Page 300 - Advanced Linear Algebra

284 Advanced Linear Algebra

This process may be continued until we reach a decomposition
= ~ span ²" ³ p Ä p span ²" ³ p <

where is symplectic as well as orthogonal. (This includes the case < ~ ¸ ¹ .)
<
Let 8 ~²" Á Ã Á " ³ .

<
<
9
If char²-³ £ , then is totally degenerate. Thus, if is a basis for , then the
union 8 9 r is an orthogonal basis for = . If char ² - ³ ~ , then
>
<~ > p rad ²<³, where is hyperbolic and so
= ~ span ²" ³pÄp span ²" ³p > p rad ²<³

where rad²<³ is totally degenerate and the " are nonisotropic. If
9 > ~²% Á & Á Ã Á % Á & ³ is a hyperbolic basis for and : ~²' Á Ã Á ' ³ is an

ordered basis for rad²<³ , then the union
; 8 ~ 9 r r : ~ ²" ÁÃÁ" Á% Á& ÁÃÁ% Á& Á' ÁÃÁ' ³

is an ordered orthogonal basis for = . However, we can do better (in some
sense).
The following lemma says that when char²-³ ~ , a pair of isotropic basis
, can be replaced by a pair of nonisotropic basis vectors,
vectors, such as %Á &
.
when coupled with a nonisotropic basis vector, such as "
Lemma 11.20 Suppose that char²-³ ~ . Let > be a three-dimensional
orthogonal geometry of the form
> ~ span ²#³ p span ²%Á &³
where is nonisotropic and / ~ span ² % Á & ³ is a hyperbolic plane. Then
#
> ~ span ²# ³ p span ²# ³ p span ²# ³

where each is nonisotropic.
#
Proof. It is straightforward to check that if º#Á #» ~ , then the vectors
# ~ "b%b&

#~ " b %

# ~ " b ² c ³% b &

are linearly independent and mutually orthogonal. Details are left to the
reader.
Using the previous lemma, we can replace the vectors ¸" Á% Á& ¹ with the

nonisotropic vectors ¸# Á# b Á# b ¹ , while retaining orthogonality. Moreover,
the replacement process can continue until the isotropic vectors are absorbed,
leaving an orthogonal basis of nonisotropic vectors.