I need to prove that a self-adjoint operator $T \in \mathcal{B}(\mathcal{H})$ is an orthogonal projection if $\sigma(T) = \{0,1\}$. I know this means I have to prove $T$ is idempotent, meaning $T = T^2$ (unless there's an equivalent definition I don't know about). Here's what I know:
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$T = T^*$.
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$T$, $I – T$, and $T – T^2$ are not invertible.
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Since $T$ is self-adjoint and therefore normal, it follows that the above operators are not bounded below.
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$r(T) = ||T|| = 1$.
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$\langle T^2x,x \rangle = \langle Tx,Tx \rangle = ||T||^2$.
I've been fiddling with these facts trying to get the pieces to fit together, but they're not clicking. What am I not seeing? I need a hint.
Best Answer
If $T$ is selfadjoint and its spectrum is $\{0,1\}$, then by the Spectral Mapping Theorem $$ \sigma(T-T^2)=\{\lambda-\lambda^2:\ \lambda\in\{0,1\}\}=\{0\}. $$ So $T-T^2$ is a selfadjoint operator with spectrum $\{0\}$. Then $$\|T-T^2\|=\operatorname{r}(T-T^2)=0.$$ and so $T-T^2=0$.