Let $\mathcal{A}$ be a C*-algebra and $(H,\pi)$ an irreducible representation of $\mathcal{A}$.
I want to prove the statement: all $\xi \in H$ are cyclic or $\pi(\mathcal{A})=\{0\}$ and $H=\mathbb{C}$.
How can one approach this problem?
Suppose there is a $\hat{\xi}\in H$ that is not cyclic, then $\{a\hat\xi:a\in \pi(\mathcal{A})\}$ is not dense in H, i.e. $\{a\hat\xi:a\in \pi(\mathcal{A})\}^{\perp}\neq \{0\}$. From here on, I don't know how to conclude that $\pi(\mathcal{A})=\{0\}$ and $H=\mathbb{C}$.
The statement to be proven should be slightly modified as follows.
Proof
If each $ \xi \in \mathcal{H} \setminus \{ 0_{\mathcal{H}} \} $ is already a cyclic vector, then there is nothing to show.
Hence, suppose that there is a $ \xi \in \mathcal{H} \setminus \{ 0_{\mathcal{H}} \} $ that is not a cyclic vector. By definition, this means that $$ \mathcal{H}' ~ \stackrel{\text{def}}{=} ~ \overline{\{ [\pi(a)](\xi) ~|~ a \in \mathcal{A} \}}^{\| \cdot \|_{\mathcal{H}}} $$ is a closed and proper linear subspace of $ \mathcal{H} $. Clearly, $ \mathcal{H}' $ is also an invariant subspace of $ (\mathcal{H},\pi) $. As $ (\mathcal{H},\pi) $ is irreducible, we have $ \mathcal{H}' = \{ 0_{\mathcal{H}} \} $, which means that $ [\pi(a)](\xi) = 0_{\mathcal{H}} $ for each $ a \in \mathcal{A} $. It follows that the one-dimensional linear subspace $ \mathbb{C} \cdot \xi \subseteq \mathcal{H} $ is an invariant subspace of $ (\mathcal{H},\pi) $; to see this, simply observe that \begin{align} \forall a \in \mathcal{A}: \quad [\pi(a)][\mathbb{C} \cdot \xi] &= \mathbb{C} \cdot [\pi(a)](\xi) \\ &= \mathbb{C} \cdot 0_{\mathcal{H}} \\ &= \{ 0_{\mathcal{H}} \} \\ &\subseteq \mathbb{C} \cdot \xi. \end{align} By the irreducibility of $ (\mathcal{H},\pi) $ again, we thus obtain $$ \mathcal{H} = \mathbb{C} \cdot \xi \cong \mathbb{C}. $$ Therefore, $ [\pi(a)][\mathcal{H}] = \{ 0_{\mathcal{H}} \} $ for each $ a \in \mathcal{A} $, which immediately yields $$ \pi[\mathcal{A}] = \{ 0_{B(\mathcal{H})} \}. \quad \blacksquare $$