Assignment 5 Solution

Assignment 5 Solution - We have Ax = x . (a) A 2 x = A ( Ax...

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CS3251, Homework 5 Solution 1. Problem Set 4.4, p. 228, # 6 Let Q 1 and Q 2 be two orthogonal matrices. Then ( Q 1 Q 2 ) T Q 1 Q 2 = Q T 2 Q T 1 Q 1 Q 2 = Q T 2 IQ 2 = I and, therefore, Q 1 Q 2 is orthogonal. 2. Problem Set 4.4, p. 230, # 22 a = (1 , 1 , 2) T , b = (1 , - 1 , 0) T and c = (1 , 0 , 4). We use Gram-Schmidt orthogonaliza- tion. We set A = a , B = b - P A ( b ) = b - aa T b/ k a k 2 = b and C = c - P A ( c ) - P B ( c ) = c - aa T c/ k a k 2 - bb T c/ k b k 2 , where P z ( x ) denotes the projection of vector x to vector y . So we have C = 1 0 4 - 9 6 1 1 2 - 1 2 1 - 1 0 = - 1 - 1 1 . 3. Problem Set 5.1, p. 240, # 3 (a) False. Example: A = ± 1 1 0 1 ² Then | I + A | = 4 6 = 2 = 1 + | A | . (b) True. | ABC | = | ( AB ) C | = | AB | · | C | = | A | · | B | · | C | . (c) False. Using linearity we have | 4 A | = 4 n | A | , where n is the number or rows of the matrix. (d) False. Example: A = ± 1 1 1 0 ² , B = ± 0 1 1 1 ² and AB - BA = ± 0 2 - 2 0 ² . Clearly, | AB - BA | = 4 6 = 0. 4. Problem Set 6.1, p. 284, # 9
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Unformatted text preview: We have Ax = x . (a) A 2 x = A ( Ax ) = A ( x ) = Ax = 2 x . (b) Ax = x multiply both sides by A-1 to get A-1 Ax = A-1 ( x ) or, equivalently, x = A-1 x . Hence, -1 x = A-1 x . (c) ( A + I ) x = Ax + Ix = x + x = ( + 1) x . 5. Problem Set 6.2, p. 298, # 4 A = S-1 S . Here is the eigenvalue matrix and S is the eigenvector matrix of A . We have A + 2 I = S S-1 + 2 SS-1 = S ( + 2 I ) S-1 . So the eigenvector matrix of A + 2 I is S and the eigenvalue matrix is + 2 I . 1...
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