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you should not actually form the covariance matrix to get there... In case anyone's wondering why: it's not only because it takes extra time. The main reason i
by Radim 13y ago
you should not actually form the covariance matrix to get there...
In case anyone's wondering why: it's not only because it takes extra time. The main reason is that computing X*X^T can bring numerical instability, where a direct SVD(X) would work just fine.
- neumann 13y agoI remember this being elegantly demonstrated here: http://math.stackexchange.com/questions/3869/what-is-the-intuitive-relationship-between-svd-and-pca http://math.stackexchange.com/questions/3869/what-is-the-int...
- conjectures 13y agoWith mean corrected observations as rows of a data matrix X, the sample covariance matrix is given by X'X not XX'. Also (X'X)g = lg (XX')Xg = lXg So with n>>p and n large, you may be able to fit X'X (p^2 entries) but not XX' (n^2 entries) in memory (and get the same eigenvalues and related eigenvectors).