# Irreducible representation of $S_3$

How can I show that this representation of $S_3$ is irreducible? $$\rho\left(e\right)=\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right),\,\,\rho\left(a_{1}\right)=\frac{1}{2}\left(\begin{array}{cc} -1 & -\sqrt{3}\\ \sqrt{3} & -1 \end{array}\right),\,\,\rho\left(a_{2}\right)=\frac{1}{2}\left(\begin{array}{cc} -1 & \sqrt{3}\\ -\sqrt{3} & -1 \end{array}\right),$$ $$\rho\left(a_{3}\right)=\left(\begin{array}{cc} -1 & 0\\ 0 & 1 \end{array}\right),\,\,\rho\left(a_{4}\right)=\frac{1}{2}\left(\begin{array}{cc} 1 & \sqrt{3}\\ \sqrt{3} & -1 \end{array}\right),\,\,\rho\left(a_{5}\right)=\frac{1}{2}\left(\begin{array}{cc} 1 & -\sqrt{3}\\ -\sqrt{3} & -1 \end{array}\right).$$

• Compute the inner product of its character with itself – Matthew Towers May 14 '16 at 16:17

## 1 Answer

Well, if it were reducible, $\rho(a_3)$ and $\rho(a_4)$ would share an eigenvector. Which is clearly not the case, because the standard basis is the only (up to scalar multiplication and permutation) basis of eigenvectors of $\rho(a_3)$.