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 and suppose , 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 such that and . But then,

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 as a series. Yes, we can write it as

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

or

, as

Here’s how it’s done.

And since for all N, we deduce that

We can also deduce that the set of all invertible operators is open, since if , 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 and , so that doesn’t pose a problem.

Then an approximation to the inverse of the matrix is

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

Ref : wikipedia

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