Let $T$ be a bounded normal operator acting from a Hilbert space $H$ to itself. I must check that if the spectrum of $T$ lies in the set defined in the title, then $T$ is an othogonal projection. I know that $T$ is an orthogonal projection iff $T=T^*=T^2$.
I have also at my disposable multiple forms of the spectral theorem, which may come in handy. I must admit though I am still trying to grasp all the notions that are usually used in the statements of the theorem. For instance, I have seen that $T$ can be written as $$T=\int_{\sigma(T)}zE(dz)$$ where $E$ represents a spectral measure, i.e. a map from a sigma algebra to the space of projections on $H$. Are these facts useful to our goal? Any hints are welcome, thank you.
Best Answer
The form of the Spectral Theorem that you wrote gives you $$ T=\int_{\sigma(T)}z\,dE(z)=0\,E(\{0\})+1\,E(\{1\})=E(\{1\}). $$ Thus $T=E(\{1\})$, a projection.
You can also conclude without the spectral theorem. The operator $S=T^2-T$ has spectrum $$ \sigma(S)=\{\lambda^2-\lambda:\ \lambda\in\{0,1\}\}=\{0\}. $$ So $S$ is normal with spectrum $\{0\}$. It is not hard to see that a normal operator satisfies $\|S\|=\operatorname{spr}(S)$. Thus $\|S\|=0$ and $S=0$. That is, $T^2=T$. Being normal with real spectrum also implies that $T$ is selfadjoint.