v2009.01.01 - Convex Optimization

6.5. EDM DEFINITION IN 11 T 461
6.5.1 Range of EDM D
FromB.1.1 pertaining to linear independence of dyad sums: If the transpose
halves of all the dyads in the sum (1093) 6.6 make a linearly independent set,
then the nontranspose halves constitute a basis for the range of EDM D .
Saying this mathematically: For D ∈ EDM N
R(D)= R([δ(V X V T X ) 1 V X ]) ⇐ rank([δ(V X V T X ) 1 V X ])= 2 + r
R(D)= R([1 V X ]) ⇐ otherwise (1106)
To prove this, we need that condition under which the rank equality is
satisfied: We know R(V X )⊥1, but what is the relative geometric 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 V X T )∝1 remains
possible (1103); this means δ(V X VX T) /∈ N(1T ) simply because it has no
negative entries. (Figure 119) If the projection of δ(V X VX T) on N(1T ) does
not belong to R(V X ), then that is a necessary and sufficient condition for
linear 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 a
δ(V X V T X ) − 1 N ‖V X ‖ 2 F 1 ≠ V X a
δ(V X V T X ) − λ
2N 1 = y ≠ V X a ⇔ {1, δ(V X V T X ), V X } is l.i.
(1107)
On the other hand when this condition is violated (when (1100) y=V X a p
for some particular a∈ R r ), then from (1099) we have
R ( ) ( )
D = y1 T + 1y T + λ N 11T − 2V X VX
T = R (VX a p + λ N 1)1T + (1a T p − 2V X )VX
T
= R([V X a p + λ N 1 1aT p − 2V X ])
= R([1 V X ]) (1108)
An example of such a violation is (1105) where, in particular, a p = 0.
Then a statement parallel to (1106) 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 ) (1109)
6.6 Identifying columns V X
∆
= [v1 · · · v r ] , then V X V T X = ∑ i
v i v T i
is also a sum of dyads.