Simple question on self-adjoint operators

Question

Let $H$ be a complex Hilbert space, $T\in H'$ and $T=T^*$. Here is where I need help: If $\sigma(T)\subset\{0,1\}$ then $T=T^2$.

Using the spectral theorem I know that $\{0,1\} \supset q(\sigma(P)) = \sigma(q(T))$ for $q:x\mapsto x^2$, thus $\sigma(T^2)=\sigma(T)$. How can I continue?

Answer 1

Or more simple: Consider the polynomial $p(x) = x^2 - x$ and let $S := p(T)$. By the spectral theorem we have $\sigma(S) = p\bigl(\sigma(T)\bigr) = \{0\}$. Also $S$ is selfadjoint since $S^* = p(T)^* = p(T^*) = p(T)$. Hence $$\|S\| = \rho(S) = \max_{\lambda\in \sigma(S)} |\lambda| = 0 $$ So $0 = S = T^2 - T$.