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I'm studying the kernelized version of the classical PCA (Principal Component Analysis). Let x_i, i=1,2,...,n be the available training points and 
an estimate of the correlation matrix, where phi maps points in X into a RKHS H. We want to perform the eigendecomposition of R, that is, 
Let's assume that the data are centered (i.e. that the sum of all the phi(x_i) is zero). By the definition of R, it can be shown that v lies in 
Indeed, 
and for lambda different from zero we can write 
Now the book says that combining this last expression with 
where 
and 
with K being the adopted kernel function. I can't see why performing the eigendecomposition of R is "equivalent" to performing that of the Gram matrix. Any help would be greatly appreciated! |
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Substituting the expression for the eigen vectors (v) in the eigen decompostion Rv = lambda v and expanding it is term of the feature space phi(x_i), you will end up at an expression that you are looking for. It is just arithmetic. Try writing down the complete matrix multiplication between Rv while expressing them in terms of phi(.)s. Max Welling's tutorial has the complete derivation I was missing a key point: I needed to multiply both members of Rv = lv by phi(x_k)^T. In the end, I can find the final expression only if K is invertible, but is it invertible?
(Aug 30 '13 at 10:26)
Kiuhnm
Yes, K is invertible. Any positive definite matrix is invertible.
(Aug 31 '13 at 09:51)
Rakesh Chalasani
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