ACM 106 a - Homework 4 Solutions prepared
نویسنده
چکیده
Assume that a1j 6= 0 for j = 2 . . . n. Then the (1, 1) element of A∗A is equal to a11+a12+. . .+a1n. The (1, 1) element of AA∗ is equal to a11. Since A is normal, a 2 11 + a 2 12 + . . . + a 2 1n = a 2 11 and, therefore, a1j = 0 for j = 2 . . . n. Now assume that a2j 6= 0 for j = 3 . . . n. Using the above proven fact that a12 = 0 we can compare the (2,2) entries of A∗A and AA∗ to show that all a2j = 0 for j = 3 . . . n. Similarly, proceeding row by row and comparing the diagonal entries of A∗A and AA∗ we can see that in order for an upper triangular A to be normal, it has to be diagonal. (b) Any n × n matrix A can be represented as A = UTU∗, where T is upper triangular and U is unitary. Assume that A is normal, then A∗A = AA∗ and we have UT ∗TU∗ = UTT ∗U∗, thus, T ∗T = TT ∗. By part (a) we know that if an upper triangular matrix T is normal, then it is diagonal. Since the diagonal entries of the Schur form T are the eigenvalues of A and T is a diagonal matrix, then A = UTU∗ gives an eigenvalue decomposition of A. Thus, A has n orthogonal eigenvectors. Now suppose A has n orthogonal eigenvectors (denote the matrix of these eigenvectors by U), then it can be represented as A = U∗DU , where D is a diagonal matrix. Then by employing the fact that any diagonal matrix has to be normal we have A∗A = U∗D∗UU∗DU = U∗D∗DU = U∗DD∗U = U∗DUU∗D∗U = AA∗. Thus, A is normal.
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