Need a confirmation Let's say I have a system of inequalities { <mtable columnalign=

fabios3

fabios3

Answered question

2022-06-19

Need a confirmation Let's say I have a system of inequalities
{ A x b x 0
A R m × n , x R n × 1 , b R m × 1 . n m Assume also that A has full rank. Is it possible to manipulates the inequalities involved such that for i = 1 mI have that x i is bounded (above or below) by x i 1 , . . . , x 0 ? Something like
a i i x i C i + j = 0 i 1 c i j x j

Answer & Explanation

jarakapak7

jarakapak7

Beginner2022-06-20Added 14 answers

The form you advocate,
a i i x i j = 0 i 1 c i j x j
has no constants anymore. Now this is not going to be the general case. As a contradictive example, consider the simple special case where the matrix A is diagonal. Then the bounds just read
a i i x i b i
which contradicts your form.
Besides that, there is also a contructive answer to your question. Your system of inequalities
{ A x b x 0 }
actually describe the feasible region of a simplex and you will find abundant literature on that, e.g. here. From the point of view of your question, the feasible region is a convex polytope which can always be described, using slack variables, in standard form as
{ A ~ x ~ = b ~ x ~ 0 }
So this answers your question (if you include slack variables) since with this theorem, you have the relation A ~ x ~ = b ~ between the variables.

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