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


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

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


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