Kernel of matrices product Let $A$ and $B$ be two matrices s.t. $\operatorname{Ker}\left(A\right)$, and $\operatorname{Ker}\left( B\right)$ are the null spaces of $A$ and $B$ respectively.
What is the $\operatorname{Ker}\left(AB\right)$?
What is the $\operatorname{Ker}\left(A+B\right)$?
What is the $\operatorname{Ker}\left(A\setminus B\right)$?
 A: I assume that $A,B$ are supposed to be square matrices.
I'm not sure what you mean by $A\backslash B$.
In general, we really can't tell anything about the nullspace of a sum of matrices just from knowing the matrices' null spaces. For example, there exist matrices $A$ and $B$ such that $\ker(A)$ and $\ker(B)$ are both trivial, but $\ker(A+B)$ is the whole space. On the other hand, there exist matrices $A,B$ such that all of $\ker(A),\ker(B),\ker(A+B)$ are trivial.
Now, we can do a little better for the kernel of $AB$, by noting that $$x\in\ker(AB)\Longleftrightarrow ABx=0\Longleftrightarrow Bx\in\ker(A).$$
Still, without knowing something more about $B$, we can't say much else.
A: In addition to @Cameron Buie answer, one can say a little more about the dimensions of $\ker(AB)$ and of $\ker(A+B)$:
1) It is not hard to prove that $$\dim\ker(AB)\geq\max\{\dim\ker(A),\dim\ker(B)\}$$ (clearly $\dim\ker(AB)\geq\dim\ker(B)$ since $\ker(B)\subseteq\ker(AB)$. Showing that $\dim\ker(AB)\geq\dim\ker(A)$ requires a little effort).
2) Since for any two matrices $A,B$ we have $\operatorname{rank}(A+B)\leq\operatorname{rank}(A)+\operatorname{rank}(B)$ and $\dim\ker(A)=n-\operatorname{rank}(A)$, one can show that:
$$\dim\ker(A+B)\geq\dim\ker(A)+\dim\ker(B)-n$$
I don't think that without more information one can get any better.
A: For the sake of completeness: One also has
$$
\dim\ker(AB) \le \dim\ker(A)+\dim\ker(B)
$$
since $x\in\ker(AB)$ implies $Bx=y\in\ker(A)\cap\mathrm{ran}(B)$. Hence you need at most $\dim\ker(A)$ basis vectors to represent $y$ and then another $\dim\ker(B)$ more to represent $x$.
