In Banach algebra $A$ if $a^2=a$ show that $\sigma_A(a)\subseteq\{0,1\}$

Question

Let $A$ be a Banach algebra with identity $e$ and let $x\in A $ be such that $a^2=a$ show that $\sigma_A(a)\subseteq\{0,1\}$ and compute the resolvet function $$R(a, \lambda)=(\lambda e-a)^{-1}$$

Any ideas or insight would be greatly appreciated

Answer 1

If $\lambda \neq 0,1$ then $\frac 1 {\lambda} (e+\frac a {\lambda -1})$ is the inverse of $\lambda e-a$. [To arrive at this you can pretend that the Banach algebra is the real line and write $\frac 1 {a-\lambda}=\frac 1 {\lambda} (a/\lambda -1)^{-1}$ and use the expansion $\frac 1 {1-x} =\sum x^{n}$. Of cousre this proves nothing but it helps us you to guess what the inverse of $ae-\lambda$ is likely to be].

Answer 2

Try to solve $(\lambda e-a)^{-1}=re+sa$. We get $$e=(\lambda e-a)(re+sa)=\lambda re+(\lambda s-r-s)a.$$ So we need $\lambda r=1$. So we get $r$ when $\lambda\ne0$. Also we need $0=-r+(\lambda-1)s$ and we get $s$ when $\lambda\ne1$.