[PDF] EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA

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EXERCISES AND SOLUTIONS

IN LINEAR ALGEBRA

Mahmut Kuzucuoglu

Middle East Technical University

matmah@metu.edu.tr

Ankara, TURKEY

March 14, 2015

ii

TABLE OF CONTENTS

CHAPTERS

0. PREFACE .................................................. 1

1. LINEAR ...................................................??

2. MAP ......................................................??

3. ............................................................??

4. ?? ......................................................??

5. ??? .........................................................??

Preface

I have given some linear algebra courses in various years. These problems are given to students from the books which I have followed that year. I have kept the solutions of exercises which I solved for the students. These notes are collection of those solutions of exercises.

Mahmut Kuzucuo˘glu

METU, Ankara

March 14, 2015

M. Kuzucuoglu

1 1956.
.............................................M E T U

DEPARTMENT OF MATHEMATICS

Math 262 Quiz IName : Answer Key

ID Number : 2360262

Signature : 0000000

Duration : 60 minutes

(05.03.2008) Show all your work. Unsupported answers will not be graded.1.)LetA=?

0- 

(a) Find the characteristic polynomial ofA. Solution.The characteristic polynomial ofAisf(x) = det(xI-A).So, f(x) =? ??????x-- 

0x+ -

 x-? = (x-)? ????x+ - x-? x+ -? = (x-)(x2-1 + 1) + (1-x-8) = (x-)(x2+ ) + (-x) = (x-)(x2+ -) = (x-)(x-1)(x+ 1) (b) Find the minimal polynomial ofA. Solution.We know that the minimal polynomial divides the characteristic polynomial and they same the same roots. Thus, the minimal polynomial forAismA(x) =f(x) = (x-)(x-1)(x+ 1). (c) Find the characteristic vectors and a basisBsuch that[A]Bis diagonal. Solution.The characteristic values ofAarec1= ,c2= 1,c3=-1.

A-I=?

?0 -

0- 

0 -1

0 0 0?

?,-x-y+ z= 0 y-z= 0?z= y x=-y

Thus,α1= (-1,1,)is a characteristic vector associated with the characteristic valuec1= .

A-I=? ?1 -

0- 

?1 -

0- 

0 0 0?

?,x+ y-z= 0 -y+ z= 0?y= k z= k x=-k

Thus,α2= (-,,)is a characteristic vector associated with the characteristic valuec2= 1.

A+I=?

0- 

0 1-1

0 0 0?

?,x+ y-z= 0 y-z= 0?x=-t y= t z= t

Thus,α3= (-,,)is a characteristic vector associated with the characteristic valuec3=-1.

Now,B={α1,α2,α3}is a basis and[A]B=?

? 0 0 0 1 0

0 0-1?

?is a diagonal matrix. (d) FindA-conductor of the vectorα= (1,1,1)into the invariant subspace spanned by(-1,1,). Solution.SetW=<(-1,1,)>and denote theA-conductor ofαintoWbyg(x).Since m A(A) = 0we havemA(A)α?W.Thus,g(x)dividesmA(x).Hence, the possibilities forg(x) arex-,x-1,x+ 1,(x-)(x-1),(x-)(x+ 1),(x-1)(x+ 1).We will try these polynomials. (Actually, the answer could be given directly.) Now, (A-I)α=? ?0 -

0- 

?1 1 1? -8? ?/?W?g(x)?=x-, (A-I)α=? ?1 -

0- 

?1 1 1? ?/?W?g(x)?=x-1, (A+I)α=?

0- 

?1 1 1? 0 ?/?W?g(x)?=x+ 1, (A-I)(A-I)α=? ?0 -

0- 

-9 -9? ?/?W?g(x)?= (x-)(x-1), (A-I)(A+I)α=? ?0 -

0- 

0 ?10 -1 ?/?W?g(x)?= (x-)(x+ 1), (A-I)(A+I)α=? ?1 -

0- 

0 ?1 -1 -0? ?=-1α1?W?g(x) =x2-1.

2.)Find a×matrix whose minimal polynomial isx2.

Solution.For the matrixA=?

?0 0 1 0 0 0

0 0 0?

?we haveA?= 0andA2= 0.Thus,Ais a×matrix whose minimal polynomial isx2.

3.)Prove that similar matrices have the sameminimalpolynomial.

Solution.LetAandBbe similar matrices, i.e.,B=P-1APfor some invertible matrixP.For anyk >0we haveBk= (P-1AP)k=P-1AkPwhich implies thatf(B) =P-1f(A)Pfor any polynomialf(x).LetfAandfBbe the minimal polynomials ofAandB,respectively. ThenfA(B) = P -1fA(A)P=P-1OP=Oimplies thatfBdividesfA.On the other hand,O=fB(B) =P-1fB(A)P gives usfB(A) =O.Hence,fAdividesfB.Therefore, we havefA=fB.

2M. KUZUCUOGLU

1. Math 262 Exercises and Solutions

(1) LetAbe a 3×3 matrix with real entries. Prove that ifAis not similar overRto a triangular matrix thenAis similar over

Cto a diagonal matrix.

Proof.SinceAis a 3×3 matrix with real entries, the characteristic polynomial,f(x),ofAis a polynomial of degree

3 with real coefficients. We know that every polynomial of

degree 3 with real coefficients has a real root, sayc1. On the other hand, sinceAis not similar overRto a tri- angular matrix, the minimal polynomial ofAis not product of polynomials of degree one. So one of the irreducible factor, h,of the minimal polynomial ofAis degree 2.Thenhhas two complex roots, one of which is the conjugate of the other. Thus, the characteristic polynomial has one real root and two complex roots,c1,λand¯λ. The minimal polynomial over complex numbers is (x- c

1)(x-λ)(x-¯λ) which implies thatAis diagonalizable over

complex numbers. (2) LetTbe a linear operator on a finite dimensional vector space over an algebraically closed fieldF.Letfbe a polynomial over F.Prove thatcis a characteristic value off(T) if and only if f(t) =cwheretis a characteristic value ofT. Proof.Lettbe a characteristic value ofTandβbe a non- zero characteristic vector associated with the characteristic valuet.Then,Tβ=tβ, T2β=T(Tβ) =T(tβ) =tTβ=t2β, and inductively we can see thatTkβ=tkβfor anyk≥1. Thus, for any polynomialf(x) we havef(T)β=f(t)βwhich means, sinceβ̸= 0,thatf(t) is a characteristic value of the linear operatorf(T). Assume thatcis a characteristic value off(T).SinceFis algebraically closed, the minimal polynomial ofTis product of linear polynomials, that is,Tis similar to a triangular op- erator. If [P-1TP]Bis triangular matrix, then [P-1f(T)P]Bis

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA3

also triangular and on the diagonal of [P-1f(T)P]Bwe have f(ci),whereciis a characteristic value ofT. (3) Letcbe a characteristic value ofTand letWbe the space of characteristic vectors associated with the characteristic value c.What is the restriction operatorT|W. Solution.Every vectorv∈Wis a characteristic vector.

Hence,Tv=cvfor allv∈W.Therefore,T|W=cI.

(4) Every matrixAsatisfyingA2=Ais similar to a diagonal matrix. Solution.A satisfies the polynomialx2-x.Thus, the min- imal polynomial,mA(x),ofAdividesx2-x,that ismA(x) =x ormA(x) =x-1 ormA(x) =x(x-1).

IfmA(x) =x, thenA= 0.

IfmA(x) =x-1, thenA=I.

IfmA(x) =x(x-1), then the minimal polynomial ofA

is product of distinct polynomials of degree one. Thus, by a Theorem, the matrixAis similar to diagonal matrix with diagonal entries consisting of the characteristic values, 0 and 1. (5) LetTbe a linear operator onV.If every subspace ofVis invariant underTthen it is a scalar multiple of the identity operator. Solution.If dimV= 1 then for any 0̸=v∈V,we have

Tv=cv,sinceVis invariant underT.Hence,T=cI.

Assume that dimV >1 and letB={v1,v2,···,vn}be a basis forV.SinceW1=⟨v1⟩is invariant underT,we have Tv

1=c1v1.Similarly, sinceW2=⟨v2⟩is invariant underT,we

haveTv2=c2v2.Now,W3=⟨v1+v2⟩is also invariant underT. Hence,T(v1+v2) =λ(v1+v2) orc1v1+c2v2=λ(v1+v2),which gives us (c1-λ)v1+ (c2-λ)v2= 0.However,v1andv2are linearly independent and hence we should havec1=c2=λ. Similarly, one can continue with the subspace⟨v1+v2+v3⟩

4M. KUZUCUOGLU

and observe thatT(v3) =λv3. So for anyvi∈ B,we have Tv i=λvi.Thus,T=λI. (6) LetVbe the vector space ofn×nmatrices overF.LetAbe a fixedn×nmatrix. LetTbe a linear operator onVdefined byT(B) =AB.Show that the minimal polynomial ofTis the minimal polynomial ofA. Solution.LetmA(x) =xn+an-1xn-1+···+a1x+a0be the minimal polynomial ofA,so thatmA(A) = 0.It is easy to see thatTk(B) =AkBfor anyk≥1.Then, for anyB∈Vquotesdbs_dbs13.pdfusesText_19