Let $A$ be a $C^*$-algebra and $\pi: A \to B(H)$ be a representation on the Hilbert space $H$, that is $\pi$ is an algebra morphism that preserves the $^*$-operation.
I'm trying to show that the following two conditions are equivalent:
$$\forall \xi \in H\setminus \{0\}: \exists a \in A: \pi(a)\xi \neq 0$$ $$\iff$$ $$\overline{\operatorname{span}\{\pi(a)\xi: a \in A, \xi \in H\}}= H$$
I'm not able to show both implications. Hints are greatly appreciated!
It is easy to prove both contrapositives.
Define $$L=\overline{\operatorname{span}\{\pi(a)\xi: a \in A, \xi \in H\}}.$$
$\implies$: Suppose that $$L\subsetneq H.$$ If $\eta\in L^\perp$ is nonzero we have, for any $\xi\in H$ and any $a\in A$, $$\tag1 \langle \pi(a)\eta,\xi\rangle=\langle \eta,\pi(a^*)\xi\rangle=0. $$ Thus $\pi(a)\eta=0$ for all $A$.
$\impliedby$: Suppose that there exists nonzero $\eta\in H$ such that $\pi(a)\eta=0$ for all $a\in A$. Again by the equality $(1)$, we have that $\eta\in L^\perp$, so $L\subsetneq H$.