Embed an equilateral triangle into $\mathbb{R}^2$ with vertices $(1,0), (\frac{-1}{2}, \frac{\sqrt{3}}{2}), (\frac{-1}{2}, -\frac{\sqrt{3}}{2})$.
Counterclockwise rotation and reflection over the $x$ axis generate $D_3$.
So we get a representation $\rho : D_6 \rightarrow GL_2(\mathbb{C})$ where counterclockwise rotation is $$\begin{bmatrix} \frac{-1}{2} &-\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& -\frac{1}{2} \end{bmatrix}$$
and reflection is $$\begin{bmatrix} 1 &0\\ 0&-1 \end{bmatrix}$$
To check this is an irreducible representation, apparently, all you need to do is check that these matrices have no common eigenvector? Why is this so? Is this something that applies generally?
Thank you for any help!
If there is a nontrivial subrepresentation, $V\subsetneq\mathbb R^2$, then $\rho (g)(v)\in V,\forall g\in D_3, v\in V$.
But, as @Tobias pointed out, we must have $\operatorname{dim} V=1$. This means $V=\operatorname{span} v$, for a common eigenvector $v$ of the generators.