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

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SUMMARY

The discussion centers on proving that for a compact operator T in a Banach space X, if a bounded linear transformation S is either left or right invertible, then S is invertible. The key equations explored are S(I-T) = I and (I-T)S = I, demonstrating that either condition leads to the conclusion that I - (I-T)^{-1} is compact. The use of the Fredholm alternative theorem is suggested as a critical tool in establishing the compactness of the operator ST.

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Homework Statement


[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.

Homework Equations


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
 
Last edited:
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Hint: Fredholm alternative.
 

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