# 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 typical way to start such a proof is to suppose you have some matrix $M$ in the left hand side, and prove that it's also an element of the right hand side. Have you considered the case where $S$ Is a singleton set, $S = \{M\}$, perhaps for some particularly simple cases like $M = I$ or $M = 0$ ? – John Hughes Aug 21 '19 at 17:05
• The infinite dimensional generalization to this is the Von Neumann bicommutant theorem – Ben Grossmann Aug 21 '19 at 17:07
• It should be noted that the reverse inclusion, $$C(C (\{A\})) \supseteq \{ p(A) | p(t) \in \mathbb C[t] \}$$ is easy to show – Ben Grossmann Aug 21 '19 at 17:09
• Some of the answers given here might be useful – Ben Grossmann Aug 21 '19 at 18:21
• This post also seems useful. I also have a vague idea that Weyr canonical form might be convenient – Ben Grossmann Aug 21 '19 at 18:27

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.

• What is your definition of $C_W (S)$ ? And what is the image of $R$ in $End (W)$ ? Are you identifying $R$ inside $End (W)$ as $r \in R$ corresponding to the map $f(w)=rw$ ? – uno Aug 23 '19 at 23:15
• $C_W(S)$ is all of the abelian group endomorphisms of $W$ commuting with $S$. When $S$ contains $\mathbb C$, then all of these will be linear transformations, reducing to the situation in your question. And indeed, the module structure of $R$ on $W$ is exactly the assignment $R \to End(W)$ which you gave (although it is not generally injective). – Joshua Mundinger Aug 23 '19 at 23:22
• What do you mean by "abelian group endomorphism" ? And what do you mean when you say an endomorphism commutes with $R$ ? Where does your $C_W(R)$ land in ? – uno Aug 24 '19 at 14:08
• Well $W$ is a module, so it’s also an abelian group. So $C_W(R) \subset End(W)$, where I mean additive endomorphisms. The endomorphism $\varphi$ commutes with $R$ if for all $r$ in $R$, $\varphi(rx) = r\varphi(x)$, i.e. that $\varphi$ is an $R$-module homomorphism. – Joshua Mundinger Aug 24 '19 at 14:12
• Okay, but then how do you define $C_W (C_W(R))$ ? – uno Aug 24 '19 at 14:28

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.

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 $$A$$.