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Unformatted text preview: Solution for HW7 November 17, 2011 2.39 (a) AB = 7 14 39 28 ( AB ) C = 21 105 98 17 285 296 (b) BC = 5 15 20 8 60 59 A ( BC ) = 21 105 98 17 285 296 2.41 (a) A T = 1 3 2 4 (b) B T = 5 6 7 (c) ( AB ) T = 7 39 14 28 (d) A T B T = 5 15 10 40 2.51 (a) A 2 = 11 15 9 14 A 3 = 67 40 24 59 . (b) f ( A ) = 50 70 42 36 , g ( A ) = 1 2.60 (a) A n = 1 2 n 1 (b) B n = 1 n n ( n 1) 2 1 n 1 2.64 A = 1 2 1 3 1 2 2.68 (a) x = 4 , y = 1 , z = 3 . (b) x = 0 , y = 6 , z = any number. 2.69 Proof: (a) ( A + A T ) T = A T +( A T ) T = A T + A = A + A T . That is, A + A T is symmetric. (b) ( A A T ) T = A T ( A T ) T = A T A = ( A A T ) , which means that ( A A T ) is skewsymmetric. (c) Put B = A + A T 2 and C = A A T 2 . We can see that B is symmetric and C is skewsymmetric and A = B + C . 2.76 A = A H gives us 3 x + 2 i yi 3 2 i 1 + zi yi 1 xi 1 = 3 3 + 2 i yi x 2 i 1 + xi yi 1 zi 1 or x = 3 , y = 0 , z = 3 . 2 2.78 Proof (a) ( A + A H ) H = A H +( A H ) H = A H + A = A + A H , which shows that A + A H is Hermitian....
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 Fall '08
 Feinberg,E

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