Let $A$ and $B$ be $m \times n$ and $n \times m$ real matrices.
I was asked to prove that if $\lambda$ is a nonzero eigenvalue of the $m \times m$ matrix $AB$ then it is also an eigenvalue of the $n \times n$ matrix $BA$ (I didn't have trouble showing that)
I need now to find an example showing that this need not be true if $\lambda = 0$
On
If $$ A = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right) \quad B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right) $$ then $$ A B = \left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \quad B A = \left( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right) $$ and $$ \mbox{eig}(A B) = ( 0, 1 , 1) $$ which includes the eigenvalue $0$, but $$ \mbox{eig}(B A) = ( 1, 1 ) $$ has only $1$ as eigenvalue.
Take $$ A= \left( \begin{matrix} 1 \\ 1 \end{matrix} \right) \ \ \ \ \ \ \ \ \mbox{ and } \ \ \ \ \ \ \ \ B= ( \begin{matrix} 1 & 1 \end{matrix} ) $$