I'm currently reading something on unbounded operators and their spectra and the author notes that the nonzero spectra of $AA^*$ and $A^*A$ are equal. $A$ here is unbounded and densely defined (on $L^2$, if that helps), but not self-adjoint. The only other useful point might be that $AA^*$ and $A^*A$ differ by a constant. This for me does not seem to help however; if $\lambda$ belongs to the resolvent of $AA^*$, i.e. $AA^* - \lambda$ is invertible, and $A^*A = AA^* + c$ where $c \neq 0$ is the constant, then $A^*A - (c + \lambda) = AA^* - \lambda$ is invertible, so $c + \lambda$ belongs to the resolvent of $A^*A$. But I guess if this was to be a proof I would expect $\lambda \in \sigma(AA^*) \setminus \{0\}$ if and only if $\lambda \in (A^*A) \setminus \{0\}$?
On the other hand, some finite dimensional tests with real matrices in Mathematica seem to suggest this result is true (i.e. that the eigenvalues of $MM^t$ are equal to the eigenvalues of $M^tM$, but the eigenvectors are not the same). I can't see why however? Do I need more hypotheses for this to be true? If not can someone provide a hint or proof? (Especially if someone can explain why zero is of particular importance too...)