Will the product of a real valued matrix and its transpose always have a real eigenvector?

Question

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:

1. Is it true that $M^TM$ must have at least a (real) eigenvector?
2. I strongly suspect this $\sigma$ they talk of is the operator norm of $M$. Is this true?
3. Is this a dead-end way to try to solve this problem. Should I try something else?

Of course, everything is real-valued here.

Answer 1

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.

Answer 2

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 $A$

Answer 3

1. yes, obvious once you realize $M^TM$ is symmetric.
2. yes, can you prove it?
3. 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.