3 ms·
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: (US
by stephencanon 9y 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'.