Page 434 - Advanced Linear Algebra
P. 434
418 Advanced Linear Algebra
As before, we need only consider one of the conditions to show that two sets are
not strongly separated. Note also that if
º5Á %» º5Á @ »
for s , then % s and @ s are stongly separated by the hyperplane
bº5Á %»
> 6 7 5Á
Separation
Now that we have the preliminaries out of the way, we can get down to some
theorems. The first is a well-known separation theorem that is the basis for
many other separation theorems. It says that if a closed convex set * s does
not contain a vector , then can be strongly separated from .
*
*
Theorem 15.5 Let be a closed convex subset of s .
5
1 * ) contains a unique vector of minimum norm, that is, there is a unique
vector 5 * for which
%
))5 ) )
for all %*Á %£5 .
)
*
2 If ¤ * , then lies in the closed half-space
2
> b 5Á )) b º5Á » 3
5
that is,
º5Á *» 5)) b º5Á » º5Á »
where is the unique vector of minimum norm in the closed convex set
5
*c ~ ¸ c *¹
Hence, and are strongly separated by the hyperplane
*
)) b º 5 Á »
5
> 5Á 9
8
Proof. For part 1), if * then this is the unique vector of minimum norm, so
we may assume that ¤* . It follows that no two distinct elements of can be
*
*
%
negative scalar multiples of each other. For if and % were in , where Á
then taking !~c °² c ³ gives
~ !% b ² c !³ % *
which is false.
