Differential Equations Solutions 20

Differential Equations Solutions 20 - by n x x true ....

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30 Chapter 5. Solutions: Matrix Factorizations (b) Substituting the SVD our equation becomes A x = V £ Σ 1 0 ¤ U x = b , where Σ 1 is n × n with the singular values on the main diagonal. Letting y = U x , we see that a particular solution is y good = · Σ 1 1 V b 0 ¸ , so x good = U · Σ 1 1 V b 0 ¸ = n X j =1 v j b σ j u j . Every solution can be expressed as x good + U 2 v for some vector v since A U 2 = 0 . CHALLENGE 5.19. (a) Since we want to minimize ( c 1 σ 1 w 1 ) 2 + ... +( c n σ n w n ) 2 + c 2 n +1 + ... + c 2 m , we set w i = c i i = u i b true i . (b) If we express x x true as a linear combination of the vectors v 1 ,... v n , then multiplying by the matrix A stretches each component by the corresponding singu- lar value. Since σ n is the smallest singular value, ± A ( x x true ) ± is bounded below
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Unformatted text preview: by n x x true . Therefore n x x true ( b b true + r ) , and the statement follows. For any matrix C and any vector z , Cz C z . Therefore, b true = Ax true A x true . (c) Using the given fact and the second statement, we see that x true 1 1 b true . Dividing the rst statement by this one gives x x true x true 1 n b b true + r b true = ( A ) b b true + r b true ....
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