NullSpace of AB if $[A,B]=0$ and $AB\neq 0$? Edited question.
Is it true that for $x\notin\mbox{null}(A)\cup\mbox{null}(B)$, then $ABx\neq0$? Alternately, for what x is it true that $ABx\neq0$?
Thanks!
 A: If $A=B$ are nilpotent but $A^2\neq0$, then you conditions are satisfied but $\ker(A)=\ker(B)\neq\ker(AB)=\ker(A^2)$. So any $x\in\ker(A^2)\setminus\ker(A)$ will show your first guess wrong. Even for commuting $A,B$ there is way to express $\ker(AB)=\ker(BA)$ in terms of other kernels.
A: First of all, referring to the comments, it is not true in general that NS$(AB)=$NS($A)\cup$ NS($B$) (NS referring to null space here). A simple counter example: \begin{equation} A=\begin{bmatrix}2 &1\\2&1\end{bmatrix},\; B=\begin{bmatrix}1 &2\\3&4\end{bmatrix},\;x=\begin{bmatrix}1 \\-2\end{bmatrix}.\end{equation}
You can verify that $x \in$ NS($A$), but $x \notin$ NS($AB$). 
Ok - so in terms of a characterization of NS($AB$) in terms of NS($A$) and NS($B$): note that $Bx$ is in the column space (I will indicate this as CS) of $B$. So if 


*

*CS($B)\cap$ NS($A)=\{0\}$ we have NS($AB)=$ NS($B$).

*CS($B)\cap$ NS($A)=$ CS($B$) we have $ABx=0$ for all $x$.


So these are the two extreme cases and there is then a range of cases in-between.
edit: ok, so I first then concluded that maybe (CS($B)\cap$ NS($A)) \cup$ NS($B)=$ NS($AB$), but as pointed out by Marc, this is also wrong since the union of subspaces is not in general a subspace (basic mistake!). My answer is based on general $A$ and $B$ and not on commuting matrices. Sorry I misunderstood the question, but hopefully there is at least some contribution in my answer. 
