Timeline for What is the dual of an semidefinitely representable (SDR) cone?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2013 at 0:47 | history | edited | Noah Stein | CC BY-SA 3.0 |
added 118 characters in body
|
| Apr 1, 2013 at 0:35 | comment | added | Noah Stein | You are right. I made a mistake in thinking that an element of the interior of $\mathcal{K}$ would make the constraint strictly positive definite, which of course it does not. I believe the argument that $\mathcal{K}^*$ iS SDR still goes through if you use Ramana's extended dual (which does not require a Slater-type constraint qualification), but this might give you slightly different operators in cases where Slater's condition does not hold. | |
| Mar 31, 2013 at 7:16 | comment | added | Alex Monras | I worked out the dual and its feasibility region gives me the $C$ cone indeed, as I expected. However, the dual only proves inclusion, and to prove equality one needs to invoke strong duality. I would get strong duality if it were clear that $\mathrm{span}\{X_i,Y_j\}$ intersects with the interior of the PSD cone. For a spectrahedron, such assumption can always be made (Lemma 2.3, arXiv:math/0306180) by restricting to the right subspace, but for an SDR I cannot see this. I am very curious to understand how you see it so clearly. You surely are invoking a more general Slater condition than mine. | |
| Mar 20, 2013 at 11:41 | comment | added | Noah Stein | There should be a } at the end of the second sentence. | |
| Mar 20, 2013 at 11:41 | comment | added | Noah Stein | They are equivalent in the homogeneous case. Define $\mathcal{K}^\circ = \{y \mid x\cdot y \leq 1\forall x\in\mathcal{K}$. If $x\in\mathcal{K}$ then $cx\in\mathcal{K}$ for all $c\geq 0$ by homogeneity. Thus for any $y\in\mathcal{K}^\circ$ and $x\in\mathcal{K}$ we actually have $cx\cdot y\leq 1$ for all $c\geq 0$. That is, $x\cdot y\leq 0$. So in your notation $\mathcal{K}^* = -\mathcal{K}^\circ$. | |
| Mar 20, 2013 at 11:14 | comment | added | Alex Monras | Thanks for the reference. Very useful indeed. I am going through it, but it seems that they use a different notion of duality $\mathcal K^\circ=\\{x'|x\cdot x'\leq 1~\forall x\in\mathcal K\\}$? Are they related? (for the homogeneous case!) | |
| Mar 19, 2013 at 18:19 | history | edited | Noah Stein | CC BY-SA 3.0 |
added 793 characters in body
|
| Mar 19, 2013 at 18:04 | history | answered | Noah Stein | CC BY-SA 3.0 |