix, that is, A ¼ A. Also, the diagonal entries of A are nonnegative real numbers. Proof Since A is a positive semidefinite matrix and ½1; 0; . . .; 0T 2 Cn , we have a11 ¼ ½1; 0; . . .; 0 A½1; 0; . . .; 0T ¼ ½1; 0; . . .; 0T A½1; 0; . . .; 0T 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

248


and hence a11 is a nonnegative real number. Similarly, a22 is a nonnegative real number, a33 is a nonnegative real number, etc. Also, ða11 þ a22 Þ þ ða12 þ a21 Þ ¼ 1ða11 1 þ a12 1Þ þ 1ða21 1 þ a22 1Þ ¼ ½1; 1; 0; . . .; 0 A½1; 1; 0; . . .; 0T ¼ ½1; 1; 0; . . .; 0T A½1; 1; 0; . . .; 0T 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so

ða11 þ a22 Þ þ Reða12 Þ þ Reða21 Þ 0 : Imða12 Þ ¼ Imða21 Þ Next, ða11 þ a22 Þ þ ðia12 ia21 Þ ¼ 1ða11 1 þ a12 iÞ þ ðiÞða21 1 þ a22 iÞ ¼ ½1; i; 0; . . .; 0 A½1; i; 0; . . .; 0T ¼ ½1; i; 0; . . .; 0T A½1; i; 0; . . .; 0T 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} so

ða11 þ a22 Þ Imða12 Þ þ Imða21 Þ 0 : Reða12 Þ ¼ Reða21 Þ Since

Reða12 Þ ¼ Reða21 Þ ; Imða12 Þ ¼ Imða21 Þ we have a12 ¼ a21 . Similarly, a13 ¼ a31 ; a23 ¼ a32 , etc. This shows that T ■ aij ¼ ai| , that is, A ¼ A . 3.3.29 Problem Let A aij be an n-square complex matrix. Suppose that A is positive semidefinite. Then the eigenvalues of A are nonnegative real numbers. Proof Let k be an eigenvalue of A. We have to show that k is a nonnegative real number. Since k is an eigenvalue of A, there exists a nonzero x 2 Cn such that Ax ¼ kx. Since A is positive semidefinite, ðkhx; xi ¼ kðx xÞ ¼ x ðkxÞ ¼ x ðAxÞ ¼Þ x Ax is a nonnegative real number, and hence khx; xi is a nonnegative real number. Since


249

x 6¼ 0, hx; xi is a positive real number. Since khx; xi is a nonnegative real number, k is a nonnegative real number. ■ 3.3.30 Problem Let A aij be an n-square complex matrix. Let A be normal. Suppose that the eigenvalues of A are nonnegative real numbers. Then A is positive semidefinite. Proof Let x be a member of Cn . We have to show that x Ax is a nonnegative real number. By 3.3.24, there exists a unitary matrix U such that 1. U AU is a diagonal matrix, 2. the eigenvalues of A are the diagonal entries of U AU. So we can write U AU ¼ diagðt1 ; . . .; tn Þ, where t1 ; . . .; tn are the eigenvalues of A. By assumption, t1 ; . . .; tn are nonnegative real numbers. Since U is unitary, we have U U ¼ UU ¼ I. It follows that U 1 ¼ U and U ðdiagðt1 ; . . .; tn ÞÞU ¼ U ðU AU ÞU ¼ ðUU ÞAðUU Þ ¼ IAI ¼ A; and hence A ¼ U ðdiagðt1 ; . . .; tn ÞÞU . Next, x Ax ¼ x ðU ðdiagðt1 ; . . .; tn ÞÞU Þx ¼ ðU xÞ ðdiagðt1 ; . . .; tn ÞÞðU xÞ; so x Ax ¼ ½y1 ; . . .; yn T ðdiagðt1 ; . . .; tn ÞÞ½y1 ; . . .; yn T ; where ½y1 ; . . .; yn T U x. We now have x Ax ¼ ½y1 ; . . .; yn T ðdiagðt1 ; . . .; tn ÞÞ½y1 ; . . .; yn T |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ½y1 ; . . .; yn ðdiagðt1 ; . . .; tn ÞÞ½y1 ; . . .; yn T ¼ ½y1 ; . . .; yn ½t1 y1 ; . . .; tn yn T ¼ y1 ðt1 y1 Þ þ þ yn ðtn yn Þ ¼ t1 jy1 j2 þ þ tn jyn j2 0; and conclude that x Ax 0. ■ 3.3.31 Note Let A aij be an n-square complex matrix. Let Mn be the collection of all n-square complex matrices. We know that Mn is a vector space over C. The collection of all C in Mn such that C has exactly one entry 1 and 0 entries elsewhere constitutes a basis of Mn . This basis has n2 members, so 2 dimðMn Þ ¼ n2 . Since I; A; A2 ; . . .; An is a collection of size ðn2 þ 1Þ ð [ dimðMn ÞÞ 2

and I; A; A2 ; . . .; An

are in Mn , it follows that I; A; A2 ; . . .; An

2

are linearly

250


dependent in Mn . Thus there exist complex numbers a0 ; a1 ; a2 ; . . .; an2 , not all zero, such that a0 I þ a1 A þ a2 A2 þ þ an2 An ¼ 0: Thus pðAÞ ¼ 0, where pðxÞ denotes the polynomial a0 þ a1 x þ a2 x2 þ þ 2 an2 xn . Clearly, degðpðxÞÞ n2 . 3.3.32 Conclusion Let A aij be an n-square complex matrix. Then there exists a polynomial pðxÞ with complex coefficients such that pðAÞ ¼ 0. In short, there exists an “annihilating polynomial” for A. 3.3.33 Problem Let A aij be an n-square complex matrix. Let A be an upper triangular matrix. Let k1 ; k2 ; . . .; kn be the diagonal entries of A. Then ðk1 I AÞ ðk2 I AÞ ðkn I AÞ ¼ 0. Proof Observe that the first column of ðk1 I AÞ is 0. So we can suppose that ðk1 I AÞ ¼ ½0; a1 ; . . .; an1 , where each ai is in Cn . Similarly, we can suppose that ðk2 I AÞ ¼ ½b1 ; 0; b2 ; . . .; bn1 , where each bi is in Cn , etc. Hence ðk1 I AÞðk2 I AÞ ¼ ½0; a1 ; . . .; an1 ½b1 ; 0; b2 ; . . .; bn1 ¼ ½½0; a1 ; . . .; an1 b1 ; ½0; a1 ; . . .; an1 0; ½0; a1 ; . . .; an1 b2 ; . . .; ½0; a1 ; . . .; an1 bn1 ¼ ½½0; a1 ; . . .; an1 b1 ; 0; ½0; a1 ; . . .; an1 b2 ; . . .; ½0; a1 ; . . .; an1 bn1 : Thus ðk1 I AÞðk2 I AÞ ¼ ½½0; a1 ; . . .; an1 b1 ; 0; ½0; a1 ; . . .; an1 b2 ; . . .; ½0; a1 ; . . .; an1 bn1 ;

ð Þ

and hence the second column of ðk1 I AÞðk2 I AÞ is 0. Since ðk2 I AÞ ¼ ½b1 ; 0; b2 ; . . .; bn1 and A is an upper triangular matrix, we have b1 ¼ ½k2 k1 ; 0; . . .; 0T 2 Cn , and hence ½0; a1 ; . . .; an1 b1 ¼ ½0; a1 ; . . .; an1 ½k2 k1 ; 0; . . .; 0T ¼ ½0; 0; . . .; 0T 2 Cn : Thus ½0; a1 ; . . .; an1 b1 ¼ 0. Now from (*), ðk1 I AÞðk2 I AÞ ¼ ½0; 0; ½0; a1 ; . . .; an1 b2 ; . . .; ½0; a1 ; . . .; an1 bn1 :


251

Thus the first and second columns of ðk1 I AÞðk2 I AÞ are 0. Similarly, the first three columns of ðk1 I AÞðk2 I AÞðk3 I AÞ are 0, etc. Finally, all the n columns of ðk1 I AÞðk2 I AÞ ðkn I AÞ are 0. Thus, ðk1 I AÞðk2 I AÞ ðkn I AÞ ¼ 0: ■ 3.3.34 Note Let A aij be an n-square complex matrix. Let k1 ; k2 ; . . .; kn be the eigenvalues of A. By 3.3.22, there exists a unitary matrix U such that 1. U AU is an upper triangular matrix, 2. k1 ; k2 ; . . .; kn are the diagonal entries of U AU. By 3.3.12, U ððk1 I AÞðk2 I AÞ ðkn I AÞÞU ¼ ðU ðk1 I AÞU ÞðU ðk2 I AÞU Þ ðU ðkn I AÞU Þ ¼ ðk1 I U AU Þðk2 I U AU Þ ðkn I U AU Þ ¼ 0 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence U ððk1 I AÞðk2 I AÞ ðkn I AÞÞU ¼ 0: Since U is unitary, we have ðk1 I AÞðk2 I AÞ ðkn I AÞ ¼ 0: It follows that pðAÞ ¼ 0, where pðxÞ denotes the polynomial ðx k1 Þ ðx k2 Þ ðx kn Þ. Observe that k1 ; k2 ; . . .; kn are the roots of the monic polynomial ðx k1 Þðx k2 Þ ðx kn Þ ð¼ pðxÞÞ. Since k1 ; k2 ; . . .; kn are the eigenvalues of A, k1 ; k2 ; . . .; kn are the roots of the monic polynomial detðkI AÞ, and since k1 ; k2 ; . . .; kn are the roots of the monic polynomial pðxÞ, we have pðkÞ ¼ detðkI AÞ: 3.3.35 Conclusion Let A aij be an n-square complex matrix. Let k1 ; k2 ; . . .; kn be the eigenvalues of A. Then pðAÞ ¼ 0, where pðkÞ ¼ detðkI AÞ. This result is known as the Cayley–Hamilton theorem. Here the polynomial detðkI AÞ is called the characteristic polynomial of A. Thus the characteristic polynomial of A is an annihilating polynomial of A. It follows that the minimal polynomial of A divides the characteristic polynomial of A.

252


Exercises 1. Let V be an n-dimensional inner product space. Let T : V ! V be a linear transformation. Let v; w1 ; w2 2 V. Suppose that u 2 V ) hu; w1 i ¼ hTðuÞ; vi ¼ hu; w2 i: Show that w1 ¼ w2 . 2. Let V be an n-dimensional inner product space. Let S1 : V ! V, S2 : V ! V, and S3 : V ! V be linear transformations. Let k; l be any complex numbers. ðS3 Þ ðS2 Þ . Show that ððkS1 þ lS2 ÞS3 Þ ¼ kðS3 Þ ðS1 Þ þ l 3. Let V be any n-dimensional vector space. Let S; T 2 AðVÞ be such that ST ¼ 0 and TS 6¼ 0. Show that T is not invertible. 4. Let V be an n-dimensional inner product space. Let T : V ! V be a normal linear transformation. Let v 2 V. Suppose that T 3 ðvÞ ¼ 0: Show that v is a member of the null space of T. 5. Let T 2 A C3 . Suppose that C is invariant under T, and C2 is invariant under T. Let p1 ðxÞ be a minimal polynomial of TjC , and p2 ðxÞ a minimal polynomial of TjC2 . Show that the least common multiple of p1 ðxÞ and p2 ðxÞ is a minimal polynomial of T. 6. Let T 2 AðCn Þ. Suppose that T is nilpotent. Show that there exist linear subspaces V1 ; V2 ; . . .; Vk of Cn such that 1. Cn ¼ V1 V2 Vk , 2. each Vi is invariant under T. Also there exist a basis v11 ; v21 ; . . .; vn11 of V1 , a basis v12 ; v22 ; . . .; vn22 of 1 2 V2 ; . . ., a basis vk ; vk ; . . .; vnk k of Vk such that the matrix of T relative to the basis v11 ; v21 ; . . .; vn11 ; v12 ; v22 ; . . .; vn22 ; . . .; v1k ; v2k ; . . .; vnk k has the canonical form 2 3 0 Mn1 0 6 7 .. 4 0 . 0 5 : 0

0

Mnk

nn

7. Let A be a nonzero 3 5 matrix with complex numbers as entries. Suppose that rankðAÞ ¼ 2. Show that there exist an invertible 3 3 matrix P and an invertible 5 5 matrix Q such that I 0 PAQ ¼ 2 : 0 0 35


253

8. Let A and B be any n-square complex matrices. Suppose that AB ¼ BA. Show that there exists a unitary matrix U such that U AU and U BU are both upper triangular matrices. 9. Let A be an n-square complex matrix. Suppose that A is a normal matrix. Show that there exists a unitary matrix U such that the diagonal entries of U AU are the eigenvalues of A. 10. Find the characteristic polynomial pðxÞ of the matrix 2 3 1 2 3 A 4 5 2 3 5: 1 1 4 Verify that pðAÞ ¼ 0.

Chapter 4

Sylvester’s Law of Inertia

Sylvester’s law characterizes an equivalence relation called congruence. This remarkable result introduces a new concept of a matrix, called its signature. It is similar to the rank of a matrix. Finally, a beautiful method of obtaining the signature of a real quadratic form is introduced.

4.1

Positive Definite Matrices

4.1.1 Theorem Let V be any n-dimensional vector space over the field F. Let T:V ! transformation. Vbe a linear Then there exists a positive integer k such that N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; and N T k1 is a proper subset of N Tk : Proof It is clear that ðfv : v 2 V and T ðvÞ ¼ 0g ¼ÞN ðT Þ N ðT T Þ ¼ N T 2 ; so N ðT Þ N ðT 2 Þ: Similarly, N ðT 2 Þ N ðT 3 Þ; etc. Since f0g N ðT Þ N T 2 N T 3 V; each “null space” N T k is a linear subspace of V, and V is a finite-dimensional vector space, the chain N ðT Þ N ðT 2 Þ N ðT 3 Þ cannot continue to increase indefinitely. Hence there exists a positive integer k such that N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; and N T k1 is a proper subset of N T k : ∎


255

256

4 Sylvester’s Law of Inertia

4.1.2 Theorem N T k is invariant under T.

Proof To show this, let us take an arbitrary v 2 N T k ; that is, T k ðvÞ ¼ 0: We have to show that T ðvÞ 2 N T k ; that is, T ð 0Þ ¼ T T k ð v Þ ¼ T k þ 1 ð v Þ ¼ T k ð T ð v Þ Þ ¼ 0 ; |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

that is, T ð0Þ ¼ 0: This is known to be true, because T : V ! V is a linear transformation. ∎ k k is a nilpotent 4.1.3 Theorem The restriction TjN ðT k Þ : N T ! N T transformation. Proof To show this, let us take an arbitrary v 2 N T k : It suffices to show that k TjN ðT k Þ ðvÞ ¼ 0: Since v 2 N T k ; we have T k ðvÞ ¼ 0: Since v 2 N T k ; we have TjN ðT k Þ ðvÞ ¼ 0: Now, k k1 ð vÞ ¼ TjN ðT k Þ TjN ðT k Þ ðvÞ TjN ðT k Þ k1 ¼ TjN ðT k Þ ð0Þ ¼ 0 ¼ RHS:

LHS ¼

4.1.4 Theorem N T

k

\ ran T

k

∎ ¼ f0g.

Proof Suppose to the contrary that there exists a nonzero v in N T k \ ran T k ; that is, v 6¼ 0; T k ðvÞ ¼ 0; and for some nonzero w 2 V; T k ðwÞ ¼ v: We seek a contradiction. Since T 2k ðwÞ ¼ T k T k ðwÞ ¼ T k ðvÞ ¼ 0; |fflfflfflfflfflffl{zfflfflfflfflfflffl} we have T 2k ðwÞ ¼ 0; and hence w 2 N T 2k : Now, since N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; we have N T k ¼ N T 2k ð3wÞ; and hence w 2 N T k : It follows that v ¼ T k ðwÞ ¼ 0; and hence v ¼ 0: This is a contradiction. ∎ |fflfflfflfflfflffl{zfflfflfflfflfflffl}

4.1 Positive Definite Matrices

257

4.1.5 Theorem V ¼ N T k ran T k :

Proof From 4.1.4, it remains to show that V ¼ N T k þ ran T k : Since N T k ; ran T k are subspaces of V, N T k þ ran T k is a subspace of V, and hence it suffices to show that dim domain of T k ¼ dimðV Þ ¼ dim N T k + ran T k |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} k k k k dim N ¼k dim N T þk dim ran T T k \ ran T k þ dim ran T dimðf0gÞ ¼ dim NT þ dim ran T 0 ¼ dim N T ¼ dim N T k þ dim ran T k ;

that is, dim domain of T k ¼ dim N T k þ dim ran T k : Since T k : V ! V is a linear transformation, we obtain from the well-known result (nullity + rank = dimension of domain) that dim N T k þ dim ran T k ¼ dim domain of T k : Thus V ¼ N T k ran T k : ∎ k k k 4.1.6 Theorem TjranðT k Þ : ran T ! ran T is a mapping, that is, ran T is invariant under T. Proof To show this, let us take an arbitrary v 2 V: We have to show that TjranðT k Þ T k ðvÞ 2 ran T k ; that is, T T k ðvÞ 2 ran T k ; that is, T k þ 1 ðvÞ 2 ∎ ran T k ; that is, T k ðT ðvÞÞ 2 ran T k : This is clearly true. k k 4.1.7 Theorem The restriction TjranðT k Þ : ran T ! ran T is invertible. Proof To show this, let us take an arbitrary v 2 V such that TjranðT k Þ T k ðvÞ ¼ 0; that is, T T k ðvÞ ¼ 0; that is, T k þ 1 ðvÞ ¼ 0; that is, v 2 N T k þ 1 : It suffices to show that T k ðvÞ ¼ 0; that is, v 2 N T k : ∎ Since v 2 N T k þ 1 ; and N T k ¼ N T k þ 1 ; we have V 2 N T k : Thus we have shown that the linear transformation TjranðT k Þ : ran T k ! ran T k is invertible. 4.1.8 Conclusion Let V be any n-dimensional vector space over the field F. Let T : V ! V be a linear transformation. Then there exists a positive integer k such that 1. N ðT Þ N ðT 2 Þ N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; 2. V ¼ N T k ran T k ; k 3. TjN ðT k Þ : N T ! N T k is a nilpotent transformation, 4. TjranðT k Þ : ran T k ! ran T k is invertible.

258


4.1.9 Theorem Let V be any n-dimensional vector space over the field F. Let T : V ! V be a linear transformation. Then there exist unique linear subspaces H and K of V such that 1. V ¼ H K; 2. TjH : H ! H is a nilpotent transformation, 3. TjK : K ! K is invertible.

1. 2. 3. 4.

Also, there exists a positive integer k such that H ¼ N T k ; and K ¼ ran T k : Proof of existence: By 4.1.1, there exists a positive integer k such that N ðT Þ N ðT 2 Þ N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; V ¼ N T k ran T k ; k TjN ðT k Þ : N T ! N T k is a nilpotent transformation, TjranðT k Þ : ran T k ! ran T k is invertible. Let us put H N T k ; and K ran T k : We get

1. V ¼ H K; 2. TjH : H ! H is a nilpotent transformation, 3. TjK : K ! K is invertible. Proof of uniqueness: Suppose that H1 and K1 are subspaces of V such that 1. V ¼ H1 K1 ; 2. TjH1 : H1 ! H1 is a nilpotent transformation, 3. TjK1 : K1 ! K1 is invertible. Suppose that H2 and K2 are subspaces of V such that 1. V ¼ H2 K2 ; 2. TjH2 : H2 ! H2 is a nilpotent transformation, 3. TjK2 : K2 ! K2 is invertible. We have to show that H1 ¼ H2 ; and K1 ¼ K2 : By 4.1.1, there exists a positive integer k such that 1. N ðT Þ N ðT 2 Þ N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; 2. V ¼ N T k ran T k ; k 3. TjN ðT k Þ : N T ! N T k is a nilpotent transformation, 4. TjranðT k Þ : ran T k ! ran T k is invertible. Since V ¼ N T k ran T k ; we have dimðV Þ ¼ dim N T k þ dim ran T k : Clearly, H1 N T k :


259

Proof To show this, let us take an arbitrary v 2 H1 : We have to show that v 2 N T k ; that is, T k ðvÞ ¼ 0: Since TjH1 : H1 ! H1 is a nilpotent transfor l mation, there exists a positive integer l such that TjH1 ¼ 0: It follows that l l1 l1 l ðT ðvÞÞ ¼ TjH1 TjH1 ðvÞ ¼ TjH1 ðvÞ ¼ 0; T ðvÞ ¼ ¼ TjH1 |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} and hence T l ðvÞ ¼ 0: Thus v 2 N T l : Since N ðT Þ N T 2 N T k ¼ N T k þ 1 ¼ N T k þ 2 ¼ ; we have v 2 N Tl N Tk ; |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence v 2 N T k : ∎ k It follows that dimðH1 Þ dim N T : We claim that dimðH1 Þ ¼ dim N T k : Suppose to the contrary that dimðH1 Þ\dim N T k : We seek a contradiction. Clearly, K1 ranðT Þ: Proof To show this, let us take an arbitrary v 2 K1 : We have to show that v 2 ranðT Þ: Since v 2 K1 , and TjK1 : K1 ! K1 is invertible, there exists w 2 K1 such that T ðwÞ ¼ TjK1 ðwÞ ¼ v ; and hence ðranðT Þ3ÞT ðwÞ ¼ v: Thus |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} v 2 ranðT Þ: Clearly, K1 ranðT 2 Þ:

∎

Proof To show this, let us take an arbitrary v 2 K1 : We have to show that 2 v 2 ranðT 2 Þ: Since TjK1 : K1 ! K1 is invertible, TjK1 : K1 ! K1 is invertible. Now, since v 2 K1 ; there exists w 2 K1 such that

2 TjK1 ðT ðwÞÞ ¼ TjK1 TjK1 ðwÞ ¼ TjK1 ðw Þ ¼ v ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

and hence TjK1 ðT ðwÞÞ ¼ v: Since w 2 K1 and TjK1 : K1 ! K1 ; we have T ðwÞ ¼ TjK1 ðwÞ 2 K1 ; and hence T ðwÞ 2 K1 : It follows that |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

260


v ¼ TjK1 ðT ðwÞÞ ¼ T ðT ðwÞÞ ¼ T 2 ðwÞ 2 ran T 2 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ∎ and hence v 2 ranðT 2 Þ: Similarly, K1 ranðT 3 Þ: etc. Thus K1 ran T k : It follows that dimðK1 Þ dim ran T k : Since V ¼ H1 K1 ; we have dimðV Þ ¼ dimðH1 Þ þ dimðK1 Þ: Similarly, dimðV Þ ¼ dimðH2 Þ þ dimðK2 Þ: Since dimðV Þ dimðH1 Þ ¼ dimðK1 Þ dim ran T k ¼ dimðV Þ dim N T k ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} we have dimðV Þ dimðH1 Þ dimðV Þ dim N T k ; and hence dim N T k dimðH1 Þ: This is a contradiction. So our claim is substantiated, that is, dimðH1 Þ ¼ dim N T k : k k Now, since H1 N T ; we have H1 ¼ N T : Similarly, H2 ¼ N T k : It follows that H1 ¼ H2 : It remains to show that K1 ¼ K2 : Since V ¼H1 K 1 ; we have k dim ðV Þ ¼ dimðH1 Þ þ dimðK1 Þ: Similarly, k þ dim ran T : Since dimðV Þ ¼ dim N T dimðV Þ dimðK1 Þ ¼ dimðH1 Þ ¼ dim N T k ¼ dimðV Þ dim ran T k ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} we have dimðV Þ dimðK1 Þ ¼ dimðV Þ dim ran T k ; and hence dimðK1 Þ ¼ dim ran T k : Now, since K1 ran T k ; we have K1 ¼ ran T k : Similarly, K2 ¼ k ran T : Hence K1 ¼ K2 : ∎ 4.1.10 Note Let V be any n-dimensional vector space. Let T : V ! V be a linear transformation.

Let v1 ; . . .; vn be any basis of V. Let A aij n n be the matrix of T relative to the basis v1 ; . . .; vn : By 3.3.22, there exists a unitary matrix U such that 1. U AU is an upper triangular matrix, 2. the eigenvalues of A are the diagonal entries of U AU: Since U is a unitary matrix, we have U U ¼ UU ¼ I; and hence U 1 ¼ U : Thus U is invertible. Also 1. U 1 AU is an upper triangular matrix, 2. the eigenvalues of T are the diagonal entries of U 1 AU:


261

Since U is invertible, by 3.1.35(b), there exists a basis w1 ; . . .; wn of V such that U 1 AU is the matrix of T relative to the basis w1 ; . . .; wn : Thus 1. the matrix of T relative to the basis w1 ; . . .; wn is upper triangular, 2. the eigenvalues of T are the diagonal entries of the matrix of T relative to the basis w1 ; . . .; wn : 4.1.11 Conclusion Let V be any n-dimensional vector space. Let T : V ! V be a linear transformation. Then there exists a basis w1 ; . . .; wn of V such that if B is the matrix of T relative to the basis w1 ; . . .; wn ; then 1. B is an upper triangular matrix, 2. the eigenvalues of T are the diagonal entries of B. 4.1.12 Note Let V be any n-dimensional vector space over the field C: Let T : V ! V be a linear transformation. Suppose that k1 ; k2 ; . . .; kp are all the distinct eigenvalues of T. Suppose that the eigenvalue k1 has multiplicity m1 ; the eigenvalue k2 has multiplicity m2 ; etc. In other words, the list of all eigenvalues of T is k1 ; k1 ; . . .; k1 ; . . .; kp ; kp ; . . .; kp ; |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} m1 in number

mp in number

where n ¼ m1 þ þ mp : m Thus the characteristic polynomial of T is ðk k1 Þm1 k kp p : Now, since the minimal polynomial of T divides the characteristic polynomial of T, we can suppose that the minimal polynomial of T is of the form l ðk k1 Þl1 k kp p ; where li mi ði ¼ 1; . . .; pÞ: Put n o V1 v : v 2 V and ðT k1 I Þl1 ðvÞ ¼ 0 ; n o V2 v : v 2 V and ðT k2 I Þl2 ðvÞ ¼ 0 ; etc: By 3.2.3, 1. 2. 3. 4.

each Vi is a nontrivial linear subspace of V, each Vi is invariant under T, V ¼ V1 V2 Vp ; for each i ¼ 1; 2; . . .; p; the minimal polynomial of TjVi is ðk ki Þli : l l From 4.1.5, ðT k1 I ÞjV1 1 ¼ TjV1 k1 I 1 ¼ 0; so ðT k1 I ÞjV1 : V1 ! V1

is a nilpotent transformation, and hence by 3.2.22, 0 is the only eigenvalue of

262


ðT k1 I ÞjV1 : It follows that k1 is the only eigenvalue of TjV1 : Similarly, k2 is the only eigenvalue of TjV2 ; etc. 4.1.13 Conclusion 4.1.13 Let V be any n-dimensional vector space over the field C: Let T : V ! V be a linear transformation. Suppose that k1 ; k2 ; . . .; kp are all the distinct eigenvalues of T. Suppose that the minimal polynomial of T is of the form l ðk k1 Þl1 k kp p : Put n o V1 v : v 2 V and ðT k1 I Þl1 ðvÞ ¼ 0 ; n o V2 v : v 2 V and ðT k2 I Þl2 ðvÞ ¼ 0 ; etc: Then 1. 2. 3. 4. 5.

each Vi is a nontrivial linear subspace of V, each Vi is invariant under T, V ¼ V1 V2 Vp ; for each i ¼ 1; 2; . . .; p; ki is the only eigenvalue of TjVi : there exists a basis B of V such that the matrix of T relative to B is of the block form 2 6 6 4

3

A1 A2

..

.

7 7 5 Ap

; n n

where A1 is a ðdim V1 Þ ðdim V1 Þ matrix of TjV1 ; A2 is a ðdim V2 Þ ðdim V2 Þ matrix of TjV2 ; etc., and all other entries are 0. Definition A square matrix of the form 2

k 60 6. 6. 6. 6. 6 .. 6 6 .. 4. 0

1 k

0 1

0 .. . .. . 0

k 0 .. . 0

..

.

3 0 07 7 7 07 7 07 7 7 15 k


263

is called a basic Jordan block belonging to k: The basic Jordan block belonging to k00 of size t t can also be written as 2

0 1 60 0 6. 6. 6. 0 kI þ 6 6 ... ... 6 6 .. .. 4. . 0 0

3 0 07 7 7 07 7 07 7 7 15 0 t t

0 1 0

..

0 .. .

.

0

or kI þ Mt ; where 2

0 60 6. 6. 6. Mt 6 6 ... 6 6 .. 4.

3 0 07 7 .. 7 0 . 07 7 : 0 07 7 7 .. . 15 0 0 t t

1 0

0 1

0 .. . .. . 0

0

4.1.14 Theorem Let V be any n-dimensional vector space over the field C: Let T : V ! V be a linear transformation. Suppose that k1 ; k2 ; . . .; kp are the distinct eigenvalues of T. Then there exists a basis B of V such that the matrix of T relative to B is of the form 2 6 6 4

3

J1 J2

..

.

7 7 5 Jp

such that for every i 2 f1; 2; . . .; pg; Ji is of the form 2 6 4

3

Bi1 Bi2

..

7 5; .

where each Bij is a basic Jordan block belonging to ki : Here a matrix of the type

264


22 66 64 6 6 6 6 6 6 6 6 6 4

B11 B12

..

3

3

7 5

7 7 7 7 7 7 7; 7 7 7 7 5

.

2 6 4

3

B21 B22

..

7 5 .

..

.

where each Bij is a basic Jordan block belonging to ki ; is called a Jordan canonical form. Thus for a given square matrix A, there exists an invertible matrix C such that C 1 AC is a Jordan canonical form. Proof By 4.1.13, there exist linear subspaces V1 ; . . .; Vp such that 1. each Vi is invariant under T, 2. V ¼ V1 V2 Vp ; 3. for each i ¼ 1; 2; . . .; p; ki is the only eigenvalue of TjVi : 4. there exists a basis B of V such that the matrix of T relative to B is of the block form 2 3 A1 6 7 A2 6 7 ; .. 4 5 . Ap n n

where A1 is a ðdim V1 Þ ðdim V1 Þ matrix of TjV1 ; A2 is a ðdim V2 Þ ðdim V2 Þ matrix of TjV2 ; etc., and all other entries are 0. Now, by 3.3.22, there exists a basis C1 of V1 such that the matrix 1 K1 ¼ ½C1 A1 ½C1 of TjV1 relative to C1 is upper triangular, and each diagonal Similarly, there exists a basis C2 of V2 such that the matrix entry of K1 is k1 : 1 K2 ¼ ½C2 A2 ½C2 of TjV2 relative to C2 is upper triangular, and each diagonal entry of K2 is k2 ; etc. Now, since V ¼ V1 V2 Vp ; there exists a basis D of V such that the matrix of T relative to D is of the block form 2 6 6 4

3

K1 K2

..

.

7 7 5 Kp

n n

:


265

Since k1 is the only eigenvalue of TjV1 ; 0 is the only eigenvalue of TjV1 k1 I ; and hence the characteristic polynomial of TjV1 k1 I is dimðTjV Þ 1 ð k 0Þ ð k 0Þ ¼ k ; and hence by the Cayley–Hamilton theorem, |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} dimðTjV Þ in number 1 dimðTjV Þ 1 ¼ 0: This shows that Tj TjV1 k1 I V1 k1 I is nilpotent. Now, by 3.2.25 , there exists a basis D1 of V1 such that the matrix ½D1 1 ðK1 k1 I Þ½D1 ¼ ½D1 1 ðK1 k1 I Þ½D1 ¼ ½D1 1 K1 ½D1 k1 I of TjV1 k1 I relative to the basis D1 has the canonical form 2

Mn1 6 0 4

0 Mn2

0

0

3 0 0 7 5: .. .

It follows that the matrix ½D1 1 K1 ½D1 ¼ ½D1 1 ½C1 1 A1 ½C1 ½D1 ¼ ð½C1 ½D1 Þ1 A1 ð½C1 ½D1 ÞÞ of TjV1 relative to the basis D1 has the canonical form 2

3 2 k1 I þ Mn1 0 6 0 0 7 5¼4 .. 0 0 . 0 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Mn1 6 0 k1 I þ 4

0 Mn2

0 k1 I þ M n 2 0

3 2 0 B11 6 0 0 7 5¼4 .. 0 .

0 B12 0

3 0 0 7 5; .. .

where B11 k1 I þ Mn1 ; B12 k1 I þ Mn2 ; etc. Clearly, each of B11 ; B12 ; . . . is a basic Jordan block belonging to k1 : Thus the matrix of TjV1 relative to the basis D1 has the canonical form 2

B11 6 0 4 0

0 B12 0

3 0 0 7 5: .. .

Similarly, there exists a basis D2 of V2 such that the matrix of TjV2 relative to the basis D2 has the canonical form 2

B21 6 0 4 0

0 B22 0

3 0 0 7 5; .. .

where each B21 ; B22 ; . . . is a basic Jordan block belonging to k2 ; etc.

266


Now, since V ¼ V1 V2 Vp ; there exists a basis B of V such that the matrix of T relative to B is of the form 2 6 6 4

3

J1 J2

..

.

7 7 5 Jp

such that for every i 2 f1; 2; . . .; pg; Ji is of the form 2 6 4

3

Bi1 Bi2

..

7 5; .

where each Bij is a basic Jordan block belonging to ki :

∎

4.1.15 Note Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Let T be normal. Suppose that k1 ; k2 ; . . .; kp are the distinct eigenvalues of T. Suppose that the minimal polynomial of T is of the form l ðk k1 Þl1 k kp p : Put n o V1 v : v 2 V and ðT k1 I Þl1 ðvÞ ¼ 0 ; n o V2 v : v 2 V and ðT k2 I Þl2 ðvÞ ¼ 0 ; etc.

1. 2. 3. 4.

By 4.1.13, each Vi is a nontrivial linear subspace of V; each Vi is invariant under T; V ¼ V1 V2 Vp ; for each i ¼ 1; 2; . . .; p; ki is the only eigenvalue of TjVi : Observe that V1 ¼ E1 [ f0g; where E1 is the set of all eigenvectors belonging to the eigenvalue k1 of T. Proof To show nonzero v 2 V1 ð¼ fv : v 2 V and this, let us take an arbitrary l1 ðT k1 I Þ ðvÞ ¼ 0gÞ: It follows that ðT k1 I Þl1 ðvÞ ¼ 0: Since T is normal, by 3.1.24, we have T ðvÞ ¼ kv: Next, since v 6¼ 0; v is an eigenvector


267

belonging to the eigenvalue k1 ; and hence v 2 E1 : Thus V1 E1 [ f0g: It suffices to show that E1 V1 : To show this, let us take an arbitrary v 2 E1 ; that is, v is an eigenvector belonging to the eigenvalue k1 of T. Hence T ðvÞ ¼ k1 v: It follows that ðT k1 I ÞðvÞ ¼ 0; and hence ðT k1 I Þl1 ðvÞ ¼ ðT k1 I Þl1 1 ððT k1 I ÞðvÞÞ ¼ ðT k1 I Þl1 1 ð0Þ ¼ 0: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus ðT k1 I Þl1 ðvÞ ¼ 0: It follows that v 2 V1 : ∎ We have shown that V1 ¼ E1 [ f0g: Similarly, V2 ¼ E2 [ f0g; where E2 is the set of all eigenvectors belonging to the eigenvalue k2 of T, etc. Since V1 is a nontrivial linear subspace of V, by 3.3.12, there exists an orthonormal basis B1 of V1 ð¼ E1 [ f0gÞ: Since a basis does not contain the zero vector, we have B1 E1 ; and hence each member of B1 is an eigenvector belonging to the eigenvalue k1 of T. Thus for every v 2 B1 ; T ðvÞ ¼ k1 v: Similarly, there exists an orthonormal basis B2 of V2 such that B2 E2 ; and for every w 2 B2 ; T ðwÞ ¼ k2 w; etc. Clearly, B1 [ B2 [ [ Bp is an orthonormal basis of V. Proof Since each Bi is an orthonormal basis of Vi , and V ¼ V1 V2 Vp ; it suffices to show that for distinct i; j 2 f1; . . .; pg; ðv 2 Bi ; w 2 Bj ) hv; wi ¼ 0Þ: To show this, let us take arbitrary i; j 2 f1; . . .; pg such that i 6¼ j: Next, let us take arbitrary v 2 Bi ; and w 2 Bj : We have to show that hv; wi ¼ 0: Since v 2 Bi ; we have T ðvÞ ¼ ki v: Since v 2 Bi ; and Bi is a basis, v is nonzero. Since i 6¼ j; and k1 ; k2 ; . . .; kp are distinct, we have ki 6¼ kj : Since w 2 Bj ; we have T ðwÞ ¼ kj w: Now, since T is normal, by 3.1.25, hv; wi ¼ 0: We have shown that B is an orthonormal basis of V, where B B1 [ B2 [ [ Bp : ∎ Since for every i 2 f1; . . .; pg; Bi Ei ; we have B ¼ B1 [ B2 [ [ Bp E1 [ E2 [ [ Ep |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðthe collection of all eigenvectors of T Þ; and hence each member of B is an eigenvector of T: Suppose that B ¼ fe1 ; e2 ; . . .; en gð V Þ: Since fe1 ; e2 ; . . .; en g is an orthonormal basis B of V we have

268


T ðe1 Þ ¼ hT ðe1 Þ; e1 ie1 ¼ hT ðe1 Þ; e1 ie1 þ 0e2 þ þ 0eP ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence T ðe1 Þ ¼ hT ðe1 Þ; e1 ie1 þ 0e2 þ þ 0eP : Similarly, T ðe2 Þ ¼ 0e1 þ hT ðe2 Þ; e2 ie2 þ 0e3 þ þ 0eP ; etc. Thus the matrix of T relative to the basis fe1 ; e2 ; . . .; en g is the diagonal matrix 2 6 4

hT ðe1 Þ; e1 i

3 hT ðe2 Þ; e2 i

..

7 5 .

:

n n

4.1.16 Conclusion Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Let T be normal. Then there exists an orthonormal basis B of V such that the matrix of T relative to the basis B is a diagonal matrix. Since every Hermitian linear transformation is normal, and every unitary linear transformation is normal, the above conclusion is also valid when either T is Hermitian or T is unitary. 4.1.17 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a normal linear transformation. Then T is Hermitian if and only if all the eigenvalues of T are real. Proof In view of 3.1.13, it remains to show that if all the eigenvalues of T are real, then T is Hermitian. So we suppose that all the eigenvalues of T are real. We have to show that T is Hermitian, that is, T ¼ T: Since T is normal, by 4.1.16, there exists an orthonormal basis fe1 ; . . .; en g of V such that the matrix of T relative to the basis fe1 ; . . .; en g is a diagonal matrix, say diagða1 ; . . .; an Þ: It follows that T ðe1 Þ ¼ a1 e1 þ 0e2 þ þ 0en ¼ a1 e1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence T ðe1 Þ ¼ a1 e1 : Since fe1 ; . . .; en g is a basis, we have e1 6¼ 0: Now, since T ðe1 Þ ¼ a1 e1 ; a1 is an eigenvalue of T. Next, by assumption, a1 is a real number. Similarly, a2 is a real number, etc. It follows that


269

ðdiagða1 ; . . .; an ÞÞ ¼ ðdiagða1 ; . . .; an ÞÞT ¼ ðdiagða1 ; . . .; an ÞÞ ¼ diagða1 ; . . .; an Þ ¼ diagða1 ; . . .; an Þ ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence ðdiagða1 ; . . .; an ÞÞ ¼ diagða1 ; . . .; an Þ: Since the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ; by 3.1.11, the matrix of T relative to the basis fe1 ; . . .; en g is ðdiagða1 ; . . .; an ÞÞ ð¼ diagða1 ; . . .; an ÞÞ; and hence the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ: Now, since the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ; the matrices of T and T relative to the basis fe1 ; . . .; en g are equal. It follows that T ðei Þ ¼ T ðei Þði ¼ 1; . . .; nÞ: Now, since T and T are linear, for every v 2 V; T ðvÞ ¼ T ðvÞ; and hence T ¼ T: ∎ 4.1.18 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a normal linear transformation. Then T is unitary if and only if the absolute value of each eigenvalue of T is 1. Proof In view of 3.1.22, it remains to show that if the absolute value of each eigenvalue of T is 1, then T is unitary. So we suppose that the absolute value of each eigenvalue of T is 1 and show that T is unitary. In view of 3.1.10, it suffices to show that T T ¼ I: Since T is normal, by 4.1.16, there exists an orthonormal basis fe1 ; . . .; en g of V such that the matrix of T relative to the basis fe1 ; . . .; en g is a diagonal matrix, say diagða1 ; . . .; an Þ: Since the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ; by 3.1.11, the matrix of T relative to the basis fe1 ; . . .; en g is ðdiagða1 ; . . .; an ÞÞ ¼ ðdiagða1 ; . . .; an ÞÞT ¼ ðdiagða1 ; . . .; an ÞÞ ¼ diagða1 ; . . .; an ÞÞ;

and hence the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ: Now, since the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ; by 3.1.33, the matrix of T T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ diagða1 ; . . .; an Þ ¼ diagða1 a1 ; . . .; an an Þ ¼ diag ja1 j2 ; . . .; jan j2 ; and hence

the

matrix 2 2 diag ja1 j ; . . .; jan j :

of T T

relative

to the

basis

fe1 ; . . .; en g is

Since the matrix of T relative to the basis fe1 ; . . .; en g is diagða1 ; . . .; an Þ; we have

270


T ðe1 Þ ¼ a1 e1 þ 0e2 þ þ 0en ¼ a1 e1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} and hence T ðe1 Þ ¼ a1 e1 : Since fe1 ; . . .; en g is a basis, we have e1 6¼ 0: Now, since T ðe1 Þ ¼ a1 e1 ; a1 is an eigenvalue of T. Next, by assumption, ja1 j ¼ 1: Similarly, ja2 j ¼ 1, etc. Now, since the matrix of T T relative to the basis fe1 ; . . .; en g is diag ja1 j2 ; . . .; jan j2 ¼ diag 12 ; . . .; 12 ¼ diagð1; . . .; 1Þ ¼ dij ;

the matrix of T T relative to the basis fe1 ; . . .; en g is dij : Also, the matrix of I relative

to the basis fe1 ; . . .; en g is dij : So, the matrices of T T and I relative to the basis fe1 ; . . .; en g are equal. It follows that ðT T Þðei Þ ¼ I ðei Þði ¼ 1; . . .; nÞ: Now, since T T and I are linear, for every v 2 V; ðT T ÞðvÞ ¼ I ðvÞ; and hence T T ¼ I: ∎ 4.1.19 Theorem Let V be any n-dimensional inner product space over the field C: Let N : V ! V be a normal linear transformation. Let T : V ! V be a linear transformation. Suppose that TN ¼ NT: Then TN ¼ N T: Proof Let us put X TN N T: We have to show that X ¼ 0: Since N is normal, we have NN ¼ N N; and hence N commutes with N : Since TN ¼ NT; N commutes with T. Since N commutes with T and N ; N commutes with ðTN N T Þð¼ X Þ; and hence N commutes with X By 3.1.7, 3.1.8, and 3.1.6, we have X ¼ ðTN N T Þ ¼ ðTN Þ ðN T Þ ¼ ðN Þ T T ðN Þ ¼ NT T N; and hence XX ¼ ðTN N T ÞðNT T N Þ ¼ ðTN N T ÞNT ðTN N T ÞT N |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ððTN N T ÞN ÞT ðTN N T ÞT N ¼ ðN ðTN N T ÞÞT ðTN N T ÞT N ¼ N ððTN N T ÞT Þ ðTN N T ÞT N ¼ N ððTN N T ÞT Þ ððTN N T ÞT ÞN ¼ NB BN; where B ðTN N T ÞT : Thus XX ¼ NB BN: Since N is normal, by 4.1.16, there exists an orthonormal basis fe1 ; . . .; en g of V such that the matrix of T relative to the basis fe1 ; . . .; en g is a diagonal matrix, say diagða1 ; . . .; an Þ: Let bij be the matrix of B relative to the basis fe1 ; . . .; en g:


271

3.1.33, the matrix of BN relative to the basis fe1 ; . . .; en g is By

bij ðdiagða1 ; . . .; an ÞÞ: Clearly, the diagonal entries of bij ðdiagða1 ; . . .; an ÞÞ are b11 a1 ; b22 a2 ; . . .; bnn an : Thus the diagonal entries of the matrix of BN relative to the basis fe1 ; . . .; en g are b11 a1 ; b22 a2 ; . . .; bnn an : By 3.1.33, the matrix of NB relative to the basis fe1 ; . . .; en g is ðdiagða1 ; . . .; an ÞÞ bij : Clearly, the diagonal entries of ðdiagða1 ; . . .; an ÞÞ bij are a1 b11 ; a2 b22 ; . . .; an bnn : Thus the diagonal entries of the matrix of NB relative to the basis fe1 ; . . .; en g are a1 b11 ; a2 b22 ; . . .; an bnn : It follows that the matrix of NB BN relative to the basis fe1 ; . . .; en g is

ðdiagða1 ; . . .; an ÞÞ bij bij ðdiagða1 ; . . .; an ÞÞ; and hence the diagonal entries of the matrix of NB BN ð¼ XX Þ relative to the basis fe1 ; . . .; en g are all 0: Thus, the diagonal entries of the matrix of XX relative to the basis fe1 ; . . .; en g are all 0:

Let xij be the matrix of X relative to the basis fe1 ; . . .; en g: By 3.1.11, the matrix of X relative to the basis fe1 ; . . .; en g is T T xij ¼ xij ¼ xij ; and hence by 3.1.33, the matrix of XX relative to the basis fe1 ; . . .; en g is T T xij xij : Here the diagonal entries of xij xij are n X

x1j x1J ;

j¼1

n X

x2j x2J ; . . .;

j¼1

n X

xnj xnJ ;

j¼1

that is, the diagonal entries of xij ½xiJ T are n X n n X X x1j 2 ; x2j 2 ; . . .; xnj 2 : j¼1

j¼1

j¼1

Hence the diagonal entries of the matrix of XX relative to the basis fe1 ; . . .; en g are n X n n X X x1j 2 ; x2j 2 ; . . .; xnj 2 : j¼1

j¼1

j¼1

Now, since the diagonal entries of the matrix of XX relative to the basis fe1 ; . . .; en g are all 0; we have

272

4 Sylvester’s Law of Inertia n n X X x1j 2 ¼ 0; x2j 2 ¼ 0; etc: j¼1

j¼1

Pn 2 Since j¼1 x1j ¼ 0; we have x1j ¼ 0ðj ¼ 1; . . .; nÞ: Similarly, x2j ¼

0 ðj ¼ 1; . . .; nÞ; etc. Thus each xij is 0, and hence the matrix xij of X relative to the ∎ basis fe1 ; . . .; en g is the zero matrix. This shows that X ¼ 0: 4.1.20 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Then T is Hermitian if and only if for every v 2 V; hT ðvÞ; vi is a real number. Proof Let T be Hermitian, that is, T ¼ T: We have to show that for every v 2 V; hT ðvÞ; vi is a real number. To do so, let us take an arbitrary v 2 V: We have to show that hT ðvÞ; vi is a real number, that is, hT ðvÞ; vi ¼ hT ðvÞ; vi: LHS ¼ hT ðvÞ; vi ¼ hv; T ðvÞi ¼ hv; T ðvÞi ¼ hT ðvÞ; vi ¼ RHS: Conversely, suppose that for every v 2 V; hT ðvÞ; vi is a real number. We have to show that T is Hermitian, that is, T ¼ T; that is, X ¼ 0; where X T T: By 3.1.1, it suffices to show that for every v 2 V; hX ðvÞ; vi ¼ 0: LHS ¼ hX ðvÞ; vi ¼ hðT T ÞðvÞ; vi ¼ hT ðvÞ T ðvÞ; vi ¼ hT ðvÞ; vi hT ðvÞ; vi ¼ hv; T ðvÞi hT ðvÞ; vi ¼ hT ðvÞ; vi hT ðvÞ; vi ¼ hT ðvÞ; vi hT ðvÞ; vi ¼ 0 ¼ RHS: Definition Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. If for every v 2 V; hT ðvÞ; vi is a nonnegative real number, then we write T 0; and we say that T is nonnegative (definite). By 4.1.20, if T 0; then T is Hermitian. Theorem 4.1.21 Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Suppose that T is nonnegative. Then all the eigenvalues of T are nonnegative. Proof To show this, let us take an arbitrary eigenvalue k of T. We have to show that k is a nonnegative real number. Since k is an eigenvalue of T, there exists a nonzero v 2 V such that T ðvÞ ¼ kv: Since T is nonnegative, ðkhv; vi ¼ hkv; vi ¼ÞhT ðvÞ; vi is a nonnegative real number, and hence khv; vi is a nonnegative real number. Since v is nonzero, hv; vi is a positive real number. Now, since khv; vi is a nonnegative real number, k is a nonnegative real number. ∎


273

Theorem 4.1.22 Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a Hermitian linear transformation. Suppose that all the eigenvalues of T are nonnegative. Then T is nonnegative. Proof To show this, let us take an arbitrary nonzero v 2 V: We have to show that hT ðvÞ; vi is a nonnegative real number. Since T is Hermitian, T is normal, and hence by 4.1.16, there exists an orthonormal basis fe1 ; . . .; en g of V such that the matrix of T relative to the basis fe1 ; . . .; en g is a diagonal matrix, say diagðt1 ; . . .; tn Þ: It follows that T ðe1 Þ ¼ t1 e1 þ 0e2 þ þ 0en ð¼ t1 e1 Þ; and hence T ðe1 Þ ¼ t1 e1 : Since fe1 ; . . .; en g is a basis, e1 is nonzero. Now, since T ðe1 Þ ¼ t1 e1 ; t1 is an eigenvalue of T. Here by assumption, t1 is a nonnegative real number. Similarly, T ðe2 Þ ¼ t2 e2 , and t2 is a nonnegative real number, etc. Since v 2 V; and fe1 ; . . .; en g is an orthonormal basis of V, we have v ¼ hv; e1 ie1 þ þ hv; en ien ; and hence * hT ðvÞ; vi ¼

T *

¼

n X

! hv; ei iei ;

i¼1 n X

¼ ¼

i¼1 n X

hv; ei iti ei ;

i¼1

hv; ei iti

+ hv; ei iei

i¼1 n X

i¼1 n X

n X

hv; ei iei

n X

v; ej ei ; ej

¼

¼ !

n X

hv; ei iti hv; ei i ¼

hv; ei iT ðei Þ; *

hv; ei iti ei ;

i¼1

¼

j¼1

n X i¼1

n X

n X i¼1

+

i¼1

*

n X

+ hv; ei iei

i¼1 n X

v; ej ej

+

j¼1

hv; ei iti

n X

v; ej dij

!

j¼1

jhv; ei ij2 ti :

i¼1

Thus hT ðvÞ; vi is a nonnegative real number.

∎

Definition Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. If for every nonzero v 2 V; hT ðvÞ; vi is a positive real number, then we write T [ 0; and we say that T is positive (definite). By 4.1.20, if T [ 0; then T is Hermitian. 4.1.23 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a Hermitian linear transformation. Suppose that T is positive. Then all the eigenvalues of T are positive. Proof To show this, let us take an arbitrary eigenvalue k of T We have to show that k is a positive real number.

274


Since k is an eigenvalue of T there exists a nonzero v 2 V such that T ðvÞ ¼ kv: Since T is positive, ðkhv; vi ¼ hkv; vi ¼ÞhT ðvÞ; vi is a positive real number, and hence khv; vi is a positive real number. Since v is nonzero, hv; vi is a positive real number. Now, since khv; vi is a positive real number, k is a positive real number. ∎ Theorem 4.1.24 Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a Hermitian linear transformation. Suppose that all the eigenvalues of T are positive. Then T is positive. Proof To show this, let us take an arbitrary nonzero v 2 V: We have to show that hT ðvÞ; vi is a positive real number. Since T is Hermitian, T is normal, and hence by 4.1.16, there exists an orthonormal basis fe1 ; . . .; en g of V such that the matrix of T relative to the basis fe1 ; . . .; en g is a diagonal matrix, say diagðt1 ; . . .; tn Þ: It follows that T ðe1 Þ ¼ t1 e1 þ 0e2 þ þ 0en ð¼ t1 e1 Þ; and hence T ðe1 Þ ¼ t1 e1 . Since fe1 ; . . .; en g is a basis, e1 is nonzero. Now, since T ðe1 Þ ¼ t1 e1 ; t1 is an eigenvalue of T. Here by assumption, t1 is a positive real number. Similarly, T ðe2 Þ ¼ t2 e2 , and t2 is a positive real number, etc. Since v 2 V; and fe1 ; . . .; en g is an orthonormal basis of V, we have v ¼ hv; e1 ie1 þ þ hv; en ien ; and hence * hT ðvÞ; vi ¼

T *

¼

n X

! hv; ei iei ;

i¼1 n X

¼ ¼

i¼1 n X

+ hv; ei iei

i¼1

hv; ei iti ei ;

n X

i¼1 n X

n X

hv; ei iei

hv; ei iti

n X

hv; eJ i ei ; ej

¼

¼ !

n X

i¼1

*

hv; ei iti ei ;

¼

n X i¼1

n X

hv; ei iT ðei Þ;

i¼1

j¼1

hv; ei iti hv; ei i ¼

n X i¼1

+

i¼1

*

n X

+ hv; ei iei

i¼1 n X

v; ej ej

+

j¼1

hv; ei iti

n X

! hv; eJ idij

j¼1

jhv; ei ij2 ti :

i¼1

Since v ¼ hv; e1 ie1 þ þ hv; en ien ; and v is nonzero, there exists j 2

2 f1; . . .; ng such that v; ej 6¼ 0; and hence v; ej [ 0: Also tj [ 0; so hT ðvÞ; vi ¼

n X

2 jhv; ei ij2 ti v; ej tj [ 0 : |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} i¼1

Thus hT ðvÞ; vi is a positive real number.

∎


275

4.1.25 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a Hermitian linear transformation. Let fe1 ; . . .; en g be an

orthonormal basis of T. Let A aij be the matrix of T relative to fe1 ; . . .; en g: Then A is a Hermitian matrix. Proof We have to show that A ¼ A: By 3.1.11, it suffices to show that aJi ¼ aij :

Since A aij is the matrix of T relative to fe1 ; . . .; en g; we have T ðe1 Þ ¼ a11 e1 þ a21 e2 þ þ an1 en ; and hence n

P hT ðe1 Þ; ei i ¼ ha11 e1 þ a21 e2 þ þ an1 en ; ei i ¼ aj1 ej ; ei |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} j¼1

¼

n P

n

P aj1 ej ; ei ¼ aj1 dji ¼ ai1 ;

j¼1

j¼1

etc. Thus aij ¼ T ej ; ei : Hence T ej ; ei is the matrix of T relative to

fe1 ; . . .; en g: Similarly, T ej ; ei is the matrix of T relative to fe1 ; . . .; en g:

Since T is Hermitian, we have T ¼ T: Since A ¼ aij ; we have AT ¼ bij ; where

bij aji ¼ T ðei Þ; ej : It follows that A ¼ AT ¼ bij ¼ biJ : Also, |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl}

aJi ¼ biJ ¼ hT ðei Þ; eJ i ¼ ej ; T ðei Þ ¼ T ej ; ei ¼ T ej ; ei ¼ aij ; so aJi ¼ aij :

∎

4.1.26 Note Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Let T be nonnegative.

Let fe1 ; . . .; en g be an orthonormal basis of T. Let A aij be the matrix of T relative to fe1 ; . . .; en g: It follows that T ðe1 Þ ¼ a11 e1 þ a21 e2 þ þ an1 en ; and hence hT ðe1 Þ; ei i ¼ ha11 e1 þ a21 e2 þ þ an1 en ; ei i ¼ ai1 ;

etc. Thus aij ¼ T ej ; ei : Hence T ej ; ei is the matrix of T relative to fe1 ; . . .; en g: Since T 0; by 4.1.20, T is a Hermitian linear transformation, and hence by 4.1.25, A is a Hermitian matrix. Since T is a Hermitian linear transformation, we have T ¼ T: Suppose that t1 ; . . .; tn are the eigenvalues of the linear transformation T.

276


Since T 0; by 4.1.21, all the eigenvalues of the linear transformation T are pffiffiffi nonnegative, that is, each ti is a nonnegative real number. Hence each ti is a nonnegative real number. Since A is a Hermitian matrix, A aij is a normal

matrix. Now, by 3.3.24, there exists a unitary matrix U uij such that 1. U AU is a diagonal matrix, 2. the eigenvalues of the matrix A (that is, the eigenvalues of the linear transformation T) are the diagonal entries of U AU: Thus U AU ¼ diagðt1 ; . . .; tn Þ: Put uj

n X

uij ei ðj ¼ 1; . . .; nÞ:

i¼1

Clearly, fu1 ; . . .; un g is an orthonormal basis of V.

Proof It suffices to show that uj ; uk ¼ dij : Since

uj ; uk ¼

n P i¼1

n n n n P P P P ulk el ¼ uij ei ; ulk el ¼ uij ulk hei ; el i i¼1 i¼1 l¼1 l¼1 l¼1 n n n n P P P P ¼ uij ulk dil ¼ uij uik ¼ uik uij ;

uij ei ;

n P

i¼1

l¼1

i¼1

i¼1

n

P

we have uj ; uk ¼ uik uij : Since uij is a unitary matrix, we have i¼1

" # n X

uki ukj ¼ ½uiJ T uij ¼ uij uij ¼ dij ; uj ; ui ¼ |fflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflffl} k¼1 and hence

uj ; ui

¼ dij : This shows that ui ; uj ¼ dij :

∎

Since U is a unitary matrix, U is invertible, and U 1 ¼ U : Let us define a linear W : V ! V as follows: for every i 2 transformation n P uki ek : It follows that the matrix of W relative to f1; . . .; ng; W ðei Þ ui ¼ k¼1

fe1 ; . . .; en g is uij :

Now, since A aij is the matrix of T relative to fe1 ; . . .; en g; by 3.1.35(a), the 1

matrix of T relative to fu1 ; . . .; un g is uij A uij ð¼ U 1 AU ¼ U AU ¼ diagðt1 ; . . .; tn ÞÞ; and hence the matrix of T relative to fu1 ; . . .; un g is diagðt1 ; . . .; tn Þ:


277

Let us define a linear transformation S : V ! V as follows: Sð ui Þ

pffiffiffi ti ui ði ¼ 1; . . .; nÞ:

pffiffiffiffi pffiffiffiffi This shows that t1 ; . . .; t1 are the eigenvalues of the linear transformation S. Also, the eigenvalues of the linear transformation S are nonnegative real numbers. Further, the matrix of the linear transformation S relative to the basis pffiffiffiffi pffiffiffiffi fu1 ; . . .; un g is diagð t1 ; . . .; tn Þ: Next, by 3.1.11, the matrix of the linear transformation S relative to the basis fu1 ; . . .; un g is pffiffiffiffi pffiffiffiffi

pffiffiffiffi pffiffiffiffi ðdiagð t1 ; . . .; tn ÞÞ ð¼ diagð t1 ; . . .; tn ÞÞ: It follows, by 3.1.33, that the matrix of the linear transformation SS relative to the basis fu1 ; . . .; un g is pffiffiffiffi pffiffiffiffi pffiffiffiffi pffiffiffiffi

diagð t1 ; . . .; tn Þðdiagð t1 ; . . .; tn ÞÞ pffiffiffiffi pffiffiffiffi pffiffiffiffi pffiffiffiffi ¼ diagð t1 ; . . .; tn Þdiagð t1 ; . . .; tn Þ ¼ diagðt1 ; . . .; tn Þ: Thus the matrix of the linear transformation SS relative to the basis fu1 ; . . .; un g is diagðt1 ; . . .; tn Þ: Now, since the matrix of T relative to fu1 ; . . .; un g is diagðt1 ; . . .; tn Þ; we have SS ¼ T: Clearly, S is a Hermitian linear transformation, that is, S ¼ S: Proof By 4.1.20, it suffices to show that for every v 2 V; hSðvÞ; vi is a real number. To this end, let us take an arbitrary v a1 u1 þ þ an un in V. Since hSðvÞ; vi ¼ hSða1 u1 þ þ an un Þ; a1 u1 þ þ an un i ¼ ha1 Sðu1 Þ þ þ an Sðun Þ; a1 u1 þ þ an un i pffiffiffiffi

pffiffiffiffi ¼ a1 t1 u1 þ þ an tn un ; a1 u1 þ þ an un pffiffiffiffi pffiffiffiffi pffiffiffiffi pffiffiffiffi ¼ a1 t1 a1 þ þ an tn an ¼ ja1 j2 t1 þ þ jan j2 tn ; hSðvÞ; vi is a real number.

∎

We have shown that S is a Hermitian linear transformation. Next, since the eigenvalues of the linear transformation S are nonnegative real numbers, by 4.1.22, S is nonnegative. 4.1.27 Conclusion Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a linear transformation. Let T be nonnegative. Then there exists a linear transformation S : V ! V such that 1. S 0; 2. SS ¼ T; 3. S2 ¼ T:

278


Definition Let A aij be an n-square complex matrix. Observe that for every x 2 Cn ; x Ax is a 1 1 matrix. By x Ax [ 0; we mean that the entry of the 1 1 matrix x Ax is a positive real number. If for every nonzero x 2 Cn ; x Ax [ 0; then we say that A is a positive definite matrix, and we write A [ 0: 4.1.28 Problem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a nonnegative linear

transformation. Let fv1 ; . . .; vn g be an orthonormal basis of V. Let A aij be the matrix of T relative to the basis fv1 ; . . .; vn g: Then A is a nonnegative definite matrix. Proof To show this, let us take an arbitrary x ½a1 ; . . .; an T 2 Cn : We have to show that x Ax

0: Since aij is the matrix of T relative to the basis fv1 ; . . .; vn g; it follows that n P aij vi : Now, T vj ¼ i¼1

T x Ax ¼ ½a1 ; . . .; an T A½a1 ; . . .; an T ¼ ½a1 ; . . .; an T A½a1 ; . . .; an T ¼ ½a1 ; . . .; an A½a1 ; . . .; an T ¼ ð½a1 ; . . .; an AÞ½a1 ; . . .; an T " # n n X X ¼ ai ai1 ; . . .; ai ain ½a1 ; . . .; an T i¼1

¼

n X

!

n n X X j¼1

n X

ai ai1 a1 þ þ

i¼1

¼

i¼1

! ai ain an

i¼1

! ai aij aj ¼

i¼1

n n X X j¼1

! ai aij aj ;

i¼1

so

x Ax ¼

n n X X j¼1

We have to show that

n n P P j¼1

! ai aij aj :

i¼1

ai aij aj 0:

i¼1

Since T : V ! V is a nonnegative linear transformation, we have hT ða1 v1 þ þ an vn Þ; ða1 v1 þ þ an vn Þi 0:


279

It suffices to show that hT ða1 v1 þ þ an vn Þ; ða1 v1 þ þ an vn Þi ¼

n n P P j¼1

LHS ¼ hT ða1 v1 þ þ an vn Þ; ða1 v1 þ þ an vn Þi

*

¼ ha1 T ðv1 Þ þ þ an T ðvn Þ; a1 v1 þ þ an vn i ¼ ¼

n X

* ai T ðvi Þ;

n X

i¼1

¼

n X i¼1

¼ ¼ ¼

ai

n X

*

n X

aki vk ;

ai aki

ai

n X

¼

aJ v k ; v j

! ai aki ak

¼

!!

n n X X i¼1

¼

aj aij ai

n n X X

¼

j¼1

n X

ai T ðvi Þ;

!

+ aj v j

aj v j *

ai aki vk ;

n X

+! aj v j

j¼1

ai aki

k¼1

n X j¼1

+

k¼1

n n X X

n X

!! aJ dkj

j¼1

ak aik ai

i¼1

n n X X

n X j¼1

i¼1

k¼1

!

aki vk ;

k¼1

aj v j

n X

:

i¼1

i¼1

j¼1

k¼1

i¼1

*

+!

j¼1

k¼1

n n X X j¼1

¼

n X i¼1

k¼1

n n X X i¼1

aj v j

j¼1

n n X X i¼1

+

ai aij aj

! ai aij aj

¼ RHS:

i¼1

∎

4.1.29 Problem Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Let T : x 7! Ax be the linear transformation from the inner product space Cn to Cn : Then T is a nonnegative linear transformation. Proof To show this, let us take an arbitrary x ½x1 ; . . .; xn T 2 Cn : We have to show that hT ð xÞ; xi 0; that is, hAx; xi 0: Since A is a nonnegative definite matrix, and x 2 Cn ; we have x Ax 0: It suffices to show that hAx; xi ¼ x Ax: By the definition of inner product of Cn ; ∎ hAx; xi ¼ x ðAxÞ ¼ x Ax; we have hAx; xi ¼ x Ax: |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl}

4.1.30 Problem Let A aij be an n-square complex matrix. Suppose that A is a

nonnegative definite matrix. Let X xij n m ¼ ½x1 ; . . .; xm be any complex matrix of size n m: (It is clear that X AX is an m m matrix.) Then X AX 0: Proof To show this, let us take an arbitrary x 2 Cm : We have to show that x ðX AX Þx 0: Since

280


x ðX AX Þx ¼ ðx X ÞAðXxÞ ¼ ðXxÞ AðXxÞ; we have to show that y Ay 0; where y Xx: Since X is a complex matrix of size n m; and x 2 Cm ; we have ðy ¼ÞXx 2 Cn ; and hence y 2 Cn : Now, since A is a nonnegative definite matrix of size n n; we have y Ay 0: ∎

4.1.31 Problem Let A aij be an n-square complex matrix. Suppose that for

every n m complex matrix X xij n m ¼ ½x1 ; . . .; xm ; X AX 0: Then A 0: Proof To show this, let us take an arbitrary x 2 Cn : We have to show that x Ax 0: Let us take x1 ¼ x; x2 ¼ 0; . . .; xm ¼ 0: By assumption, X AX 0: Observe that X AX ¼ ½x1 ; . . .; xm A½x1 ; . . .; xm ¼ ½ðx1 Þ ; . . .; ðxm Þ A½x1 ; . . .; xm T

¼ ½ðx1 Þ ; . . .; ðxm Þ ðA½x1 ; . . .; xm Þ ¼ ½ðx1 Þ ; . . .; ðxm Þ ½Ax1 ; . . .; Axm 2 3 2 3 ðx1 Þ Ax1 ðx1 Þ Ax2 ðx1 Þ Ax1 ðx1 Þ A0 6 7 6 7 ðx2 Þ Ax1 ðx2 Þ Ax2 7 ¼ 6 0 Ax1 0 A0 7 ¼6 4 5 4 5 .. .. .. .. .. .. . . . . . . 2 3

ðx1 Þ Ax1 0 6 0 0 7 ¼4 5; .. . . .. . . . T

T

so 2

ðx1 Þ Ax1

0 X AX ¼ 4 .. .

0 0 .. .

3 5: .. .

Now, since X AX 0; and ½1; 0; . . .; 0T 2 Cn ; we have

0 ½1; 0; . . .; 0T ðX AX Þ ½ 1; 0; 0 T |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 2 3 2 3 1 ðx1 Þ Ax1 0 607 0 0 56 . 7 ¼ ½ 1 0 0 4 .. . . 4 .. 5 .. . . . 0 2 3

ðx1 Þ Ax1 6 7 0 7 ¼ ðx1 Þ Ax1 ¼ x Ax; ¼ ½ 1 0 0 6 .. 4 5 . 0 and hence, 0 x Ax:

∎


281

4.1.32 Note Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix.

By 3.3.28, A is a Hermitian matrix, and hence A aij is a normal matrix. Now,

by 3.3.24, there exists a unitary matrix U uij such that 1. U AU is a diagonal matrix, say diagðk1 ; . . .; kn Þ; 2. the eigenvalues of the matrix Aare k1 ; . . .; kn : Since U is unitary, we have U U ¼ UU ¼ I; and hence U 1 ¼ U : Now, since U AU ¼ diagðk1 ; . . .; kn Þ; we have A ¼ U ðdiagðk1 ; . . .; kn ÞÞU : Let T : x 7! Ax be the linear transformation from the inner product space Cn to n C : By 4.1.29, T is a nonnegative linear transformation, and by 4.1.21, all the eigenvalues of the linear transformation T are nonnegative. Here fe1 ; . . .; en g is an orthonormal basis of V, where e1 ½1; 0; . . .; 0T ; e2 ½0; 1; 0; . . .; 0T ; etc. Since

T ðe1 Þ ¼ Ae1 ¼ aij ½1; 0; . . .; 0T ¼ ½a11 ; a21 ; . . .; an1 T ¼ a11 e1 þ a21 e2 þ þ an1 en ; we have T ðe1 Þ ¼ a11 e1 þ a21 e2 þ þ an1 en : Similarly, T ðe2 Þ ¼ a12 e1 þ a22 e2 þ þ an2 en ; P etc. Thus T ej ¼ nj¼1 aij ej : It follows that the matrix of T relative to fe1 ; . . .; en g

is aij ð¼ AÞ: Now, since the eigenvalues of the matrix A are k1 ; . . .; kn ; the eigenvalues of the linear transformation Tare also k1 ; . . .; kn : Next, since the eigenvalues of the linear transformation T are nonnegative real numbers, each ki is a nonnegative real number. Since A ¼ U ðdiagðk1 ; . . .; kn ÞÞU ¼ U ðdiagðk1 ; . . .; kn ÞÞU 1 ; we have

detð AÞ ¼ det U ðdiagðk1 ; . . .; kn ÞÞU 1 ¼ detðU Þ detðdiagðk1 ; . . .; kn ÞÞ detðU 1 Þ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ detðU Þ detðdiagðk1 ; . . .; kn ÞÞ det1ðU Þ ¼ detðdiagðk1 ; . . .; kn ÞÞ ¼ k1 k2 kn ;

and hence detð AÞ ¼ k1 k2 . . .kn : Since each ki is a nonnegative real number, detð AÞ is a nonnegative real number.

282


4.1.33 Conclusion Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Then there exists a unitary matrix U such that 1. 2. 3. 4.

A ¼ U ðdiagðk1 ; . . .; kn ÞÞU ; k1 ; . . .; kn are the eigenvalues of the matrix A, each ki is a nonnegative real number, detð AÞ is a nonnegative real number.

4.1.34 Note Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. By 4.1.33, there exists a unitary matrix U such that 1. 2. 3. 4.

A ¼ U ðdiagðk1 ; . . .; kn ÞÞU ; k1 ; . . .; kn are the eigenvalues of the matrix A, each ki is a nonnegative real number, detð AÞ is a nonnegative real number.

Since each ki is a nonnegative real number, each number. Observe that

pffiffiffiffi ki is a nonnegative real

pffiffiffiffiffi pffiffiffiffiffi 2 k ; . . .; k U

pffiffiffiffiffi U diag pffiffiffiffiffi 1 n pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn U U diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn ðU U Þ diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi

¼ Udiag k1 ; . . .; kn I diag k1 ; . . .; knU pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn diag k1 ; . . .; kn U

¼ U ðdiagðk1 ; . . .; kn ÞÞU ¼ A; so pffiffiffiffiffi pffiffiffiffiffi 2 U diag k1 ; . . .; kn U ¼ A: Thus B2 ¼ A; pffiffiffiffiffi pffiffiffiffiffi where B U diag k1 ; . . .; kn U : Since for every x ½x1 ; . . .; xn T 2 Cn ; pffiffiffiffiffi pffiffiffiffiffi x diag k1 ; . . .; kn x ffiffiffiffi ffi p pffiffiffiffiffi

¼ ½x1 ; . . .; xn T diag k1 ; . . .; kn ½x1 ; . . .; xn T pffiffiffiffiffi pffiffiffiffiffi ¼ ½x1 ; . . .; xn diag k1 ; . . .; kn ½x1 ; . . .; xn T pffiffiffiffiffi pffiffiffiffiffi ¼ ½x ; . . .; xn diag k1 ; . . .; kn ½x1 ; . . .; xn T pffiffiffiffiffi 1

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ x1 k1 ; . . .; xn kn ½x1 ; . . .; xn T ¼ x1 k1 x1 þ . . .; xn kn xn pffiffiffiffiffi 2 pffiffiffiffiffi 2 ¼ k1 jx1 j þ þ kn jxn j 0;


283

we have pffiffiffiffiffi pffiffiffiffiffi x diag k1 ; . . .; kn x 0: pffiffiffiffiffi pffiffiffiffiffi This shows that diag k1 ; . . .; kn is a nonnegative definite matrix, and hence pffiffiffiffiffi p p ffiffiffiffiffi pffiffiffiffiffi ffiffiffiffi ffi by 4.1.30, ðU Þ diag k1 ; . . .; kn ðU Þ ¼ U diag k1 ; . . .; kn U ¼ B is a nonnegative definite matrix. Thus B is a nonnegative matrix. Since pffiffiffiffiffi pffiffiffiffiffi kI B ¼ kI U diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi ¼ kUU U diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi ¼ U ðkI ÞU U diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi ¼ U kI diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi ¼ U diagðk; . . .; kÞ diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k k1 ; . . .; k kn U ; we have pffiffiffiffiffi pffiffiffiffiffi kI B ¼ U diag k k1 ; . . .; k kn U ; and hence pffiffiffiffiffi pffiffiffiffiffi detðkI BÞ ¼ det U diag k k1 ; . . .; k kn U

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} pffiffiffiffiffi pffiffiffiffiffi ¼ detðU Þ det diag k k1 ; . . .; k kn detðU Þ pffiffiffiffiffi pffiffiffiffiffi ¼ detðU Þ det diag k k1 ; . . .; k kn det U 1 pffiffiffiffiffi pffiffiffiffiffi 1 ¼ detðU Þ det diag k k1 ; . . .; k kn detðU Þ pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ det diag k k1 ; . . .; k kn ¼ k k1 k k2 k kn : Thus pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi detðkI BÞ ¼ k k1 k k2 k kn :

284


pffiffiffiffiffi Hence the characteristic polynomial of the matrix B is k k1 pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi k k2 k kn : Its roots are k1 ; . . .; kn : pffiffiffiffiffi pffiffiffiffiffi So the eigenvalues of the matrix B are k1 ; . . .; kn : Also, k1 ; . . .; kn are the eigenvalues of the matrix A. Thus the eigenvalues of the matrix B are the square roots of the eigenvalues of the matrix A.

4.1.35 Conclusion Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Let k1 ; . . .; kn be the eigenvalues of the matrix A. Then there exists a matrix B such that 1. 2. 3. 4.

B2 ¼ A; B is a nonnegative definite matrix, pffiffiffiffiffi pffiffiffiffiffi k1 ; . . .; kn are the eigenvalues of the matrix B, pffiffiffiffiffi pffiffiffiffiffi there exists a unitary matrix U such that U diag k1 ; . . .; kn U ¼ B:

4.1.36 Problem Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Let k1 ; . . .; kn be the eigenvalues of the matrix A. Then there exists a unique matrix B such that 1. B2 ¼ A; pffiffiffiffiffi pffiffiffiffiffi 2. there exists a unitary matrix U such that U diag k1 ; . . .; kn U ¼ B: Proof In view of 4.1.25, it remains to prove the uniqueness part. Uniqueness: Suppose that B is a matrix such that 1. B2 ¼ A; pffiffiffiffiffi pffiffiffiffiffi 2. there exists a unitary matrix U such that U diag k1 ; . . .; kn U ¼ B: Suppose that C is a matrix such that 1. C 2 ¼ A; pffiffiffiffiffi pffiffiffiffiffi 2. there exists a unitary matrix V such that V diag k1 ; . . .; kn V ¼ C: We have to show that B ¼ C; pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi U diag k1 ; . . .; kn U ¼ V diag k1 ; . . .; kn V ;that is,

that

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi U diag k1 ; . . .; kn ¼ V diag k1 ; . . .; kn V U; that is, pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi V U diag k1 ; . . .; kn ¼ diag k1 ; . . .; kn V U; that is,

is,


285

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ diag k1 ; . . .; kn W; W diag k1 ; . . .; kn where W V U: pffiffiffiffiffi

pffiffiffiffiffi

Suppose that W wij : Observe that diag k1 ; . . .; kn ¼ sij ; where sij pffiffiffiffi ki dij : So we have to show that

wij sij ¼ sij wij ; that is, ! n n n pffiffiffiffiffi X X pffiffiffiffi pffiffiffiffi X wik wik skj ¼ sik wkj kk dkj ¼ kj wij ¼ wij kj ¼ k¼1

k¼1

k¼1

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

that is, pffiffiffiffi pffiffiffiffi kj wij ¼ ki wij ; that is,

pffiffiffiffi pffiffiffiffi ki kj wij ¼ 0:

Thus it suffices to show that pffiffiffiffi pffiffiffiffi ki kj wij ¼ 0: Since pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi A ¼ B2 ¼ BB ¼ U diag k1 ; . . .; kn U U diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn ðU U Þ diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn I diag k1 ; . . .; kn U

pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi pffiffiffiffiffi ¼ U diag k1 ; . . .; kn diag k1 ; . . .; kn U

¼ U ðdiagðk1 ; . . .; kn ÞÞU ¼ U ðdiagðk1 ; . . .; kn ÞÞU ; we have A ¼ U ðdiagðk1 ; . . .; kn ÞÞU :

286


Similarly, A ¼ V ðdiagðk1 ; . . .; kn ÞÞV : It follows that U ðdiagðk1 ; . . .; kn ÞÞU ¼ V ðdiagðk1 ; . . .; kn ÞÞV ; that is, V U ðdiagðk1 ; . . .; kn ÞÞU ¼ ðdiagðk1 ; . . .; kn ÞÞV ; that is, ðV U Þðdiagðk1 ; . . .; kn ÞÞ ¼ ðdiagðk1 ; . . .; kn ÞÞðV U Þ; that is, W ðdiagðk1 ; . . .; kn ÞÞ ¼ ðdiagðk1 ; . . .; kn ÞÞW; that is,

wij ðdiagðk1 ; . . .; kn ÞÞ ¼ ðdiagðk1 ; . . .; kn ÞÞ wij :

Observe that diagðk1 ; . . .; kn Þ ¼ jij ; where jij ki dij : It follows that

wij jij ¼ jij wij ; and hence n X

wik jkj ¼

k¼1

n X

jik wkj :

k¼1

Now, since n X

wik jkj ¼

k¼1

n X

wik kk dkj ¼ wij kj ;

k¼1

and n X k¼1

jik wkj ¼

n X ðki dik Þwkj ¼ ki wij ¼ wij ki ; k¼1

we have wij kj ¼ wij ki ; and hence ki kj wij ¼ 0: It follows that for distinct ki and pffiffiffiffi pffiffiffiffi kj ; wij ¼ 0: Hence ki kj wij ¼ 0: ∎


287

4.1.37 Theorem Let A aij be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Then there exists a unique matrix B such that 1. B2 ¼ A; 2. B is a nonnegative definite matrix. Here the unique matrix B is denoted by nonnegative definite matrix A.

pffiffiffi A and is called the square root of the

Proof In view of 4.1.35, it remains to prove the uniqueness part. Uniqueness: Let k1 ; . . .; kn be the eigenvalues of the matrix A. Suppose that B is a matrix such that 1. B2 ¼ A; 2. B is a nonnegative definite matrix. Suppose that C is a matrix such that 1. C 2 ¼ A; 2. C is a nonnegative definite matrix. We have to show that B ¼ C: Since B is a nonnegative definite matrix, by 3.3.28, B is a Hermitian matrix, and hence by 3.3.24, there exists a unitary matrix U such that 1. U BU is a diagonal matrix, 2. the eigenvalues of B are the diagonal entries of U BU: Hence U BU ¼ diagðl1 ; . . .; ln Þ; where l1 ; . . .; ln are the eigenvalues of B. It follows that B ¼ U ðdiagðl1 ; . . .; ln ÞÞU : Now, since A ¼ B2 ¼ ðU ðdiagðl1 ; . . .; ln ÞÞU Þ2 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ ðU ðdiagðl1 ; . . .; ln ÞÞU ÞðU ðdiagðl1 ; . . .; ln ÞÞU Þ ¼ U ðdiagðl1 ; . . .; ln Þðdiagðl1 ; . . .; ln ÞÞÞU

¼ U diag ðl1 Þ2 ; . . .; ðln Þ2 U ; we have A ¼ U diag ðl1 Þ2 ; . . .; ðln Þ2 U :

288

Since

we have

and hence

Thus


kI A ¼ kI U diag ðl1 Þ2 ; . . .; ðln Þ2 U

¼ kUU U diag ðl1 Þ2 ; . . .; ðln Þ2 U

¼ U ðkI ÞU U diag ðl1 Þ2 ; . . .; ðln Þ2 U

¼ U kI diag ðl1 Þ2 ; . . .; ðln Þ2 U

¼ U diagðk; . . .; kÞ diag ðl1 Þ2 ; . . .; ðln Þ2 U

¼ U diag k ðl1 Þ2 ; . . .; k ðln Þ2 U ; kI A ¼ U diag k ðl1 Þ2 ; . . .; k ðln Þ2 U ; detðkI AÞ ¼ det U diag k ðl1 Þ2 ; . . .; k ðln Þ2 U

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ detðU Þ det diag k ðl1 Þ2 ; . . .; k ðln Þ2 detðU Þ ¼ detðU Þ det diag k ðl1 Þ2 ; . . .; k ðln Þ2 det U 1 1 ¼ detðU Þ det diag k ðl1 Þ2 ; . . .; k ðln Þ2 detðU Þ ¼ det diag k ðl1 Þ2 ; . . .; k ðln Þ2 ¼ k ðl1 Þ2 k ðl2 Þ2 k ðln Þ2 : detðkI AÞ ¼ k ðl1 Þ2 k ðl2 Þ2 k ðln Þ2 :

Hence the characteristic polynomial of the matrix A is k ðl1 Þ2 k ðl2 Þ2 k ðln Þ2 : Its roots are ðl1 Þ2 ; . . .; ðln Þ2 : So the eigenvalues of the matrix A are ðl1 Þ2 ; . . .; ðln Þ2 : Now, since the eigenvalues of the matrix A are k1 ; . . .; kn ; we can suppose that ðl1 Þ2 ¼ k1 : Since l1 is an eigenvalue of B, and B is a nonnegative definite matrix, qffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffi by 3.3.30, l1 is a nonnegative real number. It follows that k1 ¼ ðl1 Þ2 ¼ l1 ; |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} pffiffiffiffiffi pffiffiffiffiffi and hence l1 ¼ k1 : Similarly, l2 ¼ k2 ; etc.


289

Next, since pffiffiffiffiffi pffiffiffiffiffi U BU ¼ diagðl1 ; . . .; ln Þ ¼ diag k1 ; . . .; k2 ; we have U BU ¼ diag

pffiffiffiffiffi pffiffiffiffiffi k1 ; . . .; k2 ;

and hence pffiffiffiffiffi pffiffiffiffiffi B ¼ U diag k1 ; . . .; k2 U : Similarly, pffiffiffiffiffi pffiffiffiffiffi C ¼ V diag k1 ; . . .; k2 V : Thus pffiffiffiffiffi pffiffiffiffiffi 1. B2 ¼ A; 2; U diag k1 ; . . .; kn U ¼ B; where U is a unitary matrix; 3. pffiffiffiffiffi pffiffiffiffiffi C 2 ¼ A; 4. V diag k1 ; . . .; kn V ¼ C; where V is a unitary matrix. Now, by 4.1.36, B ¼ C: ∎ 4.1.38 Theorem Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a nonnegative linear transformation. Then there exists a unique nonnegative linear transformation S : V ! V such that S2 ¼ T: pffiffiffiffi Here the unique linear transformation S is denoted by T ; and is called the square root of the nonnegative linear transformation T. Proof In view of 4.1.27, it remains to prove the uniqueness part. Uniqueness: Let R : V ! V be a nonnegative linear transformation such that R2 ¼ T: Let S : V ! V be a nonnegative linear transformation such that S2 ¼ T: We have to show that R ¼ S:

Let us take an orthonormal basis fe1 ; . . .; en g of V. Let A aij be the matrix of

T relative to the basis fe1 ; . . .; en g: Let B bij be the matrix of R relative to the

basis fe1 ; . . .; en g: Let C cij be the matrix of S relative to the basis fe1 ; . . .; en g: Thus 9 n P > T ej ¼ aij ei > > > > i¼1 > = n P R ej ¼ bij ei : > i¼1 > > n > P > S ej ¼ cij ei > ; i¼1

290


It suffices to show that bij ¼ cij ði; j 2 f1; . . .; ngÞ; that is, B ¼ C: Since T is a nonnegative linear transformation, by 4.1.28, A is a nonnegative definite matrix. Similarly, B is a nonnegative definite matrix, and C is a nonnegative definite matrix. By 3.1.33, the matrix of R2 ð¼ R R ¼ T Þ relative to the basis fe1 ; . . .; en g is BBð¼ B2 Þ: Now, since the matrix of T relative to the basis fe1 ; . . .; en g is A, We have B2 ¼ A: Similarly, C2 ¼ A: Now, by 4.1.37, B ¼ C: ∎

4.2

Sylvester’s Law

4.2.1 Note Let A aij be an n-square real matrix. Since aiJ ¼ aij ; A is Hermitian if and only if A is symmetric (that is, AT ¼ A). T T Here A is unitary if and only if A is orthogonal (that is, AA ¼ A A ¼ I). Let A aij be a real symmetric matrix. It follows that A is a Hermitian matrix, and hence A is a normal matrix. Now, by 3.3.24, there exists a unitary matrix U such that 1. U AU is a diagonal matrix, 2. the eigenvalues of A are the diagonal entries of U AU: Hence U AU ¼ diagðk1 ; . . .; kn Þ; where k1 ; . . .; kn are the eigenvalues of the matrix A. It follows that diagðk1 ; . . .; kn Þ ¼ U AU ¼ U A U ¼ U A ðU Þ

¼ ðU AU Þ ¼ ðdiagðk1 ; . . .; kn ÞÞ ¼ diag k1 ; . . .; kn ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

and hence diagðk1 ; . . .; kn Þ ¼ diag k1 ; . . .; kn : It follows that ki ¼ ki ði ¼ 1; . . .; nÞ; and hence each ki is a real number.

4.2.2 Conclusion Let A aij be a real symmetric matrix. Then there exists a unitary matrix U such that U AU ¼ diagðk1 ; . . .; kn Þ; where k1 ; . . .; kn are real numbers.

4.2.3 Theorem Let A aij be an n-square real matrix. Let B bij be an nsquare real matrix. Let P be an invertible n-square complex matrix such that B ¼ P1 AP: Then there exists an invertible n-square real matrix Q such that B ¼ Q1 AQ: Proof Since P is an n-square complex matrix, we can write P ¼ P1 þ iP2 ; where P1 ; P2 are n-square real matrices. Case I: P2 ¼ 0: In this case, P ¼ P1 : Now, since B ¼ P1 AP; we have B ¼ ðP1 Þ1 AP1 ; where P1 is an n-square real matrix.

4.2 Sylvester’s Law

291

Case II: P2 6¼ 0: Since B ¼ P1 AP; we have ¼ AP ¼ AðP1 þ iP2 Þ ¼ AP1 þ iAP2 ; P1 B þ iP2 B ¼ ðP1 þ iP2 ÞB ¼ PB |fflfflfflfflffl{zfflfflfflfflffl} and hence ðP1 BÞ þ iðP2 BÞ ¼ ðAP1 Þ þ iðAP2 Þ::

ð Þ

Since P1 ; B are real matrices, P1 B is a real matrix. Similarly, P2 B; AP1 ; AP2 are real matrices. Now, from (*) P1 B ¼ AP1 : P2 B ¼ AP2

ð

Þ

Since P2 6¼ 0; detðP1 þ xP2 Þ is a polynomial in x. Suppose that fa1 ; . . .; ak g is the collection of all the roots of the polynomial detðP1 þ xP2 Þ: We can find a real number t0 62 fa1 ; . . .; ak g: It follows that detðP1 þ t0 P2 Þ 6¼ 0: Hence P1 þ t0 P2 is an invertible n-square matrix. Since P1 ; P2 are real matrices and t0 is a real number, P1 þ t0 P2 is an n-square real matrix. Thus Q is an invertible n-square real matrix, where Q P1 þ t0 P2 : It remains to show that Q1 AQ ¼ B; that is, AQ ¼ QB; that is, AðP1 þ t0 P2 Þ ¼ ðP1 þ t0 P2 ÞB; that is, AP1 þ t0 ðAP2 Þ ¼ P1 B þ t0 ðP2 BÞ: This is clearly true from (**) ∎ 4.2.4 Note Let k and l be distinct complex numbers. Observe that 2

1

6 40 0

0

0

3

2

1

0 0

3

7 6 7 0 1 5ðdiagðk; l; kÞÞ4 0 0 1 5 1 0 0 1 0 02 3 12 3 1 0 0 1 0 0 B6 7 C6 7 ¼ @4 0 0 1 5ðdiagðk; l; kÞÞA4 0 0 1 5 0 1 0 0 1 0 3 02 32 312 1 0 0 1 0 0 k 0 0 7 B6 76 7C6 ¼ @4 0 0 1 54 0 l 0 5A4 0 0 1 5 0 1 0 0 1 0 0 0 k 2 32 3 2 3 0 1 k 0 0 1 0 0 k 0 0 6 76 7 6 7 ¼ 4 0 0 k 54 0 0 1 5 ¼ 4 0 k 0 5 ¼ diag@ k; k ; l A; |{z} |{z} 2 0 l 0 0 1 0 0 0 l 1

and hence

292


2

3 2 0 1 1 5ðdiagðk; l; kÞÞ4 0 0 0 1

1 0 40 0 0 1 0

0 0 1

3 0 1 0 1 5 ¼ diag@ k; k ; l A: |{z} |{z} 0 2 1

Notation diag@ k; k ; l A is denoted by kI2 lI1 : |{z} |{z} 2

1

Thus 2

3 2 1 0 1 5ðdiagðk; l; kÞÞ4 0 0 0

1 0 40 0 0 1

3 0 1 5 ¼ kI2 lI1 : 0

0 0 1

ð Þ

Observe that 2

1 40 0

0 0 1

32 1 0 1 54 0 0 0

0 0 1

3 2 0 1 15 ¼ 40 0 0

0 1 0

3 0 0 5 ¼ I3 ; 1

so 2

1 40 0 2

3 0 0 0 15 1 0 2 1 Hermitian. Thus 4 0 0 Also, from (*) 1 Thus 4 0 0

31 2 0 0 1 0 15 ¼ 40 1 0 0

3 0 0 0 1 5: 1 0 2

is 0 0 1

1 invertible. It is clear that 4 0 0 3 0 1 5 is unitary. 0 2

1 0 diagðk; l; kÞ ¼ 4 0 0 0 1

3 0 0 0 1 5 is symmetric and 1 0

3 2 0 1 1 5ðkI2 lI1 Þ4 0 0 0

0 0 1

3 0 1 5: 0

3

2 0 1 1 5 ðkI2 lI1 Þ4 0 0 0

0 0 1

3 0 1 5: 0

Hence 2

1 diagðk; l; kÞ ¼ 4 0 0

0 0 1


293

4.2.5 Conclusion Let D be a diagonal matrix. Then D can be expressed as Q ðkIr lIs ÞQ; where k; . . .; l are the distinct members of the diagonal entries of D, and Q is a unitary matrix.

4.2.6 Note Let A aij be an n-square complex matrix. Let A be unitary (that is, A A ¼ AA ¼ I; that is, A ¼ A1 Þ. Let k be an eigenvalue of the matrix A. Then jkj ¼ 1: Proof Let T : x 7! Ax be the linear transformation from the inner product space Cn to Cn : Since k is an eigenvalue of the matrix A, we have detðkI AÞ ¼ 0; and hence there exists a nonzero x 2 Cn such that kx T ð xÞ ¼ kx Ax ¼ kIx Ax ¼ ðkI AÞx ¼ 0 : |fflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflffl} Thus kx T ð xÞ ¼ 0; and hence T ð xÞ ¼ kx; where x 6¼ 0: This shows that k is an eigenvalue of the linear transformation T. By 3.1.22, it suffices to show that T is a unitary transformation. To this end, let us take an arbitrary x 2 Cn : By 3.1.2, it suffices to show that hT ð xÞ; T ð xÞi ¼ hx; xi: LHS ¼ hT ð xÞ; T ð xÞi ¼ hAx; Axi ¼ ðAxÞ ðAxÞ ¼ ðx A ÞðAxÞ ¼ x ðA AÞx ¼ x Ix ¼ x x ¼ hx; xi ¼ RHS:

4.2.7 Note Let A aij be an n-square complex matrix. Let A be symmetric (that is, AT ¼ AÞ: Let A be unitary (that is, A A ¼ AA ¼ I; that is, A ¼ A1 Þ. Since A is unitary, A is a normal matrix, and hence by 3.3.24, there exists a unitary matrix U such that 1. U AU is a diagonal matrix, 2. the eigenvalues of A are the diagonal entries of U AU: Hence U AU ¼ diagðk1 ; . . .; kn Þ; where k1 ; . . .; kn are the eigenvalues of the matrix A. Now, since A is a unitary matrix, by 4.2.6, jk1 j ¼ 1; jk2 j ¼ 1; etc. Since U AU ¼ diagðk1 ; . . .; kn Þ; and U is a unitary matrix, we have A ¼ U ðdiagðk1 ; . . .; kn ÞÞU : Suppose that l1 ; . . .; lk are the distinct members of k1 ; . . .; kn : By 4.2.5, diagðk1 ; . . .; kn Þ can be expressed as Q ðl1 Ir1 lk Irk ÞQ; where Q is a unitary matrix. Since A ¼ U ðdiagðk1 ; . . .; kn ÞÞU ; we have

294


A ¼ U ðQ ðl1 Ir1 lk Irk ÞQÞU ¼ ðUQ Þðl1 Ir1 lk Irk ÞðUQ Þ

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ V ðl1 Ir1 lk Irk ÞV ; where V UQ : Thus A ¼ V ðl1 Ir1 lk Irk ÞV : Since l1 ; . . .; lk are the distinct members of k1 ; . . .; kn ; and each jki j ¼ 1; we have jl1 j ¼ 1; jl2 j ¼ 1; etc. Hence we can suppose that l1 eih1 ; l2 eih2 ; etc., where h1 ; h2 ; . . . are real numbers. Thus A ¼ V eih1 Ir1 eihk Irk V : Since Q is unitary, Q is unitary. Since U is unitary, ðV ¼ÞUQ is unitary, and hence V is unitary. Next, since A ¼ V eih1 Ir1 eihk Irk V ; we have

eih1 Ir1 eihk Irk ¼ V AV:

Put h1 hk S V ei 2 Ir1 ei 2 Irk V : It follows that

hk h1 ei 2 Ir1 ei 2 Irk ¼ V SV:

Here h h hk hk 1 1 S2 ¼ V ei 2 Ir1 ei 2 Irk V V ei 2 Ir1 ei 2 Irk V

h h hk hk 1 1 ¼ V ei 2 Ir1 ei 2 Irk ðV V Þ ei 2 Ir1 ei 2 Irk V

h h hk hk 1 1 ¼ V ei 2 Ir1 ei 2 Irk ei 2 Ir1 ei 2 Irk V

h 2 2 h k i 21 ¼V e Ir1 ei 2 Irk V ¼ V eih1 Ir1 eihk Irk V ¼ A; so S2 ¼ A: Clearly, S is unitary, that is, S S ¼ SS ¼ I:


295

Proof Here, h1

h1

hk hk S ¼ V ei 2 Ir1 ei 2 Irk V ¼ ðV Þ ei 2 Ir1 ei 2 Irk V

h1

h1 T hk hk ¼ V ei 2 Ir1 ei 2 Irk V ¼ V ei 2 Ir1 ei 2 Irk V

h T h1 k i 2 ¼ V e I r 1 ei 2 I r k V

h1 T h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk V ¼ V ei 2 Ir1 ei 2 Irk V ; so h1 hk S ¼ V ei 2 Ir1 ei 2 Irk V : Now, h1 h1 hk hk S S ¼ V ei 2 Ir1 ei 2 Irk V V ei 2 Ir1 ei 2 Irk V

h1 h 1 hk hk ¼ V ei 2 Ir1 ei 2 Irk ei 2 Ir1 ei 2 Irk V

h1 h1 hk hk ¼ V ei 2 ei 2 Ir1 ei 2 ei 2 Irk V ¼ V ð1Ir1 1Irk ÞV

¼ VIV ¼ VV ¼ I; so S S ¼ I: Similarly, SS ¼ I: ∎ Thus we have shown that S is unitary: . Suppose that B is any n-square complex matrix. Suppose that B commutes with A, that is, AB ¼ BA: Now clearly, V BV commutes with eih1 Ir1 eihk Irk ; that is, ih e 1 Ir1 eihk Irk ðV BV Þ ¼ ðV BV Þ eih1 Ir1 eihk Irk : Proof Here, LHS ¼ eih1 Ir1 eihk Irk ðV BV Þ ¼ ðV AV ÞðV BV Þ ¼ V AðBV Þ ¼ V ðABÞV; and RHS ¼ ðV BV Þ eih1 Ir1 eihk Irk ¼ ðV BV ÞðV AV Þ ¼ V BðAV Þ ¼ V ðBAÞV ¼ V ðABÞV; so LHS ¼ RHS:

∎

296


h hk 1 Again, it is clear that S ¼ V ei 2 Ir1 ei 2 Irk V commutes with B, that is, SB ¼ BS; that is, B ¼ S BS: Proof Here, h1 h1 hk hk RHS ¼ S BS ¼ V ei 2 Ir1 ei 2 Irk V B V ei 2 Ir1 ei 2 Irk V

h1 h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk ðV BV Þ ei 2 Ir1 ei 2 Irk V

h1 h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk ei 2 Ir1 ei 2 Irk ðV BV Þ V

h 1 h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk ei 2 Ir1 ei 2 Irk V B h1 h1 hk hk ¼ V ei 2 ei 2 Ir1 ei 2 ei 2 Irk V B ¼ V ð1Ir1 1Irk ÞV B ¼ VIV B ¼ B ¼ LHS:

Thus we have shown that if B commutes with A; then B commutes with S: ð Þ Since A is unitary, we have A A ¼ AA ¼ I; and hence the inverse of the matrix A is A : Since A is unitary, we have A A ¼ I; and hence ¼ AT ðA ÞT ¼ ðA AÞT ¼ I T ¼ I: AT A |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} ¼ I: Similarly, AA T ¼ I: Thus the inverse of the matrix AT is A: Thus AT A T Similarly, since V is unitary, the inverse of the matrix V is V; and the inverse of the matrix V is V : Since eih1 Ir1 eihk Irk V T ¼ V eih1 Ir1 eihk Irk T V T V T ¼ ðV ÞT eih1 Ir1 eihk Irk V T T ¼ V eih1 Ir1 eihk Irk V ¼ |fflfflffl{zfflfflffl} AT ¼ A ¼ V eih1 Ir1 eihk Irk V ; we have eih1 Ir1 eihk Irk V T ¼ V eih1 Ir1 eihk Irk V : V Hence eih1 Ir1 eihk Irk V T V ¼ V eih1 Ir1 eihk Irk : V


297

It follows that

eih1 Ir1 eihk Irk V T V ¼ V T V eih1 Ir1 eihk Irk :

Thus we have shown that V T V commutes with eih1 Ir1 eihk Irk : Clearly, VV T commutes with A. Proof We have to show that

VV T

V eih1 Ir1 eihk Irk V ¼ VV T A ¼ A VV T |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ih ¼ V e 1 Ir1 eihk Irk V VV T ;

that is,

V T V eih1 Ir1 eihk Irk V

¼ V eih1 Ir1 eihk Irk ðV V ÞV T ;

that is,

V T V eih1 Ir1 eihk Irk V ¼ eih1 Ir1 eihk Irk ðV V ÞV T ;

that is, T ih V V e 1 Ir1 eihk Irk V ¼ eih1 Ir1 eihk Irk IV T ; that is,

V T V eih1 Ir1 eihk Irk ¼ eih1 Ir1 eihk Irk V T V :

This is known to be true. It follows, from ð Þ; that VV T commutes with S, that is, hk i h1 V e 2 Ir1 ei 2 Irk V ¼ VV T S ¼ S VV T |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} h1 h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk V VV T ¼ V ei 2 Ir1 ei 2 Irk V T ;

VV T

that is, h1 h1 hk hk V V T V ei 2 Ir1 ei 2 Irk V ¼ V ei 2 Ir1 ei 2 Irk V T ; that is,

∎

298


h1 h1 hk hk V T V ei 2 Ir1 ei 2 Irk V ¼ ei 2 Ir1 ei 2 Irk V T ; that is,

VTV

h1 hk hk ih1 e 2 Ir1 ei 2 Irk ¼ ei 2 Ir1 ei 2 Irk V T V :

h hk 1 Thus V T V commutes with ei 2 Ir1 ei 2 Irk : Clearly, S is symmetric. Proof We have to show that h h T h h ei 21 Ir1 ei 2k Irk V T ¼ ðV ÞT ei 21 Ir1 ei 2k Irk V T V h1 T h1 hk hk ¼ V ei 2 Ir1 ei 2 Irk V ¼ S|fflfflTffl{zfflffl ¼ fflS} ¼ V ei 2 Ir1 ei 2 Irk V ;

that is, h h h h ei 21 Ir1 ei 2k Irk V T ¼ V ei 21 Ir1 ei 2k Irk V ; V that is, h h h h ei 21 Ir1 ei 2k Irk V T V ¼ V ei 21 Ir1 ei 2k Irk ; V that is,

h1

hk

ei 2 Ir1 ei 2 Irk

hk h1 V T V ¼ V T V ei 2 Ir1 ei 2 Irk :

This is known to be true. ∎ Thus S is symmetric :

4.2.8 Conclusion Let A aij be an n-square complex matrix. Let A be symmetric. Let A be unitary. Then there exists a complex matrix S such that 1. 2. 3. 4.

S2 ¼ A; S is unitary, if B commutes with A, then B commutes with S, S is symmetric.

4.2.9 Note Let A aij be an n-square real matrix. Let B aij be an n-square real matrix. Let U be a unitary complex matrix such that A ¼ UBU

ðthat is; U AU ¼ BÞ:


299

Since U is unitary, we have U U ¼ UU ¼ I; and hence the inverse of the matrix U is U : Since U is unitary, we have U U ¼ I; and hence ¼ U T ðU ÞT ¼ ðU U ÞT ¼ I T ¼ I: UT U |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl} ¼ I: Similarly, UU T ¼ I: Thus the inverse of the matrix U T is U: Thus U T U T Clearly, U U is symmetric. Proof Since

T T U T U ¼ U T U T ¼ U T U;

T

we have ðU T U Þ ¼ U T U; so U T U is symmetric. Clearly, U T U is unitary.

∎

Proof Since U is unitary, we have U U ¼ I; and hence

UT

¼ U T ðU ÞT ¼ ðU U ÞT ¼ I T ¼ I: UT ¼ UT U |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

Thus ðU T ÞðU T Þ ¼ I: Similarly, ðU T Þ ðU T Þ ¼ I: It follows that U T is unitary, 1

and hence ðU T Þ ¼ ðU T Þ : Since U is unitary, we have U 1 ¼ U : We have to 1

show that ðU T U Þ ¼ ðU T U Þ :

1 1

1 RHS ¼ U T U ¼ U U T ¼ U U T ¼ U 1 U T ¼ U T U ¼ LHS: ∎ Since U T U is unitary and symmetric, by 4.2.8, there exists a complex matrix S such that 1. 2. 3. 4.

S2 ¼ U T U; S is unitary, if C commutes with U T U; then C commutes with S, S is symmetric. Clearly, U T U commutes with B, that is, ðU T U ÞB ¼ BðU T U Þ: Proof Since A is a real matrix, we have T ¼ A ¼ UBU ; BU T ¼U BU ¼ ðUBU Þ ¼ A UBU ¼U |fflffl{zfflffl} T ¼ UBU : This shows that BU T ¼ U T ðUBU Þ; and hence and hence UBU T T ðBU ÞU ¼ U ðUBÞ: Thus

300


B U T U ¼ U T U B: ∎ Thus we have shown that U T U commutes with B. Now, by (3), B commutes with S. ð Þ Let us put Q US1 : Clearly, Q is unitary. 1

Proof We have to show that ðUS1 Þ ¼ ðUS1 Þ : Since S is unitary, we have S1 ¼ S : Now we have to show that

ðUS Þ1 ¼ US1 ¼ ðUS Þ ¼ ðS Þ U ¼ SU ¼ SU 1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} that is, ðUS Þ1 ¼ SU 1 :

1 LHS ¼ ðUS Þ1 ¼ US1 ¼ SU 1 ¼ RHS: ∎

Clearly, Q is orthogonal. Proof We have to show that QT Q ¼ I: T LHS ¼ US1 US1 ¼ ðUS ÞT US1 ¼ ðS ÞT U T US1 ¼ S U T US1 1 T 1 US ¼ ð ST Þ U T ¼ S U US ¼ S U T U S1 ¼ S1 U T U S1 ¼ S1 S2 S1 ¼ I ¼ RHS:

∎ ÞT ¼ Q ¼ Q1 ¼ QT : Since Q is orthogonal, we have QT Q ¼ I; and hence ðQ |fflfflfflfflfflffl{zfflfflfflfflfflffl} ¼ Q: Hence Q is a real matrix. ÞT ¼ QT ; and therefore Q Thus ðQ T Clearly, A ¼ QBQ : Proof Here

T ¼ QBQ ¼ QBðUS1 Þ ¼ QBðUS Þ ¼ QBðSU Þ QBQT ¼ QBQ ¼ ðQBSÞU ¼ ððUS1 ÞBSÞU ¼ U ðS1 BSÞU ; so


301

QBQT ¼ U S1 BS U : Now, since A ¼ UBU ; it suffices to show that B ¼ ðS1 BSÞ; that is, SB ¼ BS: From ð Þ; this is true. ∎ 4.2.10 Conclusion Let A and B be n-square real matrices. Let U be a unitary complex matrix such that A ¼ UBU : Then there exists a real orthogonal matrix Q such that A ¼ QBQT :

¼ AÞ. Let A be 4.2.11 Note Let A aij be an n-square real matrix (that is, A T symmetric (that is, A ¼ AÞ: ÞT ¼ AT ¼ A; and hence A ¼ A: This shows that A is It follows that A ¼ ðA Hermitian, and hence A is a normal matrix. Now, by 3.3.24, there exists a unitary matrix U such that 1. U AU is a diagonal matrix, 2. the eigenvalues of A are the diagonal entries of U AU: Hence U AU ¼ diagðk1 ; . . .; kn Þ; where k1 ; . . .; kn are the eigenvalues of the matrix A. Since A is Hermitian, by 3.3.26, k1 ; . . .; kn are real numbers. Since U AU ¼ diagðk1 ; . . .; kn Þ; and U is a unitary matrix, we have A ¼ U ðdiagðk1 ; . . .; kn ÞÞU : Also A and diagðk1 ; . . .; kn Þ are real matrices of the same size. Now, by 4.2.10, there exists a real orthogonal matrix Q such that A ¼ Qðdiagðk1 ; . . .; kn ÞÞQT : ð Þ: Since Q is orthogonal, we have Q1 ¼ QT : Now from ð Þ; A ¼ Qðdiagðk1 ; . . .; kn ÞÞQ1 : It follows that T diagðk1 ; . . .; kn Þ ¼ Q1 AQ ¼ QT AQ ¼ QT A QT ¼ PAPT ; |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} where P QT : Since Q is a real matrix, ðP ¼ÞQT is a real matrix, and hence P is a real matrix. Since Q is orthogonal, we have QQT ¼ I; and hence P1 ¼

1 ¼ Q ¼ PT : Thus P1 ¼ PT ; and hence P is orthogonal. Also, QT |fflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflffl} PAPT ¼ diagðk1 ; . . .; kn Þ: 4.2.12 Conclusion Let A be an n-square real symmetric matrix. Then there exists a real orthogonal matrix P such that PAPT ¼ diagðk1 ; . . .; kn Þ; where k1 ; . . .; kn are the eigenvalues of A.

302


In short, a real symmetric matrix can be brought to diagonal form by a real orthogonal matrix. Definition Let A be an n-square real symmetic matrix. Let B be an n-square real symmetic matrix. If there exists a real invertible matrix S such that A ¼ SBST ; then we say that A and B are congruent. 4.2.13 Note Suppose that S is an invertible matrix. It follows that S1 exists, and SS1 ¼ I: Hence

T

T T S1 ST ¼ SS1 ¼ I T ¼ I: |fflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflffl}

Thus ðS1 Þ ST ¼ I: It follows that ðST Þ

1

exists, and ðST Þ

1

T

= ðS1 Þ :

4.2.14 Conclusion If S is a real invertible matrix, then ST is a real invertible matrix, T 1 and ðST Þ ¼ ðS1 Þ : 4.2.15 Problem Congruence is an equivalence relation. Proof (i) Let us take an arbitrary real symmetric n-square matrix A. Since A ¼ IAI T ; and I is a real invertible matrix, A and A are congruent. (ii) Let us take arbitrary real symmetric n-square matrices A and B that are congruent. We have to show that B and A are congruent. Since A and B are congruent, there exists a real invertible matrix S such that T T 1 A ¼ SBST : It follows that ðST Þ = ðS1 Þ ; and B ¼ ðS1 ÞAðS1 Þ ; and hence B ¼ T 1 RAR ; where R S : Since S is a real invertible matrix, ðR ¼ÞS1 is a real invertible matrix, and hence R is a real invertible matrix. Thus B and A are congruent. (iii) Let us take any real symmetric n-square matrices A, B, C. Suppose that A and B are congruent. Suppose that B and C are congruent. We have to show that A and C are congruent. Since A and B are congruent, there exists a real invertible matrix S such that A ¼ SBST : Since B and C are congruent, there exists a real invertible matrix R such that B ¼ RCRT : It follows that A ¼ SðRCRT ÞST ¼ ðSRÞC ðSRÞT ; and hence A ¼ ðSRÞCðSRÞT : Since S,R are real invertible matrices,SR is also a real invertible matrix. Thus A and C are congruent. Hence, congruence is an equivalence relation. ∎ 4.2.16 Note Let A be an n-square real symmetric matrix.


303

ÞT ¼ AT ¼ A; we have A ¼ A: This shows that A is Hermitian, Since A ¼ ðA and hence by 3.3.26, its eigenvalues are real numbers. Let l1 ; . . .; lr be the positive |fflfflfflfflffl{zfflfflfflfflffl} r

distinct eigenvalues of A, and let lr þ 1 ; . . .; lr þ s be the negative distinct |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} s

eigenvalues of A. By 4.2.12, there exists a real orthogonal matrix P such that PAPT ¼ l1 Ik1 lr Ikr lr þ 1 Ikr þ 1 lr þ s Ikr þ s 0Inðk1 þ þ kr þ s Þ : Put 1 1 1 1 D pffiffiffiffiffi Ik1 pffiffiffiffiffi Ikr pffiffiffiffiffiffiffiffiffiffi Ikr þ 1 pffiffiffiffiffiffiffiffiffiffi Ikr þ s l1 lr lr þ 1 lr þ s 1Inðk1 þ þ kr þ s Þ : It follows that DT ¼ ¼

1 1 p1ffiffiffiffi Ik1 p1ffiffiffiffi Ikr pffiffiffiffiffiffiffi pffiffiffiffiffiffiffi l1 lr lr þ 1 Ikr þ 1 lr þ s Ikr þ s 1Inðk1 þ þ kr þ s Þ 1 1 1 1 pffiffiffiffi Ik1 pffiffiffiffi Ikr pffiffiffiffiffiffiffi Ikr þ 1 pffiffiffiffiffiffiffi Ikr þ s 1Inðk þ þ k Þ ; 1 rþs l l l l 1

r

rþ1

rþs

and hence ðDPÞAðDPÞT ¼ DðPAPT ÞDT

0

B 1 1 1 1 B ¼ Bpffiffiffiffiffi l1 pffiffiffiffiffi Ik1 pffiffiffiffiffiffiffiffiffiffi lr þ 1 pffiffiffiffiffiffiffiffiffiffi Ikr þ 1 @ l1 lr þ 1 lr þ 1 l1 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} r s ! 1 0 1Inðk1 þ þ kr þ s Þ ¼ 1Ik1 1Ikr ð1ÞIkr þ 1 ð1ÞIkr þ s 0IInðk ¼ 1Il ð1ÞIm 0Inðl þ mÞ where l k1 þ þ kr ; m kr þ 1 þ þ kr þ s : Thus RART ¼ 1Il ð1ÞIm 0Inðl þ mÞ ; where R DP: Since

1 þ þ kr þ s Þ

T

304


1 1 1 1 D ¼ pffiffiffiffiffi Ik1 pffiffiffiffiffi Ikr pffiffiffiffiffiffiffiffiffiffi Ikr þ 1 pffiffiffiffiffiffiffiffiffiffi Ikr þ s l1 lr lr þ 1 lr þ s 1Inðk1 þ þ kr þ s Þ ; D is a real invertible matrix. Since P is a real orthogonal matrix, P is a real invertible matrix, and P1 ¼ PT : Since D; P are real invertible matrices, ðR ¼ÞDP is a real R is a real invertible matrix. Since ðRART ¼Þ invertible matrix, and hence 1Il ð1ÞIm 0Inðl þ mÞ is a real symmetric matrix, RART is a real symmetric matrix. Now, since R is a real invertible matrix, A and RART ð¼ ð1Il ð1Þ Im 0Inðl þ mÞ ÞÞ are congruent, and hence A and 1Il ð1ÞIm 0Inðl þ mÞ are congruent. Next, by 4.2.15, 1Il ð1ÞIm 0Inðl þ mÞ is a member of the congruence class of A. It is clear that l þ m is the rank of A. Now we want to show that l and m are unique. To this end, suppose that 1Ir ð1ÞIs 0Inðr þ sÞ and 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ are congruent. We have to show that r ¼ r 0 : Suppose to the contrary that r\r 0 : We seek a contradiction. Since

1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ and 1Ir ð1ÞIs 0Inðr þ sÞ

are congruent, there exists a real invertible matrix S such that

1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ ¼ S 1Ir ð1ÞIs 0Inðr þ sÞ ST :

Since S is invertible, by 4.2.14, ST is invertible, and hence r 0 þ s0 ¼ rank 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ ¼ rank S 1Ir ð1ÞIs 0Inðr þ sÞ ST ¼ rank 1Ir ð1ÞIs 0Inðr þ sÞ |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ r þ s: Thus r 0 þ s0 ¼ r þ s: Now, since r\r 0 ; we have s0 \s: ð Þ Observe that the set 9 82 3T > > = < 6 7 U 40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 05 : y1 ; . . .; ys 2 R > > ; : |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} r

s

nðr þ sÞ


305

is an s-dimensional subspace of the real inner product space Rn ; where Rn denotes the collection of all n 1 column matrices with real entries. Also 9 82 3T > > = < 6 7 W 4x1 ; . . .; xr0 ; 0; . . .; 0 ; z1 ; . . .; znðr0 þ s0 Þ 5 : x1 ; . . .; xr0 ; z1 ; . . .; znðr0 þ s0 Þ 2 R |fflfflffl{zfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} > > ; : |fflfflfflfflffl{zfflfflfflfflffl} 0 0 r

s

nðr0 þ s0 Þ

is an ðn s0 Þ-dimensional subspace of the real inner product space Rn : 2 3T 6 7 Observe that, for every nonzero 40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 05 2 U; |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} r

*

nðr þ sÞ

s

2 3T 6 7 1Ir ð1ÞIs 0Inðr þ sÞ 40; . . .; 0 ; y1 ; . . .; ys ; 0; . . .; 05 ; |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} r

2

3T

6 7 40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 0 5 |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} r

02

s

nðr þ sÞ

nðr þ sÞ

s

+

3 T 1T

B6 7 C ¼ @40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 05 A |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} r

0

s

nðr þ sÞ

2 3T 1 B 6 7 C @ 1Ir ð1ÞIs 0Inðr þ sÞ 40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 05 A |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} 2

3

r

nðr þ sÞ

s

6 7 ¼ 40; . . .; 0 ; y1 ; . . .; ys ; 0; . . .; 05 |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} 0

r

nðr þ sÞ

s

2 3T 1 B 6 7 C @ 1Ir ð1ÞIs 0Inðr þ sÞ 40; . . .; 0; y1 ; . . .; ys ; 0; . . .; 05 A |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} 2

32

r

nðr þ sÞ

s

3T

6 76 7 ¼ 40; . . .; 0 ; y1 ; . . .; ys ; 0; . . .; 0540; . . .; 0; ð1Þy1 ; . . .; ð1Þys ; 0; . . .; 0 5 |fflfflffl{zfflfflffl} |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} |fflfflffl{zfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflffl{zfflfflffl} r

s

nðr þ sÞ

r

s

nðr þ sÞ

306


¼ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} þ þ 0 þ ð y1 Þ 2 þ þ ð ys Þ 2 þ 0 þ þ0 ¼ |fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl {zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl ffl } r nðr þ sÞ s ðy1 Þ2 þ þ ðys Þ2 \0; so for every nonzero u 2 U; 1Ir ð1ÞIs 0Inðr þ sÞ u; u is negative. 2 3T 6 7 Observe that, for every 4x1 ; . . .; xr0 ; 0; . . .; 0; z1 ; . . .; znðr0 þ s0 Þ 5 2 W; |fflfflfflfflffl{zfflfflfflfflffl} |fflfflffl{zfflfflffl} |fflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflffl} r0

*

s0

nðr0 þ s0 Þ

2

3T

6 7 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ 4x1 ; ; 0; ; z1 ; 5 ; |fflfflffl{zfflfflffl} |ffl{zffl} |fflffl{zfflffl} 2

3T

6 7 4x1 ; ; 0; ; 0; z1 ; 5 |fflffl{zfflffl} |fflfflfflffl{zfflfflfflffl} |fflffl{zfflffl} r0

02

s0

r0

s0

nðr0 þ s0 Þ

+

nðr0 þ s0 Þ

3 T 1T

B6 7 C ¼ @4x1 ; ; 0; ; z1 ; 5 A |fflffl{zfflffl} |ffl{zffl} |fflffl{zfflffl} r0

0

s0

nðr 0 þ s0 Þ

2 3T 1 B 6 7 C @ 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ 4x1 ; ; 0; z1 ; 5 A |fflffl{zfflffl} |ffl{zffl} |fflffl{zfflffl} 2

r0

3

s0

nðr0 þ s0 Þ

6 7 ¼ 4x1 ; ; 0; ; z1 ; 5 |fflffl{zfflffl} |fflffl{zfflffl} |fflffl{zfflffl} 0

r0

s0

nðr 0 þ s0 Þ

2 3T 1 B 6 7 C @ 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ 4x1 ; ; 0; ; z1 ; 5 A |fflffl{zfflffl} |ffl{zffl} |fflffl{zfflffl} r0

2

32

s0

nðr0 þ s0 Þ

3T

6 76 7 ¼ 4x1 ; ; xr0 ; 0; ; 0; z1 ; ; znðr0 þ s0 Þ 54x1 ; ; xr0 ; 0; ; 0; 0; ; 0 5 |fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflfflfflffl{zfflfflfflffl} |fflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflffl} |fflfflfflfflfflffl{zfflfflfflfflfflffl} |fflfflfflffl{zfflfflfflffl} |fflfflfflffl{zfflfflfflffl} r0

s0

2

nðr0 þ s0 Þ

r0

2

¼ ðx1 Þ þ þ ðxr0 Þ þ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} þ þ 0 þ 0|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl} þ þ0 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} s0

r0

2

2

¼ ðx1 Þ þ þ ðxr0 Þ 0; so for every w 2 W;

nðr 0 þ s0 Þ

s0

nðr0 þ s0 Þ


307

1Ir ð1ÞIs 0Inðr þ sÞ ST w ; ST w T 1Ir ð1ÞIs 0Inðr þ sÞ ST w ¼ ST w ¼ wT S 1Ir ð1ÞIs 0Inðr þ sÞ ST w

¼ S 1Ir ð1ÞIs 0Inðr þ sÞ ST w; w ¼ 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ w; wi 0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

Thus for every v 2 fST w : w 2 W g;

1Ir ð1ÞIs 0Inðr þ sÞ v; v 0:

Further, we have seen that for every nonzero u 2 U;

1Ir ð1ÞIs 0Inðr þ sÞ u; u 0:

It follows that fST w : w 2 W g \ U ¼ f0g: Clearly, fST w : w 2 W g is an ðn s0 Þ-dimensional real vector space. Proof Since S is invertible, ST is invertible. The map T : x 7! ST x from the real vector space Rn to Rn is a linear transformation. Since ST is invertible, T is one-to-one, and hence dim ST w : w 2 W ¼ dimðW Þ ¼ n s0 : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus we have shown that fST w : w 2 W g is an ðn s0 Þ-dimensional real vector space. ∎ Next, since dimðU Þ ¼ s; we have n ¼ dimðRn Þ dim ST w : w 2 W [ U ¼ dim ST w : w 2 W þ dimðU Þ dim ST w : w 2 W \ U ¼ dim ST w : w 2 W þ dimðU Þ dimðf0gÞ ¼ dim ST w : w 2 W þ dimðU Þ 0 T ¼ dim S w : w 2 W þ dimðU Þ ¼ ðn s0 Þ þ dimðU Þ ¼ ðn s0 Þ þ s ¼ n þ ðs s0 Þ;

and hence n n þ ðs s0 Þ: It follows that s s0 : This contradicts ð Þ: Thus we have shown that r ¼ r 0 : Finally, we have to show that s ¼ s0 :

308


Since 1Ir ð1ÞIs 0Inðr þ sÞ and 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ are congruent, there exists a real invertible matrix S such that

1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ ¼ S 1Ir ð1ÞIs 0Inðr þ sÞ ST :

Since S is invertible, by 4.2.14, ST is invertible, and hence r 0 þ s0 ¼ rank 1Ir0 ð1ÞIs0 0Inðr0 þ s0 Þ ¼ rank S 1Ir ð1ÞIs 0Inðr þ sÞ ST ¼ rank 1Ir ð1ÞIs 0Inðr þ sÞ ¼ r þ s: |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus r 0 þ s0 ¼ r þ s: Now, since r ¼ r 0 ; we have s ¼ s0 :

∎

4.2.17 Conclusion Let A be an n-square real symmetric matrix. There exist a real invertible matrix R, and two nonnegative integers r and s such that RART ¼ 1Ir ð1ÞIs 0Inðr þ sÞ : Also, r and s are unique. Further, rankð AÞ ¼ r þ s: The integer ðr sÞ is called the signature of A, and is denoted by sgð AÞ: Thus there exists a real invertible matrix R such that RAR ¼ 1IrankðAÞ þ sgðAÞ ð1ÞIIrankðAÞsgðAÞ 0Inrankð AÞ : T

2

2

This result is known as Sylvester’s law. 4.2.18 Theorem Let V be an n-dimensional real inner product space. Let S : V ! V be a linear transformation. Let v 2 V: Then there exists a unique w 2 V such that u 2 V ) u; w ¼ hSðuÞ; vi: We denote w by ST ðvÞ: Thus ST : V ! V; and for every u; v 2 V; hu; ST ðvÞi ¼ hSðuÞ; vi: Also, ST : V ! V is linear. Proof Existence: Since V is an n-dimensional real inner product space, there exists an orthonormal basis fu1 ; . . .; un g of V. Put w hSðu1 Þ; viu1 þ þ hSðun Þ; viun : Let us fix an arbitrary u

Pn i¼1

ai ui : We have to show that


309

*

n n P P S uj ; v uj ¼ S ai ui ; v : i¼1 j¼1 i¼1 * + n n

P ij P P P ai ui ; S uj ; v uj ¼ ai S uj ; v ui ; uj ¼ ai S uj ; v d LHS ¼ i;j i;j i¼1 j¼1 n n n P P P ¼ ai hSðui Þ; vi ¼ ai Sðui Þ; v ¼ S ai ui ; v ¼ RHS: n P

i¼1

a i ui ;

i¼1

i¼1

Uniqueness: Suppose that there exist w1 ; w2 2 V such that u 2 V ) hu; w1 i ¼ hSðuÞ; vi; and hu; w2 i ¼ hSðuÞ; vi: We have to show that w1 ¼ w2 ; that is, hw1 w2 ; w1 w2 i ¼ 0: Here u 2 V ) hu; w1 i ¼ hu; w2 i; so for every u 2 V; hu; w1 w2 i ¼ 0: It follows that hw1 w2 ; w1 w2 i ¼ 0: Linearity: Let us take arbitrary v1 ; v2 2 V: Let a; b be arbitrary real numbers. We have to show that ST ðav1 þ bv2 Þ ¼ aST ðv1 Þ þ bST ðv2 Þ: It suffices to show that for every u 2 V;

u; ST ðav1 þ bv2 Þ ¼ u; aST ðv1 Þ þ bST ðv2 Þ :

To this end, let us fix an arbitrary u 2 V: We have to show that hu; ST ðav1 þ bv2 Þi ¼ hu; aST ðv1 Þ þ bST ðv2 Þi : LHS ¼ hu; S ðav1 þ bv2 Þi ¼ hSðuÞ; av1 þ bv2 i ¼ ahSðuÞ; v1 i þ bhSðuÞ; v2 i ¼ ahu; ST ðv1 Þi þ bhu; ST ðv2 Þi ¼ hu; aST ðv1 Þ þ bST ðv2 Þi ¼ RHS: T

∎ Definition Let V be an n-dimensional real inner product space. Let S : V ! V be a linear transformation. By 4.2.18, ST : V ! V is a linear transformation such that for all u; v 2 V; hu; ST ðvÞi ¼ hSðuÞ; vi: Here ST is called the transpose of S. 4.2.19 Theorem Let V be an n-dimensional real inner product space. Let S : V ! T V be a linear transformation. Then ðST Þ ¼ S: Proof Let us take an arbitrary v 2 V: We have to show that

310


T T S ðvÞ ¼ SðvÞ: To this end, let us take an arbitrary u 2 V: It suffices to show that D

E T u; ðST Þ ðvÞ ¼ hu; SðvÞi :

D E T LHS ¼ u; ðST Þ ðvÞ ¼ hST ðuÞ; vi ¼ hv; ST ðuÞi ¼ hSðvÞ; ui ¼ hu; SðvÞi ¼ RHS: ∎ 4.2.20 Theorem Let V be an n-dimensional real inner product space. Let R : V ! V and S : V ! V be linear transformations. Let k; l be any real numbers. Then ðkR þ lSÞT ¼ kRT þ lST : Proof Let us take an arbitrary v 2 V: We have to show that ðkR þ lSÞT ðvÞ ¼ kRT þ lST ðvÞ; that is, ðkR þ lSÞT ðvÞ ¼ kRT ðvÞ þ lST ðvÞ: To this end, let us take an arbitrary u 2 V: It suffices to show that

u; ðkR þ lSÞT ðvÞ ¼ hu; kRT ðvÞ þ lST ðvÞi :

LHS ¼ u; ðkR þ lSÞT ðvÞ ¼ hðkR þ lSÞðuÞ; vi ¼ hkRðuÞ þ lSðuÞ; vi ¼ khRðuÞ; v þ lSðuÞ; vi

¼ k u; RT ðvÞ þ l u; ST ðvÞ ¼ u; kRT ðvÞ þ lST ðvÞ ¼ RHS: ∎ 4.2.21 Theorem Let V be an n-dimensional real inner product space. Let R : V ! V and S : V ! V be linear transformations. Then ðRSÞT ¼ ST RT : Proof Let us take an arbitrary v 2 V: We have to show that ðRSÞT ðvÞ ¼ ST RT ðvÞ; that is, ðRSÞT ðvÞ ¼ ST RT ðvÞ : To this end, let us take an arbitrary u 2 V: It suffices to show that


311

u; ðRSÞT ðvÞ ¼ hu; ST ðRT ðvÞÞi :

LHS ¼ u; ðRSÞT ðvÞ ¼ hðRSÞðuÞ; v ¼ RðSðuÞÞ; vi

¼ SðuÞ; RT ðvÞ ¼ u; ST RT ðvÞ ¼ RHS: ∎ 4.2.22 Theorem Let V be an n-dimensional real inner product space. Let S : V !

V be a linear transformation. Let fv1 ; . . .; vn g be an orthonormal basis of V. Let aij be the matrix of S relative to the basis fv1 ; . . .; vn g; in the sense that n P Sðv1 Þ ¼ a11 v1 þ a21 v2 þ þ an1 vn ¼ ai1 vi ; i¼1

Sðv2 Þ ¼ a12 v1 þ a22 v2 þ þ an2 vn ; .. . Sðvn Þ ¼ a1n v1 þ a2n v2 þ þ ann vn :

P In short, S vj ¼ ni¼1 aij vi :

Then the matrix of ST relative to the basis fv1 ; . . .; vn g is bij ; where, bij ¼ aji : P In short, ST vj ¼ ni¼1 bij vi : Proof By the proof of 4.2.18, ST ðv1 Þ ¼ hSðv1 Þ; v1 iv1 þ þ hSðvn Þ; v1 ivn ; ST ðv2 Þ ¼ hSðv1 Þ; v2 iv1 þ þ hSðvn Þ; v2 ivn ; .. . ST ðvn Þ ¼ hSðv1 Þ; vn iv1 þ þ hSðvn Þ; vn ivn :

Since ST ð v 1 Þ ¼ ¼ ¼

n X i¼1 n X i¼1 n X

hSðvi Þ; v1 ivi ¼

n X

ha1i v1 þ a2i v2 þ þ ani vn ; v1 ivi

i¼1

ða1i hv1 ; v1 i þ a2i hv2 ; v1 i þ þ ani hvn ; v1 iÞvi ða1i 1 þ a2i 0 þ þ ani 0Þvi ¼

i¼1

n X

a1i vi

i¼1

¼ a11 v1 þ a12 v2 þ þ a1n vn ; we have S ðv1 Þ ¼ a11 v1 þ a12 v2 þ þ a1n vn ¼ T

n X i¼1

! a1i vi :

312


Similarly, ST ðv2 Þ ¼ a21 v1 þ a22 v2 þ þ a2n vn ; P etc. In short, ST vj ¼ ni¼1 aji vi : If the matrix of ST relative to the basis

∎ fv1 ; . . .; vn g is bij ; then bij ¼ aji : Definition Let V be an n-dimensional real inner product space. Let S : V ! V be a linear transformation. If ST ¼ S; then we say that S is symmetric. 4.2.23 Theorem Let V be an n-dimensional real inner product space. Let S : V ! V be a linear transformation. Let S be symmetric. Let fe1 ; . . .; en g be any orthonormal basis of V. Let aij be the matrix of S relative to fe1 ; . . .; en g; that is, P S e ¼ ni¼1 aij ei : Let Q : v 7! hSðvÞ; vi be a function from V to R: Let v Pnj i¼1 xi ei : Then ! n n X X Q xi ei ¼ a11 ðx1 Þ2 þ þ ann ðxn Þ2 þ 2 aij xi xj : i\j

i¼1

P Here a11 ðx1 Þ2 þ þ ann ðxn Þ2 þ 2 ni\j aij xi xj is called the real quadratic form of S.

Proof Since S is symmetric and aij is the matrix of S relative to fe1 ; . . .; en g; by 4.2.22, aji ¼ aij : It follows that ! * ! + n n n X X X xi ei ¼ QðvÞ ¼ hSðvÞ; vi ¼ S xi ei ; xi ei Q i¼1

* ¼

n X i¼1

¼

n X i¼1

¼

¼

i¼1

*

n X

n X

xi Sðei Þ;

i¼1

+ xj ej

j¼1

! ! n n n X X X

xj Sðei Þ; ej xi xj Sðei Þ; ej ¼ j¼1

xi xj

*

i¼1

n X k¼1

n n X X

n X

xi xj

j¼1

n n X X i¼1

¼

x j ej

j¼1

i¼1

i¼1

+

j¼1

n n X X i¼1

¼

xi

xi Sðei Þ;

n X

j¼1

aki ek ; ej

¼

!! aki dkj

!

¼

n n X X i¼1

¼

n n X X i¼1

j¼1

!

aij xi xj

j¼1

¼ a11 ðx1 Þ2 þ þ ann ðxn Þ2 þ 2

n X i\j

xi xj

j¼1

n n X X i¼1

k¼1

aji xi xj

j¼1

+!

aij xi xj ;

xi xj aji

n X

!

k¼1

aki ek ; ej

!!


313

and hence Q

n X

! xi ei

¼ a11 ðx1 Þ2 þ þ ann ðxn Þ2 þ 2

n X

aij xi xj :

i\j

i¼1

∎ 4.2.24 Example Let us consider the following real quadratic form: ðx1 Þ2 þ 2ðx2 Þ2 þ 2ðx3 Þ2 þ x1 x2 þ 2x1 x3 þ 4x2 x3 : Here ðx1 Þ2 þ2ðx2Þ2 þ 2ðx3 Þ2 þ x1 x2 þ 2x1x3 þ4x2 x3 2 ¼ ðx1 Þ þ 12 x1 x2 þ x1 x3 þ 12 x1 x2 þ 2ðx2 Þ2 þ 2x2 x3 þ x1 x3 þ 2x2 x3 þ 2ðx3 Þ2 1 x2 þ ðx1 þ 2x2 þ 2x3 Þx3 ¼ x1 þ 12 x2 þ x2 x þ 2x 3 x1 þ 2 x13þ 2 3 2 3 x1 1 12 1 ¼ ½x1 x2 x3 4 12 2 2 54 x2 5 ¼ ½x1 x2 x3 A½x1 x2 x3 T ; x3 1 2 2 where 2

a12 a22 a32

a11 A 4 a21 a31

3 2 a13 1 12 a23 5 ¼ 4 12 2 a33 1 2

3 1 2 5: 2

Observe that 2

1

41 2 1

1 2

2 2

3 2 1 R2 !R2 1R1 1 2 ! 40 25 2 1

1 2 7 4

1 3 2

2 2

2

3 5

C2 !C2 12C1

!

1 40 1

0

1

7 4 3 2

3 2

2

0

1

7 4 3 2

3 2

3 5:

Thus 2

1 41 2 0

32 0 0 1 1 0 54 12 0 1 1

1 2

2 2

32 1 1 2 54 0 0 2

3 2 1 12 0 1 05 ¼ 40 1 0 1

3 5;

2

or 2

1 41 2 0

32 0 0 1 1 0 54 12 0 1 1

1 2

2 2

32 1 1 2 54 12 0 2

3T 2 1 0 0 0 1 0 5 ¼ 4 0 74 1 32 0 1

1 3 2

2

3 5:

314


Since 0 0 1 0 ¼ 1 6¼ 0; 0 1

0 1 0 ¼ 0 1 0

1 0 1 2 1 0 0 it follows that

1 0 1 2 1 0 0

0 0 1

2

3

is invertible. Observe that 2

1 40 1

0 7 4 3 2

1

3

1 3 5 R3 !R3 R1 4 0 ! 2 2 0

0 1

2

1 3 5 C3 !C3 C1 4 0 ! 2 1 0

7 4 3 2

0

0

7 4 3 2

3 2

3 5:

1

Thus 2

1 4 0 1

32 1 0 0 1 0 54 0 1 0 1

0

1

32

7 4 3 2

1 3 54 0 2 2 0

0

1

3 2 1 0 1 0 5 ¼ 4 0 74 0 32 1

0 1 0

0 3 2

3 5;

1

or 2

1 4 0 1

32 1 0 0 1 0 54 0 1 0 1

7 4 3 2

32

1 3 54 0 2 2 1

0 1 0

3T 2 1 0 0 0 5 ¼ 4 0 74 0 32 1

Since 1 0 1

0 1 0

0 1 0 ¼ 0 1 0

0 1 0

0 0 ¼ 1 6¼ 0; 1

it follows that 1 0 1 is invertible.

0 1 0

0 0 1

0 3 2

1

3 5:


315

Observe that 2

1 40 0

0 0 7 4 3 2

3

2

1 0 35 40 7 ! 2 2 0 32 1 R2 !2R2

2 3 0 1 C2 !2C2 35 ! 40 1 0

0 7 3

3 0 3 5: 1

Thus 2

1 40 0

32 1 0 0 2 0 54 0 0 0 1

32

0

0

7 4 3 2

3 2

1 54 0 1 0

0

0

7 4 3 2

3 2

0 2 0

3 2 0 1 05 ¼ 40 1 0

3 0 0 7 3 5; 3 1

0 2 0

3T 2 0 1 05 ¼ 40 1 0

3 0 0 7 3 5: 3 1

or 2

1 40 0

32 1 0 0 5 4 0 2 0 0 0 1

32

1 54 0 1 0

Since 1 0 0

0 2 0

1 0 0 0 ¼ 2 0 1 0 0 1

0 0 ¼ 2 6¼ 0; 1

it follows that 1 0 0

0 2 0

0 0 1


1 40 0

3 2 0 0 1 R2 !R2 2R3 7 35 ! 40 3 1 0

3 2 0 0 1 C2 !C2 2C3 1 15 ! 40 3 1 0

3 0 0 1 1 5: 1 1

Thus 2

1 40 0 or

0 1 0

32 0 1 2 54 0 1 0

0 7 3

32 0 1 3 54 0 0 1

0 1 2

3 2 0 1 05 ¼ 40 1 0

0 1 1

3 0 1 5; 1

316


2

1 40 0

0 1 0

32 0 1 2 54 0 1 0

32 0 1 3 54 0 1 0

0 1 0

3T 2 0 1 2 5 ¼ 4 0 1 0

0 1 2 ¼ 0 1 0

0 1 0

0 0 ¼ 1 6¼ 0; 1

0 7 3

3 0 1 5: 1

0 1 1

Since 1 0 0 1 0 0 it follows that 1 0 0

0 1 0

0 2 1


1 0 4 0 1 0 1

3 2 0 1 R3 !R3 þ R2 15 ! 40 1 0

0 1 0

3 2 0 1 C3 !C3 þ C2 15 ! 40 2 0

0 1 0

3 0 0 5: 2

Thus 2

1 40 0

0 1 1

32 0 1 0 54 0 1 0

0 1 1

32 0 1 1 54 0 1 0

0 1 0

3 2 0 1 15 ¼ 40 1 0

0 1 0

3 0 0 5; 2

0 1 1

32 0 1 0 54 0 1 0

0 1 1

32 0 1 1 54 0 1 0

0 1 1

3T 2 0 1 05 ¼ 40 1 0

0 1 0

3 0 0 5: 2

or 2

1 40 0 Since

1 0 0 it follows that

0 1 1

0 1 0 0 ¼ 0 1 1 0 0

0 0 ¼ 1 6¼ 0; 1


317

1 0 0

0 1 1

0 0 1


1 40 0

0 1 0

3 2 0 R3 !p1ffi R3 1 2 05 ! 40 2 0

0 1 0

3 2 0 C3 !p1ffi C3 1 2 5 ! 40 p0ffiffiffi 2 0

3 0 0 1 0 5: 0 1

Thus 2

1 40 0

0 1 0

32 0 1 0 54 0 p1ffiffi 0

0 1 0

32 1 0 5 4 0 0 0 2

0 1 0

3 2 0 1 0 5 ¼ 40 p1ffiffi 0

0 1 0

3 0 0 5; 1

0 1 0

32 0 1 0 54 0 p1ffiffi 0

0 1 0

32 1 0 0 54 0 0 2

0 1 0

3 2 0 T 1 0 5 ¼ 40 p1ffiffi 0

0 1 0

3 0 0 5: 1

2

2

or 2

1 40 0

2

2

Since 1 0 0

0 1 0

1 0 1 0 ¼ pffiffiffi 0 20 p1ffiffi 2

0 0 1 1 0 ¼ pffiffiffi 6¼ 0; 2 0 1

it follows that 1 0 0 1 0 0

0 0 p1ffiffi 2


1 40 0

0 1 0

3 2 0 1 R23 0 5 !4 0 1 0

3 2 0 0 1 C23 0 1 5 !4 0 1 0 0

0 1 0

3 0 0 5: 1

318


Thus 2

1 40 0

0 0 1

32 0 1 1 54 0 0 0

0 1 0

32 0 1 0 54 0 1 0

0 0 1

3 2 0 1 15 ¼ 40 0 0

3 0 0 1 0 5; 0 1

0 0 1

32 0 1 1 54 0 0 0

0 1 0

32 0 1 0 54 0 1 0

0 0 1

3T 2 0 1 15 ¼ 40 0 0

3 0 0 1 0 5: 0 1

1 0 0 0 0 1

1 0 0 1 ¼ 0 1 0 0 0

or 2

1 40 0 Since

0 0 ¼ 1 6¼ 0; 1

it follows that 1 0 0

0 0 1

0 1 0

is invertible. From 2

1 0 40 1 0 0

3 0 0 5; 1

we find that r ¼ 2 and s ¼ 1: Hence the signature of the real quadratic form is r sð¼ 2 1 ¼ 1Þ: If we collect the above results, we get 2

1 RAR ¼ 4 0 0 T

0 1 0

3 0 0 5; 1


319

where R stands for 32 3 32 32 32 32 32 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 4 0 0 1 54 0 1 0 54 0 1 0 54 0 1 2 54 0 2 0 54 0 1 0 54 1 1 0 5 2 0 0 p1ffiffi2 0 0 1 0 1 0 0 1 1 0 0 1 0 0 1 1 0 1 2 32 3 32 32 32 32 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 ¼ 4 0 0 pffiffi2 54 0 1 0 54 0 1 2 54 0 2 0 54 0 1 0 54 2 1 0 5 0 30 1 0 312 1 0 0 321 0 302 1 1 302 1 0 12 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 ¼ 4 0 p1ffiffi2 p1ffiffi2 54 0 1 2 54 0 2 0 54 0 1 0 54 12 1 0 5 0 30 1 0 032 1 0 302 1 1 302 1 0 12 0 1 0 0 1 0 0 1 0 0 1 0 0 1ffiffi 54 0 2 0 54 0 1 0 54 12 1 0 5 ¼ 4 0 p1ffiffi2 p 2 0 30 1 0 302 1 1 302 1 0 21 2 1 p0ffiffiffi 0 1 0 0 1 0 0 1ffiffi 54 2 p 0 1 0 54 12 1 0 5 ¼ 40 2 0 2 2 2 32 1 0 31 2 0 0 1 3 1 p0ffiffiffi 0 1 p0ffiffiffi 0 1 0 0 1ffiffi 54 1ffiffi 5 1 5 ¼ 40 2 p 2 p : 1 0 ¼ 4 p1ffiffi2 2 2 2 0 0 1 2 2 2 1 2 2 2

Thus 2

1 40 1

p0ffiffiffi 2 2

2

1 4 Clearly, 0 1 2

1

61 42 1

32 0 1 1ffiffi 54 1 p 2 2 1 2

2 2

2

2

1

1 6 ¼ 40 1 and hence

2 2

32 1 1 4 5 0 2 2 1

p0ffiffiffi 2 2

2 3 0 T 1 0 1ffiffi 5 40 1 p ¼ 2 0 0 2

3 0 0 5: 1

3 0 p0ffiffiffi 1 2 pffiffi2 5ð¼RÞ is invertible. It follows that 2 2

3 2 1 1 7 60 25 ¼ 4

1 2

1 2

3 2 0 1 1 1ffiffi 7 6 p 0 2 5 4 0 2 2 31 2 0 0 1 pffiffiffi 1 7 6 2 pffiffi2 5 4 0

0 pffiffiffi 2

2

2

0

0 1 0 0 1 0

302 1 0 7B6 0 0 5@4 1 1 302 1 0 7B6 0 5@4 0 1 1

0 pffiffiffi 2 2 0 pffiffiffi 2 2

3 11 0 T 1ffiffi 7 C p 2 5 A 2 3 1T 0 1 1ffiffi 7 C p 5 A ; 2

2

320


2

3 1 ðx1 Þ2 þ 2ðx2 Þ2 þ 2ðx3 Þ2 þ x1 x2 þ 2x1 x3 þ 4x2 x3 ¼ ½x1 x2 x3 4 12 2 2 5½x1 x2 x3 T 1 2 2 3 02 2 31 2 3 1T 1 p0ffiffiffi 0 1 1 p0ffiffiffi 0 1 0 0 1ffiffi 5 4 1ffiffi 5 A 2 p 2 p 0 1 0 5 @4 0 ¼ ½ x1 x2 x3 4 0 ½ x1 x2 x3 T 2 2 0 0 1 1 2 2 1 2 2 0 30 2 31 12 2 3 1T 1 p0ffiffiffi 0 1 p0ffiffiffi 0 1 1 0 0 1ffiffi 5 A4 1ffiffi 5 A 2 p 2 p 0 1 0 5@½x1 x2 x3 4 0 ¼ @½ x1 x2 x3 4 0 2 2 1 2 2 1 2 2 2 0 0 1 3 1 0 0 ¼ ½y1 y2 y3 4 0 1 0 5½y1 y2 y3 T ; 0 0 1 1

1 2

where 2

3 1 p0ffiffiffi 0 1 1 2 pffiffi2 5 : ½ y1 y2 y3 ½ x1 x2 x3 4 0 1 2 2 It follows that 2

1 ½x1 x2 x3 ¼ ½y1 y2 y3 4 0 1

p0ffiffiffi 2 2

3 0 pffiffiffi 1 1ffiffi 5 p p ffiffi ffi 2 y þ y þ 2y 2y ¼ y y ; 1 3 2 3 2 3 2 2 2

or 9 x1 ¼ pffiffiyffi1 þ y3 = x2 ¼ 2y2 þ 2y3 : 1ffiffi x3 ¼ p y 2y3 ; 2 2 Also, 2

1 ðx1 Þ þ 2ðx2 Þ þ 2ðx3 Þ þ x1 x2 þ 2x1 x3 þ 4x2 x3 ¼ ½y1 y2 y3 4 0 0 ¼ ðy1 Þ2 þ ðy2 Þ2 ðy3 Þ2 ; 2

that is,

2

2

0 1 0

3 0 0 5½y1 y2 y3 T 1


321

ðx1 Þ2 þ 2ðx2 Þ2 þ 2ðx3 Þ2 þ x1 x2 þ 2x1 x3 þ 4x2 x3 ¼ ðy1 Þ2 þ ðy2 Þ2 ðy3 Þ2 ; where

9 x1 ¼ pffiffiyffi1 þ y3 = x2 ¼ 2y2 þ 2y3 : 1ffiffi x3 ¼ p y 2y3 ; 2 2

Verification: Here LHS ¼ ðx1 Þ2 þ 2ðx2 Þ2 þ 2ðx3 Þ2 þ x1 x2 þ 2x1 x3 þ 4x2 x3 2 pffiffiffi 2 pffiffiffi 1 ¼ ðy1 þ y3 Þ2 þ 2 2y2 þ 2y3 þ 2 pffiffiffi y2 2y3 þ ðy1 þ y3 Þ 2y2 þ 2y3 2 pffiffiffi 1 1 þ 2ðy1 þ y3 Þ pffiffiffi y2 2y3 þ 4 2y2 þ 2y3 pffiffiffi y2 2y3 2 2 ¼ ðy1 Þ2 þ ðy2 Þ2 ð4 þ 1 4Þ þ ðy3 Þ2 ð1 þ 8 þ 8 þ 2 4 16Þ pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi pffiffiffi þ y1 y2 2 2 þ y1 y3 ð2 þ 2 4Þ þ y2 y3 8 2 þ 4 2 þ 2 2 12 2 ¼ ðy1 Þ2 þ ðy2 Þ2 ðy3 Þ2 ¼ RHS:

Verified.

4.2.25 Conclusion Let A aij be an n-square real symmetric matrix. Let /ðx1 ; . . .; xn Þ

X

aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T

i;j

be a real quadratic form. Then there exists a real invertible matrix C cij such that the transformation 9 x1 ¼ c11 y1 þ þ c1n yn > = .. . > ; xn ¼ cn1 y1 þ þ cnn yn reduces the form

P i;j

aij xi xj to the “normal form”

ðy1 Þ2 þ þ ðyr Þ2 ðyr þ 1 Þ2 ðyr þ s Þ2 : Definition If r ¼ 0; then the normal form becomes ðy1 Þ2 ðys Þ2 ; and /ðx1 ; . . .; xn Þ 0 for every real xi ði ¼ 1; . . .; nÞ: In this case, we say that / is negative definite.

322


If s ¼ 0; then the normal form becomes ð y1 Þ 2 þ þ ð yr Þ 2 ; and /ðx1 ; . . .; xn Þ 0 for every real xi ði ¼ 1; . . .; nÞ: In this case, we say that / is positive definite. By / is definite we mean that either / is negative definite or / is positive definite. If / is not definite, then we say that / is indefinite. In the above example, the quadratic form is indefinite. Definition Let P / P aij xi xj w bij xi xj be a pair of real quadratic forms. For a parameter k 2 C; the quadratic form

P aij kbij xi xj is denoted by / kw: By the discriminant of / we mean det aij :

Similarly, the discriminant of w is det bij ; and the discriminant of / kw is

det aij kbij :

Clearly, det aij kbij is a polynomial in k: The polynomial equation

det aij kbij ¼ 0 is called the k-equation of the pair of quadratic forms

P

aij xi xj and

P

bxi xj :

4.2.26 Theorem Let P / Paij xi xj w bij xi xj

be a pair of real quadratic forms. Let bij be invertible. Then all the roots of the kequation of / and w are real.

Proof Let us denote aij by A; and bij by B. Now, aij kbij ¼ A kB; and the k-equation of / and w becomes detðA kBÞ ¼ 0:

Since bij is invertible, B1 exists, and detðBÞ 6¼ 0: Now, A kB ¼ AB1 kI B; and hence


323

detðA kBÞ ¼ det AB1 kI B ¼ det AB1 kI detðBÞ: Thus

detðA kBÞ ¼ det AB1 kI detðBÞ:

Since detðBÞ 6¼ 0; every root of the k-equation of / and w is an eigenvalue of AB1 : Since B is real and symmetric, B1 is real and symmetric. Since A is real and

symmetric, the product AB1 is real and symmetric, and hence ðAB1 Þ ¼ ðAB1 Þ: 1 This shows that ðAB Þ is Hermitian, and hence by 3.3.26, all the eigenvalues of ðAB1 Þ are real. Now, since every root of the k-equation of / and w is an eigen∎ value of AB1 ; every root of the k-equation of / and w is real.

4.3

Application to Riemannian Geometry

4.3.1 Note Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ

X


be a real quadratic form. Suppose that a11 6¼ 0: Pn 2 Clearly, /ðx1 ; . . .; xn Þ a111 is a real quadratic form independent of x1 : i¼1 a1i xi Proof Observe that 1 /ðx1 ; . . .; xn Þ a11

n X

!2 a1i xi

i¼1

1 ¼ /ðx1 ; . . .; xn Þ a11

n X i¼1

! a1i xi

n X j¼1

! a1j xj

! n n X 1 X ¼ /ðx1 ; . . .; xn Þ ða1i xi Þ a1j xj a11 i¼1 j¼1 ! ! n n n n X X X 1 X ¼ aij xi xj a1i xi a1j xj a11 i¼1 j¼1 i¼1 j¼1 ! n n X X a11 aij a1i a1j ¼ xi xj ; a11 i¼1 j¼1 so

324

4 Sylvester’s Law of Inertia n X

1 /ðx1 ; . . .; xn Þ a11 where cij

a11 aij a1i a1j a11

!2 ¼

a1i xi

X

cij xi xj ;

i¼1

: Here

a11 aji a1j a1i a11 aji a1i a1j a11 aij a1i a1j ¼ ¼ ¼ cij ; a11 a11 a11

so cji ¼ cij : Thus cij is a symmetric n-square real matrix, and hence cji ¼

n X

1 /ðx1 ; . . .; xn Þ a11

!2 a1i xi

i¼1

is a real quadratic form. It suffices to show that c1j ¼ 0: Here LHS ¼ c1j ¼

a11 a1j a11 a1j ¼ 0 ¼ RHS: a11

Thus we have shown that /ðx1 ; . . .; xn Þ

1 a11

n P

2 a1i xi

is a real quadratic form

i¼1

independent of x1 : We can denote it by /1 ðx2 ; . . .; xn Þ: Thus n X

1 /ðx1 ; . . .; xn Þ a11

!2 þ /1 ðx2 ; . . .; xn Þ:

a1i xi

i¼1

Put 2

a11 6 0 ½y1 ; . . .; yn T 6 4 ... 0

a12 1 .. .

.. .

3 a1n 0 7 ½x1 ; . . .; xn T : .. 7 . 5

0 1

Since 2

a11 6 0 det6 4 ... 0

a12 1 .. .

.. .

3 a1n 0 7 ¼ a11 6¼ 0; .. 7 . 5

0 1

4.3 Application to Riemannian Geometry

2

a11 6 0 Q6 4 ... 0

a12 1 .. .

.. .

325

3 a1n 0 7 is invertible, and hence .. 7 . 5

0 1

½x1 ; . . .; xn T 7! Q½x1 ; . . .; xn T from R3 to R3 is a one-to-one linear transformation. Since n X

1 /ðx1 ; . . .; xn Þ a11

!2 þ /1 ðx2 ; . . .; xn Þ;

a1i xi

i¼1

we have /ðx1 ; . . .; xn Þ

1 ðy1 Þ2 þ /1 ðy2 ; . . .; yn Þ: a11

4.3.2 Conclusion Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ

X


be a real quadratic form. Suppose that a11 6¼ 0: Then the one-to-one linear transformation 2

a11 6 0 ½y1 ; . . .; yn T ¼ 6 4 ... 0

a12 1 .. .

.. .

3 a1n 0 7 ½x1 ; . . .; xn T .. 7 . 5

0 1

reduces /ðx1 ; . . .; xn Þ to a111 ðy1 Þ2 þ /1 ðy2 ; . . .; yn Þ; where /1 ðy2 ; . . .; yn Þ is a quadratic form. This result is known as Lagrangian reduction.

4.3.3 Note Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ

X


be a real quadratic form. Suppose that a11 ¼ 0; a22 ¼ 0; and a12 6¼ 0: Observe that

326


X /ðx1 ; . . .; xn Þ aij xi xj ¼ a11 ðx1 Þ2 þ a22 ðx2 Þ2 þ a12 x1 x2 þ a21 x2 x1 þ ð2a13 x1 x3 þ 2a14 x1 x4 þ þ 2a1n x1 xn Þ n n X X

þ ð2a23 x2 x3 þ 2a24 x2 x4 þ þ 2a2n x2 xn Þ þ

i¼3

þ 2x2 ða23 x3 þ a24 x4 þ þ a2n xn Þ þ

i¼3

¼ 2 a12 x1 x2 þ x1

n X

a1i xi þ x2

n X

i¼3

¼

2 a12

þ ða12 x2 Þ a12 x2 þ

n X

!

a1i xi

a12 x2 þ

!

!

n X

a1i xi

i¼3

2 ¼ a12

a12 x2 þ

þ

2 a12

n X

þ

a2i xi

n X

!

n X

!

a2i xi

þ

i¼3

!

n X

j¼3

! aij xi xj

!

a21 x1 þ a2i xi i¼3 !! ! n n n n P P P P a1i xi a2j xj aij xi xj þ a1i xi

i¼3

i¼3

j¼3

aij xi xj

j¼3

n n X X

j¼3

n X

!

i¼3

! a2j xj

!

n n X X

n X

a21 x1 þ

a1i xi

aij xi xj

j¼3

a2i xi

i¼3

!

!

i¼3

!!

i¼3

a12 x2 þ

n n X X

i¼3

n 2 X þ a1i xi a12 i¼3

2 ¼ a12

a21 x1 þ

a1i xi

a2i xi

i¼3

i¼3 n X

þ

a2i xi

!

i¼3

i¼3 n X

aij xi xj

n X

!!

n X

aij xi xj

j¼3

!

j¼3

a12 x1 þ

a1i xi

!

a1i xi

i¼3

!

i¼3

2 ¼ a12

n X

i¼3 n X

i¼3

n n X X

þ

a2i xi

aij xi xj

n n X X

i¼3

!

i¼3

2 ¼ a12

þ

a2i xi

!

j¼3

i¼3

ða12 x1 Þða12 x2 Þ þ ða12 x1 Þ n X

!

aij xi xj

j¼3

¼ 2a12 x1 x2 þ 2x1 ða13 x3 þ a14 x4 þ þ a1n xn Þ n n X X

!

i¼3

j¼3


2 ¼ a12

a12 x2 þ

n X

327

! a21 x1 þ

a1i xi

n X

i¼3

! a2i xi

i¼3

! !! n n n n X X X 2 X þ a1i xi a2j xj aij xi xj þ a12 i¼3 j¼3 i¼3 j¼3 ! ! n n X X 2 ¼ a12 x2 þ a1i xi a2i xi a21 x1 þ a12 i¼3 i¼3 ! n n X X 2a1i a2j þ þ aij xi xj ; a12 i¼3 j¼3

so 2 /ðx1 ; . . .; xn Þ a12 where cij

2a1i a2j a12

a12 x2 þ

n X

! a21 x1 þ

a1i xi

i¼3

n X

! a2i xi

þ

i¼3

n n X X i¼3

! cij xi xj ;

j¼3

þ aij : Now, since

2a1j a2i 2a2i a1j 2a2i a1j þ aji ¼ þ aji ¼ þ aij ¼ cij ; a12 a12 a12 Pn Pn we have cji ¼ cij ; and hence c x x is a real quadratic form indei¼3 j¼3 ij i j P Pn c x x pendent of x1 and x2 : So we can denote the real quadratic form ni¼3 j¼3 ij i j cji ¼

by /1 ðx3 ; . . .; xn Þ: Thus 2 /ðx1 ; . . .; xn Þ a12

a12 x2 þ

n X

! a21 x1 þ

a1i xi

i¼3

n X

! a2i xi

þ /1 ðx3 ; . . .; xn Þ:

i¼3

Put 2

0 6 a21 6 0 ½y1 ; . . .; yn T 6 6 . 4 .. 0

a12 0 0 .. . 0

3 a1n a2n 7 7 0 7½x1 ; . . .; xn T : .. 7 . 5 0 1

a13 a23 1 .. .

.. .

328


Since 2

0 6 a21 6 0 det6 6 . 4 .. 0

a12 0 0 .. . 0

3 a1n 2 a12 a2n 7 6 0 7 0 7 ¼ a21 6 . 4 .. .. 7 . 5 0 0 1

a13 a23 1 .. .

.. .

2

0 6 a21 6 0 Q6 6 . 4 ..

a12 0 0 .. .

a13 a23 1 .. .

.. .

3 a1n 0 7 ¼ ða12 Þ2 6¼ 0; .. 7 . 5

0 1

.. .

3 a1n a2n 7 7 0 7 .. 7 . 5

0 1

0

0

a13 1 .. .

is invertible, and hence ½x1 ; . . .; xn T 7! Q½x1 ; . . .; xn T from R3 to R3 is a one-to-one linear transformation. Since 2 /ðx1 ; . . .; xn Þ a12

a12 x2 þ

n X

! a21 x1 þ

a1i xi

i¼3

n X

! a2i xi

i¼3

we have 2 y1 y2 þ /1 ðy3 ; . . .; yn Þ: a12

/ðx1 ; . . .; xn Þ Put 2

1 61 6 0 ½z1 ; . . .; zn T 6 6. 4 ..

1 1 0 .. .

0 0 1 .. .

.. .

3 0 07 7 0 7½y1 ; . . .; yn T : .. 7 .5

0 0 0 1

Since 2

1 61 6 0 det6 6. 4 ..

1 1 0 .. .

0 0 1 .. .

.. .

3 0 07 7 0 7 ¼ 2 6¼ 0; .. 7 .5

0 0 0 1

þ /1 ðx3 ; . . .; xn Þ;


2

1 61 6 0 R6 6. 4 ..

1 1 0 .. .

0 0 1 .. .

.. .

329

3 0 07 7 0 7 is invertible, and hence .. 7 .5

0 0 0 1

½y1 ; . . .; yn T 7! R½y1 ; . . .; yn T from R3 to R3 is a one-to-one linear transformation. Since /ðx1 ; . . .; xn Þ

2 y1 y2 þ /1 ðy3 ; . . .; yn Þ; a12

we have 2 1 1 /ðx1 ; . . .; xn Þ ðz1 þ z2 Þ ðz1 z2 Þ þ /1 ðz3 ; . . .; zn Þ a12 2 2 1 1 ðz1 Þ2 þ ðz2 Þ2 þ /1 ðz3 ; . . .; zn Þ; 2a12 2a12 and hence 1 1 ðz1 Þ2 þ ðz2 Þ2 þ /1 ðz3 ; . . .; zn Þ: 2a12 2a12

4.3.4 Conclusion (I) Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ ¼

/ðx1 ; . . .; xn Þ

X


be a real quadratic form. Suppose that a11 ¼ 0; a22 ¼ 0; and a12 6¼ 0: Then the one-to-one linear transformation 02

1 B6 1 B6 60 ½z1 ; . . .; zn T B B6 . @4 ..

1 1 0 .. .

0 0 1 .. .

.. .

32 0 0 0 76 a21 76 0 76 0 6 . .. 7 . 54 ..

0 0 0 1

0

a12 0 0 .. . 0

a13 a23 1 .. .

.. .

31 a1n a2n 7C 7C 0 7C½x1 ; . . .; xn T C .. 7 . 5A

0 1

1 reduces /ðx1 ; . . .; xn Þ to 2a112 ðz1 Þ2 þ 2a ðz2 Þ2 þ /1 ðz3 ; . . .; zn Þ; where /1 ðz3 ; . . .; zn Þ 12 is a quadratic form. This result is also known as Lagrangian reduction. By repeated application of Lagrangian reduction, we get the following result.

330


4.3.5 Conclusion (II) Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ

X


be a real quadratic form. Then there exists a one-to-one linear transformation ½y1 ; . . .; yn T Q½x1 ; . . .; xn T such that /ðx1 ; . . .; xn Þ reduces to a form ½y1 ; . . .; yn ðdiagðc1 ; . . .; cn ÞÞ½x1 ; . . .; xn T :

4.3.6 Theorem Let A aij be a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ

X


be a real quadratic form. Let A be invertible. Let / be a definite form. Then each aii ði ¼ 1; . . .; nÞ is nonzero. Proof Suppose to the contrary that there exists a diagonal entry of A that is 0. We seek a contradiction. For simplicity, suppose that a11 ¼ 0: Case I: / is a positive definite form. Since A is a symmetric n-square real matrix and / ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T is a definite form, there exists, by 4.2.17, a real invertible matrix R such that RART ¼ In : It follows that 1 T A ¼ R1 RT ¼ R1 R1 : Since X

aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T T ¼ ½x1 ; . . .; xn R1 R1 ½x1 ; . . .; xn T T ¼ ½x1 ; . . .; xn R1 ½x1 ; . . .; xn R1

/ðx1 ; . . .; xn Þ

¼ ½y1 ; . . .; yn ½y1 ; . . .; yn T ; where ½y1 ; . . .; yn ½x1 ; . . .; xn R1 ; : it follows that


331

/ðx1 ; . . .; xn Þ ¼ ½y1 ; . . .; yn ½y1 ; . . .; yn T ¼ ðy1 Þ2 þ þ ðyn Þ2 : Since R is invertible, we have that ½x1 ; . . .; xn 7! ½x1 ; . . .; xn R1 from R3 to R3 is a one-to-one linear transformation, and hence ½1; 0; . . .; 0R1 6¼ ½0; 0; . . .; 0: T

1 1 P It follows that ð½1; 0; . . .; 0R Þð½1; 0; . . .; 0R Þ [ 0: Since /ðx1 ; . . .; xn Þ aij xi xj ; we have

T 0\ ½1; 0; . . .; 0R1 ½1; 0; . . .; 0R1 ¼ /ð1; 0; . . .; 0Þ ¼ a11 1 1 þ 0 þ þ 0 |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} ¼ a11 ¼ 0: Thus we have obtained a contradiction. Case II: / is a negative definite form. This case is similar to Case I. ∎

4.3.7 Theorem Let B bij be a symmetric n-square real matrix. Let B be invertible. Let wðx1 ; . . .; xn Þ

X

bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T

be a real quadratic form. Let w be positive definite. Let P be a real orthogonal nsquare real matrix. Then the (1,1)-entry in PBPT is nonzero. Similarly, the (2,2)entry in PBPT is nonzero, etc. Proof Suppose to the contrary that the (1,1)-entry in PBPT is 0. We seek a contradiction.

Since B bij is a symmetric n-square real matrix and wðx1 ; . . .; xn Þ

X


is a positive definite form, by 4.2.17, there exists matrix C such that a real invertible

CBC T ¼ In : It follows that B ¼ C 1 ðC T Þ

1

¼ C 1 ðC1 Þ

T

: Now,

T T 1 1 T PC ; P ¼ PC PBPT ¼ P C1 C1 T

so PBPT ¼ ðPC 1 ÞðPC 1 Þ : Since P and C are invertible, PC 1 is invertible.

332


2

Suppose that PC 1

c11 6 .. 4 . cn1

.. .

2

c11 6 .. T PBP ¼ 4 . 2 cn1 c11 6 . ¼ 4 .. cn1

3 c1n .. 7 . 5; where each cij is a real number. Now, cnn 32 c1n c11 .. 76 .. .. . . 54 . cnn 32 cn1 c1n c11 .. 76 .. .. . . 54 . c1n cnn

3T c1n . 7 .. . .. 5 cnn 3 cn1 . 7 .. . .. 5: cnn

Since the (1,1)-entry in 2

c11 6 .. 4 . cn1

32 c1n c11 .. 76 .. .. . . 54 . c1n cnn

.. .

3 cn1 .. 7 T . 5 ¼ PBP cnn

is ðc11 Þ2 þ þ ðc1n Þ2 ; the (1,1)-entry. in PBPT is ðc11 Þ2 þ þ ðc1n Þ2 : By assumption, the (1,1)-entry. in PBPT is 0, so ðc11 Þ2 þ þ ðc1n Þ2 ¼ 0: Since each cij is a real number, we have c1i ¼ 0ði ¼ 1; . . .; nÞ: It follows that detðPC 1 Þ ¼ 0; and hence PC 1 is not invertible. This is a contradiction. ∎

4.3.8 Note Let A aij be a symmetric n-square real matrix. Let B bij be a symmetric n-square real matrix. Let B be invertible. Let X /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T be a real quadratic form. Let wðx1 ; . . .; xn Þ

X


be a real quadratic form. Let w be a positive definite form. Let k1 be a root of the kequation of / and w; that is, det½A k1 B ¼ 0: Since det½A k1 B ¼ 0; the characteristic equation det½ðA k1 BÞ kIn ¼ 0 of ðA k1 BÞ is satisfied by k ¼ 0; and hence 0 is an eigenvalue of ðA k1 BÞ: Since A; B are symmetric n-square real matrices, ðA k1 BÞ is also a symmetric n-square real matrix. It follows, by 4.2.12, that there exists a real orthogonal matrix P such that PðA k1 BÞPT ¼ diagðl1 ; . . .; ln Þ;


333

where l1 ; . . .; ln are the eigenvalues of ðA k1 BÞ: Since 0 is an eigenvalue of ðA k1 BÞ; one of the li s is 0. For simplicity, suppose that l1 ¼ 0: Thus ðA k1 BÞ ¼ PT ðdiagð0; l2 ; . . .; ln ÞÞP: It follows that ½x1 ; . . .; xn ðA k1 BÞ½x1 ; . . .; xn T ½x1 ; . . .; xn PT ðdiagð0; l2 ; . . .; ln ÞÞP ½x1 ; . . .; xn T |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} T ½x1 ; . . .; xn PT ðdiagð0; l2 ; . . .; ln ÞÞ ½x1 ; . . .; xn PT ½y1 ; . . .; yn ðdiagð0; l2 ; . . .; ln ÞÞ½y1 ; . . .; yn T ;

where ½y1 ; . . .; yn ½x1 ; . . .; xn PT : Since P is invertible, ½x1 ; . . .; xn T 7! P½x1 ; . . .; xn T from R3 to R3 is a one-to-one linear transformation. Clearly, ½y1 ; . . .; yn ðdiagð0; l2 ; . . .; ln ÞÞ½y1 ; . . .; yn T is a real quadratic form independent of y1 ; so we can denote ½y1 ; . . .; yn ðdiagð0; l2 ; . . .; ln ÞÞ½y1 ; . . .; yn T by /1 ðy2 ; . . .; yn Þ: Thus ½x1 ; . . .; xn ðA k1 BÞ½x1 ; . . .; xn T /1 ðy2 ; . . .; yn Þ: Hence in the reduced form /1 ðy2 ; . . .; yn Þ of /ðx1 ; . . .; xn Þ k1 wðx1 ; . . .; xn Þ; the coefficient of ðy1 Þ2 is zero. By 4.3.7, the ð1; 1Þ-entry in PBPT is nonzero. It follows that the coefficient of ðy1 Þ2 in the real quadratic form ðwðx1 ; . . .; xn Þ ¼ ½x1 ;. . .; xn B½x1 ; . . .; xn T ð½y1 ; . . .; yn PÞB ½y1 ; . . .; yn PÞT ½y1 ; . . .; yn ðPBPT Þ½y1 ; . . .; yn T is nonzero. Thus in the reduced form, say w1 ðy1 ; . . .; yn Þ; of wðx1 ; . . .; xn Þ; the coefficient of ðy1 Þ2 is nonzero. Now we can suppose that

w1 ðy1 ; . . .; yn Þ ¼ ½y1 ; . . .; yn dij ½y1 ; . . .; yn T ; where d11 6¼ 0: Next, by 4.3.2 the one-to-one linear transformation

334


2

d11 6 0 ½z1 ; . . .; zn T ¼ 6 4 ... 0

d12 1 .. .

.. .

3 d1n 0 7 ½y1 ; . . .; yn T .. 7 . 5

0 1

reduces w1 ðy1 ; . . .; yn Þ to d111 ðz1 Þ2 þ /2 ðz2 ; . . .; zn Þ; where /2 ðz2 ; . . .; zn Þ is a quadratic form. Here we can write 9 z1 ¼ d11 y1 þ d12 y2 þ þ d1n yn > > > > z 2 ¼ y2 > = z 3 ¼ y3 ; > .. > > . > > ; z n ¼ yn or y1 ¼ d111 z1 þ

d12 d11 z2 þ y2 ¼ z 2

y3 ¼ z 3 .. .

yn ¼ z n

þ

d1n d11 zn

9 > > > > > = > > > > > ;

:

Hence /ðx1 ; . . .; xn Þ k1 wðx1 ; . . .; xn Þ ½x1 ; . . .; xn ðA k1 BÞ½x1 ; . . .; xn T /1 ðy2 ; . . .; yn Þ /1 ðz2 ; . . .; zn Þ : |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} Thus /ðx1 ; . . .; xn Þ k1 wðx1 ; . . .; xn Þ /1 ðz2 ; . . .; zn Þ: Since wðx1 ; . . .; xn Þ reduces to w1 ðy1 ; . . .; yn Þ; and w1 ðy1 ; . . .; yn Þ reduces to ðz1 Þ2 þ /2 ðz2 ; . . .; zn Þ; it follows that wðx1 ; . . .; xn Þ reduces to d111 ðz1 Þ2 þ /2 ðz2 ; . . .; zn Þ: Thus

1 d11

wðx1 ; . . .; xn Þ

1 ðz1 Þ2 þ /2 ðz2 ; . . .; zn Þ: d11


335

It follows that /ðx1 ; . . .; xn Þ k1

2 1 d11 ðz1 Þ þ /2 ðz2 ; . . .; zn Þ þ /1 ðz2 ; . . .; zn Þ k1 d111 ðz1 Þ2 þ /3 ðz2 ; . . .; zn Þ;

where /3 ðz2 ; . . .; zn Þ k1 /2 ðz2 ; . . .; zn Þ þ /1 ðz2 ; . . .; zn Þ:

4.3.9 Conclusion Let A aij ; and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 be a root of the kequation of / and w: Then there exists a one-to-one linear transformation ½x1 ; . . .; xn 7! ½z1 ; . . .; zn such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðz1 Þ2 þ /1 ðz2 ; . . .; zn Þ ; wðx1 ; . . .; xn Þ c1 ðz1 Þ2 þ w1 ðz2 ; . . .; zn Þ where c1 is a nonzero real number.

Definition Let B bij be a symmetric n-square real matrix. Let wðx1 ; . . .; xn Þ

n n X X

!

bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Suppose that ðc1 ; . . .; cn Þ 6¼ ð0; . . .; 0Þ; where each ci is

real. If bij ½c1 ; . . .; cn T ¼ ½0; . . .; 0T ; then we say that ðc1 ; . . .; cn Þ is a vertex of wðx1 ; . . .; xn Þ:

4.3.10 Let B bij be a symmetric n-square real matrix. Let B be invertible. Let wðx1 ; . . .; xn Þ

n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be an indefinite form. Since wðx1 ; . . .; xn Þ is an indefinite form, wðx1 ; . . .; xn Þ is neither positive definite nor negative definite. If follows, by 4.2.25, that there exists a real invertible matrix

C cij such that the one-to-one transformation

336


½x1 ; . . .; xn T ¼ C ½y1 ; . . .; yn T reduces the form wðx1 ; . . .; xn Þ to the normal form ðy1 Þ2 þ þ ðyr Þ2 ðyr þ 1 Þ2 ðyn Þ2 ; where 1 r\n: Put ½a1 ; . . .; an T ¼ C ½1; 0; . . .; 0T : Now, since C is invertible, ða1 ; . . .; an Þ 6¼ ð0; . . .; 0Þ: Also, wða1 ; . . .; an Þ ¼ 12 þ 02 þ þ 02 02 02 ¼ 1 [ 0; so wða1 ; . . .; an Þ is positive. Similarly, there exist nonzero ðb1 ; . . .; bn Þ such that wðb1 ; . . .; bn Þ is negative.

4.3.11 Conclusion Let B bij be a symmetric n-square real matrix. Let B be invertible. Let wðx1 ; . . .; xn Þ

n n X X

!

bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be an indefinite form. Then there exist nonzero real points ða1 ; . . .; an Þ and ðb1 ; . . .; bn Þ such that wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative.

4.3.12 Let B bij be a symmetric n-square real matrix. Let B be invertible. Let wðx1 ; . . .; xn Þ

n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be an indefinite form. Let ða1 ; . . .; an Þ and ðb1 ; . . .; bn Þ be nonzero real points such that wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative. Observe that


337

wða1 þ kb1 ; . . .; an þ kbn Þ ¼ ½a1 þ kb1 ; . . .; an þ kbn bij ½a1 þ kb1 ; . . .; an þ kbn T

¼ ð½a1 ; . . .; an þ k½b1 ; . . .; bn Þ bij ð½a1 ; . . .; an þ k½b1 ; . . .; bn ÞT

¼ ½a1 ; . . .; an bij ½a1 ; . . .; an T þ ð½a1 ; . . .; an bij ½b1 ; . . .; bn T

þ ½b1 ; . . .; bn bij ½a1 ; . . .; an Þk! þ ½b1 ; . . .; bn bij ½b!1! ; . . .; bn T k2 n n n n P P P P ¼ ðwðb1 ; . . .; bn ÞÞk2 þ bij ai bj þ bij bi aj k þ wða1 ; . . .; an Þ i¼1 j¼1 i¼1 j¼1 ! !! n n n n P P P P 2 bij ai bj þ bji bi aj ¼ ðwðb1 ; . . .; bn ÞÞk þ k þ wða1 ; . . .; an Þ i¼1 j¼1 ! i¼1 j¼1 !! n n n n P P P P bij ai bj þ bji aj bi ¼ ðwðb1 ; . . .; bn ÞÞk2 þ k þ wða1 ; . . .; an Þ i¼1 j¼1 ! ! i¼1 j¼1 n n n n P P P P bij ai bj þ bij ai bj ¼ ðwðb1 ; . . .; bn ÞÞk2 þ k þ wða1 ; . . .; an Þ i¼1 j¼1 j¼1 i¼1 ! !! n n n n P P P P 2 bij ai bj þ bij ai bj ¼ ðwðb1 ; . . .; bn ÞÞk þ k þ wða1 ; . . .; an Þ i¼1 j¼1 i¼1 j¼1 ! n n P P bij ai bj k þ wða1 ; . . .; an Þ; ¼ ðwðb1 ; . . .; bn ÞÞk2 þ 2 i¼1

j¼1

so wða1 þ kb1 ; . . .; an þ kbn Þ ðwðb1 ; . . .; bn ÞÞk þ 2 2

n n X X i¼1

! bij ai bj k þ wða1 ; . . .; an Þ:

j¼1

Since wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative, the discriminant of ðwða1 þ kb1 ; . . .; an þ kbn Þ ¼Þðwðb1 ; . . .; bn ÞÞk2 ! n n X X þ2 bij ai bj k þ wða1 ; . . .; an Þ i¼1

j¼1

is positive, and hence there exist two distinct real numbers k1 and k2 such that wða1 þ k1 b1 ; . . .; an þ k1 bn Þ ¼ 0 : wða1 þ k2 b1 ; . . .; an þ k2 bn Þ ¼ 0 Clearly k1 k2 is negative, and hence k1 ; k2 are of opposite signs. Also, wðx1 ; . . .; xn Þ vanishes at the two real points ða1 þ k1 b1 ; . . .; an þ k1 bn Þ ð¼ ða1 ; . . .; an Þ þ k1 ðb1 ; . . .; bn ÞÞ and ða1 þ k2 b1 ; . . .; an þ k2 bn Þð¼ ða1 ; . . .; an Þ þ k2 ðb1 ; . . .; bn ÞÞ: Since k1 and k2 are distinct real numbers and ðb1 ; . . .; bn Þ are

338


nonzero, ða1 ; . . .; an Þ þ k1 ðb1 ; . . .; bn Þ and ða1 ; . . .; an Þ þ k2 ðb1 ; . . .; bn Þ are distinct points.

Clearly, bij ½a1 þ k1 b1 ; . . .; an þ k1 bn T 6¼ ½0; . . .; 0T :

Proof Suppose to the contrary that bij ½a1 þ k1 b1 ; . . .; an þ k1 bn T ¼ ½0; . . .; 0T : We seek a contradiction.

Since bij ½a1 þ k1 b1 ; . . .; an þ k1 bn T ¼ ½0; . . .; 0T ; and bij is invertible, we have ½a1 þ k1 b1 ; . . .; an þ k1 bn ¼ ½0; . . .; 0; and hence ða1 ; . . .; an Þ þ k1 ðb1 ; . . .; bn Þ ¼ ð0; . . .; 0Þ: It follows that ða1 ; . . .; an Þ ¼ ðk1 b1 ; . . .; k1 bn Þ: Since wða1 ; . . .; an Þ is positive, wðk1 b1 ; . . .; k1 bn Þ

¼ ½k1 b1 ; . . .; k1 bn bij ½k1 b1 ; . . .; k1 bn T ¼ ðk1 Þ2 wðb1 ; . . .; bn Þ is positive, and hence ðk1 Þ2 wðb1 ; . . .; bn Þ is positive. Since k1 is real and wðb1 ; . . .; bn Þ is negative, ðk1 Þ2 wðb1 ; . . .; bn Þ 0: This is a contradiction. ∎

T T Thus we have shown that bij ½a1 þ k1 b1 ; . . .; an þ k1 bn 6¼ ½0; . . .; 0 : Hence ða1 ; . . .; an Þ þ k1 ðb1 ; . . .; bn Þ is different from the origin and is not a vertex of wðx1 ; . . .; xn Þ: Similarly, ða1 ; . . .; an Þ þ k2 ðb1 ; . . .; bn Þ is different from the origin and is not a vertex of wðx1 ; . . .; xn Þ:

4.3.13 Conclusion Let B bij be a symmetric n-square real matrix. Let B be invertible. Let wðx1 ; . . .; xn Þ

n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be an indefinite form. Let ða1 ; . . .; an Þ and ðb1 ; . . .; bn Þ be nonzero real points such that wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative. Then there exist two distinct real numbers k1 and k2 such that 1. 2. 3. 4.

k1 ; k2 are of opposite signs, wða1 þ k1 b1 ; . . .; an þ k1 bn Þ ¼ 0; wða1 þ k2 b1 ; . . .; an þ k2 bn Þ ¼ 0; ða1 þ k1 b1 ; . . .; an þ k1 bn Þ and ða1 þ k2 b1 ; . . .; an þ k2 bn Þ are points different from origin and the vertices of wðx1 ; . . .; xn Þ:


339


n n X X

!

bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Suppose that for every real point ðc1 ; . . .; cn Þ that is different from the origin and the vertices of wðx1 ; . . .; xn Þ; wðc1 ; . . .; cn Þ 6¼ 0: Then wðx1 ; . . .; xn Þ is definite. Proof Suppose to the contrary that wðx1 ; . . .; xn Þ is indefinite. We seek a contradiction. By 4.3.11, there exist nonzero real points ða1 ; . . .; an Þ and ðb1 ; . . .; bn Þ such that wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative. By 4.3.13, there exist two distinct real numbers k1 and k2 such that 1. 2. 3. 4.


Since ða1 þ k1 b1 ; . . .; an þ k1 bn Þ is a point different from origin and the vertices of wðx1 ; . . .; xn Þ; by assumption, wða1 þ k1 b1 ; . . .; an þ k1 bn Þ 6¼ 0: This is a contradiction. ∎

4.3.15 Theorem Let B bij be a symmetric n-square real matrix. Let B be not invertible. Let wðx1 ; . . .; xn Þ

n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Then there exists a real point ða1 ; . . .; an Þ different from the origin such that wða1 ; . . .; an Þ ¼ 0: Proof Since B is not invertible, rankðBÞ\n: By 4.2.17, there exists a real invertible matrix R such that 0

1

RBRT ¼ diag@1; . . .1; 1; . . .; 1; 0; . . .; 0A: |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n

It follows that

340


0

0

11

1 B ¼ R1 @diag@1; . . .1; 1; . . .; 1 ; 0; . . .; 0AA RT |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl} 0 0 11 T ¼ R1 @diag@1; . . .1; 1; . . .; 1; 0; . . .; 0AA R1 ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n

and hence wðx1 ; . . .; xn Þ ½x1 ; . . .; xn B½x1 ; . . .; xn T 0 0 0

11

1 T ½x1 ; . . .; xn @R1 @diag@1; . . .1; 1; . . .; 1 ; 0; . . .; 0AA R1 A ½x1 ; . . .; xn T |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n 0 1

½y1 ; . . .; yn diag@1; . . .1; 1; . . .; 1; 0; . . .; 0A½y1 ; . . .; yn T ; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n

where ½y1 ; . . .; yn ½x1 ; . . .; xn R1 : Since R is invertible, ½0; . . .; 0; 1R is nonzero. Put ½a1 ; . . .; an ½0; . . .; 0; 1R: Thus ða1 ; . . .; an Þ 6¼ ð0; . . .; 0Þ: Also, wða1 ; . . .; an Þ

0

1 T ¼ ½a1 ; . . .; an @R1 @diag@1; . . .1; 1; . . .; 1; 0; . . .; 0AA R1 A½a1 ; . . .; an T |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n 0 0 11 0

0

11

¼ ½0; . . .; 0; 1@diag@1; . . .1; 1; . . .; 1 ; 0; . . .; 0AA½0; . . .; 0; 1T ¼ 0; |fflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflffl} \n

so wða1 ; . . .; an Þ ¼ 0:

∎


n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be an indefinite form. Then there exists a real point ðc1 ; . . .; cn Þ different from the origin such that wðc1 ; . . .; cn Þ ¼ 0:


341

Proof By 4.3.11, there exist nonzero real points ða1 ; . . .; an Þ and ðb1 ; . . .; bn Þ such that wða1 ; . . .; an Þ is positive and wðb1 ; . . .; bn Þ is negative. By 4.3.13, there exist two distinct real numbers k1 and k2 such that 1. 2. 3. 4.


Let us take ðc1 ; . . .; cn Þ ða1 þ k1 b1 ; . . .; an þ k1 bn Þ: Now, wðc1 ; . . .; cn Þ ¼ 0; and ðc1 ; . . .; cn Þ is a real point different from the origin. ∎


n n X X

! bij xi xj

n X

xi

½x1 ; . . .; xn bij ½x1 ; . . .; xn T i¼1

j¼1

i¼1

n X

!! bij xj

j¼1

be a real quadratic form. Let wðx1 ; . . .; xn Þ be a definite form. Let ða1 ; . . .; an Þ be a real point that is different from the origin. Then wða1 ; . . .; an Þ 6¼ 0: Proof Case I: wðx1 ; . . .; xn Þ is a positive definite form. Since wðx1 ; . . .; xn Þ is a definite form, rankðBÞ ¼ n: By 4.2.17, there exists a real invertible matrix R such that 0 1 RBRT ¼ diag@1; . . .; 1 A: |fflfflffl{zfflfflffl} n

It follows that 0

0

11

0 0 11 T 1 B ¼ R1 @diag@1; . . .; 1AA RT ¼ R1 @diag@1; . . .; 1AA R1 ; |fflfflffl{zfflfflffl} |fflfflffl{zfflfflffl} n n |fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

and hence wðx1 ; . . .; xn Þ ½x1 ; . . .; xn B½x1 ; . . .; xn T 0 0 0

11

1 T ½x1 ; . . .; xn @R1 @diag@1; . . .; 1 AA R1 A ½x1 ; . . .; xn T |fflfflffl{zfflfflffl} n

½y1 ; . . .; yn ½y1 ; . . .; yn T ;

342


where ½y1 ; . . .; yn ½x1 ; . . .; xn R1 : Since R is invertible and ða1 ; . . .; an Þ 6¼ ð0; . . .; 0Þ; ½a1 ; . . .; an R1 is nonzero. Put ½b1 ; . . .; bn ½a1 ; . . .; an R1 : Thus ðb1 ; . . .; bn Þ 6¼ ð0; . . .; 0Þ: It follows that ðb1 Þ2 þ þ ðbn Þ2 6¼ 0: Also, 0

0

0

11

1 1 T A

wða1 ; . . .; an Þ ¼ ½a1 ; . . .; an @R1 @diag@1; . . .; 1 AAðR Þ |fflfflffl{zfflfflffl} n 0 0 11

½a1 ; . . .; an T

¼ ½b1 ; . . .; bn @diag@1; . . .; 1AA½b1 ; . . .; bn T ¼ ðb1 Þ2 þ þ ðbn Þ2 6¼ 0; |fflfflffl{zfflfflffl} n

so wða1 ; . . .; an Þ 6¼ 0: Case II: wðx1 ; . . .; xn Þ is a negative definite form. This case is similar to Case I. ∎

4.3.18 Note Let A aij , and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 be a root of the kequation of / and w: Then by 4.3.9, there exists a one-to-one linear transformation ½x1 ; . . .; xn 7! ½z1 ; . . .; zn such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðz1 Þ2 þ /1 ðz2 ; . . .; zn Þ ; wðx1 ; . . .; xn Þ c1 ðz1 Þ2 þ w1 ðz2 ; . . .; zn Þ where c1 is a nonzero real number. Clearly, the matrix associated with w1 is invertible. Proof Suppose to the contrary that the matrix associated with w1 is not invertable. We seek a contradiction. By 4.3.15, there exists a real point ðd2 ; . . .; dn Þ 6¼ ð0; . . .; 0Þ such that w1 ðd2 ; . . .; dn Þ ¼ 0: Suppose that ð½x1 ; . . .; xn ¼Þ½a1 ; . . .; an 7! ½0; d2 ; . . .; dn ð¼ ½z1 ; . . .; zn Þ: Since ð0; d2 ; . . .; dn Þ 6¼ ð0; . . .; 0Þ; and the linear transformation ½x1 ; . . .; xn 7! ½z1 ; . . .; zn is one-to-one, ða1 ; . . .; an Þ is nonzero. Also, wða1 ; . . .; an Þ ¼ c1 ð0Þ2 þ w1 ðd2 ; . . .; dn Þ ¼ 0; so wða1 ; . . .; an Þ ¼ 0: Since ða1 ; . . .; an Þ is nonzero and w is positive definite, by 4.3.17, wða1 ; . . .; an Þ 6¼ 0: This is a contradiction. ∎


343

Clearly, w1 is definite. Proof Suppose to the contrary that w1 is indefinite. We seek a contradiction. By 4.3.16, there exists a real point ðd2 ; . . .; dn Þ 6¼ ð0; . . .; 0Þ such that w1 ðd2 ; . . .; dn Þ ¼ 0: Suppose that ð½x1 ; . . .; xn ¼Þ½a1 ; . . .; an 7! ½0; d2 ; . . .; dn ð¼ ½z1 ; . . .; zn Þ: Since ð0; d2 ; . . .; dn Þ 6¼ ð0; . . .; 0Þ; and the linear transformation ½x1 ; . . .; xn 7! ½z1 ; . . .; zn is one-to-one, ða1 ; . . .; an Þ is nonzero. Also, wða1 ; . . .; an Þ ¼ c1 ð0Þ2 þ w1 ðd2 ; . . .; dn Þ ¼ 0; so wða1 ; . . .; an Þ ¼ 0: Since ða1 ; . . .; an Þ is nonzero and w is positive definite, by 4.3.17, wða1 ; . . .; an Þ 6¼ 0: This is a contradiction. ∎

4.3.19 Conclusion Let A aij ; and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 be a root of the kequation of / and w: Let ½x1 ; . . .; xn 7! ½z1 ; . . .; zn be a one-to-one linear transformation such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðz1 Þ2 þ /1 ðz2 ; . . .; zn Þ ; wðx1 ; . . .; xn Þ c1 ðz1 Þ2 þ w1 ðz2 ; . . .; zn Þ where c1 is a nonzero real number. Then 1. the matrix associated with w1 is invertible, 2. w1 is definite.

4.3.20 Theorem Let A aij ; and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 ; . . .; kn be the roots of the k-equation of / and w: Let ½z1 ; . . .; zn ¼ ½x1 ; . . .; xn Q be a one-to-one linear transformation such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðz1 Þ2 þ /1 ðz2 ; . . .; zn Þ ; wðx1 ; . . .; xn Þ c1 ðz1 Þ2 þ w1 ðz2 ; . . .; zn Þ where c1 is a nonzero real number and Q is an invertible n-square real matrix. Then k2 ; . . .; kn are the roots of the k-equation of /1 and w1 :

344


Proof Let A1 be the ðn 1Þ-square real symmetric matrix associated with the quadratic form /1 ðz2 ; . . .; zn Þ: Let B1 be the ðn 1Þ-square real symmetric matrix associated with the quadratic form w1 ðz2 ; . . .; zn Þ: Clearly, ½x1 ; . . .; xn A½x1 ; . . .; xn T k 1 c1 0 ¼ ½z1 ; . . .; zn ½z1 ; . . .; zn T 0 A1 k1 c 1 0 ¼ ð½x1 ; . . .; xn QÞ ð½x1 ; . . .; xn QÞT 0 A1 k1 c 1 0 ¼ ½x1 ; . . .; xn Q QT ½x1 ; . . .; xn T ; 0 A1 so k c A¼Q 1 1 0

0 QT : A1

Similarly,

c B¼Q 1 0

0 QT : B1

Since Q is invertible, detðQÞ is a nonzero real number. Since k1 ; . . .; kn are the roots of the k-equation of / and w; we have detðA kBÞ ¼ ðk1 kÞðk2 kÞ ðkn kÞ: It suffices to show that detðA1 kB1 Þ ¼ ðnonzero constantÞðk2 kÞ ðkn kÞ: Since ðk1 kÞðk2 kÞ ðkn kÞ ¼ detðA kBÞ c1 0 k1 c 1 0 ¼ det Q QT kQ QT 0 A1 0 B1 k1 c1 kc1 0 ¼ det Q QT 0 A1 kB1 k1 c1 kc1 0 ¼ detðQÞ det det QT 0 A1 kB1


345

k1 c1 kc1 0 ¼ detðQÞ det detðQÞ 0 A1 kB1 k1 c1 kc1 0 ¼ ðdetðQÞÞ2 det 0 A1 kB1 ¼ ðdetðQÞÞ2 ðk1 c1 kc1 Þ detðA1 kB1 Þ ¼ ðdetðQÞÞ2 ðk1 kÞc1 detðA1 kB1 Þ; we have detðA1 kB1 Þ ¼

1 c1 ðdetðQÞÞ2

ðk2 kÞ ðkn kÞ:

∎

4.3.21 Note Let A aij and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 ; . . .; kn be the roots of the k-equation of / and w: Let ½z1 ; . . .; zn ¼ ½x1 ; . . .; xn Q be a one-to-one linear transformation such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðz1 Þ2 þ /1 ðz2 ; . . .; zn Þ ; wðx1 ; . . .; xn Þ c1 ðz1 Þ2 þ w1 ðz2 ; . . .; zn Þ where c1 is a nonzero real number and Q is an invertible n-square real matrix. By 4.3.19, 1. the matrix associated with w1 is invertible, 2. w1 is definite. Next, by 4.3.20, 3. k2 ; . . .; kn are the roots of the k-equation of /1 and w1 : Again, by repeating the same procedure, there exists a one-to-one linear transformation ½z1 ; z2 ; . . .; zn 7! ½w1 ; w2 ; . . .; wn such that z1 ¼ w1 , and the pair’s reduced forms are /1 ðz2 ; . . .; zn Þ k2 c2 ðw2 Þ2 þ /2 ðw3 ; . . .; wn Þ ; w1 ðz2 ; . . .; zn Þ c2 ðw2 Þ2 þ w2 ðw3 ; . . .; wn Þ where c2 is a nonzero real number. Also,

346


1. the matrix associated with w2 is invertible, 2. w2 is definite, 3. k3 ; . . .; kn are the roots of the k-equation of /2 and w2 : It follows that /ðx1 ; . . .; xn Þ k1 c1 ðw1 Þ2 þ k2 c2 ðw2 Þ2 þ /2 ðw3 ; . . .; wn Þ : wðx1 ; . . .; xn Þ c1 ðw1 Þ2 þ c2 ðw2 Þ2 þ w2 ðw3 ; . . .; wn Þ On repeating the above procedure, we get a one-to-one linear transformation ½x1 ; . . .; xn 7! ½v1 ; . . .; vn such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðv1 Þ2 þ þ kn cn ðvn Þ2 ; wðx1 ; . . .; xn Þ c1 ðv1 Þ2 þ þ cn ðvn Þ2 where each ci is a nonzero real number.

4.3.22 Conclusion Let A aij ; and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 ; . . .; kn be the roots of the k-equation of / and w: Then there exists a one-to-one linear transformation ½x1 ; . . .; xn 7! ½v1 ; . . .; vn such that the pair’s reduced forms are /ðx1 ; . . .; xn Þ k1 c1 ðv1 Þ2 þ þ kn cn ðvn Þ2 ; wðx1 ; . . .; xn Þ c1 ðv1 Þ2 þ þ cn ðvn Þ2 where each ci is a nonzero real number.

4.3.23 Theorem Let A aij ; and B bij be symmetric n-square real matrices. Let B be invertible. Let P /ðx1 ; . . .; xn Þ aij xi xj ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T P wðx1 ; . . .; xn Þ bij xi xj ¼ ½x1 ; . . .; xn B½x1 ; . . .; xn T be a pair of real quadratic forms. Let w be positive definite. Let k1 ; . . .; kn be the roots of the k-equation of / and w: Then there exists a one-to-one linear transformation ½x1 ; . . .; xn 7! ½y1 ; . . .; yn such that the pair’s reduced forms are 9 /ðx1 ; . . .; xn Þ k1 ðy1 Þ2 þ þ kn ðyn Þ2 = : ; wðx1 ; . . .; xn Þ ðy1 Þ2 þ þ ðyn Þ2


347

Proof By 4.3.22, there exists a one-to-one linear ½x1 ; . . .; xn 7! ½v1 ; . . .; vn such that the pair’s reduced forms are

transformation

/ðx1 ; . . .; xn Þ k1 c1 ðv1 Þ2 þ þ kn cn ðvn Þ2 ; wðx1 ; . . .; xn Þ c1 ðv1 Þ2 þ þ cn ðvn Þ2 where each ci is a nonzero real number. On applying the one-to-one linear transformation 9 v1 ¼ p1ffiffiffiffiffi y1 > > jc1 j > = .. ; . > > 1 > vn ¼ pffiffiffiffiffi yn ; jcn j

we get the following reduced forms: /ðx1 ; . . .; xn Þ k1 jcc11 j ðy1 Þ2 þ þ kn jccnn j ðyn Þ2 wðx1 ; . . .; xn Þ jcc11 j ðy1 Þ2 þ þ

ci jci j

)

2 cn jcn j ðyn Þ

:

Here each jccii j is equal to 1 or −1. So if wðx1 ; . . .; xn Þ is positive definite, then each is equal to 1, and hence /ðx1 ; . . .; xn Þ k1 ðy1 Þ2 þ þ kn ðyn Þ2 : wðx1 ; . . .; xn Þ ðy1 Þ2 þ þ ðyn Þ2 Similarly, if wðx1 ; . . .; xn Þ is negative definite, then 9 /ðx1 ; . . .; xn Þ k1 ðy1 Þ2 þ þ kn ðyn Þ2 = : ; wðx1 ; . . .; xn Þ ðy1 Þ2 þ þ ðyn Þ2 ∎

Exercises 1. Let V be any n-dimensional vector space over the field F. Let T : V ! V be a linear transformation. Show that there exists a positive integer k such that i k ) N Tk ¼ N Ti ; and N T k1 is a proper subset of N T k : 2. Let V be any n-dimensional vector space over the field F. Let T : V ! V be a linear transformation. Show that ranðT 3 Þ is invariant under T.

348


3. Let V be any n-dimensional inner product space over the field C: Let T : V ! V be a normal linear transformation. Suppose that all the eigenvalues of T are real. Show that T is Hermitian. 4. Let A be a 6-square complex matrix. Suppose that A is a nonnegative definite matrix. Show that there exists a unitary 6 6 matrix U such that a. b. c. d.

A ¼ U ðdiagðk1 ; . . .; k6 ÞÞU ; k1 ; . . .; k6 are the eigenvalues of the matrix A, each ki is a nonnegative real number, detð AÞ is a nonnegative real number.

5. Let A be a 6-square complex matrix. Let A be symmetric and unitary. Show that there exists a symmetric unitary complex matrix S such that S2 ¼ A:

6. Let aij be an n-square real symmetric matrix. Let /ðx1 ; . . .; xn Þ

X

aij xi xj

i;j

be a real quadratic form. Show that there exists a real invertible matrix cij such that the transformation 9 x1 ¼ c11 y1 þ þ c1n yn > = .. . > ; xn ¼ cn1 y1 þ þ cnn yn reduces the form

P i;j

aij xi xj to the form

ðy1 Þ2 þ þ ðyr Þ2 ðyr þ 1 Þ2 ðyr þ s Þ2 :

7. Let A aij ; and B bij be symmetric 5-square real matrices. Let B be invertible. Let /ðx1 ; . . .; x5 Þ ½x1 ; . . .; x5 A½x1 ; . . .; x5 T wðx1 ; . . .; xn Þ ½x1 ; . . .; x5 B½x1 ; . . .; x5 T

be a pair of real quadratic forms. Let w be positive definite. Let k1 ; . . .; k5 be the roots of the k-equation of / and w: Show that there exists a one-to-one linear transformation ½x1 ; . . .; x5 7! ½v1 ; . . .; v5 such that the pair’s reduced forms are /ðx1 ; . . .; x5 Þ k1 c1 ðv1 Þ2 þ þ k5 c5 ðv5 Þ2 ; wðx1 ; . . .; x5 Þ c1 ðv1 Þ2 þ þ c5 ðv5 Þ2 where each ci is a nonzero real number.


349

8. Let A be an n-square complex matrix. Suppose that A is a nonnegative definite matrix. Show that the square root of A exists. 9. Let V be any n-dimensional vector space over the field F. Let T : V ! V be a linear transformation. Show that there exists a positive integer k such that V ¼ N T k ran T k :

10. Suppose that A aij is a symmetric n-square real matrix. Let /ðx1 ; . . .; xn Þ ¼ ½x1 ; . . .; xn A½x1 ; . . .; xn T be a real quadratic form. Suppose that a11 6¼ 0: Show that the one-to-one linear transformation 2

a11 6 0 T ½y1 ; . . .; yn ¼ 6 4 ... 0

a12 1 .. .

.. .

3 a1n 0 7 ½x1 ; . . .; xn T .. 7 5 .

0 1

reduces /ðx1 ; . . .; xn Þ to 1 ðy1 Þ2 þ /1 ðy2 ; . . .; yn Þ; a11 where /1 ðy2 ; . . .; yn Þ is a quadratic form.

Bibliography

1. 2. 3. 4. 5. 6.

M. Artin, Algebra (Prentice Hall, 2008) P.R. Halmos, Finite-Dimensional Vector Spaces (Springer, 2011) I.N. Herstein, Topics in Algebra, 2nd edn. (Wiley-India, 2008) N. Jacobson, Lectures in Abstract Algebra (D. Van Nostrand Company, Inc., 1965) I.S. Luthar, I.B.S. Passi, Field Theory (Narosa, 2008) F. Zhang, Matrix Theory (Springer, 1999)

© Springer Nature Singapore Pte Ltd. 2020 R. Sinha, Galois Theory and Advanced Linear Algebra, https://doi.org/10.1007/978-981-13-9849-0

351

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