Non strictly-singular operators and complemented subspaces 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 $Y$?
 A: Thanks for the email, Markus.
Let’s agree that “space” means “infinite dimensional Banach space” so that subspaces are always infinite dimensional.
A Banach space $X$ is decomposable if it is the direct sum of two subspaces; in other words, if there is a (bounded, linear) projection $P$ on $X$ s.t. $PX$ and $(I-P)X$ are both infinite dimensional. The first indecomposable Banach space was constructed by Gowers and Maurey; in fact, their space is hereditarily indecomposable. Now we know that indecomposable spaces are very common; see [AFHORSZ] and references therein. In particular, $\ell_p$, $1<p<\infty$, is a subspace of a separable indecomposable space. 
For an example that gives a negative answer to Markus’ problem, take an indecomposable space $X$ that contains a decomposable subspace $Y$ ($Y$ can be a Hilbert space). Take a projection $P$ on $Y$ that has infinite dimensional range and infinite dimensional kernel.  Extend $P$ to an operator $T$ from $X$ into some injective space that contains $X$.  $T$ is obviously not strictly singular since $T$ is the identity on $PY$. Also,  the kernel of $T$, being infinite dimensional, has infinite dimensional intersection with  every finite codimensional subspace.  But since $X$ is indecomposable, all complemented subspaces of $X$ are finite codimensional.
I could not have answered this natural and basic (though I never thought of it until reading this post) question a few years ago.
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[AFHORSZ] Argyros, S. A.(GR-ATHN); Freeman, D.(1-TX); Haydon, R.(4-OXBR); Odell, E.(1-TX); Raikoftsalis, Th.(GR-ATHN); Schlumprecht, Th.(1-TXAM); Zisimopoulou, D.(GR-ATHN)
Embedding uniformly convex spaces into spaces with very few operators. (English summary) 
J. Funct. Anal. 262 (2012), no. 3, 825–849. 
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EDIT: After I posted this, user19038 gave a reference in a comment above that shows that the OP's question was raised by Vitali Milman in a 1970 paper and solved in the linked paper.  The example involves only classical spaces; it is the inclusion mapping from $L \log^\lambda L$ into $L_1$ with $\lambda < 1/2$.
A: 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]. 
