An idempotent bounded linear operator has eigenvalues $0,1$ I am thinking of the following problem: suppose $T$ is an idempotent bounded linear operator on a Banach space $X$ over the complex field. Of course, suppose $T$ is not zero map or identity map to avoid triviality. I have shown that every complex number not equal to $0$ or $1$ are all regular values of $T$; however I found it extremely difficult to show they are both spectral values. Of course if $X$ is finite dimensional, then $T$ is a projection and the conclusion is trivial. Also by the spectral theorem, either $0$ or $1$ must be a spectral value. 
 A: We have
$T^2 = T \tag{1}$
by idempotence;  thus
$(T - I)T = T(T -I) = T^2 - T = 0; \tag{2}$
by hypothesis, $T \ne 0$;  thus there exists $x \in X$ with $Tx \ne 0$; and thus by (2),
$(T - I)Tx = 0, \tag{3}$
which, since $Tx \ne 0$, affirms that $Tx$ is an eigenvector of $T$ corresponding to eigenvalue $1$.
Likewise $T \ne I$ affirms that $T -I \ne 0$; thus again we have
$(T - I)x \ne 0 \tag{4}$
for some $x \in X$; then again via (2),
$T(T - I)x = 0; \tag{5}$
(4) and (5) imply $(T - I)x$ is an eigenvector associated with $0$; we have shown both $0$ and $1$ are eigenvalues of $T$.  $0$ and $1$ being eigenvalues, neither $T - 1I = T - I$ nor $T -0 = T$ are invertible since each annihilates a non-zero vector; $0$ and $1$ are both thus spectral values of $T$.  QED.
A: They're not necessarily both spectral values. $I$ has spectrum $\{1\}$ and $0$ has spectrum $\{0\}$.
Every $x \in X$ can be written as $x = (I-T)x+Tx$. And, $y=(I-T)x$, $z=Tx$ satisfy $Ty=0$ and $(T-I)z=0$. If $T$ does not have a non-zero vector with eigenvalue $0$, then $(I-T)x=0$ for all $x$, which means $T=I$. If $T$ does not have a non-zero vector with eigenvalue $1$, then $Tx=0$ for all $x$, which means $T=0$.
