Existence of a G-invariant matrix Let $\phi: G \to GL(\mathbb{R}^n)$ be a homomorphism, $G$ finite. Prove that there is a positive-definite matrix $M$ such that $\phi(g)^tM \phi(g) =M$ $\forall g \in G $. 
This looks really interesting. Any idea is appreciated.
 A: Here is an explanation on my comment. Often times one wants some inner product $\langle -,-\rangle$ on your vector space $V$ ($V$ is the vector space for which you have your representation on: $G\to\text{GL}(V)$) which is "$G$-invariant".By this I mean that $G$ actually maps into the unitary group of $(V,\langle-,-\rangle)$ or, in other words, $\langle gv,gu\rangle=\langle v,u\rangle$ for all $v,u\in V$. 
The most common technique to achieve this is to take any inner product $\langle -,-\rangle_0$ on $V$ and "average" it. Namely, define $\langle -,-\rangle$ as follows:
$$\langle v,u\rangle=\frac{1}{|G|}\sum_{g\in G}\langle gv,gu\rangle_0$$
You can quickly check that $\langle -,-\rangle$ is indeed a $G$-invariant bilinear form. 
Consider then the Gram matrix associated to $\langle-,-\rangle$. Namely, fix a basis $\{e_1,\ldots,e_n\}$ for $V$ and define the matrix $M=[a_{i,j}]$ by $a_{i,j}=\langle e_i,e_j\rangle$. You can quickly check then that for all $u,v\in V$ one has that 
$$\langle v,u\rangle=v^\top M u$$
where this right hand-side is the dot-product, once we identify $V$ with $\mathbb{R}^n$ by the basis $\{e_1,\ldots,e_n\}$. 
Now, the fact that $\langle -,-\rangle$ is an inner product says precisely that $M$ is positive definite and symmetric. Moreover, we see that for all $g\in G$ and all $u,v\in V$ we have 
$$v^\top\rho(g)^\top M\rho(g)u=(gv)^\top M(gu)=\langle gv,gu\rangle=\langle v,u\rangle=v^\top M u$$
Thus, we can see that $\rho(g)^\top M\rho(g)=M$ as desired.
A: I think this will work:
Let $A$ be the matrix defined by
$A = \sum_{h \in G} \rho(h)^t \rho(h); \tag{1}$
the sum here ranges over all $h \in G$; since $G$ is finite, $A$ is well-defined.  For any $g \in G$ we have
$\rho(g)^t A \rho(g) = \sum_{h \in G} \rho(g)^t \rho(h)^t \rho(h) \rho(g). \tag{2}$
In (2), since $\rho$ is a homomorphism, $\rho(h) \rho(g) = \rho(hg)$ and furthermore
$\rho(g)^t \rho(h)^t = (\rho(h) \rho(g))^t = \rho(hg)^t \tag{3};$
thus (2) becomes
$\rho(g)^t A \rho(g) = \sum_{h \in G} \rho(hg)^t \rho(hg). \tag{4}$
As $h$ ranges over $G$, so does $hg$, since multiplication by a fixed element (in this case $g$) merely induces a permutation on the elements of the group (in this case $G$); the map $G \to G$ given by $h \to hg$ for all $h \in G$ is thus a "permutation of the indices" in the sum (1); every term of (1) in fact also occurs in (4), and vice versa; the sums are in fact the same.  Thus
$A = \sum_{h \in G} \rho(h)^t \rho(h) = \sum_{h \in G} \rho(hg)^t \rho(hg) = \rho(g)^t A \rho(g). \tag{5}$
(5) shows that $A$ as defined in (1) satisfies $A = \rho(g)^t A \rho(g)$; it remains to be seen that $A$ is positive definite symmetric.  To see that $A$ is symmetric, note that each summand matrix occurring in (1) is in fact symmetric:
$(\rho(h)^t \rho(h))^t = \rho(h)^t (\rho(h)^t)^t = \rho(h)^t \rho(h), \tag{6}$
which implies the sum (1) is also symmetric, being itself a sum of symmetric matrices.  Each term in (1) is also positive definite; to see this, note that since $\rho$ is a homomorphism,
$\rho(h) \rho(h^{-1}) = \rho(hh^{-1}) = I \in \Bbb G \Bbb L_n, \tag{7}$
which shows that $\rho(h)$ is nonsingular.  The positive definiteness of $\rho(h)^t \rho(h)$ follows easily from this fact, since for any vector $\mathbf x \ne 0$ we have
$\langle \mathbf x, \rho(h)^t \rho(h) \mathbf x \rangle = \langle \rho(h) \mathbf x,  \rho(h) \mathbf x \rangle > 0, \tag{8}$
strict inequality holding since $\rho(h) \mathbf x \ne 0$ by the nonsingularity of $\rho(h)$.  The sum of positive definite matrices is also positive definite: since $\langle \mathbf x, P_i \mathbf x \rangle > 0$ for a collection of matrices $P_i$ yields
$\langle \mathbf x, (\sum_i P_i) \mathbf x \rangle = \sum_i \langle \mathbf x, P_i \mathbf x \rangle > 0, \tag{9}$
which shows the sum in (1) is in fact positive definite symmetric.  This fact, coupled with (5), completes the proof that $A$ has the requisite properties. QED
Hope this helps.  A Cheery Yule Season to One and All,
and of course, whether it be Christmas Trees or Candles,
Fiat Lux!!!
