Constructing Copositive Matrices from Interior Matrices∗
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Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 17, pp. 9-20, January 2008 ELA CONSTRUCTING COPOSITIVE MATRICES FROM INTERIOR MATRICES∗ CHARLES R. JOHNSON† AND ROBERT REAMS‡ Abstract. Let A be an n by n symmetric matrix with real entries. Using the l1-norm for vectors + n and letting S1 = {x ∈ R |||x||1 =1,x ≥ 0},thematrixA is said to be interior if the quadratic form T + x Ax achieves its minimum on S1 in the interior. Necessary and sufficient conditions are provided for a matrix to be interior. A copositive matrix is referred to as being exceptional if it is not the sum of a positive semidefinite matrix and a nonnegative matrix. A method is provided for constructing exceptional copositive matrices by completing a partial copositive matrix that has certain specified overlapping copositive interior principal submatrices. Key words. Copositive matrix, Interior matrix, Quadratic form, Almost positive semidefinite, Exceptional copositive matrix, Extreme copositive matrix. AMS subject classifications. 15A18, 15A48, 15A57, 15A63. T n 1. Introduction. We will call a vector v =(v1,...,vn) ∈ R nonnegative, n×n denoted v ≥ 0, if vi ≥ 0, for all i,1≤ i ≤ n. Similarly, a matrix A ∈ R will be called nonnegative, in the event that all its entries are nonnegative. A symmetric matrix A ∈ Rn×n will be called positive semidefinite (positive definite)ifxT Ax ≥ 0 for all x ∈ Rn (xT Ax > 0 for all x ∈ Rn,x= 0). A symmetric matrix A ∈ Rn×n will be called copositive (strictly copositive)ifxT Ax ≥ 0 for all x ∈ Rn,x≥ 0(xT Ax > 0 for all x ∈ Rn,x ≥ 0,x = 0). The vector in Rn of all ones will be denoted by e,so T n n×n that e =(1, 1,...,1) ∈ R .ThematrixEij ∈ R denotes the matrix with a 1 in n the (i, j) position and zeroes elsewhere, while ei ∈ R denotes the vector with a 1 in the ith position and zeroes elsewhere. A symmetric matrix A ∈ Rn×n will be called almost positive semidefinite (almost positive definite)ifxT Ax ≥ 0 for all x ∈ Rn such that xT e =0(xT Ax > 0 for all x ∈ Rn, x = 0, such that xT e = 0)[13]. Note that if a matrix is copositive then all its principal submatrices are copositive. We will say that a matrix is partial copositive if it has some entries that are specified, and some that are not, and for every principal submatrix which has all its entries specified, this submatrix is copositive. It is easy to show that a 2 × 2 copositive matrix is ∗Received by the editors on August 8, 2007. Accepted for publication on January 9, 2008 Handling Editor: Daniel Hershkowitz. †Department of Mathematics, College of William and Mary, P.O. Box 8795, Williamsburg, VA 23187, USA ([email protected]). ‡Department of Mathematics, Virginia Commonwealth University, 1001 West Main Street, Rich- mond, VA 23284, USA ([email protected]). 9 Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 17, pp. 9-20, January 2008 ELA 10 C.R. Johnson and R. Reams either positive semidefinite or nonnegative. Diananda [7] showed that every 3 × 3or 4 × 4 copositive matrix is the sum of a positive semidefinite matrix and a nonnegative matrix. We will call a copositive matrix exceptional if it is not the sum of a positive semidefinite matrix and a nonnegative matrix. Thus, n × n exceptional copositive matrices can only occur for n ≥ 5. Horn [8] gave an example of an exceptional matrix (see Section 3 of this paper). By an extreme copositive matrix Q we mean that if Q = Q1 + Q2,whereQ1 and Q2 are copositive then Q1 = aQ,andQ2 =(1− a)Q, for some a such that 0 ≤ a ≤ 1. n + We will denote the vector l1-norm [12] by ||x||1 = i=1 |xi|,andbyS1 the n portion of the unit 1-sphere S1 = {x ∈ R |||x||1 =1} in the nonnegative orthant, + n n×n thus S1 = {x ∈ R |||x||1 =1,x ≥ 0}.LetA ∈ R be symmetric. The set + T S1 is compact so a minimum for x Ax is achieved on this set. We will say that A T is interior if such a minimum, namely min + x Ax, is achieved in the interior of x∈S1 + S1 .NotethatA is not necessarily copositive in this definition, although A being copositive is our primary interest. This definition does not preclude the possibility that the minimum is also achieved on the boundary, just that it will be achieved in the interior in any case. For example, with n =3and 1 −11 A = −11−1 , 1 −11 the matrix A is interior, since the minimum of the quadratic form xT Ax is achieved + 1 T in the interior of S1 at the vector 4 (1, 2, 1) . Although the minimum is also achieved 1 T n×n on the boundary at 2 (1, 1, 0) . For a symmetric matrix A ∈ R , if the minimum T + of x Ax is not achieved in the interior of S1 then the minimum is achieved at a vector u with some zero components. Let us call u ∈ Rk,where1≤ k<n,the vector consisting of the positive components of u,sothatu is of the form u = u ,...,u T S+ {x ∈ Rk|||x|| ,x ≥ } ( i1 ik ) , and is in the interior of 1 = p =1 0 ,and where i1,...,ik are determined by deleting the zero components of the minimizer u. T T Then minx∈S+ x Ax =min + x A x ,whereA is the principal submatrix of A 1 x ∈S1 obtained by deleting the ith row and column of A, for every i such that i ∈{i1,...,ik}. Thus A is an interior matrix. To be consistent with when n = 1 we will say that the matrix A =(a)isinterior, + T even though S1 = {1} has no interior or boundary. In this case, min + x Ax = x∈S1 2 minx=1 ax = a. Note that for a 1 × 1matrixA, being copositive and not strictly copositive implies that A = (0). For another example, this time with n =2,take 21 T T 2 2 A = .Thenwithx =[x1 x2 ] we have x Ax = x +(x1 + x2) ,so 11 1 T T min + x Ax = 1 and the minimum is achieved at u =[0 1] (in the notation of x∈S1 the preceding paragraph) and A = (1) is an interior matrix and u =1. Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 17, pp. 9-20, January 2008 ELA Constructing Copositive Matrices from Interior Matrices 11 In Section 2 we provide necessary and sufficient conditions for a matrix to be interior. In Section 3 we employ interior principal submatrices to construct coposi- tive matrices that are exceptional. In particular, we will show how to construct an exceptional copositive matrix by performing a completion of a matrix that has certain overlapping principal submatrices specified, which are copositive and interior. Section 4 contains examples that demonstrate this method of construction. + 2. Interior matrices. Theorem 1 below minimizes the quadratic form on S1 . + A variation of Theorem 1 in which the quadratic form is maximized on S1 ,and xT Ax has no maxima on the boundary, was proved by Sidorenko in [17], whose proof relies on [5] and [6]. Our interest lies in minimizing the quadratic form because with copositive matrices, which are not strictly copositive, the value of that minimum on the nonnegative orthant is zero. We are also interested in the location of the minimizer, which may move from the boundary to the interior, or vice versa, on changing from maximizing to minimizing. Our proofs are self-contained and different than Sidorenko’s. Theorem 1. Let A ∈ Rn×n be symmetric. Then A is interior if and only if (i) and (ii) hold. (i) There exists u>0, ||u||1 =1,andµ ∈ R such that Au = µe; (ii) yT Ay ≥ 0 for all y ∈ Rn such that yT e =0(i.e. A is almost positive semidefinite). T T If A is interior then µ =min + x Ax = u Au. x∈S1 If (i) holds and yT Ay > 0 for all y =0 such that yT e =0(i.e. A is almost positive + definite), then u>0 is the only minimizer in S1 . + If A is interior and u is not the only minimizer in S1 then A is singular, and Ay =0 for some y =0 such that yT e =0. Proof. Suppose that A is interior. Since there is a minimizer u for xT Ax in S+ xT Ax the interior of 1 we may use Lagrange Multipliers and minimize subject to T n ||x||1 =1.Thus∇(x Ax)=λ∇( i=1 xi), when x = u, which implies 2(Au)i = λ, λ T T or Au = µe,whereµ = . So (i) holds. Evidently, µ =min + x Ax = u Au. 2 x∈S1 Let y ∈ Rn be any y =0besuchthat yT e =0.Thenforsomec =0, u + cy ≥ 0, (u+cy)T (u+cy) T u A ≥ u Au ||u cy||1 and since is a minimizer we have ||u+cy||1 ||u+cy||1 .But + = (u + cy)T e =1,sowehaveuT Au +2cyT Au + c2yT Ay ≥ uT Au,orc2yT Ay ≥ 0, proving (ii). + Conversely, suppose (i) and (ii) hold. Let z be any vector in S1 . Since the vector u together with a basis for {y|yT e =0} spans Rn we can write z = νu + y,where Electronic Journal of Linear Algebra ISSN 1081-3810 A publication of the International Linear Algebra Society Volume 17, pp.