$AB$ and $BA$ have same set of eigenvalues.

If $0$ is an eigenvalue of $AB$ then clearly, $\det(AB)=0$, which as a result (without loss of generality) gives $\det(A)=0$. Hence $\det(BA)=0$, as a result $0$ is an eigenvalue of $BA$.

Let $\lambda\neq 0$ be an eigenvalue of $AB$ with $x\neq 0$ and $ABx=\lambda \cdot x$. Let $Bx=y$. If $y=0$ then $\lambda \cdot x=ABx=Ay=0$. Hence $\lambda=0$ or $x=0$, a contradiction. So we have  $y\neq 0$.

$\begin{eqnarray*} BAy &=& BABx\\&=& B\lambda \cdot x \\&=& \lambda Bx\\&=& \lambda BAy. \end{eqnarray*}$

That proves $\lambda$ is an eigenvalue of $BA$ with $y=Bx$ a corresponding eigenvector.

One can prove the same fact in another way as follows, if (without loss of generality) $A$ is invertible.

Note that, $AB = ABAA^{-1}$, which proves $AB$ and $BA$ are similar matrices and any two similar matrices have the same eigenvalues.