solq12 - Solutions to quiz # 12 (December 2) 1. First, we...

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Unformatted text preview: Solutions to quiz # 12 (December 2) 1. First, we find the eigenvalues of A. The sum of the eigenvalues is 3 + 4 = 7, the product is 4 · 3 − 2 · 1 = 10, so the eigenvalues are 2 and 5. Next, we find eigenvectors. An eigenvector with eigenvalue 2 is found by solving the system 21 21 so |0 21 −→ |0 00 −1 is an eigenvector. An eigenvector with eigenvalue 5 is found by solving the system 2 −1 1 2 −2 so |0 , |0 1 1 | | 0 1 −1 −→ 0 00 |0 , |0 is an eigenvector. Next, we write 4 −1 1 =x +y , 13 2 1 that is, x −1 2 y 1 1 1 | 4 −→ 0 | 13 −1 3 | −4 1 −→ | 21 0 0| 1| 3 , 7 so x = 3 and y = 7. Finally, A k v = 3A k Answer: Ak v = 1 −1 7 · 5k − 3 · 2k 1 −1 . = + 7 · 5k = 3 · 2k + 7A k 1 7 · 5k + 3 · 2k+1 2 1 2 7 · 5k − 3 · 2k . 7 · 5k + 3 · 2k+1 2. First, we find the eigenvalues from the equation 1−λ a 0 0 = det 0 1−λ b = (1 − λ)2 (2 − λ), 0 0 2−λ so the eigenvalues are λ = 1 and λ = 2. To diagonalize the matrix, we need three linearly independent eigenvectors. We get exactly one of these eigenvectors from the eigenvalue λ = 2, so we need two linearly independent eigenvectors with eigenvalue λ = 1. That is, we need 0a0 dim ker 0 0 b = 2, 001 1 that is, we need 0 1 = rank 0 0 a0 0a0 0 b = rank 0 0 0 . 01 001 In other words, we need a = 0. Answer: The matrix is diagonalizable if and only if a = 0 (the value of b doesn’t matter). 2 ...
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This note was uploaded on 12/26/2011 for the course MATH 214 taught by Professor Conger during the Fall '08 term at University of Michigan.

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