v2007.11.26 - Convex Optimization

6.5. EDM DEFINITION IN 11 T 419then 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 (1033)To prove this, we need that condition under which the rank equality issatisfied: We know R(V X )⊥1, but what is the relative geometric orientationof δ(V X VX T) ? δ(V X V X T)≽0 because V X V X T ≽0, and δ(V X V X T )∝1 remainspossible (1030); this means δ(V X VX T) /∈ N(1T ) simply because it has nonegative entries. (Figure 105) If the projection of δ(V X VX T) on N(1T ) doesnot belong to R(V X ), then that is a necessary and sufficient condition forlinear independence (l.i.) of δ(V X VX T) with respect to R([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.(1034)On the other hand when this condition is violated (when (1027) y=V X a pfor some particular a∈ R r ), then from (1026) 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 ]) (1035)An example of such a violation is (1032) where, in particular, a p = 0. Then a statement parallel to (1033) is, for D ∈ EDM N (Theorem 5.7.3.0.1)rank(D) = r + 2 ⇔ y /∈ R(V X ) ( ⇔ 1 T D † 1 = 0 )rank(D) = r + 1 ⇔ y ∈ R(V X ) ( ⇔ 1 T D † 1 ≠ 0 ) (1036))