v2006.03.09 - Convex Optimization

298 CHAPTER 4. EUCLIDEAN DISTANCE MATRIXtheorem (A.3.1.0.4) it is sufficient to prove all d ij are nonnegative, alltriangle inequalities are satisfied, and det(−VN TDV N) is nonnegative. WhenN = 4, in other words, that nonnegative determinant becomes the fifth andlast Euclidean metric requirement for D ∈ EDM N . We now endeavor toascribe geometric meaning to it.4.14.2.1 Nonnegative determinantBy (519) when D ∈EDM 4 , −VN TDV N is equal to inner product (514),⎡√ √ ⎤d 12 d12 d 13 cosθ 213 d12 √d12 √d 14 cosθ 214Θ T Θ = ⎣ d 13 cosθ 213 d 13 d13 √d12d 14 cosθ 314⎦√(704)d 14 cosθ 214 d13 d 14 cosθ 314 d 14Because Euclidean space is an inner-product space, the more conciseinner-product form of the determinant is admitted;det(Θ T Θ) = −d 12 d 13 d 14(cos(θ213 ) 2 +cos(θ 214 ) 2 +cos(θ 314 ) 2 − 2 cos θ 213 cosθ 214 cos θ 314 − 1 )The determinant is nonnegative if and only if(705)cos θ 214 cos θ 314 − √ sin(θ 214 ) 2 sin(θ 314 ) 2 ≤ cos θ 213 ≤ cos θ 214 cos θ 314 + √ sin(θ 214 ) 2 sin(θ 314 ) 2⇔cos θ 213 cos θ 314 − √ sin(θ 213 ) 2 sin(θ 314 ) 2 ≤ cos θ 214 ≤ cos θ 213 cos θ 314 + √ sin(θ 213 ) 2 sin(θ 314 ) 2⇔cos θ 213 cos θ 214 − √ sin(θ 213 ) 2 sin(θ 214 ) 2 ≤ cos θ 314 ≤ cos θ 213 cos θ 214 + √ sin(θ 213 ) 2 sin(θ 214 ) 2which simplifies, for 0 ≤ θ i1l ,θ l1j ,θ i1j ≤ π and all i≠j ≠l ∈{2, 3, 4}, to(706)cos(θ i1l + θ l1j ) ≤ cos θ i1j ≤ cos(θ i1l − θ l1j ) (707)Analogously to triangle inequality (621), the determinant is 0 upon equalityon either side of (707) which is tight. Inequality (707) can be equivalentlywritten linearly as a “triangle inequality”, but between relative angles[261,1.4];|θ i1l − θ l1j | ≤ θ i1j ≤ θ i1l + θ l1jθ i1l + θ l1j + θ i1j ≤ 2π(708)0 ≤ θ i1l ,θ l1j ,θ i1j ≤ πGeneralizing this: