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Abstract

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.