Computing the resolvent of a rank one projector

Question

I'm reading through this paper in dynamical systems, that's using a bit of perturbation theory and operator theory. The authors make the following claim:

Now, since $P_0$ is a projection of rank one, it is easy to compute explicitly its resolvent, namely $$ R_0(ζ) = (P_0 − ζ)^{−1} = \frac{P_0}{1−ζ} − \frac{I−P_0}{ζ} $$

I have no background in operator theory, but I've skimmed through Kato's Perturbation Theory for Linear Operators and done a quick Google search and it's not at all obvious to me where this formula comes from. Does anyone have a source for the resolvent of a rank one projection?

Answer 1

A geometric series expansions gives you what you want, at least for $|\zeta| > 1$: \begin{align} (P_0-\zeta I)^{-1} &=\left(\zeta\left(\frac{1}{\zeta}P_0-I\right)\right)^{-1} \\ &= -\frac{1}{\zeta}\left(I-\frac{1}{\zeta}P_0\right)^{-1} \\ &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}P_0+\frac{1}{\zeta^2}P_0^2+\frac{1}{\zeta^3}P_0^3+\cdots\right) \\ &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}\left(1+\frac{1}{\zeta}+\frac{1}{\zeta^2}+\cdots \right)P_0\right) \\ &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta}\frac{1}{1-1/\zeta}P_0\right) \\ &= -\frac{1}{\zeta}\left(I+\frac{1}{\zeta-1}P_0\right) \\ &=-\frac{1}{\zeta}I-\left(\frac{1}{\zeta-1}-\frac{1}{\zeta}\right)P_0 \\ &=-\frac{1}{\zeta}(I-P_0)-\frac{1}{\zeta-1}P_0. \end{align} Then you can check that the right-hand side is the inverse for all $\zeta\ne 0,1$.

Answer 2

For nonzero $a,b$, $$(aP + b(I-P))^{-1} = \frac{P}{a} + \frac{I-P}{b} .$$ Setting $a = 1-\zeta$ and $b = -\zeta$ gives the result.