Let $A\in\operatorname{M}_n(F)$. How to prove the following identity. $$\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}.$$ Here $m_A(x)$ the minimal polynomial of $A$ and $C(A)$ is the centralizer of $A$ and $C(X)$ is the centralizer of $X$. We can view $F^n$ as an $F[x]$-module via $A$, which is well known, denote this module by $F^n_A$, we have $$C(A)=\operatorname{End}_{F[x]}(F^n_A,F^n_A)$$ Since $$F^n_A\simeq\bigoplus_{i=1}^t\frac{F[x]}{(d_i(x))},\quad d_1\mid d_2\mid\cdots\mid d_t\quad\text{(invariant factors)}.$$ We have $$C(A)=\bigoplus_{i}\bigoplus_{j}\operatorname{Hom}_{F[x]} \left(\frac{F[x]}{(d_i(x))},\frac{F[x]}{(d_j(x))}\right) \simeq\bigoplus_{i=1}^{t}\left(\frac{F[x]}{(d_i(x))}\right)^{2(t-i)+1}.$$ For $X\in C(A)$, what is $C(X)$? Can we write the structure of $C(X)$ explicitly(using the above isomorphism)?
I saw one proof of this result, but the proof is over algebraic closed field and they use Jordan canonical form. Could anyone explain this for me and give a purely algebraic proof. Use matrix method or the module structrue, but must be on the ground field $F$. Thanks in advance.
By the way, are there any relations between this and the double centralizer theorem of central simple algebra?