v2010.10.26 - Convex Optimization

506 CHAPTER 6. CONE OF DISTANCE MATRICES6.4.1 Range of EDM DFromB.1.1 pertaining to linear independence of dyad sums: If the transposehalves of all the dyads in the sum (1185) 6.5 make a linearly independent set,then the nontranspose halves constitute a basis for the range of EDM D .Saying this mathematically: For D ∈ EDM NR(D)= R([δ(V X V T X ) 1 V X ]) ⇐ rank([δ(V X V T X ) 1 V X ])= 2 + rR(D)= R([1 V X ]) ⇐ otherwise (1198)6.4.1.0.1 Proof. We need that condition under which the rankequality is satisfied: We know R(V X )⊥1, but what is the relativegeometric orientation of δ(V X VX T) ? δ(V X V X T)≽0 because V X V X T ≽0,and δ(V X VX T)∝1 remains possible (1195); this means δ(V X V X T) /∈ N(1T )simply because it has no negative entries. (Figure 139) If the projection ofδ(V X VX T) on N(1T ) does not belong to R(V X ) , then that is a necessary andsufficient condition for linear independence (l.i.) of δ(V X VX T ) with respect toR([1 V X ]) ; id est,V δ(V X V T X ) ≠ V X a for any a∈ R r(I − 1 N 11T )δ(V X V T X ) ≠ V X aδ(V X V T X ) − 1 N ‖V X ‖ 2 F 1 ≠ V X aδ(V X V T X ) − λ2N 1 = y ≠ V X a ⇔ {1, δ(V X V T X ), V X } is l.i.(1199)When this condition is violated (when (1192) y=V X a p for some particulara∈ R r ), on the other hand, then from (1191) we haveR ( D = y1 T + 1y T + λ N 11T − 2V X V T X)= R((VX a p + λ N 1)1T + (1a T p − 2V X )V T X= R([V X a p + λ N 1 1aT p − 2V X ])= R([1 V X ]) (1200)An example of such a violation is (1197) where, in particular, a p = 0. )6.5 Identifying columns V X [v 1 · · · v r ] , then V X V T X = ∑ iv i v T iis also a sum of dyads.