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lottin
3218 days ago
Actually it's eigendecomposition of A'A and the SVD of A, is it not?
1 comments
stephencanon
3218 days ago
The SVD of A'A
is
its eigendecomposition (since it is symmetric semi-definite, the two factorizations are the same).
It is closely related to the SVD of A: (USV')'USV' = VSU'USV' = VS^2V'.
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It is closely related to the SVD of A: (USV')'USV' = VSU'USV' = VS^2V'.