Let $H$ be a Hilbert space and $A:H \longrightarrow H$ a symmetric operator, that is, $(Au,v)_H=(u,Av)_H$ for all $u,v \in H$. Then, $\ker(A)^\perp=\operatorname{Im}(A)$.
My attempt:
$``\ker(A)^\perp \supset\operatorname{Im}(A)"$
Let $v \in \operatorname{Im}(A)$, then $v=Au$ for some $u \in H$. Thus, \begin{eqnarray} (v,w)_{H}&=&(Au,w)\\ &=&(u,Aw)\\ &=&0, \end{eqnarray} for all $w \in \ker(A)$. Therefore, $v \in \ker(A)^\perp$.
Unfortunately, I could not prove the other inclusion.