13.07.2015 Views

v2006.03.09 - Convex Optimization

v2006.03.09 - Convex Optimization

v2006.03.09 - Convex Optimization

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

5.6. CORRESPONDENCE TO PSD CONE S N−1+ 3275.6.0.0.1 Proof. Case r = 1 is easily proved: From the nonnegativitydevelopment in4.8.1, extreme direction (751), and Schoenberg criterion(479), we need show only sufficiency; id est, proverankV T N DV N =1D ∈ S N h ∩ R N×N+}⇒D ∈ EDM ND is an extreme directionAny symmetric matrix D satisfying the rank condition must have the form,for z,q ∈ R N and nonzero z ∈ N(1 T ) ,because (4.6.2.1, conferE.7.2.0.2)D = ±(1q T + q1 T − 2zz T ) (761)N(V N (D)) = {1q T + q1 T | q ∈ R N } ⊆ S N (762)Hollowness demands q = δ(zz T ) while nonnegativity demands choice ofpositive sign in (761). Matrix D thus takes the form of an extreme direction(751) of the EDM cone. The foregoing proof is not extensible in rank: An EDMwith corresponding affine dimension r has the general form, for{z i ∈ N(1 T ), i=1... r} an independent set,( r)∑ T ( r)∑ ∑D = 1δ z i zi T + δ z i ziT 1 T − 2 r z i zi T ∈ EDM N (763)i=1i=1i=1The EDM so defined relies principally on the sum ∑ z i ziT having positivesummand coefficients (⇔ −VN TDV N ≽0) 5.6 . Then it is easy to find asum incorporating negative coefficients while meeting rank, nonnegativity,and symmetric hollowness conditions but not positive semidefiniteness onsubspace R(V N ) ; e.g., from page 280,⎡−V ⎣0 1 11 0 51 5 05.6 (⇐) For a i ∈ R N−1 , let z i =V †TN a i .⎤⎦V 1 2 = z 1z T 1 − z 2 z T 2 (764)

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!