# Positive-definite symmetric matrix satisfying a certain property

1. Dec 5, 2013

### Convergence

1. The problem statement, all variables and given/known data
We have a finite group $G$ and a homomorphism $\rho: G \rightarrow \mathbb{GL}_n(\mathbb{R})$ where $n$ is a positve integer. I need to show that there's an $n\times n$ positive definite symmetric matrix that satisfies $\rho(g)^tA\rho(g)=A$ for all $g \in G$, where $t$ means transpose

2. Relevant equations

3. The attempt at a solution
Well I've basically just written down definitions and tried to go from there. So for all non-zero vectors $v$ with $n$ real entries, $v^tAv>0$, and I know $A=A^t$ Also, I know $A$ is positive-definite if and only if there exists an invertible $n\times n$ matrix $P$ such that $A=P^tP$. But I'm not really sure where to go with that. I was wondering maybe I can start with $\rho(g)^tA\rho(g)=A$ and then show that $A$ is positive-definite, but I don't really see that going anywhere. If $\rho(g)$ was orthogonal, then maybe that would help, but again I'm not to sure. I'd appreciate any hints. Also, this is from a group theory course that only requires one quarter of linear algebra

2. Dec 6, 2013

### brmath

What would help me is knowing the definition of $GL_n(R)$.

3. Dec 6, 2013

### Office_Shredder

Staff Emeritus
brmath, it's the group of nxn invertible matrices over the real numbers.

Convergence, if the determinant of $\rho(g)$ is not 1, then
$$\det \left( \rho(g^n) \right) = \det \left( \rho(g)^n \right) = \det\left( \rho(g) \right)^n$$
takes on infinitely many values as n goes to infinity which is a contradiction as G is finite. So $\rho(g)$ has determinant 1 for all g.

I mention this only because you said it might help if $\rho(g)$ was orthogonal; knowing that G is finite does give you information about what the image looks like. I don't know if that helps though.

EDIT: Actually that's a red herring I think... I would suggest focusing on A = PtP, and also remember that the sum of two positive definite matrices is still positive definite!

4. Dec 6, 2013

### brmath

Office-Shredder -- if G is finite $g^n = g$ for some n, so I'm not clear where you got inf many values for the determinants.

5. Dec 6, 2013

### Office_Shredder

Staff Emeritus
Yes, which is why if $$\det(\rho(g))^n$$ takes on more than finitely many values you get a contradiction (since eventually it should become $\det(\rho(g))$ again). For example if the determinant was 2 then you would get that the determinant of $\rho(g^n)$ is 2n which is impossible because eventually it has to be 2 again.

Although I realize in my above post I missed the obvious possibility that the determinant could be -1 as well.