Relaxing LMI domination matricially

Given linear matrix inequalities (LMIs) L<inf>1</inf> and L<inf>2</inf> in the same number of variables it is natural to ask: (Q<inf>1</inf>) 1) does one dominate the other, that is, does L<inf>1</inf>(X) ≽ 0 imply L<inf>2</inf>(X) ≽ 0? 2) are they mutually dominant, that is, do they have the same solution set? Such problems can be NP-hard. We describe a natural relaxation of an LMI, based on substituting matrices for the variables x<inf>j</inf>. With this relaxation, the domination questions (Q<inf>1</inf>) and (Q<inf>2</inf>) have elegant answers, indeed reduce to semidefinite programs (SDP) which we show how to construct. For our “matrix variable” relaxation a positive answer to (Q<inf>1</inf>) is equivalent to the existence of matrices V<inf>j</inf> such that L<inf>2</inf>(x) = V<inf>1</inf>*L<inf>1</inf>(x)V<inf>1</inf> + … + V<inf>μ</inf>*L<inf>1</inf>(x)V<inf>μ</inf>. (A<inf>1</inf>) As for (Q<inf>2</inf>) we show that L<inf>1</inf> and L<inf>2</inf> are mutually dominant if and only if, up to certain redundancies described in the paper, L<inf>1</inf> and L<inf>2</inf> are unitarily equivalent. An observation at the core of the paper is that the relaxed LMI domination problem is equivalent to a classical problem. Namely, the problem of determining if a linear map τ from a subspace of matrices to a matrix algebra is “completely positive”.

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