We have a matrix $A_{m \times n}$. I know that the two matrices $A^*A$ and $AA^*$ share nonzero eigenvalues. This comes from multiplying $A^*Ax=\lambda x$ on both sides by $A$ ($\lambda$ is an eigenvalue of $A^*A$, and $x$ is an eigenvector of $A^*A$). Then, we have:
$$AA^*(Ax)=\lambda (Ax)$$
So now $\lambda$ is also an eigenvalue of $AA^*$, but now the eigenvector is $Ax$ (for $x \notin KerA$, so that $Ax\neq0$).
All other $\lambda$ not common to both matrices $AA^*$ and $A^*A$ is equal to zero.
My question is: when will the multiplicity of the zero eigenvalues be the same for both matrices $(AA^*)_{m\times m}$ and $(A^*A)_{n\times n}$?
Intuition tells me that this will only happen when both matrices $AA^*$ and $A^*A$ have the same dimensions, which implies $A_{m \times n}$ being square $(m=n)$. Otherwise, one matrix will always have more eigenvalues than the other (or more repetitions of eigenvalues, so higher multiplicity), the extra eigenvalues being equal to zero.
Is this correct? and how would you prove this more rigorously?