[gabrielgoh]

6 word answer

PCA is the SVD of A'A

[lottin]

Actually it's eigendecomposition of A'A and the SVD of A, is it not?

[stephencanon]

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'.