v2010.10.26 - Convex Optimization

v2010.10.26 - Convex Optimization v2010.10.26 - Convex Optimization

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644 APPENDIX B. SIMPLE MATRICESwhere b is a vector. Then because H is nonsingular (A.3.1.0.5) [184,3][ A 0−V DV = −H0 T 0]H ≽ 0 ⇔ −A ≽ 0 (1653)and affine dimension is r = rankA when D is a Euclidean distance matrix.B.4.2Schoenberg auxiliary matrix V N1. V N = 1 √2[ −1TI2. V T N 1 = 03. I − e 1 1 T = [ 0 √ 2V N]4. [ 0 √ 2V N]VN = V N5. [ 0 √ 2V N]V = V]∈ R N×N−16. V [ 0 √ 2V N]=[0√2VN]7. [ 0 √ 2V N] [0√2VN]=[0√2VN]8. [ 0 √ 2V N] †=[ 0 0T0 I]V9. [ 0 √ 2V N] †V =[0√2VN] †10. [ 0 √ ] [ √ ] †2V N 0 2VN = V11. [ 0 √ [ ]] † [ √ ] 0 0T2V N 0 2VN =0 I12. [ 0 √ ] [ ]0 02V TN = [ 0 √ ]2V0 IN[ ] [ ]0 0T [0 √ ] 0 0T13.2VN =0 I0 I

B.4. AUXILIARY V -MATRICES 64514. [V N1 √21 ] −1 =[ ]V†N√2N 1T15. V † N = √ 2 [ − 1 N 1 I − 1 N 11T] ∈ R N−1×N ,16. V † N 1 = 017. V † N V N = I18. V T = V = V N V † N = I − 1 N 11T ∈ S N(I −1N 11T ∈ S N−1)19. −V † N (11T − I)V N = I ,(11 T − I ∈ EDM N)20. D = [d ij ] ∈ S N h (915)tr(−V DV ) = tr(−V D) = tr(−V † N DV N) = 1 N 1T D 1 = 1 N tr(11T D) = 1 NAny elementary matrix E ∈ S N of the particular form∑d iji,jE = k 1 I − k 2 11 T (1654)where k 1 , k 2 ∈ R , B.7 will make tr(−ED) proportional to ∑ d ij .21. D = [d ij ] ∈ S Ntr(−V DV ) = 1 N∑i,ji≠jd ij − N−1N∑d ii = 1 N 1T D 1 − trDi22. D = [d ij ] ∈ S N htr(−V T N DV N) = ∑ jd 1j23. For Y ∈ S NV (Y − δ(Y 1))V = Y − δ(Y 1)B.7 If k 1 is 1−ρ while k 2 equals −ρ∈R , then all eigenvalues of E for −1/(N −1) < ρ < 1are guaranteed positive and therefore E is guaranteed positive definite. [301]

B.4. AUXILIARY V -MATRICES 64514. [V N1 √21 ] −1 =[ ]V†N√2N 1T15. V † N = √ 2 [ − 1 N 1 I − 1 N 11T] ∈ R N−1×N ,16. V † N 1 = 017. V † N V N = I18. V T = V = V N V † N = I − 1 N 11T ∈ S N(I −1N 11T ∈ S N−1)19. −V † N (11T − I)V N = I ,(11 T − I ∈ EDM N)20. D = [d ij ] ∈ S N h (915)tr(−V DV ) = tr(−V D) = tr(−V † N DV N) = 1 N 1T D 1 = 1 N tr(11T D) = 1 NAny elementary matrix E ∈ S N of the particular form∑d iji,jE = k 1 I − k 2 11 T (1654)where k 1 , k 2 ∈ R , B.7 will make tr(−ED) proportional to ∑ d ij .21. D = [d ij ] ∈ S Ntr(−V DV ) = 1 N∑i,ji≠jd ij − N−1N∑d ii = 1 N 1T D 1 − trDi22. D = [d ij ] ∈ S N htr(−V T N DV N) = ∑ jd 1j23. For Y ∈ S NV (Y − δ(Y 1))V = Y − δ(Y 1)B.7 If k 1 is 1−ρ while k 2 equals −ρ∈R , then all eigenvalues of E for −1/(N −1) < ρ < 1are guaranteed positive and therefore E is guaranteed positive definite. [301]

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