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Same eigenvalues can be shown easily $XX'v=\lambda v \implies X'X(X'v)=\lambda (X'v)$ and similarly for $XX'$ case.

But it is hard to see immediately how engienspace of each $\lambda $have same dimensionality for $XX', X'X$.

I know this is true from SVD but I don't want to use it here because I look forward to proving SVD using this as a corollary.

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    $\begingroup$ I don't believe this is true, the two matrices may not even be the same size. (Take $X$ to be an $n \times 1$ matrix, for example.) $\endgroup$ – angryavian Oct 12 '19 at 3:11
  • $\begingroup$ @angryavian I don't see how is that a counter-example. $\endgroup$ – Daniel Li Oct 12 '19 at 3:16
  • $\begingroup$ Surely your computation shows that a reasonable candidate map from the $\lambda$-espace of $XX'$ to that of $X'X$ is $v \mapsto X'v$. The only question is whether it has a kernel that intersects the $\lambda$-eigenspace. The tough case will probably be $\lambda = 0$, but maybe not. Anyhow, have you tried looking at that map? $\endgroup$ – John Hughes Oct 12 '19 at 3:24
  • $\begingroup$ @JohnHughes Yes I did. I can't quite make sense of it. If $X$ is invertible or something this would be trivial. $\endgroup$ – Daniel Li Oct 12 '19 at 3:32
  • $\begingroup$ @DanielLi Say, $X = (1,0,\ldots, 0)^\top$. Then $X^\top X = 1$ which has no zero eigenvalues, but $XX^\top$ has zero as an eigenvalue with high multiplicity. Anyway, just trying to get you to notice that your claim should be adjusted somehow, for example by focusing on nonzero eigenvalues... $\endgroup$ – angryavian Oct 12 '19 at 3:35

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