avis-kaluzny

avis-kaluzny - since A is Noetherian(see[1 Exercise 7 p 126...

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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 , Addison-Wesley, Reading, MA, 1969. Fabrizio Zanello, Department of Mathematics and Statistics, Queen’s University, Kingston, Ontario, K7L 3N6 Canada [email protected] 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 criss-cross 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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with 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 leFt-hand 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 leFt-hand side, and the others on the right-hand 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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avis-kaluzny - since A is Noetherian(see[1 Exercise 7 p 126...

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