is there a connection between the following? Assume $A$ is $m \times n$ and $B$ is $m \times n$.
Is there a connection between the eigenvalues of $AB'$ and the eigenvalues of $B'A$?
One is an $m \times m$ and the other is $n \times n$.
($B'$ stands for the transpose of $B$)

It seems easier for me to assume that $B$ is an $n \times m$ matrix. In that case, a classical argument shows that $AB$ and $BA$ have the same nonzero eigenvalues, not counting multiplicity. The case that these eigenvalues are distinct is dense in the general case, so $AB$ and $BA$ have the same nonzero eigenvalues counting multiplicity. Of course one of them has $|n-m|$ more zero eigenvalues than the other. 
