# tut6 - MATH 1104F Solutions to Tutorial Assignment 6 Total...

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MATH 1104F - Solutions to Tutorial Assignment 6 - March 16, 2010 Total: 10 points 1. Diagonalize A = - 1 0 1 3 0 - 3 1 0 - 1 , if possible. Is A invertible? Why or why not? Solution: The characteristic polynomial is - λ 2 ( λ + 2). Hence the eigenvalues are λ 1 = 0 with algebraic multiplicity 2 and λ 2 = - 2 with algebraic multiplicity 1. The eigenspace corresponding to λ 1 = 0 is E 0 = span 0 1 0 , 1 0 1 . The eigenspace corresponding to λ 2 = - 2 is E - 2 = span - 1 3 1 . Therefore A = PDP - 1 where P = 0 1 - 1 1 0 3 0 1 1 and D = 0 0 0 0 0 0 0 0 - 2 . A is not invertible because 0 is an eigenvalue of A . 2. Let A be a square matrix with eigenvalue λ and corresponding eigenvector v . Show the following statements by using that Av = λv . (a) For any positive integer n , λ n is an eigenvalue of A n with corresponding eigenvector v . (Hint: what is A 2 v ? A 3 v ?) (b) If A is invertibe, then 1 is an eigenvalue of A - 1 with corresponding eigenvector v . (Hint: what is A - 1 v ?)

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