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I was discussing applying Principal Component Analysis to a covariance matrix versus applying PCA to the corresponding correlation matrix with a collegue. This led me to think about the following question.

Given a covariance matrix $\Sigma \in \mathbb{R}^{n\times n}$ of random variables $X_1, \ldots, X_n$. From this covariance matrix we can calculate the correlation matrix $R \in \mathbb{R}^{n\times n}$ through $$R_{ij} = \frac{\Sigma_{ij}}{\sigma_{X_i}\sigma_{X_j}}$$ where $\sigma_{X_i}$ is the standard deviation of $X_i$.

What can we say about the eigenspace of $\Sigma$ in terms of $R$? Is there any way to relate the eigenvectors of $\Sigma$ to those of $R$?

If this is a trivial question then I apologize, a Google search hasn't given me any answers.

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    $\begingroup$ As you see, $\Sigma = D R D$, where $D = \text{diag}(\sigma_{X_1}, \ldots \sigma_{X_n})$. $\endgroup$
    – Piotr Migdal
    Feb 27, 2013 at 15:04
  • $\begingroup$ Ah, thank you. Now I feel silly for not seeing that. I suppose deriving the transformed eigenvectors is straightforward through the singular value decomposition? I'd be willing to accept your comment as the answer if you submit it as such. $\endgroup$
    – Stijn
    Feb 27, 2013 at 17:12
  • $\begingroup$ Well, it's not the result yet. I don't know how one can squeeze out of it, besides for looking at smallest and largest $\sigma_{X_i}$. $\endgroup$
    – Piotr Migdal
    Mar 2, 2013 at 3:04

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