v1 v2 vk v2 v1 vk v1 linearly independent; aﬃnely ...

Affine sets î�� |9�½+Ë Linear Inequality Systems

Definition 6.12

The vectors v1, v2, . . ., vk are affinely independent if v2 − v1, . . ., vk − v1 islinearly independent; affinely dependent, otherwise.

We first check the subtracting vector is irrelevant in the definition: v2 − v1,v3 − v1, . . ., vk − v1 are linearly independent ⇔ v1 − v2, v3 − v2, . . ., vk − v2 arelinearly independent ⇔ · · · . (Check it.) It is also easy to see that affineindependence is invariant with a translation.

Proposition 6.13

The vectors v1, v2, . . ., vk are affinely independent only λ = 0 satisfies thefollowing. The converse is also true.

λ1v1 + · · ·+ λkvk = 0λ1 + · · ·+ λk = 0.

(6.15)

(6.15) is equivalent to that the following vectors are linearly independent.[v1

1

],

[v2

1

], · · · ,

[vk

1

].

Optimization Lab. 25th March 2018 42 / 50


Exercise 6.14

Linearly independent vectors are also affinely independent.

If we translate, by w /∈ S, a basis of a subspace S, and add w to it, thenthe resulting set is a set of affinely independent vectors. Therefore, themaximum number of affinely independent vectors from S + w is ≥dim(S) + 1. But it can not exceed dim(S) + 1 (why?).

Proposition 6.15

The maximum number of affinely independent vectors in S + w is dim S+ 1.



FYI �ÃÐ�¦

Let L and L− w with w ∈ L, respectively, be an affine space and its subspace.Then {v1, v2, . . ., vk} ⊆ L ⇔ {v2 − v1, . . ., vk − v1} ⊆ L− w.

Example 6.16

In the figure, v1, v2, v3 are linearly as well as affinely idependent. Also 0,v2 − v1, v3 − v1 is affinely independent but not linearly independent.



FYI �ÃÐ�¦

Proposition 6.17

If an affine space does not contain the origin, its affinely independent vectors arelinearly independent. (Therefore, their maximum number is the same as thedimension of the subspace plus 1.)

Proof: Let’s consider a subspace S and an affine space S + w for some w /∈ S.Assume v1, v2, . . ., vk are affinely independent vectors in S + w and α1v1 +α2v2 + · · · + αkvk = 0. Then

α1(v1 − w) + α2(v2 − w) + · · ·+ αk(vk − w) = −(α1 + · · ·+ αk)w. (6.16)

While v1 − w, . . ., vk − w are in S, w is not. Therefore (6.16) is possible only ifα1 + · · · + αk = 0. Since vi’s are affinely independent, we have, by Exercise6.13, α1 = · · · = αk = 0.

Definition 6.18

The dimension of an affine space L is defined to be the maximum number ofaffinely independent vectors from L minus 1.



FYI �ÃÐ�¦

Suppose w1, · · · , wk are affinely dependent vectors. Then there is β = (β1, . . .,βk) 6= 0:

β1w1 + β1w1 + · · ·+ βkwk = 0, β1 + β2 + · · ·+ βk = 0. (6.17)

We may assume β1 6= 0. Then, if we write µi := −βi/β1, (6.17) becomes

w1 = µ2w2 + · · ·+ µk+1wk+1,µ2 + · · ·+ µk+1 = 1

(6.18)

Hence w1 is an affine combination of w2, . . ., wk. We can see that

Proposition 6.19

The vectors w1, · · · , wk are affinely dependent if and only there is i such that wi

is an affine combination of the others.



FYI �ÃÐ�¦

A subspace S is the set of linear combinations of a maximal linearly independentset of vectors it contains. Likewise, an affine set L is the set of affinecombinations of a maximal affinely independent set of vectors it has.

Proposition 6.20

A k-dimensional affine set is the set of affine combinations of its (k + 1) affinelyindependent vectors.

Definition 6.21

The dimension of a convex set is defined to be the the maximum number ofits affinely independent vectors minus 1.


Valid inequalities Linear Inequality Systems

Definition 7.1

Valid inequalities (��{©�ô�Ç ÂÒ1pxd��) If the half space {x : aT x ≥ b} includesS, aT x ≥ b is called a valid inequality for S.

Valid inequality and supporting hyperplane

Definition 7.2

Supporting hyperplanes (]X�î��) If aT x ≥ b is valid for S and S ∩{x : aT x = b} 6= ∅, H = {x : aT x = b} is called a supporting hyperplane of S.



Definition 7.3

Separating hyperplane (ì�ro��íî��) H = {x : aT x = b} is called aseparating hyperplane of a set S ⊆ Rn and a point z ∈ Rn − S if aT x ≥ b isvalid for S but not for z.

A separating hyperplane of S and z.



Theorem 7.4

Separating heperplane theorem (ì�ro��íî��&ño�) If S is a closed set and z /∈S, then there is a separating hyperplane of z and S.

The theorem implies that any closed convex set C is the intersection of the halfspaces separating C and the points not in C.

We can prove Farkas Lemma by using Separating hyperplane theorem.


Polyhedra,

Polyhedra

Optimization Lab.

IE departmentSeoul National University

23rd April 2018

Optimization Lab. 23rd April 2018 1 / 38

Polyhedra Polyhedra,

Definition 1.1

We call the feasible solution set of a linear system Ax ≥ b (A ∈ Rm×n) apolyhedron. In other words, a polyhedron is the intersection of a finite numberof half spaces.

A =

0 3 20 −3 22 0 32 0 −33 2 0−3 2 00 −3 −20 3 −2−2 0 −3−2 0 3−3 −2 03 −2 0

, b =

565455656546

.

Dodecahedron ( 12 ��) - from “Therepresentation of polyhedra by polynomialinequalities” M Grotschel and M. Henk,(2004).



Regular dodecahedron (&ñ 12 ��)

φ = 1+√

52 , α > 0,

±φ2x± φy ≤ α,±φ2y ± φz ≤ α,±φ2z ± φx ≤ α.



A polyhedron P is said to be bounded if it does not contain any half line(or unbounded ray), or equivalently if P has vector lower and upperbounds: there are l, u ∈ Rn such that P ⊆ {x : l ≤ x ≤ u}.

If every point of P = {x|Ax ≥ b}satisfies an inequality Ai·x ≥ bi forsome i with equality, we call it animplicit equality. The subsystemof implicit equalities of Ax ≥ b iscalled the implicit equality subsys-tem and denoted by A=x ≥ b=.

Exercise 1.2

There is a point of P satisfying each of the remaining subsystem withstrict inequality, “>.”



Lemma 1.3

If an affine set L ⊆ Rn has a point u satisfying DT u > d, then L and L ∩ {x :DT x ≥ d} has the same dimension.

Proof: By the assumption, there is ε > 0 suchthat every point of L ∩ Bε(u) satisfies DT x ≥ d.(The figure shows the case when Dx ≥ d isaT x ≥ β.) Let v1, v2, . . ., vk be affinely inde-pendent vectors in L. Since L is affine and henceclosed under translation between its points, thefollowing vectors are also in L: u, u+ ε

2v2−v1‖v2−v1‖ ,

. . ., u + ε2

vk−v1‖vk−v1‖ . They are also affinely inde-

pendent vectors each contained in L ∩ Bε(u).This implies L and L ∩ {x : DT x ≥ d} havethe same dimension.

Proposition 1.4

P has the same dimension as the affine set A=x = b=: dim(P ) = n− r(A=).


Faces Polyhedra,

Definition 2.1

By a face of a polyhedron P , we mean its intersection with asupporting hyperplane.

If there is an implicit equality aT x ≥ β, H = {x| aT x = β} is asupporting hyperplane which includes P . Hence P itself is a face. By aproper face, we mean a face other than P .

Proposition 2.2

Face-subsystem proposition (��-ÂÒì�rr�Û¼%7� $í|9�) Let F be a face ofP = {x : Ax ≥ b}. Then there is a subsystem A◦x ≥ b◦ of Ax ≥ b suchthat F = {x ∈ P : A◦x = b◦}. Conversely, if P ∩ {x : A◦x = b◦} 6= ∅ fora subsystem A◦x ≥ b◦, F := P ∩H is a face of P .

(Will call A◦x = b◦} in the proposition a face subsystem of F .)


Faces Polyhedra,

Proof: Let F be a face of P and cT x = d be the supporting hyperplane.I.e. F = {x ∈ P : cT x = d} and cT x ≥ d for every x ∈ P . This meansexactly that F is the set of optimal solutions of LP min{cT x : Ax ≥ b}.By strong duality, there is dual optimal solution y ≥ 0. Let A◦x ≥ b◦ be asubsystem of Ax ≥ b corresponding to positive components of y. Then bycomplementary slackness theorem, an element x of P is optimal if andonly if A◦x = b◦. Hence F = {x ∈ P : A◦x = b◦}.Conversely, suppose there is a subsystem A◦x ≥ b◦ such that F := P ∩{x : A◦x = b◦} 6= ∅. Suppose we have taken the sum of the equations ofA◦x◦ = b◦ to get equation cT x = δ. The equation is satisfied by everyelement of F and hence F ⊆ P ∩ {x : cT x = δ}.For the reverse inclusion, consider any x ∈ P − F . Then by definition atleast one inequality from A◦x ≥ b◦ is satisfied with >. Thus any point ofP satisfying cT x = δ is also in F : F ⊇ P ∩ {x : cT x = δ}.Assume c 6= 0 so that cT x = δ is a hyperplane. Then by definition F is aface of P . For the case when c = 0, see Exercise 2.3.


Faces Polyhedra,

Exercise 2.3

Complete the proof for the case c = 0 (so that δ = 0). (Hint: P is a face.)

A face subsystem is, however, not unique in general. Furthermore eventheir ranks can be different. In the left figure, the subsystems {a1x ≥ b1,a2x ≥ b2} and {a3x ≥ b3} determine the same face F . In the right, the0-dimensional face F has the determining subsystems of rank 1, 2, and 3.


Faces Polyhedra,

Let A◦x = b◦ and A◦◦x = b◦◦ be two face subsystems of F : F = {x ∈P : A◦x = b◦} = {x ∈ P : A◦◦x = b◦◦}. Then F = {x ∈ P : A◦x = b◦,A◦◦x = b◦◦}. The merged subsystem also a face subsystem of F .

Definition 2.4

Maximum face subsystem The subsystem A◦x ≥ b◦ obtained bymerging all the face subsystems of a face F is called the maximum facesubsystem of F .

By definition, for any inequality not in its maximum face subsystem, F hasa point satisfying it with strict inequality. Therefore, the dimension of F isthe same as the dimension of the affine set A◦x = b◦: dim(F ) = n −r(A◦).


Faces Polyhedra,

Example 2.5

The maximum subsystem of the top extreme point has five inequalities,each set of three inequalities from which is also a face subsystem of it.


Faces Polyhedra,

Assume Ax = b has a solution. Recall Exercise 9.2 that if cT increases therank of A, the system Ax = b, cT x = d has a solution. Otherwise, Ax = b,cT x = d has either no solution or the same solution set as Ax = b.

Let F and F ′ be the faces of P having A◦x ≥ b◦ and A◦◦x ≥ b◦◦,respectively, as their maximum face subsystems.

Suppose F ′ ⊆ F . Then A◦◦x = b◦◦ should contain every inequality ofA◦x = b◦. If, in addition, F ( F ′, then A◦◦x = b◦◦ has more inequalitiesthan A◦x ≥ b◦. And we should have r(A◦) < r(A◦◦) since, otherwise, thesolution set does not change or is empty. Thus we have proved thefollowing proposition.

Proposition 2.6

Suppose two faces F and F ′ of a polyhedron satisfy F ( F ′. Thendim(F ) < dim(F ′).


Some “typical” arguments Faces Polyhedra,

1 If a face F of P (or P itself) has at the same time the points satisfying itsconstraint aT x ≥ β with = and >, then F ′ := F ∩ {aT x = β} is a face ofP such that F ′ ( F .

2 Consider a face F (F can be P itself) and x any point of F . Take anystraight line passing through x. Suppose there is the last point z of P onthe line. Then there should be a constraint aT x ≥ β which z satisfies withequality and the later points on the line will violate.

If x 6= z, we have aT x > β (sinceif aT x = β, the hyperplane aT x =β includes the whole line.)Therefore, from 1, we get a face F ′

:= F ∩ {aT x = β} ( F .

Therefore, if a face F (or P )contains no face among its propersubsets, it should include every linepassing through two distinct pointof it. Namely, F (or P ) is an affineset.


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