126 a at b at t ab a b we can easily see what the

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126 ˜ A = AT ˜ B = AT - t . AB = ˜ A ˜ B .
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We can easily see what the solution is by requiring that A'A and B'B are diagonal. Then and A and B are the left and right singular vectors corresponding with the r largest singular values. At the optimum 127 X A = B Λ , XB = A Λ , σ ( A , B ) = min( n , m ) s = r + 1 λ 2 s .
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Missing Data We can use the low-rank approximation problem to illustrate a familiar imputation approach to analyzing matrices in which there are missing data . Define where w ij is 1 if x ij is observed and w ij is 0 if x ij is missing. 128 σ ( A , B ) = n i = 1 m j = 1 w i j ( x i j - a i b j ) 2 ,
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We tackle this problem by using augmentation . Define and define Then 129 σ ( A , B , Z ) = Z - AB 2 , Z = { Z | z i j = x i j if w i j = 1 } . σ ( A , B ) = min Z Z σ ( A , B , Z ) .
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This leads to the following alternating least squares algorithm. We start with a Iteration k consists of two steps, that are alternated, the singular value step and the imputation step . Since clearly the algorithm produces a decreasing, and thus convergent, sequence of loss function values. 130 Z (0) Z . ( A ( k ) , B ( k ) ) argmin A , B Z ( k ) - AB 2 , Z ( k + 1) = argmin Z Z Z - A ( k ) { B ( k ) } 2 . σ ( A ( k ) , B ( k ) , Z ( k + 1) ) < σ ( A ( k ) , B ( k ) , Z ( k ) ) < σ ( A ( k - 1) , B ( k - 1) , Z ( k ) )
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