Spectrum of idempotent element

Question

Let $A$ be some unitary algebra over $\mathbb{C}$. If $a^2=a$ and $0\ne a\ne 1$ then $\{0,1\}\subset \sigma_A(a)$ ($\sigma_A(a)$ is the spectrum if $a$). I believe that also $\sigma_A(a)\subset \{0,1\}$. For examlpe, if $A=\mathbb{C}\times\mathbb{C}\times...\times\mathbb{C}$ with pointwise multiplication then each idempotent is $(\epsilon_1,\epsilon_2,...,\epsilon_n)$ where $\epsilon_i\in \{0,1\}$ and $(\epsilon_1-\lambda,\epsilon_2-\lambda,...,\epsilon_n-\lambda)$ is invertible iff $\epsilon_i\ne \lambda$ i.e. $\lambda\notin \{0,1\}$. Also, if $K\subset \mathbb{R}^2$ is a compact set then the space of continuous functions $A=C(K,\mathbb{C})$ is also a $\mathbb{C}$-algebra and $f$ is idempotent iff $f(K)\subset \{0,1\}$. But $\sigma_A(f)=f(K)$.
First, I am stuck on to find some easy trick to prove that the spectrum of an idempotent element (not $0$ and not $1$) is $\{0,1\}$. Secondly, this statement is easy for $\mathbb{C}^S$ (this is the algebra of all functions and $\sigma(f)=f(S)$), perhaps there is a nice way to reduce our problem to the case of $\mathbb{C}^S$?

Answer 1

The following argument works for any idempotent $a$ (not necessarily a projection, i.e., not necessarily selfadjoint), and shows that $\sigma(a)\subset\{0,1\}$.

Suppose that $\lambda\not\in\{0,1\}$. Then $$a(a-\lambda)=a^2-\lambda a=a-\lambda a=(1-\lambda)a. $$ Similarly, $$ (1-a)(a-\lambda)=(1-a)a-(1-a)\lambda=-(1-a)\lambda. $$ Let $$b=\frac1{1-\lambda}\,a-\frac1\lambda\,(1-a).$$ Then $$ b(a-\lambda)=ba-\lambda b=\frac1{1-\lambda}\,a-\frac\lambda{1-\lambda}\,a+1-a=1, $$ and one can also check that $(a-\lambda)b=1$. So $a-\lambda$ is invertible, which shows that $\lambda\not\in\sigma(a)$.