(2) Show that two symmetric matrices are similar if and only if they have the same characteristic polynomials.

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1 () Which of the following statements are true and which are false? Justify your answer. (a) The product of two orthogonal n n matrices is orthogonal. Solution. True. Let A and B be two orthogonal matrices of the same size. Then A T A B T B I. It follows that (AB) T (AB) B T (A T A)B B T B I. So AB is orthogonal. (b) Two matrices with the same characteristic polynomials are similar. Solution. False. Let A and B Then det(λi A) det(λi B) (λ ) 2. But A and B are not similar. Otherwise, B P AP for an invertible matrix P. But P AP P P I and B I. Contradiction. (c) For two n n invertible matrices A and B, AB and BA are similar. Solution. True since AB B (BA)B. (d) rank(a) rank(a 2 ) for every square matrix A. Solution. False. Let A Then rank(a) but A 2 and rank(a 2 ). (e) For a Householder matrix A and two vectors u and v, if Au v, then Av u. Solution. True. Since A is a Householder matrix, A is symmetric and orthogonal. Therefore, A 2 A T A I. Consequently, Au v A 2 u Av u Av (f) If A is diagonalizable, so is A T. Solution. True. Since A is diagonalizable, there is an invertible matrix P such that P AP D is diagonal. It follows that D T (P AP ) T (P ) T A T P T (P T ) A T P T

2 2 Since D D T is diagonal, A T is diagonalizable. (2) Show that two symmetric matrices are similar if and only if they have the same characteristic polynomials. Proof. Let A and B be two symmetric matrices. Suppose that A and B are similar. Then A P BP for an invertible matrix P. Then det(λi A) det(λi P BP ) det(p (λi B)P ) det(p ) det(λi B) det(p ) det(λi B) Therefore, A and B have the same characteristic polynomials. On the other hand, suppose that A and B have the same characteristic polynomials. Let det(λi A) det(λi B) (λ λ )(λ λ 2 )...(λ λ n ) where λ, λ 2,..., λ n are the eigenvalues of A and B. Since A and B are symmetric, they are diagonalizable. There exist invertible matrices P and Q such that P AP λ λ 2... λ n and QBQ λ λ 2... λ n Hence P AP QBQ A P QBQ P (P Q)B(P Q) So A and B are similar. (3) Let u (,, ) and v (,, ) be two vectors in R 3. (a) Find the hyperplane W {a x + a 2 x 2 + a 3 x 3 } such that u is the reflection of v with respect to W. Solution. We have W w where w u v (, 2, 2). So W {x + 2x 2 + 2x 3 }. (b) Let T : R 3 R 3 be the reflection with respect to the hyperplane obtained in part (a). Find the matrix [T ] B representing T with respect to the standard basis B.

3 3 Solution. The corresponding Householder matrix is [T ] B I 2 w ( ) T w I 2 2 [ 2 2 ] w w 9 2 I /9 4/9 4/ /9 /9 8/ /9 8/9 /9 (4) Let Q(x, x 2, x 3 ) (x + x 2 + x 3 ) 2 + (x x 2 ) 2 be a quadratic form in the three variables x, x 2 and x 3. (a) Find a 3 3 symmetric matrix A such that Q(x, x 2, x 3 ) x T Ax, where x x. x 2 x 3 Solution. We have Q(x, x 2, x 3 ) 2x 2 + 2x x x x 3 + 2x 2 x 3 x T 2 2 x (b) Find a 3 3 orthogonal matrix P such that P T AP is diagonal. Solution. Note that (,, ) and (, ) are orthogonal. Therefore, P T / 3 / 3 / 3 / 2 / 2 a b c Since P T is orthogonal, a+b+c a b and a 2 +b 2 +c 2. Therefore, / 3 / 3 / 3 P T / 2 / 2 / / 2/ and hence / 3 / 2 / P / 3 / 2 / / 3 2/

4 4 (c) Let x P y. What is Q in terms of y, y 2 and y 3, where y y? y 2 y 3 Solution. We have Q(x, x 2, x 3 ) x T Ax y T P T AP y 3y 2 + 2y 2 2 (d) Find the maximum and minimum values of Q(x, x 2, x 3 ) subject to the constraint x 2 + x x 2 3 and where these extreme values are achieved. (5) Let A Solution. We have 3y 2 + 2y 2 2 3(y 2 + y y 2 3) Therefore, Q min when y y 2 and y 3 ±, i.e., x x 2 P / ± / x 3 ± 2/ and Q max 3 when y ± and y 2 y 3, i.e., x x 2 P ± / 3 ± / 3 x 3 / 3. (a) Find an invertible matrix P such that P AP is diagonal. Solution. The characteristic polynomial of A is det(λi A) λ(λ 2). For λ, the corresponding eigenspace is [ null( A) Span{ } ] For λ, the corresponding eigenspace is [ null(2i A) Span{ } ] Let P 2 /2 /2 /2 /2

5 5 Then P AP (b) Find A 29. [ /2 /2 /2 /2 Solution. Since ] [ P A 29 P (P AP ) 29 we have ] [ ] 2 29 A 29 P 2 29 P /2 / /2 / Alternatively, we may apply the Cayley-Hamilton theorem: A 2 2A. It follows that A n 2 n A. Therefore, 2 A A () Let P 2 be the vector space of polynomials in x of degree at most 2 and let T : P 2 P 2 be the map given by T (f(x)) f( x). (a) Show that T is a linear transformation. Proof. For all f(x) and g(x) in P 2, T (f(x)+g(x)) (f +g)( x) f( x)+g( x) T (f(x))+t (g(x)) For all f(x) P 2 and λ R, T (λf(x)) λf( x) λt (f(x)) Therefore, T is a linear transformation.

6 (b) Find the matrix [T ] S representing T with respect to the basis S {, x, x 2 }. Solution. We have T (), T (x) x and T (x 2 ) ( x) 2. Therefore, [T ()] S [] S, [T (x)] [ x] S and [T (x 2 )] S [( x) 2 ] S [ 2x + x 2 ] S 2 It follows that [T ] S 2 (c) Let B {, x, (x ) 2 }. Find the transition matrix P S B and the matrix [T ] B representing T with respect to B. Solution. Since and [] B, [x] B [ + (x )] B [x 2 ] B [( + (x )) 2 ] B [ + 2(x ) + (x ) 2 ] B 2 we have [P S B ] 2 Since [T ()] B [] B, [T (x )] [ x] B [ (x )] B

7 and [T ((x ) 2 )] B [x 2 ] B [ + 2(x ) + (x ) 2 ] B 2 we have [T ] B 2 (d) Does there exist a basis B such that [T ] B is diagonal? If yes, find such a basis B. If no, justify your answer. Solution. Such a basis exists if and only if [T ] S is diagonalizable. The matrix [T ] S has two eigenvalues and. For λ, the corresponding eigenspace is null(i [T ] S ) Span{, } For λ, the corresponding eigenspace is null( I [T ] S ) Span{ 2 } Therefore, [T ] S has three linearly independent eigenvectors and is hence diagonalizable. The corresponding basis B is {, x x 2, 2x}. (7) Do the following: (a) Find the least squares solution of x 8 x 2 Solution. The associated normal system is x 8 [ 3 x 8 x 2 x 2 ] So the least squares solution is { x 4 x 2 4 7

8 8 (b) Express w (8,, ) in the form w w + w 2, where w lies in the subspace W of R 3 spanned by the vectors u (,, ) and u 2 (,, ) and w 2 is orthogonal to W. Solution. Note that where By part (a), and w Proj W w Proj col(a) w A w x u + x 2 u 2 (4,, 4) w 2 w w (4,, 4)

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