Asymptotically nilpotent matrices

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.

• 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. Oct 16, 2022 at 9:50
• 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. Oct 16, 2022 at 9:58
• 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.
– YCor
Oct 16, 2022 at 10:07

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 $$i. Choose $$(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.

• Thank You. I asked the second part of the question here mathoverflow.net/questions/432547/…. Oct 16, 2022 at 10:59
• 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? Oct 21, 2022 at 8:41
• @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 $(1/a)*b$.
– YCor
Oct 21, 2022 at 9:30