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Unformatted text preview: Partial Solution Set, Leon 6.5 6.5.1 We are to show first of all that A and A T have the same nonzero singular values, then to describe the relationship between the singular value decompositions of A and A T . So begin by assuming that is a nonzero singular value for A . By definition of singular value, we know that = 2 is a positive eigenvalue of A T A . Let x be an eigenvector for A T A belonging to . Then A T A x = x . So ( A x ) = A ( x ) = A ( A T A x ) = AA T ( A x ) , so A x is an eigenvector for AA T , also belonging to . Conversely, suppose that is a nonzero singular value for A T . Then = 2 is a positive eigenvalue for AA T , with eigenvector x , i.e., AA T x = x . Then ( A T x ) = A T ( x ) = A T ( AA T x ) = A T A ( A T x ) , so A T x is an eigenvector for A T A , also belonging to . Thus AA T and A T A have the same positive eigenvalues, hence the same nonzero singular values. 2 How, then, are the singular value decompositions for A and A T related? This is more easily answered: if A = U V T , then A T = ( U V T ) T = V T U T . 6.5.2 We are to find the singular value decompositions of several matrices, using the method outlined in the text. Here are two of the solutions....
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 Spring '08
 Anshelvich
 Linear Algebra, Algebra

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