Will the product of a real valued matrix and its transpose always have a real eigenvector? I'm trying to solve this really interesting problem, taken from "Berkeley Problmes in Mathematics" by Souza and Silva
https://www.amazon.com/Berkeley-Problems-Mathematics-Problem-Books/dp/0387204296
Trying to make headway, I've noticed that if the last two statements are correct, then $M^T(Mu)=\sigma^2 u$, so in other words, there exists an eigenvector $u$ with the eigenvalue $\sigma^2$.
Therefore, I have two questions, and I would dearly love a proof or counter-example:

*

*Is it true that $M^TM$ must have at least a (real) eigenvector?

*I strongly suspect this $\sigma$ they talk of is the operator norm of $M$. Is this true?

*Is this a dead-end way to try to solve this problem. Should I try something else?

Of course, everything is real-valued here.
 A: The matrix $M^\top M$ is called the Gramian of $M$.
One important property of the Gramian is that it is symmetric. Indeed, one easily checks that
$$
(M^\top M)^\top
= M^\top (M^\top)^\top
= M^\top M
$$
This means that the spectral theorem applies to $M^\top M$. 
The spectral theorem says that every symmetric matrix $S$ can be factored as $S=QDQ^\top$ where $D$ is real-diagonal and $Q$ is orthogonal. This means that every $n\times n$ symmetric matrix has real eigenvalues and that there is an orthonormal basis of $\Bbb R^n$ consisting of eigenvectors of $S$. 
Applying the spectral theorem to $M^\top M$ immediately implies that $M^\top M$ has real eigenvalues and that there is an orthonormal basis of $\Bbb R^n$ consisting of eigenvectors of $M^\top $M.
One can actually say a little more. The Gramian $M^\top M$ is positive semidefinite so its eigenvalues are nonnegative. If $M$ is full column rank, then $M^\top M$ is positive definite, so its eigenvalues are all positive.
A: The spectral theorem: if $A$ is real valued matrix then $A^TA$ is symmetric. Such matrices must have a whole ON system of eigenvectors with strictly non negative real eigenvalues. It follows for example directly from for example Singular Value Decomposition (SVD) which is a more powerful result regarding $A$
A: *

*yes, obvious once you realize $M^TM$ is symmetric.

*yes, can you prove it?

*I think this is a good way to do it. Be somewhat careful in finding condition (2) since $v$ is a unit vector. I have not ideas besides doing this by construction from condition (3). have fun.  

