Let $A$ and $B$ matrices of dimensions $d \times n$. We know that the non-zero eigenvalues of $AB^{\top}$ and $B^{\top} A$ are the same. Let $m$ be the number of non-zero eigenvalues.

Is there any connection between the top $m$ singular vectors of $AB^{\top}$ and $B^{\top} A$ when performing singular value decomposition on $AB^{\top}$ and $B^{\top} A$?

This is a simplification question of the following problem that I would like to solve:

Same $A$ and $B$. Let $C = AB^{\top}$.

We also know that $C = I \mathrm{diag}(\gamma) J$ for some matrices $I$ and $J$ and vector $\gamma$ of length $m$, $m < \min(d,n)$ (i.e. $I$ is of dimension $d \times m$ and $J$ is of dimension $m \times d$).

Using *only* the matrix $B^{\top}A$ (and not $AB^{\top}$, $I$, $J$ or $\gamma$), 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 calculated. You can apply any decomposition or extract any information you need from $B^{\top} A$.