The problem bothers me for a long time.


Suppose, we have two matrix $A$ and $B$, where $A$ is a $m$ by $n$ complex matrix while $B$ is a $n$ by $m$ complex matrix.


Apparently, $AB$ and $BA$ have the same non-zero eigenvalues.
However, can we predict the non-zero eigenvalues just based on the information of singular values of $A$, $B$, $AB$ and $BA$.


Is there any relation between the singular values and eigenvalues?



The relation between eigenvalue and singular value of non-symmetric square matrix
We certainly can't predict phases: e.g. multiply $A$ by a complex number $\omega$ with $|\omega|=1$, and you don't change any of the singular values but you multiply the eigenvalues of $A$ and $AB$ by $\omega$.


Consider the case $A = I$, $B$ a unitary $n \times n$ matrix. Then the singular values of $A$, $AB$, $BA$ are all $1$, but the eigenvalues of $B$ could be any $n$ points on the unit circle.


One relation you do have: the product of the eigenvalues of $AB$ (counted by algebraic multiplicity) is the determinant of $AB$, and its absolute value is the product of the singular values of $AB$.


