I writing my Bachelorthesis and I am doubting a proof i made for following statement
Let $A$ be a C*-algebra and $\pi: A \rightarrow B(\mathcal{H})$ a *-representation. If $\pi$ is irreducible (i.e. $\mathcal{H}$ has no $\pi(A)$-invariant subspaces), then $\pi$ is non-degenerate (i.e. $\pi(A)\mathcal{H}$ is dense in $\mathcal{H}$).
My proof: the closure of $\pi(A)\mathcal{H}$ is an invariant subspace. The only problem is if $\pi(A)\mathcal{H} = \{ 0 \}$, but then $\pi(A)\mathcal{H} = \{ zero-operator \}$, then every closed subspace of $\mathcal{H}$ is invariant under $\pi(A)$, thus $\pi$ not irreducible.
I doubt my last statement $\pi(A)\mathcal{H} = \{ 0 \}$, is this pathelogical case excluded, or should i begin my statement with "$\pi$ is a non-trivial *-repr. ?
Thanks in advance!
You have that $\pi : \mathcal{A} \to \mathcal{B}(H)$ is a nonzero representation. So $\exists x \in \mathcal{A}$ such that $\pi(x) \neq 0$. So $\exists \gamma \in H$ such that $\pi(x) \gamma \neq 0$. By assumption, $\pi$ is irreducible so the only invariant closed subspaces it has are the trivial subspaces, namely $H$ and $\{0\}$.
As you then state in your proof $\overline{\pi(\mathcal{A}) H }$ is closed invariant subspace. By the first line this has a non-zero vector so what can you conclude?
You can even further conclude that $\gamma$ is a cyclic vector for your representation namely that $\overline{\pi(\mathcal{A}) \gamma}= H$.