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.



Equality of spectra of products of operators
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 $\dim E\_\lambda(A^\*A)= \dim E\_\lambda(AA^\*)$. 


