Linear Programming Lecture Notes - Penn State Personal Web Server

Theorem 4.31. Let P ⊆ R n be a polyhedral set and suppose P is defined as:
(4.24) P = {x ∈ R n : Ax ≤ b}
where A ∈ R m×n and b ∈ R m . A point x0 ∈ P is an extreme point of P if and only if x0 is
the intersection of n linearly independent hyperplanes from the set defining P .
Remark 4.32. The easiest way to see this as relevant to linear programming is to assume
that
(4.25) P = {x ∈ R n : Ax ≤ b, x ≥ 0}
In this case, we could have m < n. In that case, P is composed of the intersection of n + m
half-spaces. The first m are for the rows of A and the second n are for the non-negativity
constraints. An extreme point comes from the intersection of n of the hyperplanes defining
these half-spaces. We might have m come from the constraints Ax ≤ b and the other n − m
from x ≥ 0.
Proof. (⇐) Suppose that x0 is the intersection of n hyperplanes. Then x0 lies on n
hyperplanes. By way of contradiction of suppose that x0 is not an extreme point. Then
there are two points x, ˆx ∈ P and a scalar λ ∈ (0, 1) so that
x0 = λx + (1 − λ)ˆx
If this is true, then for some G ∈ R n×n whose rows are drawn from A and a vector g whose
entries are drawn from the vector b, so that Gx0 = g. But then we have:
(4.26) g = Gx0 = λGx + (1 − λ)Gˆx
and Gx ≤ g and Gˆx ≤ g (since x, ˆx ∈ P ). But the only way for Equation 4.26 to hold is if
(1) Gx = g and
(2) Gˆx = g
The fact that the hyper-planes defining x0 are linearly independent implies that the solution
to Gx0 = g is unique. (That is, we have chosen n equations in n unknowns and x0 is the
solution to these n equations.) Therefore, it follows that x0 = x = ˆx and thus x0 is an
extreme point since it cannot be expressed as a convex combination of other points in P .
(⇒) By Lemma 4.30, we know that any extreme point x0 lies on the boundary of P and
therefore there is at least one row Ai· such that Ai·x0 = bi (otherwise, clearly x0 does not
lie on the boundary of P ). By way of contradiction, suppose that x0 is the intersection of
r < n linearly independent hyperplanes (that is, only these r constraints are binding). Then
there is a matrix G ∈ R r×n whose rows are drawn from A and a vector g whose entries are
drawn from the vector b, so that Gx0 = g. Linear independence of the hyperplanes implies
that the rows of G are linearly independent and therefore there is a non-zero solution to
the equation Gd = 0. To see this, apply Expression 3.51 and choose solution in which d is
non-zero. Then we can find an ɛ > 0 such that:
(1) If x = x0 + ɛd, then Gx = g and all non-binding constraints at x0 remain nonbinding
at x.
(2) If ˆx = x0 − ɛd, then Gˆx = g and all non-binding constraints at x0 remain nonbinding
at ˆx.
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