$n$ by $n$ matrices such that $Ax=0$ implies $Bx=0$, then what can we say about $A$ and $B$?

Question

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.

Answer 1

(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 $x\in\mathbb{R}^n$.

Answer 2

Correct answer: (c)

$Ax=0$ is equivalent to "$x$ is perpendicular to all the rows of $A$". Hence, the statement $$ \forall x\in\mathbb R^n (Ax=0\Longrightarrow Bx=0) $$ tells us that the space spanned by the rows of $B$ is a subspace of the space spanned by the rows of $A$. Hence, each row $b_i$, $i=1,\ldots,n$, the of $B$ is a linear combination of the rows of $A$. Thus, $b_i=c_iA$, where $c_i$ is a vector of the same dimension.

Therefore, $B=CA$, where $C$ is the matrix with rows $c_1,\ldots,c_n$.