Page 290 - Advanced Linear Algebra
P. 290
274 Advanced Linear Algebra
and we refer to as a hyperbolic extension of . If is nonsingular, we
<
<
<
say that is a hyperbolic extension of itself.
<
)
Proof. For part 1 , the nonsingularity of = implies that : ~ : . Hence,
#¤: ~: and so there is an %: for which º#Á %»£ . If = is
symplectic, then all vectors are isotropic and so we can take ' ~ ² °º#Á %»³% . If
= ' is orthogonal, let ~ # b % . The conditions defining ² # Á ' ³ as a hyperbolic
pair are since is isotropic)
(
#
~ º#Á '» ~ º#Á # b %» ~ º#Á %»
and
~ º'Á'» ~ º # b %Á # b %» ~ º#Á%» b º%Á%» ~ b º%Á%»
Since º#Á %» £ , the first of these equations can be solved for and since
char²-³ £ , the second equation can then be solved for . Thus, in either case,
there is a vector ' : for which / ~ span ²#Á '³ : is hyperbolic. Hence,
: : / and since / is nonsingular, that is, / q / ~ ¸ ¹, we have
/q : ~ ¸ ¹ and so /p : exists.
Part 2) is proved by induction on . Note first that all of the vectors are
#
)
isotropic. If ~ , then span ²# ³ p > exists and so part 1 implies that there is
a hyperbolic plane /~ span ²# Á '³ for which / p > exists.
Assume that the result is true for independent sets of size less than . Since
span²# ³ p span²# ÁÃÁ# ³ p >
exists, part 1) implies that there exists a hyperbolic plane /~ span ²# Á ' ³ for
which
/ p ²# ÁÃÁ# ³ p > span
are in the radical of span ²#Á Ã Á # ³ p > , the inductive
exists. Since #Á Ã Á #
with
hypothesis implies that there is a hyperbolic space /p Ä p /
hyperbolic basis ²# Á' ÁÃÁ# Á' ³ for which the orthogonal direct sum
/p Ä p / p >
exists. Hence, /p Ä p / p > also exists.
We can now prove that the hyperbolic extensions of are precisely the minimal
<
nonsingular extensions of .
<
Theorem 11.11 (Nonsingular extension theorem ) Let be a subspace of a
<
nonsingular finite-dimensional metric vector space = . The following are
equivalent:
)
1 ;~ > p > is a hyperbolic extension of <
2 ; ) is a minimal nonsingular extension of <
