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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 6 . 2 2 ) S = 1 1 0 1 is an elementary matrix with S 1 = 1 1 1 so S Λ S 1 = 1 1 0 1 2 0 0 5 1 1 1 = 2 5 0 5 1 1 1 = 2 3 0 5 . 12 ) (a) False. Consider A = 3 1 16 5 . This matrix is invertible and has (1 , 4) is an eigenvector with eigenvalue 1. It has no other eigenvectors. Ok where did this come from? We want a counterexample. The matrix 1 1 0 1 is in vertible and has both eigenvalues 1, but only one eigenvector (1 , 0). So we set S = 1 1 4 5 . It’s columns form a basis, the first being the eigenvector we want, and then compute S 1 1 0 1 S 1 . (b) True. An eigenvector is missing, so we have a repeated eigenvalue. (c) True. Diagonalizable n × n matrices have n independent eigenvectors. 18 ) A = 2 1 1 2 has as its eigenvalues the roots of λ 2 4 λ + 3 which are 3 and 1. An eigenvector for eigenvalue 3 is (1 , 1) and an eigenvector for eigenvalue 1 is (1 , 1). Set S = 1 1 1 1 so S 1 = 1 2 1 1 1 1 . Then A k = ( S Λ S 1 ) k = S...
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 Math, Linear Algebra, Algebra, Matrices, xT ABx

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