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v2009.01.01 - Convex Optimization

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2.13. DUAL CONE & GENERALIZED INEQUALITY 171<br />

hence N ≤ n (X skinny 2.64 -or-square full-rank). In other words, fat full-rank<br />

matrix X is prohibited by uniqueness because of the existence of an infinity<br />

of right-inverses;<br />

polyhedral cones whose extreme directions number in excess of the<br />

ambient space dimension are precluded in biorthogonal expansion.<br />

2.13.8.1 x ∈ K<br />

Suppose x belongs to K ⊆ R n . Then x =Xa for some a≽0. Vector a is<br />

unique only when {Γ i } is a linearly independent set. 2.65 Vector a∈ R N can<br />

take the form a =Bx if R(B)= R N . Then we require Xa =XBx = x and<br />

Bx=BXa = a . The pseudoinverse B =X † ∈ R N×n (E) is suitable when X<br />

is skinny-or-square and full-rank. In that case rankX =N , and for all c ≽ 0<br />

and i=1... N<br />

a ≽ 0 ⇔ X † Xa ≽ 0 ⇔ a T X T X †T c ≥ 0 ⇔ Γ T i X †T c ≥ 0 (359)<br />

The penultimate inequality follows from the generalized inequality and<br />

membership corollary, while the last inequality is a consequence of that<br />

corollary’s discretization (2.13.4.2.1). 2.66 From (359) and (347) we deduce<br />

K ∗ ∩ aff K = cone(X †T ) = {X †T c | c ≽ 0} ⊆ R n (360)<br />

is the vertex-description for that section of K ∗ in the affine hull of K because<br />

R(X †T ) = R(X) by definition of the pseudoinverse. From (280), we know<br />

2.64 “Skinny” meaning thin; more rows than columns.<br />

2.65 Conic independence alone (2.10) is insufficient to guarantee uniqueness.<br />

2.66<br />

a ≽ 0 ⇔ a T X T X †T c ≥ 0 ∀(c ≽ 0 ⇔ a T X T X †T c ≥ 0 ∀a ≽ 0)<br />

∀(c ≽ 0 ⇔ Γ T i X†T c ≥ 0 ∀i) <br />

Intuitively, any nonnegative vector a is a conic combination of the standard basis<br />

{e i ∈ R N }; a≽0 ⇔ a i e i ≽0 for all i. The last inequality in (359) is a consequence of the<br />

fact that x=Xa may be any extreme direction of K , in which case a is a standard basis<br />

vector; a = e i ≽0. Theoretically, because c ≽ 0 defines a pointed polyhedral cone (in fact,<br />

the nonnegative orthant in R N ), we can take (359) one step further by discretizing c :<br />

a ≽ 0 ⇔ Γ T i Γ ∗ j ≥ 0 for i,j =1... N ⇔ X † X ≥ 0<br />

In words, X † X must be a matrix whose entries are each nonnegative.

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