v2010.10.26 - Convex Optimization

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

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442 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXbecause the projection of −D/2 on S N c (1998) can be 0 if and only ifD ∈ S N⊥c ; but S N⊥c ∩ S N h = 0 (Figure 129). Projector V on S N h is thereforeinjective hence uniquely invertible. Further, −V S N h V/2 is equivalent to thegeometric center subspace S N c in the ambient space of symmetric matrices; asurjection,S N c = V(S N ) = V ( S N h ⊕ S N⊥h) ( )= V SNh(997)because (72)V ( ) ( ) (S N h ⊇ V SN⊥h = V δ 2 (S N ) ) (998)Because D(G) on S N c is injective, and aff D ( V(EDM N ) ) = D ( V(aff EDM N ) )by property (127) of the affine hull, we find for D ∈ S N hid est,orD(−V DV 1 2 ) = δ(−V DV 1 2 )1T + 1δ(−V DV 1 2 )T − 2(−V DV 1 2 ) (999)D = D ( V(D) ) (1000)−V DV = V ( D(−V DV ) ) (1001)S N h = D ( V(S N h ) ) (1002)−V S N h V = V ( D(−V S N h V ) ) (1003)These operators V and D are mutual inverses.The Gram-form D ( )S N c (903) is equivalent to SNh ;D ( S N c)= D(V(SNh ⊕ S N⊥h) ) = S N h + D ( V(S N⊥h ) ) = S N h (1004)because S N h ⊇ D ( V(S N⊥h ) ) . In summary, for the Gram-form we have theisomorphisms [90,2] [89, p.76, p.107] [7,2.1] 5.30 [6,2] [8,18.2.1] [2,2.1]and from bijectivity results in5.6.1,S N h = D(S N c ) (1005)S N c = V(S N h ) (1006)EDM N = D(S N c ∩ S N +) (1007)S N c ∩ S N + = V(EDM N ) (1008)5.30 In [7, p.6, line 20], delete sentence: Since G is also...not a singleton set.[7, p.10, line 11] x 3 =2 (not 1).

5.6. INJECTIVITY OF D & UNIQUE RECONSTRUCTION 4435.6.2 Inner-product form bijectivityThe Gram-form EDM operator D(G)= δ(G)1 T + 1δ(G) T − 2G (903) is aninjective map, for example, on the domain that is the subspace of symmetricmatrices having all zeros in the first row and columnS N 1 = {G∈ S N | Ge 1 = 0}{[ ] [ 0 0T 0 0T= Y0 I 0 I]| Y ∈ S N }(2002)because it obviously has no nullspace there. Since Ge 1 = 0 ⇔ Xe 1 = 0 (905)means the first point in the list X resides at the origin, then D(G) on S N 1 ∩ S N +must be surjective onto EDM N .Substituting Θ T Θ ← −VN TDV N (970) into inner-product form EDMdefinition D(Θ) (958), it may be further decomposed:[0D(D) =δ ( −VN TDV )N] [1 T + 1 0 δ ( −VN TDV ) ] TN[ ] 0 0T− 20 −VN TDV N(1009)This linear operator D is another flavor of inner-product form and an injectivemap of the EDM cone onto itself. Yet when its domain is instead the entiresymmetric hollow subspace S N h = aff EDM N , D(D) becomes an injectivemap onto that same subspace. Proof follows directly from the fact: linear Dhas no nullspace [85,A.1] on S N h = aff D(EDM N )= D(aff EDM N ) (127).5.6.2.1 Inversion of D ( −VN TDV )NInjectivity of D(D) suggests inversion of (confer (908))V N (D) : S N → S N−1 −V T NDV N (1010)a linear surjective 5.31 mapping onto S N−1 having nullspace 5.32 S N⊥c ;V N (S N h ) = S N−1 (1011)5.31 Surjectivity of V N (D) is demonstrated via the Gram-form EDM operator D(G):Since S N h = D(S N c ) (1004), then for any Y ∈ S N−1 , −VN T †TD(VN Y V † N /2)V N = Y .5.32 N(V N ) ⊇ S N⊥c is apparent. There exists a linear mappingT(V N (D)) V †TN V N(D)V † N = −V DV 1 2 = V(D)

5.6. INJECTIVITY OF D & UNIQUE RECONSTRUCTION 4435.6.2 Inner-product form bijectivityThe Gram-form EDM operator D(G)= δ(G)1 T + 1δ(G) T − 2G (903) is aninjective map, for example, on the domain that is the subspace of symmetricmatrices having all zeros in the first row and columnS N 1 = {G∈ S N | Ge 1 = 0}{[ ] [ 0 0T 0 0T= Y0 I 0 I]| Y ∈ S N }(2002)because it obviously has no nullspace there. Since Ge 1 = 0 ⇔ Xe 1 = 0 (905)means the first point in the list X resides at the origin, then D(G) on S N 1 ∩ S N +must be surjective onto EDM N .Substituting Θ T Θ ← −VN TDV N (970) into inner-product form EDMdefinition D(Θ) (958), it may be further decomposed:[0D(D) =δ ( −VN TDV )N] [1 T + 1 0 δ ( −VN TDV ) ] TN[ ] 0 0T− 20 −VN TDV N(1009)This linear operator D is another flavor of inner-product form and an injectivemap of the EDM cone onto itself. Yet when its domain is instead the entiresymmetric hollow subspace S N h = aff EDM N , D(D) becomes an injectivemap onto that same subspace. Proof follows directly from the fact: linear Dhas no nullspace [85,A.1] on S N h = aff D(EDM N )= D(aff EDM N ) (127).5.6.2.1 Inversion of D ( −VN TDV )NInjectivity of D(D) suggests inversion of (confer (908))V N (D) : S N → S N−1 −V T NDV N (1010)a linear surjective 5.31 mapping onto S N−1 having nullspace 5.32 S N⊥c ;V N (S N h ) = S N−1 (1011)5.31 Surjectivity of V N (D) is demonstrated via the Gram-form EDM operator D(G):Since S N h = D(S N c ) (1004), then for any Y ∈ S N−1 , −VN T †TD(VN Y V † N /2)V N = Y .5.32 N(V N ) ⊇ S N⊥c is apparent. There exists a linear mappingT(V N (D)) V †TN V N(D)V † N = −V DV 1 2 = V(D)

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