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Unformatted text preview: Math 235 Assignment 9 Due: Wednesday, Dec 1st 1. For any θ ∈ R , let R θ = cos θ sin θ sin θ cos θ . a) Diagonalize R θ over C . Solution: The characteristic polynomial is C ( λ ) = det( R θ λI ) = cos θ λ sin θ sin θ cos θ λ = λ 2 2 cos θ + 1 Hence, by the quadratic formula, we have λ = 2 cos θ ± √ 4 cos 2 θ 4 2 = cos θ ± p sin 2 θ = cos θ ± i sin θ Observe, that if sin θ = 0, then R θ is diagonal, so we could just take P = I . So, we now assume that sin θ 6 = 0. For λ 1 = cos θ + i sin θ , we get R θ λ 1 I = i sin θ sin θ sin θ i sin θ ∼ 1 i We get that an eigenvector of λ 1 is i 1 . Similarly, for λ 2 = cos θ i sin θ , we get R θ λ 2 I = i sin θ sin θ sin θ i sin θ ∼ 1 i 0 0 We get that an eigenvector of λ 1 is i 1 . Thus, P = i i 1 1 and D = cos θ + i sin θ cos θ i sin θ . b) Verify your answer in a) is correct, by showing the matrix P and diagonal matrix D from part a) satisfy P 1 R θ P = D for θ = 0 and θ = π 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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