Differential Equations Solutions 74

Differential Equations Solutions 74 - f nonzero and the...

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84 Chapter 14. Solutions: Case Study: Blind Deconvolution: Errors its eigenvalues). We can use the fact that trace( AB ) = trace( BA ). (See Challenge 12.3.) It can also be proven just from the deFnition of the ±robenius norm and the fact that ± Ux ± 2 = ± x ± 2 for all vectors x and orthogonal matrices U . Using this fact, and letting a i be the i th column of A , we see that ± UA ± 2 F = n X i =1 ± Ua i ± 2 2 = n X i =1 ± a i ± 2 2 = ± A ± 2 F . Similarly, letting b a T i be the i th row of A , ± AV ± 2 F = m X i =1 ± b a T i V ± 2 2 = n X i =1 ± b a i ± 2 2 = ± A ± 2 F , and the conclusion follows. (c) ±rom the constraint £ K + Eg + r ¤ · f 1 ¸ = 0 , we see that ˜ U T £ K + Eg + r ¤ ˜ V ˜ V T · f 1 ¸ = 0 , so ( ˜ Σ + ˜ E ) ˜ f = 0 , where ˜ E = ˜ U T [ E , r ] ˜ V and ˜ f = ˜ V T · f 1 ¸ . ±rom part (b) we know that minimizing ± [ E , r ] ± F is the same as minimizing ± ˜ E ± F . Therefore, to solve our problem, we want to make the smallest change to ˜ Σ that makes the matrix ˜ Σ + ˜ E rank deFcient, so that the constraint can be satisFed by a nonzero ˜ f . Changing the ( n +1 ,n + 1) element of ˜ Σ from ˜ σ n +1 to 0 certainly makes the constraint feasible (by setting the last component of ˜
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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 n +1 , and since we veriFed in part (a) that our solution has this value, we are Fnished. If you dont 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 decom-position of B that the robenius norm of B is just the square root of the sum of the squares of its singular values. If ( + E ) f = , then f = E f ....
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