How to prove $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$? I have a question regarding Functional Analysis:
Assume that $X$ is a Banach space and that $P \in B(X)$ is a projection with $ran P \neq {[0]}$.
Then I have easily shown that $\lambda = 0$ and $\lambda = 1$ are eigenvalues of $P$.
I have troubles proving that if $\lambda \notin\{0,1\}$, then it follows that we have $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$.
Any help would be grateful.
 A: You can derive the inverse for $|\lambda| > 1$ using a geometric series:
$$
       (\lambda I-P)^{-1}=\frac{1}{\lambda}(I-\frac{1}{\lambda}P)^{-1}=\frac{1}{\lambda}\sum_{n=0}^{\infty}\frac{1}{\lambda^n}P^n=\frac{1}{\lambda}(I+\sum_{n=1}^{\infty}\frac{1}{\lambda^n}P) \\
       = \frac{1}{\lambda}(I+\frac{1/\lambda}{1-1/\lambda}P) = \frac{1}{\lambda}I+\frac{1}{\lambda(\lambda-1)}P.
$$
Both sides are holomorphic in $\lambda$ except at $\lambda=0,1$. So the expressions are equal for all such $\lambda$.
A: There are a few ways to get this answer depending on how lazy you want to be about showing it. In order of decreasing laziness:

*

*The first way is just to multiply your proposed inverse by $P-\lambda I$ and show you get the identity.

*The second way is to note that since $P^2 = P$, it stands to reason that the inverse should be of the form $aI + bP$ and then solve $(aI + bP)(P-\lambda I) = I$ for $a$ and $b$. (This is informed by 3.)

*The third way is to write $(P - \lambda I)^{-1}$ as a geometric series. You have some cases to consider here based on the nature of $\lambda$ (i.e. $|\lambda|>1$, $|\lambda| < 1$, and $|\lambda| = 1$). If $|\lambda| > 1$, you'll need to factor out $\lambda^{-1}$ in that term and then expand to ensure convergence. The conditional convergence for $|\lambda| = 1$ is a bit tricky. Note that in the series expansion for the inverse, $P^n = P$ for $n\ge 1$ which informs the ansatz in 2.

A: $$\begin{aligned}
(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I\quad
&\Leftrightarrow\quad I = (P-\lambda)\left(\frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I\right)\\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}P^2 - \frac{1}{\lambda}P-\frac{1}{(1 - \lambda)}P \\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}P^2 - \frac{1}{\lambda(1 - \lambda)}P \\
&\Leftrightarrow\quad 0 = \frac{1}{\lambda(1 - \lambda)}\left(P^2 - P\right)\\
&\Leftrightarrow\quad P^2=P \\
\end{aligned}$$
