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since
A
is Noetherian (see [
1
, Exercise 7, p. 126]). Thus the height of
M
is exactly 2,
as required by (i). The proof of the theorem is complete.
ACKNOWLEDGMENTS.
I wish to express my warm gratitude to Professor Ram Murty for an interesting
discussion that led to the idea of writing this paper.
REFERENCE
1.
M. F. Atiyah and I. G. Macdonald,
Introduction to Commutative Algebra
, AddisonWesley, Reading, MA,
1969.
Fabrizio Zanello, Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, K7L 3N6
Canada
fabrizio@mast.queensu.ca
Solving Inequalities and Proving
Farkas’s Lemma Made Easy
David Avis and Bohdan Kaluzny
1. INTRODUCTION.
Every college student has learned how to solve a system of
linear equations, but how many would know how to solve
Ax
≤
b
for
x
≥
0orshow
that there is no solution? Solving a system of linear inequalities has traditionally been
taught only in higher level courses and is given an incomplete treatment in introductory
linear algebra courses. For example, the text of Strang [
4
] presents linear programming
and states Farkas’s lemma. It does not, however, include any proof of the ±niteness of
the simplex method or a proof of the lemma. Recent developments have changed the
situation dramatically. Re±nements of the simplex method by Bland [
1
] in the 1970s
led to simpler proofs of its ±niteness, and Bland’s original proof was simpli±ed further
by several authors. In this paper we use a variant of Bland’s pivot rule to solve a system
of inequalities directly, without any need for introducing linear programming. We give
a simple proof of the ±niteness of the method, based on ideas contained in the paper
of Fukuda and Terlaky [
3
] on the related crisscross method. Finally, if the system is
infeasible, we show how the termination condition of the algorithm gives a certi±cate
of infeasibility, thus proving the Farkas lemma. Terminology and notation used here
follows that of Chv´atal’s linear programming book [
2
].
We consider the following problem: given a matrix
A
=[
a
ij
]
in
R
m
×
n
and a column
vector
b
in
R
m
,±nd
x
=
(
x
1
,
x
2
,...,
x
n
)
T
that satis±es the following linear system, or
prove that no such vector
x
exists:
≤
b
,
x
≥
0
.
(1)
We illustrate a simple method for doing this with an example:
−
x
1
−
2
x
2
+
x
3
≤−
1
x
1
−
3
x
2
−
x
3
≤
2(
2
)
−
x
1
−
2
x
2
+
2
x
3
2
152
c
°
THE MATHEMATICAL ASSOCIATION OF AMERICA
[Monthly 111
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x
i
≥
0
(
i
=
1
,
2
,
3
)
. We frst convert this system oF inequalities into a system oF
equations by introducing a new nonnegative
slack
variable For each inequality. This
slack variable represents the diFFerence between the right and leFthand sides oF the
inequality. In our example, we need three new variables, which we label
x
4
,
x
5
,and
x
6
.
Putting these variables on the leFthand side, and the others on the righthand side we
have the Following system:
x
4
=−
1
+
x
1
+
2
x
2
−
x
3
x
5
=
2
−
x
1
+
3
x
2
+
x
3
(3)
x
6
2
+
x
1
+
2
x
2
−
2
x
3
It is easy to see that any nonnegative solution oF (2) then extends to a nonnegative
solution oF (3) by assigning the slack variables values via their respective equations.
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 Spring '09
 MeinolfSellman

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