Differential Equations Solutions 73 - σ i ˜ u i ˜ v T i so using the formula for £ E r ¤ we see that £ K g ¤ £ E r ¤ = n X i =1 ˜ σ i ˜ u i

Chapter 14 Solutions: Case Study: Blind Deconvolution: Errors, Errors Everywhere CHALLENGE 14.1. See the posted program problem1 and 3.m . The program is not diﬃcult, but it is important to make sure that you do the SVD only once (at a cost of O ( mn 3 )) and then form each of the trial solutions at a cost of O ( n 2 ). This requires using the associative law of multiplication. In fact, it is possible to form each solution by updating a previous one (by adding the newly added term) at a cost of O ( n ), and this would be an even better algorithm, left as an exercise. CHALLENGE 14.2. (a) We know that

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Unformatted text preview: σ i ˜ u i ˜ v T i , so using the formula for £ E r ¤ we see that £ K g ¤ + £ E r ¤ = n X i =1 ˜ σ i ˜ u i ˜ v T i . Now, since ˜ v n +1 is orthogonal to ˜ v i for k = 1 , . . . , n , it follows that ( £ K g ¤ + £ E r ¤ ) · f − 1 ¸ = − n X i =1 ˜ σ i ˜ u i ˜ v T i 1 ˜ v n +1 ,n +1 ˜ v n +1 = . Note that ± [ E , r ] ± F = ˜ σ n +1 . (b) This can be proven using the fact that ± A ± 2 F = trace( A T A ) where trace( B ) is the trace of the matrix B , equal to the sum of its diagonal elements (or the sum of 83...
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