Let $A_{\theta}$ be the rotation C$^{*}$-algebra with rotation $\theta$. I.e., $A_{\theta}=C^{*}(u,v)$, where $vu=e^{2\pi i \theta}uv$. Suppose that $\theta=p/q$, where $p$ and $q$ are non-zero positive integers that are relatively prime.
I am trying to show that $A_{\theta}$ is not simple.
Consider the C$^{*}$-algebra $M_{q}(\mathbb{C})$ and the two unitary matrices $$ U = \begin{pmatrix}0 & 1 & 0 & \cdots & 0\\ 0 & 0 & 1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & 1\\ 1 & 0 & 0 & \cdots & 0 \end{pmatrix} \quad\text{and}\quad V=\begin{pmatrix} 1 & 0 & 0 & \cdots & 0\\ 0 & e^{2\pi i \theta} & 0 & \cdots & 0\\ 0 & 0 & e^{4\pi i \theta} & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \cdots & e^{2(q-1)\pi i\theta} \end{pmatrix} $$ in $M_{q}(\mathbb{C})$. It is easy to check that $VU=e^{2\pi i \theta}UV$. It follows by the universal property of $A_{\theta}$ that there is a $*$-homomorphism $\pi\colon A_{\theta}\to M_{q}(\mathbb{C})$ such that $\pi(u)=U$ and $\pi(v)=V$. If $\pi$ is not injective, then we are finished, since $\ker\pi$ is a non-zero proper ideal in $A_{\theta}$. So, we may suppose $\pi$ is injective. Now note that $\pi$ is an irreducible representation. Otherwise, it would decompose into a direct sum of smaller irreducible ones, which is impossible. But then it must be that $\pi(A_{\theta})=M_{q}(\mathbb{C})$ since irreducible representations acting on a finite-dimensional space are surjective.
Thus, we can conclude that $A_{\theta}=C^{*}(u,v)=C^{*}(U,V)=M_{q}(\mathbb{C})$. I am looking for a straight-forward way to get a contradiction here.
As you can read in the first paragraph of Chapter VI in Davidson, you can realize unitaries $u,v$ with $uv=e^{2\pi i \theta}vu$ by taking $u,v\in B(L^2(\mathbb T))$ where $u$ is multiplication by $z$, i.e. the bilateral shift.
The C$^*$-algebra generated by $u$ is $C^*(u)=C(\mathbb T)$, so by restricting your $\pi$ to $C^*(u)$ you get a $*$-homomorphism $$\pi:C(\mathbb T)\to M_n(\mathbb C).$$ If $\pi$ were injective, you would have an infinite-dimensional subalgebra of $M_n(\mathbb C)$.