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enter image description hereMy question could appear naive but I am reading up representation theory by myself. I am at a loss to understand the way how the action of the matrix happens for $S_3$. For example how does each of these representation work on the vectors $(e_1 -e_2)$ and $(e_2 -e_3)$ is unclear.

The text says that $$(1, 2) (e_1-e_2) = -(e_1 - e_2)$$ and $$(1, 2) (e_2 - e_3) = (e_1 - e_3).$$ How does one derive from the matrix action on the vector $(e_1, -e_2)$ and $(e_2, -e_3)$.

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  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Jan 16 '18 at 10:16
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    $\begingroup$ Where did you take this from? It doesn't look correct. For instance, the action of $(1\ \ 2)$ on $e_1-e_2$ should give $-2e_1-e_2$, not $-e_1+e_2$. $\endgroup$ – José Carlos Santos Jan 16 '18 at 10:21
  • $\begingroup$ I will add the link of the place where I got from. Please look at pg 6 and pg 7 of the text. jdc.math.uwo.ca/M9140a-2012-summer/Diaconis.pdf $\endgroup$ – user2714795 Jan 16 '18 at 10:35
  • $\begingroup$ @JoséCarlosSantos Why? It seems that elements of $S_3$ act by permuting the basis, so, indeed, $(1 2)(e_1 - e_2) = (1 2)e_1 - (1 2)e_2 - e_2 - e_1 = - (e_1 - e_2)$. $\endgroup$ – lisyarus Jan 16 '18 at 10:55
  • $\begingroup$ @lisyarus I am surely missing something. Where did you read that $(1\ \ 2)$ permutes the basis? What I read is that $(1\ \ 2)e_1=-e_1$ and that $(1\ \ 2)e_2=e_1+e_2$. $\endgroup$ – José Carlos Santos Jan 16 '18 at 11:05
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Since$$(1\ \ 2)e_1=e_2,\ (1\ \ 2)e_2=e_1\text{, and }(1\ \ 2)e_3=e_3,$$it is clear that$$(1\ \ 2)(e_1-e_2)=e_2-e_1=-(e_1-e_2)\text{ and that }(1\ \ 2)(e_2-e_3)=e_1-e_3.$$So, if we define $w_1$ and $w_2$ as $e_1-e_2$ and as $e_2-e_3$ respectively, we get that $(1\ \ 2)w_1=-w_1$ and that$$(1\ \ 2)w_2=e_1-e_3=e_1-e_2+e_2-e_3=w_1+w_2.$$Therefore, the matrix of the action of $(1\ \ 2)$ on $\bigl\{(x,y,z)\in\mathbb{R}^3\,|\,x+y+z=0\bigr\}$ with respect to the basis $\{w_1,w_2\}$ is indeed$$\begin{pmatrix}-1&1\\0&1\end{pmatrix}.$$

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