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Unformatted text preview: 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 = span 1 , 1 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 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 /λ is an eigenvalue of...
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This note was uploaded on 03/22/2010 for the course MATH 1104 taught by Professor Unknown during the Winter '10 term at Carleton.
 Winter '10
 Unknown
 Math, Linear Algebra, Algebra

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