Page 289 - Advanced Linear Algebra

Metric Vector Spaces: The Theory of Bilinear Forms 273

(
for > as a hyperbolic basis . In the symplectic case, the usual term is
symplectic basis.)
Note that any hyperbolic space is nonsingular.
>
In the orthogonal case, hyperbolic planes can be characterized by their degree of
(
isotropy, so to speak. In the symplectic case, all spaces are totally isotropic by
)
definition. Indeed, we leave it as an exercise to prove that a two-dimensional
nonsingular orthogonal geometry = is a hyperbolic plane if and only if =
contains exactly two one-dimensional totally isotropic equivalently, totally
(
)
degenerate subspaces. Put another way, the cone of isotropic vectors is the
union of two one-dimensional subspaces of .
=
Nonsingular Completions of a Subspace
Let be a subspace of a nonsingular metric vector space . If is singular, it
=
<
<
is of interest to find a minimal nonsingular subspace of containing .
<
=
Definition Let be a nonsingular metric vector space and let be a subspace
=
<
of = . A subspace of = : for which <  : is called an extension of < . A
nonsingular completion of is an extension of that is minimal in the family
<
<
<
of all nonsingular extensions of .
Theorem 11.10 Let be a nonsingular finite-dimensional metric vector space
=
over . We assume that char ² - ³ £ when is orthogonal.
-
=
)
1 Let be a subspace of . If is isotropic and the orthogonal direct sum
=
#
:
span²#³ p :
exists, then there is a hyperbolic plane /~ span ²#Á '³ for which
/p :
exists. In particular, if is isotropic, then there is a hyperbolic plane
#
containing .
#
)
2 Let be a subspace of and let
<
=
< ~ span ²# ÁÃÁ# ³ p >

where > is nonsingular and ¸ # Á Ã Á # ¹ are linearly independent in
~ / p Ä p / with
rad²<³. Then there is a hyperbolic space >
hyperbolic basis ²# Á' ÁÃÁ# Á' ³ for which

<~ > p >
is a nonsingular proper extension of < . If ¸ # Á Ã Á # ¹ is a basis for
rad²<³, then
dim²<³ ~ dim²<³ b dim²rad ²<³³