# When is the orbit of a vector a minimal sequence? When does an operator have a minimal orbit vector?

Let $X$ be a banach space. We say a sequence $(x_n)$ is minimal if for each $k$, $x_k\notin [x_n]_{n\neq k}$, where $[x,y,z,\cdots]$ is the closed linear span of the vectors.

For $T\in\mathcal{L}(X)$, a bounded linear operator on $X$, $Orb(T,x)=\{x,Tx,T^2x,\cdots\}$ is the orbit of $x$ under $T$.

The question I am asking is that when an operator has a vector whose orbit is minimal. Since the existence of such an operator would imply the existence of almost invariant half spaces for a large class of operators, including all quasi nilpotent ones in particular, such a condition would be very interesting.

I am not quite sure where to look. But any conditions on either the operator or the vector would be helpful.

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What is the simplest case that you cannot already do? An important part of approaching such general questions is to break them down into smaller ones. – user16299 Jun 9 '12 at 22:46
@YemonChoi I totally agree with you. But the only cases I know are that i) when the operator is triangularizable and injective, and ii) when the operator is a weighted shift with nonzero weights decreasing to 0. I am not sure they are easy or not because they are dealt with very different techniques. – Hui Yu Jun 9 '12 at 23:30