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Re: Principal component analysis by several decomposition
From: |
Andreas Stahel |
Subject: |
Re: Principal component analysis by several decomposition |
Date: |
Wed, 12 May 2021 01:39:45 -0500 (CDT) |
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
--
Sent from: https://octave.1599824.n4.nabble.com/Octave-General-f1599825.html