pith. sign in

arxiv: 1205.3575 · v3 · pith:WZVGBPIInew · submitted 2012-05-16 · 🧮 math.FA

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

classification 🧮 math.FA
keywords mathbbsupercyclicstronglydensedimensionalfinitefraclfloor
0
0 comments X
read the original abstract

We prove that on $\mathbb{R}^n$, there is no $N$-supercyclic operator with $1\leq N< \lfloor \frac{n+1}{2}\rfloor$ i.e. if $\mathbb{R}^n$ has an $N$ dimensional subspace whose orbit under $T$ is dense in $\mathbb{R}^n$, then $N$ is greater than $\lfloor\frac{n+1}{2}\rfloor$. Moreover, this value is optimal. We then consider the case of strongly $N$-supercyclic operators. An operator $T$ 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 dimension.

This paper has not been read by Pith yet.

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.