Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?
If it's not true, then under what conditions is it true or not true?
If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn & Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.
A trivial proof would be OK, but a citation is better.

Notice that $\lambda$ being an eigenvalue of $AB$ implies that $\det(AB-\lambda I)=0$ which implies that $$\det(A^{-1})\det(AB-\lambda I)\det(B^{-1})=0=\det(A^{-1}(AB-\lambda I)B^{-1})=\det((B-\lambda A^{-1})B^{-1}) $$ $$=\det(I-\lambda A^{-1}B^{-1}) = 0.$$
This further implies that $$\det(BA)\det(I-\lambda A^{-1}B^{-1})=\det(BA(I-\lambda A^{-1}B^{-1}))=\det(BA-\lambda I)=0,$$
i.e., $\lambda$ is an eigenvalue of $BA$.
This proof holds only for invertible matrices $A$ and $B$ though. For singular matrices you can show that $0$ is a common eigenvalue, but I can't think of a way to show that the rest of the eigenvalues are equal.  
