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 $B^{\top} A$.

EDIT: I managed to refine the question.

Let $\sigma(D)$ be the non-zero eigenvalues of a square matrix $D$ and let $s(C)$ be the singular values of a matrix $C$.

We know that if $C = AB^{\top}$, then:

$s^2(C) = \sigma(C C^{\top}) = \sigma(AB^{\top} B A^{\top}) = \sigma(A^{\top} A B^{\top} B)$

I am also assuming that I can compute $A^{\top} A$ and $B^{\top} B$, which means that $s^2(C)$ is computable.

I think $U$ and $V$ that I am looking for could come from the right and left singular vectors of $C$, but I don't know how to compute these singular vectors based on $A^{\top} A$, $B^{\top} B$ and $B^{\top} A$. Worst even, even if I had these $U$ and $V$, it is not clear how to compute $UA$ and $VB$ without explicitly representing $A$ and $B$.