Let $\Sigma\subseteq\mathrm{Sym}(n)$ be a permutation group on $N:=\{1,...,n\}$. My goal is to determine the irreducible invariant subspaces of the permutation action of $\Sigma$ on $\Bbb R^n$, and I wonder whether the following technique always works.
Consider the element-wise action of $\Sigma$ on the subsets $\{i,j\}\subseteq N$ for $i,j\in N$. Let $o(i,j)$ denote the orbit of $\{i,j\}$ w.r.t. this action, and let $\mathcal O(\Sigma)$ denote the set of all these orbits.
Now let $R:= \Bbb R[\,x_o\!\mid\! o\in\mathcal O(\Sigma)\,]$ be the polynomial ring over $\Bbb R$ with one variable $x_o$ for each orbit, and consider the matrix $M\in R^{n\times n}$ with components $ M_{ij} = x_{o(i,j)}$.
Question: Is is true that that the eigenspaces of this matrix $M$ are exactly the irreducible invariant subspaces of the permutation action of $\Sigma$ on $\Bbb R^n$?
The matrix is symmetric, and thus the eigenspaces are real and mutually orthogonal. Each eigenspace is indeed $\Sigma$-invariant, but is it also irreducible?
I have applied this idea computationally and it never failed me. In practice, I do not use variables $x_o$, but some algebraically independent number (or just some random numbers). But does it work always?
