Matrix that commutes with this particular $A\in M_n(\mathbb{Q})$ must be a polynomial in $A$

Question

Recently, I received a "basic" linear algebra question, but it seems like I highly underestimated its difficulty. The question stated:

Consider $A,B\in M_n(\mathbb{Q})$ with $A^n=0$ but $A^{n-1}\neq 0$. Show \begin{align*} AB=BA \implies B= F(A)\text{ for some }F \in \mathbb{Q}[x]\text{ such that }\operatorname{deg}(F)<n \end{align*}

This question is basically asking us to prove the centralizer of $A$ must be $\mathbb{Q}[A]$, which I guess now is highly nontrivial. One of my attempts is to argue that the dimension of the centralizer of $A$ is $n$, since the Jordan form of $A$ is pretty nice, but I fail to prove how commutativity is preserved under conjugation, and I also fail to compute the centralizer of Jordan block. Anyway, that was just my attempt. It may now really work, so if you have another idea, it is very likely yours is better.

Thanks for any help in any form in advance.

Answer 1

Since $A^n=0$ but $A^{n-1}\neq0$, the minimal polynomial of $A$ must be $x^n$. In particular, no smaller power of $A$ vanishes. Because the characteristic polynomial also has degree $n$, it too must be $x^n$. In other words, $A$ is nilpotent of index exactly $n$, and its minimal and characteristic polynomials coincide—so $A$ is “non-derogatory.”

Equivalently, over $\mathbb{Q}$ one can put $A$ into a single $n\times n$ Jordan block for the eigenvalue $0$. Concretely, there is an invertible $P\in GL_n(\mathbb{Q})$ so that

$$ A \;=\; P\,J\,P^{-1}, \quad J \;=\; \begin{pmatrix} 0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & & \ddots & \ddots & \vdots\\ 0 & 0 & \cdots & 0 & 1\\ 0 & 0 & \cdots & 0 & 0 \end{pmatrix}. $$

Because $AB=BA$, if we set $B' = P^{-1}BP$, then $J\,B' = B'\,J$. Writing $B'=(b_{ij})$, one checks that this forces

$$ b_{i,j-1} \;=\; b_{\,i+1,j} \quad \text{and}\quad b_{k1}=0 \text{ for }k>1,\quad b_{n,j-1}=0\text{ for }j>1. $$

In plain language, each super-diagonal of $B'$ has a constant entry, and every entry strictly below the main diagonal must be zero. Thus

$$ B' \;=\; \begin{pmatrix} c_0 & c_1 & c_2 & \cdots & c_{n-1}\\ 0 & c_0 & c_1 & \cdots & c_{n-2}\\ 0 & 0 & c_0 & \cdots & c_{n-3}\\ \vdots & \vdots & & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & c_0 \end{pmatrix}, $$

with each $c_k\in\mathbb{Q}$. But that is exactly

$$ c_0 I + c_1 J + c_2 J^2 + \cdots + c_{n-1} J^{\,n-1}. $$

Since $J^n=0$, the polynomial

$$ F(x) \;=\; c_0 + c_1 x + \cdots + c_{n-1}x^{\,n-1} $$

satisfies $B' = F(J)$. Going back to the original basis,

$$ B \;=\; P\,B'\,P^{-1} \;=\; P\,F(J)\,P^{-1}. $$

But $J = P^{-1}AP$, so $F(J) = P^{-1}F(A)\,P$. Hence

$$ B \;=\; F(A), $$

and $F$ has rational coefficients of degree at most $n-1$.

Once you realize $A$ is a single $n$-block Jordan nilpotent, any matrix commuting with it must be a polynomial in $J$; conjugating back shows $B$ is that same polynomial in $A$. Therefore

$$ B \;=\; F(A), \quad F(x)\in \mathbb{Q}[x],\;\deg F < n, $$

as claimed.