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v2009.01.01 - Convex Optimization

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7.3. THIRD PREVALENT PROBLEM: 529<br />

7.3.1.4 Dual interpretation, projection on EDM cone<br />

FromE.9.1.1 we learn that projection on a convex set has a dual form. In<br />

the circumstance K is a convex cone and point x exists exterior to the cone<br />

or on its boundary, distance to the nearest point Px in K is found as the<br />

optimal value of the objective<br />

‖x − Px‖ = maximize a T x<br />

a<br />

subject to ‖a‖ ≤ 1 (1892)<br />

a ∈ K ◦<br />

where K ◦ is the polar cone.<br />

Applying this result to (1287), we get a convex optimization for any given<br />

symmetric matrix H exterior to or on the EDM cone boundary:<br />

maximize 〈A ◦ , H〉<br />

minimize ‖D − H‖ 2 A ◦<br />

F<br />

D<br />

subject to D ∈ EDM ≡ subject to ‖A ◦ ‖ N F ≤ 1 (1298)<br />

A ◦ ∈ EDM N◦<br />

Then from (1894) projection of H on cone EDM N is<br />

D ⋆ = H − A ◦⋆ 〈A ◦⋆ , H〉 (1299)<br />

Critchley proposed, instead, projection on the polar EDM cone in his 1980<br />

thesis [78, p.113]: In that circumstance, by projection on the algebraic<br />

complement (E.9.2.2.1),<br />

which is equal to (1299) when A ⋆ solves<br />

D ⋆ = A ⋆ 〈A ⋆ , H〉 (1300)<br />

maximize 〈A , H〉<br />

A<br />

subject to ‖A‖ F = 1<br />

(1301)<br />

A ∈ EDM N<br />

This projection of symmetric H on polar cone EDM N◦ can be made a convex<br />

problem, of course, by relaxing the equality constraint (‖A‖ F ≤ 1).

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