ALS with missing entries

Hi,

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?

Thanks

als.JPG

Hello,
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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