Removing points from the discrete spectrum of a self-adjoint operator

Question

I'm currently following the chapter 6 of Gerald Teschl's book "Mathematical Methods in Quantum Mechanics" and got stuck in something that seems irrelevant but is killing me anyway.

As a example of how inestable is the discrete spectrum, it is stated in the book that: Given an self-adjoint operator $A$ and $\lambda_0\in \sigma_{dis}(A)$, we can easily remove this eigenvalue with a finite rank perturbation of arbitrarily small norm. In fact, consider $$A+\varepsilon P_{\{\lambda_0\}}(A).$$

While it is easy to see that such perturbation is of finite rank and its norm is arbitrarily small, understanding why $\lambda_0$ can not be an eigenvalue anymore is not clear at all.

I'd appreciate any help with this.

Answer 1

Let $V$ be the eigenspace of $A$ for eigenvalue $\lambda_0$. If $v \in V$, then $(A + \epsilon P_{\lambda_0}(A)) v = (\lambda_0 + \epsilon) v$. An eigenvector $w$ of $A + \epsilon P_{\lambda_0}(A)$ for any eigenvalue other than $\lambda_0 + \epsilon$ must be orthogonal to $V$, so $(A + \epsilon P_{\lambda_0}(A)) w = A w$, and thus $w$ would be an eigenvector of $A$ for some eigenvalue other than $\lambda_0$.