v2009.01.01 - Convex Optimization

v2009.01.01 - Convex Optimization v2009.01.01 - Convex Optimization

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274 CHAPTER 4. SEMIDEFINITE PROGRAMMING 4.3.3.0.1 Example. Aδ(X) = b . This academic example demonstrates that a solution found by rank reduction can certainly have rank less than Barvinok’s upper bound (245): Assume a given vector b∈ R m belongs to the conic hull of columns of a given matrix A∈ R m×n ; ⎡ ⎤ ⎡ ⎤ A = ⎢ ⎣ −1 1 8 1 1 −3 2 8 1 2 −9 4 8 1 4 1 3 1 9 Consider the convex optimization problem ⎥ ⎦ , b = ⎢ ⎣ 1 1 2 1 4 ⎥ ⎦ (658) minimize trX X∈ S 5 subject to X ≽ 0 Aδ(X) = b (659) that minimizes the 1-norm of the main diagonal; id est, problem (659) is the same as minimize X∈ S 5 ‖δ(X)‖ 1 subject to X ≽ 0 Aδ(X) = b (660) that finds a solution to Aδ(X)=b. Rank-3 solution X ⋆ = δ(x M ) is optimal, where (confer (618)) ⎡ 2 ⎤ 128 0 x M = 5 ⎢ 128 ⎥ (661) ⎣ 0 ⎦ 90 128 Yet upper bound (245) predicts existence of at most a (⌊√ ⌋ ) 8m + 1 − 1 rank- = 2 2 (662) feasible solution from m = 3 equality constraints. To find a lower rank ρ optimal solution to (659) (barring combinatorics), we invoke Procedure 4.3.1.0.1:

4.3. RANK REDUCTION 275 Initialize: C =I , ρ=3, A j ∆ = δ(A(j, :)) , j =1, 2, 3, X ⋆ = δ(x M ) , m=3, n=5. { Iteration i=1: ⎡ Step 1: R 1 = ⎢ ⎣ √ 2 128 0 0 0 0 0 0 √ 5 0 128 0 0 √ 0 0 0 90 128 ⎤ . ⎥ ⎦ find Z 1 ∈ S 3 subject to 〈Z 1 , R T 1A j R 1 〉 = 0, j =1, 2, 3 (663) A nonzero randomly selected matrix Z 1 having 0 main diagonal is feasible and yields a nonzero perturbation matrix. Choose, arbitrarily, Z 1 = 11 T − I ∈ S 3 (664) then (rounding) ⎡ B 1 = ⎢ ⎣ Step 2: t ⋆ 1= 1 because λ(Z 1 )=[−1 −1 ⎡ X ⋆ ← δ(x M ) + B 1 = ⎢ ⎣ 0 0 0.0247 0 0.1048 0 0 0 0 0 0.0247 0 0 0 0.1657 0 0 0 0 0 0.1048 0 0.1657 0 0 2 ] T . So, ⎤ ⎥ ⎦ 2 0 0.0247 0 0.1048 128 0 0 0 0 0 0.0247 0 5 128 0 0.1657 0 0 0 0 0 0.1048 0 0.1657 0 90 128 ⎤ (665) ⎥ ⎦ (666) has rank ρ ←1 and produces the same optimal objective value. }

4.3. RANK REDUCTION 275<br />

Initialize:<br />

C =I , ρ=3, A j ∆ = δ(A(j, :)) , j =1, 2, 3, X ⋆ = δ(x M<br />

) , m=3, n=5.<br />

{<br />

Iteration i=1:<br />

⎡<br />

Step 1: R 1 =<br />

⎢<br />

⎣<br />

√<br />

2<br />

128<br />

0 0<br />

0 0 0<br />

0<br />

√<br />

5<br />

0<br />

128<br />

0 0<br />

√<br />

0<br />

0 0<br />

90<br />

128<br />

⎤<br />

.<br />

⎥<br />

⎦<br />

find Z 1 ∈ S 3<br />

subject to 〈Z 1 , R T 1A j R 1 〉 = 0, j =1, 2, 3<br />

(663)<br />

A nonzero randomly selected matrix Z 1 having 0 main diagonal<br />

is feasible and yields a nonzero perturbation matrix. Choose,<br />

arbitrarily,<br />

Z 1 = 11 T − I ∈ S 3 (664)<br />

then (rounding)<br />

⎡<br />

B 1 =<br />

⎢<br />

⎣<br />

Step 2: t ⋆ 1= 1 because λ(Z 1 )=[−1 −1<br />

⎡<br />

X ⋆ ← δ(x M<br />

) + B 1 =<br />

⎢<br />

⎣<br />

0 0 0.0247 0 0.1048<br />

0 0 0 0 0<br />

0.0247 0 0 0 0.1657<br />

0 0 0 0 0<br />

0.1048 0 0.1657 0 0<br />

2 ] T . So,<br />

⎤<br />

⎥<br />

⎦<br />

2<br />

0 0.0247 0 0.1048<br />

128<br />

0 0 0 0 0<br />

0.0247 0<br />

5<br />

128<br />

0 0.1657<br />

0 0 0 0 0<br />

0.1048 0 0.1657 0<br />

90<br />

128<br />

⎤<br />

(665)<br />

⎥<br />

⎦ (666)<br />

has rank ρ ←1 and produces the same optimal objective value.<br />

}

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