What is the resolvent and spectrum of the projection operator? What is the resolvent and spectrum of the projection operator?
Well, i am reading about this subject and for one projection operator $P:X\to X$ ($P^2=P$) the excercise ask me to find the resolvent and spectrum. (The excersice do not say who is $X$. I suppose that $X$ is a Banach space)
My attempt. I note that If $\lambda x=Px$ implies that $P(\lambda x)=P^2(x)=P(x)$ then
$\lambda=1$. Therefore, for me the spectrum is $\{1\}$ and the resolvent is $\mathbb C-\{1\}.$
Am i fine or there is other elements? Please somebody can to help me?  Thank you.
Best
 A: Most projections will have a kernel, so $0$ is also an eigenvalue. So $$\sigma(P)\subset \{0,1\}.$$
All possibilities can occur, though. If $P=0$, then $\sigma(P)=\{0\}$, if $P=I_X$ then $\sigma(P)=\{1\}$, and if $P$ is any other projection then $\sigma(P)=\{0,1\}$.
But for this you need to show that no other $\lambda$ is in the spectrum: , if $\lambda\not\in \{0,1\}$, you can check explicitly that $P-\lambda I$ is invertible by noting that
$$
(P-\lambda I_X)^{-1}=\frac1{\lambda(1-\lambda)}P-\frac1\lambda \,I_X. 
$$
A: If $P^2=P$, then
\begin{align}
    \lambda(\lambda-1)I&=\lambda(\lambda-1)I-P(P-I) \\
          &=\lambda^2I-\lambda I-P^2+P \\
          &=(\lambda^2 I-P^2)-(\lambda I-P) \\
          &=(\lambda I-P)(\lambda I+P)-(\lambda I-P) \\
          &=(\lambda I-P)((\lambda-1)I+P)
\end{align}
Therefore, the resolvent of $P$ is defined for $\lambda\notin\{0,1\}$ by
\begin{align}
          (\lambda I-P)^{-1}=\frac{1}{\lambda(\lambda-1)}((\lambda-1)I+P)
\end{align}
The spectrum can be $\{0\},\{1\},\{0,1\}$.
NOTE: This technique works for any operator $P$ such that $p(P)=0$ for some polynomial $p$, which includes all $n\times n$ matrices because the characteristic polynomial is such a polynomial. But it also covers operators in infinite dimensions that are annihilated by a polynomial $p(\lambda)$, such as a projection. The technique for finding the resolvent $(\lambda I-A)^{-1}$ amounts to looking at the annihilating polynomial $p$ and writing
$$
                p(\lambda)-p(\mu)=(\lambda-\mu)q(\lambda,\mu)
$$
and substituting $\mu=A$, and using $p(A)=0$:
$$
                    p(\lambda)I = (\lambda I-A)q(\lambda,A) \\
               I=(\lambda I-A)\left[\frac{1}{p(\lambda)}q(\lambda,A)\right]=\left[\frac{1}{p(\lambda)}q(\lambda,A)\right](\lambda I-A)
$$
This works for all $\lambda$ for which $p(\lambda)\ne 0$ and gives the two-sided inverse:
$$
             (\lambda I-A)^{-1}=\frac{1}{p(\lambda)}q(\lambda,A).
$$
This also works for all matrices over $\mathbb{C}$ by setting $p$ equal to the characteristic or minimal polynomial.
