Quadratic Forms? [Sylvester's Theorem(signature) and Fourier's Theorem

[Orizom]

How do I prove that given Q indefinite quadratic form it has at least an isotropic vector?
I think I should use the signature (Sylvester's Theorem) and Fuorier's Theorem?
Like I know that sgn(Q)=(p, r-p) [0<p<r] but how do I prove that r<n (with n equal to the dimension of the ambient space)
Apologies for my poor english.

 

[blamocur]

What is ambient space here? The space in which Q is defined? If so, then [imath]r<n[/imath] because there is at least one isotropic vector, and thus there are only [imath]n-1[/imath] dimensions left for the signatures.

 

[Orizom]

What is ambient space here? The space in which Q is defined? If so, then [imath]r<n[/imath] because there is at least one isotropic vector, and thus there are only [imath]n-1[/imath] dimensions left for the signatures.

Thank you a lot for the answer. But I think I'm just missing something very easy... there's at least one isotropic vector because of the trivial one? You know I basically have to prove the quadratic forms-correspondent of the Bolzano's zeros theorem... Because practically speaking I know that if I have a Q defined in R^5 with sgn(Q)=(2,1) this implies I have 3 isotropic vectors, but I think I'm missing something for that r<<n for every indefinite Q, thank you really again

 

[blamocur]

Thank you a lot for the answer. But I think I'm just missing something very easy... there's at least one isotropic vector because of the trivial one? You know I basically have to prove the quadratic forms-correspondent of the Bolzano's zeros theorem... Because practically speaking I know that if I have a Q defined in R^5 with sgn(Q)=(2,1) this implies I have 3 isotropic vectors, but I think I'm missing something for that r<<n for every indefinite Q, thank you really again

"...in R^5 with sgn(Q)=(2,1) this implies I have 3 isotropic vectors,..." -- are you sure? Are we using the same definitions of signatures and isotropic vectors (e.g., https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia, https://en.wikipedia.org/wiki/Null_vector, https://encyclopediaofmath.org/wiki/Isotropic_vector)?

 

[Orizom]

"...in R^5 with sgn(Q)=(2,1) this implies I have 3 isotropic vectors,..." -- are you sure? Are we using the same definitions of signatures and isotropic vectors (e.g., https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia, https://en.wikipedia.org/wiki/Null_vector, https://encyclopediaofmath.org/wiki/

"...in R^5 with sgn(Q)=(2,1) this implies I have 3 isotropic vectors,..." -- are you sure? Are we using the same definitions of signatures and isotropic vectors (e.g., https://en.wikipedia.org/wiki/Sylvester's_law_of_inertia, https://en.wikipedia.org/wiki/Null_vector, https://encyclopediaofmath.org/wiki/Isotropic_vector)?

Ooops sorry I meant 2 isotropic since I would have an associated matrix with 2 positive values and a negative one (rk=3) and the last two values would be 0s? Does it make sense? Thank you for your patience but I'm still wrapping my head around these things and for sure I'm missing something very easy for the previous demonstration... I was also thinking to demonstrate it using Fourier's Theorem (span(v) directsum v(ortho subspace)= Ambient) but it seems so much easier, maybe I'm missing some corollary of Sylvester's theorem that states that the sum of the two components of the signature in an indefinite Q is << dimAmbient... thanks again

 

[blamocur]

Agree about 2 isotropic vectors.
"...the sum of the two components of the signature in an indefinite Q is << dimAmbient..." -- the two components of the signature are the numbers of positive and negative elements on the diagonal of a "Silvester-equivalent" matrix, but the total number of the elements on the diagonal is the dimension of the whole ("ambient"?) space. And if there is a zero on the diagonal (corresponding to an isotropic vector) then that sum must be strictly less then the whole dimension. Does this make sense?
Also, I am not sure what Fourier theorem you mean -- the one I am familiar with does not seem to be relevant here.

 

[Orizom]

Agree about 2 isotropic vectors.
"...the sum of the two components of the signature in an indefinite Q is << dimAmbient..." -- the two components of the signature are the numbers of positive and negative elements on the diagonal of a "Silvester-equivalent" matrix, but the total number of the elements on the diagonal is the dimension of the whole ("ambient"?) space. And if there is a zero on the diagonal (corresponding to an isotropic vector) then that sum must be strictly less then the whole dimension. Does this make sense?
Also, I am not sure what Fourier theorem you mean -- the one I am familiar with does not seem to be relevant here.

Yes, that does make sense, the thing is that theoretically I'm missing something to prove it, I mean like something, because this demonstration exercise on my book is like the correspondent of the Bolzano's zeros' theorem, like if I know that the signature is (p, r-p) 0<p<r for sure there must be an isotropic vector non trivial

 

[blamocur]

I've just realized that I misunderstood your original post. I assumed that the existence of an isotropic vector is a given, whereas you post says that it has to be proven -- sorry about this confusion.

I also apologize for a false statement I've made earlier ("If so, then [imath]r<n[/imath] because there is at least one isotropic vector"). A counter example would be [imath]Q(x,y) = x^2-y^2[/imath] for which [imath]r=2[/imath], but it has an isotropic vector (1,1).

Now I believe that the correct statement of the problem is "Prove that any indefinite quadratic form [imath]Q[/imath] has an isotropic vector". This can be proven by considering vectors [imath]\bar u[/imath] and [imath]\bar v[/imath] such that [imath]Q(\bar u) \geq 0[/imath] and [imath]Q(\bar v) \leq 0[/imath]. Their existence is guaranteed because [imath]Q[/imath] is indefinite. Is this enough of a hint for you to construct a proof?

Again, sorry for my earlier confusion

 

[Orizom]

I've just realized that I misunderstood your original post. I assumed that the existence of an isotropic vector is a given, whereas you post says that it has to be proven -- sorry about this confusion.

I also apologize for a false statement I've made earlier ("If so, then [imath]r<n[/imath] because there is at least one isotropic vector"). A counter example would be [imath]Q(x,y) = x^2-y^2[/imath] for which [imath]r=2[/imath], but it has an isotropic vector (1,1).

Now I believe that the correct statement of the problem is "Prove that any indefinite quadratic form [imath]Q[/imath] has an isotropic vector". This can be proven by considering vectors [imath]\bar u[/imath] and [imath]\bar v[/imath] such that [imath]Q(\bar u) \geq 0[/imath] and [imath]Q(\bar v) \leq 0[/imath]. Their existence is guaranteed because [imath]Q[/imath] is indefinite. Is this enough of a hint for you to construct a proof?

Again, sorry for my earlier confusion

Don't worry, as I said it wasn't well written... thank you for the patience!