The k-th derivative of the resolvent set

Question

I want to prove $$\frac{d^{k}}{dz^{k}}(zI-A)^{-1}=(-1)^{k}k!(zI-A)^{-k-1}$$ I have the resolvent equation $(zI-A)^{-1}-(\lambda I-A)^{1}=(\lambda-z)(zI-A)^{-1}(\lambda I-A)^{-1}$, i.e. $$\begin{aligned}(zI-A)^{-1}&=(\lambda-z)(zI-A)^{-1}(\lambda I-A)^{-1}+(\lambda I-A)^{-1} \\ &=(\lambda I-A)^{-1}[(\lambda-z)(zI-A)^{-1}+1] \\ \end{aligned}$$

$$\implies \frac{d^{k}}{dz^{k}}(zI-A)^{-1}=(\lambda I-A)^{-1}\frac{d^{k}}{dz^{k}}[(\lambda-z)(zI-A)^{-1}]$$ But I don't see how the final equality is derived from this.

Answer 1

The ring generated by $z$, $I$ and $A$ is commutative, thus the derivative of $(zI-A)^{-1}$ can be computed as if $A$ were a scalar.

Or by the geometric/Neumann series $$ ((z+h)I-A)^{-1}=(zI-A)^{-1}(I+h((zI-A)^{-1})^{-1}\\ =-(zI-A)^{-1}\sum_{k=0}^\infty (-h)^k(zI-A)^{-k} $$ from where you can read off the derivatives by comparing with the Taylor series.