$n$ by $n$ matrices such that $Ax=0$ implies $Bx=0$, then what can we say about $A$ and $B$? Ok, this is one of the interesting questions I encountered in my GRE Maths exam 2 weeks ago. Fix an $x\in \mathbb{R}^n$ (doesn't have to be $0$), and $A$, $B$ are $n$ by $n$ matrices such that $Ax=0$ implies $Bx=0$, then what can we say about $A$ and $B$? I think the choices were: $\exists C$ an $n$ by $n$ matrices such that: 
a) $A=BC$, b) $A=CB$, c) $B=CA$, d) $B=AC$, e) $B=C^{-1}AC$.
Well, the set of matrices $D$ such that $Dx=0$ form an left ideal, but I only know what bi-ideals look like in this case. Also the wording the this question is funny, we dont know if $A$ is in the ideal or not, we just know if $A$ is in the ideal then $B$ is in the ideal.
Edit: As we discussed below, I think I misunderstood the question in the exam (although I don't have the actually paper now). $x$ is probably not fixed.

(c) is the answer. This is a standard result. 
We have $Ker A \subseteq Ker B$.  Then we may define $C$ on $Im A$ by:
$$
C: Im A \rightarrow \mathbb{R}^n, \ \ C( Ax) = Bx.
$$
This is well defined because $Ax=Ay$ implies $Bx=By$. 
We may extend $C$ to the whole $\mathbb{R}^n$.  Then we have $B=CA$ for all $x\in\mathbb{R}^n$. 
