v2010.10.26 - Convex Optimization

v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization

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178 CHAPTER 2. CONVEX GEOMETRY2.13.4.2.4 Example. Boundary membership to proper polyhedral cone.For a polyhedral cone, test (330) of boundary membership can be formulatedas a linear program. Say proper polyhedral cone K is specified completelyby generators that are arranged columnar inX = [ Γ 1 · · · Γ N ] ∈ R n×N (280)id est, K = {Xa | a ≽ 0}. Then membership relationc ∈ ∂K ⇔ ∃ y ≠ 0 〈y , c〉 = 0, y ∈ K ∗ , c ∈ K (330)may be expressedfinda , yy ≠ 0subject to c T y = 0X T y ≽ 0Xa = ca ≽ 0This linear feasibility problem has a solution iff c∈∂K .(371)2.13.4.3 smallest face of pointed closed convex coneGiven nonempty convex subset C of a convex set K , the smallest face of Kcontaining C is equivalent to intersection of all faces of K that contain C .[307, p.164] By (309), membership relation (330) means that each andevery point on boundary ∂K of proper cone K belongs to a hyperplanesupporting K whose normal y belongs to dual cone K ∗ . It follows that thesmallest face F , containing C ⊂ ∂K ⊂ R n on boundary of proper cone K ,is the intersection of all hyperplanes containing C whose normals are in K ∗ ;whereF(K ⊃ C) = {x ∈ K | x ⊥ K ∗ ∩ C ⊥ } (372)When C ∩ int K ≠ ∅ then F(K ⊃ C)= K .C ⊥ {y ∈ R n | 〈z, y〉=0 ∀z∈ C} (373)2.13.4.3.1 Example. Finding smallest face of cone.Suppose polyhedral cone K is completely specified by generators arrangedcolumnar inX = [ Γ 1 · · · Γ N ] ∈ R n×N (280)

2.13. DUAL CONE & GENERALIZED INEQUALITY 179To find its smallest face F(K ∋c) containing a given point c∈K , by thediscretized membership theorem 2.13.4.2.1, it is necessary and sufficient tofind generators for the smallest face. We may do so one generator at atime: 2.67 Consider generator Γ i . If there exists a vector z ∈ K ∗ orthogonal toc but not to Γ i , then Γ i cannot belong to the smallest face of K containingc . Such a vector z can be realized by a linear feasibility problem:find z ∈ R nsubject to c T z = 0X T z ≽ 0Γ T i z = 1(374)If there exists a solution z for which Γ T i z=1, thenΓ i ̸⊥ K ∗ ∩ c ⊥ = {z ∈ R n | X T z ≽0, c T z=0} (375)so Γ i /∈ F(K ∋c) ; solution z is a certificate of null membership. If thisproblem is infeasible for generator Γ i ∈ K , conversely, then Γ i ∈ F(K ∋c)by (372) and (363) because Γ i ⊥ K ∗ ∩ c ⊥ ; in that case, Γ i is a generator ofF(K ∋c).Since the constant in constraint Γ T i z =1 is arbitrary positively, then bytheorem of the alternative there is correspondence between (374) and (348)admitting the alternative linear problem: for a given point c∈Kfinda∈R N , µ∈Ra , µsubject to µc − Γ i = Xaa ≽ 0(376)Now if this problem is feasible (bounded) for generator Γ i ∈ K , then (374)is infeasible and Γ i ∈ F(K ∋c) is a generator of the smallest face thatcontains c .2.13.4.3.2 Exercise. Finding smallest face of pointed closed convex cone.Show that formula (372) and algorithms (374) and (376) apply more broadly;id est, a full-dimensional cone K is an unnecessary condition. 2.68 2.67 When finding a smallest face, generators of K in matrix X may not be diminished innumber (by discarding columns) until all generators of the smallest face have been found.2.68 Hint: A hyperplane, with normal in K ∗ , containing cone K is admissible.

2.13. DUAL CONE & GENERALIZED INEQUALITY 179To find its smallest face F(K ∋c) containing a given point c∈K , by thediscretized membership theorem 2.13.4.2.1, it is necessary and sufficient tofind generators for the smallest face. We may do so one generator at atime: 2.67 Consider generator Γ i . If there exists a vector z ∈ K ∗ orthogonal toc but not to Γ i , then Γ i cannot belong to the smallest face of K containingc . Such a vector z can be realized by a linear feasibility problem:find z ∈ R nsubject to c T z = 0X T z ≽ 0Γ T i z = 1(374)If there exists a solution z for which Γ T i z=1, thenΓ i ̸⊥ K ∗ ∩ c ⊥ = {z ∈ R n | X T z ≽0, c T z=0} (375)so Γ i /∈ F(K ∋c) ; solution z is a certificate of null membership. If thisproblem is infeasible for generator Γ i ∈ K , conversely, then Γ i ∈ F(K ∋c)by (372) and (363) because Γ i ⊥ K ∗ ∩ c ⊥ ; in that case, Γ i is a generator ofF(K ∋c).Since the constant in constraint Γ T i z =1 is arbitrary positively, then bytheorem of the alternative there is correspondence between (374) and (348)admitting the alternative linear problem: for a given point c∈Kfinda∈R N , µ∈Ra , µsubject to µc − Γ i = Xaa ≽ 0(376)Now if this problem is feasible (bounded) for generator Γ i ∈ K , then (374)is infeasible and Γ i ∈ F(K ∋c) is a generator of the smallest face thatcontains c .2.13.4.3.2 Exercise. Finding smallest face of pointed closed convex cone.Show that formula (372) and algorithms (374) and (376) apply more broadly;id est, a full-dimensional cone K is an unnecessary condition. 2.68 2.67 When finding a smallest face, generators of K in matrix X may not be diminished innumber (by discarding columns) until all generators of the smallest face have been found.2.68 Hint: A hyperplane, with normal in K ∗ , containing cone K is admissible.

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