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Prove: For T Compact, left or right invertible implies invertible 
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#1
Mar3112, 03:05 PM

P: 62

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(IT)=I[/itex] if and only if [itex](IT)S=I[/itex] The question also asks to show that either of these equalities implies that [itex]I(IT)^{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(IT)=I\Rightarrow SST=I \Rightarrow S=I+ST[/itex] [itex]I(IT)^{1} = IS = I(I+ST) = ST [/itex] And ST is compact since T is compact 


#2
Apr412, 12:01 PM

Sci Advisor
HW Helper
P: 2,020

Hint: Fredholm alternative.



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