2
$\begingroup$

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:enter image description here

  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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $A$

$\endgroup$
0
$\begingroup$
  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.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.