v2010.10.26 - Convex Optimization

5.4. EDM DEFINITION 419equality constraints on that Gram matrix while the sensor is constrainedonly by dimensioning to lie in R 2 . Although the objective trG ofminimization 5.15 insures a solution on the boundary of positive semidefinitecone S 4 + , for this problem, we claim that the set of feasible Gram matricesforms a line (2.5.1.1) in isomorphic R 10 tangent (2.1.7.1.2) to the positivesemidefinite cone boundary. (5.4.2.3.6, confer4.2.1.3)By Schur complement (A.4,2.9.1.0.1), any feasible G and X provideG ≽ X T X (937)which is a convex relaxation of the desired (nonconvex) equality constraint[ ] [ ] I X I [I X ]X T =G X T (938)expected positive semidefinite rank-2 under noiseless conditions.by (1491), the relaxation admitsBut,(3 ≥) rankG ≥ rankX (939)(a third dimension corresponding to an intersection of three spheres,not circles, were there noise). If rank of an optimal solution equals 2,[ ]I Xrank⋆X ⋆T G ⋆ = 2 (940)then G ⋆ = X ⋆T X ⋆ by Theorem A.4.0.1.3.As posed, this localization problem does not require affinely independent(Figure 27, three noncollinear) anchors. Assuming the anchors exhibitno rotational or reflective symmetry in their affine hull (5.5.2) andassuming the sensor x 1 lies in that affine hull, then sensor position solutionx ⋆ 1= X ⋆ (:, 1) is unique under noiseless measurement. [320] This preceding transformation of trilateration to a semidefinite programworks all the time ((940) holds) despite relaxation (937) because the optimalsolution set is a unique point.5.15 Trace (tr G = 〈I , G〉) minimization is a heuristic for rank minimization. (7.2.2.1)It may be interpreted as squashing G which is bounded below by X T X as in (937);id est, G−X T X ≽ 0 ⇒ trG ≥ tr X T X (1489). δ(G−X T X)=0 ⇔ G=X T X (A.7.2)⇒ tr G = tr X T X which is a condition necessary for equality.