Page 437 - Advanced Linear Algebra
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Positive Solutions to Linear Systems: Convexity and Separation 421
º5Á » b º5Á » ~ º5Á b » 5))
for all : and * . But if º5Á »£ for some : , then we can replace
by an appropriate scalar multiple of in order to make the left side of this
inequality negative, which is impossible. Hence, º5Á » ~ for all : , that
is, 5 : and
º5Á *» 5))
We can now prove (15.1) and (15.2).
:
Theorem 15.7 Let be a subspace of s .
1 : q s ) ~J if and only if : q s £J
b b b
2 : q s ) bb ~J if and only if : q s b £J
Proof. In both cases, one direction is easy. It is clear that there cannot exist
vectors " s bb and # s b that are orthogonal. Hence, : q s b and
:q s bb cannot both be nonempty and so :q s bb £ J implies
:q s b ~ J. Also, :q s b b and : q s b cannot both be nonempty and so
: q s £J implies that : q s ~J.
b b b
For the converse in part 1), to prove that
:q s b ~ J ¬ : q s b b £ J
a good candidate for an element of :q s bb would be a normal to a
:
hyperplane that separates from a subset of s . Note that our separation
b
theorems do not allow us to separate from s b , because s b is not compact. So
:
"
consider instead the convex hull of the standard basis vectors ÁÃÁ in
s :
b
" ~ ¸! b Ä b ! ' ! Á !~ ¹
which is compact. Moreover, " s b implies that " q : ~ J and so Theorem
15.6 implies that there is a nonzero vector 5 ~ ² ÁÃÁ ³ : such that
º5Á » 5))
for all " À Taking ~ gives
~ º5Á » 5))
and so 5 : q s bb , which is therefore nonempty.
)
To prove part 2 , again we note that there cannot exist orthogonal vectors
" s bb and # s b : q s and so bb : q s and b cannot both be nonempty.
Thus, : q s b £J implies that : q s b b ~J .
