v2009.01.01 - Convex Optimization
v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization
120 CHAPTER 2. CONVEX GEOMETRY Created by means of Geršgorin discs, K always belongs to the positive semidefinite cone for any nonnegative value of p ∈ R m + . Hence any point in K corresponds to some positive semidefinite matrix A . Only the extreme directions of K intersect the positive semidefinite cone boundary in this dimension; the four extreme directions of K are extreme directions of the positive semidefinite cone. As p 1 /p 2 increases in value from 0, two extreme directions of K sweep the entire boundary of this positive semidefinite cone. Because the entire positive semidefinite cone can be swept by K , the system of linear inequalities [ Y T svec A = ∆ p1 ±p 2 / √ ] 2 0 0 ±p 1 / √ svec A ≽ 0 (224) 2 p 2 when made dynamic can replace a semidefinite constraint A≽0 ; id est, for given p where Y ∈ R m(m+1)/2×m2m−1 but K = {z | Y T z ≽ 0} ⊂ svec S m + (225) svec A ∈ K ⇒ A ∈ S m + (226) ∃p Y T svec A ≽ 0 ⇔ A ≽ 0 (227) In other words, diagonal dominance [176, p.349,7.2.3] A ii ≥ m∑ |A ij | , ∀i = 1... m (228) j=1 j ≠ i is only a sufficient condition for membership to the PSD cone; but by dynamic weighting p in this dimension, it was made necessary and sufficient. In higher dimension (m > 2), the boundary of the positive semidefinite cone is no longer constituted completely by its extreme directions (symmetric rank-one matrices); the geometry becomes complicated. How all the extreme directions can be swept by an inscribed polyhedral cone, 2.40 similarly to the foregoing example, remains an open question. 2.40 It is not necessary to sweep the entire boundary in higher dimension.
2.9. POSITIVE SEMIDEFINITE (PSD) CONE 121 2.9.2.5.3 Exercise. Dual inscription. Find dual proper polyhedral cone K ∗ from Figure 42. 2.9.2.6 Boundary constituents of the positive semidefinite cone 2.9.2.6.1 Lemma. Sum of positive semidefinite matrices. For A,B ∈ S M + rank(A + B) = rank(µA + (1 −µ)B) (229) over the open interval (0, 1) of µ . Proof. Any positive semidefinite matrix belonging to the PSD cone has an eigen decomposition that is a positively scaled sum of linearly independent symmetric dyads. By the linearly independent dyads definition inB.1.1.0.1, rank of the sum A +B is equivalent to the number of linearly independent dyads constituting it. Linear independence is insensitive to further positive scaling by µ . The assumption of positive semidefiniteness prevents annihilation of any dyad from the sum A +B . 2.9.2.6.2 Example. Rank function quasiconcavity. (confer3.3) For A,B ∈ R m×n [176,0.4] that follows from the fact [287,3.6] rankA + rankB ≥ rank(A + B) (230) dim R(A) + dim R(B) = dim R(A + B) + dim(R(A) ∩ R(B)) (231) For A,B ∈ S M + rankA + rankB ≥ rank(A + B) ≥ min{rankA, rankB} (1370) that follows from the fact N(A + B) = N(A) ∩ N(B) , A,B ∈ S M + (146) Rank is a quasiconcave function on S M + because the right-hand inequality in (1370) has the concave form (578); videlicet, Lemma 2.9.2.6.1. From this example we see, unlike convex functions, quasiconvex functions are not necessarily continuous. (3.3) We also glean: ⋄
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2.9. POSITIVE SEMIDEFINITE (PSD) CONE 121<br />
2.9.2.5.3 Exercise. Dual inscription.<br />
Find dual proper polyhedral cone K ∗ from Figure 42.<br />
<br />
2.9.2.6 Boundary constituents of the positive semidefinite cone<br />
2.9.2.6.1 Lemma. Sum of positive semidefinite matrices.<br />
For A,B ∈ S M +<br />
rank(A + B) = rank(µA + (1 −µ)B) (229)<br />
over the open interval (0, 1) of µ .<br />
Proof. Any positive semidefinite matrix belonging to the PSD cone<br />
has an eigen decomposition that is a positively scaled sum of linearly<br />
independent symmetric dyads. By the linearly independent dyads definition<br />
inB.1.1.0.1, rank of the sum A +B is equivalent to the number of linearly<br />
independent dyads constituting it. Linear independence is insensitive to<br />
further positive scaling by µ . The assumption of positive semidefiniteness<br />
prevents annihilation of any dyad from the sum A +B . <br />
2.9.2.6.2 Example. Rank function quasiconcavity. (confer3.3)<br />
For A,B ∈ R m×n [176,0.4]<br />
that follows from the fact [287,3.6]<br />
rankA + rankB ≥ rank(A + B) (230)<br />
dim R(A) + dim R(B) = dim R(A + B) + dim(R(A) ∩ R(B)) (231)<br />
For A,B ∈ S M +<br />
rankA + rankB ≥ rank(A + B) ≥ min{rankA, rankB} (1370)<br />
that follows from the fact<br />
N(A + B) = N(A) ∩ N(B) , A,B ∈ S M + (146)<br />
Rank is a quasiconcave function on S M + because the right-hand inequality in<br />
(1370) has the concave form (578); videlicet, Lemma 2.9.2.6.1. <br />
From this example we see, unlike convex functions, quasiconvex functions<br />
are not necessarily continuous. (3.3) We also glean:<br />
⋄