Perhaps this question is too easy for mathoverflow, at least this is how it seems, but I got no answer on stackexchange. 


Suppose $T$ is a bounded linear operator on $l\_2$ and $x\in l\_2$ is a vector such that the orbit $(T^{n}x)$ is linearly independent. 


Can one find an $\epsilon>0$ such that for all $||y-x||<\epsilon$ the orbit $(T^{n}y)$ is also linearly independent?


This is probably not true in general. Does then a weaker requirement hold: that for all $\epsilon>0$ we can find a $y\in l\_2$ with $||y-x||<\epsilon$ and the orbit $(T^{n}x)$ is linearly independent?


What if *linearly independent* is replace by *not linearly independent* or by *basic sequence*? 



Behavior of orbits under small perturbations
The first version does not work, as you suspected. For example, if $Te\_n= (1/n)e\_{n+1}$ for $n\ge 0$ on $H=\ell^2(\mathbb Z)$, then the $T^n e\_0$, $n\ge 0$ are linearly independent. However, we can easily arrange matters in such a way that $T^N(e\_0+\delta e\_{-1})=0$ for a small $\delta>0$: $T$ will map $e\_{-1}$ to $e\_{-2}$ and then to $e\_{-3}$ etc. We wait until $T^Ne\_0$ has the same size as $\delta$, and then we finally map $e\_{-N}$ to $-e\_N$. 


Then we just continue in this way: we define $Te\_j$ for $j<-N$ such that $T^{M} (e\_0+\delta' e\_{-N-1})=0$ for an even smaller $\delta'>0$ etc.


The second claim is true. In fact, the set of vectors $y$ for which $\{T^ny\}$ is l.i. is a dense $G\_{\delta}$ set. This follows from the observation that
$$
A\_N = \{ y\in H: y,Ty,\ldots , T^N y \:\:\textrm{are l.i.}\}
$$
is a dense open set. It's easy to see that $A\_N$ is open, and to prove that it's dense, suppose that we had a $y\in H$ and an $N\ge 1$ such that $y+z\notin A\_N$ for all $\|z\|<r$, for some $r>0$. Then, if $N$ was minimal with this property, we have that
$$
T^Nz \in L(y,Ty, \ldots, T^{N-1}y, z, Tz, \ldots T^{N-1} z ) \quad\quad\quad\quad (1)
$$
for all $z$ of small norm. However, since $T$ is linear, (1) then holds for all $z\in H$. Moreover, $T^n z$ for $n\ge N$ lies in the same (at most) $2N$ dimensional space. This contradicts our assumption that there is an $x\in\bigcap A\_n$.


