# Asymptotically nilpotent Lie sets of matrices

A matrix $$A\in\textbf{Mat}_n(\mathbb{R})$$ is called asymptotically nilpotent if for each vector $$v$$, $${\lim}_{k\to\infty}A^k(v) = 0$$.

Question 1. Assume that $$\mathcal{A}$$ is the subset of $$\textbf{Mat}_n(\mathbb{R})$$ with only asymptotically nilpotent matrices and which is closed under the Lie bracket operation (i.e. $$[X, Y] = XY - YX\in \mathcal{A},\forall X, Y\in\mathcal{A}$$). Is it true that for some $$P\in\text{GL}_n(\mathbb{R})$$, $$P\mathcal{A}P^{-1}$$ is a subset of all matrices with coefficients of absolute value $$<1$$ on and below the diagonal?

A set $$S\subseteq \textbf{Mat}_n(\mathbb{R})$$ is called asymptotically nilpotent if for each sequence $$A_1, A_2, A_3,\ldots\in S$$ and each vector $$v$$, $$\lim_{k\to\infty}A_kA_{k - 1}\ldots A_2A_1(v) = 0$$.

Question 2. Assume that $$S$$ is an asymptotically nilpotent set of matrices. Is it true that for some $$P\in\text{GL}_n(\mathbb{R})$$, $$PSP^{-1}$$ is a subset of all matrices with coefficients of absolute value $$<1$$ on and below the diagonal?

Question 3. Let $$\mathbf{Q}_p$$ be the field of $$p$$-adic numbers. Assume that $$S\subset \text{Mat}_n(\mathbf{Q}_p)$$ is an asymptotically nilpotent finite set of matrices. Is it true that for some $$P\in\text{GL}_n(\mathbf{Q}_p)$$, $$PSP^{-1}$$ is a subset of all matrices with coefficients of absolute value $$<1$$ on and below the diagonal?

I will state some personal ideas.

In the conditions of the Question 2 it is possible to prove the next simpler theorem

Theorem 1. Assume that $$S$$ is a finite asymptotically nilpotent set of matrices. Then there is an operator $$P\in\text{GL}_n(\mathbb{R})$$ such that $$PSP^{-1}$$ is a subset of matrices $$(a_{ij})_{i, j\leq n}$$ where $$a_{i1}<1$$ for $$i\leq n$$.

Proof. Consider the open set of vectors $$U = \{(a_1, a_2,\ldots, a_n)\in\mathbb{R}^n\:|\: a_1 < 1\}$$. Since $$S$$ is asymptotically nilpotent there is a vector $$e_1\in\mathbb{R}^n\setminus U$$ such that $$A(e_1)\in U$$ for each $$A\in \langle S\rangle$$ ($$\langle S\rangle$$ is the semigroup generated with $$S$$). Now for a natural $$N$$ let $$e_1^N = e_1\in\mathbb{R}^n$$ and $$e_i^N = (\underbrace{0,\ldots, 0}_{i -1 }, N,0,\ldots, 0)\in\mathbb{R}^n$$ for $$2\leq i\leq n$$. Let $$P_N\in \text{GL}_n(\mathbb{R})$$ denote the operator with $$P_N(e_i^N) = (\underbrace{0,\ldots, 0}_{i -1 }, 1,0,\ldots, 0)$$ for $$1\leq i\leq N$$. Then for $$N$$ large enough $$P_NSP_N^{-1} =(a_{ij})_{i, j\leq n}$$ where $$a_{i1}<1$$ for $$i\leq n$$.

Question 3 in the case $$n=2$$ follows from the next Theorem 2.

Theorem 2. Let $$\mathbf{Q}_p$$ be the field of $$p$$-adic numbers. Assume that $$S\subset \text{Mat}_2(\mathbf{Q}_p)$$ is an asymptotically nilpotent finite set of matrices. Then for some $$P\in\text{GL}_2(\mathbf{Q}_p)$$, $$PSP^{-1}$$ is a subset of all matrices with coefficients of absolute value $$<1$$ on and below the diagonal.

Proof. Consider $$P\in\text{GL}_2(\mathbf{Q}_p)$$ as in the Theorem 1. Thus for each $$A\in S$$, $$PAP^{-1} = (a_{ij})$$ with $$|a_{11}|_p, |a_{21}|_p<1$$. $$\text{tr}(A) = \text{tr}(PAP^{-1}) = a_{11} + a_{22} = \lambda_1^A + \lambda_2^A$$ where $$\lambda_i^A$$ are eigenvalues of $$A$$. So $$|a_{11} + a_{22}|_p = |\lambda_1^A + \lambda_2^A|_p < 1$$ and since $$|a_{11}|_p<1$$, then in the case $$|a_{22}|_p\not= |a_{11}|_p$$, $$\max(|a_{11}|_p, |a_{22}|_p) = |a_{11} + a_{22}|_p < 1$$ and $$|a_{22}|_p < 1$$ as well.

• Follow-up question of mathoverflow.net/questions/432543
– YCor
Oct 16, 2022 at 11:01
• The joint-spectral radius is one of many generalizations of the notion of the spectral radius to multiple operators. en.wikipedia.org/wiki/Joint_spectral_radius Oct 17, 2022 at 16:38
• Gian-Carlo Rota AND W. Gilbert Strang have shown in the 1960 paper A Note On The Joint Spectral Radius that the joint-spectral radius $\rho_\infty(\mathcal{A})$ of a collection of $d\times d$-matrices $\mathcal{A}$ is just $\inf_{N\in\mathfrak{N}}\sup_{A\in\mathcal{A}}N(A)$ where $\mathfrak{N}$ is the collection of all multiplicative $d\times d$-norms. Oct 17, 2022 at 17:29