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 A

_{S}:= A + A(I-A)S?

That is, S is OK if either 0 or 1 is not in the spectrum of A_{S}, 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(A_{S}) 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 $A\in \mathcal{A}$.

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