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

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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- 2 ( + 2). Hence the eigenvalues are 1 = 0 with algebraic multiplicity 2 and 2 =- 2 with algebraic multiplicity 1. The eigenspace corresponding to 1 = 0 is E = span 1 , 1 1 . The eigenspace corresponding to 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 and corresponding eigenvector v . Show the following statements by using that Av = v . (a) For any positive integer n , 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...
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tut6 - MATH 1104F - Solutions to Tutorial Assignment 6 -...

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