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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1If your design matrix has $n$ rows & $p$ columns (& let's say $p
– gung - Reinstate Monica Sep 11 '19 at 15:38 -
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. – qualiaMachine Sep 11 '19 at 15:41