Suppose that $a \in \mathbb F^n$ is nonzero and $V = M_{m \times n}(\mathbb F)$. Let $S=\{ X \in V : Xa =0 \}$ be a subspace of $V$. What is the dimension of $S$?
My Solution: $m(n-1)$;
Supposing a is not zero for all elements $a_{i}$ and X=$\{b_{ij}\}$. then multiplying $Xa=0$ WLOG i=1
$b_{11}a_{1}+b_{12}a_{2}+...+b_{1n}a_{n}=0$ which has $(n-1)$ independent vectors and since there are $m$ rows I get $m(n-1)$ as a dimension. Is this right?
With your notation and given symbols, define
$$\phi: V\to F^m\,,\,\,\phi X:=Xa$$
It reasonably easy to check $\;\text{Im}\,\phi=F^m\;$, from where $\;\dim\text{Im}\,\phi=m\;$ and thus from the dimensions theorem, $\;\dim\ker\phi=mn-m=m(n-1)\;$ .
This is more or less, I guess, what you wrote...but, imo, this is clearer and easier to grasp.