Generator operator of $v$?

Question

Let $\text{U}_+$ be an associative $\mathbb{C}$-algebra with two generators $E$, $H$, and one defining relation $HE - EH = 2E$. Let $M$ be an $\text{U}_+$-module. If $v \in M$ is a nonzero eigenvector $H: M \to M$, then what can we say about $E(v)$?

Answer 1

There exists some $\lambda \in \Bbb C$ such that $Hv = \lambda v$. It follows that $$ (HE - EH)v = 2Ev \implies\\ H[E(v)] - \lambda [E(v)] = 2[E(v)] \implies\\ H[E(v)] = (2 + \lambda)E(v) $$ In other words, $E(v)$ is either $0$ or an eigenvector of $H$ associated with the eigenvalue $2 + \lambda$.