Every matrix of the centralizer of the centralizer of a matrix is a polynomial in that matrix Let $V=M(n,\mathbb C)$. For a subset $S \subseteq V$, let $C(S):=\{A \in V | AB=BA, \forall B \in S \}$ . 
How to prove that for every $A\in V$, we have $C(C (\{A\})) \subseteq \{ p(A) | p(t) \in \mathbb C[t] \}$ ?
My thoughts (motivated by Omnomnomnom's comment ): the reverse inclusion holds. Since both the sets are vector subspaces of $V$, enough to show their dimensions are equal. Now the dimension of the polynomial set is the degree of the minimal polynomial of $A$. So enough to show that dim$C(C(A)) \le $ degree of minimal polynomial of $A$. 
 A: Let's say that an $R$-module $W$ has the double commutant property if $C_W(C_W(R))$ equals the image of $R$ in $End(W)$. 
I'll give a proof via Jordan decomposition. One consequence of Jordan decomposition is: if we view $V$ as a $\mathbb{C}[x]$-module, then there exists a direct sum decomposition
$$ V = V_1 \oplus \cdots \oplus V_n$$ 
into cyclic $\mathbb{C}[x]$-modules, i.e. for all $i$ there exists $v_i \in V_i$ such that $V_i = \mathbb{C}[x]v_i$.

Lemma 1: If $W$ is a cyclic $R$-module and $R$ is a commutative ring, then $C_W(R)$ is the image of $R$ in $End(W)$.

Proof: say $\varphi: W \to W$ intertwines $R$. If $v$ is the cyclic vector, then $\varphi(w) = rw$ for some $r \in R$. Then $\varphi(sw) = s\varphi(w) = srw = rsw$, 
so since $W = Rw$, we conclude that $\varphi$ is left-multiplication by $r$. Since $R$ is commutative, the reverse holds.

Lemma 2: If $W_1$ and $W_2$ are $R$-modules, then for $V = W_1 \oplus W_2$, $$C_V(C_V(R)) \subseteq C_{W_1}(C_{W_1}(R)) \oplus C_{W_2}(C_{W_2}(R)).$$

Proof: Suppose that $T: V \to V$ lies in $C_V(C_V(R))$. We can write $T$ uniquely as $T = T_{11} + T_{12} + T_{21} + T_{22}$ where $T_{ij} \in Hom(W_j,W_i)$. We aim to show that $T_{12} = 0$, $T_{21} = 0$, and that $T_{11}$ and $T_{22}$ lie in their respective double commutants.
Let $V = W_1 \oplus W_2$. By definition of direct sum, the structure maps
$$ \pi_1: W_1 \oplus W_2 \to W_1 \to W_1 \oplus W_2 \qquad \pi_2: W_1 \oplus W_2 \to W_2 \to W_1 \oplus W_2$$
lie in $C_V(R)$. Then $\pi_1 T = T\pi_1$ since $T$ lies in the double commutant. Expanding into the direct sum $Hom(V,V) = \oplus_{ij} Hom(W_j,W_i)$ gives that $T_{11} + T_{12} = T_{11} + T_{21}$, implying that $T_{12} = 0$ and $T_{21} = 0$.
We may also observe that $C_{W_i}(R) \subseteq \oplus_{ij} Hom(W_j,W_i)$ is also contained in $C_V(R)$. If $\varphi_i \in C_{W_i}(R)$, then as $T$ is in the double commutant of $V$, $\varphi_iT_{ii} = T_{ii}\varphi_i$. Hence $T_{ii}$ is in the double commutant of $W_i$, as desired.
Now we can prove the desired result. Jordan decomposition states more than that we have a decomposition of $V$ into cyclic $\mathbb{C}[x]$-modules, but that we may in fact decompose into Jordan blocks, i.e. modules of the form $\mathbb{C}[x]/(x-\lambda)^n$. We will use these blocks to refine the inclusion in Lemma 2.

Lemma 3: If $V$ is a direct sum of Jordan blocks with the same eigenvalue, then $V$ has the double commutant property.

Proof: By replacing $x$ with $x -\lambda$, we may assume that the Jordan blocks have eigenvalue $0$, i.e. that $x$ acts nilpotently. Say that
$V = \oplus_i V_i$ where $V_i = \mathbb{C}[x]/x^{n_i}$, and without loss of generality that $n_i \geq n_{i+1}$ for all $i$. Suppose that $T \in C_{V}(C_{V}(\mathbb{C}[x]))$; by Lemma 2, we may write $T = \oplus T_{ii}$ for $T_{ii} \in C_{V_i}(C_{V_i}(\mathbb{C}[x]))$. 
Now since $n_i \geq n_{i+1}$ for all $i$, we have $\mathbb{C}[x]$-module surjections
$$ s_{i1}: \mathbb{C}[x]/x^{n_1} \to \mathbb{C}[x]/x^{n_i},$$
which may be viewed as maps $s_{i1}: V \to V$ by composing with the structure maps for the direct sum. Then $s_{i1} \in C_V(\mathbb{C}[x])$, so we have $s_{1i}T = Ts_{1i}$ for all $i$. Hence, $s_{i1}T_{11} = T_{ii}s_{i1}$.  By Lemma 1, $C_{V_1}(C_{V_1}(\mathbb{C}[x])) = \mathbb{C}[x]/x^{n_1}$. So if $T_{11} = p(x)$, then
$$ T_{ii}s_{i1} = s_{i1}T_{11} = s_{i1}p(x) = p(x)s_{i1}.$$
Since $s_{i1}$ is surjective, we conclude $T_{ii} = p(x)$ as well (as an endomorphism of $V_i$). As this holds for all $i$, $T = p(x)$ (as an endomorphism of $V$).

Lemma 4: If $W_1$ and $W_2$ are $\mathbb{C}[x]$-modules with the double commutant property and the minimal polynomials $r_1$ and $r_2$ of $x$ on $W_1$ and $W_2$ are coprime, then $W_1 \oplus W_2$ has the double commutant property.

Proof: Let $V = W_1 \oplus W_2$. By Lemma 2, if $\pi_i: W_1 \oplus W_2 \to W_i \to W_1 \oplus W_2$ are the structure maps, then
$$ C_V(C_V(\mathbb{C}[x])) \subseteq  \mathbb{C}[x]\pi_1 \oplus  \mathbb{C}[x]\pi_2.$$
It suffices to show that any transformation $V \to V$ of the form $p_1(x) \pi_1 + p_2(x) \pi_2$ is in the image of $\mathbb{C}[x]$ in $End(V)$. So it suffices to show that $\pi_1$ and $\pi_2$ are in the image of $\mathbb{C}[x]$.
Since $r_1$ and $r_2$ are coprime, there exist polynomials $f_1, f_2$ such that
$$f_1r_1 + f_2r_2 = 1.$$
Then $f_1r_1(x)= 1 - f_2r_2(x)$ acts as zero on $V_1$ and as the identity on $V_2$, so it acts as $\pi_2$ on $V$. This shows that $\pi_2$ and thus $\pi_1 = 1 - \pi_2$ are in the image of $\mathbb{C}[x]$, as desired.

Theorem: Any finite-dimensional $\mathbb{C}[x]$-module $V$ has the double commutant property.

Proof: By Jordan decomposition, we may decompose $V$ into summands $V_1\oplus\cdots \oplus V_n$ where $x$ acts with a single eigenvalue on each block. By Lemma 3, each $V_i$ has the double commutant property. Then induct via Lemma 4 to show that $V_1 \oplus \cdots \oplus V_k$ has the double commutant property for all $k$, as the minimal polynomial of $x$ on $V_1 \oplus \cdots \oplus V_{k-1}$ and $V_k$ have no roots in common.
A: It is sufficient to prove this result for a matrix similar to $A$, and so we can suppose that $A$ is in Jordan canonical form. Furthermore, let $\lambda$ be an eigenvalue of $A$, then $C(A)=C(A-\lambda I)$ and so we can replace $A$ by $A-\lambda I$ in order that $A$ has a zero eigenvalue.
If $A$ consists of a single block, then  direct calculation shows that $C(A)$ is itself equal to $\{p(A) | p(t) \in \mathbb C[t] \}$ and hence the required result for $C(C(A))$.
So we can suppose that there are at least two blocks. 
Case 1.  If at least two blocks have different eigenvalues 
We can suppose $A$ is \begin{pmatrix}U&0\\0&V\end{pmatrix} where all  the blocks corresponding to one particular eigenvalue are in $U$. 
Let $M=$\begin{pmatrix}I&0\\0&0\end{pmatrix} where $I$ is the identity matrix of the same size as $U$.Then $M$ is in C(A) and $C(M)$ consists of matrices of the form \begin{pmatrix}R&0\\0&S\end{pmatrix} where $R$ has the same size as $I$. 
A matrix $N$ in $C(C(A))$ is in $C(M)$ and so has this form and then, by induction,  $N$  is  \begin{pmatrix}f(U)&0\\0&g(V)\end{pmatrix} for some polynomials $f$ and $g$. 
The characteristic polynomials $p_U$ and $p_V$ of $U$ and $V$ are coprime and so there are polynomials $u$ and $v$ such that $up_U+vp_V=1$. The matrix $N$ is then $h(A)$, where $$h=f+(g-f)up_U.$$
Case 2.  If all blocks have eigenvalue $0$
Each block of $A$ now has $1$s on the super-diagonal and $0$s elsewhere. We shall first consider the case where $A$ has just two blocks $U$ and $V$ where we can suppose that dim$(V)$ is no greater than dim$(U)$. As in Case 1, any matrix in $C(C(A))$ has the form \begin{pmatrix}f(U)&0\\0&g(V)\end{pmatrix} for some polynomials $f$ and $g$. Let $h=g-f$ and then the matrix 
$T=$\begin{pmatrix}0&0\\0&h(V)\end{pmatrix} is also in $C(C(A))$. 
Let $I$ be an identity matrix of same dimension as $V$ and let $I^*$ be the matrix with $\dim(U)-\dim(V)$ rows of $0$s added underneath $I$. Then \begin{pmatrix}0&{I^*}\\0&0\end{pmatrix} is in $C(A)$. For $T$ to commute with this matrix we require $I^*h(V)=0$ and therefore $h(V)=0$. Then the matrix of $C(C(A))$ is $f(A)$. 
For any number of blocks, this argument can be used with the the block of largest dimension and any other block to complete the proof. 
A: If $A$ is a linear map from a finite dimensional $F$ vector space $V$ to itself then, we get a $R:=F[x]$ module structure on $V$, where $x$ acts by $A$. 
Let $p(x)$ be the minimal polynomial of $A$, then it is well-known from cyclic decomposition theorem that there is a vector $v\in V$ such that its minimal annihilating polynomial is $p(x)$ and if $W$ is the cyclic $R$-submodule generated by $v$, then there is an $R$-submodule $W'$ such that $V=W\oplus W'$. Let $e:V\rightarrow V$ be the idempotent associated to the projection onto $W$. Let $B$ be in $C(C(A))$, then $B$ commutes with $e\in C(A)=End_R(V)$ and hence $B(W)=B(eW)=e(B(W))\subseteq W$.
This implies that $Bv=f(A)v$ for some $f\in R$. We claim that $Bx=f(A)x$ for any $x\in V$. Define $e':V\rightarrow V$ to be the $R$-endomorphism that is zero on $W'$ and on $W$ is defined by sending $g(A)v$ to $g(A)x$ for any $g\in R$. This is well-defined since if $g(A)v=h(A)v$ then $g-h$ is a multiple of the minimal polynomial $p$ and hence $g(A)x=h(A)x$. Now $e'Bv=e'f(A)v=f(A)e'v=f(A)x$ and on the other hand $e'Bv=Be'v=Bx$. So we have proved that $B=f(A)$ is a polynomial in $A$.
