J-class abelian semigroups of matrices on ℝ

Let M_n(ℝ) be the set of all square matrices over ℝ of order n ≥ 1 and by GL(n, ℝ) the group of invertible matrices of M_nℝ. Let G be a finitely generated abelian sub-semigroup of M_n(ℝ). By a sub-semigroup of M_n(ℝ), we mean a subset which is stable under multiplication and contains the identity matrix. For a vector v ∈ ℂⁿ, we consider the orbit of G through v: G(v) = {Av: A ∈ G}⊂ ℝⁿ. A subset E ⊂ ℝⁿ is called G-invariant if A(E) ⊂ E for any A ∈ G. The orbit G(v)⊂ ℝⁿ is dense in ℝⁿ if $\bar{G (v)} = ℝ^{n}$ $\overline{G(v)}= {\mathbb{R}}^{n}$ , where Ē denotes the closure of a subset E⊂ ℝⁿ. The semigroup G is called hypercyclic if there exists a vector v ∈ℝⁿ such that G(v) is dense in ℝⁿ. We refer the reader to the recent books [4] and [9] for a thorough account on hypercyclicity. In [7], [8], Costakis and Manoussos localized the notion of hypercyclicity through the use of the J-sets, firstly for a single bounded linear operator T acting on a complex Banach space X, and secondly extended to that of semigroup. Recall that T is called a J-class operator if for every non-zero vector x in X and for every open neighborhood U ⊂ X of x and every non-empty open set V ⊂ X, there exists a positive integer n such that Tⁿ(U) ∩ V = ∅. Recently, there are much research that study or use J-class operators, We mention for instance, the series of papers by Azimi and Müller [3], Nasseri [10], Chan and Seceleanu [5]. Among properties, there are locally hypercyclic, non hypercyclic operators and that finite dimensional Banach spaces do not admit locally hypercyclic operators (see [7]).

In [8], Costakis and Manoussos extend the notion of J-class operator to that of a J-class of semigroup G as follows: Suppose that G is generated by p matrices A₁,…, A_p(p ≥ 1) then we define the extended limit set J_G(x) of x under G to be the set of y ∈ ℝⁿ for which there exists a sequence (x_m)_m ⊂ ℝⁿ and sequences of non-negative integers ${k_{m}^{(j)} : m \in ℕ}$ $\{k^{(j)}_{m}:\ m\in \mathbb{N}\}$ for j = 1, 2,…, p with $k_{m}^{(1)} + k_{m}^{(2)} + \dots + k_{m}^{(p)} \to + \infty$ $k^{(1)}_{m}+k^{(2)}_{m}+\dots + k^{(p)}_{m}\to+\infty$ such that x_m → x and $A_{1}^{k_{m}^{(1)}} A_{2}^{k_{m}^{(2)}} \dots A_{p}^{k_{m}^{(p)}} x_{m} \to y$ $A_{1}^{k^{(1)}_{m}}A_{2}^{k^{(2)}_{m}}\dots A_{p}^{k^{(p)}_{m}}x_{m}\to y$ .

This describes the asymptotic behavior of the orbits of nearby points to x. Note that the condition $k_{m}^{(1)} + k_{m}^{(2)} + \dots + k_{m}^{(p)} \to + \infty$ $\ k^{(1)}_{m} + k^{(2)}_{m}+\dots+ k^{(p)}_{m}\to+\infty$ is equivalent to having at least one of the sequences ${k_{m}^{(j)} : m \in ℕ}$ $\{k^{(j)}_{m}:\ m\in\mathbb{N}\}$ for j = 1, 2,…, p containing an increasing subsequence tending to +∞. We say that G is locally hypercyclic (or J-class) if there exists a vector v ∈ ℝⁿ\{0} such that J_G(v) = ℝⁿ. This notion is a “localization” of the concept of hypercyclicity, this can be justified by the following: J_G(x) = ℝⁿ if and only if for every open neighborhood U_x ⊂ ℝⁿ of x and every nonempty open set V ⊂ ℝⁿ there exists A ∈ G such that A(U_x)∩ V≠∅.

As we have mentioned above, in ℂⁿ or ℝⁿ, no matrix can be locally hypercyclic. However, what is rather remarkable is that in ℂⁿ or ℝⁿ, a pair of commuting matrices exists which forms a locally hypercyclic, non-hypercyclic semigroup (see [8]).

In the present work, we show that G is hypercyclic if and only if there exists a vector v in an open set V, defined according to the structure of G, such that J_G(v) = ℝⁿ (Theorem 1). This answer the question 1 raised by the author in [2]. Furthermore, we construct for every n ≥ 2, a locally hypercyclic abelian semigroup G generated by matrices A₁,…, A_n+1 which is non-hypercyclic whenever J_G(u_k) = ℝⁿ, k = 1,…, n, for a basis (u₁,…, u_n) of ℝⁿ (Theorem 4), this answers negatively the question raised by Costakis and Manoussos in [8]. However, we prove that the question is true (see Proposition 5) for any abelian semigroup G consisting of lower triangular matrices on ℝⁿ with all diagonal elements equal.

Before stating our main results, let introduce the following notations and definitions.

Set ℕ be the set of non negative integers and write ℕ₀ = ℕ\{0}. Let n ∈ ℕ₀ be fixed. Denote by:

ℬ; = (e₁,..., e₁) the canonical basis of ℝⁿ.

I_n the identity matrix on ℝⁿ.

For each m = 1, 2,…, n, denote by:

$T_{m} (ℝ)$ $\mathbb{T}_{m}(\mathbb{R})$ the set of matrices over ℝ of the form:

$[\begin{matrix} μ & 0 \\ a_{2, 1} & ⋱ \\ ⋮ & ⋱ & ⋱ \\ a_{m, 1} & \dots & a_{m, m - 1} & μ \end{matrix}]$ $$ \begin{equation}\left[\begin{array}{cccc} \mu & \ & \ & 0 \\ a_{2,1} & \ddots & \ & \ \\ \vdots & \ddots & \ddots & \ \\ a_{m,1} & \dots & a_{m,m-1} & \mu \end{array} \right] \end{equation} $$

𝕊; the set of matrices over ℝ of the form

$[\begin{matrix} α & β \\ - β & α \end{matrix}]$ $$ \begin{equation}\begin{bmatrix} \alpha & \beta \\ -\beta & \alpha \\ \end{bmatrix} \end{equation} $$

For each $1 \leq m \leq \frac{n}{2}$ $1\leq m\leq \frac{n}{2}$ , denote by

$B_{m} (ℝ)$ $\mathbb{B}_{m}(\mathbb{R})$ the set of matrices of M_2m(ℝ) of the form

$[\begin{matrix} C & 0 \\ C_{2, 1} & C \\ ⋮ & ⋱ & ⋱ \\ C_{m, 1} & \dots & C_{m, m - 1} & C \end{matrix}] : C, C_{i, j} \in S, 2 \leq i \leq m, 1 \leq j \leq m - 1$ $$ \begin{equation}\begin{bmatrix} C & & & 0 \\ C_{2,1} & C & & \\ \vdots & \ddots & \ddots & \\ C_{m,1} & \dots & C_{m,m-1} & C \end{bmatrix}: \ C, \ C_{i,j}\in \mathbb{S}, \ 2\leq i\leq m, 1\leq j\leq m-1 \end{equation} $$

Let r, s ∈ ℕ. By a partition of n we mean a finite sequence of positive integers

$η = {\begin{matrix} (n_{1}, \dots, n_{r}; m_{1}, \dots, m_{s}) & if r s \neq 0, \\ (m_{1}, \dots, m_{s}) & if r = 0, \\ (n_{1}, \dots, n_{r}) & if s = 0 \end{matrix}$ $$ \begin{equation}\eta = \begin{cases} (n_{1},\dots,n_{r};\ m_{1},\dots,m_{s}) & \textrm{if} rs\neq 0, \\ (m_{1},\dots,m_{s}) & \textrm{if} r= 0, \\ (n_{1},\dots,n_{r}) & \textrm{if} s=0 \end{cases} \end{equation} $$

such that (n₁+…+ n_r) +2(m₁+…+ m_s) = n. In particular, we have r + 2s ≤ n. The number r + 2s will be called the length of the partition.

Given a partition η = (n₁,…, n_r; m₁,…, m_s), denote by:

$K_{η} (ℝ) : = T_{n_{1}} (ℝ) \oplus \dots \oplus T_{n_{r}} (ℝ) \oplus B_{m_{1}} (ℝ) \oplus \dots \oplus B_{m_{s}} (ℝ)$ $\mathscr{K}_{\eta}(\mathbb{R}): = \mathbb{T}_{n_{1}}(\mathbb{R})\oplus\dots \oplus \mathbb{T}_{n_{r}}(\mathbb{R})\oplus \mathbb{B}_{m_{1}}(\mathbb{R})\oplus\dots \oplus \mathbb{B}_{m_{s}}(\mathbb{R})$ .

$K_{η}^{*} (ℝ) : = K_{η} (ℝ) \cap GL (n, ℝ)$ $\mathscr{K}^{*}_{\eta}(\mathbb{R}): = \mathscr{K}_{\eta}(\mathbb{R})\cap \textrm{GL}(n, \ \mathbb{R})$ , it is a sub-semigroup of GL(n, ℝ). In particular:

If r = 1, s = 0 then $K_{η} (ℝ) = T_{n} (ℝ)$ $\mathscr{K}_{\eta}(\mathbb{R}) = \mathbb{T}_{n}(\mathbb{R})$ and η = (n).

If r = 0, s = 1 then $K_{η} (ℝ) = B_{m} (ℝ)$ $\mathscr{K}_{\eta}(\mathbb{R}) = \mathbb{B}_{m}(\mathbb{R})$ and η = (m), n = 2m.

If r = 0, s = 1 then $K_{η} (ℝ) = B_{m_{1}} (ℝ) \oplus \dots \oplus B_{m_{s}} (ℝ)$ $\mathscr{K}_{\eta}(\mathbb{R}) = \mathbb{B}_{m_{1}}(\mathbb{R})\oplus\dots\oplus\mathbb{B}_{m_{s}}(\mathbb{R})$ and η = (m₁,…, m_s).

For a row vector v ∈ ℝⁿ, we will be denoting by v^T the transpose of v.

u_η = [e_η,1,…,e_η,r; f_η,1,…,f_η,s]^T ∈ ℝⁿ, where for every k = 1,…, r; l = 1,…, s, e_η,k = [1,0,…, 0]^T ∈ ℝ^n_k, $f_{η, l} = {[1, 0, \dots, 0]}^{T} \in ℝ^{2 m_{l}}$ $f_{\eta,l} = [1,0,\dots, 0]^{T}\in \mathbb{R}^{2m_{l}}$ .

We let

$U : = \underset{k = 1}{\prod^{r}} (ℝ^{*} \times ℝ^{n_{k} - 1}) \times \underset{l = 1}{\prod^{s}} ((ℝ^{2} \ {(0, 0)}) \times ℝ^{2 m_{l} - 2}) .$ $$ \begin{equation}U := \underset{k=1}{\overset{r}{\prod}}(\mathbb{R}^{*}\times\mathbb{R}^{n_{k}-1})\times \underset{l=1}{\overset{s}{\prod}}\left((\mathbb{R}^{2}\backslash\{(0,0)\})\times\mathbb{R}^{2m_{l}-2}\right). \end{equation} $$

It is plain that U is open and dense in ℝⁿ.

Let G be an abelian sub-semigroup of M_n(ℝ), we have the following “normal form of G”:

There exists a P ∈ GL(n, ℝ) such that $P^{- 1} G P \subset K_{η} (ℝ)$ $P^{-1}GP\subset \mathscr{K}_{\eta}(\mathbb{R})$ for some partition η of n (see Proposition 6).

Given two integers r, s ∈ ℕ, we shall say that the semigroup G has “a normal form of length r + 2s” if G has a normal form in 𝒦_η(ℝ) for some partition η with length r + 2s. For such a choice of matrix P, we let:

v_η = Pu_η.

V = P(U), it is a dense open set in ℝⁿ.

Our principal results are the following:

Theorem 1

LetG be a finitely generated abelian semigroup of matrices on ℝⁿ. If J_G(v) = ℝⁿfor somev ∈ Vthen $\bar{G (v_{η})} = ℝ^{n}$ $\overline{G(v_{\eta})} = \mathbb{R}^{n}$ .

Corollary 2

Under the hypothesis of Theorem 1, the following are equivalent:

G is hypercyclic.

J_G(v_η) = ℝⁿ.

$\bar{G (v_{η})} = ℝ^{n}$ $\overline{G(v_{\eta})} = \mathbb{R}^{n}$ .

Corollary 3

Under the hypothesis of Theorem 1, assume that G is not hypercyclic. Then

$E : = {x \in ℝ^{n} : J_{G} (x) = ℝ^{n}} \subset \underset{k = 1}{\overset{r}{\cup}} H_{k} \cup \underset{l = 1}{\overset{s}{\cup}} F_{l}, (r + 2 s \leq n)$ $$ \begin{equation}E: = \{x\in \mathbb{R}^{n}: J_{G}(x)= \mathbb{R}^{n}\}\subset \underset{k=1}{\overset{r}{\cup}}H_{k}\cup \underset{l=1}{\overset{s}{\cup}}F_{l}, \ (r+2s\leq n) \end{equation} $$

, where H_k (resp. F_l) are G-invariant vector subspaces of ℝⁿof dimension n − 1 (resp. n − 2).

Remark

If n = 1, then v_η = 1 and by Corollary 2, a sub-semigroup G of ℝ is hypercylic if and only if it is locally hypercyclic.

Theorem 4

Let n ≥ 2 be an integer. Then there exists an abelian semigroup G generated by diagonal matricesA₁, …,A_n+1 ∈ GL(n,ℝ) which is not hypercyclic such that J_G(e_k) = for every k = 1,…, n.

Proposition 5

Let G be a finitely generated abelian sub-semigroup of $T_{n} (ℝ)$ $\mathbb{T}_{n}(\mathbb{R})$ . If there exists a basis (e′₁,…,e′_nof ℝⁿsuch that J_G(e′_k) = ℝⁿfor everyk = 1,…, n, thenGis hypercyclic.

Notation and some results

Recall first the following proposition.

Proposition 6

([1], Proposition 2.2) Let G be an abelian sub-semigroup of M_n(ℝ). Then there exists a P ∈ GL(n, ℝ) such thatP⁻¹GPis an abelian sub-semigroup of𝒦_η(ℝ), for some partitionηofn.

Notation.

If G is a sub-semigroup of $T_{n} (K)$ $\mathbb{T}_{n}(\mathbb{K})$ $(K = ℝ$ $(\mathbb{K} = \mathbb{R}$ or ℂ, denote by $F_{G} = vect ({(B - μ_{B} I_{n}) e_{i} \in K^{n} : 1 \leq i \leq n - 1, B \in G})$ $$F_{G} = \textrm{vect}\big(\{(B - \mu_{B}I_{n})e_{i}\in \mathbb{K}^{n}: \ 1\leq i\leq n-1, \ B\in G\}\big)$$

the vector subspace generated by the family of vectors {(B − μ_BI_n)e_i ∈ K;ⁿ : 1 ≤ i ≤ n − 1 , B ∈ G}.

- rank(F_G) the rank of F_G. We have rank(F_G) ≤ n − 1.

If G is an abelian sub-semigroup of $B_{m} (ℝ)$ $\mathbb{B}_{m}(\mathbb{R})$ , (n = 2m), denote by:

$F_{G} = vect ({(\tilde{B} - C I_{n}) e_{i} \in ℝ^{n} : 1 \leq i \leq n - 1, \tilde{B} \in G})$ $$ F_{G} = \textrm{vect}\left(\{(\widetilde{B} - CI_{n})e_{i}\in \mathbb{R}^{n}: \ 1\leq i\leq n-1, \ \widetilde{B}\in G\}\right)$$

where

$\tilde{B} = [\begin{matrix} C & 0 \\ C_{2, 1} & C \\ ⋮ & ⋱ & ⋱ \\ C_{m, 1} & \dots & C_{m, m - 1} & C \end{matrix}] : C, C_{i, j} \in S, i = 2, \dots, m, j = 1, \dots, m - 1.$ $$ \begin{equation}\widetilde{B} = \begin{bmatrix} C & & & 0 \\ C_{2,1} & C & & \\ \vdots & \ddots & \ddots & \\ C_{m,1} & \dots & C_{m,m-1} & C \end{bmatrix}: \ C, \ C_{i,j}\in \mathbb{S}, \ i=2, \dots, m, j=1, \dots, m-1. \end{equation} $$

If G is an abelian sub-semigroup of 𝒦_n(ℝ), then for every B ∈ 𝒦_η(ℝ), we have $B = diag (B_{1}, \dots, B_{r}; {\tilde{B}}_{1}, \dots, {\tilde{B}}_{s})$ $B = \textrm{diag}(B_{1},\dots, B_{r}; \ \widetilde{B_{1}},\dots,\widetilde{B_{s}})$ with $B_{k} \in T_{n_{k}} (ℝ)$ $B_{k}\in\mathbb{T}_{n_{k}}(\mathbb{R})$ , ${\tilde{B}}_{l} \in B_{m_{l}} (ℝ)$ $\widetilde{B_{l}}\in\mathbb{B}_{m_{l}}(\mathbb{R})$ , k = 1,…, r, l = 1,…, s. Denote by:

G_k = {B_k : B ∈ G}, k = 1,…, r, it is an abelian sub-semigroup of $T_{n_{k}} (ℝ)$ $\mathbb{T}_{n_{k}}(\mathbb{R})$ .

${\tilde{G}}_{l} = {{\tilde{B}}_{l} : B \in G}$ $\widetilde{G}_{l} = \{\widetilde{B_{l}}\ : \ B\in G\}$ , l = 1,…, s, it is an abelian sub-semigroup of $B_{m_{l}} (ℝ)$ $\mathbb{B}_{m_{l}}(\mathbb{R})$ .

For every $x = {[x_{1}, \dots, x_{r}; {\tilde{x}}_{1}, \dots, {\tilde{x}}_{s}]}^{T} \in ℝ^{n}$ $x = [x_{1},\dots,x_{r}; \widetilde{x_{1}},\dots,\widetilde{x_{s}}]^{T}\in \mathbb{R}^{n}$ , where $x_{k} = {[a_{k, 1}, \dots, a_{k, n_{k}}]}^{T} \in ℝ^{*} \times ℝ^{n_{k} - 1}$ $x_{k} = [a_{k,1},\dots, a_{k,n_{k}}]^{T}\in\mathbb{R}^{*}\times\mathbb{R}^{n_{k}-1}$ , ${\tilde{x}}_{l} = {[b_{l, 1}, \dots, b_{l, m_{l}}]}^{T} \in (ℝ^{2} \ {(0, 0)}) \times ℝ^{2 m_{l} - 2}$ $\widetilde{x_{l}} = [b_{l,1},\dots, b_{l,m_{l}}]^{T}\in(\mathbb{R}^{2}\backslash\{(0,0)\})\times\mathbb{R}^{2m_{l}-2}$ , we let:

$H_{x_{k}} = ℝ x_{k} + F_{G_{k}}$ $H_{x_{k}} = \mathbb{R}x_{k}+F_{G_{k}}$ , k = 1,…, r.

${\tilde{H}}_{{\tilde{x}}_{l}} = C {\tilde{x}}_{l} + F_{{\tilde{G}}_{l}}$ $\widetilde{H}_{\widetilde{x_{l}}} = \mathcal{C}\widetilde{x_{l}} + F_{\widetilde{G}_{l}}$ , l = 1,…, s, where $C = {diag (C, \dots, C) : C \in S}$ $\mathcal{C}= \{\textrm{diag}(C,\dots, C): \ C\in \mathbb{S}\}$ .

$H_{x} = \underset{k = 1}{\overset{r}{\oplus}} H_{x_{k}} \oplus \underset{l = 1}{\overset{s}{\oplus}} {\tilde{H}}_{{\tilde{x}}_{l}}$ $H_{x} = \underset{k=1}{\overset{r}{\bigoplus}}H_{x_{k}} \bigoplus \underset{l=1}{\overset{s}{\bigoplus}}\widetilde{H}_{\widetilde{x_{l}}}$ .

We start with the following lemmas:

Lemma 7

LetGbe an abelian sub-semigroup of 𝒦_η(ℝ). Under the notations above, for everyx ∈ ℝⁿ, H_x is G-invariant.

Proof

It suffices to prove that $H_{x_{k}}$ $H_{x_{k}}$ is G_k-invariant (resp. ${\tilde{H}}_{{\tilde{x}}_{l}}$ $\widetilde{H}_{\widetilde{x_{l}}}$ is ${\tilde{G}}_{l}$ $\widetilde{G}_{l}$ -invariant): write for short G = G_k and x = x_k (resp. $G = {\tilde{G}}_{l}$ $G = \widetilde{G}_{l}$ and $x = {\tilde{x}}_{l})$ $x = \widetilde{x_{l}})$ . One has H_x = ℝ_x (resp. ${\tilde{H}}_{x} = C x + F_{G})$ $\widetilde{H}_{x} = \mathcal{C}x + F_{G})$ .

- If w = [w₁,…, w_n]^T ∈ H_x and B ∈ G with eigenvalue μ ∈ ℝ, then $B w = μ w + (B - μ I_{n}) w = μ w + \underset{i = 1}{\sum^{n - 1}} w_{k} (B - μ I_{n}) e_{i} \in H_{x}$ $Bw = \mu w+(B-\mu I_{n})w = \mu w+\underset{i=1}{\overset{n-1}{\sum}}w_{k}(B-\mu I_{n})e_{i}\in H_{x}$ (since w, (B−μ I_n)e_i ∈ H_x and H_x is a vector space).

- If $\tilde{w} = {[{\tilde{w}}_{1}, \dots, {\tilde{w}}_{s}]}^{T} \in {\tilde{H}}_{x}$ $\widetilde{w} = [\widetilde{w_{1}},\dots,\widetilde{w_{s}}]^{T}\in \widetilde{H}_{x}$ and B ∈ G, then we have $B \tilde{w} = C \tilde{w} + (B - C I_{n}) \tilde{w} = C \tilde{w} + \underset{i = 1}{\sum^{n - 1}} {\tilde{w}}_{k} (B - C I_{n}) e_{i} \in {\tilde{H}}_{x}$ $B\widetilde{w} = C\widetilde{w}+ (B- C I_{n})\widetilde{w} = C \widetilde{w} + \underset{i=1}{\overset{n-1}{\sum}}\widetilde{w_{k}}(B- C I_{n})e_{i}\in \widetilde{H}_{x}$ (since $C \tilde{w}, (B - C I_{n}) e_{i} \in {\tilde{H}}_{x}$ $C\widetilde{w}, \ (B-C I_{n})e_{i}\in \widetilde{H}_{x}$ and ${\tilde{H}}_{x}$ $\widetilde{H}_{x}$ is a vector space).

Proposition 8

Let G be an abelian sub-semigroup of 𝒦_η(ℝ). If J_G(u) = ℝⁿfor someu ∈ U, then rank(F_{G_k}) = n_k − 1 and $rank (F_{{\tilde{G}}_{l}}) = 2 m_{l} - 2$ $\mathrm{rank}(F_{\widetilde{G}_{l}}) = 2m_{l}-2$ , for everyk = 1,…, r; l = 1,…, s.

Proof

Let $u = {[u_{1}, \dots, u_{r}; {\tilde{u}}_{1}, \dots, {\tilde{u}}_{s}]}^{T} \in ℝ^{n}$ $u = [u_{1},\dots,u_{r}; \ \widetilde{u_{1}},\dots, \widetilde{u_{s}}]^{T}\in \mathbb{R}^{n}$ , where $u_{k} \in ℝ^{n_{k}}$ $u_{k}\in \mathbb{R}^{n_{k}}$ , ${\tilde{u}}_{l} \in ℝ^{2 m_{l}}$ $\widetilde{u_{l}}\in \mathbb{R}^{2m_{l}}$ , for every k = 1,…, r; l = 1,…, s.

i) First, we will show that $J_{G_{k}} (u_{k}) = ℝ^{n_{k}}$ $J_{G_{k}}(u_{k}) = \mathbb{R}^{n_{k}}$ and $J_{{\tilde{G}}_{l}} ({\tilde{u}}_{l}) = ℝ^{2 m_{l}}$ $J_{\widetilde{G}_{l}}(\widetilde{u_{l}}) = \mathbb{R}^{2m_{l}}$ . For this, let $x_{k} \in ℝ^{n_{k}}$ $x_{k}\in \mathbb{R}^{n_{k}}$ , ${\tilde{x}}_{l} \in ℝ^{2 m_{l}}$ $\widetilde{x_{l}}\in \mathbb{R}^{2m_{l}}$ and set $y = {[y_{1}, \dots, y_{r}; {\tilde{y}}_{1}, \dots, {\bar{y}}_{s}]}^{T} \in ℝ^{n}$ $y = [y_{1},\dots,y_{r};\widetilde{y_{1}},\dots,\widetilde{y_{s}}]^{T}\in \mathbb{R}^{n}$ such that

$y_{i} = {\begin{matrix} 0 \in ℝ^{n_{i}}, & if i \neq k \\ x_{k}, & if i = k . \end{matrix}$ $$ y_{i} = \begin{cases} 0\in \mathbb{R}^{n_{i}}, & \textrm{if } \ i\neq k\\ x_{k}, & \textrm{if } \ i= k. \end{cases}$$

and

$y_{i} = {\begin{matrix} 0 \in ℝ^{n_{i}}, & if i \neq k \\ x_{k}, & if i = k . \end{matrix}$ $$ \widetilde{y_{l}} = \begin{cases} 0\in \mathbb{R}^{2m_{l}}, & \textrm{if } \ i\neq l\\ \widetilde{x_{l}}, & \textrm{if } \ i= l. \end{cases}$$

As J_G(u) = ℝⁿ, there exist two sequences (z_m)_m ⊂ ℝⁿ and (B_m)_m⊂ G such that

$\lim_{m \to + \infty} z_{m} = u and \lim_{m \to + \infty} B_{m} z_{m} = y .$ $$ \begin{equation}\underset{m\to +\infty}\lim z_{m} = u\ \ \ \textrm{and} \ \ \ \underset{m\to +\infty}\lim B_{m}z_{m} =y.\ \ \ \ \ \ \ \end{equation} $$(2)

Write $z_{m} = {[z_{m, 1}, \dots, z_{m, r}; \tilde{z_{m, 1}}, \dots, \tilde{z_{m, s}}]}^{T}$ ${z_m} = {[{z_{m,1}}, \ldots ,{z_{m,r}};\;\widetilde {{z_{m,1}}}, \ldots ,\,\widetilde {{z_{m,s}}}]^T}$ with $z_{m, k} \in ℝ^{n_{k}} \tilde{z_{m, l}} \in ℝ^{2 m_{l}}$ \[{z_{m,k}} \in {\mathbb{R}^{{n_k}}}{\text{ }}\widetilde {{z_{m,l}}} \in {\mathbb{R}^{2{m_l}}}\] , and $B_{m} = diag (B_{m, 1}, \dots, B_{m, r}; \tilde{B_{m, 1}}, \dots, \tilde{B_{m, s}})$ $B_{m} = \textrm{diag}(B_{m,1},\dots, B_{m,r}; \ \widetilde{B_{m,1}},\dots,\widetilde{B_{m,s}})$ with $B_{m, k} \in T_{n_{k}} (ℝ) \tilde{B_{m, l}} \in B_{m_{l}} (ℝ) k = 1, \dots, r, l = 1, \dots, s$ $B_{m,k}\in\mathbb{T}_{n_{k}}(\mathbb{R}) \widetilde{B_{m,l}}\in\mathbb{B}_{m_{l}}(\mathbb{R}) k=1,\dots,r, l=1,\dots,s$ .

By (2), we have

$\lim_{m \to + \infty} z_{m, k} = u_{k}, \lim_{m \to + \infty} \tilde{z_{m, l}} = \tilde{u_{l}}$ $$\underset{m\to +\infty}\lim z_{m,k} = u_{k}, \ \underset{m\to +\infty}\lim \widetilde{z_{m,l}} = \widetilde{u_{l}}$$

and

$\lim_{m \to + \infty} B_{m, k} z_{m, k} = y_{k} = x_{k}, \lim_{m \to + \infty} \tilde{B_{m, l}} \tilde{z_{m, l}} = \tilde{y_{l}} = \tilde{x_{l}} .$ $$\underset{m\to +\infty}\lim B_{m,k}z_{m,k}=y_{k}=x_{k}, \ \underset{m\to +\infty}\lim \widetilde{B_{m,l}} \widetilde{z_{m,l}} = \widetilde{y_{l}}= \widetilde{x_{l}}.$$

Therefore x_k ∈ J_{G_k}(u_k) and $\tilde{x_{l}} \in J_{\tilde{G_{l}}} (\tilde{u_{l}})$ $\widetilde{x_{l}}\in J_{\widetilde{G}_{l}}(\widetilde{u_{l}})$ . It follows that J_{G_k}(u_k) = ℝ^n_k and $J_{\tilde{G_{l}}} (\tilde{u_{l}}) = ℝ^{2 m_{l}}$ $J_{\widetilde{G}_{l}}(\widetilde{u_{l}})= \mathbb{R}^{2m_{l}}$.

ii) Second, one can then suppose that G ⊂ 𝕋_n(ℝ) and u ∈ ℝ^* × ℝ^{n – 1} (resp. G ⊂ 𝔹_m(ℝ) and u ∈ (ℝ²\{(0, 0)}) × ℝ^{2m – 2}. It is clear that u ∉ F_G.

Assume that G ⊂ 𝕋_n(ℝ). By Lemma 7, H_e₁ = ℝ_e₁ + F_G is G-invariant.

Suppose that ℝⁿ\H_e₁ ≠ Ø, so there exist y ∈ ℝⁿ\H_e₁ and two sequences (x_m)_m ⊂ ℝⁿ and (B_m)_m ⊂ D such that $\lim_{m \to + \infty} x_{m} = e_{1}$ $\underset{m\to +\infty}\lim x_{m}=e_{1}$ and $\lim_{m \to + \infty} B_{m} x_{m} = y$ $\underset{m\to +\infty}\lim B_{m}x_{m}=y$ . Let H_{x_m} = ℝ_{x_m} + F_G for every m ∈ ℕ. By Lemma 7, H_{x_m} is G-invariant, so B_mx_m, for every m ∈ ℕ. Write B_mx_m = α_mx_m + z_mα_m ∈ ℝ and z_m ∈ F_G. We distinguish two cases:

If (α_m)_m is bounded, one can suppose by passing to a subsequence, that (α_m)α_{m ≥ 1} is convergent, say $\lim_{m \to + \infty} α_{m} = a \in ℝ$ $\underset{m\to +\infty}\lim \alpha_{m} = a\in\mathbb{R}$ . It follows that $\lim_{m \to + \infty} z_{m} = y - a u \in F_{G}$ $\underset{m\to +\infty}\lim z_{m} = y-au\in F_{G}$ and so y ∈ H_e₁, a contradiction.

If (α_m)_m is not bounded, one can suppose by passing to a subsequence, that $\lim_{m \to + \infty} | α_{m} | = + \infty$ $\underset{m\to +\infty}\lim|\alpha_{m}| = +\infty$ , then $\lim_{m \to + \infty} \frac{1}{α_{m}} z_{m} = - u \in F_{G}$ $\underset{m\to +\infty}\lim \frac{1}{\alpha_{m}}z_{m} = -u\in F_{G}$ , a contradiction. We conclude that H_e₁ = ℝⁿ and so dim (F_G) = n – 1.

Assume that G ⊂ 𝔹_m(ℝ) (n = 2m). By Lemma 7, H_e₁ = 𝒞_e₁ + F_G is G-invariant. Suppose that ℝⁿ\H_e₁ ≠ Ø and let y ∈ ℝⁿ\H_e₁. Then there exist two sequences (x_k)_k ⊂ ℝⁿ and (B_k)_k ⊂ G such that $\lim_{k \to + \infty} x_{k} = e_{1}$ $\underset{k\to +\infty}\lim x_{k} = e_{1}$ and $\lim_{k \to + \infty} B_{k} x_{k} = y$ $\underset{k\to +\infty}\lim B_{k}x_{k}=y$ . Let H_{x_k} = 𝒞_{x_k} + F_G for every k ∈ ℕ. By Lemma 7, H_{x_k} is G-invariant, so B_kx_k ∈ H_{x_k}, for every k ∈ ℕ. Write B_kx_k = C_kx_k + z_k, with C_k = diag (R_k,..., R_k), $R_{k} = [\begin{matrix} α_{k} & β_{k} \\ - β_{k} & α_{k} \end{matrix}],$ $R_{k}=\begin{bmatrix} \alpha_{k} & \beta_{k} \\ -\beta_{k} & \alpha_{k} \\ \end{bmatrix},$ , and z_k ∈ F_G. Take ||C_k|| = ||R_k|| = sup(|α_k|, |β_k|). We distinguish two cases:

If (||C_k||)_k is bounded, one can suppose by passing to a subsequence, that (α_k)_k and (β_k)_k converge, say $\lim_{k \to + \infty} α_{k} = α \in ℝ$ $\underset{k\to +\infty}\lim \alpha_{k} = \alpha\in\mathbb{R}$ and $\lim_{k \to + \infty} β_{k} = β \in ℝ$ $\underset{k\to +\infty}\lim \beta_{k} = \beta\in\mathbb{R}$ . As $\lim_{k \to + \infty} C_{k} e_{1} = α e_{1} - β e_{2} = C e_{1}$ $\underset{k\to +\infty}\lim C_{k}e_{1}= \alpha e_{1}-\beta e_{2}=Ce_{1}$ , where C = diag(R,..., R) with $R = [\begin{matrix} α & β \\ - β & α \end{matrix}],$ $R=\begin{bmatrix} \alpha& \beta \\ -\beta & \alpha \\ \end{bmatrix},$ and as B_kx_k = C_kx_k = C_k(x_k – e₁) + C_ke₁ + z_k, then $\lim_{k \to + \infty} z_{k} = y - C e_{1} \in F_{G}$ $\underset{k\to +\infty}\lim z_{k} = y-Ce_{1}\in F_{G}$ and therefore y ∈ Ce₁ + F_G ⊂ H_e₁, a contradiction.

If (||C_k||)_k is not bounded, one can suppose by passing to a subsequence, that $\lim_{k \to + \infty} ‖ C_{k} ‖ = + \infty$ $\underset{k\to +\infty}\lim\|C_{k}\| = +\infty$ , then C_k is invertible for k large, so from B_kx_k = C_kx_k = C_k(x_k – e₁) + C_ke₁ + z_k, we have $\frac{1}{‖ C_{k} ‖} C_{k} e_{1} + \lim_{k \to + \infty} \frac{1}{‖ C_{k} ‖} z_{k} = 0$ $ \frac{1}{\|C_{k}\|}C_{k}e_{1}+ \underset{k\to +\infty}\lim \frac{1}{\|C_{k}\|}z_{k} = 0$ . In particular, we get $\lim_{k \to + \infty} \frac{α_{k}}{‖ C_{k} ‖} = \lim_{k \to + \infty} \frac{β_{k}}{‖ C_{k} ‖} = 0$ $\underset{k\to +\infty}\lim \frac{\alpha_{k}}{\|C_{k}\|} = \underset{k\to +\infty}\lim \frac{\beta_{k}}{\|C_{k}\|}=0$ , this is a contradiction with ||C_k|| (|α_k|, |β_k|).

We conclude that H_e₁ = ℝⁿ and so dim (F_G) = n – 1.

Proof of Theorem 1, Corollaries 2 and 3

Lemma 9.

LetGbe an abelian sub-semigroup of 𝕋_n(𝕂) (𝕂= ℝ or 𝕔) such that rank (F_G) = n – 1. Let u, ∈ 𝕂^* × 𝕂^{n – 1}. If two sequences (u_m}_m∈ℕin 𝕂^* × 𝕂^{n – 1} and (B_m)_m∈ℕinGsuch that $\lim_{m \to + \infty} u_{m} = u$ $\underset{m\to+\infty}\lim u_{m} = u$ and $\lim_{m \to + \infty} B_{m} u_{m} = v$ $\underset{m\to+\infty}\lim B_{m}u_{m} = v$ then (B_m)_m∈ℕis bounded.

Proof

The proof is down in ([2], Lemma 3.4) if $K = ℂ$ $\mathbb{K}= \mathbb{C}$ and the same proof works if $K = ℝ$ $\mathbb{K}= \mathbb{R}$ .

Now consider the following basis change:

Assume that n = 2m, m ∈ ℕ₀. For every k = 1, …, m, we let:

$e_{k}^{'} = \frac{e_{2 k - 1} - i e_{2 k}}{2}$ $e^{\prime}_{k} = \frac{e_{2k-1}-ie_{2k}}{2}$ and $C_{0} = (e_{1}^{'}, \dots, e_{m}^{'}, \bar{e_{1}^{'}}, \dots, \bar{e_{m}^{'}})$ $\mathscr{C}_{0} = (e^{\prime}_{1},\dots,e^{\prime}_{m}, \ \overline{e^{\prime}_{1}},\dots,\overline{e^{\prime}_{m}})$ , where $\bar{u} = {[\bar{z_{1}}, \dots, \bar{z_{m}}]}^{T}$ $\overline{u}=[\overline{z_{1}},\dots,\overline{z_{m}}]^{T}$ is the conjugate of u = [z₁,…, z_m]^T. Then 𝒞₀ is a basis of ℂ^2m. Denote by Q ∈ GL(2m, ℂ) the matrix of basis change from ℬ₀ to 𝒞₀. Then a simple calculation yields that:

Lemma 10

Under the notation above, for every $B \in B_{m} (ℝ)$ $B\in\mathbb{B}_{m}(\mathbb{R})$ , $Q^{- 1} B Q = diag (B_{1}^{'}, \bar{B_{1}^{'}})$ $Q^{-1}BQ = \mathrm{diag}(B^{\prime}_{1},\ \overline{B^{\prime}_{1}})$ , where $B_{1}^{'} \in T_{m} (ℂ)$ $B^{\prime}_{1}\in \mathbb{T}_{m}(\mathbb{C})$ .

Lemma 11

Let G be an abelian sub-semigroup of $B_{m} (ℝ)$ $\mathbb{B}_{m}(\mathbb{R})$ such that rank(F_G) = 2m − 2. Letv ∈ (ℝ²\{(0,0)}) × ℝ^2m−2. If two sequences (u_m)_m∈ℕ ⊂ (ℝ²\{(0,0)}) × ℝ^2m−2and (B_m)_m∈ℕ ⊂ Gsuch that $\lim_{m \to + \infty} u_{m} = u$ $\underset{m\to+\infty}\lim u_{m} = u$ and $\lim_{m \to + \infty} B_{m} u_{m} = v$ $\underset{m\to+\infty}\lim B_{m}u_{m} = v$ then (B_m)_m∈ℕis bounded.

Proof

For B ∈ G, we have $Q^{- 1} B Q = diag ({B^{'}}_{1}, \bar{{B^{'}}_{1}})$ $Q^{-1}BQ = \textrm{diag}(B^{\prime}_{1},\ \overline{B^{\prime}_{1}})$ , where ${B^{'}}_{1} \in T_{m} (ℂ)$ $B^{\prime}_{1}\in \mathbb{T}_{m}(\mathbb{C})$ . Write G′₁ = {B′₁ : B ∈ G} and $\bar{{G^{'}}_{1}} = {\bar{{B^{'}}_{1}} : B \in G}$ $\overline{G^{\prime}_{1}} = \{\overline{B^{\prime}_{1}}: B\in G\}$ . Then G′₁ (resp. $\bar{{G^{'}}_{1}}$ $\overline{G^{\prime}_{1}}$ ) is an abelian sub-semigroup of $T_{m} (ℂ)$ $\mathbb{T}_{m}(\mathbb{C})$ .

First we prove that $rank (F_{{G^{'}}_{1}}) = m - 1$ $\textrm{rank}(F_{G^{\prime}_{1}}) = m-1$ . Write C = diag(R,…,R) with $R = [\begin{matrix} α & β \\ - β & α \end{matrix}]$ $R = \begin{bmatrix} \alpha& \beta \\ -\beta & \alpha \\ \end{bmatrix}$ and μ = α + iβ. One has

$\begin{matrix} Q^{- 1} (F_{G}) & = vect ({(Q^{- 1} B Q - Q^{- 1} C Q I_{2 m}) Q^{- 1} e_{i} : 1 \leq i \leq n - 1, B \in G}) \\ = vect ({(diag ({B^{'}}_{1}, \bar{{B^{'}}_{1}}) - diag (μ I_{m}, \bar{μ} I_{m})) {e^{'}}_{1} : 1 \leq i \leq n - 1, {B^{'}}_{1} \in {G^{'}}_{1}}) \\ = F_{{G^{'}}_{1}} \oplus F_{\bar{{G^{'}}_{1}}} \end{matrix}$ $$ \begin{align*} Q^{-1}(F_{G}) & = \textrm{vect}\left(\{(Q^{-1}BQ-Q^{-1}CQ I_{2m})Q^{-1}e_{i}: 1\leq i\leq n-1, \ B\in G\}\right) \\ & = \textrm{vect}\left(\{(\textrm{diag}(B^{\prime}_{1},\ \overline{B^{\prime}_{1}})-\textrm{diag}(\mu I_{m}, \overline{\mu}I_{m}))e_{i}^{\prime}: 1\leq i\leq n-1, \ B^{\prime}_{1}\in G_{1}^{\prime}\}\right) \\ & = F_{G_{1}^{\prime}}\oplus F_{\overline{G_{1}^{\prime}}} \end{align*} $$

As rank(F_G) = 2m−2 then rank (Q⁻¹(F_G))= 2m−2 = 2rank(F_G′₁). So rank(F_G′₁) = m − 1.

Second, we let u′_m = Q^-1_{u_m} = [u′_m,1, u′_m,2]^T], u′ = Q^-1u = [u′₁, u′₂]^T and v′ = Q^-1v = [v′₁,v′₂]^T, where u′_m,i, u′_i, v′_i ∈ ℂ^m, i = 1, 2. As u, v, u_m ∈ (ℝ²\{(0,0)}) × ℝ^2_m−2 then u′₁, v′₁, u′_m,1 ∈ ℂ* × ℂ^m−1. We have $\lim_{m \to + \infty} {u^{'}}_{m} = u^{'}$ $\underset{m\to+\infty}\lim u_{m}^{\prime} = u^{\prime}$ and so $\lim_{m \to + \infty} {u^{'}}_{m, 1} = {u^{'}}_{1}$ $\underset{m\to+\infty}\lim u_{m,1}^{\prime} = u_{1}^{\prime}$ . Take B′_m = Q^-1B_mQ. By Lemma 10, ${B^{'}}_{m} = diag ({B^{'}}_{m, 1}, \bar{{B^{'}}_{m, 1}})$ $B_{m}^{\prime} = \textrm{diag}(B_{m,1}^{\prime},\ \overline{B_{m,1}^{\prime}})$ , where ${B^{'}}_{m, 1} \in T_{m} (ℂ)$ $B_{m,1}^{\prime}\in \mathbb{T}_{m}(\mathbb{C})$ . Then B′_m,1 ∈ G′₁ and $\lim_{m \to + \infty} {B^{'}}_{m} {u^{'}}_{m} = v^{'}$ $\underset{m\to+\infty}\lim B_{m}^{\prime}u_{m}^{\prime} = v^{\prime}$ . So $\lim_{m \to + \infty} {B^{'}}_{m, 1} {u^{'}}_{m, 1} = {v^{'}}_{1}$ $\underset{m\to+\infty}\lim B_{m,1}^{\prime}u_{m,1}^{\prime} = v^{\prime}_{1}$ . By Lemma 9, (B_m,1)_m∈ℕ is bounded, so is (B′_m)_m∈ℕ. We conclude that (B_m)_m∈ℕ bounded.

It follows from Lemmas 9 and 11, the following:

Corollary 12

Let G be a finitely generated abelian sub-semigroup of 𝒦_η(ℝ). Suppose that rank(F_{G_k}) = n_k−1, $rank (F_{{\tilde{G}}_{l}}) = 2 m_{l} - 2$ $\mathrm{rank}(F_{\widetilde{G}_{l}})= 2m_{l}-2$ , k = 1,…, r; l = 1,…, s. Ifx, y ∈ Uand two sequences (B_m)_m∈ℕ ⊂ Gand (x_m)_m∈ℕ ⊂ ℝⁿsuch that $\lim_{m \to + \infty} x_{m} = x$ $\underset{m\to+\infty}\lim x_{m}=x$ and $\lim_{m \to + \infty} B_{m} x_{m} = y$ $\underset{m\to+\infty}\lim B_{m}x_{m} = y$ then (B_m)_m∈ℕis bounded.

Proposition

Let G be a finitely generated abelian sub-semigroup ofM_n(ℝ). Then G is hypercyclic if and only ifJ_G(x) = ℝⁿfor everyx∈ℝⁿ.

Proof

[Proof of Theorem 1] One can assume, by Proposition 6, that G is a sub-semigroup of 𝒦_η(ℝ). Suppose that J_G(u) = ℝⁿ. Then by Proposition 8, rank(F_{G_k}) = n_k−1 and $rank (F_{{\tilde{G}}_{l}}) = 2 m_{l} - 2$ $\mathrm{rank}(F_{\widetilde{G}_{l}})= 2m_{l}-2$ , for every k = 1,…, r; l = 1,…, s. Let y ∈ U, then there exist two sequences (B_m)_m∈ℕ ⊂ G and (x_m)_m∈ℕ ⊂ ℝⁿ satisfying:

$\lim_{m \to + \infty} x_{m} = u and \lim_{m \to + \infty} B_{m} x_{m} = y .$ $$ \begin{equation}\underset{m\to+\infty}\lim x_{m}=u\ \ \ \textrm{and}\ \ \ \underset{m\to+\infty}\lim B_{m}x_{m}=y. \end{equation} $$

So by Corollary 12, (B_m)_m∈ℕ is bounded: ||B_m|| ≤ M for some M > 0, where ||·|| is the Euclidean norm on ℝⁿ. Then

$\begin{matrix} ∥ B_{m} u - y ∥ & = ∥ B_{m} u - B_{m} x_{m} + B_{m} x_{m} - y ∥ \\ \leq ∥ B_{m} u - B_{m} x_{m} ∥ + ∥ B_{m} x_{m} - y ∥ \\ \leq ∥ B_{m} ∥ ∥ u - x_{m} ∥ + ∥ B_{m} x_{m} - y ∥ \\ \leq M ∥ u - x_{m} ∥ + ∥ B_{m} x_{m} - y ∥ \end{matrix}$ $$ \begin{align*} \|B_{m}u-y\|& =\|B_{m}u-B_{m}x_{m}+B_{m}x_{m}-y\|\\ \ & \leq\|B_{m}u-B_{m}x_{m}\| + \|B_{m}x_{m}-y\|\\ \ & \leq \|B_{m}\| \|u-x_{m}\| + \|B_{m}x_{m}-y\|\\ \ & \leq M \|u-x_{m}\| + \|B_{m}x_{m}-y\| \end{align*} $$

Thus $\lim_{m \to + \infty} B_{m} u = y$ $\underset{m\to+\infty}\lim B_{m}u=y$ and so $y \in \bar{G (u)}$ $y\in \overline{G(u)}$ . It follows that $U \subset \bar{G (u)}$ $U\subset\overline{G(u)}$ . Since $\bar{U} = ℝ^{n}$ $\overline{U} = \mathbb{R}^{n}$ , we get $\bar{G (u)} = ℝ^{n}$ $\overline{G(u)} = \mathbb{R}^{n}$ . □

Proof of Corollary 2. (i)⟹(ii) follows from Proposition 13.

(ii)⟹(iii) results from Theorem 1. (iii)⟹(i) is clear.

Proof of Corollary 3. If G is not hypercyclic then by Theorem 1,

J_G(v) ≠ ℝⁿ for any v ∈ V. Thus V ∩ E = ∅ and therefore $E \subset ℝ^{n} \ V = P (\underset{k = 1}{\overset{r}{\cup}} H_{k} \cup \underset{l = 1}{\overset{s}{\cup}} F_{l})$ $E\subset\mathbb{R}^{n}\backslash V = P(\underset{k=1}{\overset{r}{\bigcup}}H_{k} \cup \underset{l=1}{\overset{s}{\bigcup}}F_{l})$ , where

$H_{k} := (u = [u_{1}, \dots, u_{r}; {\tilde{u}}_{1}, \dots, {\tilde{u}}_{s}]^{T} \in R^{n}; u_{j} \in R^{n_{j}}, u_{k} \in {0} \times R^{n_{k} - 1}, 1 \leq j \neq k \leq r)$ $$ \begin{equation} H_{k}: = \left\{u = [u_{1},\dots,u_{r};\ \widetilde{u}_{1},\dots,\widetilde{u}_{s}]^{T}\in \mathbb{R}^{n}; \ u_{j}\in \mathbb{R}^{n_{j}}, u_{k}\in\{0\}\times\mathbb{R}^{n_{k}-1}, \ 1\leq j\neq k\leq r \right\} \end{equation} $$

and $F_{l} := (u = [u_{1}, \dots, u_{r}; {\tilde{u}}_{1}, \dots, {\tilde{u}}_{s}]^{T} \in R^{n} : {\tilde{u}}_{l} \in {(0, 0)} \times R^{2 m_{l} - 2}, 1 \leq l \leq s)$ $F_{l}: =\left\{u=[u_{1},\dots,u_{r};\ \widetilde{u}_{1},\dots,\widetilde{u}_{s}]^{T}\in \mathbb{R}^{n}: \ \widetilde{u}_{l}\in\{(0,0)\}\times\mathbb{R}^{2m_{l}-2}, \ 1\leq l\leq s\right\}$ . □

Proof of Theorem 4 and Proposition 5

Lemma 14

[6], Lemma 2.6]. Leta, b ∈ ℝ with −1 < a < 0, b > 1 and $\frac{\log | a |}{\log b}$ $\dfrac{\log|a|}{\log b}$ is irrational. Then the set {a^kb^l : k, l ∈ ℕ} is dense in ℝ.

Proof

[Proof of Theorem 4) Consider the abelian sub-semigroup G of GL(n, ℝ) generated by B, A₁,…, A_n, where B = bI_n and $A_{k} = diag (\underset{(k - 1) - t e r m s}{\underset{}{\underset{︸}{1, \dots \dots, 1}}}, a, 1 \dots, 1), k = 1, \dots, n$ $A_{k} = \textrm{diag}(\underset{(k-1)-terms}{\underbrace{1,\dots\dots,1}},a,1\dots,1),\ k=1,\dots, n$ . Then G is a sub-semigroup of $K_{η}^{*} (ℝ)$ $\mathscr{K}^{*}_{\eta}(\mathbb{R})$ with r = n and η = (1,…, 1). One has u_η = [1,…, 1]^T.

First, we will show that G is not hypercyclic: for this, it is equivalent to prove, by Corollary 2, that $\bar{G (u_{η})} \neq ℝ^{n}$ $\overline{G(u_{\eta})}\neq \mathbb{R}^{n}$ : We have

$G (u_{η}) = {{[b^{m} a^{k_{1}}; b^{m} a^{k_{2}}; \dots \dots; b^{m} a^{k_{n}}]}^{T} : m, k_{1}, \dots, k_{n} \in ℕ}$ $$ \begin{equation}G(u_{\eta}) = \left\{[b^{m}a^{k_{1}};\ b^{m}a^{k_{2}};\dots\dots;\ b^{m}a^{k_{n}}]^{T}:~ m,k_{1},\dots, k_{n}\in\mathbb{N}\right\} \end{equation} $$

Observe that for every x = [x₁,…, x_n]^T ∈ G(u_η), we have $\frac{x_{1}}{x_{2}} = a^{k_{1} - k_{2}}$ $\dfrac{x_{1}}{x_{2}}= a^{k_{1}-k_{2}}$ . Since the set ${a^{p} : p \in ℤ}$ $\left\{a^{p} :\ \ p\in \mathbb{Z}\right\}$ is not dense in ℝ, the orbit G(u_η) cannot be dense in ℝⁿ.

Second, we will show that J_G(e₁) = ℝⁿ (the other e_k work in the same way). Fix a vector y = [y₁, …, y_{n]^T ∈ ℝⁿ such that y₁ ≠ 0. By Lemma 14, choose two sequences of positive integers (i_m)_{m ∈ℕ} and (j_m)_m∈ℕ} with i_m, j_m → +∞ such that $\lim_{m \to + \infty} a^{i_{m}} b^{j_{m}} = y_{1}$ $\underset{m\to +\infty}\lim a^{i_{m}}b^{j_{m}}= y_{1}$ . Let $x^{(m)} = (1, x_{2}^{(m)}, \dots, x_{n}^{(m)})$ $x^{(m)} = (1, x^{(m)}_{2},\dots, x^{(m)}_{n})$ with $x_{k}^{(m)} = \frac{y_{k}}{a^{i_{m}} b_{j_{m}} a^{m}}$ $x^{(m)}_{k} = \frac{y_{k}}{a^{i_{m}}b_{j_{m}}a^{m}}$ , k = 2,…, n. Since −1 < a < 0 and y₁ ≠ 0, we see that $\lim_{m \to + \infty} x^{(m)} = e_{1}$ $\underset{m\to+\infty}{\lim}x^{(m)}=e_{1}$ . On the other hand, consider $B_{m} : = B^{j_{m}} A_{1}^{i_{m}} A_{2}^{i_{m} + m} A_{3}^{i_{m} + m} \dots A_{n}^{i_{m} + m}$ $B_{m}: = B^{j_{m}}A_{1}^{i_{m}}A_{2}^{i_{m}+m}A_{3}^{i_{m}+m}\dots A_{n}^{i_{m}+m}$ . Then $B_{m} x^{(m)} = {[a^{i_{m}} b^{j_{m}}, y_{2}, \dots, y_{n}]}^{T}$ $B_{m}x^{(m)} = \left[a^{i_{m}}b^{j_{m}},y_{2},\dots,y_{n}\right]^{T}$ and so, $\lim_{m \to + \infty} B_{m} x^{(m)} = y$ $\underset{m\to+\infty}{\lim}B_{m}x^{(m)}=y$ . We conclude that y ∈ J_G(e₁) and therefore J_G(e₁) = ℝⁿ.

Proof of Proposition 5. Since (e′₁,…,e′_n) is a basis of ℝⁿ, there exists i₀ ∈ {1,…, n} such that ℝ* × ℝⁿ⁻¹. As V = U = ℝ* × ℝⁿ⁻¹ and $J_{G} ({e^{'}}_{i_{0}}) = ℝ^{n}$ ${\rm J}_{G}(e^{\prime}_{i_{0}}) = \mathbb{R}^{n}$ then by Theorem 1, $\bar{G (e_{i_{0}}^{'})} = ℝ^{n}$ $\overline{G(e^{\prime}_{i_{0}})} = \mathbb{R}^{n}$ and hence G is hypercyclic.

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J-class abelian semigroups of matrices on ℝn

Publicado en línea: 08 dic 2017

Páginas: 519 - 528

Recibido: 07 may 2017

Aceptado: 08 dic 2017

DOI: https://doi.org/10.21042/AMNS.2017.2.00043

Palabras claveHypercyclic semigroup, locally hypercyclic, J-class operator, extended limit set, dense orbit, semigroup

© 2017 Habib Marzougui published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

J-class abelian semigroups of matrices on ℝⁿ

Palabras clave
Hypercyclic semigroup, locally hypercyclic, J-class operator, extended limit set, dense orbit, semigroup