Write $D_n = \langle r, s \mid s^2 = r^n = 1, srs= r^{-1} \rangle$ with $n \geq 3$. Define a representation $\rho$ on the generators as $\rho(r) = \begin{pmatrix} \omega & 0 \\ 0 & \omega^{-1} \end{pmatrix}$ and $\rho(s) = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$. Note that $\omega$ is an $n$th root of unity. How do you show that this extends to an irreducible representation of $D_n$?
I know that I need to show that there is no non-zero subspace of $\mathbf{C}^2$ which gets mapped to itself, but I'm having trouble making progress with this definition.
Call your basis vectors (of $\mathbb{C}^2$) $\bf{x}$ and $\bf{y}$, so e.g. $\rho(r)({\bf x}) = \omega \bf{x}$ and $\rho(r)({\bf y}) = \omega^{-1}\bf{y}$. Now, a non-zero proper subspace of $\mathbb{C}^2$ must be 1-dimensional, so it must be spanned (as a vector space) by some single vector, say ${\bf v} = \lambda {\bf x} + \mu {\bf y}$.
For the subrepresentation generated by ${\bf v}$ to be 1-dimensional, $\bf v$ would need to be an eigenvector of every matrix in $\rho(D_n)$. Show that it can't be.