# Showing that $\lambda$ is an eigenvalue iff $\mathbb{1}_{\left\{\lambda\right\}} (T)$ is non-zero

Let $$T$$ be a bounded self-adjoint operator. I already showed that the operator $$\mathbb{1}_{U} (T)$$ is an orthogonal projection, where $$\mathbb{1}_{U}$$ is the characteristic function of the Borel set $$U \subset \mathbb{R}$$. (To make sense of this, one can think of $$\mathbb{1}_U(T)$$ as restricted to the spectrum $$\sigma(T)$$.)

However, now I wish to show that $$\mathbb{1}_{\left\{\lambda \right\}} (T)$$ is non-zero if and only if $$\lambda$$ is an eigenvalue of $$T$$.

My attempt is this: I think we have the identity $$(x- \lambda) \mathbb{1}_{\left\{ \lambda \right\}}(x) = 0$$ for all $$x \in U$$. Then applying the star homomorphism $$\Phi$$ to this, we get $$(T - \lambda) \mathbb{1}_{ \left\{ \lambda \right\} } (T) = 0.$$ Does this prove that $$\lambda$$ is an eigenvalue of $$T$$?

Yes. If $$1_{\{\lambda\}}(T)\ne0$$, there exists nonzero $$v\in H$$ with $$1_{\{\lambda\}}(T)v=v$$. Then $$Tv=T1_{\{\lambda\}}(T)v=\lambda 1_{\{\lambda\}}(T)v=\lambda v,$$ so $$\lambda$$ is an eigenvalue. Conversely, if $$\lambda$$ is an eigenvalue then $$Tv=\lambda v$$ for some nonzero $$v$$, and so you can deduce that $$f(T)v=f(\lambda)v$$ for all bounded Borel functions $$f$$ (start with monomials, then polynomials, then take limits). So $$1_{\{\lambda\}}(T)v=1_{\{\lambda\}}(\lambda)v=v.$$