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

  1. Mar 31, 2012 #1
    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
     
    Last edited: Mar 31, 2012
  2. jcsd
  3. Apr 4, 2012 #2

    morphism

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    Hint: Fredholm alternative.
     
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