I asked the following question on Math Stack Exchange, but no people reply. I know MO is more professional and it is for mathematicians to discuss research problems. Maybe this question is unsuitable for here. I truly appreciated if someone who could give me an explanation.


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?



Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$
This follows from the Frobenius normal form, see e.g. Theorem 39.3 in Prasolov's book *Problems and Theorems in Linear Algebra*. They same argument appears in [Lagerstrom's paper](http://dx.doi.org/10.1090/S0002-9904-1945-08386-4). As the identity holds without any semisimplicity assumption on $A$, it is conceivable that there is a cleaner proof too. But I do not see one.


