v2010.10.26 - Convex Optimization
v2010.10.26 - Convex Optimization 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)
5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 4575.8.3.1.1 Example. Small completion problem, II.Applying the inequality for λ 1 in (1072) to the small completion problem onpage 398 Figure 116, the lower bound on √ d 14 (1.236 in (884)) is tightenedto 1.289 . The correct value of √ d 14 to three significant figures is 1.414 .5.8.4 Affine dimension reduction in two dimensions(confer5.14.4) The leading principal 2×2 submatrix T of −VN TDV N haslargest eigenvalue λ 1 (1060) which is a convex function of D . 5.43 λ 1 cannever be 0 unless d 12 = d 13 = d 23 = 0. Eigenvalue λ 1 can never be negativewhile the d ij are nonnegative. The remaining eigenvalue λ 2 is a concavefunction of D that becomes 0 only at the upper and lower bounds of triangleinequality (1061a) and its equivalent forms: (confer (1063))| √ d 12 − √ d 23 | ≤ √ d 13 ≤ √ d 12 + √ d 23 (a)⇔| √ d 12 − √ d 13 | ≤ √ d 23 ≤ √ d 12 + √ d 13 (b)⇔| √ d 13 − √ d 23 | ≤ √ d 12 ≤ √ d 13 + √ d 23 (c)(1073)In between those bounds, λ 2 is strictly positive; otherwise, it would benegative but prevented by the condition T ≽ 0.When λ 2 becomes 0, it means triangle △ 123 has collapsed to a linesegment; a potential reduction in affine dimension r . The same logic is validfor any particular principal 2×2 submatrix of −VN TDV N , hence applicableto other triangles.5.43 The largest eigenvalue of any symmetric matrix is always a convex function of itsentries, while the ⎡ smallest ⎤ eigenvalue is always concave. [61, exmp.3.10] In our particulard 12case, say d ⎣d 13⎦∈ R 3 . Then the Hessian (1760) ∇ 2 λ 1 (d)≽0 certifies convexityd 23whereas ∇ 2 λ 2 (d)≼0 certifies concavity. Each Hessian has rank equal to 1. The respectivegradients ∇λ 1 (d) and ∇λ 2 (d) are nowhere 0.
- Page 405 and 406: 5.4. EDM DEFINITION 4055.4.2 Gram-f
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- Page 411 and 412: 5.4. EDM DEFINITION 411is the fact:
- Page 413 and 414: 5.4. EDM DEFINITION 413Figure 120:
- Page 415 and 416: 5.4. EDM DEFINITION 415is found fro
- Page 417 and 418: 5.4. EDM DEFINITION 417one less dim
- Page 419 and 420: 5.4. EDM DEFINITION 419equality con
- Page 421 and 422: 5.4. EDM DEFINITION 421How much dis
- Page 423 and 424: 5.4. EDM DEFINITION 423105ˇx 4ˇx
- Page 425 and 426: 5.4. EDM DEFINITION 425now implicit
- Page 427 and 428: 5.4. EDM DEFINITION 427by translate
- Page 429 and 430: 5.4. EDM DEFINITION 429Crippen & Ha
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- Page 433 and 434: 5.4. EDM DEFINITION 433because (A.3
- Page 435 and 436: 5.5. INVARIANCE 4355.5.1.0.1 Exampl
- Page 437 and 438: 5.5. INVARIANCE 437x 2 x 2x 3 x 1 x
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- Page 445 and 446: 5.7. EMBEDDING IN AFFINE HULL 4455.
- Page 447 and 448: 5.7. EMBEDDING IN AFFINE HULL 447Fo
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- Page 473 and 474: 5.11. EDM INDEFINITENESS 473So beca
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- Page 485 and 486: 5.13. RECONSTRUCTION EXAMPLES 485Th
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- Page 497 and 498: Chapter 6Cone of distance matricesF
- Page 499 and 500: 6.1. DEFINING EDM CONE 4996.1 Defin
- Page 501 and 502: 6.2. POLYHEDRAL BOUNDS 501This cone
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- Page 505 and 506: 6.4. EDM DEFINITION IN 11 T 505N(1
5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 4575.8.3.1.1 Example. Small completion problem, II.Applying the inequality for λ 1 in (1072) to the small completion problem onpage 398 Figure 116, the lower bound on √ d 14 (1.236 in (884)) is tightenedto 1.289 . The correct value of √ d 14 to three significant figures is 1.414 .5.8.4 Affine dimension reduction in two dimensions(confer5.14.4) The leading principal 2×2 submatrix T of −VN TDV N haslargest eigenvalue λ 1 (1060) which is a convex function of D . 5.43 λ 1 cannever be 0 unless d 12 = d 13 = d 23 = 0. Eigenvalue λ 1 can never be negativewhile the d ij are nonnegative. The remaining eigenvalue λ 2 is a concavefunction of D that becomes 0 only at the upper and lower bounds of triangleinequality (1061a) and its equivalent forms: (confer (1063))| √ d 12 − √ d 23 | ≤ √ d 13 ≤ √ d 12 + √ d 23 (a)⇔| √ d 12 − √ d 13 | ≤ √ d 23 ≤ √ d 12 + √ d 13 (b)⇔| √ d 13 − √ d 23 | ≤ √ d 12 ≤ √ d 13 + √ d 23 (c)(1073)In between those bounds, λ 2 is strictly positive; otherwise, it would benegative but prevented by the condition T ≽ 0.When λ 2 becomes 0, it means triangle △ 123 has collapsed to a linesegment; a potential reduction in affine dimension r . The same logic is validfor any particular principal 2×2 submatrix of −VN TDV N , hence applicableto other triangles.5.43 The largest eigenvalue of any symmetric matrix is always a convex function of itsentries, while the ⎡ smallest ⎤ eigenvalue is always concave. [61, exmp.3.10] In our particulard 12case, say d ⎣d 13⎦∈ R 3 . Then the Hessian (1760) ∇ 2 λ 1 (d)≽0 certifies convexityd 23whereas ∇ 2 λ 2 (d)≼0 certifies concavity. Each Hessian has rank equal to 1. The respectivegradients ∇λ 1 (d) and ∇λ 2 (d) are nowhere 0.