Clarifying a corollary about Quadratic Forms

In summary, the conversation discusses the corollary of a theorem about symmetric bilinear forms on a vector space over the field of real or complex numbers. The corollary states that for some values of p and q, there exists a basis for the vector space where the quadratic form can be written as a sum of squares with p positive terms and q negative terms. This is shown by substituting appropriate values for the basis vectors, which can be chosen arbitrarily.
  • #1
Euler2718
90
3
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?
 
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  • #2
Euler2718 said:
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.
 
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  • #3
fresh_42 said:
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?
 
  • #4
Euler2718 said:
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:

Euler2718 said:
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.
 
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  • #5
fresh_42 said:
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).
 
  • #6
Euler2718 said:
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.
 
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  • #7
fresh_42 said:
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.
 

FAQ: Clarifying a corollary about Quadratic Forms

1. What is a corollary about quadratic forms?

A corollary about quadratic forms is a statement that follows as a consequence of a previously proven theorem. In this context, it refers to a statement related to the properties and characteristics of quadratic forms, which are mathematical expressions involving variables raised to the second power.

2. How do you clarify a corollary about quadratic forms?

To clarify a corollary about quadratic forms, you would need to carefully examine the original theorem and its proof to understand how the corollary follows from it. You may also need to use mathematical techniques and logical reasoning to break down the corollary and explain it in simpler terms.

3. What are some common examples of corollaries about quadratic forms?

Some common examples of corollaries about quadratic forms include statements about the symmetry and definiteness of quadratic forms, as well as the relationship between quadratic forms and matrix operations.

4. How does understanding corollaries about quadratic forms benefit scientific research?

Understanding corollaries about quadratic forms can help scientists in fields such as mathematics, physics, and engineering to better analyze and solve problems involving quadratic equations. This knowledge can also aid in the development of new theories and applications in these fields.

5. Are there any limitations to corollaries about quadratic forms?

Yes, there are limitations to corollaries about quadratic forms as they are only applicable to specific types of equations and may not hold true for all cases. Additionally, corollaries may be affected by certain assumptions made in the original theorem, so they should be carefully evaluated before being applied in scientific research.

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