Page 436 - Advanced Linear Algebra
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420 Advanced Linear Algebra
>2 )) b º5Á » 35Á ² ° ³
5
Thus, we need only prove part 2) for ~ , that is, we need only prove that
º5Á *» 5))
If there is a nonzero %* for which
º5Á %» 5))
then ))5 ) ) and
%
º5Á %» ~ 5)) c
for some . Then for the open line segment ²!³ ~!5 b ² c !³% with
! , we have
P ²!³P ~ !5 b ² c !³%)
)
~ ! P5P b !² c !³º5Á %» b ² c !³ P%P
² ! c! ³P5P c !² c!³ b² c!³ P%P
))
~ ²cP5P b b % ³! b ² 5) )) ) c c P%P ³! b %
Let ²!³ denote the final expression above, which is a quadratic in . It is easy to
!
see that ²!³ has its minimum at the interior point of the line segment ´5Á %µ
corresponding to
c5 b b %) )
))
!~
))
c5 b b %) )
and so ) ) ²! ³ ²! ³ ² ³ ~ 5 ) , which is a contradiction.
)
The next result brings us closer to our goal by replacing a single vector with a
subspace disjoint from . However, we must also require that be bounded,
*
*
:
and therefore compact.
:
Theorem 15.6 Let be a compact convex subset of s and let be a subspace
*
of s such that *q : ~ J . Then there exists a nonzero 5 : such that
º5Á %» 5))
>
:
for all % * . Hence, the hyperplane ²5Á 5)) ° ³ strongly separates and
*.
Proof. Theorem 15.2 implies that the set :b * is closed and convex.
Furthermore, *q : ~ J implies that ¤ : b * and so Theorem 15.5 implies
that :b * can be strongly separated from the origin. Hence, there is a nonzero
5 s such that
