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Consider a square matrix $A$ over the real numbers that is not orthogonal. Let $v \neq 0$ be a non-zero vector such that $A^T A v = A A^T v = v$. Is it true that $v$ must be an eigenvector of $A$?

It's not hard to see that if $v$ is an eigenvector, then its eigenvalue must be on the unit circle. But I can't determine whether or not $v$ is necessarily an eigenvector.

Somehow this looks like it would be a consequence of the spectral theorem, but I can't see a crisp proof. Thanks!

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  • $\begingroup$ Sorry, I missed the $=v$ at the end. So my answer didn't work as written. $\endgroup$ – Arthur Jul 2 '20 at 14:04
  • $\begingroup$ Choose $v=0.$ Then $v$ isn't an eigenvector (unless you've defined $0$ as an eigenvector). $\endgroup$ – Chickenmancer Jul 2 '20 at 14:07
  • $\begingroup$ @Chickenmancer Thanks. What about non-zero vectors? $\endgroup$ – Leo Jul 2 '20 at 14:09
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No. Let $Q\ne\pm I$ be any $n\times n$ orthogonal matrix, $u\in\mathbb R^n$ be any vector that is not an eigenvector of $Q$, and $a\ne\pm1$. Let $$ A=\pmatrix{Q&0\\ 0&a},\ v=\pmatrix{u\\ 0}. $$ Then $A$ is not orthogonal and $AA^Tv=A^TAv=v$, but $v$ is not an eigenvector of $A$.

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In general $v$ will not be an eigenvector. For instance let $A$ be a Jordan block matrix with respect to the eigenvalue $0$. (That is $A$ has 1 over the main diagonal and all other entries are zero). Let $e_i$ denote the column vector with the i'th entry equal to 1 and the other entries zero. Then $Ae_1 = 0$ and $Ae_i = e_{i-1}$ for $i = 2, \ldots, n$. And $A^Te_n = 0$ and $A^Te_i = e_{i+1}$ for $i = 1, \ldots, n-1$. Hence $AA^Te_i = A^TAe_i = e_i$ for $i = 2,\ldots,n-1$.

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