To answer this lets deal with real symmetric

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Unformatted text preview: σ2 , . . . , σr ) contains the nonzero singular values of A, then V∗ A∗ AV = D2 0 0 0 , (7.5.9) 2 and this means that (σi , vi ) for i = 1, 2, . . . , r is an eigenpair for A∗ A. In other words, the nonzero singular values of A are precisely the positive square roots of the nonzero eigenvalues of A∗ A, and right-hand singular vectors vi of A are particular eigenvectors of A∗ A. Note that this establishes the uniqueness of the σi ’s (but not the vi ’s), and pay attention to the fact that the number of zero singular values of A need not agree with the number of zero eigenvalues of A∗ A —e.g., A1×2 = (1, 1) has no zero singular values, but A∗ A has one zero eigenvalue. The same game can be played with AA∗ in place of A∗ A to argue that the nonzero singular values of A are the positive square roots of Copyright c 2000 SIAM Buy online from SIAM http://www.ec-securehost.com/SIAM/ot71.html Buy from AMAZON.com 554 Chapter 7 Eigenvalues and Eigenvectors http://www.amazon.com/exec/obidos/ASIN/0898714540 It is illegal to print, duplicate, or distribute this material Please report violations to meyer@ncsu.edu the nonzero eigenvalues of AA∗ , and left-hand singular vectors ui of A are particular eigenvectors of AA∗ . Caution! The statement that right-hand singular vectors vi of A are eigenvectors of A∗ A and left-hand singular vectors ui of A are eigenvectors of AA∗ is a one-way street—it doesn’t mean that just any orthonormal sets of eigenvectors for A∗ A and AA∗ can be used as respective right-hand and left-hand singular vectors for A. The columns vi of any unitary matrix V that diagonalizes A∗ A as in (7.5.9) can serve as right-hand singular vectors for A, but corresponding left-hand singular vectors ui are constrained by the relationships Avi = σi ui , i = 1, 2, . . . , r u∗ A = 0, i =⇒ i = r + 1, . . . , m =⇒ D E Avi Avi = , i = 1, 2, . . . , r, σi Avi 2 span {ur+1 , ur+2 , . . . , um } = N (A∗ ). ui = T H In other words, t...
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