v2010.10.26 - Convex Optimization

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

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450 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXFor all practical purposes, (1041)max{0, rank(D) − 2} ≤ r ≤ min{n, N −1} (1050)5.8 Euclidean metric versus matrix criteria5.8.1 Nonnegativity property 1When D=[d ij ] is an EDM (891), then it is apparent from (1035)2VNX T T XV N = −VNDV T N ≽ 0 (1051)because for any matrix A , A T A≽0 . 5.36 We claim nonnegativity of the d ijis enforced primarily by the matrix inequality (1051); id est,−V T N DV N ≽ 0D ∈ S N h}⇒ d ij ≥ 0, i ≠ j (1052)(The matrix inequality to enforce strict positivity differs by a stroke of thepen. (1055))We now support our claim: If any matrix A∈ R m×m is positivesemidefinite, then its main diagonal δ(A)∈ R m must have all nonnegativeentries. [159,4.2] Given D ∈ S N h−VN TDV N =⎡⎤1d 12 2 (d 112+d 13 −d 23 )2 (d 11,i+1+d 1,j+1 −d i+1,j+1 ) · · ·2 (d 12+d 1N −d 2N )12 (d 112+d 13 −d 23 ) d 13 2 (d 11,i+1+d 1,j+1 −d i+1,j+1 ) · · ·2 (d 13+d 1N −d 3N )1⎢2 (d 11,j+1+d 1,i+1 −d j+1,i+1 )2 (d .1,j+1+d 1,i+1 −d j+1,i+1 ) d .. 11,i+1 2 (d 14+d 1N −d 4N )⎥⎣..... . .. . ⎦12 (d 112+d 1N −d 2N )2 (d 113+d 1N −d 3N )2 (d 14+d 1N −d 4N ) · · · d 1N= 1 2 (1D 1,2:N + D 2:N,1 1 T − D 2:N,2:N ) ∈ S N−1 (1053)5.36 For A∈R m×n , A T A ≽ 0 ⇔ y T A T Ay = ‖Ay‖ 2 ≥ 0 for all ‖y‖ = 1. When A isfull-rank skinny-or-square, A T A ≻ 0.

5.8. EUCLIDEAN METRIC VERSUS MATRIX CRITERIA 451where row,column indices i,j ∈ {1... N −1}. [312] It follows:−V T N DV N ≽ 0D ∈ S N h}⎡⇒ δ(−VNDV T N ) = ⎢⎣⎤d 12d 13⎥.d 1N⎦ ≽ 0 (1054)Multiplication of V N by any permutation matrix Ξ has null effect on its rangeand nullspace. In other words, any permutation of the rows or columns of V Nproduces a basis for N(1 T ); id est, R(Ξ r V N )= R(V N Ξ c )= R(V N )= N(1 T ).Hence, −VN TDV N ≽ 0 ⇔ −VN TΞT rDΞ r V N ≽ 0 (⇔ −Ξ T c VN TDV N Ξ c ≽ 0).Various permutation matrices 5.37 will sift the remaining d ij similarlyto (1054) thereby proving their nonnegativity. Hence −VN TDV N ≽ 0 isa sufficient test for the first property (5.2) of the Euclidean metric,nonnegativity.When affine dimension r equals 1, in particular, nonnegativity symmetryand hollowness become necessary and sufficient criteria satisfying matrixinequality (1051). (6.5.0.0.1)5.8.1.1 Strict positivityShould we require the points in R n to be distinct, then entries of D off themain diagonal must be strictly positive {d ij > 0, i ≠ j} and only those entriesalong the main diagonal of D are 0. By similar argument, the strict matrixinequality is a sufficient test for strict positivity of Euclidean distance-square;−V T N DV N ≻ 0D ∈ S N h5.8.2 Triangle inequality property 4}⇒ d ij > 0, i ≠ j (1055)In light of Kreyszig’s observation [227,1.1 prob.15] that properties 2through 4 of the Euclidean metric (5.2) together imply nonnegativityproperty 1,5.37 The rule of thumb is: If Ξ r (i,1) = 1, then δ(−V T N ΞT rDΞ r V N )∈ R N−1 is somepermutation of the i th row or column of D excepting the 0 entry from the main diagonal.

450 CHAPTER 5. EUCLIDEAN DISTANCE MATRIXFor all practical purposes, (1041)max{0, rank(D) − 2} ≤ r ≤ min{n, N −1} (1050)5.8 Euclidean metric versus matrix criteria5.8.1 Nonnegativity property 1When D=[d ij ] is an EDM (891), then it is apparent from (1035)2VNX T T XV N = −VNDV T N ≽ 0 (1051)because for any matrix A , A T A≽0 . 5.36 We claim nonnegativity of the d ijis enforced primarily by the matrix inequality (1051); id est,−V T N DV N ≽ 0D ∈ S N h}⇒ d ij ≥ 0, i ≠ j (1052)(The matrix inequality to enforce strict positivity differs by a stroke of thepen. (1055))We now support our claim: If any matrix A∈ R m×m is positivesemidefinite, then its main diagonal δ(A)∈ R m must have all nonnegativeentries. [159,4.2] Given D ∈ S N h−VN TDV N =⎡⎤1d 12 2 (d 112+d 13 −d 23 )2 (d 11,i+1+d 1,j+1 −d i+1,j+1 ) · · ·2 (d 12+d 1N −d 2N )12 (d 112+d 13 −d 23 ) d 13 2 (d 11,i+1+d 1,j+1 −d i+1,j+1 ) · · ·2 (d 13+d 1N −d 3N )1⎢2 (d 11,j+1+d 1,i+1 −d j+1,i+1 )2 (d .1,j+1+d 1,i+1 −d j+1,i+1 ) d .. 11,i+1 2 (d 14+d 1N −d 4N )⎥⎣..... . .. . ⎦12 (d 112+d 1N −d 2N )2 (d 113+d 1N −d 3N )2 (d 14+d 1N −d 4N ) · · · d 1N= 1 2 (1D 1,2:N + D 2:N,1 1 T − D 2:N,2:N ) ∈ S N−1 (1053)5.36 For A∈R m×n , A T A ≽ 0 ⇔ y T A T Ay = ‖Ay‖ 2 ≥ 0 for all ‖y‖ = 1. When A isfull-rank skinny-or-square, A T A ≻ 0.

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