Unformatted text preview: f nonzero and the other components zero). Any other change gives a bigger ± ˜ E ± F . Thus the smallest value of the minimization function is ˜ σ n +1 , and since we veriFed in part (a) that our solution has this value, we are Fnished. If you don’t Fnd that argument convincing, we can be more precise. We use a fact found in the Frst pointer in Chapter 2: for any matrix B and vector z for which Bz is deFned: ± Bz ± 2 ≤ ± B ± 2 ± z ± 2 , where ± B ± 2 is deFned to be the largest singular value of B . Therefore, • ± B ± 2 ≤ ± B ± F , since we can see from part (b) and the singular value decomposition of B that the ±robenius norm of B is just the square root of the sum of the squares of its singular values. • If ( ˜ Σ + ˜ E ) ˜ f = , then ˜ Σ ˜ f = − ˜ E ˜ f ....
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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.
 Fall '11
 Dr.Robin

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