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
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