Show that there is a symmetric matrix A such that $\forall g\in G,\rho(g)^tA\rho(g)=A$.

Question

Let $G$ be a finite group, $n$ a positive integer and $\rho: G \to GL(n;\mathbb{R} ) $ a homomorphism. Show that there is a positive definite symmetric $n\times n$ matrix A such that $\forall g\in G, \rho(g)^tA\rho(g)=A $.

Answer 1

An attack that works is the following:

1. Let $(,)$ be the usual inner product on $\Bbb{R}^n$. Define a "new" inner product $\langle,\rangle$ by the formula (for all $x,y\in\Bbb{R}^n$) $$ \langle x, y\rangle=\sum_{g\in G}(\rho(g)x,\rho(g)y). $$ Verify that this is, indeed, an inner product.
2. Verify that $\langle x, y\rangle=\langle \rho(g)x, \rho(g)y\rangle$ for all $x,y\in\Bbb{R}^n$ and all $g\in G$.
3. Recall from linear algebra that any inner product on $\Bbb{R}^n$ is of the form $$ \langle x,y\rangle=x^tAy $$ for some positive definite symmetric matrix $A$.

Hmm. I guess I'll stop here.