The Weyl-Minkowski Theorems - Eaton.math.rpi.edueaton.math.rpi.edu/faculty/Mitchell/courses/matp6640/... · 2020. 2. 3. · Minkowski for Cones Corollary (Minkowski’s Theorem) Any

The Weyl-Minkowski Theorems

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

February 2018

Mitchell The Weyl-Minkowski Theorems 1 / 21

Introduction

There are two fundamentally different ways to look at polyhedra:As constrained polyhedra: P = {x ∈ IRn : Ax ≥ b}.As finitely generated polyhedra:Q = {x ∈ IRn : x = Dy , with possible additional

linear constraints on y}.The columns of D are the generators of Q.

The Weyl and Minkowski Theorems show that any polyhedron that isrepresented in one form can also be represented in the other form.

The theorems are first stated and proved for the case of convex conesand then the results are extended out to more general polyhedra.


Introduction

There are two fundamentally different ways to look at polyhedra:As constrained polyhedra: P = {x ∈ IRn : Ax ≥ b}.As finitely generated polyhedra:Q = {x ∈ IRn : x = Dy , with possible additional

linear constraints on y}.The columns of D are the generators of Q.

The Weyl and Minkowski Theorems show that any polyhedron that isrepresented in one form can also be represented in the other form.

The theorems are first stated and proved for the case of convex conesand then the results are extended out to more general polyhedra.


Example

{x ∈ IR2 : x =

[4 2 5 3 11 2 1 2 8

]y , y ∈ IR5, y ≥ 0, y3 + y4 + y5 = 1

}

=

{x ∈ IR2 : x1 + 2x2 ≥ 7, 3x1 + x2 ≥ 11,

−x1 + 4x2 ≥ −1, 2x1 − 2x2 ≥ −14

}


Cones

Constrained and finitely generated convex cones have the forms:Constrained: K = {x ∈ IRn : Ax ≥ 0} for some matrix A.Finitely generated: K = {x ∈ IRn : x = Dy for some y ≥ 0}.


Cones

Constrained and finitely generated convex cones have the forms:Constrained: K = {x ∈ IRn : Ax ≥ 0} for some matrix A.Finitely generated: K = {x ∈ IRn : x = Dy for some y ≥ 0}.


Duals of sets

Recall that the dual of a set M ⊆ IRn is

M+ := {y ∈ IRn : yT x ≥ 0 ∀ x ∈ M}.

The next proposition is left as an exercise:

Proposition

Let S ⊆ IRn. We have S = S++ if and only if S is a constrained cone.


Examples of dual cones

Find dual cones to:IRn.IRn+.

Cone of symmetric positive semidefinite matrices.










Weyl Thm for Cones

Theorem (Weyl)Any finitely generated convex cone is finitely constrained.

Proof.Let K be a finitely generated cone. Then for some matrix D we have

K = {x ∈ IRn : x = Dy for some y ≥ 0}= {x ∈ IRn : the system x − Dy = 0, y ≥ 0 is consistent in x , y}= {x ∈ IRn : Bx ≥ 0} for some matrix B,

by using Fourier-Motzkin elimination to eliminate y .


Weyl Thm for Cones

Theorem (Weyl)Any finitely generated convex cone is finitely constrained.

Proof.Let K be a finitely generated cone. Then for some matrix D we have

K = {x ∈ IRn : x = Dy for some y ≥ 0}= {x ∈ IRn : the system x − Dy = 0, y ≥ 0 is consistent in x , y}= {x ∈ IRn : Bx ≥ 0} for some matrix B,

by using Fourier-Motzkin elimination to eliminate y .


Duals of standard cones

Corollary

Given A ∈ IRm×n, let K := {Ax : x ≥ 0} ⊆ IRm andL := {y ∈ IRm : AT y ≥ 0}. Then K+ = L and L+ = K .

Proof.We first show K+ = L. We have

K+ = {z ∈ IRm : zT w ≥ 0 ∀w ∈ K}= {z ∈ IRm : zT Ax ≥ 0∀x ≥ 0}= {z ∈ IRm : (AT z)T x ≥ 0 ∀x ≥ 0}

which holds if and only if AT z ≥ 0, so K+ = L.


Duals of standard cones

Corollary


Proof.We first show K+ = L. We have

K+ = {z ∈ IRm : zT w ≥ 0 ∀w ∈ K}= {z ∈ IRm : zT Ax ≥ 0∀x ≥ 0}= {z ∈ IRm : (AT z)T x ≥ 0 ∀x ≥ 0}

which holds if and only if AT z ≥ 0, so K+ = L.


Proof of second half

Corollary


Proof.Now we show L+ = K . By Weyl’s Theorem, the cone K can also beexpressed as a constrained cone, so there is a matrix C withK = {x ∈ IRn : Cx ≥ 0}. From Proposition 1, we haveK = K++ = (K+)+ = L+, as required.


Proof of second half

Corollary


Proof.Now we show L+ = K . By Weyl’s Theorem, the cone K can also beexpressed as a constrained cone, so there is a matrix C withK = {x ∈ IRn : Cx ≥ 0}. From Proposition 1, we haveK = K++ = (K+)+ = L+, as required.


FarkasCorollary (Farkas Lemma)Let A ∈ IRm×n and b ∈ IRm. Exactly one of the following systems has asolution:

(I) Ax = b, x ≥ 0.(II) AT y ≤ 0, bT y > 0.

Proof.Define K and L as in Corollary 2.First, assume (I) fails. Then b 6∈ K . By Weyl’s Theorem andProposition 1, we have b 6∈ K++ = (K+)+. So there exists y ∈ K+ withbT y < 0. From Corollary 2, we have K+ = {y ∈ IRm : AT y ≥ 0}, so (II)is consistent.Now, assume (II) holds, so ∃y ∈ L with bT y < 0. So b 6∈ L+ = K byCorollary 2, so (I) fails.



(I) Ax = b, x ≥ 0.(II) AT y ≤ 0, bT y > 0.




(I) Ax = b, x ≥ 0.(II) AT y ≤ 0, bT y > 0.




(I) Ax = b, x ≥ 0.(II) AT y ≤ 0, bT y > 0.



Minkowski for Cones

Corollary (Minkowski’s Theorem)Any finitely constrained cone is nonempty and finitely generated.

Proof.Define K and L as in Corollary 2. We need to show that the genericfinitely constrained cone L is finitely constrained and nonempty.Note first that the origin is a point in L, so L is nonempty.Now we need to show that L is finitely generated. We know fromCorollary 2 that L+ = K , a finitely generated cone. By Weyl’s Theorem,K is also finitely constrained, so K = {x ∈ IRn : Bx ≥ 0} for somematrix B. Again from Corollary 2, we have L = K+ = {BT z : z ≥ 0}, soL is indeed finitely generated.


Minkowski for Cones

Corollary (Minkowski’s Theorem)Any finitely constrained cone is nonempty and finitely generated.

Proof.Define K and L as in Corollary 2. We need to show that the genericfinitely constrained cone L is finitely constrained and nonempty.Note first that the origin is a point in L, so L is nonempty.Now we need to show that L is finitely generated. We know fromCorollary 2 that L+ = K , a finitely generated cone. By Weyl’s Theorem,K is also finitely constrained, so K = {x ∈ IRn : Bx ≥ 0} for somematrix B. Again from Corollary 2, we have L = K+ = {BT z : z ≥ 0}, soL is indeed finitely generated.


Example of homogenizationThe proof of the affine Weyl Theorem uses a technique calledhomogenization, where a polyhedron is embedded in a higherdimensional cone.

ExampleThe polyhedron{x1 ∈ IR1 : x1 ≥ 3} = {x1 ∈ IR1 : x1 = 5y + 3z, y ≥ 0, z ≥ 0, z = 1} canbe regarded as the slice of the cone

{x ∈ IR2 : x1 = 5y + 3z, x2 = z, y , z ≥ 0}= {x ∈ IR2 : x1 − 3x2 ≥ 0, x2 ≥ 0} ⊆ IR2

with x2 = 1.

Note that the homogenized version is the closure of the conic hull ofthe slice corresponding to the polyhedron.


Affine Weyl

Theorem (Affine Weyl Theorem)Finitely generated polyhedra are finitely constrained:Let

P := {x ∈ IRn : x = By + Cz, y ≥ 0, z ≥ 0q∑

i=1

zi = 1}

with B ∈ IRn×p and C ∈ IRn×q. Then there exists a matrix A ∈ IRm×n anda vector b ∈ IRm such that

P = {x ∈ IRn : Ax ≥ b}.


Proof of Affine Weyl

If P = ∅, we can take m = 1, A to be composed of zeroes, and b = 1.Otherwise, if P 6= ∅:Define

P ′ :={[

xxn+1

]:

[xxn+1

]=

[B C0 eT

] [yz

], y ≥ 0, z ≥ 0

}where e ∈ IRq is a vector of ones. Note that P corresponds to points inP ′ with xn+1 = 1. The set P ′ is a finitely generated cone, so by Weyl’sTheorem it is also finitely constrained:

P ′ ={[

xxn+1

]: [A

... − b][

xxn+1

]≥ 0

}


Proof of Affine Weyl 2

Then

P =

{x :

[x1

]∈ P ′

}=

{x : [A

... − b][

x1

]≥ 0

}= {x : Ax ≥ b}

as required. QED


Affine Minkowski

Theorem (Affine Minkowski Theorem)Finitely constrained polyhedra are finitely generated:Suppose P = {x ∈ IRn : Ax ≥ b} where A ∈ IRm×n and b ∈ IRm. Thenthere exist matrices B ∈ IRn×p and C ∈ IRn×q such that

P = {x ∈ IRn : x = By + Cz, y ≥ 0, z ≥ 0q∑

i=1

zi = 1}.


Proof of Affine Minkowski

This is also proved using homogenization.First take care of the trivial case: If P = ∅, can take p = q = 0, so Band C are vacuous.Now the general case, with P 6= ∅. Let

P ′ :={[

xxn+1

]:

[A −b0 1

] [xxn+1

]≥ 0

}

so again x ∈ P ⇔[

x1

]∈ P ′. By Minkowski’s Theorem, we have

P ′ :={[

xxn+1

]:

[xxn+1

]= Dw , w ≥ 0

}for some D ∈ IR(n+1)×m.


Proof of Affine Minkowski 2

We can rearrange the columns of D if necessary so that

P ′ :={[

xxn+1

]:

[xxn+1

]=

[B C0 6= 0 . . . 6= 0

] [yz

], y , z ≥ 0

}.

In order to get the desired form, we need to pin down the nonzeroes inthe last row of D. Note that if any of the nonzeroes are negative thenwe could choose the corresponding zj > 0 and y and all othercomponents of z equal to zero. This would give a point with xn+1 < 0,which is not in P ′. So all the nonzeroes in the last row must be positive.


Proof of Affine Minkowski 3

By scaling the last q columns of D, we can rewrite as

P ′ :={[

xxn+1

]:

[xxn+1

]=

[B C0 eT

] [yz

], y ≥ 0, z ≥ 0

}and

P =

{x :

[x1

]∈ P ′

}= {x ∈ IRn : x = By + Cz, y ≥ 0, z ≥ 0

q∑i=1

zi = 1}

as desired. QED


Goldman Resolution Theorem

Together, the affine Weyl and the affine Minkowski Theorems areknown as the Double Decomposition Theorem. They can be extendedto the following theorem:

Theorem (Goldman Resolution Theorem)Suppose P = {x ∈ IRn : Ax ≥ b} 6= ∅ where A ∈ IRm×n and b ∈ IRm.Then P = S + K + Q, where

S = {x ∈ IRn : Ax = 0}S + K = {x ∈ IRn : Ax ≥ 0}K is a pointed coneK + Q is a pointed polyhedronQ is a polytope given by the convex hull

of extreme points of K + Q


ExampleAn example where S 6= {0} and K and Q are not unique:

P = {x ∈ IR2 : x1 + x2 ≥ 3}

=

{x ∈ IR2 : x =

[12

]+ s

[1−1

]+ t[

10

], t ≥ 0

}=

{x ∈ IR2 : x =

[5−2

]+ s

[1−1

]+ t[

14

], t ≥ 0

}For example,

x =

[49

]=

[12

]+ (−7)

[1−1

]+ 10

[10

]

=

[5−2

]+ (−3)

[1−1

]+ 2

[14

].


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The Weyl-Minkowski Theorems - Eaton.math.rpi.edueaton.math.rpi.edu/faculty/Mitchell/courses/matp6640/... · 2020. 2. 3. · Minkowski for Cones Corollary (Minkowski’s Theorem) Any