Let $m$ and $n$ be integers, with $n>0$ and $\gcd(m,n)=1$. Let $\theta=m/n$ and let $A_{\theta}$ be the rational rotation C$^{*}$-algebra generated by two unitaries $u$ and $v$, satisfying the relation $vu=e^{2\pi i \theta}uv$. I am working on an exercise in Davidson's C$^{*}$-algebra and my goal is to
Find all irreducible representations of $A_{\theta}$ and show that they lie in $M_{n}(\mathbb{C})$.
I was able to show that $u^{n}$ and $v^{n}$ lie in the center of $A_{\theta}$. Thus, if $\pi\colon A_{\theta}\to B(H)$ is an irreducible representation of $A_{\theta}$, it must be that $\pi(u^{n})$ and $\pi(v^{n})$ are scalar multiples of the identity. Using this I can show that the set $S=\{\pi(u)^{j}\pi(v)^{k}:0\leq j,k\leq n-1\}$ linearly spans $\pi(A_{\theta})$. Thus, $\dim(\pi)\leq n$. But I don't know how to rule out the case that $\dim(\pi)< n$ or how to go about finding all of the irreps.
Suppose that $\pi(A_\theta)\subset M_k(\mathbb C)$ is a unitary, where $k<n$. Then we have matrices $U,V$ such that $VU=e^{2\pi i\theta}UV$. In particular, $$ V=U^*(e^{2\pi i \theta} V)U. $$ So $V$ is unitarily equivalent to $e^{2\pi i \theta}V$. Take $\lambda\in\sigma(V)$. Then $$ \lambda, e^{2\pi i \theta}\lambda, e^{2\pi i 2\theta}\lambda, \ldots, e^{2\pi i (n-1)\theta}\lambda $$ are $n$ eigenvalues of $V$. But $V$ has at most $n-1$ distinct eigenvalues, so there exist integers $r,s$ with $0\leq r,s\leq n-1$ and $e^{2\pi i r\theta}\lambda=e^{2\pi i s\theta}\lambda$. This implies $e^{2\pi i (r-s)\theta}=1$, a contradiction since $|r-s|<n$.