Let $A$ and $B$ matrices of dimensions $d \times n$ and $n \times d$ respectively. We know that the non-zero eigenvalues of $AB$ and $BA$ are the same.
Is there any connection between the top $m$ singular values of AB' and B'A when performing SVD decomposition on $AB$ and $BA$? ($m$ is the number of non-zero eigenvalues of AB' or $B'A$)
EDIT: What I wrote up there is a simplification of the following problem that I need to solve:
Same $A$ and $B$. We know that $C = BA^{\top}$.
We also know that $C = I \mathrm{diag}(\gamma) J$ for some matrices $I$ and $J$ and vector $\gamma$ of length $m$, $m < \min(d,n)$.
Using only the matrix $A^{\top}B$ (and not $BA^{\top}$), I want to find $U$ and $V$ of dimensions $m \times d$ such that $U I$ and $V J$ are invertible and $U A$ and $V B$ can be computed, or at least $V B A^{\top} U^{\top}$ is computable. You can apply any decomposition or extract any information you need from $A^{\top} B$.