v2010.10.26 - Convex Optimization

244 CHAPTER 3. GEOMETRY OF CONVEX FUNCTIONSconvex. Consider Schur-form (1511) fromA.4: for t ∈ R[ ] A zG(A, z , t) =z T ≽ 0t⇔z T (I − AA † ) = 0t − z T A † z ≥ 0A ≽ 0(562)Inverse image of the positive semidefinite cone S n+1+ under affine mappingG(A, z , t) is convex by Theorem 2.1.9.0.1. Of the equivalent conditions forpositive semidefiniteness of G , the first is an equality demanding vector zbelong to R(A). Function f(A, z)=z T A † z is convex on convex domainS+× n R(A) because the Cartesian product constituting its epigraphepif(A, z) = { (A, z , t) | A ≽ 0, z ∈ R(A), z T A † z ≤ t } = G −1( )S n+1+ (563)is convex.3.5.0.0.3 Exercise. Matrix product function.Continue Example 3.5.0.0.2 by introducing vector variable x and makingthe substitution z ←Ax . Because of matrix symmetry (E), for all x∈ R nf(A, z(x)) = z T A † z = x T A T A † Ax = x T Ax = f(A, x) (564)whose epigraph isepif(A, x) = { (A, x, t) | A ≽ 0, x T Ax ≤ t } (565)Provide two simple explanations why f(A, x) = x T Ax is not a functionconvex simultaneously in positive semidefinite matrix A and vector x ondomf = S+× n R n .3.5.0.0.4 Example. Matrix fractional function. (confer3.7.2.0.1)Continuing Example 3.5.0.0.2, now consider a real function of two variableson domf = S n +×R n for small positive constant ǫ (confer (1870))f(A, x) = ǫx T (A + ǫI) −1 x (566)