Prove: For T Compact, left or right invertible implies invertible

CornMuffin

1. The problem statement, all variables and given/known data
[itex]X[/itex] is a Banach space
[itex]S\in B(X)[/itex] (Bounded linear transformation from X to X)
[itex]T\in K(X)[/itex] (Compact bounded linear transformation from X to X)

[itex]S(I-T)=I[/itex] if and only if [itex](I-T)S=I[/itex]




The question also asks to show that either of these equalities implies that [itex]I-(I-T)^{-1}[/itex] is compact.



2. Relevant equations



3. The attempt at a solution
I have tried using the adjoint, cause S is invertible if and only if S* is invertible. but that didn't get me anywhere.

If there happens to be a theorem that says ST = TS, then it would be easy, but i couldn't find anything like that.


For the second part:
[itex]S(I-T)=I\Rightarrow S-ST=I \Rightarrow S=I+ST[/itex]
[itex]I-(I-T)^{-1} = I-S = I-(I+ST) = ST [/itex]
And ST is compact since T is compact

morphism

Hint: Fredholm alternative.