TITLE: Can spectrum "specify" an operator?
QUESTION [22 upvotes]: Given a bounded operator $A$ on a Banach space $X$, one may find the spectrum $\sigma(A)\subset{\bf C}$. 
Here are my questions:


Given some set in the complex plane, say, $S\subset{\bf C}$, can one find an operator $A$ such that $\sigma(A)=S$? 
Is there a "big picture" for this kind of questions?

REPLY [6 votes]: In the case of Hilbert spaces, there is something like a big picture concerning related questions. Briefly this goes as follows:
let $X$ be a compact subset of the complex plane $\mathbb{C}$. 
Denote by $\mathrm{Ext}(X)$ the set of equivalence classes of essentially normal operators with spectrum equal to $X$.
Then this set $\mathrm{Ext}(X)$ is an abelian group and moreover one model for the $K$-Homology of $X$, the homology theory dual to Atiyah-Hirzebruch $K$-Theory. All of this is part of the socalled Brown Douglas Filmore Theorem which was one of the starting points of $K$-Homology.
So classifying certain operators (essentially normal ones up to equivalence) with fixed spectrum $X \subset \mathbb{C}$ has a connection to classical algebraic topology 
A great reference for this is the book "Analytic K-Homology" by Nigel Higson and John Roe.