Linear Programming and Polyhedral Combinatorics

This is a special case of a projection onto a linear space (here, we consider only coordinate projection). By repeatedly projecting, we can eliminate any subset of coordinates. We claim that Pk is also a polyhedron and this can be proved by giving an explicit description of Pk in terms of linear inequalities. For this purpose, one uses Fourier-Motzkin elimination. Let P = {x : Ax ≤ b} and let • S+ = {i : aik > 0}, • S− = {i : aik < 0}, • S0 = {i : aik = 0}. Clearly, any element in Pk must satisfy the inequality a T i x ≤ bi for all i ∈ S0 (these inequalities do not involve xk). Similarly, we can take a linear combination of an inequality in S+ and one in S− to eliminate the coefficient of xk. This shows that the inequalities: