Differential Equations Solutions 20

Differential Equations Solutions 20 - by σ n ± x − x...

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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 σ n ± x − x true ± . Therefore σ 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 σ 1 ± b true ± . Dividing the ±rst statement by this one gives ± x − x true ± ± x true ± ≤ σ 1 σ n ± b − b true + r ± ± b true ± = κ ( A ) ± b − b true + r ± ± b true ± ....
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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