v2007.09.17 - Convex Optimization

412 CHAPTER 6. EDM CONE6.6.0.0.1 Proof. Case r = 1 is easily proved: From the nonnegativitydevelopment in5.8.1, extreme direction (1015), and Schoenberg criterion(728), we need show only sufficiency; id est, proverankV T N DV N =1D ∈ S N h ∩ R N×N+}⇒D ∈ EDM ND is an extreme directionAny symmetric matrix D satisfying the rank condition must have the form,for z,q ∈ R N and nonzero z ∈ N(1 T ) ,because (5.6.2.1, conferE.7.2.0.2)D = ±(1q T + q1 T − 2zz T ) (1025)N(V N (D)) = {1q T + q1 T | q ∈ R N } ⊆ S N (1026)Hollowness demands q = δ(zz T ) while nonnegativity demands choice ofpositive sign in (1025). Matrix D thus takes the form of an extremedirection (1015) of the EDM cone.The foregoing proof is not extensible in rank: An EDMwith corresponding affine dimension r has the general form, for{z i ∈ N(1 T ), i=1... r} an independent set,( r)∑ T ( r)∑ ∑D = 1δ z i zi T + δ z i ziT 1 T − 2 r z i zi T ∈ EDM N (1027)i=1i=1i=1The EDM so defined relies principally on the sum ∑ z i ziT having positivesummand coefficients (⇔ −VN TDV N ≽0) 6.8 . Then it is easy to find asum incorporating negative coefficients while meeting rank, nonnegativity,and symmetric hollowness conditions but not positive semidefiniteness onsubspace R(V N ) ; e.g., from page 362,⎡−V ⎣0 1 11 0 51 5 06.8 (⇐) For a i ∈ R N−1 , let z i =V †TN a i .⎤⎦V 1 2 = z 1z T 1 − z 2 z T 2 (1028)