- #1

Euler2718

- 90

- 3

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?