Linear Programming Lecture Notes - Penn State Personal Web Server

a T x2 = b
Then we have:
(4.10) a T x = a T [λx1 + (1 − λ)x2] = λa T x1 + (1 − λ)a T x2 = λb + (1 − λ)b = b
Thus, x ∈ H and we see that H is convex. This completes the proof.
Lemma 4.15. Every half-space is convex.
Proof. Let a ∈ R n and b ∈ R. Without loss of generality, consider the half-space Hl
defined by a and b. For arbitrary x1 and x2 in Hl we have:
a T x1 ≤ b
a T x2 ≤ b
Suppose that a T x1 = b1 ≤ b and a T x2 = b2 ≤ b. Again let x = λx1 + (1 − λ)x2. Then:
(4.11) a T x = a T [λx1 + (1 − λ)x2] = λa T x1 + (1 − λ)a T x2 = λb1 + (1 − λ)b2
Since λ ≤ 1 and 1 − λ ≤ 1 and λ ≥ 0 we know that λb1 ≤ λb, since b1 ≤ b. Similarly we
know that (1 − λ)b2 ≤ (1 − λ)b, since b2 ≤ b. Thus:
(4.12) λb1 + (1 − λ)b2 ≤ λb + (1 − λ)b = b
Thus we have shown that a T x ≤ b. The case for Hu is identical with the signs of the
inequalities reversed. This completes the proof.
Using these definitions, we are now in a position to define polyhedral sets, which will be
the subject of our study for most of the remainder of this chapter.
Definition 4.16 (Polyhedral Set). If P ⊆ R n is the intersection of a finite number
of half-spaces, then P is a polyhedral set. Formally, let a1, . . . , am ∈ R n be a finite set of
constant vectors and let b1, . . . , bm ∈ R be constants. Consider the set of half-spaces:
Then the set:
m
(4.13) P =
Hi = {x|a T i x ≤ bi}
i=1
Hi
is a polyhedral set.
It should be clear that we can represent any polyhedral set using a matrix inequality.
The set P is defined by the set of vectors x satisfying:
(4.14) Ax ≤ b,
where the rows of A ∈ R m×n are made up of the vectors a1, . . . , am and b ∈ R m is a column
vector composed of elements b1, . . . , bm.
Theorem 4.17. Every polyhedral set is convex.
Exercise 41. Prove Theorem 4.17. [Hint: You can prove this by brute force, verifying
convexity. You can also be clever and use two results that we’ve proved in the notes.]
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