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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 $A^{\top} B$.

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    Hi kloop! I have merged your other account into your current account. If you have trouble logging in, or if you accidentally create duplicate accounts, simply flag one of your own questions for moderator attention, and we will help out. Registering your account should help to avoid problems.2012-01-29

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