Comput Spectrum of Idempotent

Question

Let A be a unital banach algebra and a in A if a is idmepotent and a do not equal to 0 and 1

then

the spectrum of a = {0,1}??

Answer 1

Hint: $a^2 - a = 0$ and the spectral mapping theorem strongly restrict the possible members of the spectrum of $a$.

Answer 2

Since $a = a^{2}$, by induction it can be shown that $a = a^{n}$ for all $n \in \mathbb{N}$. Then the spectral radius formula yields $$spr(a) = \lim\limits_{n \to \infty}\| a^{n} \|^{\frac{1}{n}} = \lim\limits_{n \to \infty} \| a \|^{\frac{1}{n}} = 1$$

Hence $\sigma(a) \subset B[0,1]$.

From the spectral mapping theorem for polynomials we get that $$\sigma(a) = \sigma(a^{n}) = (\sigma(a))^{n} ~~~~\forall~n $$

Hence since $\sigma(a) \subset B[0,1] \implies \sigma(a)= \{0,1\}$.