275A-Solutions-1

# In which case px 0 x and ? 0 6 let a be hermitian a h

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, in which case Px = 0 · x and λ = 0. 6. Let A be hermitian, A H = A . (a) Note that x H Ax = ( x H Ax ) H = x H A H x = x H Ax , and therefore x H Ax is real. Let x be any eigenvector of A with eigenvalue λ . Then from Ax = λx we obtain, x H Ax = λx H x = λ x 2 . (3) Therefore λ must be real. (b) From equation (3). we see that if x H Ax is nonegative, then the eigenvalue λ must be nonegative. (c) If A has multiple eigenvectors for the same eigenvalue λ (in this case we say that these vectors span the eigenspace associated with λ ), we can apply the Gram-Schmidt procedure to orthogonalize within this eigenspace. Now suppose that x 1 and x 2 are normalized eigenvectors associated with the two distinct eigenvalues λ 1 and λ 2 respectively. Then, using the fact that the eigenvalues are real, λ 1 x 1 , x 2 = λ 1 x H 1 x 2 = ( Ax 1 ) H x 2 = x H 1 A H x 2 = x H 1 Ax 2 = λ 2 x H 1 x 2 = λ 2 x 1 , x 2 . 2

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Since λ 1 ̸ = λ 2 it must be the case that x 1 , x 2 = 0. This shows that the eigenspaces of A are orthogonal. 3
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