Non strictly singular operators Let $X$ be a separable Banach space and let $T:X\to X$ be a bounded operator that is not strictly singular. Can we always find an infinite dimensional, closed, and complemented subspace $Y$ of $X$ such that $T:Y\to T(Y)$ is an isomorphism?
I cannot show this is true, and I cannot think of a counterexample. The only idea of counterexample I had was looking at HI spaces, where every operator is of the form $\lambda I+S$, where $S$ is strictly singular. However, as far as I know, such operators (when $\lambda\neq 0$) are isomorphisms on a closed, finite-codimensional subspace, thus complemented. 

This was answered in the negative in this post at MathOverflow. 
See the first answer there and Bill Johnson's comment to the second answer.
