Do $AB$ and $BA$ have the same eigenvalues? Suppose $A$ and $B$ are $n\times n$ matrices and they are both invertible. Can we say that $AB$ and $BA$ have the same eigenvalues?

A more general result:
If $A$ and $B$ are $n \times n$ matrices, then we have for a scalar $\lambda \ne 0$:
$ \lambda$ is an eigenvalue of $AB$ iff $ \lambda$ is an eigenvalue of $BA$ .
Proof: if $ \lambda$ is an eigenvalue of $AB$ , then there is $x \ne 0$ such that 
$(*)$  $ABx= \lambda x$. 
Let $y:=Bx$. Then $y \ne 0$  (otherwise we would get from $(*)$ that $ \lambda =0$ or $x=0$).
Now we have
$$ BAy=BABx=B(ABx)=B( \lambda x)=\lambda Bx = \lambda y.$$
It follows that $\lambda$ is an eigenvalue of $BA$.
