If $\lambda$ is an eigenvalue of a normal operator $A$, then $\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$ is well-defined and normal as well

Question

Let $H$ be a $\mathbb R$-Hilbert space, $A$ be a normal$^1$ linear operator on $H$ and $\lambda\in\mathbb R$ be an eigenvalue of $A$.

Are we able to show that ${\mathcal N(\lambda-A)}^\perp\subseteq\mathcal D(A)$ and $$A_\lambda:=\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}$$ is normal well?

The claim is easily seen to be true when $\mathcal D(A)=H$ and $A$ is bounded.

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$^1$ i.e. $A$ is densely-defined, $\mathcal D(A)=\mathcal D(A^\ast)$ and $$\left\|Ax\right\|_H=\left\|A^\ast x\right\|_H\;\;\;\text{for all }x\in\mathcal D(A).\tag1$$