# Dimension of commutator of an $n \times n$ matrix $A$ is $n$ implies that min poly and ch poly of $A$ coincides.

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.

https://en.wikipedia.org/wiki/Frobenius_normal_form

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.

https://en.wikipedia.org/wiki/Companion_matrix

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. $$\square$$

• +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 $\dim(\ker(p_k(A)))$. – Ewan Delanoy Oct 16 '19 at 10:11
• @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. – user91684 Oct 17 '19 at 13:43