Let
- $H$ be a $\mathbb R$-Hilbert space
- $A$ be a compact and self-adjoint bounded linear operator on $H$
- $I:=\left\{n\in\mathbb N:n\le\operatorname{rank}A\right\}$
By the Hilbert-Schmidt theorem, there is a $(\lambda_i)_{i\in I}\subseteq\mathbb R\setminus\left\{0\right\}$ with $$Ae_i=\lambda_ie_i\;\;\;\text{for all }i\in I$$ for some orthonormal basis $(e_i)_{i\in I}$ of $\overline{AH}$.
Now, if $A$ is nonnegative (i.e. $\langle Ax,x\rangle_H\ge0$ for all $x\in H$) and has finite trace, why can we conclude that $\overline{AH}=H$?