v2007.09.13 - Convex Optimization

522 APPENDIX B. SIMPLE MATRICESB.1.1.1 Biorthogonality condition, Range and Nullspace of SumDyads characterized by a biorthogonality condition W T S =I areindependent; id est, for S ∈ C M×k and W ∈ C N×k , if W T S =I thenrank(SW T )=k by the linearly independent dyads theorem because(conferE.1.1)W T S =I ⇔ rankS=rankW =k≤M =N (1406)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). (1387)(⇒) 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.5.1) 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 ) ⇐ ∃ B BS = IR(SW T ) = R(S) ⇐ ∃ Z W T Z = I(1407)⋄Proof. (⇒) N(SW T )⊇ N(W T ) and R(SW T )⊆ R(S) are obvious.(⇐) Assume the existence of a left inverse B ∈ R k×N and a right inverseZ ∈ R N×k . B.5N(SW T ) = {x | SW T x = 0} ⊆ {x | BSW T x = 0} = N(W T ) (1408)R(SW T ) = {SW T x | x∈ R N } ⊇ {SW T Zy | Zy ∈ R N } = R(S) (1409)B.4 Left inverse is not unique, in general.B.5 By counter-example, the theorem’s converse cannot be true; e.g., S = W = [1 0].