v2010.10.26 - Convex Optimization

454 CHAPTER 5. EUCLIDEAN DISTANCE MATRIX5.8.2.1.1 Shore. The columns of Ξ r V N Ξ c hold a basis for N(1 T )when Ξ r and Ξ c are permutation matrices. In other words, any permutationof the rows or columns of V N leaves its range and nullspace unchanged;id est, R(Ξ r V N Ξ c )= R(V N )= N(1 T ) (898). Hence, two distinct matrixinequalities can be equivalent tests of the positive semidefiniteness of D onR(V N ) ; id est, −VN TDV N ≽ 0 ⇔ −(Ξ r V N Ξ c ) T D(Ξ r V N Ξ c )≽0. By properlychoosing permutation matrices, 5.39 the leading principal submatrix T Ξ ∈ S 2of −(Ξ r V N Ξ c ) T D(Ξ r V N Ξ c ) may be loaded with the entries of D needed totest any particular triangle inequality (similarly to (1053)-(1061)). Becauseall the triangle inequalities can be individually tested using a test equivalentto the lone matrix inequality −VN TDV N ≽0, it logically follows that the lonematrix inequality tests all those triangle inequalities simultaneously. Weconclude that −VN TDV N ≽0 is a sufficient test for the fourth property of theEuclidean metric, triangle inequality.5.8.2.2 Strict triangle inequalityWithout exception, all the inequalities in (1061) and (1062) can be madestrict while their corresponding implications remain true. The thenstrict inequality (1061a) or (1062) may be interpreted as a strict triangleinequality under which collinear arrangement of points is not allowed.[223,24/6, p.322] Hence by similar reasoning, −VN TDV N ≻ 0 is a sufficienttest of all the strict triangle inequalities; id est,δ(D) = 0D T = D−V T N DV N ≻ 0⎫⎬⎭ ⇒ √ d ij < √ d ik + √ d kj , i≠j ≠k (1064)5.8.3 −V T N DV N nestingFrom (1058) observe that T =−VN TDV N | N←3 . In fact, for D ∈ EDM N ,the leading principal submatrices of −VN TDV N form a nested sequence (byinclusion) whose members are individually positive semidefinite [159] [202][331] and have the same form as T ; videlicet, 5.405.39 To individually test triangle inequality | √ d ik − √ d kj | ≤ √ d ij ≤ √ d ik + √ d kj forparticular i,k,j, set Ξ r (i,1)= Ξ r (k,2)= Ξ r (j,3)=1 and Ξ c = I .5.40 −V DV | N←1 = 0 ∈ S 0 + (B.4.1)