Page 435 - Advanced Linear Algebra
P. 435
Positive Solutions to Linear Systems: Convexity and Separation 419
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We first show that contains a vector 5 of minimum norm. Recall that the
Euclidean norm (distance) is a continuous function. Although need not be
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compact, if we choose a real number such that the closed ball
))
) ² ³ ~¸' s ' ¹
intersects , then that intersection * Z ~ * q ) ² ³ is both closed and bounded
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and so is compact. The norm function therefore achieves its minimum on * Z ,
Z
say at the point 5 * * . It is clear that if # 5) for some # * , then
)
))
Z
#* , in contradiction to the minimality of 5. Hence, 5 is a vector of
minimum norm in .
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We establish uniqueness first for closed line segments ´"Á #µ in s . If " ~ #
where , then
)
(
) )!" b ² c !³# ~ ! b ² c !³ # )
(
is smallest when !~ for and !~ for . Assume that and are
#
"
not scalar multiples of each other and suppose that %£& in ´"Á #µ have
minimum norm . If ' ~ ²% b &³° then since and are also not scalar
%
&
multiples of each other, the Cauchy-Schwarz inequality is strict and so
P'P ~ ) % b & )
~ ²P%P b º%Á &» b P&P ³
² b % ))) ) ³
&
~
which contradicts the minimality of . Thus, ´ " Á # µ has a unique point of
minimum norm.
5
%
Finally, if %* also has minimum norm, then and are points of minimum
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norm in the line segment ´5Á %µ * and so % ~ 5 . Hence, has a unique
element of minimum norm.
For part 2), suppose the result is true when ¤* . Then ¤* implies that
¤* c and so if 5 * c has smallest norm, then
º5Á * c » 5))
Therefore,
º5Á *» 5)) b º5Á » º5Á »
and so and are strongly separated by the hyperplane
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