To simplify the series of matrices

Question

Let $A$ be a square matrix in the form $A=B+O(h^2)$, where $B$ is a fixed matrix, and $O(h^2)$ is a matrix with very small elements. Assume that: $$(I-A)^{-1}=I+A+A^2+A^3+...$$ How can I esimate the right hand side of the equality above ? Is that right $(I-B)^{-1}+O(h^2) ?$

Answer 1

Edit: the below answer is incorrect - the binomial theorem doesn't apply because $O(h^2)$ doesn't necessarily commute past $B$.

A rather slapdash (i.e. non-rigourous) way of doing it is just to substitute $A = B + O(h^2)$ in the right-hand side, and notice that by the binomial theorem we have $$(B+O(h^2))^n = B^n +nO(h^2)B^{n-1} = B^n + O(h^2)$$

This gives you $I + B + B^2 + \ldots + O(h^2) = (I-B)^{-1} + O(h^2)$, as you say.

Answer 2

Let $A=B-K$, where $K$ is a small variable matrix. Then, by the Taylor's formula $(I-A)^{-1}=(I-B+K)^{-1}=(I-B)^{-1}-(I-B)^{-1}K(I-B)^{-1}+O(||K||^2)$. If $||K||=O(h^2)$, then $||(I-B)^{-1}K(I-B)^{-1}||=O(h^2)$ and we are done.