How to prove $(P-\lambda)^{-1} = \frac{1}{\lambda(1 - \lambda)}P - \frac{1}{\lambda}I$?

Question

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.

Answer 1

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 $\lambda$.

Answer 2

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:

1. The first way is just to multiply your proposed inverse by $P-\lambda I$ and show you get the identity.
2. 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.)
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.

Answer 3

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