Spectrum of an unbounded operator Consider a densely defined unbounded operator $A_0:D(A_0)(\subset{H})\to H$ which has the following properties:
1- Symmetric, $\langle A_0x,y\rangle=\langle x,A_0y \rangle$
2- Positive, $\langle A_0x,x\rangle \ge0$
3- $A_0$ is surjective and $A_0^{-1}$ is a compact operator
I am wondering whether these are sufficient ground for saying that:
$A_0$ admits an infinite set of eigenvalues which are positive and strictly increasing; furthermore, the corresponding eigenfunctions form an orthonormal basis of $H$.  
 A: Yes, it is enough. Because $A_0$ is positive, symmetric and surjective, then $A_0$ is densely-defined, injective and selfadjoint. Therefore, $A_0^{-1}$ is compact, selfadjoint with trivial null space. So $A_{0}^{-1}$ has an orthnormal basis of eigenfunctions $\{e_n \}$ with corresponding eigenvalue sequence of positive numbers
$$
       \lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \cdots \ge \lambda_n \ge \cdots \rightarrow 0
$$
Thus $A_0e_n = \lambda_n^{-1}e_n$ and $\lambda_n^{-1}\rightarrow\infty$.
A: You are implicitly assuming that $A_0$ is injective. 
Let $y\in H$. Then there exists $x\in D(A_0)$ with $y=A_0x$. Then
$$
\langle A_0^{-1}y,y\rangle=\langle x,A_0x\rangle\geq0. 
$$
Thus $A_0^{-1}$ is positive, and so there exists an orthonormal basis $\{e_n\}$ of eigenvectors. Since $A_0^{-1}$ is compact, its eigenvalues $\lambda_n$ satisfy $\lambda_n\searrow0$. 
Also, from $A_0^{-1}e_n=\lambda_ne_n$ and $A_0^{-1}$ bijective $H\to D(A_0)$, we get that $e_n\in D(A_0)$ and that $\lambda_n>0$ for all $n$. So 
$$
A_0e_n=\frac1{\lambda_n}\,e_n,
$$
and $\lambda_n\nearrow\infty$. 
