Matrices for action wrt basis

Question

Consider the permutation representation where $G=S_3$ on $\mathbb{C^3}$ with the action:

$\pi(g)e_i=e_{g(i)}$

$W=\{ \lambda_i e_i ; \sum \lambda_i=0 \}$ is an invariant subsoace of vector space $V$

matrices for the action on $W$ wrt basis $v_1=e_1+we_2+w^2e_3$ and $v_2=we_1+e_2+w^2e_3$ of $W$ where $w$ is a primitive third root of unity are given for two group elements as follows:

$\pi((12))=\begin{bmatrix} 0&1\\1&0 \end{bmatrix}$

$\pi((23))=\begin{bmatrix} 0&w\\w^2&0 \end{bmatrix}$

Can someone please guide me through how these are calculated?

Answer 1

To determine the matrix representation of the action for $\pi((12))$ for example, you just let $\pi((12))$ act on each of the basis vectors $v_1$ and $v_2$. So you have $$\pi((12))v_1 = \pi((12))e_1 + w\pi((12))e_2 + w^2\pi((12))e_3 = e_2 + we_1 + w^2e_3 = v_2$$ and thus with respect to the basis $\{v_1,v_2\}$ the first column of your matrix is $\begin{pmatrix}0\\1\end{pmatrix}$ indicating that it simply switches the two basis vectors. Consequently the second column will be $\begin{pmatrix}1\\0\end{pmatrix}$ giving the matrix $$\pi((12)) = \begin{pmatrix}0&1\\1&0\end{pmatrix}$$. I will leave the second computation (of the matrix $\pi((23))$) up to you.

Answer 2

$$ \pi((12))(e_1+we_2+w^2e_3) = we_1+e_2+w^2e_3 $$$$ \pi((12))(we_1+e_2+w^2e_3) = e_1+we_2+w^2e_3 $$

$$ \pi((23))(e_1+we_2+w^2e_3) = e_1+w^2e_2+we_3 = w^2(we_1+e_2+w^2e_3) $$$$ \pi((23))(we_1+e_2+w^2e_3) = we_1+w^2e_2+e_3 = w(e_1+w^2e_2+we_3) $$