v2009.01.01 - Convex Optimization

176 CHAPTER 2. CONVEX GEOMETRY
2.13.9.4.1 Exercise. Conically independent columns and rows.
We suspect the number of conically independent columns (rows) of X to
be the same for X †T , where † denotes matrix pseudoinverse (E). Prove
whether it holds that the columns (rows) of X are c.i. ⇔ the columns (rows)
of X †T are c.i.
2.13.9.4.2 Example. Monotone nonnegative cone.
[53, exer.2.33] [306,2] Simplicial cone (2.12.3.1.1) K M+ is the cone of all
nonnegative vectors having their entries sorted in nonincreasing order:
K M+ ∆ = {x | x 1 ≥ x 2 ≥ · · · ≥ x n ≥ 0} ⊆ R n +
= {x | (e i − e i+1 ) T x ≥ 0, i = 1... n−1, e T nx ≥ 0}
= {x | X † x ≽ 0}
(383)
a halfspace-description where e i is the i th standard basis vector, and where 2.70
X †T = ∆ [ e 1 −e 2 e 2 −e 3 · · · e n ] ∈ R n×n (384)
For any vectors x and y , simple algebra demands
n∑
x T y = x i y i = (x 1 − x 2 )y 1 + (x 2 − x 3 )(y 1 + y 2 ) + (x 3 − x 4 )(y 1 + y 2 + y 3 ) + · · ·
i=1
+ (x n−1 − x n )(y 1 + · · · + y n−1 ) + x n (y 1 + · · · + y n ) (385)
Because x i − x i+1 ≥ 0 ∀i by assumption whenever x∈ K M+ , we can employ
dual generalized inequalities (291) with respect to the self-dual nonnegative
orthant R n + to find the halfspace-description of the dual cone K ∗ M+ . We can
say x T y ≥ 0 for all X † x ≽ 0 [sic] if and only if
y 1 ≥ 0, y 1 + y 2 ≥ 0, ... , y 1 + y 2 + · · · + y n ≥ 0 (386)
id est,
where
x T y ≥ 0 ∀X † x ≽ 0 ⇔ X T y ≽ 0 (387)
X = [e 1 e 1 +e 2 e 1 +e 2 +e 3 · · · 1 ] ∈ R n×n (388)
2.70 With X † in hand, we might concisely scribe the remaining vertex- and
halfspace-descriptions from the tables for K M+ and its dual. Instead we use dual
generalized inequalities in their derivation.