# Neumann Series

So, I have a gut feeling this result might be important enough to remember.

Let X be a Banach space, a space where every Cauchy sequence converges, and let L(X) be the set of all bounded operators from X to X.

If we have an operator $T\in L(X)$ and suppose $||I-T||<1$, then we can say a couple of interesting stuff about the operator, with my favourite being that then T is invertible. That isn’t so difficult to see.

Supposing that T is not invertible is equivalent to saying that there is an $x\in X$ such that $||x||=1$ and $Tx=0$. But then,

$||Ix-Tx||\leq ||I -T||\cdot ||x||\Rightarrow 1\leq ||I-T||$

which is a contradiction to what we have already said above. What was a surprise for me (and I think that it must have greater implications for some other area of mathematics) is that then we can write $T^{-1}$ as a series. Yes, we can write it as

$T^{-1}=\sum_{n=0}^{+\infty}(I-T)^{n}$

Again, the proof is fairly simple. We just need to show that

$T\left (\sum_{n=0}^{+\infty}(I-T)^{n}\right)=I$

or

$T\left (\sum_{n=0}^{N}(I-T)^{n}\right)\to I$, as $N\to \infty$

Here’s how it’s done.

$T\left (\sum_{n=0}^{N}(I-T)^{n}\right)=T\left (I + (I-T) + (I-T)^2 +\ldots +(I-T)^N\right)=T\left (\frac{I-(I-T)^{N+1}}{T}\right)$

And since $||(I-T)^{N}||\leq ||(I-T)||^{N}<1$ for all N, we deduce that

$\frac{I-(I-T)^{N+1}}{T}\to \frac{1}{T}$

We can also deduce that the set of all invertible operators is open, since if $||S-T||<||S^{-1}||^{-1}$, we can use the idea here to show it.

This way, you can also built approximations to the inverse of a matrix (being a representation of a linear operator and all that).  If A is invertible, then we can find a B such that $||B||<1$ and $B=c\cdot A,\ c\in\mathbb{R}^+$, so that doesn’t pose a problem.

Then an approximation to the inverse of the matrix is

$B^{-1}\simeq I+(I-B)$

I guess I will revisit this fact as soon as I learn more about the spectrum of an operator.

Ref : wikipedia