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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 definition of the Frobenius 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 i =1 Ua i 2 2 = n i =1 a i 2 2 = A 2 F . Similarly, letting a T i be the i th row of A , AV 2 F = m i =1 a T i V 2 2 = n i =1 a i 2 2 = A 2 F , and the conclusion follows. (c) From the constraint K + E g + r f 1 = 0 , we see that ˜ U T K + E g + 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 . From 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 deficient, so that the constraint can be satisfied 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 ˜ f
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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 ˜ 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 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 ( ˜ Σ + ˜ E ) ˜ f = , then ˜ Σ ˜ f = − ˜ E ˜ f ....
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