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Unformatted text preview: Math 235 Assignment 9 Due: Wednesday, Dec 1st 1. For any R , let R = cos  sin sin cos . a) Diagonalize R over C . Solution: The characteristic polynomial is C ( ) = det( R  I ) = cos   sin sin cos  = 2 2 cos + 1 Hence, by the quadratic formula, we have = 2 cos 4 cos 2  4 2 = cos p sin 2 = cos i sin Observe, that if sin = 0, then R is diagonal, so we could just take P = I . So, we now assume that sin 6 = 0. For 1 = cos + i sin , we get R  1 I = i sin  sin sin  i sin 1 i We get that an eigenvector of 1 is i 1 . Similarly, for 2 = cos  i sin , we get R  2 I = i sin  sin sin i sin 1 i 0 0 We get that an eigenvector of 1 is i 1 . Thus, P = i i 1 1 and D = cos + i sin cos  i sin . b) Verify your answer in a) is correct, by showing the matrix P and diagonal matrix D from part a) satisfy P 1 R P = D for = 0 and = 4 . Solution: We have R = 1 0 0 1 which is already diagonal, so R is diagonalized by P = I as we stated in a)....
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 Fall '08
 CELMIN
 Linear Algebra, Algebra, Quadratic Formula

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