Do $A$ and $B$ have the same eigenvalues?

Question

There is a property that I've been told by my quantum physics teacher that states:

Being $A$ and $B$ Hermitian, if $AB-BA=0$ then $A$ and $B$ have the same eigenvalues.

I do not even know how to start the proof for this. I believe my teacher but I just want to understand where that comes from. Would the property also be valid for eigenvectors? I mean, if $AB-BA=0$, would not only the eigenvalues but also the eigenvectors be the same for both $A$ and $B$?

Answer 1

If this were true, then $A$ and $A^2$ would have the same eigenvalues, for any Hermitian matrix $A$. SO it is false.

It is true that if both products $AB$ and $BA$ are defined, then $AB$ and $BA$ have the same non-zero eigenvalues with the same multiplities.

Answer 2

Let $A=[3]$ and $B=[2]$ then both $A$ and $B$ has different eigenvalues but $AB-BA=0$.