0

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.

Daniel Li
  • 3,200
  • 1
    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.) – angryavian Oct 12 '19 at 03:11
  • @angryavian I don't see how is that a counter-example. – Daniel Li Oct 12 '19 at 03:16
  • 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? – John Hughes Oct 12 '19 at 03:24
  • @JohnHughes Yes I did. I can't quite make sense of it. If $X$ is invertible or something this would be trivial. – Daniel Li Oct 12 '19 at 03:32
  • @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... – angryavian Oct 12 '19 at 03:35
  • Shouldn't the q. be restricted to square matrices? – Wlod AA Oct 12 '19 at 03:50
  • Otherwise, as @angryavian have suggested, vector-matrix $\ [1\ 0\ 0\ 0\ 0 \ 0\ 0]\ $ (or simply $\ [1\ 0])$ is a counterexample. – Wlod AA Oct 12 '19 at 03:53
  • The "mark as duplicate" by @user1551 is mistaken; that related question is for square matrices, while this one is not. (As it happens, some answers in that question address the non-square case, however.) – John Hughes Oct 12 '19 at 12:06
  • @JohnHughes While the question is about square matrices, the answers are not. See this answer for instance. – user1551 Oct 12 '19 at 12:22
  • Uh...that's exactly what I said. But you marked this as a duplicate question, which I believe is a mistake. I suppose in the greater scheme, it doesn't really matter. OP has the answer at this point from several comments. – John Hughes Oct 12 '19 at 12:27

0 Answers0