$AA'$ and $A' A$ have same set of eigenvalues.

Suppose $\lambda$ is an eigenvalue of $AA'$. Hence
$\begin{eqnarray*} AA' \cdot x &=& \lambda\cdot x\\A'A(A' \cdot x) &=& \lambda A'\cdot x\\ \end{eqnarray*}$
This proves, $\lambda$ is an eigenvalue of $A'A$.