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Unformatted text preview: Solutions for Homework 5 Numerical Linear Algebra 1 Fall 2010 Problem 5.1 Let A R n k have rank k . The pseudoinverse for rectangular full columnrank matrices behaves much as the inverse for nonsingular matrices. To see this show the following identities are true (Stewart 73): 5.1.a . AA A = A 5.1.b . A AA = A 5.1.c . A A = ( A A ) T 5.1.d . AA = ( AA ) T 5.1.e . If A R n k has orthonormal columns then A = A T . Why is this important for consistency with simpler forms of least squares problems that we have discussed? Solution: AA A = A ( A T A ) 1 A T A = A ( A T A ) 1 ( A T A ) = AI = A A AA = ( A T A ) 1 A T A ( A T A ) 1 A T = ( A T A ) 1 ( A T A )( A T A ) 1 A T = I ( A T A ) 1 A T = A A A = ( A T A ) 1 A T A = ( A T A ) 1 ( A T A ) = I = A T A ( A T A ) 1 = A T A ( A T A ) T = ( A A ) T 1 AA = A ( A T A ) 1 A T = A ( A T A ) T A T = ( A ( A T A ) 1 A T ) T = ( AA ) T Since A has orthonormal columns we have A T A = I and therefore A = ( A T A ) 1 A T = IA T = A T as desired. Recall we have the problem of minimizing k b Ax k 2 with A R n k with n k having full column rank. This splits into several cases when we consider consistency. If n = k then A is nonsingular and x = A 1 b . We have shown earlier that in this case A = A 1 . If n > k then we want x = A b to be such that Ax = Pb where P is the projector from R n onto R ( A ). We showed this in the notes based on the orthogonality of the residual. This case was also discussed in simplified form earlier in the notes when first considering orthonormal bases for a space, i.e., a basis for a subspace such that Q T Q = I and the columns of Q are the basis. In this case we had for the case above with n > k that x = Q T b . The exercise shows this is consistent with the definition of A when A T A = I ....
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 Fall '06
 gallivan

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