# No common eigenvectors then representation irreducible

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!

• Note that the representation is $2$-dimensional, so any non-trivial proper subrepresentation would need to be $1$-dimensional. – Tobias Kildetoft Nov 8 '18 at 19:25

## 1 Answer

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.

• But in general the non-existence of a common eigenvector is not enough to show that a representation is irreducible. I think that was the main question. – Tobias Kildetoft Nov 8 '18 at 19:39
• I see. That was the point of your comment above. Here reducible implies common eigenvector. I did the converse. – Chris Custer Nov 8 '18 at 19:51
• Why does it have to be that specific vector $v$? I understand that $\rho(g)(v)$ must be in $V$, which is one dimensional, hence has basis of one element ${\{v}\}$. You're saying this $v$ must be the common eigenvector. Why? Thank you. – the man Nov 8 '18 at 20:27
• I think I get it now. Thank you! – the man Nov 8 '18 at 20:37