Let $A$ be a linear operator between two Hilbert spaces. Let $A^*$ be its adjoint.
Question. Under what conditions the non-zero spectra of $A^*A$ and $AA^*$ coincide counting multiplicities?
In my situation $A$ is an elliptic differential operator between two complex line bundles over a compact smooth manifold.
Remark. If all Hilbert spaces are finite dimensional then the result is true.
If $A:C^\infty(M,F_1)\to C^\infty(M,F_2)$ is an elliptic operator on a closed manifold, then both $A^*A$ and $AA^*$ are self-adjoint elliptic operators. Hence they have discrete spectrum. For $\lambda\in \sigma(A^*A)$ let $E_\lambda(A^*A)$ and $E_\lambda(AA^*)$ denote the eigenspaces of $A^*A$ and $AA^*$ with eigenvalue $\lambda$. Then $L^2(M,F_1)=\oplus_{\lambda\in \sigma(A^*A)}E_\lambda(A^*A)$ and a similar decomposition holds for $L^2(M,F_2)$. Now if $v\in E_\lambda(A^*A)$ then $A^*Av=\lambda v$ and, hence, $$AA^*(Av)= A(A^*Av)= A(\lambda v)= \lambda Av.$$ Hence, $Av\in E_\lambda(AA^*)$. If $\lambda\not=0$ then $A^*Av\not=0$ which implies $Av\not=0$. Thus $A$ defines an injective map $A:E_\lambda(A^*A)\to E_\lambda(AA^*)$. Similarly, the map $A^*:E_\lambda(AA^*)\to E_\lambda(A^*A)$ is injective. It follows that $\dim E_\lambda(A^*A)= \dim E_\lambda(AA^*)$.
For $\lambda\neq 0$, it is not difficult to show that $A^*A-\lambda I$ bijective if and only if so is $AA^*-\lambda I$. Moreover the kernels of $A^*A-\lambda I$ and $AA^*-\lambda I$ have the same dimensión, etc.
This is true for bounded operators between Banach spaces $A:X\to Y$ and $B:Y\to X$. If $AB-I$ is invertible, then $B(AB-I)^{-1}A+I$ is an inverse of $BA-I$.
Using $(I-BA)Bx =B(I-AB)x$ you get that the kernels of $I-AB$ and $I-BA$ have the same dimension.
There are more relations between the properties of $BA-\lambda I$ and $AB-\lambda I$, and they are not difficult to prove. I can put references tomorrow.
