v2009.01.01 - Convex Optimization

B.1. RANK-ONE MATRIX (DYAD) 581
range of summation is the vector sum of ranges. B.3 (Theorem B.1.1.1.1)
Under the assumption the dyads are linearly independent (l.i.), then the
vector sums are unique (p.708): for {w i } l.i. and {s i } l.i.
( k∑
)
R s i wi
T = R ( ) ( )
s 1 w1 T ⊕ ... ⊕ R sk wk T = R(s1 ) ⊕ ... ⊕ R(s k ) (1507)
i=1
B.1.1.0.1 Definition. Linearly independent dyads. [182, p.29, thm.11]
[294, p.2] The set of k dyads
{si
wi T | i=1... k } (1508)
where s i ∈ C M and w i ∈ C N , is said to be linearly independent iff
(
)
k∑
rank SW T =
∆ s i wi
T = k (1509)
i=1
where S ∆ = [s 1 · · · s k ] ∈ C M×k and W ∆ = [w 1 · · · w k ] ∈ C N×k .
△
As defined, dyad independence does not preclude existence of a nullspace
N(SW T ) , nor does it imply SW T is full-rank. In absence of an assumption
of independence, generally, rankSW T ≤ k . Conversely, any rank-k matrix
can be written in the form SW T by singular value decomposition. (A.6)
B.1.1.0.2 Theorem. Linearly independent (l.i.) dyads.
Vectors {s i ∈ C M , i=1... k} are l.i. and vectors {w i ∈ C N , i=1... k} are
l.i. if and only if dyads {s i wi T ∈ C M×N , i=1... k} are l.i.
⋄
Proof. Linear independence of k dyads is identical to definition (1509).
(⇒) Suppose {s i } and {w i } are each linearly independent sets. Invoking
Sylvester’s rank inequality, [176,0.4] [344,2.4]
rankS+rankW − k ≤ rank(SW T ) ≤ min{rankS , rankW } (≤ k) (1510)
Then k ≤rank(SW T )≤k which implies the dyads are independent.
(⇐) Conversely, suppose rank(SW T )=k . Then
k ≤ min{rankS , rankW } ≤ k (1511)
implying the vector sets are each independent.
B.3 Move of range R to inside the summation depends on linear independence of {w i }.