# 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:

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.

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.

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 $$A$$

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.