v2009.01.01 - Convex Optimization

v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization

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458 CHAPTER 6. CONE OF DISTANCE MATRICES The problem dual to maximum variance unfolding problem (1092) (less the Gram rank constraint) has been called the fastest mixing Markov process. Explored in [295], that dual has simple interpretations in graph and circuit theory and in mechanical and thermal systems. Optimal Gram rank turns out to be tightly bounded above by minimum multiplicity of the second smallest eigenvalue of a dual optimal variable. 6.5 EDM definition in 11 T Any EDM D corresponding to affine dimension r has representation D(V X ) ∆ = δ(V X V T X )1 T + 1δ(V X V T X ) T − 2V X V T X ∈ EDM N (1093) where R(V X ∈ R N×r )⊆ N(1 T ) = 1 ⊥ , VX T V X = δ 2 (VX T V X ) and V X is full-rank with orthogonal columns. (1094) Equation (1093) is simply the standard EDM definition (798) with a centered list X as in (880); Gram matrix X T X has been replaced with the subcompact singular value decomposition (A.6.2) 6.4 V X V T X ≡ V T X T XV ∈ S N c ∩ S N + (1095) This means: inner product VX TV X is an r×r diagonal matrix Σ of nonzero singular values. Vector δ(V X VX T ) may me decomposed into complementary parts by projecting it on orthogonal subspaces 1 ⊥ and R(1) : namely, P 1 ⊥( δ(VX V T X ) ) = V δ(V X V T X ) (1096) P 1 ( δ(VX V T X ) ) = 1 N 11T δ(V X V T X ) (1097) Of course δ(V X V T X ) = V δ(V X V T X ) + 1 N 11T δ(V X V T X ) (1098) ∆ 6.4 Subcompact SVD: V X VX T = Q √ Σ √ ΣQ T ≡ V T X T XV . So VX T is not necessarily XV (5.5.1.0.1), although affine dimension r = rank(VX T ) = rank(XV ). (937)

6.5. EDM DEFINITION IN 11 T 459 by (821). Substituting this into EDM definition (1093), we get the Hayden, Wells, Liu, & Tarazaga EDM formula [159,2] where D(V X , y) ∆ = y1 T + 1y T + λ N 11T − 2V X V T X ∈ EDM N (1099) λ ∆ = 2‖V X ‖ 2 F = 1 T δ(V X V T X )2 and y ∆ = δ(V X V T X ) − λ 2N 1 = V δ(V XV T X ) (1100) and y=0 if and only if 1 is an eigenvector of EDM D . Scalar λ becomes an eigenvalue when corresponding eigenvector 1 exists. 6.5 Then the particular dyad sum from (1099) y1 T + 1y T + λ N 11T ∈ S N⊥ c (1101) must belong to the orthogonal complement of the geometric center subspace (p.671), whereas V X V T X ∈ SN c ∩ S N + (1095) belongs to the positive semidefinite cone in the geometric center subspace. Proof. We validate eigenvector 1 and eigenvalue λ . (⇒) Suppose 1 is an eigenvector of EDM D . Then because it follows V T X 1 = 0 (1102) D1 = δ(V X V T X )1T 1 + 1δ(V X V T X )T 1 = N δ(V X V T X ) + ‖V X ‖ 2 F 1 ⇒ δ(V X V T X ) ∝ 1 (1103) For some κ∈ R + δ(V X V T X ) T 1 = N κ = tr(V T X V X ) = ‖V X ‖ 2 F ⇒ δ(V X V T X ) = 1 N ‖V X ‖ 2 F1 (1104) so y=0. (⇐) Now suppose δ(V X VX T)= λ 1 ; id est, y=0. Then 2N D = λ N 11T − 2V X V T X ∈ EDM N (1105) 1 is an eigenvector with corresponding eigenvalue λ . 6.5 e.g., when X = I in EDM definition (798).

6.5. EDM DEFINITION IN 11 T 459<br />

by (821). Substituting this into EDM definition (1093), we get the<br />

Hayden, Wells, Liu, & Tarazaga EDM formula [159,2]<br />

where<br />

D(V X , y) ∆ = y1 T + 1y T + λ N 11T − 2V X V T X ∈ EDM N (1099)<br />

λ ∆ = 2‖V X ‖ 2 F = 1 T δ(V X V T X )2 and y ∆ = δ(V X V T X ) − λ<br />

2N 1 = V δ(V XV T X )<br />

(1100)<br />

and y=0 if and only if 1 is an eigenvector of EDM D . Scalar λ becomes<br />

an eigenvalue when corresponding eigenvector 1 exists. 6.5<br />

Then the particular dyad sum from (1099)<br />

y1 T + 1y T + λ N 11T ∈ S N⊥<br />

c (1101)<br />

must belong to the orthogonal complement of the geometric center subspace<br />

(p.671), whereas V X V T X ∈ SN c ∩ S N + (1095) belongs to the positive semidefinite<br />

cone in the geometric center subspace.<br />

Proof. We validate eigenvector 1 and eigenvalue λ .<br />

(⇒) Suppose 1 is an eigenvector of EDM D . Then because<br />

it follows<br />

V T X 1 = 0 (1102)<br />

D1 = δ(V X V T X )1T 1 + 1δ(V X V T X )T 1 = N δ(V X V T X ) + ‖V X ‖ 2 F 1<br />

⇒ δ(V X V T X ) ∝ 1 (1103)<br />

For some κ∈ R +<br />

δ(V X V T X ) T 1 = N κ = tr(V T X V X ) = ‖V X ‖ 2 F ⇒ δ(V X V T X ) = 1 N ‖V X ‖ 2 F1 (1104)<br />

so y=0.<br />

(⇐) Now suppose δ(V X VX T)= λ 1 ; id est, y=0. Then<br />

2N<br />

D = λ N 11T − 2V X V T X ∈ EDM N (1105)<br />

1 is an eigenvector with corresponding eigenvalue λ . <br />

6.5 e.g., when X = I in EDM definition (798).

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