Page 299 - Advanced Linear Algebra
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Metric Vector Spaces: The Theory of Bilinear Forms 283
only if it is totally degenerate. Furthermore, we have seen that if char²-³ £ ,
then all orthogonal symplectic geometries are totally degenerate and so all such
geometries have orthogonal bases. But if char²-³ ~ , then there are orthogonal
symplectic geometries that are not totally degenerate and therefore do not have
orthogonal bases.
Thus, if we exclude orthogonal symplectic geometries when char²-³ ~ , we
can say that every orthogonal geometry has an orthogonal basis.
If a metric vector space has an orthogonal basis, the natural next step is to
=
look for an orthonormal basis. However, if is singular, then there is a nonzero
=
vector #= and such a vector can never be a linear combination of vectors
from an orthonormal basis ¸" ÁÃÁ" ¹ , since the coefficients in such a linear
combination are º#Á " » ~ .
However, even if is nonsingular, orthonormal bases do not always exist and
=
the question of how close we can come to such an orthonormal basis depends on
the nature of the base field. We will examine this issue in three cases:
algebraically closed fields, the field of real numbers and finite fields.
We should also mention that even when has an orthogonal basis, the Gram–
=
Schmidt orthogonalization process may not apply to produce such a basis,
because even nonsingular orthogonal geometries may have isotropic vectors,
and so division by º"Á "» is problematic.
For example, consider an orthogonal hyperbolic plane /~ span ²"Á #³ and
"
assume that char²-³ £ . Thus, and are isotropic and º"Á #» ~ . The
#
vector cannot be extended to an orthogonal basis using the Gram–Schmidt
"
process, since ¸"Á " b #¹ is orthogonal if and only if ~ . However, does
/
have an orthogonal basis, namely, ¸" b#Á " c#¹ .
Orthogonal Bases
Let = be an orthogonal geometry. As we have discussed, if = is also
symplectic, then has an orthogonal basis if and only if it is totally degenerate.
=
Moreover, when char²-³ £ , all orthogonal symplectic geometries are totally
degenerate and so all orthogonal symplectic geometries have an orthogonal
basis.
= = " ,
If is orthogonal but not symplectic, then contains a nonisotropic vector
the subspace span²" ³ is nonsingular and
=~ span ²" ³ p =
where =~ span ²" ³ . If is not symplectic, then we may decompose it to get
=
= ~ span ²" ³ p span ²" ³ p =
