Expanding $(I+A)^{-1}$

Question

I think that there is an expansion of $(I+A)^{-1}$ when $A$ is a matrix with a norm smaller than 1, but I cannot seem to recall this expansion. Any suggestions?

http://en.wikipedia.org/wiki/Neumann_series – Nikolaj-K Mar 21 '13 at 13:45 — Nikolaj-K, Mar 21 '13 at 13:45
Do you know how to example $(I-A)^{-1}$? – Thomas Andrews Mar 21 '13 at 13:46 — Thomas Andrews, Mar 21 '13 at 13:46
A related problem. – Mhenni Benghorbal Mar 21 '13 at 15:07 — Mhenni Benghorbal, Mar 21 '13 at 15:07

score 3 · Accepted Answer · answered Mar 21 '13 at 14:03

Theorem: If $|X|<1$, then matrix $X=I+A$ is Invertible. And $X^{-1}=1-A+A^2-A^3+...$ .

Proof:

Step 1. Let's Prove that the series is convergent when $|X|<1$.

(using the inequalities $|A+B|\leq|A|+|B|$ and $|AB|\leq|A||B|$)

$$|A^m-A^{m+1}+A^{m+2}-...\pm A^{m+k-1}|\leq|A^m|\cdot|1+|A|+...+|A^{k-1}|=|A^m|\frac{1-|X|^k}{1-|X|}$$

So sequence of partial sums is Cauchy sequence when $|X|<1$ so the series converges.

Step 2 $$XX^{-1}=(1+A)(1-A+A^2-A^3+...)=1-X+X^2-X^3+...+X-X^2+X^3-...=1$$

$\blacksquare$

score 2 · Answer 2 · answered Mar 21 '13 at 13:50

2

It is $I-A+A^2-A^3.....=\sum_{i=0}^{\infty}(-A)^i$

It is just like the expansion of $(1+x)^{-1}$ with $|x|<1$

answered Mar 21 '13 at 13:50

Abhra Abir Kundu

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Expanding $(I+A)^{-1}$

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