The restriction of a self-adjoint operator $A$ to ${\mathcal N(\lambda-A)}^\perp$ is symmetric, but is it even self-adjoint? Let $H$ be a $\mathbb R$-Hilbert space, $A$ be a densely-defined self-adjoint linear oprator on $H$, $\lambda\in\mathbb R$ and $A_\lambda:=\left.A\right|_{\mathcal D(A)\:\cap\:{\mathcal N(\lambda-A)}^\perp}$. It's easy to see that $$A_\lambda\left(\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp\right)\subseteq{\mathcal N(\lambda-A)}^\perp\tag1$$ and hence $A_\lambda$ is a linear operator on ${\mathcal N(\lambda-A)}^\perp$. Since $\mathcal D(A)$ is dense in $H$, $\mathcal D(A_\lambda)=\mathcal D(A)\cap{\mathcal N(\lambda-A)}^\perp$ is dense in ${\mathcal N(\lambda-A)}^\perp$, i.e. $A_\lambda$ is densely-defined.

Now $$\langle A_\lambda x,y\rangle_H=\langle Ax,y\rangle_H=\langle x,A^\ast y\rangle_H\;\;\;\text{for all }x,y\in\mathcal D(A_\lambda)\tag2$$ and hence $\mathcal D(A_\lambda)\subseteq\mathcal D(A_\lambda^\ast)$ and $\left.A_\lambda^\ast\right|_{\mathcal D(A_\lambda)}=\left.A^\ast\right|_{\mathcal D(A_\lambda)}$, i.e. $A_\lambda$ is symmetric. Are we able to show that $A_\lambda$ is even self-adjoint?

 A: This is true. It might be true under more general assumptions on the space $\mathcal N(\lambda-A)^\perp$, but the fact that this is the complement of an eigenspace makes the relevant calculation very simple here.
Let $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp}^*)$, that is $z\in\mathcal N(\lambda-A)^\perp$ so that $|\langle z , Ax\rangle|≤\|x\|\,C_z$ for all $x\in\mathcal D(A)\cap \mathcal N(\lambda -A)^\perp$. We want to show that $z\in\mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$. This, together with symmetry, shows self-adjointness of $A\lvert_{\mathcal N(\lambda-A)^\perp}$.
Now for any $y\in\mathcal D(A)$ you have $y=x+v$ with $v\in \mathcal D(A)\cap \overline{\mathcal N(\lambda-A)}= \mathcal N(\lambda -A)$, that is $v$ an eigenvector of $A$ to the eigenvalue $\lambda$, and $x\in \mathcal D(A)\cap\mathcal N(\lambda-A)^\perp$. Note that $\|x+v\|^2=\|x^2\|+\|v\|^2$ since $x$ and $v$ are perpendicular. It follows that:
$$\langle z, Ay\rangle = \langle z,Ax\rangle +\lambda\langle z,v\rangle = \langle z,Ax\rangle$$
since $z$ is perpendicular to the eigenvectors to $\lambda$ of $A$. Hence
$$|\langle z, Ay\rangle| ≤ \|x\|\,C_z≤\|y\|\,C_z$$
and $z\in\mathcal D(A^*)$. But $\mathcal D(A^*)=\mathcal D(A)$, hence $z\in \mathcal D(A\lvert_{\mathcal N(\lambda-A)^\perp})$ and $A\lvert_{\mathcal N(\lambda -A)^\perp}$ is self-adjoint.
