Adjoint of a symmetric operator acting on the complement of the graph Let $A:\mathcal D(A)\to\mathcal H$ be a closed symmetric (densely defined) operator on a Hilbert space. In an online lecture the lecturer remarks that the following is trivial:
Viewing $\mathrm{Gr}(A)$ as a subspace of the graph of $A^*$, the map $\sigma : \mathrm{Gr}(A)^\perp\to\mathrm{Gr}(A)^\perp, x\mapsto -iA^*x$ is a self-adjoint involution.
How can one see this? I can see that if $x$ lies in the orthogonal complement of the graph of $A$ (as a subspace of $\mathrm{Gr}(A^*)$ ) that then $x$ also lies in $\mathcal D(A^*)$. But I do not see any further steps, which is dispiriting given the remark that this is supposedly trivial.
The context is that this step was apparently an important insight by von Neumann about the existence of self-adjoint extensions of symmetric operators.

The orthogonal complement of $\mathcal{G}(A)$ in $\mathcal{G}(A^*)$ consists of all $(y,A^*y)\in\mathcal{G}(A^*)$ such that
$$
         \langle (x,Ax),(y,A^*y)\rangle_{X\times X} = 0, \;\;\; x\in\mathcal{D}(A) \\
          (x,y)_{X}+(Ax,A^*y)_{X} = 0,\;\;\; x\in\mathcal{D}(A) \\
           (Ax,A^*y)_{X} = -(x,y)_{X},\;\;\; x\in\mathcal{D}(A).
$$
The last relation is equivalent to $A^*y\in\mathcal{D}(A^*)$ and $(A^*)^2y=-y$. That is, $y\in\mathcal{D}((A^*)^2)$ and $(A^*)^2y=-y$. So,
$$
           \mathcal{G}(A)^{\perp}\cap\mathcal{G}(A^*)=\{ (y,A^*y)\in X\times X : y\in\mathcal{N}((A^*)^2+I)\}.
$$
You can see how applying $A^*$ to both coordinates of $(y,A^*y)\in\mathcal{G}(A)^{\perp}\cap\mathcal{G}(A^*)$ defines to $J$ such that $J^2=-I$:
$$
          J(y,A^*y) = (A^*y, (A^*)^2y)=(A^*y,-y) \\
           J(A^*y,-y) = ((A^*)^2y,-A^*y) = (-y,-A^*y)=-(y,A^*y) \\
          \therefore J^2 = -I.
$$
You can check that $iJ$ is then selfadjoint.
