Linear Algebra Solutions 11

# Linear Algebra Solutions 11 - If 1 = 2 we see that(1 2)kn...

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If λ 1 6 = λ 2 , we see that ( λ 1 - λ 2 ) k n = ( λ 1 - λ 2 )( λ n - 1 1 + λ n - 2 1 λ 2 + ··· + λ 1 λ n - 2 2 + λ n - 1 2 ) = λ n 1 + λ n - 1 1 λ 2 + ··· + λ 1 λ n - 1 2 - ( λ n - 1 1 λ 2 + ··· + λ 1 λ n - 1 2 + λ n 2 ) = λ n 1 - λ n 2 . Hence k n = λ n 1 - λ n 2 λ 1 - λ 2 . We have to prove A n = k n A - λ 1 λ 2 k n - 1 I 2 . * n=1: A 1 = A ; also k 1 A - λ 1 λ 2 k 0 I 2 = k 1 A - λ 1 λ 2 0 I 2 = A. Let n 1 and assume equation * holds. Then A n +1 = A n · A = ( k n A - λ 1 λ 2 k n - 1 I 2 ) A = k n A 2 - λ 1 λ 2 k n - 1 A. Now A 2 = ( a + d ) A - ( ad - bc ) I 2 = ( λ 1 + λ 2 ) A - λ 1 λ 2 I 2 . Hence A n +1 = k n ( λ 1 + λ 2 ) A - λ 1 λ 2 I 2 - λ 1 λ 2 k n - 1 A = { k n ( λ 1 + λ 2 ) - λ 1 λ 2 k n - 1 } A - λ 1 λ 2 k n I 2 = k n +1 A - λ 1 λ 2 k n I 2 , and the induction goes through. 8. Here λ 1 2 are the roots of the polynomial x 2 - 2 x - 3 = ( x - 3)(
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## This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.

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