v2009.01.01 - Convex Optimization

3.1. CONVEX FUNCTION 215
3.1.7.0.2 Example. Matrix pseudofractional function.
Consider a real function of two variables
f(A, x) : S n × R n → R = x T A † x (498)
on domf = S+× n R(A). This function is convex simultaneously in both
variables when variable matrix A belongs to the entire positive semidefinite
cone S n + and variable vector x is confined to range R(A) of matrix A .
To explain this, we need only demonstrate that the function epigraph is
convex. Consider Schur-form (1410) fromA.4: for t ∈ R
[ ] A z
G(A, z , t) =
z T ≽ 0
t
⇔
z T (I − AA † ) = 0
t − z T A † z ≥ 0
A ≽ 0
(499)
Inverse image of the positive semidefinite cone S n+1
+ under affine mapping
G(A, z , t) is convex by Theorem 2.1.9.0.1. Of the equivalent conditions for
positive semidefiniteness of G , the first is an equality demanding vector z
belong to R(A). Function f(A, z)=z T A † z is convex on convex domain
S+× n R(A) because the Cartesian product constituting its epigraph
epi f(A, z) = { (A, z , t) | A ≽ 0, z ∈ R(A), z T A † z ≤ t } = G −1( S n+1
+
is convex.
)
(500)
3.1.7.0.3 Exercise. Matrix product function.
Continue Example 3.1.7.0.2 by introducing vector variable x and making
the substitution z ←Ax . Because of matrix symmetry (E), for all x∈ R n
f(A, z(x)) = z T A † z = x T A T A † Ax = x T Ax = f(A, x) (501)
whose epigraph is
epif(A, x) = { (A, x, t) | A ≽ 0, x T Ax ≤ t } (502)
Provide two simple explanations why f(A, x) = x T Ax is not a function
convex simultaneously in positive semidefinite matrix A and vector x on
domf = S+× n R n .