If $\lambda$ is an eigenvalue of an operator $A$, then $\left(\lambda-\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}\right)^\ast$ is injective

Question

Let $H$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H)$, $\sigma(A)\subseteq\mathbb R$ denote the spectrum of $A$, $\lambda\in\sigma(A)$ be an eigenvalue of $A$ and $$A_\lambda:=\left.A\right|_{{\mathcal N(\lambda-A)}^\perp}.$$

Note that $\lambda-A_\lambda$ is injective. In the comments below this answer it is claimed that $(\lambda-A_\lambda)^\ast$ is injective as well.

EDIT: I doubt that the claim is true in general. However, if $A$ is normal, then $A_\lambda$ should be normal as well and we know that for any normal $B\in\mathcal L(H)$ we've got $\mathcal N(B^\ast)=\mathcal N(B)$. Do we need this assumption?

Remark: The desired goal is to show $\mathcal{R}(\lambda - A_\lambda)^\perp=\{0\}$.

Answer 1

This is false in general. For instance, let $B:H_0\to H_0$ be any operator that is injective but whose image is not dense (say, $B$ could be an isometry to a proper subspace of $H_0$). Let $H=H_0\oplus\mathbb{R}$ and let $A:H\to H$ be $B$ on the first coordinate and $0$ on the second. Then $\lambda=0$ is an eigenvalue of $A$, and $A_0$ is just $B$. But $B^*$ is not injective since the image of $B$ is not dense.