Linear Algebra Solutions 73

Linear Algebra Solutions 73 - A ch A = f f f f f f...

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Hence the eigenspace consists of vectors X = [ x, y, z ] t satisfying x = - 2 z and y = z , with z arbitrary. Hence X = - 2 z z 0 = z - 2 1 0 and we can take X 3 = [ - 2 , 1 , 1] t . Hence we can take P = 1 - 1 - 2 1 1 1 1 0 1 . (c) A = P diag(1 , 0 , 1 4 ) P - 1 so A n = P diag(1 , 0 , 1 4 n ) P - 1 . Hence A n = 1 - 1 - 2 1 1 1 1 0 1 1 0 0 0 0 0 0 0 1 4 n 1 3 1 1 1 0 3 - 3 - 1 - 1 2 = 1 3 1 0 - 2 4 n 1 0 1 4 n 1 0 1 4 n 1 1 1 0 3 - 3 - 1 - 1 2 = 1 3 1 + 2 4 n 1 + 2 4 n 1 - 4 4 n 1 - 1 4 n 1 - 1 4 n 1 + 2 4 n 1 - 1 4 n 1 - 1 4 n 1 + 2 4 n = 1 3 1 1 1 1 1 1 1 1 1 + 1 3 · 4 n 2 2 - 4 - 1 - 1 2 - 1 - 1 2 . 10. Let A = 5 2 - 2 2 5 - 2 - 2 - 2 5 . (a) We Frst determine the characteristic polynomial ch
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Unformatted text preview: A ( λ ). ch A ( λ ) = f f f f f f λ-5-2 2-2 λ-5 2 2 2 λ-5 R 3 → R 3 + R 2 = f f f f f f λ-5-2 2-2 λ-5 2 λ-3 λ-3 f f f f f f = ( λ-3) f f f f f f λ-5-2 2-2 λ-5 2 1 1 f f f f f f 76...
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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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