# is there a connection between singular vectors of $AB$ and $BA$?

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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kloop: Is the user who edited the question just now the same as the one who posted it? If yes, then I can flag the moderators to merge the two accounts. – Srivatsan Jan 28 '12 at 16:39
yes, that's the same user. thanks! – harmonic Jan 28 '12 at 19:43
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. – Zev Chonoles Jan 29 '12 at 5:24