Spectrum in Hilbert space

Question

Let $H$ be a Hilbert space and $T\in B(H)$ be normal. Want to show that if $\lambda$ is an isoloated point in $\sigma(T)$ then $\lambda$ is an eigenvalue of $T$.

I have no idea how to do this one. should I try contradiciton? Please help. Thank you.

Answer 1

Note that $C^* (T) \cong C (\sigma(T))$, where $T$ corresponds to the identity function $id(z) =z$ on $\sigma(T)$. Define $f \in C(\sigma(T))$ as $f(z) = z - \lambda$. It corresponds to $K = T - \lambda$ in $C^*(T)$.

Now consider the non-zero function $g$ on $\sigma(T)$ where $g(z) = 1$ if $z =\lambda$ and $g(z) =0$ otherwise. Since $g \in C(\sigma(T)$, it corresponds to a non-zero element $L\in C^*(T)$.

Note that $KL =0$.