I have an $n\times n$ matrix
$M=1-\sigma(g)$
where $\sigma:S_n\to\mathbb{C}^n$ is a representation (reducible) of the symmetric group. I want to find another matrix $A(g)$ such that
$MA(g)=0.$
I've written $g$ here in $A(g)$ because $A$ will explicitly depend on the group element. So this matrix obviously satisfies
$\sigma(g)A(g)=A(g)$,
so $A(g)$ consists of elements fixed by $\sigma (g)$. Since I am thinking of a specific representation, I could go ahead and calculate the matrix explicitly; something like
$A(g)_{ij}=\delta_{ij}-(\delta_{ik}\sigma(g)_{kj}-\delta_{jk}\sigma(g)_{ik})$
(Summation over $k$ implied - what I think that does is subtract from the identity any rows or column that is moved by $g$...). However, it'd be nice to have some representation-independent way to describe the matrix $A(g)$.
I guess $A(g)$ is not actually $in$ the representation $\sigma$ so that's a problem with this approach, but I was wondering if anyone had any tricks. The only reason why I think there might be something tricky here is that I am apparently looking for some kind of projector, and projectors are usually nice and tricky!
The matrix $A\left(g\right)$ is not unique. Since you are working over $\mathbb C$ (actually, any characteristic-$0$ field would do here), you can take $A\left(g\right) = \sum\limits_{k=0}^{n!-1} \sigma\left(g^k\right)$; then you can easily check (using $\sigma\left(g^{n!}\right) = \sigma\left(e\right) = \operatorname*{id}$) that not only $MA\left(g\right)$, but also the image of $A\left(g\right)$ is exactly the kernel of $M$.