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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?

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Let $E$ be the spectral measure of $T$, then we have, as $\lambda^2 = \lambda$ for $\lambda \in \sigma(T)$, that $T^2 = \int_{\sigma(T)} \lambda^2\, dE_\lambda = \int_{\sigma(T)}\lambda\, dE_\lambda = T$. – martini Dec 1 '12 at 10:37

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 $0 = S = T^2 - T$.

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the last equation should be $T^2 - T$ not $T^2 = T$. – mland Dec 1 '12 at 12:22
@mland Thx. Corrected it. – martini Dec 1 '12 at 12:32

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