ALS with missing entries


I wanted to confirm the inverse is taken after the sum right? it's not a sum of inverses?

Also I am not sure how do we compute complexity for such a step?



The corrected solution has been uploaded. Taking inverse after the sum is correct indeed

thank you! and how would should we go about calculating the complexity?

Each step(updating W or Z) is similar to computing solution for a similar Linear Regression(Ridge in case of regularized)

I have a question about the sign of the gradients: shouldn't we have \(dL/dw_i = - \Sigma z_n(x_{in}-(WZ^T)_{in}) \) for the first one for example?? :'(

Also, since \(w_i\) and \(z_n\) are rows, the dimensions of \(w_i^T z_n\) are not correct for every entry of \( (WZ^T)_{in} \), it should be \(w_i z_n^T\), shouldn't it ? What do you guys think ?

Individually, w_i when written is thought of as a column vector.

Yes that actually makes sense after checking the dimensions of the final result (for example, the inverse was hinting at the fact that the product resulted in a matrix, not a scalar). What about the minus sign ? Of course, I know it doesn't change the result, but just for rigour

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