Finding idempotent elements of a $C^{*}$-algebra Let $T \in B(\mathcal{H})$ (bounded operators on a complex Hilbert space $\mathcal{H}$) and suppose $T$ is normal. Suppose furthermore $\sigma(T)=\{-1\}\cup[2,3]$.
Then I want to find a complete list of all idempotent elements of $\mathcal{A}=C^{*}(I,T)$ (the $C^*$-algebra generated by $I$ and $T$) and which of these corresponds to eigenvalues of $T$.

I am interested in finding out what the method is for a problem like this. I suspect it has something to do with the "Spectral Theorem for normal bounded operators" and from this theorem I have already shown that isolated points of the spectrum are eigenvalues. So $-1$ is a an eigenvalue but it need not be the only one.
Clearly $I$ is one idempotent element. But I struggle to find a systematic way of identifying the others.
Any help is appreciated!
 A: Since the spectrum is real the operator is self -adjoint. The point $-1$ is an eigenvalue. The $C^*$-algebra generated by $T$ and $I$ is isometrically isomorphic to $C(\sigma (T)).$ Therefore the are exactly $2$ nontrivial idempotents.  Namely, the orthogonal projection $P$ onto the eigenspace corresponding to $-1$ and $I-P.$ The operator $T$ may have more eigenvalues, even a dense subset of $[2,3],$ but the projections on the corresponding eigenspaces do not belong to the $C^*$-algebra.
For example, assume that $2\le \lambda\le 3 $ is an eigenvalue of $T$ and let $P_\lambda$ denote the orthogonal projection on the corresponding eigenspace. We thus have
$TP_\lambda=\lambda P_\lambda.$ Assume, by contradiction that $P_\lambda$ belongs to the $C^*$-algebra generated by $T$ and $I.$ Then $P_\lambda=f(T)$ for a continuous function on $\sigma(T)$ and $Tf(T)=\lambda f(T).$ Thus
$(T-\lambda I)f(T)=0.$ This implies $(x-\lambda)f(x)=0$ for $x\in \sigma(T).$ Thus $f(x)=0$ for $x\neq \lambda.$ Hence $f=0,$ a contradiction.
Concerning the number $-1,$ let $p_n(x)$ denote the sequence of polynomials such that $p_n(x)$ tends uniformly to $0$ for $2\le x\le 3$ and $p_n(-1)\to 1.$ The existence is guarranteed by the Weierstrass Theorem, although we can give the formula explicitly
$$p_n(x)={(3-2x)^n\over 5^n}.$$ Then $p_n(T) $ is convergent to an operator $P.$  Moreover  $P=P^*$ and $P^2=P.$  Let $Tv=-v.$ Then
$$Pv=\lim_np_n(T)v=\lim_np_n(-1)v= v.$$ On the other hand $$TP=\lim_nTp_n(T)=-P$$
Hence $P$ is the orthogonal projection onto the eigenspace corresponding to $-1.$
