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A10_soln

# A10_soln - Math 235 Assignment 10 Solutions 1 Let A be...

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Math 235 Assignment 10 Solutions 1. Let A be skew-Hermitian. Prove that all the eigenvalues of A are purely imaginary. Solution: If A is skew-Hermitian, then A * = - A . Hence ( iA ) * = - iA * = iA so iA is Hermitian. Then U * ( iA ) U = D and so U * AU = iD . 2. Let A be normal and invertible. Prove that B = A * A - 1 is unitary. Solution: We have A * A = AA * . So B * B = ( A * A - 1 ) * ( A * A - 1 ) = ( A - 1 ) * AA * A - 1 = ( A - 1 ) * A * AA - 1 = I. 3. Let A and B be Hermitian matrices. Prove that AB is Hermitian if and only if AB = BA . Solution: If AB is Hermitian, then ( AB ) = ( AB ) * = B * A * = BA, since A and B are Hermitian. If AB = BA , then ( AB ) * = B * A * = BA = AB. 4. Unitarily diagonalize the following matrices. a) A = a b - b a Solution: We have C ( λ ) = λ 2 - 2 + a 2 + b 2 so by the quadratic formula we get eigenvalues λ = a ± bi . For λ = a + bi we get A - λI = - bi b - b - bi ~ z 1 = 1 i For λ = a - bi we get A - λI = bi b - b bi ~ z 2 = - 1 i Hence we have D = a + bi 0 0 a - bi and U = " 1 2 - 1 2 i 2 i 2 # . 1

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2 b) B = 4 i 1 + 3 i - 1 + 3 i i Solution: We have C ( λ ) = λ 2 - 5 + 6 so by the quadratic formula we get eigenvalues λ = 6 i and λ = - i . For λ = 6 i we get B - λI = - 2 i 1 + 3 i - 1 + 3 i - 5 i ~ z 1 = 3 - i 2 For λ = - i we get B - λI = 5 i 1 + 3 i - 1 + 3 i 2 i ~ z 2 = 1 + 3 i - 5 i Hence we have D = 6 i 0 0 - i and U = " 3 - i 14 - 1+3 i 35 2 14 - 5 i 35 # .
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A10_soln - Math 235 Assignment 10 Solutions 1 Let A be...

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