v2007.09.13 - Convex Optimization

v2007.09.13 - Convex Optimization v2007.09.13 - Convex Optimization

convexoptimization.com
from convexoptimization.com More from this publisher
10.07.2015 Views

156 CHAPTER 2. CONVEX GEOMETRY2.13.5.1.1 Example. Linear matrix inequality. (confer2.13.2.0.3)Consider a peculiar vertex-description for a closed convex cone defined overthe positive semidefinite cone (instead of the nonnegative orthant as indefinition (83)): for X ∈ S n given A j ∈ S n , j =1... m⎧⎡⎨K = ⎣⎩⎧⎡⎨= ⎣⎩〈A 1 , X〉.〈A m , X〉⎤ ⎫⎬⎦ | X ≽ 0⎭ ⊆ Rm⎤ ⎫svec(A 1 ) T⎬. ⎦svec X | X ≽ 0svec(A m ) T ⎭∆= {A svec X | X ≽ 0}(324)where A∈ R m×n(n+1)/2 , and where symmetric vectorization svec is definedin (47). K is indeed a convex cone because by (144)A svec X p1 , A svec X p2 ∈ K ⇒ A(ζ svec X p1 +ξ svec X p2 ) ∈ K for all ζ,ξ ≥ 0(325)since a nonnegatively weighted sum of positive semidefinite matrices must bepositive semidefinite. (A.3.1.0.2) Although matrix A is finite-dimensional,K is generally not a polyhedral cone (unless m equals 1 or 2) becauseX ∈ S n + . Provided the A j matrices are linearly independent, thenrel int K = int K (326)meaning, the cone interior is nonempty implying the dual cone is pointedby (268).If matrix A has no nullspace, on the other hand, then (by2.10.1.1 andDefinition 2.2.1.0.1) A svec X is an isomorphism in X between the positivesemidefinite cone and R(A). In that case, convex cone K has relativeinteriorrel int K = {A svec X | X ≻ 0} (327)and boundaryrel ∂K = {A svec X | X ≽ 0, X ⊁ 0} (328)

2.13. DUAL CONE & GENERALIZED INEQUALITY 157Now consider the (closed convex) dual cone:K ∗ = {y | 〈A svec X , y〉 ≥ 0 for all X ≽ 0} ⊆ R m= { y | 〈svec X , A T y〉 ≥ 0 for all X ≽ 0 }= { y | svec −1 (A T y) ≽ 0 } (329)that follows from (322) and leads to an equally peculiar halfspace-descriptionK ∗ = {y ∈ R m |m∑y j A j ≽ 0} (330)j=1The summation inequality with respect to the positive semidefinite cone isknown as a linear matrix inequality. [44] [101] [191] [270]When the A j matrices are linearly independent, function g(y) ∆ = ∑ y j A jon R m is a linear bijection. The inverse image of the positive semidefinitecone under g(y) must therefore have dimension m . In that circumstance,the dual cone interior is nonemptyint K ∗ = {y ∈ R m |m∑y j A j ≻ 0} (331)j=1having boundary∂K ∗ = {y ∈ R m |m∑y j A j ≽ 0,j=1m∑y j A j ⊁ 0} (332)j=12.13.6 Dual of pointed polyhedral coneIn a subspace of R n , now we consider a pointed polyhedral cone K given interms of its extreme directions Γ i arranged columnar in X ;X = [ Γ 1 Γ 2 · · · Γ N ] ∈ R n×N (240)The extremes theorem (2.8.1.1.1) provides the vertex-description of apointed polyhedral cone in terms of its finite number of extreme directionsand its lone vertex at the origin:

2.13. DUAL CONE & GENERALIZED INEQUALITY 157Now consider the (closed convex) dual cone:K ∗ = {y | 〈A svec X , y〉 ≥ 0 for all X ≽ 0} ⊆ R m= { y | 〈svec X , A T y〉 ≥ 0 for all X ≽ 0 }= { y | svec −1 (A T y) ≽ 0 } (329)that follows from (322) and leads to an equally peculiar halfspace-descriptionK ∗ = {y ∈ R m |m∑y j A j ≽ 0} (330)j=1The summation inequality with respect to the positive semidefinite cone isknown as a linear matrix inequality. [44] [101] [191] [270]When the A j matrices are linearly independent, function g(y) ∆ = ∑ y j A jon R m is a linear bijection. The inverse image of the positive semidefinitecone under g(y) must therefore have dimension m . In that circumstance,the dual cone interior is nonemptyint K ∗ = {y ∈ R m |m∑y j A j ≻ 0} (331)j=1having boundary∂K ∗ = {y ∈ R m |m∑y j A j ≽ 0,j=1m∑y j A j ⊁ 0} (332)j=12.13.6 Dual of pointed polyhedral coneIn a subspace of R n , now we consider a pointed polyhedral cone K given interms of its extreme directions Γ i arranged columnar in X ;X = [ Γ 1 Γ 2 · · · Γ N ] ∈ R n×N (240)The extremes theorem (2.8.1.1.1) provides the vertex-description of apointed polyhedral cone in terms of its finite number of extreme directionsand its lone vertex at the origin:

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!