Dimension of commutator space of matrices (again)
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.
Sure. The hard step is to prove that any matrix $A$ can be written (up to conjugation) as the direct sum of matrices $A_i$ whose characteristic and minimal polynomials coincide. Once you know this, it suffices to prove the statement for each $A_i$, but it's obvious because in this case the dimension of $\text{span}(1, A_i, A_i^2, ... A_i^{k-1})$ (where $A_i$ is a $k \times k$ matrix) is $k$ and this is clearly contained in $C_{A_i}$.
The hard step follows from e.g. the structure theorem for finitely-generated modules over a PID (here $F[x]$) which I think is linear algebra just as much as Jordan decomposition is. Anyway, it tells you that $F^n$, as a $F[x]$-module with $x$ acting by $A$, decomposes as a direct sum $$\bigoplus_i F[x]/p_i(x)^{d_i}$$
where $p_i$ is an irreducible polynomial. The restriction of $A$ to each summand has the property that the characteristic and minimal polynomials coincide and we are done.