294 VII The Arithmetic of Quadratic Forms
Proposition 1If a quadratic space V contains a vector u such that(u,u)= 0 ,then
V=U⊥U⊥,where U=〈u〉is the one-dimensional subspace spanned by u.
Proof For any vectorv∈V, putv′=v−αu,whereα=(v,u)/(u,u). Then (v′,u)=
0 and hencev′∈U⊥.SinceU∩U⊥={ 0 }, the result follows.
A vector space basisu 1 ,...,unof a quadratic spaceVis said to be anorthogonal
basisif(uj,uk)=0 wheneverj=k.
Proposition 2Any quadratic space V has an orthogonal basis.
Proof IfVhas dimension 1, there is nothing to prove. SupposeV has dimension
n>1 and the result holds for quadratic spaces of lower dimension. If(v,v)=0for
allv∈V, then any basis is an orthogonal basis, by (1). Hence we may assume that
Vcontains a vectoru 1 such that(u 1 ,u 1 )=0. IfU 1 is the 1-dimensional subspace
spanned byu 1 then, by Proposition 1,
V=U 1 ⊥U 1 ⊥.By the induction hypothesisU 1 ⊥has an orthogonal basisu 2 ,...,un,andu 1 ,u 2 ,...,un
is then an orthogonal basis forV.
Proposition 2 says that any symmetic matrixAis congruent to a diagonal matrix,
or that the corresponding quadratic formfis equivalent overFto a diagonal form
δ 1 ξ 12 +···+δnξn^2. Evidently detf =δ 1 ···δnandf is non-singular if and only if
δj= 0 ( 1 ≤j≤n).IfA=0 then, by Propositions 1 and 2, we can takeδ 1 to be any
element ofF×which is represented byf.
Hereγ∈F×is said to berepresentedby a quadratic spaceVover the fieldFif
there exists a vectorv∈Vsuch that(v,v)=γ.
As an application of Proposition 2 we prove
Proposition 3If U is a non-singular subspace of the quadratic space V , then
V=U⊥U⊥.
Proof Letu 1 ,...,umbe an orthogonal basis forU.Then(uj,uj)= 0 ( 1 ≤j≤m),
sinceUis non-singular. For any vectorv∈V,letu=α 1 u 1 +···+αmum,where
αj=(v,uj)/(uj,uj)for eachj.Thenu∈Uand(u,uj)=(v,uj)( 1 ≤j≤m).
Hencev−u∈U⊥.SinceU∩U⊥={ 0 }, the result follows.
It may be noted that ifUis a non-singular subspace andV =U⊥Wfor some
subspaceW, then necessarilyW = U⊥. For it is obvious thatW ⊆ U⊥and
dimW=dimV−dimU=dimU⊥, by Proposition 3.
Proposition 4Let V be a non-singular quadratic space. Ifv 1 ,...,vmare linearly
independent vectors in V then, for anyη 1 ,...,ηm∈F , there exists a vectorv∈V
such that(vj,v)=ηj( 1 ≤j≤m).
Moreover, if U is any subspace of V , then