2

Given $A,B\in\mathbb{C}^{n\times n}$. Suppose $A^kB=BA^k$ for some $k\in \mathbb{N}^*$ with $A-E$ nilpotent, where $E$ is the identity. Show that $AB=BA$

I have a primary idea. Suppose $A=E+N$ with $N$ nilpotent, say $N^m=0$ and $N^{m-1}\neq 0$. It is easy to conclude from $A^k B= BA^k$ that for all $f(x)\in \mathbb{C}[x]$ we have $$f(A^k)=f\left( E+kN+\cdots+ \binom{m-1}{k} N^{m-1} \right)\in \mathbb{C}(B)$$

It follows from there is a specific $g(x)$ such that $g(A^k)=E+N=A$.

New Progress

From Jordan-Chevalley Theorem, we can find a polynomial $p(x)$ and $q(x)$ such that $p(A^k)= E$ and $q(A^k)=kN+ r(N)$, where $\deg r(x)\geq 2$ . Notice that we can get $N^{m-1}$ from $q^{m-1}(A^k)$. By analogy, we can get all $N^{i},i=1,2,\dots,m-1$ by linear combination of $q^{l}(A^k)$. Hence we can find the poynormial $g(x)$ such that $g(A^k)=E+N$.

William Smith
  • 81
  • 2
    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Dec 24 '24 at 09:02
  • Yes, it suffices to find a polynomial $g$ such that $g(A^k)=A$. – user1551 Dec 24 '24 at 09:51
  • Yes, I just thought a aprroach. But I look forward to a better way. – William Smith Dec 24 '24 at 10:35
  • While this question has a very different setting from https://math.stackexchange.com/q/4774695 , the techniques in user8675309’s answer and mine there are applicable (as illustrated by our answers below). So, I close this question as a duplicate. – user1551 Dec 30 '24 at 17:28
  • @user1551 I suppose the reconciliation is we can assume WLOG $B\in GL_n(\mathbb C)$ -- if not consider $B':= B + \delta I$-- and the OP says $\big(B^{-1}AB\big)^k=B^{-1}A^k B = A^k$ for some $k$ and since $\big(B^{-1}AB\big)$ and $A$ each have only positive spectra, the duplicate link says $\big(B^{-1}AB\big) =A$ – user8675309 Dec 31 '24 at 19:55

2 Answers2

3

All you need is to express $A$ as $g(A^k)$ for some polynomial $g$. By assumption, the minimal polynomial of $A$ is $(x-1)^m$ for some $m\ge1$. So, it suffices to find $g$ such that $(x-1)^m$ divides $f(x)=g(x^k)-x$. This gives rise to a system of equations $f(1)=f'(1)=\cdots=f^{(m-1)}(1)=0$. Since each $r$-th order derivative of $f$ is a $\mathbb C[x]$-linear combination of $g,g',\ldots,g^{(r)}$, by solving the system of equations $f(1)=f'(1)=\cdots=f^{(m-1)}(1)=0$ recursively, the problem boils down to finding $g$ such that $g^{(r)}(1)=c_r$ for $r=0,1,\ldots,m-1$, where $c_0,c_1,\ldots,c_{m-1}$ are some constants that depend on $m$ and $k$ only.

(E.g. since \begin{align*} f(x)&=g(x^k)-x,\\ f'(x)&=g'(x^k)kx^{k-1}-1,\\ f''(x)&=g''(x^k)(kx^{k-1})^2+g'(x^k)k(k-1)x^{k-2},\\ \end{align*} $f(1)=f'(1)=f''(1)=0$ if and only if $g(1)=c_0:=1,\,g'(1)=c_1:=\frac{1}{k}$ and $g''(1)=c_2:=-\frac{g'(1)k(k-1)}{k^2}=\frac{1}{k}-1$.)

So, $g$ always exists because we can pick the Taylor polynomial $$ g(x)=\sum_{r=0}^{m-1}\frac{c_r}{r!}(x-1)^r. $$

user1551
  • 149,263
  • Terrific solution! Thanks! – William Smith Dec 24 '24 at 12:52
  • I think this question is in some non-standard sense a duplicate of https://math.stackexchange.com/questions/4774695/let-a-b-in-m-n-mathbbc-suppose-that-all-of-the-eigenvalues-of-a-and/ ... in particular your argument of writing $A=C=p\big(A^k\big)$ works verbatim and I posted my argument under "generalizing" essentially verbatim – user8675309 Dec 24 '24 at 17:02
3

define $T:M_n\big(\mathbb C\big)\longrightarrow M_n\big(\mathbb C\big)$ given by $T(v) = A^{-1}vA$
$T$ has all eigenvalues equal to one since $A$ is unipotent. Thus with $w:=B$
$T^k(w) =w\implies T(w)=w$

i.e. let $\lambda$ be a primitive $k$th root of unity. We have
$\prod_{r=1}^k \big(T-\lambda^r \cdot\text{id}\big)w= T^k(w) -w =\mathbf 0=\big(T-\text{id}\big)w$
where the right hand side follows since $\big(T-\lambda^r \cdot\text{id}\big)$ is invertible for $1\leq r\leq k-1$

user8675309
  • 12,193