Show that $(\lambda A+B)^{-1}B=1-\lambda\left((\lambda A+B)^{-1}A\right)$

Question

Let $A,B\in\mathbb C^{n\times n}$ and suppose there exists $\lambda\in\mathbb C$ such that $\lambda A+B$ invertible. How does one deduce that:

$$(\lambda A+B)^{-1}B=1-\lambda\left((\lambda A+B)^{-1}A\right)?$$

I feel that this should be quite easy, and no doubt it is - my apprehension is that $(\lambda A+B)^{-1}$ need not have an immediately obvious form.

Answer 1

Begin with the simple observation that

$B = (\lambda A + B) - \lambda A \tag 1$

and, exploiting the given hypothesis that $\lambda A + B$ is invertible, multiply through by $(\lambda A + B)^{-1}$:

$(\lambda A + B)^{-1}B$ $= (\lambda A + B)^{-1} (\lambda A + B) - (\lambda A + B)^{-1} \lambda A$ $= I - \lambda ((\lambda A + B)^{-1} A), \tag 2$

thus obtaining the requisite result.

Answer 2

$$(\lambda A+B)^{-1}B=I-\lambda\left((\lambda A+B)^{-1}A\right)$$ $$\iff(\lambda A+B)^{-1}B+\lambda\left((\lambda A+B)^{-1}A\right)=I$$ $$\iff(\lambda A+B)^{-1}(B+\lambda A)=I$$ which is true simce $\lambda A+B$ was presumed invertible.