I want to know why the following claim is true: Let $A$ be a $C^*$-algebra. $p\in A$ is a projection (that means $p^2=p^*=p$) iff p is normal and $\sigma (p)\subseteq \{0,1\}$.
"$\implies$" why p normal, it is clear. I know that $\sigma(p)\subseteq [-1,1]$, since $\sigma(p)\subseteq [-\|p\|,\|p\|]$ and $p$ projection, therefore: $\|p\|=1$. But its not enough. (If $A$ is a vector space and $p$ an endomorphism, then i know how to prove it, i consider $p^2-p=p(p-1)$ ). Maybe i have to use the continuous functional calculus, but i dont know how. Can anyone help me? And can you give me a hint how to do "$\impliedby$" ? Regards
Edit: Maybe i could use $f(x)=x^2-x$ and functional calculus to prove "$\impliedby$". But I don't know how to do this in detail.
If $p$ is normal and $\sigma(p)\subseteq \{ 0, 1\}$, then the spectral mapping theory gives $\sigma(p(p-1))=\{0\}$. Because $p$ is normal, then the spectral radius and the norm are the same, which gives $p(p-1)=0$ or $p^{2}=p$. A normal operator with spectrum $\sigma(p)\subset\mathbb{R}$ must be selfadjoint.
Okay, the opposite direction. If $p^{2}=p^{\star}=p$, then $p$ is normal because it is selfadjont. To show that $\sigma(p)\subseteq\{0,1\}$, you can directly determine the resolvent for $\lambda \notin \{0,1\}$. Write $$ \lambda 1- p = \lambda(1-p+p)-p= \lambda(1-p)+(\lambda-1) p. $$ Then the inverse is easily spotted because $p(1-p)=0$, $p^{2}=p$, and $(1-p)^{2}=1-p$: $$ (\lambda 1-p)^{-1} = \frac{1}{\lambda}(1-p)+\frac{1}{\lambda-1}p. $$ So $\sigma(p)\subseteq\{0,1\}$ because $(\lambda 1-p)^{-1}$ exists for all other $\lambda$.