# Geometric interpretation of spectra equality of matrix product

Following this question, let $A$ and $B$ be square $n\times n$ matrices. It is a well known fact (e.g., 12.3 at [1]) that $AB$ and $BA$ have the same eigenvalues (spectra):

The proof is trivial for $\lambda=0$ in case it is an eigenvalue. If $v$ is an eigenvector of $AB$ which belongs to $\lambda\neq 0$ then $A(Bv)=\lambda v\neq\vec{0}$, and so $Bv\neq \vec{0}$. It is then easy to see that $Bv$ is an eigenvector of $BA$ which belongs to the very same $\lambda$.

While the proof is straightforward, why should one expect this to be true? I'd be glad to have a geometric explanation to this fact.

[1] Golan, Jonathan S., The linear algebra. A beginning graduate student ought to know, Dordrecht: Springer (ISBN 978-94-007-2635-2/pbk; 978-94-007-2636-9/ebook). xii, 497 p. (2012). ZBL1237.15002.

• In the case that $A$ is invertible, the operation $A^{-1}(AB)A = BA$ can be interpreted geometrically as a change of basis. This doesn't quite account for the case in which both matrices are singular, though. – Omnomnomnom Oct 9 '17 at 13:18

It is easy to see that $AB$ and $BA$ have the same non-zero eigenvalues. Say $ABv = \lambda v$, with $v\ne 0$. Then $B(AB)v = \lambda Bv$. Assume that $\lambda \ne 0$. Then from the first equation we get $Bv \ne 0$. So now we got $Bv$ a $\lambda$ eigenvector for $BA$. This trick unfortunately does not account for multiplicities.
• Yes, that's exactly how Golan proves that $AB$ and $BA$ have the same spectra. But that does not answer my questions. – Shlomi A Oct 9 '17 at 20:33