I have the following statement i want to prove and i'm unsure about my proof:
Let $A$ be a C*-algebra and $\pi: A \rightarrow B(\mathcal{H})$ an irreducible representation. For any $\eta \neq 0$ in $\mathcal{H}$, the set $\{ \pi(a)\eta \mid a \in A \}$ is not equal to $\{ 0 \}$.
Proof: Let $\mathbb{C} \eta$ be the subspace generated by $\eta $ in $\mathcal{H}$.
If $\{ \pi(a)\eta \mid a \in A \} = \{ 0\} \subseteq \mathbb{C}\eta$, then $\mathbb{C} \eta$ invariant subspace of $\pi$. As $\pi$ irreducible, we have $\mathbb{C}\eta $ is either whole $\mathcal{H}$ or $\{0\}$(second option is not possible due to $\eta \neq 0$.)
As $\mathcal{H} = \mathbb{C} \eta$ and $\{ \pi(a)\eta \mid a \in A \} = \{ 0\}$, we have $\pi(A) = \{ zero-operator \}$, but then is $\pi$ the zero representatioon, which is not irreducible.
I know the proof is a bit too spread out, but is this correct?
thanks in advance!
All non trivial irreducible representations are nondegenerate. Suppose for sake of contradiction that $\exists 0 \neq \gamma \in H$ such that the orbit of $\gamma$ under the representation was zero, i.e, $\{\pi(x) \gamma : x \in \mathcal{A} \} = \{0\}$. What does this tell you?
This says that $\gamma \in \{\delta \in H : \pi(x) \delta = 0 \text{ } \forall x \in \mathcal{A} \}$. But, this set is zero for a nondegenerate representation.
Suppose that $\overline{\pi(\mathcal{A})H}=H$. Let $0 \neq \delta \in H$ then $\exists x \in \mathcal{A}$ and $\gamma \in H$ such that $0 < \frac{||\delta||^2}{2} \leq (\pi(x^*) \gamma , \delta) = (\gamma, \pi(x) \delta)$. This implies that $\pi(x) \delta \neq 0$, i.e, $\{\delta \in H : \pi(x) \delta = 0 \text{ } \forall x \in \mathcal{A} \} = \{0\}$. This is one direction for the equivalence, can you show the other?