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Assume $A$ is $m \times n$ and $B$ is $m \times n$.

Is there a connection between the eigenvalues of $AB'$ and the eigenvalues of $B'A$?

One is an $m \times m$ and the other is $n \times n$.

($B'$ stands for the transpose of $B$)

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Have you tried a small case, say, $m=1$, $n=2$, to see what happens? – Gerry Myerson Jan 27 '12 at 22:06
Using block matrices, we can see that $\sigma(AB^t)\setminus \{0\}=\sigma(B^tA)\setminus \{0\}$. We have a relationship of the form $\det(XI_m-AB^t)X^n=\det(XI_n-B^tA)$. – Davide Giraudo Jan 27 '12 at 22:06
@Gerry I am embarrassed to say I have just tried it a bit, and the singular values seem to be the same, like Davide claims, I think, for the general case. Is there a way to see it (for the general case)? – harmonic Jan 27 '12 at 22:09
I missed a $X^m$ in the last comment. – Davide Giraudo Jan 27 '12 at 22:15
up vote 4 down vote accepted

It seems easier for me to assume that $B$ is an $n \times m$ matrix. In that case, a classical argument shows that $AB$ and $BA$ have the same nonzero eigenvalues, not counting multiplicity. The case that these eigenvalues are distinct is dense in the general case, so $AB$ and $BA$ have the same nonzero eigenvalues counting multiplicity. Of course one of them has $|n-m|$ more zero eigenvalues than the other.

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