1) yes, I agree it is a typo in this case it should be N instead of D. In general it is the smallest of the two.
2) It is a direct implication of the identity \(X^\top X = V S^\top S V^\top\) and the fact the columns of \(V\) are orthonormal.
To be more precise \(X^\top X = V S^\top S V^\top\) means \(X^\top X V = V S^\top S\) which translates to \(X^\top X (v_1,\ldots,v_N) = V S^\top S(v_1,\ldots,v_N) = V (s_1^2 e_1,\ldots,s_N^2e_N)=(s_1^2 v_1,\ldots,s_N^2 v_N)\)
Where \((e_i)_{i=1}^N\) is the canonical basis of \(\mathcal{R}^N\).
Lab 12 Problem 1 Compute U, S efficiently
1) possible typo In the solution
"Let \( v_i \), i = 1, ..., N, denote the columns of V" instead of "i = 1, ..., D" since V is NxN matrix
2) I don't get how identity (1) is established, in particular why \( VS^TSV^Tv_j = s_j^2v_j \)
Thank you!
1) yes, I agree it is a typo in this case it should be N instead of D. In general it is the smallest of the two.
2) It is a direct implication of the identity \(X^\top X = V S^\top S V^\top\) and the fact the columns of \(V\) are orthonormal.
To be more precise \(X^\top X = V S^\top S V^\top\) means \(X^\top X V = V S^\top S\) which translates to \(X^\top X (v_1,\ldots,v_N) = V S^\top S(v_1,\ldots,v_N) = V (s_1^2 e_1,\ldots,s_N^2e_N)=(s_1^2 v_1,\ldots,s_N^2 v_N)\)
Where \((e_i)_{i=1}^N\) is the canonical basis of \(\mathcal{R}^N\).
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