The following question came to me while working on a technical matter about transversality in infinite dimension, and I'm really curious to know whether it has an affirmative answer at least under extra hypotheses.



> 
> Let A be a bounded linear operator on
>  a Banach space E. Does it exist a
>  bounded linear operator S such that 0
>  and 1 do not belong to the same
>  connected component of the spectrum
>  of the operator AS:= A + A(I-A)S?
> 
> 
> 


That is, S is OK if either 0 or 1 is not in the spectrum of AS, or if they both are in the spectrum, they should belong to different connected component of it. Thus it may be assumed that 0 and 1 belong to the same component of spec(A), otherwise S=0 trivially solves the problem.


The first idea is to look for S of the form *f*(A), but this can't work if *f* is continuous,
since then spec(AS) is the continuous image of spec(A) with a map that fixes 0 and 1. However, if A admits a discontinuous functional calculus (e.g. a normal operator on Hilbert space), the trick does work. 


I do not know the answer to the question even on Hilbert spaces. In a general Banach space the problem seems even harder, due the difficulty of building operators.


I'd very grateful of any suggestion! (Pietro Majer).


**edit (17/11/2011).** 


Here are a few more or less trivial facts that I know.


* For a Banach space $X$, the set $\mathcal{A}$ of all $A\in L(X)$ such that there exists $S\in L(X)$ such that no connected component of $\operatorname{spec}(A\_S)$ contains both $0$ and $1$, is an open set;
* If $A\in \mathcal{A}$, then there is $S$ such that $A\_S$ is even a linear projector (thus satifying the condition on the spectrum *ad abundantiam*);
* If $A\in L(X)$ and $A\_S\in \mathcal{A}$ for some $S\in L(X)$, then $A$ itself is in $\mathcal{A}$;
* $A\in \mathcal{A}$ if and only if there are closed subspaces $V$ and $W$ of $X$ such that $V\times W\ni (v,w) \mapsto Av + (I-A)w \in X$ is bijective;
* if $AX$ is a closed subspace and $(I-A)^{-1}(AX)$ is a complemented subspace of $X$, then $A\in \mathcal{A}$.



Perturbations of an operator that disconnect the spectrum 
For Hilbert spaces, the conjecture follows from fact 4 and the answer to question [Complement of a subspace which is a cartesian product](https://mathoverflow.net/questions/81652/complement-of-a-subspace-which-is-a-cartesian-product) applied to the kernel of the map $H\times H\ni (v,w) \mapsto Av + (I-A)w \in X$ . 


