Countable weighted shift has no invariant subspace.

Question

Suppose I have

$T(e_n)=w_ne_{n+1}$

where $w_n>0$ (and are bounded) and $\{e_n\}$ denotes the canonical basis of $l^{2}(\mathbb{N})$. I would like to prove that the only (closed) invariant subspaces of $T$ are the whole of $l^{2}(\mathbb{N})$ and the space $\{0\}$. Any hints without giving the game away?

Answer 1

False. The closed span of $e_k$ for $k \ge n$ is invariant.