Let $A$ be a C*-algebra and $(H,\pi)$ an irreducible representation such that $\dim(H)<\infty$. Then how do I show that $\pi(A)=B(H)$?
Here is what I tried: Since $(H,\pi)$ is irreducible, we know that $\pi(A)'=\mathbb{C}1_{B(H)}$ (cf. Murphy, Theorem 5.1.5), where $\pi(A)'$ is the commutant of $\pi(A)$. Hence $\pi(A)''=B(H)$. Furthermore, I know that $\pi(A)$ is norm-closed in $B(H)$, since $B(H)\cong M_{\dim(H)}(\mathbb{C})$ is finite-dimensional and $\pi(A)\subset B(H)$ is a vector-subspace.
I think that I have to implement Neumann's double commutant theorem somewhere. This theorem says that a $*$-algebra $N$ acting on $H$ (and $1_{B(H)}\in N$) is strongly closed if and only if $N''=N$ (cf. Murphy, Theorem 4.1.5). Also, I don't see how I can relate the strong closure and norm closure of $\pi(A)$.
Any suggestions are greatly appreciated. Thanks in advance!
Every finite-dimensional subspace of a topological vector space is closed, hence $\pi(A)$ is strongly closed. Using this, you have everything you need to finish the proof.