This question was asked in my quiz of spectral theory and I was not able to make much progress on this question. So, I am asking for help here.
Question : Let H be be a Hilbert Space and let $T\in L(H)$ be auto -adjoint. Show that if $\lambda \in \sigma(T)$ , then $T-\lambda I$ is not surjective.
I am clear with the definitions and related results but still I was not able to make much progress on this question in the class. I tried it again but I am still struck. Can you please give me a couple of hints?
Thanks!
We want to show that $\lambda \in \sigma(T) \implies T - \lambda I \text{ not surjective}$. We can equivalently try to prove the contrapositive: $T - \lambda I \text{ surjective} \implies \lambda \notin \sigma(T)$. Since $\lambda \notin \sigma(T) \iff T-\lambda I \text{ bijective}$ this is equivalent to showing that $T - \lambda I \text{ surjective} \implies T-\lambda I \text{ injective}$.
So, suppose $T- \lambda I$ is surjective but not injective. Then there is some $x \in H$ such that $x \neq 0$ and $(T - \lambda I)x = 0$.
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Thus $\lambda \in \mathbb{R}$, so $T - \lambda I$ is also self-adjoint. Since $T - \lambda I$ is surjective, there is some $y \in H$ such that $(T - \lambda I)y = x$.
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Thus, $\langle x, x \rangle = 0$, so $x = 0$. Contradiction!