Re: Principal component analysis by several decomposition

onewayenzyme wrote
> Thank you for your reply.
> 
> My major issue now is:
> 
> by eigendecomposition of the covariance matrix, says Xm in my example, I
> obtain PCs from
> 
> W=V'*Xm' (ordering the eigenvectors in V according to eigenvalues); 
> 
> however according to svd
> 
> [U S T]=svd(Xm);
> PC=T';
> 
> but W does not match with the PCs obtained after SVD.
> 
> Which are the relationships among the matrices obatined by
> eigendecomposition and SVD? In particular how can I get the same PCs and
> coefficients from the two decomposition in such a way thay are equal?
> 
>  
> 
> 
> 
> 
> 
> --
> Sent from:
> https://octave.1599824.n4.nabble.com/Octave-General-f1599825.html

For symmetric matrices there is a simple relation between eigenvectors and
the SVD, and it applies in the context of PCA.
Have a look at the draft of my lecture notes at
https://web.sha1.bfh.science/NumMethods/NumMethods.pdf
Find the connection between eigenvectors and SVD on page 149.
On pages 151-161 the path from Gaussian distributions to eigenvalues to PCA,
incuding SVD, is spelled out.

I hope it helps

Andreas




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