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Positive-definite symmetric matrix satisfying a certain property

  1. Dec 5, 2013 #1
    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. jcsd
  3. Dec 6, 2013 #2
    What would help me is knowing the definition of ##GL_n(R)##.
     
  4. Dec 6, 2013 #3

    Office_Shredder

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

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

    I mention this only because you said it might help if [itex] \rho(g) [/itex] 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!
     
  5. Dec 6, 2013 #4
    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.
     
  6. Dec 6, 2013 #5

    Office_Shredder

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    Yes, which is why if [tex] \det(\rho(g))^n [/tex] takes on more than finitely many values you get a contradiction (since eventually it should become [itex] \det(\rho(g)) [/itex] again). For example if the determinant was 2 then you would get that the determinant of [itex] \rho(g^n) [/itex] 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.
     
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