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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 $T^2 = \int_{\sigma(T)} \lambda^2\, dE_\lambda = \int_{\sigma(T)}\lambda\, dE_\lambda = T$.2012-12-01

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