Proving Indefiniteness of a Quadratic Form on a Subspace | V -> R

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Discussion Overview

The discussion revolves around the properties of a quadratic form defined on a vector space, specifically addressing the conditions under which the quadratic form is considered indefinite. Participants explore the implications of a subspace defined by non-negative values of the quadratic form.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a problem regarding proving that a quadratic form q is indefinite, given a subspace T defined by non-negative values of q.
  • Another participant suggests that to prove q is indefinite, one must leverage the properties of quadratic forms and the structure of the subspace T.
  • A participant corrects their initial claim, stating they need to prove that q is not indefinite, indicating a misunderstanding of the problem's requirements.
  • Another participant proposes a method involving an arbitrary element of T and suggests proving that certain combinations of elements lead to contradictions, ultimately aiming to show that q(v) must equal zero for all v in T.

Areas of Agreement / Disagreement

Participants express differing interpretations of the problem, with some arguing for the proof of indefiniteness while others explore the possibility of proving that q is not indefinite. The discussion remains unresolved with multiple competing views on the approach to take.

Contextual Notes

There are limitations in the understanding of quadratic forms among some participants, which may affect the clarity of the arguments presented. The discussion also reflects uncertainty regarding the implications of the subspace T on the properties of the quadratic form.

TTob
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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.
 
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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.
 
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 ?
 
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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