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A matrix $A\in\textbf{Mat}_n(\mathbb{R})$ is called asymptotically nilpotent if for each vector $v$, ${\lim}_{k\to\infty}A^k(v) = 0$. Assume that $\mathcal{A}, \mathcal{B}$ be maximal (under inclusion) among those subsets of $\textbf{Mat}_n(\mathbb{R})$ with only asymptotically nilpotent matrices and which are closed under the Lie bracket operation (i.e. $\forall X, Y\in\mathcal{A}, [X, Y] = XY - YX\in \mathcal{A}$ and similarly for $\mathcal{B}$).

Is it true that for some $P\in\text{GL}_n(\mathbb{R})$, $P\mathcal{A}P^{-1} = \mathcal{B}$?

Further variation of this question is asked here.

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    $\begingroup$ So as long as $n$ is fixed then Jordan Normal form tells you that all asymptotically nilpotent matrices are alle the matrices with eigenvalues of absolute value stricltly less than 1. $\endgroup$
    – Frederik Ravn Klausen
    Commented Oct 16, 2022 at 9:50
  • $\begingroup$ My conjecture is that after the suitable conjugation we can assume that $\mathcal{A}$ is a Lie algebra formed with matrices $X + Y$, where $X$ is a matrix with all entries less than $1$ and $Y$ is an upper triangular matrix. $\endgroup$
    – solver6
    Commented Oct 16, 2022 at 9:58
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    $\begingroup$ A subalgebra (or even a linear subspace) consists of asymptotically nilpotent matrices iff it consists of nilpotent matrices. Because if $M$ is as. nilp. but not nilp., then $tM$ for $t$ large enough is not as. nilp. $\endgroup$
    – YCor
    Commented Oct 16, 2022 at 10:07

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The answer to your first question is no: they're not all conjugate.

Indeed, let $A$ be the set of all upper triangular matrices of absolute value $<1$ on the diagonal. Then $A$ consists of asymptotically nilpotent matrices (clear) and is maximal for this property (1).

Let $B_0$ be the set of all $d\times d$ matrices with all coefficients of absolute value $<1/2d^2$. Then $B_0$ is stable under taking brackets and consists of asympotically nilpotent matrices. Let $B$ be maximal for these properties and containing $B_0$. Then $B$ is not conjugate to $A$, since $0$ is in the interior of $B$ and not of $A$.

(1) Let $A'$ be a larger subset with the given properties, and let by contradiction $M\in A'$ be a matrix outside $A$. If $M$ is upper triangular, then $M$ has a diagonal coefficient of absolute value $\ge 1$ and hence is not asymptotically nilpotent. Otherwise $M_{ji}\neq 0$ for some $i<j$. Choose $(i,j)$ with $j-i$ maximal for this property, and then among the possible choices, choose $i$ minimal. Write $M'=[M,E_{ij}]$. Then by stability, $[M,tE_{ij}]=tM'$ belongs to $A'$, for every scalar $t$. By the choices, $M'$ is upper triangular, and $M'_{ii}\neq 0$. This is a contradiction.

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    $\begingroup$ Thank You. I asked the second part of the question here mathoverflow.net/questions/432547/…. $\endgroup$
    – solver6
    Commented Oct 16, 2022 at 10:59
  • $\begingroup$ For my understanding: $1/2d^2$ is $1/(2d^2)$, right, not $(1/2)d^2$? Or is there some simple-in-retrospect miracle that I am overlooking? $\endgroup$
    – Vincent
    Commented Oct 21, 2022 at 8:41
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    $\begingroup$ @Vincent yes, by $1/ab$ I mean $1/(ab)$. This is quite common usage although I'm aware that a software would interpret $1/a*b$ as $(1/a)*b$. $\endgroup$
    – YCor
    Commented Oct 21, 2022 at 9:30

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