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Quadratic form question

  1. Aug 8, 2008 #1
    I have a problem with this question:
    let q : V -> R is quadratic form and suppose T = {v|q(v) [tex]\geq[/tex] 0} is subspace of V. prove that q is indefinite.

    Thanks in advance.
     
  2. jcsd
  3. Aug 8, 2008 #2

    CompuChip

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    What do you need to show to prove that q is indefinite?
    Then you will have to try and prove this, using that T is a vector subspace. In particular, if you have some element in it, this tells you that you also have several other elements. With the right element you can use some properties of quadratic forms to complete the proof.
     
  4. Aug 8, 2008 #3
    I made a mistake in my post. I need to prove that q is not indefinite, it means it's impossible that exist a,b[tex]\neq[/tex]0 such that q(a) > 0 and q(b) < 0.

    suppose that exist a,b[tex]\neq[/tex]0 such that q(a) > 0 and q(b) < 0. so a is in T. so what ?

    how can I reach a contradiction ?
     
  5. Aug 8, 2008 #4

    morphism

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    I know nothing about quadratic forms, so this is probably not the most elegant way to deal with this problem, but here goes anyway...

    Suppose that there exists a b such that q(b) < 0, and let v be an arbitrary element of T.

    (1) Prove q(v+b) and q(v-b) < 0.
    (2) Deduce that q(v) + q(b) < 0, and hence that q(v) < |q(b)|.

    Now use this to conclude that q(v)=0 for all v in T.
     
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