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The title says it all. If $X$ is a design matrix (columns containing variables, rows containing observations), I have observed that eigs($X'X$)=eigs($XX'$). I actually found this by accident when I was trying to compute eigenvalues of a covariance matrix in Matlab. Why is this the case? Can someone provide me some intuition, proofs, and/or reading materials?

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    $\begingroup$ If your design matrix has $n$ rows & $p$ columns (& let's say $p<n$ WLG), you mean that the first $p$ eigenvalues are the same, right? Otherwise $eig(XX')$ has more eigenvalues than $eig(X'X)$. $\endgroup$ – gung - Reinstate Monica Sep 11 '19 at 15:38
  • $\begingroup$ Technically, yes. However, I believe the ranks will be the same. I guess I should clarify by asking why all non-zero eigenvalues are the same. $\endgroup$ – qualiaMachine Sep 11 '19 at 15:41