TITLE: Olympiad linear algebra problem
QUESTION [6 upvotes]: This is a problem from an olympiad I took today. I tried but couldn't solve it.
Let $A$ and $B$ rectangular matrices with real entries, of dimensions $k\times n$ and $m\times n$ respectively. Prove that if for every $n\times l$ matrix $X$ ($l\in \mathbb{Z}^+$) the equality $AX=0$ implies $BX=0$, then there exists a matrix $C$ such that $B=CA$.
I tried to define a commutative diagram, but failed. Anyhow, I'd prefer a solution that works with the matrices explicitly. But I'll welcome and appreciate any solutions.

REPLY [3 votes]: $AX=0$ means that the rows of $A$ are all orthogonal to the columns of $X$, and likewise for $BX=0$. $B=CA$ means that the rows of $B$ are linear combinations of the rows of $A$.
Assume that there is a row of $B$ that is not a linear combination of the rows of $A$. Orthogonalize that row to the rows of $A$, and choose $X$ as the column vector corresponding to this orthogonalized row. Then $AX=0$, but $BX\neq0$. Hence the rows of $B$ are linear combinations of the rows of $A$.