Let $A$ be an $n \times n$ matrix over a field $\mathbb{F}.$ Then consider the subspace of $M_n(\mathbb{F})$ given by $W=\{ B \in M_n(\mathbb{F}): AB =BA\}.$ Then I know $\text{dim}(W) \geq n$ from my earlier question (Dimension of centraliser of an $n \times n$ matrix is atleast n.)

How can I show that $\text{dim}(W)=n$ will imply minimal polynomial equals the characteristic polynomial of $A.$

So enough to show that $I,A,A^2,\ldots,A^{n-1}$ are linearly independent. Does it follow immediately ? I cannot see it.

Also what can we say about the converse ? i.e., minimal polynomial of $A$=characteristic poly of $A$ imply that $\text{dim}(W)=n$ ?

I really need some help. Thank you.


Up to a change of basis in $F^n$, we may assume that $A$ is in Frobenius normal form, cf.


there are monic polynomials $(p_i)_{i\leq k}\in F[x]$ s.t.

i) $p_1|p_2|\cdots|p_k$, $p_k$ is the minimal polynomial of $A$ and $p_1\cdots p_k$ is the characteristic polynomial of $A$.

ii) $A=diag(C_{p_1},\cdots,C_{p_k})$, where $C_p$ is the companion matrix of the polynomial $p$. cf.


Note that $C_p$ is cyclic because $e_1,f(e_1),\cdots,f^{(degree(p)-1)}(e_1)$ is a basis of $F^n$. Therefore, it is not difficult to see that

a) the minimal polynomial of $C_p$ is $p$.

b) the commutant of $C_p$ is $F[C_p]$ (the matrices that are polynomials in $C_p$), and then, has dimension $degree(p)$.

Now, we show the equivalence of the two hypotheses considered by the OP.

$\Leftarrow$ If the minimal polynomial has degree $n$, then $p_k=p_1\cdots p_k$ and $k=1$, that is $A=C_{p_1}$ is cyclic; according to b), $dim(W)=n$.

$\Rightarrow$ Assume that $dim(W)=n$ and $k>1$; we look for a contradiction. Let $Z=\{B=diag(B_1,\cdots,B_k);C_{p_i}B_i=B_iC_{p_i}\}$; according to b) $dim(Z)=degree(p_1)+\cdots +degree(p_k)=n$ and $W=Z$. Considering a matrix in the form $B=diag(U_{degree(p_1)+degree(p_2)},I_{n-degree(p_1)-degree(p_2)})\in W$, we reduce the problem to the case $k=2$, $A=diag(C_p,C_q)$ where $p|q$.

Let $B=\begin{pmatrix}0&X\\0&0\end{pmatrix}$; $B\in W$ iff $C_pX-XC_q=0$. This Sylvester equation has a non-zero solution iff $C_p$ and $C_g$ have a common eigenvalue, that is, iff $p$ and $q$ have a common root, that is true because $p|q$. We obtain a matrix $B$ which is in $W\setminus Z$, that is a contradiction and we are done. $\square$

  • $\begingroup$ +1 I was going to post a near-identical answer, but you beat me to it by a few minutes ... I wonder if the dimension of the commutant can be expressed in terms of the degrees of the $p_k$. All I could obtain so far is that the dimension of the commutant is the sum of the $\dim(\ker(p_k(A)))$. $\endgroup$ – Ewan Delanoy Oct 16 '19 at 10:11
  • $\begingroup$ @Ewan Delanoy , it's funny; I asked myself the same question. If $A$ has a sole eigenvalue, then the Jordan blocks and the Frobenius blocks have same dimensions. Otherwise, I don't know. However, there must be a formula since if we know the multiplicities of the roots of the $(p_i)$, then we can determine the dimensions of the Jordan blocks. $\endgroup$ – user91684 Oct 17 '19 at 13:43

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