This is part of an exercise in Serre's representation theory book I am self-studying, but mostly it is about manipulation of symmetric polynomials.
Let $\rho$ be a representation of a finite group $G$ and define $\sigma_T(\chi) = \Sigma \chi_{\sigma}^kT^k \in R(G)[T]$ where $\chi_{\sigma}^k$ is the $k$-th symmetric power of the representation. Show that $\sigma_T(\chi)(s) = 1/det(1-\rho(s)T)$
Attempt
Well we should show that $\sigma_T(\chi)(s)det(1-\rho(s)T) =1$. Let $\lambda_i$ denote the eigenvalues of $\rho(s)$. We have that $$det(1-\rho(s)T) = det(\rho)det(\rho(s)^{-1} - T) = \lambda_1...\lambda_n \cdot \Sigma T^n + a_{n-1}T^{n-1} + ... + a_0$$
where $a_i = (-1)^i sym^i(\lambda_j^{-1})$ where $sym^i$ denote the $i$-th symmetric polynomial. So it suffices to show that the coefficients of $T^n$, $n >0$ in the product $\sigma_T(\chi)(s)det(1-\rho(s)T)$ is zero and this is explicitly: $$\rho(s) \Sigma_{0 \leq k \leq n} \chi^k_{\sigma}(s)a_{n-k}$$ and we know that $\chi_{\sigma}^k(s)$ is equal to the sum of all possible degree $k$ monomials in the $\lambda_i$'s. Somehow I don't see how the cancellations occur. Help/conceptual explanation would be greatly appreciated!
Suppose an $n\times n$ matrix $X$ has eigenvalues $\lambda_1,\ldots,\lambda_n$.
Then $$\det(1 - XT)^{-1} = (1-\lambda_1 T)^1 \cdots (1 - \lambda_n T)^{-1}$$ $$= (1 + \lambda_1 T + \lambda_1^2 T^2 + \cdots) \cdots (1 + \lambda_n T + \lambda_n^2 T^2 + \cdots )$$ $$ = \sum_{i = 0}^{\infty} S_i(\lambda_1,\ldots,\lambda_n) T^i,$$ where $S_i(\lambda_1,\ldots,\lambda_n)$ is a polynomial in the $\lambda_i$ which you can easily calculate, and which you can check coincides with the trace of the $i$th symmetric power of $X$.