For a square matrix $A\in F^{n\times n}$ over a field $F$, define the commutator subspace $C_A = \{ B\in F^{n\times n} \vert AB = BA\}$ of matrices which commute with $A$. This other question by RiaD asks for a proof that $\dim C_A\geq n$ for any $A$. The answer by Johannes Kloos uses Jordan Normal Form, and so yields the same result over any algebraically closed field $F$ in place of $\mathbb{C}$.
Now let's write $C_{F,A}$ in place of $C_A$ to keep track of the field. Note that $C_{F,A}$ is the kernel of the linear map $B\mapsto AB-BA$. The dimension of a kernel doesn't change in a field extension because row reduction proceeds in the same way regardless of field, so if $K$ is a field extension of $F$ then $\dim_F C_{F,A} = \dim_K C_{K,A}$. Since any field $F$ is contained in an algebraically closed field $K$ we can combine these facts to get $\dim_F C_{F,A} = \dim_K C_{K,A}\geq n$, so the original result holds for all fields $F$, not just algebraically closed fields.
The question: is there a purely linear algebraic proof of this? By this I mean one which works for all fields and does not involve passing to a field extension.