Show that if $AX=BX$ not necessarily $A=B$ without using an inverse matrix

Question

I'm trying to solve this problem:

If $A, B$ and $X$ are squared matrices with $X\neq 0$ and $AX=BX$, not necessarily $A=B$. What is the condition for A = B?

I know that X must be non-singular because

$$ AXX^{-1}=BXX^{-1} $$ $$ \Rightarrow AI=BI \Rightarrow A=B.$$

My question is: Is there another way to show this?

Answer 1

You have successfully argued that if $X$ is non-singular, then $A=B.$ However, you have not yet ruled out the possibility that there exists a singular $X$ and yet it is still the case that $A$ must equal $B$

$AX = BX\\ (A-B)X = 0$

If $X$ is singular it has a non-trivial left null space. There exist matrices $A,B$ with $A\ne B$ such that $(A-B)X = \mathbf 0$

If X is non-singular the left null-space is trivial and it must be the case that $A=B$

Answer 2

I would suggest something along these lines:

matrix multiplication is basically a set of inner products for each element, which in your case results in $\bf a_i \dot x_j = b_i \dot x_j$ While the equivalence of the dot product itself does not imply that $\bf a_i = b_i$ For a single $\bf x_j$, see if you can make an argument that if it must hold for arbitrary $j$, then $\bf a_i = b_i$ if $X$ is not the null matrix.