A linear operator $T$ on a (complex) separable Hilbert space $H$ is said to be a weighted shift operator if there is some orthogonal basis $\{e_n\}_n$ and weight sequence $\{w_n\}_n$ such that $$Te_n=w_n e_{n+1}, \forall n$$ I have a remark I want to show, that is
$T$ is injective if and only if none of the weights is zero.
My attempt is as follows:
$ker(T)={0}$ if and only if $$Tx=T(\sum \alpha_n e_n)=\sum \alpha_n T(e_n)=\sum \alpha_n w_n e_{n+1}=0 \text{ implies } x=\sum \alpha_n e_n=0$$ if and only if $$\alpha_n w_n=0 \text{ implies } \alpha_n=0, \forall n$$ How do I go ahead from here.