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Suppose we have a $n\times n$ real symmetric positive definite matrix $\Sigma$, and $V=(v_1,...,v_n)$ whose columns are the eigenvectors corresponding to the $n$ eigenvalues $\lambda_1\geq \lambda_2 ...\geq \lambda_n$. Let $\Delta$ be a diagonal matrix $diag(\delta_1,...,\delta_n)$, where $\delta_i>0$ for $i=1,...,n$. If $M=\Delta V$, then I want to ask that whether the columns of $M$ correspond to the eigenvectors of some matrix, saying $\tilde{\Sigma}$?

Many thanks!

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$V$ is the eigenvector matrix, $D$ is the diagonal matrix.

$VD$ is still an eigenvector matrix. Multiplying a diagonal matrix from the right, scales the colums. Columns are eigenvectors. Scaled eigenvectors are still eigenvectors (of the same matrix that $V$ belongs to).

$DV$, is not. Multiplying a diagonal matrix from the left, scales the rows. This destroys the proportions of all the eigenvectors.

Regarding your last sentence, all vectors are the eigenvectors of the identity matrix, so that statement means very little.

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  • $\begingroup$ Thanks! I understand that $VD$ should be still the eigenvector matrix of the same matrix $\Sigma$. I am just confused... let's say $\Sigma=VUV'$, where $U$ is the diagonal matrix of eigenvalues. Then we have $\Sigma_1=DVUV'D$. Does this mean $DV$ is the eigenvectors of $\Sigma_1$? $\endgroup$ – Eridk Poliruyt May 29 '15 at 22:23
  • $\begingroup$ Was that D in the end supposed to be D inverse? $\endgroup$ – grdgfgr May 29 '15 at 22:34
  • $\begingroup$ If that is the case, then it is called a similarity transformation. It has the same eigenvalues as $\Sigma$ and different eigenvectors. $D$ doesn't have to be diagonal either. It could be any invertible matrix. $\endgroup$ – grdgfgr May 29 '15 at 22:38
  • $\begingroup$ Why should it be $D$ inverse? $\endgroup$ – Eridk Poliruyt May 29 '15 at 22:45
  • $\begingroup$ I meant to say was it supposed to be $\Sigma_1=DVUV'D'$ and $DV$ is the eigenvectors of $\Sigma_1$ $\endgroup$ – grdgfgr May 29 '15 at 22:47

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