The Copositive-Plus Matrix Completion Problem, Xuhua Liu

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The Copositive-Plus Matrix Completion Problem, Xuhua Liu Alabama Journal of Mathematics ISSN 2373-0404 39 (2015) The Copositive-Plus Matrix Completion Problem Xuhua Liu Ronald L. Smith Department of Mathematics Department of Mathematics The University of Tennessee at Chattanooga The University of Tennessee at Chattanooga Chattanooga, TN 37403, USA Chattanooga, TN 37403, USA We derive a useful characterization of 3-by-3 copositive-plus matrices in terms of their entries, and use it to solve the copositive-plus matrix completion problem in the case of specified diagonal by showing that a partial copositive-plus matrix with graph G has a copositive-plus completion if and only if G is a pairwise disjoint union of complete subgraphs. Introduction (cf. Hiriart-Urruty and Seeger (2010)). See recent surveys Hiriart-Urruty and Seeger (2010) and Ikramov and Savel’eva In this article, the superscript “>” denotes transposi- n (2000) on copositive matrices and references therein. tion and R the set of all real n-vectors. A vector x = A partial matrix is one in which some entries are spec- > 2 n (x1;:::; xn) R is said to be nonnegative, denoted by ified, while the remaining entries are unspecified and free x > 0, if xi > 0 for all i = 1;:::; n. to be chosen. A completion of a partial matrix is a choice × Let S n denote the set of all n n real symmetric matrices. of values for the unspecified entries. A matrix completion 2 A matrix A S n is said to be problem asks which partial matrices have completions with + ∗ > n a desired property. A partial C (C, C , resp.) matrix is a (1) (real) positive semidefinite if x Ax > 0 for all x 2 R ; real symmetric partial matrix such that every fully specified > + ∗ ∗ (2) copositive if x Ax > 0 for all x > 0; principal submatrix is C (C, C , resp.). The C and C ma- trix completion problems were solved in Hogben, Johnson, (3) copositive-plus if A is copositive and if x>Ax = 0 with and Reams (2005) and Hogben (2007), but the C+ matrix x > 0 implies Ax = 0; completion problem has remained open. + > Our main interest here is in the following C matrix com- (4) strictly copositive if x Ax > 0 for all nonzero x > 0. pletion problem: Under the assumption that the main diago- Positive semidefinite matrices are copositive by definition. nal is specified, which patterns for the specified entries of a + Moreover, if A 2 S is positive semidefinite and x 2 n, partial C matrix ensure that each partial matrix with one of n R + then x>Ax = 0 if and only if Ax = 0 (Horn & Johnson, these patterns can be completed to a C matrix? 1985, p.400, Problem 1). A parallel comparison of positive In Section 2, we give a new characterization of 3 × 3 semidefinite matrices and copositive-plus matrices was made copositive-plus matrices in terms of their entries. In Sec- + in Cottle, Habetler, and Lemke (1970b). tion 3, this result is used to solve the C matrix completion Following Cottle, Habetler, and Lemke (1970a), we de- problem listed above. note the class of copositive (copositive-plus, strictly copos- A characterization of copositive-plus matrices of order itive, resp.) matrices by C (C+, C∗, resp.). Obviously, n 3 C∗ ⊂ C+ ⊂ C. Each of the three classes of copositive matri- 6 ces has three important properties: inheritance, closure un- We begin with the following characterization of 2 × 2 der permutation similarity, and closure under positive diag- copositive-plus matrices in terms of their entries. We omit onal congruence, i.e., if S denotes any class of copositive the proof since it follows from straightforward computation. matrices and A 2 S , then every principal submatrix of A is in " # > a b S , P AP 2 S for any permutation matrix P, and DAD 2 S Lemma 1. The real symmetric matrix is for any positive diagonal matrix D. b d Copositive matrices have applications in control theory, copositive-plus if and only if the following conditions are sat- optimization modeling, linear complementarity problems, isfied: and many other branches of pure and applied mathematics (1)a > 0 and d > 0, (2a)b = 0 when ad = 0, p Corresponding Author Email: [email protected] (2b)b > − ad when ad > 0. 1 2 XUHUA LIU & RONALD L. SMITH Moreover,p it is strictly copositive if and only if a > 0, d > 0, If α + β = 0, then Q(v) = 0 if and only if and b > − ad. αx + y − z = 0 and (1 − α2)x = 0 Combining (Hadeler, 1983, Theorem 4) and (Simpson & i.e., if and only if Spector, 1983, Theorem 2.2), we have the following charac- 8 terization of 3 × 3 copositive matrices, which will be useful <>x + α(y − z) = 0 if α = ±1 > in characterizing 3 × 3 copositive-plus matrices in terms of :>x = 0 and y = z if α , ±1 their entries. i.e., if and only if Av = 0. Hence, A 2 C+. 2 3 6 a b s 7 On the other hand, if α + β , 0, then Q(v) = 0 and 6 7 > > Lemma 2. The real symmetric matrix A = 6 b d c 7 is Av = (α + β, 0; 0) for v = (0; 1; 1) , hence A < C+. Note 46 57 s c e that α + β = 0 and α, β > −1 imply −1 6 α, β 6 1. copositive if and only if the following conditions are satisfied: Case 2: α, β, γ > −1. We claim that A 2 C+ if and only if 2 ∗ (1)a 0, d 0, e 0, (1) holds. Since A C if and only if strict inequality in (1) > > > 2 + p p p holds, it remains to show that A C if − − − p (2)b > ad, c > de, s > ae, 1 + α + β + γ + 2(1 + α)(1 + β)(1 + γ) = 0; (2) p p p p (3) ade + b e + c a + s d + which implies that 1 + α + β + γ 0. Since we assume q p p p 6 α, β, γ > −1, it follows that + 2(b + ad)(c + de)(s + ae) > 0. − 1 < α, β, γ < 1: (3) Moreover, A is strictly copositive if and only if all thep above conditionsp p are satisfiedp with strict inequality. If ade + Moreover, we have det A = 0 by Lemma 2. Now that b e + c a + s d 0, Condition (3) is equivalent to 6 1 + 2αβγ − α2 − β2 − γ2 = det A = 0; (4) det A > 0. we then have The following characterization of 3 × 3 copositive-plus q matrices with unit diagonal will be crucial in the proof of αγ − β = (1 − α2)(1 − γ2) > 0; (5) the main results. q αβ − γ = (1 − α2)(1 − β2) 0; (6) 2 1 α β 3 > 6 7 q Lemma 3. The matrix A = 6 α 1 γ 7 is copositive-plus 6 7 βγ − α = (1 − β2)(1 − γ2) 0: (7) 4 β γ 1 5 > if and only if By solving the linear equation Av = 0, we know that every solution is a scalar multiple of (αγ − β, αβ − γ; 1 − α2)>, i.e., 2 1 α −α 3 p p p 6 7 a scalar multiple of ( 1 − γ2; 1 − β2; 1 − α2)>. (1) it is permutation similar to 6 α 1 −1 7 with 6 7 Now we write 4 −α −1 1 5 2 2 2 −1 6 α 6 1, or Q(v) = x + y + z + 2αxy + 2βxz + 2γyz 0 12 0 12 (2) α, β, γ > −1 and B x y C B x z C p = p B p − p C + q B p − p C 1 + α + β + γ + 2(1 + α)(1 + β)(1 + γ) > 0. @ 1 − γ2 1 − β2 A @ 1 − γ2 1 − α2 A 0 12 Proof. Since C+ ⊂ C, by Lemma 2 we may assume that B y z C α, β, γ −1 and + r B p − p C > @ 1 − β2 1 − α2 A p 2 2 2 1 + α + β + γ + 2(1 + α)(1 + β)(1 + γ) > 0: (1) =: p(x ˆ − yˆ) + q(x ˆ − zˆ) + r(ˆy − zˆ) ; > > where by (4), (5), (6), and (7) Denote Q(v) =: v Av for v = (x; y; z) > 0. We consider the following two cases: q 1 + α2 − β2 − γ2 p = −α (1 − β2)(1 − γ2) = α(α − βγ) = ; Case 1: at least one of α, β, γ is −1. Then after applying 2 permutation similarity, we can assume γ = −1. Then (1) (8) + reduces to α + β > 0. We claim that A 2 C if and only if q 1 + β2 − α2 − γ2 α + β = 0. Note that q = −β (1 − α2)(1 − γ2) = β(β − αγ) = ; 2 (9) Q(v) = x2 + y2 + z2 + 2αxy + 2βxz − 2yz q 1 + γ2 − α2 − β2 = x2 + (y − z)2 + 2αx(y − z) + 2(α + β)xz r = −γ (1 − α2)(1 − β2) = γ(γ − αβ) = : 2 = (αx + y − z)2 + (1 − α2)x2 + 2(α + β)xz: (10) THE COPOSITIVE-PLUS MATRIX COMPLETION PROBLEM 3 2 3 From (2) and (3), we see that at most one of α, β, γ is non- a b s 6 7 negative. If −1 < α, β, γ 0, then p; q; r 0 and hence Proof. Let A = 6 b d c 7 be real symmetric with non- 6 > 46 57 Q(v) = 0 if and only ifx ˆ = yˆ = zˆ, i.e., Av = 0, hence A 2 C+. s c e If exactly one of α, β, γ is positive, say −1 < β, γ < 0 < α < 1 negative diagonal entries.
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