2
$\begingroup$

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

$\endgroup$
  • 3
    $\begingroup$ Compute the inner product of its character with itself $\endgroup$ – Matthew Towers May 14 '16 at 16:17
1
$\begingroup$

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 $\rho(a_3)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.