Let $X$ be a banach space and $P\in B(X)$ a projection. Show that $\sigma(P) \subset {0,1}$.
In which case $\sigma(P)={0}$? In which case $\sigma(P)={1}$?
The definition of spectrum:
For a banach space $X$ and $T\in B(X)$: $\sigma(T)={\lambda \in F : \lambda*I-T}$ is not invertible}.
A projection $P$ satisfy that $P^2=P$.
I know that there is a theorem regarding Hilbert spaces and a projection yhat belongs to it, which states that if P is a projection then it is also normal and self-adjoint and $||P||=1$. But here my space is banach (not a hipbert space), so how can this works?
If $\lambda$ is neither $0$ nor $1$ then from algebraic calculation (using only the fact that $P^2 = P$) we have $$ \begin{align*} (\lambda I - P)\left(\frac{1}{\lambda} I - \frac{1}{\lambda(1-\lambda)} P\right ) & = I - \frac{1}{(1-\lambda)} P - P\left(\frac{1}{\lambda} I - \frac{1}{\lambda(1-\lambda)} P\right )\\ & = I - \frac{1}{(1-\lambda)} P - \frac{1}{\lambda} P + \frac{1}{\lambda(1-\lambda)} P \\ & = I, \end{align*} $$ because $1/(x(1-x)) = 1/x + 1/(1-x)$ for any complex $x$ that is neither $0$ nor $1$. A similar computation (or just the observation that $I$ and $P$ commute with $I$ and $P$) shows that the same operator is a left inverse of $\lambda I - P$, so that $\lambda I - P$ is invertible for all $\lambda \not \in \{0, 1\}$.
This shows that if $P^2 = P$ then $\sigma(P) \subseteq \{0,1\}$. We have not used the completeness of $X$ here, just the operator identity $P^2 = P$.
If there are no vectors $v$ for which $Pv$ is nonzero then $\sigma(P) = \{0\}$, clearly; $P$ is just the zero operator.
If there is a vector $v$ for which $Pv$ is nonzero (note that $v$ is necessarily nonzero), then $P(Pv) = Pv$ shows that $I-P$ is not invertible (because there is a nonzero vector $Pv$ in its kernel), so that $1 \in \sigma(P)$.
If there is a vector $v$ for which $Pv \neq v$ then $(I - P)v$ is a nonzero vector in the kernel of $P$ and hence $0 \in \sigma(P)$.
In sum:
The spectrum of the zero operator is $\{0\}$.
If the projection is not zero then $1$ will be in the spectrum of $P$.
$0$ will also be in the spectrum of $P$ unless $Pv = v$ holds for all $v$, in which case $P = I$.