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# soln5 - MATH 245 Linear Algebra 2 Solutions to Assignment 5...

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MATH 245 Linear Algebra 2, Solutions to Assignment 5 1: (a) Let A = 1 2 1 2 4 2 1 2 1 . Find an orthogonal matrix P and a diagonal matrix D such that P * AP = D . Solution: The characteristic polynomial of A is f A ( λ ) = det( A - λI ) = det 1 - λ 2 1 2 4 - λ 2 1 2 1 - λ = (1 - 2 λ + λ 2 )(4 - λ ) + 8 + 8( λ - 1) + ( λ - 4) = 4 - 9 λ + 6 λ 2 - λ 3 - 4 + 9 λ = - ( λ 3 - 6 λ 2 ) = - λ 2 ( λ - 6) so the eigenvalues of A are λ 1 = λ 2 = 0 and λ 3 = 6. For λ = 0 we have A - λI = 1 2 1 2 4 2 1 2 1 1 2 1 0 0 0 0 0 0 so a basis for the eigenspace E 0 (which is the same as Null( A )) is given by { v 1 , v 2 } where v 1 = - 1 0 1 and v 2 = - 2 1 0 . We perform the Gram Schmidt Procedure, letting w 1 = v 1 , and w 2 = v 2 - v 2 . w 1 | w 1 | 2 w 1 = - 2 1 0 - 2 2 - 1 0 1 = - 1 1 - 1 and then normalize to get u 1 = w 1 | w 1 | = 1 2 - 1 0 1 and u 2 = w 2 | w 2 | = 1 3 - 1 1 - 1 . For λ = 6 we have A - λI = - 5 2 1 2 - 2 2 1 2 - 5 1 2 - 5 1 - 1 1 - 5 2 1 1 2 - 5 0 3 - 6 0 12 - 24 1 2 - 5 0 1 - 2 0 1 - 2 1 0 - 1 0 1 - 2 0 0 0 so a unit eigenvector is u 3 = 1 6 1 2 1 . Thus we can take P = ( u 1 , u 2 , u 3 ) = - 1 2 - 1 3 1 6 0 1 3 2 6 1 2 - 1 3 1 6 and D = λ 1 0 0 0 λ 2 0 0 0 λ 3 = 0 0 0 0 0 0 0 0 6 .

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(b) Let A = 2 i - 2 + i 1 - i 3 . Find a unitary matrix P and an upper-triangular matrix T so that P * AP = T . Solution: The characteristic polynomial is f A ( λ ) = det( A + λI ) = det 2 i - λ - 2 + i 1 - i 3 - λ = λ 2 - (3 + 2 i ) λ + 6 i - ( - 1 + 3 i ) = λ 2 - (3 + 2 i ) + (1 + 3 i ) so the eigenvalues are λ = (3 + 2 i ) ± p (3 + 2 i ) 2 - 4(1 + 3 i ) 2 = (3 + 2 i ) ± p (5 + 12 i ) - (4 + 12 i ) 2 = (3 + 2 i ) ± 1 2 = 2 + i, 1 + i . When λ = λ 1 = 2 + i we have A - λI = - 2 + i - 2 + i 1 - i 1 - i 1 1 0 1 , so u 1 = 1 2 - 1 1 is a unit eigenvector. We choose u 2 = 1 2 1 1 to extend { u 1 } to the orthonormal basis U = { u 1 , u 2 } for C 2 . Thus we can take P = ( u 1 , u 2 ) = 1 2 - 1 1 1 1 , and T = P * A P = 1 2 - 1 1 1 1 2 i - 2 + i 1 - i 3 - 1 1 1 1 = 1 2 - 1 1 1 1 - 2 - i - 2 + 3 i 2 + i 4 - i = 1 2 4 + 2 i 6 - 4 i 0 2 + 2 i = 2 + i 3 - 2 i 0 1 + i .
2: (a) Let U be a (possibly infinite-dimensional) inner product space over R or C , and let L : U U be linear. Show that if L * = L then all of the eigenvalues of L are real, and any two eigenvectors associated to distinct eigenvalues are orthogonal. Solution: Suppose that L has an adjoint and L * = L . We claim that every eigenvalue of L is real. Let λ be an eigenvalue of L , say L ( u ) = λu where 0 6 = u U . Then λ h u, u i = h λu, u i = h L ( u ) , u i = h u, L * ( u ) i = h u, L ( u ) i = h u, λu i = λ h u, u i . Since u 6 = 0, we can divide both sides by h u, u i to get λ = λ . Thus λ R . Next we claim that two eigenvectors associated to two distinct eigenvalues are orthogonal. Let λ 1 and λ 2 be two distinct eigenvalues and let u 1 and u 2 be corresponding eigenvectors. Then λ 1 h u 1 , u 2 i = h λ 1 u 1 , u 2 i = h L ( u 1 ) , u 2 i = h u 1 , L * ( u 2 ) i = h u 1 , L ( u 2 ) i = h u 1 , λ 2 u 2 i = λ 2 h u 1 , u 2 i = λ 2 h u 1 , u 2 i where at the final step we used the fact that λ 2 R . Since λ 1 6 = λ 2 , this implies that h u 1 , u 2 i = 0.

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soln5 - MATH 245 Linear Algebra 2 Solutions to Assignment 5...

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