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Bilinear forms.

  1. Jun 30, 2007 #1
    let f:VxV->R be an antisymmetric billinear form in real vector space V, there exists an operator that satisfies J:V->V J^2=-I.
    i need to prove that the form q:VxV->R, for every a,b in V q(a,b)=f(a,J(b)) is symmetric and definite positive.
    i tried to show that it's symmetric with its definition and that J^2=-I, but with no success, any hints here how to approach this question?
     
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  3. Jun 30, 2007 #2

    HallsofIvy

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    If a and b are two vectors, you know, from the formula what q(a,b) are. What is q(b,a), also using the formula. For the positive definite part, you will need to look at q(v,v) for v any vector.
     
  4. Jun 30, 2007 #3

    matt grime

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    Aren't you omitting something key about f?
     
  5. Jul 1, 2007 #4
    yes, but i don't how to say it english.
    i know that it's equivalent to that the matrix of f is invertible, i.e has an inverse.
    does this help to resolve the question?
     
  6. Jul 1, 2007 #5
    halls it doesnt help, i tried it this way but got stuck.
     
  7. Jul 1, 2007 #6

    matt grime

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    Google for symplectic forms - that's what this is. The point about f is that you can pick an element, v, and using f and its non-degeneracy, find a w so that f(v,w)=-1...
     
  8. Jul 1, 2007 #7
    yes, f is non degenrate.
    but i don't see how does it help, i need to show that q(a,b)=q(b,a)
    wait, a minute if f is antisymmetric then it means that f(x,y)=-f(y,x) then f(a,a)=0 always, but i don't see how can i find a w such that f(v,w)=-1, i mean if J^2=-I then if we write w=J(v), then J^2(v)=J(w)=-v
    then f(-J(w),J(v))=f(v,J(v))=-f(J(v),v)=f(J(v),J(w))
    f(J(v-w),J(v+w))=0, but how procceed from here?
     
  9. Jul 1, 2007 #8

    matt grime

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    it's a non-degenerate bilinear form, the very definition means that you can find such a w given a non-zero v. That is almost precisely what non-degenerate means.

    Ah, wait - I misread your question - I thought you had to prove that J existed. The point is that that J is not any old J, is it? It is a very particular J, and is gotten by taking the decomposition of V using the symplectic form.
     
    Last edited: Jul 1, 2007
  10. Jul 1, 2007 #9
    yes ofcourse that J is the symplectic form matrix, (and to be quite frank with you, i had to prove it before the question iv'e given here, which i ofcourse did).
    i know that to be non degenerates means that for every v different than zero, there exists w in V such that f(v,w) is different than zero, anyway i can choose w to be w=(1,0,0,...,0) and v=(0,1,0,..,0) then f(v,w)=-1, but how does this helps me to prove that q is symmetric and definite positive?
     
  11. Jul 1, 2007 #10

    matt grime

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    If all else fails, just use the natural symplectic basis.
     
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