Does the matrix equation $XA = XB$ always have a solution?

Question

Let $A,B$ be two $n \times m$ matrices with entries in $\mathbb{R}$. Is it possible to always find a $m \times n$ matrix $X$ such that

$$XA = XB$$

Answer 1

$$XA=XB\to X(A-B)=0$$

So, you must have $X_{m\times n}$ matrix where the line space is orthogonal to the column space of $A-B$.

The line space of $X$ is inside $\Bbb R^n$, and we have $m$ vectors in it. The same is true for the column space of $A-B$.

So we have that $\dim[\text{column space } (A-B)]\le m$.

So it is only possible to find such line space if we have

$$n-\dim[\text{column space } (A-B)]\ge m$$

If that happen we can find a set of $m$ vectors in $\Bbb R^n$ such that it is orthogonal to the column space of $A-B$.