Clarifying a corollary about Quadratic Forms

[Euler2718]

The question comes out of a corollary of this theorem:
Let [itex] B [/itex] be a symmetric bilinear form on a vector space, V, over a field [itex]\mathbb{F}= \mathbb{R}[/itex] or [itex]\mathbb{F}= \mathbb{C}[/itex]. Then there exists a basis [itex]v_{1},\dots, v_{n}[/itex] such that [tex] B(v_{i},v_{j}) = 0 [/tex] for [itex]i\neq j[/itex] and such that for all [itex]i=1,\dots , n[/itex]: [tex] B(v_{i},v_{i}) = \begin{cases} 0,1 & F=\mathbb{C} \\ 0, \pm 1 & F=\mathbb{R}\end{cases} [/tex]
(that is to say, there is a choice of basis so that [itex]\beta[/itex] (symmetric [itex]n\times n[/itex] matrix representing the form)is diagonal with diagonal entires 0,1 if [itex]\mathbb{F} = \mathbb{C}[/itex] and 0,1,-1, if [itex]\mathbb{F} = \mathbb{R}[/itex]

Then the corollary is presented: Let [itex]V[/itex] and [itex]B[/itex] be as above and let [itex]Q(\mathbf{v}) = B(\mathbf{v},\mathbf{v})[/itex]. Then (I'll just consider the case over the field of reals): for some [itex]p,q[/itex] with [itex]p+q\leq n[/itex] there exists a basis such that: [tex]Q(\mathbf{v}) = Q(z_{1}\mathbf{v}_{1} + \dots + z_{n}\mathbf{v}_{n}) = \sum_{i=1}^{p} z_{i}^{2} - \sum_{i=p+1}^{p+q}z_{i}^{2} = z_{1}^{2} + z_{2}^{2} +\dots + z_{p-1}^{2} + z_{p}^{2} - z_{p+1}^{2} - z_{p+2}^{2} - \dots -z_{p+q}^{2} [/tex]

I feel like it is probably obvious, but how does: [itex]\displaystyle \sum_{i=1}^{p} z_{i}^{2} - \sum_{i=p+1}^{p+q}z_{i}^{2}[/itex] describe the quadric? The proof I have seen is: choice an appropriate basis satisfying the theorem, then [tex] Q(z_{1}\mathbf{v}_{1} + \dots + z_{n}\mathbf{v}_{n}) = B(z_{1}\mathbf{v}_{1} + \dots + z_{n}\mathbf{v}_{n},z_{1}\mathbf{v}_{1} + \dots + z_{n}\mathbf{v}_{n}) = \sum_{i,j=1}^{n} z_{i}z_{j}B(\mathbf{v}_{i},\mathbf{v}_{j}) = \sum_{i=1}^{n}z_{i}^{2}B(\mathbf{v}_{i},\mathbf{v}_{i})[/tex], where [itex]B(\mathbf{v}_{i},\mathbf{v}_{i})[/itex] are 0,1,or -1 as appropriate.
So the question then becomes how does [itex]\sum_{i=1}^{n}z_{i}^{2}B(\mathbf{v}_{i},\mathbf{v}_{i})[/itex] imply the result I have in the corollary? Is it just a manipulation of the series? Or is it simply my lack of experience showing its true colors?

 

[fresh_42]

Is it just a manipulation of the series? Or is it simply my lack of experience showing its true colors?

I think neither nor. It's probably a case of math blindness. You just have to substitute
$$B(v_1,v_1)=1, \ldots , B(v_p,v_p) = 1\, , \,B(v_{p+1},v_{p+1})=-1, \ldots , B(v_{p+q},v_{p+q})=-1\; , \; \\
B(v_{p+q+1},v_{p+q+1})=0, \ldots ,B(v_n,v_n)=0 $$
which you have from the theorem after eventually renumbering the basis vectors.

 

[Euler2718]

I think neither nor. It's probably a case of math blindness. You just have to substitute
$$B(v_1,v_1)=1, \ldots , B(v_p,v_p) = 1\, , \,B(v_{p+1},v_{p+1})=-1, \ldots , B(v_{p+q},v_{p+q})=-1\; , \; \\
B(v_{p+q+1},v_{p+q+1})=0, \ldots ,B(v_n,v_n)=0 $$
which you have from the theorem after eventually renumbering the basis vectors.

So I break this up[itex]\displaystyle Q(\mathbf{v}) = \sum_{i=1}^{n}z_{i}^{2}B(\mathbf{v}_{i},\mathbf{v}_{i}) = z_{1}^{2}B(\mathbf{v}_{1},\mathbf{v}_{1}) + z_{2}^{2}B(\mathbf{v}_{2},\mathbf{v}_{2}) + z_{3}^{2}B(\mathbf{v}_{3},\mathbf{v}_{3}) + \dots + z_{n-1}^{2}B(\mathbf{v}_{n-1},\mathbf{v}_{n-1}) + z_{n}^{2}B(\mathbf{v}_{n},\mathbf{v}_{n}) = z_{1}^{2}B(\mathbf{v}_{1},\mathbf{v}_{1}) + \dots + z_{p}^{2}B(\mathbf{v}_{p},\mathbf{v}_{p}) +z_{p+1}^{2}B(\mathbf{v}_{p+1},\mathbf{v}_{p+1})+\dots +z_{p+q-1}^{2}B(\mathbf{v}_{p+q-1},\mathbf{v}_{p+q-1})+z_{p+q}^{2}B(\mathbf{v}_{p+q},\mathbf{v}_{p+q})+z_{p+q+1}^{2}B(\mathbf{v}_{p+q+1},\mathbf{v}_{p+q+1})+\dots + z_{n}^{2}B(\mathbf{v}_{n},\mathbf{v}_{n})[/itex], which I think can see now where the summations in the corollary come from. But whose to say, for instance, that [itex]1,\dots ,p[/itex] indexed terms can't be -1 or 0 also? Is it just a convenience to make those substitutions or is there something governing this?

 

[fresh_42]

So I break this up[itex]\displaystyle Q(\mathbf{v}) = \sum_{i=1}^{n}z_{i}^{2}B(\mathbf{v}_{i},\mathbf{v}_{i}) = z_{1}^{2}B(\mathbf{v}_{1},\mathbf{v}_{1}) + z_{2}^{2}B(\mathbf{v}_{2},\mathbf{v}_{2}) + z_{3}^{2}B(\mathbf{v}_{3},\mathbf{v}_{3}) + \dots + z_{n-1}^{2}B(\mathbf{v}_{n-1},\mathbf{v}_{n-1}) + z_{n}^{2}B(\mathbf{v}_{n},\mathbf{v}_{n}) = z_{1}^{2}B(\mathbf{v}_{1},\mathbf{v}_{1}) + \dots + z_{p}^{2}B(\mathbf{v}_{p},\mathbf{v}_{p}) +z_{p+1}^{2}B(\mathbf{v}_{p+1},\mathbf{v}_{p+1})+\dots +z_{p+q-1}^{2}B(\mathbf{v}_{p+q-1},\mathbf{v}_{p+q-1})+z_{p+q}^{2}B(\mathbf{v}_{p+q},\mathbf{v}_{p+q})+z_{p+q+1}^{2}B(\mathbf{v}_{p+q+1},\mathbf{v}_{p+q+1})+\dots + z_{n}^{2}B(\mathbf{v}_{n},\mathbf{v}_{n})[/itex], which I think can see now where the summations in the corollary come from. But whose to say, for instance, that [itex]1,\dots ,p[/itex] indexed terms can't be -1 or 0 also? Is it just a convenience to make those substitutions or is there something governing this?

You wrote:

Then (I'll just consider the case over the field of reals): for some ##p,q## with ##p+q\leq n## there exists a basis such that:

This means, we can enumerate the basis vectors as we like, so we start with the ##1's## - say there are ##p## many - then continue with the ##-1's## - say ##q## many of them - and finally the ##0's## which must be all the rest. All other terms ##B(v_i,v_j)## for ##i \neq j## are zero anyway.

 

[Euler2718]

This means, we can enumerate the basis vectors as we like, so we start with the 1′s1's - say there are pp many - then continue with the −1′s-1's - say qq many of them - and finally the 0′s0's which must be all the rest. All other terms B(vi,vj)B(v_i,v_j) for i≠ji \neq j are zero anyway.

Ah yes, I think it is clear now. Thank you. Now I can proceed on what I'm supposed to do (proving p and q are independent of choice of basis).

 

[fresh_42]

Ah yes, I think it is clear now. Thank you. Now I can proceed on what I'm supposed to do (proving p and q are independent of choice of basis).

Yes, if I remember correctly, it's called the signature of the quadratic form. Note that only the numbers ##p## and ##q## are independent of the basis, not the ordering or any other linear combination of basis elements. ##B## itself can have many appearances depending on which basis is chosen, however, ##p,q## are invariant.

 

[Euler2718]

Yes, if I remember correctly, it's called the signature of the quadratic form. Note that only the numbers ##p## and ##q## are independent of the basis, not the ordering or any other linear combination of basis elements. ##B## itself can have many appearances depending on which basis is chosen, however, ##p,q## are invariant.

Thank you again, I will keep this in mind.