v2009.01.01 - Convex Optimization

v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization

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174 CHAPTER 2. CONVEX GEOMETRY These results are summarized for a pointed polyhedral cone, having linearly independent generators, and its ordinary dual: 2.13.9.2 Simplicial case Cone Table 1 K K ∗ vertex-description X X †T , ±X ⊥ halfspace-description X † , X ⊥T X T When a convex cone is simplicial (2.12.3), Cone Table 1 simplifies because then aff coneX = R n : For square X and assuming simplicial K such that rank(X ∈ R n×N ) = N ∆ = dim aff K = n (373) we have Cone Table S K K ∗ vertex-description X X †T halfspace-description X † X T For example, vertex-description (372) simplifies to K ∗ = {X †T b | b ≽ 0} ⊂ R n (374) Now, because dim R(X)= dim R(X †T ) , (E) the dual cone K ∗ is simplicial whenever K is. 2.13.9.3 Cone membership relations in a subspace It is obvious by definition (270) of the ordinary dual cone K ∗ in ambient vector space R that its determination instead in subspace M⊆ R is identical to its intersection with M ; id est, assuming closed convex cone K ⊆ M and K ∗ ⊆ R (K ∗ were ambient M) ≡ (K ∗ in ambient R) ∩ M (375) because { y ∈ M | 〈y , x〉 ≥ 0 for all x ∈ K } = { y ∈ R | 〈y , x〉 ≥ 0 for all x ∈ K } ∩ M (376)

2.13. DUAL CONE & GENERALIZED INEQUALITY 175 From this, a constrained membership relation for the ordinary dual cone K ∗ ⊆ R , assuming x,y ∈ M and closed convex cone K ⊆ M y ∈ K ∗ ∩ M ⇔ 〈y , x〉 ≥ 0 for all x ∈ K (377) By closure in subspace M we have conjugation (2.13.1.1): x ∈ K ⇔ 〈y , x〉 ≥ 0 for all y ∈ K ∗ ∩ M (378) This means membership determination in subspace M requires knowledge of the dual cone only in M . For sake of completeness, for proper cone K with respect to subspace M (confer (294)) x ∈ int K ⇔ 〈y , x〉 > 0 for all y ∈ K ∗ ∩ M, y ≠ 0 (379) x ∈ K , x ≠ 0 ⇔ 〈y , x〉 > 0 for all y ∈ int K ∗ ∩ M (380) (By conjugation, we also have the dual relations.) Yet when M equals aff K for K a closed convex cone x ∈ rel int K ⇔ 〈y , x〉 > 0 for all y ∈ K ∗ ∩ aff K , y ≠ 0 (381) x ∈ K , x ≠ 0 ⇔ 〈y , x〉 > 0 for all y ∈ rel int(K ∗ ∩ aff K) (382) 2.13.9.4 Subspace M = aff K Assume now a subspace M that is the affine hull of cone K : Consider again a pointed polyhedral cone K denoted by its extreme directions arranged columnar in matrix X such that rank(X ∈ R n×N ) = N ∆ = dim aff K ≤ n (366) We want expressions for the convex cone and its dual in subspace M=aff K : Cone Table A K K ∗ ∩ aff K vertex-description X X †T halfspace-description X † , X ⊥T X T , X ⊥T When dim aff K = n , this table reduces to Cone Table S. These descriptions facilitate work in a proper subspace. The subspace of symmetric matrices S N , for example, often serves as ambient space. 2.69 2.69 The dual cone of positive semidefinite matrices S N∗ + = S N + remains in S N by convention, whereas the ordinary dual cone would venture into R N×N .

174 CHAPTER 2. CONVEX GEOMETRY<br />

These results are summarized for a pointed polyhedral cone, having<br />

linearly independent generators, and its ordinary dual:<br />

2.13.9.2 Simplicial case<br />

Cone Table 1 K K ∗<br />

vertex-description X X †T , ±X ⊥<br />

halfspace-description X † , X ⊥T X T<br />

When a convex cone is simplicial (2.12.3), Cone Table 1 simplifies because<br />

then aff coneX = R n : For square X and assuming simplicial K such that<br />

rank(X ∈ R n×N ) = N ∆ = dim aff K = n (373)<br />

we have<br />

Cone Table S K K ∗<br />

vertex-description X X †T<br />

halfspace-description X † X T<br />

For example, vertex-description (372) simplifies to<br />

K ∗ = {X †T b | b ≽ 0} ⊂ R n (374)<br />

Now, because dim R(X)= dim R(X †T ) , (E) the dual cone K ∗ is simplicial<br />

whenever K is.<br />

2.13.9.3 Cone membership relations in a subspace<br />

It is obvious by definition (270) of the ordinary dual cone K ∗ in ambient vector<br />

space R that its determination instead in subspace M⊆ R is identical to<br />

its intersection with M ; id est, assuming closed convex cone K ⊆ M and<br />

K ∗ ⊆ R<br />

(K ∗ were ambient M) ≡ (K ∗ in ambient R) ∩ M (375)<br />

because<br />

{<br />

y ∈ M | 〈y , x〉 ≥ 0 for all x ∈ K<br />

}<br />

=<br />

{<br />

y ∈ R | 〈y , x〉 ≥ 0 for all x ∈ K<br />

}<br />

∩ M<br />

(376)

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