v2007.09.13 - Convex Optimization
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v2007.09.13 - Convex Optimization

v2007.09.13 - Convex Optimization v2007.09.13 - Convex Optimization

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342 CHAPTER 5. EUCLIDEAN DISTANCE MATRIX5.8.2 Triangle inequality property 4In light of Kreyszig’s observation [165,1.1, prob.15] that properties 2through 4 of the Euclidean metric (5.2) together imply property 1,the nonnegativity criterion (857) suggests that the matrix inequality−V T N DV N ≽ 0 might somehow take on the role of triangle inequality; id est,δ(D) = 0D T = D−V T N DV N ≽ 0⎫⎬⎭ ⇒ √ d ij ≤ √ d ik + √ d kj , i≠j ≠k (861)We now show that is indeed the case: Let T be the leading principalsubmatrix in S 2 of −VN TDV N (upper left 2×2 submatrix from (858));[T =∆1d 12 (d ]2 12+d 13 −d 23 )1(d 2 12+d 13 −d 23 ) d 13(862)Submatrix T must be positive (semi)definite whenever −VN TDV N(A.3.1.0.4,5.8.3) Now we have,is.−V T N DV N ≽ 0 ⇒ T ≽ 0 ⇔ λ 1 ≥ λ 2 ≥ 0−V T N DV N ≻ 0 ⇒ T ≻ 0 ⇔ λ 1 > λ 2 > 0(863)where λ 1 and λ 2 are the eigenvalues of T , real due only to symmetry of T :(λ 1 = 1 d2 12 + d 13 + √ )d23 2 − 2(d 12 + d 13 )d 23 + 2(d12 2 + d13)2 ∈ R(λ 2 = 1 d2 12 + d 13 − √ )d23 2 − 2(d 12 + d 13 )d 23 + 2(d12 2 + d13)2 ∈ R(864)Nonnegativity of eigenvalue λ 1 is guaranteed by only nonnegativity of the d ijwhich in turn is guaranteed by matrix inequality (857). Inequality betweenthe eigenvalues in (863) follows from only realness of the d ij . Since λ 1always equals or exceeds λ 2 , conditions for the positive (semi)definiteness ofsubmatrix T can be completely determined by examining λ 2 the smaller ofits two eigenvalues. A triangle inequality is made apparent when we expressT eigenvalue nonnegativity in terms of D matrix entries; videlicet,

5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 343T ≽ 0 ⇔ detT = λ 1 λ 2 ≥ 0 , d 12 ,d 13 ≥ 0 (c)⇔λ 2 ≥ 0(b)⇔| √ d 12 − √ d 23 | ≤ √ d 13 ≤ √ d 12 + √ d 23 (a)(865)Triangle inequality (865a) (confer (768) (877)), in terms of three rootedentries from D , is equivalent to metric property 4√d13 ≤ √ d 12 + √ d 23√d23≤ √ d 12 + √ d 13√d12≤ √ d 13 + √ d 23(866)for the corresponding points x 1 ,x 2 ,x 3 from some length-N list. 5.345.8.2.1 CommentGiven D whose dimension N equals or exceeds 3, there are N!/(3!(N − 3)!)distinct triangle inequalities in total like (768) that must be satisfied, of whicheach d ij is involved in N−2, and each point x i is in (N −1)!/(2!(N −1 − 2)!).We have so far revealed only one of those triangle inequalities; namely, (865a)that came from T (862). Yet we claim if −VN TDV N ≽ 0 then all triangleinequalities will be satisfied simultaneously;| √ d ik − √ d kj | ≤ √ d ij ≤ √ d ik + √ d kj , i

5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 343T ≽ 0 ⇔ detT = λ 1 λ 2 ≥ 0 , d 12 ,d 13 ≥ 0 (c)⇔λ 2 ≥ 0(b)⇔| √ d 12 − √ d 23 | ≤ √ d 13 ≤ √ d 12 + √ d 23 (a)(865)Triangle inequality (865a) (confer (768) (877)), in terms of three rootedentries from D , is equivalent to metric property 4√d13 ≤ √ d 12 + √ d 23√d23≤ √ d 12 + √ d 13√d12≤ √ d 13 + √ d 23(866)for the corresponding points x 1 ,x 2 ,x 3 from some length-N list. 5.345.8.2.1 CommentGiven D whose dimension N equals or exceeds 3, there are N!/(3!(N − 3)!)distinct triangle inequalities in total like (768) that must be satisfied, of whicheach d ij is involved in N−2, and each point x i is in (N −1)!/(2!(N −1 − 2)!).We have so far revealed only one of those triangle inequalities; namely, (865a)that came from T (862). Yet we claim if −VN TDV N ≽ 0 then all triangleinequalities will be satisfied simultaneously;| √ d ik − √ d kj | ≤ √ d ij ≤ √ d ik + √ d kj , i

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