v2010.10.26 - Convex Optimization

638 APPENDIX B. SIMPLE MATRICESProof. Linear independence of k dyads is identical to definition (1624).(⇒) Suppose {s i } and {w i } are each linearly independent sets. InvokingSylvester’s rank inequality, [202,0.4] [393,2.4]rankS+rankW − k ≤ rank(SW T ) ≤ min{rankS , rankW } (≤ k) (1625)Then k ≤rank(SW T )≤k which implies the dyads are independent.(⇐) Conversely, suppose rank(SW T )=k . Thenk ≤ min{rankS , rankW } ≤ k (1626)implying the vector sets are each independent.B.1.1.1Biorthogonality condition, Range and Nullspace of SumDyads characterized by biorthogonality condition W T S =I are independent;id est, for S ∈ C M×k and W ∈ C N×k , if W T S =I then rank(SW T )=k bythe linearly independent dyads theorem because (conferE.1.1)W T S =I ⇒ rankS=rankW =k≤M =N (1627)To see that, we need only show: N(S)=0 ⇔ ∃ B BS =I . B.4(⇐) Assume BS =I . Then N(BS)=0={x |BSx = 0} ⊇ N(S). (1607)(⇒) If N(S)=0[ then]S must be full-rank skinny-or-square.B∴ ∃ A,B,C [S A] = I (id est, [ S A] is invertible) ⇒ BS =I .CLeft inverse B is given as W T here. Because of reciprocity with S , itimmediately follows: N(W)=0 ⇔ ∃ S S T W =I . Dyads produced by diagonalization, for example, are independent becauseof their inherent biorthogonality. (A.5.0.3) The converse is generally false;id est, linearly independent dyads are not necessarily biorthogonal.B.1.1.1.1 Theorem. Nullspace and range of dyad sum.Given a sum of dyads represented by SW T where S ∈C M×k and W ∈ C N×kN(SW T ) = N(W T ) ⇐R(SW T ) = R(S) ⇐B.4 Left inverse is not unique, in general.∃ B BS = I∃ Z W T Z = I(1628)⋄