MATB61 Feb. 8th notes
1. Show one of the following linear systems has a solution if and only if the
other one has no solution.
2. We want to prove above by using the following facts.
One of the following linear systems has a solution if and only if the other
one has no solution.
A~x
=
~
b, ~x
≥
0
(3)
and
A
T
~
y
≥
0
, ~
y
T
~
b <
0
(4)
Proof.
We notice that there is no sign constraint in system (1). Thus,
we should rewrite the (1) to an equivalent system which dose have such
a constraint.
Let
~x
=
~u

~v
and
~u,~v
≥
0
.
Rewrite (1) in terms
~u,~v
as the following.
A~x
=
~
b
⇔
A
(
~u

~v
) =
~
b
⇔
A~u

A~v
=
~
b
⇔
h
A

A
i
"
~u
~v
#
=
~
b,
(5)
where you can view
M
=
h
A

A
i
is a bigger matrix and
~
c
=
"
~u
~v
#
≥
0
.
Now you may compare system (5) with (3) and you should see they are
equivalent. ( ie, (5) gives you
M~
c
=
~
b,~
c
≥
0.)
Also note that (5) is
same as (1).
Apply the fact in our case, you should get the following.