TITLE: Non strictly-singular operators and complemented subspaces
QUESTION [7 upvotes]: If $T$ is a bounded operator which is not strictly singular, acting on a separable Banach space $X$, can one always find an infinite dimensional, closed and complemented, subspace $Y$ such that $T$ restricted to $Y$ is an isomorphism on $Y$?

REPLY [5 votes]: I think that the correct Markus' question concerns operators in L(X).In this case the answer is positive for separable C(K). I do not know what happens in the case of spaces with an unconditional basis.
 Moreover the following seems interesting.
 Let X be a separable reflexive space and T in L(X). Does there exist indecomposable Y containing isomorphically X and S in L(Y) that extends T?
  I also do not know what is the answer if in the previous question if we replace the indecomposable
   Y by the space C[o,1].