No abstract

2016-12-26T15:11:00

Given linear matrix inequalities (LMIs) L1 and L2 it is natural to ask:(Q1) when does one dominate the other, that is, does L1(X) 0 imply L2(X) 0? (Q2) when are they mutually dominant, that is, when do they have the same solution set?The matrix cube problem of Ben-Tal and Nemirovski [B-TN02] is an example of LMI domination. Hence such problems can be NP-hard. This paper describes a natural relaxation of an LMI, based on substituting matrices for the variables xj. With this relaxation, the domination questions (Q1) and (Q2) have elegant answers, indeed reduce to constructible semidefinite programs. As an example, to test the strength of this relaxation we specialize it to the matrix cube problem and obtain essentially the relaxation given in . Thus our relaxation could be viewed as generalizing it.Assume there is an X such that L1(X) and L2(X) are both positive definite, and suppose the positivity domain of L1 is bounded. For our "matrix variable" relaxation a positive answer to (Q1) is equivalent to the existence of matrices Vj such thatAs for (Q2) we show that L1 and L2 are mutually dominant if and only if, up to certain redundancies described in the paper, L1 and L2 are unitarily equivalent. Algebraic certificates for positivity, such as (A1) for linear polynomials, are typically called Positivstellensätze. The paper goes on to derive a Putinar-type Positivstellensatz for polynomials with a cleaner and more powerful conclusion under the stronger hypothesis of positivity on an underlying bounded domain of the form {X | L(X) 0}.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". Complete positivity is one of the main techniques of modern operator theory and the theory of operator algebras. On one hand it provides tools for studying LMIs and on the other hand, since completely positive maps are not so far from representations and generally are more tractable than their merely positive counterparts, the theory of completely positive maps provides perspective on the difficulties in solving LMI domination problems.

A theorem of Beurling Lax Halmos represents a subspace M of H 2 C (D) the Hardy space of analytic functions with values in the Hilbert space E and square summable power series invariant for multiplication by z as 8H 2 F , where F is a subspace of E and 8 is an inner function with values in L(F, E). When the Hardy space is replaced by the Hilbert space H(k) determined by a Nevanlinna Pick kernel k, such as the Dirichlet kernel or the row contraction kernel on the ball in C d , the BLH Theorem survives with F an auxiliary Hilbert space and 8 a L(F, E) valued function which is inner in the sense that the operator M 8 of multiplication by 8 is a partial isometry. Under mild additional hypotheses, when E=C, M z , the operator of multiplication by z, is cellularly indecomposable and has the codimension one property; however, if M is invariant for M z , M M z M need not be a cyclic subspace for M z restricted to M. Academic Press

Abstract. A non-commutative polynomial which is positive on a bounded semi-algebraic set of operators has a weighted sum of squares representation. This Positivstellensatz parallels similar results in the commutative case.A broader issue is, to what extent does real semi-algebraic geometry extend to non-commutative polynomials? Our "strict" Positivstellensatz is positive news, on the opposite extreme from strict positivity would be a Real Nullstellensatz. We give an example which shows that there is no non-commutative Real Nullstellensatz along certain lines. However, we include a successful type of non-commutative Nullstellensatz proved by George Bergman.