EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA Mahmut Kuzucuoğlu Middle East Technical University matmah@metu.edu.tr Ankara, TURKEY March 14, 015

ii TABLE OF CONTENTS CHAPTERS 0. PREFACE.................................................. 1 1. LINEAR...................................................??. 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 Kuzucuoğlu METU, Ankara March 14, 015 M. Kuzucuoğlu 1

........ 1956 M E T U DEPARTMENT OF MATHEMATICS Math 6 Quiz I (05.03.008) Name : Answer Key ID Number : 3606 Signature : 0000000 Duration : 60 minutes Show all your work. Unsupported answers will not be graded. 1.) Let A = 4 0 4 3 3 7 4. (a) Find the characteristic polynomial of A. Solution. The characteristic polynomial of A is f(x) = det(xi A). So, x 4 f(x) = 0 x + 4 3 3 7 x 4 x + 4 3 = (x ) 7 x 4 + 3 4 x + 4 3 (b) Find the minimal polynomial of A. = (x )(x 16 + 1) + 3(1 x 8) = (x )(x + 5) + 3(4 x) = (x )(x + 5 6) = (x )(x 1)(x + 1) Solution. We know that the minimal polynomial divides the characteristic polynomial and they same the same roots. Thus, the minimal polynomial for A is m A (x) = f(x) = (x )(x 1)(x + 1). (c) Find the characteristic vectors and a basis B such that [A] B is diagonal. Solution. The characteristic values of A are c 1 =, c = 1, c 3 = 1. 0 4 3 7 3x 7y + z = 0 A I = 0 6 3 0 1, y z = 0 z = y x = y 3 7 0 0 0 Thus, α 1 = ( 1, 1, ) is a characteristic vector associated with the characteristic value c 1 =. 1 4 1 4 y = 3k x + 4y z = 0 A I = 0 5 3 0 5 3, 5y + 3z = 0 z = 5k 3 7 3 0 0 0 x = k Thus, α = (, 3, 5) is a characteristic vector associated with the characteristic value c = 1. 3 4 3 4 x = t 3x + 4y z = 0 A + I = 0 3 3 0 1 1, y z = 0 y = 3t 3 7 5 0 0 0 z = 3t

Thus, α 3 = (, 3, 3) is a characteristic vector associated with the characteristic value c 3 = 1. 0 0 Now, B = {α 1, α, α 3 } is a basis and [A] B = 0 1 0 is a diagonal matrix. 0 0 1 (d) Find A -conductor of the vector α = (1, 1, 1) into the invariant subspace spanned by ( 1, 1, ). Solution. Set W =< ( 1, 1, ) > and denote the A -conductor of α into W by g(x). Since m A (A) = 0 we have m A (A)α W. Thus, g(x) divides m A (x). Hence, the possibilities for g(x) are x, 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, 0 4 1 (A I)α = 0 6 3 1 = 3 / W g(x) x, (A I)α = 3 7 1 1 4 1 0 5 3 1 3 7 3 1 3 4 1 (A + I)α = 0 3 3 1 3 7 = = 8 3 / W g(x) x 1, 7 3 7 5 1 5 0 4 3 (A I)(A I)α = 0 6 3 (A I)(A + I)α = (A I)(A + I)α = 0 4 0 6 3 3 7 1 4 0 5 3 3 7 3 7 5 0 5 0 5 5 0 5 / W g(x) x + 1, = = = 6 9 9 10 15 / W g(x) (x )(x 1), / W g(x) (x )(x + 1), 5 15 15 = 15α 1 W g(x) = x 1..) Find a 3 3 matrix whose minimal polynomial is x. 0 0 1 Solution. For the matrix A = 0 0 0 we have A 0 and A = 0. Thus, A is a 3 3 matrix whose minimal polynomial is x. 0 0 0 3.) Prove that similar matrices have the same minimal polynomial. Solution. Let A and B be similar matrices, i.e., B = P 1 AP for some invertible matrix P. For any k > 0 we have B k = (P 1 AP ) k = P 1 A k P which implies that f(b) = P 1 f(a)p for any polynomial f(x). Let f A and f B be the minimal polynomials of A and B, respectively. Then f A (B) = P 1 f A (A)P = P 1 OP = O implies that f B divides f A. On the other hand, O = f B (B) = P 1 f B (A)P gives us f B (A) = O. Hence, f A divides f B. Therefore, we have f A = f B. 30

M. KUZUCUOĞLU 1. Math 6 Exercises and Solutions (1) Let A be a 3 3 matrix with real entries. Prove that if A is not similar over R to a triangular matrix then A is similar over C to a diagonal matrix. Proof. Since A is a 3 3 matrix with real entries, the characteristic polynomial, f(x), of A is a polynomial of degree 3 with real coefficients. We know that every polynomial of degree 3 with real coefficients has a real root, say c 1. On the other hand, since A is not similar over R to a triangular matrix, the minimal polynomial of A is not product of polynomials of degree one. So one of the irreducible factor, h, of the minimal polynomial of A is degree. Then h has two complex roots, one of which is the conjugate of the other. Thus, the characteristic polynomial has one real root and two complex roots, c 1, λ and λ. The minimal polynomial over complex numbers is (x c 1 )(x λ)(x λ) which implies that A is diagonalizable over complex numbers. () Let T be a linear operator on a finite dimensional vector space over an algebraically closed field F. Let f be a polynomial over F. Prove that c is a characteristic value of f(t ) if and only if f(t) = c where t is a characteristic value of T. Proof. Let t be a characteristic value of T and β be a nonzero characteristic vector associated with the characteristic value t. Then, T β = tβ, T β = T (T β) = T (tβ) = tt β = t β, and inductively we can see that T k β = t k β for any k 1. Thus, for any polynomial f(x) we have f(t )β = f(t)β which means, since β 0, that f(t) is a characteristic value of the linear operator f(t ). Assume that c is a characteristic value of f(t ). Since F is algebraically closed, the minimal polynomial of T is product of linear polynomials, that is, T is similar to a triangular operator. If [P 1 T P ] B is triangular matrix, then [P 1 f(t )P ] B is

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 3 also triangular and on the diagonal of [P 1 f(t )P ] B we have f(c i ), where c i is a characteristic value of T. (3) Let c be a characteristic value of T and let W be the space of characteristic vectors associated with the characteristic value c. What is the restriction operator T W. Solution. Every vector v W is a characteristic vector. Hence, T v = cv for all v W. Therefore, T W = ci. (4) Every matrix A satisfying A = A is similar to a diagonal matrix. Solution. A satisfies the polynomial x x. Thus, the minimal polynomial, m A (x), of A divides x x, that is m A (x) = x or m A (x) = x 1 or m A (x) = x(x 1). If m A (x) = x, then A = 0. If m A (x) = x 1, then A = I. If m A (x) = x(x 1), then the minimal polynomial of A is product of distinct polynomials of degree one. Thus, by a Theorem, the matrix A is similar to diagonal matrix with diagonal entries consisting of the characteristic values, 0 and 1. (5) Let T be a linear operator on V. If every subspace of V is invariant under T then it is a scalar multiple of the identity operator. Solution. If dim V = 1 then for any 0 v V, we have T v = cv, since V is invariant under T. Hence, T = ci. Assume that dim V > 1 and let B = {v 1, v,, v n } be a basis for V. Since W 1 = v 1 is invariant under T, we have T v 1 = c 1 v 1. Similarly, since W = v is invariant under T, we have T v = c v. Now, W 3 = v 1 +v is also invariant under T. Hence, T (v 1 +v ) = λ(v 1 +v ) or c 1 v 1 +c v = λ(v 1 +v ), which gives us (c 1 λ)v 1 + (c λ)v = 0. However, v 1 and v are linearly independent and hence we should have c 1 = c = λ. Similarly, one can continue with the subspace v 1 + v + v 3

4 M. KUZUCUOĞLU and observe that T (v 3 ) = λv 3. So for any v i B, we have T v i = λv i. Thus, T = λi. (6) Let V be the vector space of n n matrices over F. Let A be a fixed n n matrix. Let T be a linear operator on V defined by T (B) = AB. Show that the minimal polynomial of T is the minimal polynomial of A. Solution. Let m A (x) = x n + a n 1 x n 1 + + a 1 x + a 0 be the minimal polynomial of A, so that m A (A) = 0. It is easy to see that T k (B) = A k B for any k 1. Then, for any B V we have m A (T )B = (T n + a n 1 T n 1 + + a 1 T + a 0 I)B = T n (B) + a n 1 T n 1 (B) + + a 1 T (B) + a 0 B = A n B + a n 1 A n 1 B + + a 1 AB + a 0 B = (A n + a n 1 A n 1 + + a 1 A + a 0 I)B = m A (A)B = 0. Thus, we obtain m A (T ) = 0, which means that m T (x) divides m A (x). Now, let m T (x) = x m + c m 1 x m 1 + + c 1 x + c 0 be the minimal polynomial of T, so that m T (T ) = 0. Then, for any B V we have m T (A)B = (A m + c m 1 A m 1 + + c 1 A + c 0 I)B = A m B + c m 1 A m 1 B + + c 1 AB + c 0 B = T m (B) + c m 1 T m 1 (B) + + c 1 T (B) + c 0 B = (T m + c m 1 T m 1 + + c 1 T + c 0 I)B = m T (T )B = 0, which leads to m T (A) = 0, meaning that m A (x) divides m T (x). Since, monic polynomials dividing each other are the same we have m T (x) = m A (x).

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 5 (7) If E is a projection and f is a polynomial, then show that f(e) = ai + be. What are a and b in terms of the coefficients of f? Solution. Let f(x) = c 0 + c 1 x + + c n x n. Then, f(e) = c 0 I + c 1 E + + c n E n. Since E is a projection, (E = E), we have E k = E for any k 1. Then, f(e) = c 0 I + c 1 E + + c n E n = c 0 I + c 1 E + + c n E = c 0 I + (c 1 + + c n )E. Thus, a is the constant term of f and b is the sum of all other coefficients. (8) Let V be a finite dimensional vector space and let W 1 be any subspace of V. Prove that there is a subspace W of V such that V = W 1 W. Proof. Let B W1 = {β 1,, β k } be a basis for W 1. We may extend B W1 to a basis B V of V, say B V = {β 1,, β k, β k+1,, β n }. Let W be the subspace spanned by β k+1,, β n. Then, as they are linearly independent in V, we have B W = {β k+1,, β n }. Clearly W 1 + W = V as W 1 + W contains a basis of V and so spans V. Let β W 1 W. Then, β W 1 implies that β = c 1 β 1 + + c k β k, and β W implies that β = c k+1 β k+1 + + c n β n. The last two equalities give us c 1 β 1 + +c k β k c k+1 β k+1 +c n β n = 0, but since β i s are linearly independent, we obtain c i = 0 for all i = 1,, n which means that β = 0. That is W 1 W = {0}, and hence V = W 1 W. (9) Let V be a real vector space and E be an idempotent linear operator on V, that is a projection. Prove that I + E is invertible. Find (I + E) 1. Proof. Since E is an idempotent linear operator it is diagonalizable by Question 4. So there exists a basis of V consisting of characteristics vectors of E corresponding to the characteristic values 0 and 1. That is, there exists a basis

6 M. KUZUCUOĞLU ([I + E] B ) 1 = B = {β 1,, β n } such that Eβ i = β i for i = 1,, k, and Eβ i = 0 for i = k + 1,, n. Then (I + E)β i = β i for i = 1,, k and (I + E)β i = β i for i = k + 1,, n, that is, [I + E] B = [ I 1 0 0 I where I 1 stands for k k identity matrix, I is (n k) (n k) identity matrix and each 0 represents the zero matrix of appropriate dimension. It is now easy to see that [I + E] B is invertible, since det(i + E) = k 0. To find the inverse of (I + E), we note that [ 1 I 1 0 0 I ] = [ I 1 0 0 I ] + [ ], 1 I 1 0 0 0 ] = I 1 [E] B. Therefore, (I + E) 1 = I 1 E. (You may verify that really this is the inverse, by showing that (I + E)(I 1 E) = (I E)(I + E) = I.) 1 (10) Let T be a linear operator on V which commutes with every projection operator on V. What can you say about T? Solution. Let B be a basis for V and β i B, i I where I is some index set. We can write V as a direct sum V = W i U where W i = β i. Then there exists a projection E i of V onto the subspace W i for each i I. Note that E i v W i for all v V, and E i β i = β i. Now, by assumption, the linear operator T commutes with E i for all i I, that is, T E i = E i T. Then, for β i W i, we have T E i β i = E i T β i W i implies that T β i = T (E i β i ) = c i β i for some constant c i F. Thus, β i is a characteristic vector of T. Hence, V has a basis consisting of characteristic vectors of T. It follows that T is a diagonalizable linear operator on V. (11) Let V be the vector space of continuous real valued functions on the interval [ 1, 1] of the real line. Let W e be the space

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 7 of even functions, f( x) = f(x), and W o be the space of odd functions, f( x) = f(x). a) Show that V = W e W o. b) If T is the indefinite integral operator (T f)(x) = are W e and W o invariant under T? Solution. a) Let f V. Then, we may write x 0 f(t)dt, f(x) = f(x) + f( x) + f(x) f( x) = f(x) + f( x) f(x) f( x) +. f(x) + f( x) Observe that f e (x) = is a continuous even function and f o (x) = is a continuous odd function. f(x) f( x) Hence, f = f e + f o, that is V = W e + W o. To show that V = W e W o, we need to show that W e W o = {0}. To see this, let g W e W o. Then, g W e implies that g( x) = g(x), and g W o implies that g( x) = g(x). Thus, we have g(x) = g(x) or g(x) = 0 for all x [ 1, 1], which means that g = 0. b) For f(x) = x W o, we have (T f)(x) = x / / W o, and for g(x) = x W e, we have (T g)(x) = x 3 /3 / W e. Thus, neither W e nor W o are invariant under T. (1) Let V be a finite dimensional vector space over the field F, and let T be a linear operator on V, such that rank(t ) = 1. Prove that either T is diagonalizable or T is nilpotent, but not both. Proof. Since rank(t ) = dim(im(t )) = 1, we have dim(ker(t )) = n 1. Let 0 β Im(T ). So, Im(T ) = β. Since β Im(T ), there exists a vector α 0 V such that T α 0 = β. Let {α 1, α,, α n 1 } be a basis for Ker(T ). Then, B = {α 0, α 1, α,, α n 1 } is a basis for V. We have T α i = 0 for all i = 1,,, n 1.

8 M. KUZUCUOĞLU If T α 0 Ker(T ), then T α 0 = c 1 α 1 + + c n 1 α n 1 and 0 0 0 0 c 1 0 0 0 [T ] B = c 0 0 0....... c n 1 0 0. and it is easily seen that T = 0 meaning that T is nilpotent. Note that at least one of c i s is nonzero, since otherwise, α 0 would be in Ker(T ) which contradicts with the choice of B. If T α 0 / Ker(T ), then T β Im(T ) and T β = c 0 β. In this case we construct a new basis B = {β, α 1, α,, α n 1 } and c 0 0 0 0 0 0 [T ] B =...... 0 0 0 which means that T is diagonalizable. (13) Let T be a linear operator on the finite dimensional vector space V. Suppose T has a cyclic vector. Prove that if U is any linear operator which commutes with T, then U is a polynomial in T. Proof. Let B = {α, T α,, T n 1 α} be a basis for V containing the cyclic vector α and let m(x) = x n + a n 1 x n 1 + + a 1 x + a 0 be the minimal polynomial of T. Since Uα is in V, it can be written as a linear combination of basis vectors. Then, Uα = b 0 α+b 1 T α+ +b n 1 T n 1 α where b 0, b 1,, b n 1 are elements of the field F. That is, (b 0 I+b 1 T + +b n 1 T n 1 U)α = 0. Now, since U and T commute, we have UT (α) = T U(α) = T (b 0 α + b 1 T α + + b n 1 T n 1 α) = b 0 T α + b 1 T α + + b n 1 T n α = (b 0 I + b 1 T + + b n 1 T n 1 )T α

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 9 which means that (b 0 I + b 1 T + + b n 1 T n 1 U)T α = 0. Similarly, we can show that (b 0 I + b 1 T + + b n 1 T n 1 U)T i α = 0 for all i =, 3,, n 1. Since the transformation b 0 I + b 1 T + + b n 1 T n 1 U maps each basis vector to the zero vector, it is identically equal to zero on the whole space. Thus, we obtain U = b 0 I + b 1 T + + b n 1 T n 1. (14) Give an example of two 4 4 nilpotent matrices which have the same minimal polynomial but which are notsimilar. 0 0 0 0 0 0 0 0 1 0 0 0 Solution. Let A = 0 0 0 0 and B = 1 0 0 0 0 0 0 0. 0 0 0 0 0 0 1 0 It is easy to see that m A (x) = m B (x) = x but they are not similar since, A has 3 distinct characteristic vectors corresponding to the characteristic value zero, but B has only two characteristic vectors corresponding to the characteristic value zero. (15) Show that if N is a nilpotent linear operator on an n dimensional vector space V, then the characteristic polynomial for N is x n. Solution. Recall that N is nilpotent, if N k = 0 for some k N +. Since, N is a nilpotent linear operator on V, the minimal polynomial for N is of the form x m for some m n. Then, all characteristic values of N are zero. Since the minimal polynomial is a product of linear polynomials, N is a triangulable operator. It follows that there exists a basis B of

10 M. KUZUCUOĞLU V such that 0 0 0 0 0 [N] B =..... 0 since, similar matrices have the characteristic polynomial, it follows that the characteristic polynomial of N is x n where n = dimv. (16) Let T be a linear operator on R 3 which is represented in the standard ordered basis by the matrix 0 0 0 0. 0 0 1 Prove that T has no cyclic vector. What is the T cyclic subspace generated by the vector β = (1, 1, 3)? Solution. Assume that T has a cyclic vector α = (a 1, a, a 3 ). Then B = {α, T α, T α} will be a basis for R 3. That is, the vectors α = (a 1, a, a 3 ), T α = (a 1, a, a 3 ), T α = (4a 1, 4a, a 3 ) must be linearly independent, or the matrix a 1 a a 3 a 1 a a 3 4a 1 4a a 3 must be invertible. Applying elementary row operations, we obtain R 1 + R a 1 a a 3 R + R 3 a 1 a a 3 0 0 3a 3 0 0 a 3 4R 1 + R 3 0 0 3a 3 1R 3 0 0 0 a 1 a a 3 a 1 a a 3 4a 1 4a a 3 which is not invertible. Hence, T has no cyclic vector. To find the cyclic subspace generated by β, it is enough to check if β and T β are independent since we have already shown that the set {α, T α, T α} can not be linearly independent for any α R 3. Clearly, β = (1, 1, 3) and T β = (,, 3) are

EXERCISES AND SOLUTIONS IN LINEAR ALGEBRA 11 linearly independent since, otherwise, one of them would be a multiple of the other one which is not the case here. Thus, the cyclic subspace generated by β is Z(β; T ) = (1, 1, 3), (,, 3) = {λ(1, 1, 3)+µ(,, 3) : λ, µ R}. (17) Find the minimal polynomial and rational form of the matrix c 0 1 T = 0 c 1. 1 1 c Solution. The characteristic polynomial of T is x c 0 1 f T (x) = det(xi T ) = 0 x c 1 1 1 x c x c 1 = (x c) 1 x c + 0 1 x c 1 = (x c)((x c) 1) (x c) = (x c)((x c) ) = (x c)(x c )(x c + ). Since the characteristic polynomial and the minimal polynomial have the same roots and the minimal polynomial divides the characteristic polynomial we have m T (x) = f T (x) = (x c)((x c) ) = (x c) 3 (x c) = x 3 + ( 3c)x + (3c )x + ( c 3 + c). Thus the rational form of T is R = 0 0 c 3 c 1 0 3c + 0 1 3c.