How do I prove $\textit{V}_\textit{n}$ is irreducible symmetric group representation?

Question

$\textit{V}_\textit{n}=\left\lbrace (x_1,x_2,...,x_n)\in\mathbb{R}^\textit{n}|\sum_{i=1}^{n}x_i=0\right\rbrace $,we define $ \textit{S}_\textit{n} $ act on $ \textit{V}_\textit{n} $ by $ \textit{S}_\textit{n}\cdotp x :=\sigma\cdotp(x_1, x_2,...,x_\textit{n})=(x_{\sigma\cdotp 1}, x_{\sigma\cdotp 2},...,x_{\sigma\cdotp (\textit{n})})$, this sapce is invariant when the symmetric group acts on $\textit{V}_\textit{n}$ because we permute the coordinates of any such vector, its sum still be zero. let $\alpha_\textit{i}=\textit{e}_\textit{i}-\textit{e}_\textit{n}, \textit{i} \in 1,2,...,\textit{n}-1$ is the basis of $\textit{V}_\textit{n}$, where $ \left\lbrace \textit{e}_\textit{i}\right\rbrace $ is the standard basis for $ \mathbb{R}^\textit{n} $, therefore $\textbf{w}=\alpha_\textit{1}+\alpha_\textit{2}+...+\alpha_\textit{n-1}=(1,1,...,1-n) \in \textit{V}_\textit{n}$, But I don't know how to prove this representation is irreducible. I tried to let $ \textit{S}_\textit{n} $ acts on a subspace of $\textit{V}_\textit{n}$, such as $\textit{P}$=$(\alpha_1, \alpha_2,\alpha_3)$, when $ \textit{S}_\textit{n} $ act on $\textit{P}$ its still invariant.

Answer 1

First off, your definition of the action is probably wrong. There is a left linear action of $S_n$ given by

$$ \sigma e_i:=e_{\sigma i} \implies \sigma(x_1,\cdots,x_n)=(x_{\sigma^{-1}1},\cdots,x_{\sigma^{-1}n}). $$ This is because $\sigma\big(\sum x_i e_i\big)=\sum x_i e_{\sigma i}=\sum x_{\sigma^{-1}i} e_i$. Or, you can say there's a right linear action

$$ (x_1,\cdots,x_n)\sigma=(x_{\sigma1},\cdots,x_{\sigma n}). $$

In general, if $g\cdot x$ is a left action, then $x\cdot g:=g^{-1}\cdot x$ is a right action, and vice-versa.

---

Anyway, suppose $W$ is a nonzero subrepresentation of $S_n$'s standard representation. Pick any $w\in W$, expressible as $w=(w_1,\cdots,w_n)$. The coordinates cannot all be equal, so there are some $w_i$ and $w_j$ that are not equal. Then $w-(ij)w=(w_i-w_j)(e_i-e_j)$, so $e_i-e_j\in W$. Can you go from there?