$n$-supercyclic and strongly $n$-supercyclic operators in finite dimensions

Romuald Ernst¹

¹ Clermont Université, UniversitéBlaise Pascal Laboratoire de Mathématiques bp 10448, F-63000 Clermont-Ferrand, France and CNRS, UMR 6620 Laboratoire de Mathématiques F-63177 Aubière, France

Studia Mathematica, Tome 220 (2014) no. 1, pp. 15-53

Résumé

We prove that on $\mathbb{R}^N$, there is no $n$-supercyclic operator with $1\leq n \lfloor {(N+1)}/{2}\rfloor$, i.e. if $\mathbb{R}^N$ has an $n$-dimensional subspace whose orbit under $T\in\mathcal{L}(\mathbb{R}^N)$ is dense in $\mathbb{R}^N$, then $n$ is greater than $\lfloor{(N+1)}/{2}\rfloor$. Moreover, this value is optimal. We then consider the case of strongly $n$-supercyclic operators. An operator $T\in\mathcal{L}(\mathbb{R}^N)$ is strongly $n$-supercyclic if $\mathbb{R}^N$ has an $n$-dimensional subspace whose orbit under $T$ is dense in $\mathbb{P}_n(\mathbb{R}^N)$, the $n$th Grassmannian. We prove that strong $n$-supercyclicity does not occur non-trivially in finite dimensions.

Détail

DOI : 10.4064/sm220-1-2

Keywords: prove mathbb there n supercyclic operator leq lfloor rfloor mathbb has n dimensional subspace whose orbit under mathcal mathbb dense mathbb greater lfloor rfloor moreover value optimal consider strongly n supercyclic operators operator mathcal mathbb strongly n supercyclic mathbb has n dimensional subspace whose orbit under dense mathbb mathbb nth grassmannian prove strong n supercyclicity does occur non trivially finite dimensions