v2009.01.01 - Convex Optimization

4.2. FRAMEWORK 261
When equality is attained in (610)
rankX ⋆ + rankS ⋆ = n (612)
there are no coinciding main diagonal zeros in Λ X ⋆Λ S ⋆ , and so we have what
is called strict complementarity. 4.11 Logically it follows that a necessary and
sufficient condition for strict complementarity of an optimal primal and dual
solution is
X ⋆ + S ⋆ ≻ 0 (613)
4.2.3.1 solving primal problem via dual
The beauty of Corollary 4.2.3.0.1 is its conjugacy; id est, one can solve
either the primal or dual problem in (584) and then find a solution to
the other via the optimality conditions. When a dual optimal solution is
known, for example, a primal optimal solution belongs to the hyperplane
{X | 〈S ⋆ , X〉=0}.
4.2.3.1.1 Example. Minimum cardinality Boolean. [81] [31,4.3.4] [300]
(confer Example 4.5.1.2.1) Consider finding a minimum cardinality Boolean
solution x to the classic linear algebra problem Ax = b given noiseless data
A∈ R m×n and b∈ R m ;
minimize ‖x‖ 0
x
subject to Ax = b
x i ∈ {0, 1} ,
i=1... n
(614)
where ‖x‖ 0 denotes cardinality of vector x (a.k.a, 0-norm; not a convex
function).
A minimum cardinality solution answers the question: “Which fewest
linear combination of columns in A constructs vector b ?” Cardinality
problems have extraordinarily wide appeal, arising in many fields of science
and across many disciplines. [276] [185] [144] [145] Yet designing an efficient
algorithm to optimize cardinality has proved difficult. In this example, we
also constrain the variable to be Boolean. The Boolean constraint forces
4.11 distinct from maximal complementarity (4.1.1.1).