Non square matrices and spectrum of $AB$

Question

There is a known theorem (true in the general context of Banach algebras) that if $a,b \in A$ (where $A$ is a Banach algebra) then $\sigma(ab) \cup \{0\}=\sigma(ba) \cup \{0\}$ where $\sigma$ is the spectrum. I wonder whether the same result is true in the specific context of finite dimensional spaces, i.e. $a,b$ are matrices but we don't assume that they are square matrices. We only assume that $a,b$ are such that $ab$ and $ba$ make sense, say $a$ is $m \times n$ and $b$ is $n \times m$ matrix.

Answer 1

Yes. In fact, if $A$ is $m \times n$ and $B$ is $n \times m$, $m \ge n$, the characteristic polynomial of $AB$ is $\lambda^{m-n}$ times the characteristic polynomial of $BA$.