v2010.10.26 - Convex Optimization

456 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXλ 2 ≤ d 12 ≤ λ 1 (1067)where d 12 is the eigenvalue of submatrix (1065a) and λ 1 , λ 2 are theeigenvalues of T (1065b) (1058). Intertwining in (1067) predicts that shouldd 12 become 0, then λ 2 must go to 0 . 5.42 The eigenvalues are similarlyintertwined for submatrices (1065b) and (1065c);γ 3 ≤ λ 2 ≤ γ 2 ≤ λ 1 ≤ γ 1 (1068)where γ 1 , γ 2 ,γ 3 are the eigenvalues of submatrix (1065c). Intertwininglikewise predicts that should λ 2 become 0 (a possibility revealed in5.8.3.1),then γ 3 must go to 0. Combining results so far for N = 2, 3, 4: (1067) (1068)γ 3 ≤ λ 2 ≤ d 12 ≤ λ 1 ≤ γ 1 (1069)The preceding logic extends by induction through the remaining membersof the sequence (1065).5.8.3.1 Tightening the triangle inequalityNow we apply Schur complement fromA.4 to tighten the triangle inequalityfrom (1057) in case: cardinality N = 4. We find that the gains by doing soare modest. From (1065) we identify:[ ] A BB T −V TCNDV N | N←4 (1070)A T = −V T NDV N | N←3 (1071)both positive semidefinite by assumption, where B=ν(4) (1066), andC = d 14 . Using nonstrict CC † -form (1511), C ≽0 by assumption (5.8.1)and CC † =I . So by the positive semidefinite ordering of eigenvalues theorem(A.3.1.0.1),−V T NDV N | N←4 ≽ 0 ⇔ T ≽ d −114 ν(4)ν(4) T ⇒{λ1 ≥ d −114 ‖ν(4)‖ 2λ 2 ≥ 0(1072)where {d −114 ‖ν(4)‖ 2 , 0} are the eigenvalues of d −114 ν(4)ν(4) T while λ 1 , λ 2 arethe eigenvalues of T .5.42 If d 12 were 0, eigenvalue λ 2 becomes 0 (1060) because d 13 must then be equal tod 23 ; id est, d 12 = 0 ⇔ x 1 = x 2 . (5.4)