fa.functional analysis - rules for operator commutativity?

One obvious but important observation is that, for operators on a $n$-dimensional vector space over a field, if $1 < n < infty$, we have $AB neq BA$ generically. In other words, consider the commutativity locus $mathcal{C}_n$ of all pairs of $n times n$ matrices $A,B$ such that $AB = BA$ as a subset of $mathbb{A}^{n^2}$. This is clearly a Zariski closed set -- i.e., defined by the vanshing of polynomial equations. It is also proper: take e.g.
$A = left[ begin{array}{cc} 1 & 1 \ 0 & 1 end{array} right] oplus 0_{n-2}$ and $B =
left[ begin{array}{cc} 0 & 1 \ 0 & 1 end{array} right] oplus 0_{n-2}$. Since $mathbb{A}^{n^2}$ is an irreducible variety, $mathcal{C}_N$ therefore has dimension less than $N^2$. This implies that over a field like $mathbb{R}$ or $mathbb{C}$ where such things make sense, $mathcal{C}_N$ has measure zero, thus giving a precise meaning to the idea that two matrices, taken at random, will not commute.

One could ask for more information about the subvariety $mathcal{C}_N$: what is its dimension? is it irreducible? and so forth. (Surely someone here knows the answers.)

I would guess it is also true that for a Banach space $E$ (over any locally compact, nondiscrete field $k$, say) of dimension $> 1$, the locus $mathcal{C}_E$ of all commuting pairs of bounded linear operators is meager (in the sense of Baire category) in the space $B(E,E) times B(E,E)$ of all pairs of bounded linear operators on $E$.

Kevin Buzzard has enunciated a principle that without further constraints, the optimal answer to a question "What is a necessary and sufficient condition for $X$ to hold?" is simply "X". This seems quite applicable here: I don't think you'll find a necessary and sufficient condition for two linear operators to commute which is nearly as simple and transparent as the beautiful identity $AB = BA$.

Still, you could ask for useful sufficient conditions. Diagonalizable operators with the same eigenspaces, as mentioned by Jonas Meyer above, is one. Another is that if $A$ and $B$ are both polynomials in the same operator $C$: this shows up for instance in the Jordan decomposition.