Proving $(\lambda^{-1}-A^{-1})^{-1}=\lambda-\lambda^2(\lambda-A)^{-1}$

Question

Let $X$ be a Banach space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, let $A\subset X\oplus X$ be a linear relation and let $\lambda\in\mathbb{K}\setminus\{0\}$

I want to prove that $(\lambda^{-1}-A^{-1})^{-1}=\lambda-\lambda^2(\lambda-A)^{-1}$.

Note that $A=\lambda-R(\lambda,A)^{-1}=\lambda-(\lambda -A)$; hence

$\lambda-\lambda^2(\lambda-A)^{-1}=\lambda-\lambda^2(\lambda-\lambda+(\lambda-A))=\lambda-\lambda^2(\lambda-A)$

which would imply that $(\lambda-A)^{-1}=(\lambda -A)$, but that can't be correct. I would appreciate some hints on solving this problem

Answer 1

Factor out $\lambda^{-1}A^{-1}$ and follow your nose: \begin{align} (\lambda^{-1}-A^{-1})^{-1} &= \{ \lambda^{-1}A^{-1}(A-\lambda))^{-1} \\ &= (A-\lambda)^{-1}\lambda A \\ &= (A-\lambda)^{-1}(\lambda (A-\lambda)+\lambda) \\ &= \lambda-\lambda^2(A-\lambda)^{-1}. \end{align}

Answer 2

It works in the reals without commuting multiplication, so it should work here. $$\begin {align}\lambda-\lambda^2(\lambda-A)^{-1} &=[(\lambda^2-\lambda A)-\lambda)^2](\lambda-A)^{-1}\\ &=-\lambda A(\lambda - A)^{-1}\\ &=-(A^{-1}-\lambda^{-1})^{-1}\\ &=(\lambda^{-1}-A^{-1})^{-1}\end {align}$$