Is there a matrix $X$ possible such that $AXB=O$?

Question

Let $A$ be a $9\times4$ matrix and $B$ be a $7\times3$ matrix. Is there a $4\times 7$ matrix $X$ possible such that $X\neq O$ and $AXB=O$?

My approach: I tried to prove that there is a nonzero solution to $AX=O$ by proving that the number of pivots in $A$ is less than $4$ but that depends on the matrix $A$ so I can't figure it out.

Answer 1

Consider 28 entries of matrix X as variable and then put AXB =0. So we have 27 equations and 28 variable.Thus there is a non trivial solution.

Answer 2

It might be helpful to think about it this way: in the expression $$ AXB, $$

we have a composition of three maps. First, $B$ is a map from $3$-dimensional space ($\mathbb{R}^3$) to $7$-dimensional space ($\mathbb{R}^7$). That means the output of $B$, the range, must only be at most a $3$-dimensional subspace of $\mathbb{R}^7$ (in particular there are lots of "unused" dimensions in $\mathbb{R}^7$ that don't occur in the output of $B$). Next, $X$ is a map from $\mathbb{R}^7$ to $\mathbb{R}^4$, and $A$ is a map from $\mathbb{R}^4$ into $\mathbb{R}^9$.

In order to force the whole product to be $0$, what should $X$ be? It should send the output of $B$ to $0$, as that way any vector that we apply $AXB$ to will end up $0$ ($v$ gets sent to $Bv$, which then gets sent to $0$, and $A$ applied to $0$ is still $0$.) But the output of $B$ is at most $3$ dimensions out of $7$ that we have to work with; so we can have $X$ do something else on the other $4$ dimensions.

Therefore, the answer is yes, this is possible with $X \ne 0$: just pick $X$ so that it sends the output of $B$ to $0$, but sends the other $4$ dimensions to something nonzero.