Page 283 - Advanced Linear Algebra
P. 283
Metric Vector Spaces: The Theory of Bilinear Forms 267
Proof. It is clear that orthogonality is symmetric if the form is symmetric or
alternate, since in the latter case, the form is also skew-symmetric.
For the converse, assume that orthogonality is symmetric. For convenience, let
% & mean that º%Á &» ~ º&Á %» and let % = mean that º%Á #» ~ º#Á %» for all
#= . If %= for all % = , then is orthogonal and we are done. So let us
=
examine vectors with the property that % \ = .
%
We wish to show that
\
%= ¬ % is isotropic and ²%& ¬%&³ (11.1)
Note that if the second conclusion holds, then since %% , it follows that is
%
isotropic. So suppose that %& . Since %= , there is a ' = for which
\
º%Á'» £ º'Á%» and so % & if and only if
º%Á &»²º%Á '» c º'Á %»³ ~
Now,
º%Á &»²º%Á '» c º'Á %»³ ~ º%Á &»º%Á '» c º%Á &»º'Á %»
~ º&Á %»º%Á '» c º%Á &»º'Á %»
~ º%Á º&Á %»' c &º'Á %»»
But reversing the coordinates in the last expression gives
ºº&Á%»' c &º'Á%»Á%» ~ º&Á%»º'Á%» c º&Á%»º'Á%» ~
and so the symmetry of orthogonality implies that the last expression is and so
we have proven (11.1).
Let us assume that = is not orthogonal and show that all vectors in = are
isotropic, whence = is symplectic. Since = is not orthogonal, there exist
"Á #= for which "# and so "= and #= . Hence, the vectors " and #
\
\
\
are isotropic and for all & = ,
& " ¬ & "
& # ¬ & #
\
Since all vectors for which $ = are isotropic, let $= . Then $ " and
$
$# and so $ " and $#. Now write
$~²$ c "³ b "
where $c" " , since is isotropic. Since the sum of two orthogonal
"
isotropic vectors is isotropic, it follows that is isotropic if c $ " is isotropic.
$
But
º$ b "Á#» ~ º"Á#» £ º#Á"» ~ º#Á$ b "»
