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If $A$ is an $r$ x $s$ matrix and B is an $s$ x $r$ matrix.

$E_\mu(C)$ is the eigenspace of square matrix C with eigenvalue $\mu ≠ 0$.

Proof:

$dimE_\mu(AB) = dimE_\mu(BA)$

I imagine one would need to find the basis of eigenvectors for both spaces, but I'm not sure how...

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Hint: If $v$ is an eigenvector for $AB$, then $Bv$ is an ...

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    $\begingroup$ Eigenvector of $BA$. Does that imply that the spaces have the same number of eigenvectors? $\endgroup$ – user79572 May 26 '13 at 19:20
  • $\begingroup$ I still don't understand... sigh $\endgroup$ – user79572 May 26 '13 at 19:58
  • $\begingroup$ You have an invertible linear map taking the eigenspace for one onto the eigenspace for the other. Such maps take a basis to a basis, and thus preserve dimension. $\endgroup$ – Robert Israel May 27 '13 at 3:10

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