If $A^T A v = A A^T v = v$ with $A$ *not* orthogonal. Must $v$ be an eigenvector?

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!

• Sorry, I missed the $=v$ at the end. So my answer didn't work as written. – Arthur Jul 2 '20 at 14:04
• Choose $v=0.$ Then $v$ isn't an eigenvector (unless you've defined $0$ as an eigenvector). – Chickenmancer Jul 2 '20 at 14:07
• @Chickenmancer Thanks. What about non-zero vectors? – Leo Jul 2 '20 at 14:09

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 $$A$$.
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 $$i = 2,\ldots,n-1$$.