# solq13 - Solutions to quiz # 13 (December 9) 1. The sum of...

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Unformatted text preview: Solutions to quiz # 13 (December 9) 1. The sum of the eigenvalues is 2 + 0 = 2 while the product is 2 · 0 − (1) · (−5) = 5, so the eigenvalues are 1 ± 2i. We ﬁnd an eigenvector with eigenvalue 1 + 2i from a solution to the system 2 − (1 + 2i) −5 1 − 2i 0 −→ 1 −1 + 2i where so 1 0 1 | 0 − (1 + 2i) | |0 , |0 1 0 +i −1 2 = S= Answer: A = 0 2 1 −1 0 1 − 2i −→ 0 −5 1 2 1 −1 − 2i |0 |0 is an eigenvector. Therefore, we can write A = SDS −1 , 01 2 −1 −2 1 and 01 2 −1 D= 1 2 −2 . 1 −1 . 2. We write √ √ √ cos α 1/√10 −3/ 10 √ = 10 B = 12 + (−3)2 3/ 10 1/ 10 sin α 1 3 where cos α = √ and sin α = √ . 10 10 − sin α , cos α Therefore, B k = (10)k/2 and B k v = (10)k/2 Answer: B k v = 10k/2 cos kα sin kα cos kα sin kα − sin kα cos kα − sin kα cos kα 0 − sin kα = 10k/2 . 1 cos kα √ − sin kα , where α = arccos 1/ 10. cos kα 1 ...
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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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