M. Seetharama Gowda. J. Tao. G. Ravindran

Some complementarity properties of Z and Lyapunov-like transformations on symmetric cones M. Seetharama Gowda Department of Mathematics and Statistics, University of Maryland, Baltimore County, Baltimore, MD 2250, USA email: gowda@math.umbc.edu J. Tao Department of Mathematical Sciences, Loyola University Maryland, Baltimore, MD 220, USA email: jtao@loyola.edu G. Ravindran Indian Statistical Institute, Chennai, India email: gomatamr@hotmail.com This article deals with some complementarity properties of Z and Lyapunov-like transformations on a symmetric cone. Similar to the results proved for Lyapunov and Stein transformations on Herm(R n n ), we show that for Lyapunov-like transformations, the P and S properties are equivalent, and that for a Z- transformation, the S-property implies the P-property of a transformation that is (quadratically) similar to the given transformation. In addition, we show that a Lyapunov-like transformation is the sum of a Lyapunov transformation and a derivation. Key words: Euclidean Jordan algebra; symmetric cone; complementarity problems; Z and Lyapunov-like transformations; P and GUS properties MSC2000 Subject Classification: Primary: 90C33, 7C55; Secondary: 5A48 OR/MS subject classification: Primary: Programming; Secondary: Mathematics. Introduction Given a closed convex set K in a (finite dimensional) real Hilbert space (H,, ), a function f : K H, and an element q H, the variational inequality problem VI(f, K, q) is to find an x H such that x K and f(x ) + q, x x 0 for all x K. (.) This problem has been well studied in the literature and appears in numerous applications, see [6], [9], [8]. When K is a closed convex cone, the above problem reduces to a cone complementarity problem: Find x H such that x K, f(x ) + q K, and x, f(x ) + q = 0, (.2) where K := {x H : x, y 0, y K} is the dual of K. This further reduces to the so-called cone linear complementarity problem (cone LCP) when f is linear and to the nonlinear complementarity problem (NCP) when H = R n with the usual inner product and K = R+. n Two important cone LCPs are the standard LCP (this is NCP with f linear) and semidefinite LCP (when H is the space of all n n real symmetric matrices and K is the cone of positive semidefinite matrices in H). One of the most basic and longstanding problems in this area is to characterize (or describe conditions on) f and K so that VI(f, K, q) has a unique solution for all q. Three well known uniqueness results/conditions are the following: () f is strongly monotone and K is arbitrary (Theorem 2.3.3, [6]). (2) In the setting of standard LCP, f(x) = Mx, where M is a P-matrix, that is, all principal minors of M are positive, or equivalently, x Mx 0 x = 0 [8], [0]. (3) K is polyhedral, f is linear and coherently oriented [20]. 0

Gowda, Tao, Ravindran: Complementarity properties Going beyond polyhedral sets/cones, in recent times, some effort has been made to understand this uniqueness problem over symmetric cones in Euclidean Jordan algebras, in particular over the semidefinite cone and the Lorentz (or second order) cone. Item (2) above has the following analogue in the symmetric cone setting [4]: Let V be a Euclidean Jordan algebra with the corresponding symmetric cone K, L : V V be linear. Then VI(L, K, q), which is written as LCP(L, K, q), has a unique solution for all q V in which case, we say that L has the globally uniquely solvable (GUS) property if and only if L has the P-property, that is, [x and L(x) operator commute, and x L(x) 0 ] x = 0; L has the cross-commutative property, that is, for any q and any two solutions x and x 2 of LCP(L, K, q), x operator commutes with L(x 2 ) + q and x 2 operator commutes with L(x ) + q. Although the above cross-commutative property is obvious in the case of the Euclidean Jordan algebra R n and for monotone L [4], it is not easily verifiable in the general case. A uniqueness result avoiding the cross-commutative property is yet to be obtained and, at present, the work has been limited to understanding the P-property for certain classes of linear transformations on symmetric cones. The present paper continues this line of work; we investigate the P-property for the classes of Z and Lyapunov-like transformations. A linear transformation L : V V is said to be a Z-transformation (or said to have the Z-property) if x 0, y 0, and x, y = 0 L(x), y 0, (.3) and Lyapunov-like if both L and L are Z-transformations. The Z-transformations, the negative of which are called cross-positive transformations in [22], are generalizations of Z-matrices and have numerous complementarity properties, see the recent paper [5]. In particular, it was shown in [5] that for a Z-transformation L, the following are equivalent, and moreover, they hold when, additionally, L has the P-property: (a) L has the Q-property, that is, for all q V, LCP(L, K, q) has a solution. (b) L has the S-property, that is, there exists d > 0 such that L(d) > 0. (c) L has the positive stable property, that is, all eigenvalues of L have positive real parts. (d) L is invertible with L (K) K. This paper is concerned with the converse issue: Whether these conditions imply the P-property for Z-transformations. Our motivation comes from the following particular cases: (i) On the algebra R n, a Z-matrix has the P-property if and only if it has the S-property, see Theorem 3..0, [2]. (ii) On the Jordan spin algebra L n, a Z-transformation has the P-property if and only if it has the S-property, see Theorem 3, [5]. (iii) On the algebra Herm(R n n ) of all n n real symmetric matrices, the Lyapunov transformation L A, defined by L A (X) = AX + XA T (for a given n n real matrix A), has the P-property if and only if it has the S-property [2]. (We note that L A has the Lyapunov-like property and the S-property is related to the asymptotic stability of the continuous linear dynamical system ẋ + Ax = 0.) (iv) On the algebra Herm(R n n ), the Stein transformation S A, defined by S A (X) = X AXA T (for a given n n real matrix A), has the P-property if and only if it has the S-property []. (We note that S A has the Z-property and the S-property is related to the asymptotic stability of the discrete linear dynamical system x(k + ) = Ax(k), k =, 2,...)

2 Gowda, Tao, Ravindran: Complementarity properties (v) For a Lyapunov-like transformation L on a Euclidean Jordan algebra, L has the GUSproperty if and only if L is positive stable and positive semidefinite, see Theorem 7., [3]. In this paper, we show that for Lyapunov-like transformations, the P and S properties are equivalent, and that for a Z-transformation, the S-property implies the P-property of transformation (quadratically) similar to the given transformation. In [3], Damm has shown that on Herm(R n n ) (Herm(C n n )), a Lyapunov-like transformation is of the form L A for some A R n n (respectively, A C n n ). Based on Lie algebraic ideas, we show that Lyapunov-like transformations on a general Euclidean Jordan algebra are of the form L = L a + D, where a V and D is a derivation on V. This will allow us to extend Damm s result to Herm(H n n ) (the algebra of all n n Hermitian quaternionic matrices) and describe Lyapunov-like transformations on the Jordan spin algebra L n. Here is an outline of our paper. In Section 2, we briefly recall concepts and notations from the Euclidean Jordan algebra theory. Section 3 deals with the P-property of a Z-transformation. In Section 4, we characterize Lyapunov-like transformations and describe them on various simple Euclidean Jordan algebras. Finally Section 5 deals with the characterization of the P-property of a Lyapunov-like transformation. 2. Euclidean Jordan Algebras Throughout this paper, we let (V,,, ) denote a Euclidean Jordan algebra of rank r. Recall that V is a finite dimensional real Hilbert space with the bilinear Jordan product - written x y - satisfying the following properties for all x, y, z V : (i) x y = y x, (ii) x (x 2 y) = x 2 (x y), and (iii) x y, z = y, x z. We refer to Faraut and Korányi [7], Gowda, Sznajder, and Tao [4], Schmieta and Alizadeh [9] for basic definitions, concepts, and results. In V we write x y if x, y = 0. The symmetric cone of V is the (closed convex self-dual) cone of squares K := {x 2 : x V }. We use the notation x 0 (x > 0) when x K (respectively, x interior(k)). As is well known, any Euclidean Jordan algebra is a product of simple Euclidean Jordan algebras and every simple algebra is isomorphic to one of the following: () The Jordan spin algebra L n ; (2) The algebra Herm(R n n ) of n n real symmetric matrices; (3) The algebra Herm(C n n ) of all n n complex Hermitian matrices; (4) The algebra Herm(H n n ) of all n n quaternion Hermitian matrices; (5) The algebra Herm(O 3 3 ) of all 3 3 octonion Hermitian matrices. Here, the symbols R, C, H, and O, denote, respectively, the set of all real numbers, complex numbers, quaternions, and octonions. In all of the above matrix algebras, the inner product is given by X, Y = Re tr (XY ) and the Jordan product is given by X Y = 2 (XY + Y X). Recall that a set {e, e 2,...,e r } is a Jordan frame in V when each e i is a primitive idempotent, e i e j = δ ij, and r e i = e, where e is the identity element in V. Given any a V, we let L a and P a denote the corresponding Lyapunov transformation and quadratic representation of a on V : L a (x) := a x and P a (x) := 2a (a x) a 2 x. These are linear and self-adjoint on V. In addition, P a (K) K for any a. We say that elements a and b in V operator commute if L a L b = L b L a. It is known that a and b operator commute if and only if they have their spectral decompositions with respect to a common Jordan frame. We recall the spectral and Peirce decompositions of an element in V.

Gowda, Tao, Ravindran: Complementarity properties 3 Theorem 2. (Spectral decomposition; Theorem III..2, [7]) Let V be a Euclidean Jordan algebra with rank r. Then for every x V, there exists a Jordan frame {e,..., e r } and real numbers λ,...,λ r such that x = λ e + + λ r e r. (2.) Here, the numbers λ i are the eigenvalues of x and the expression λ e + + λ r e r is the spectral decomposition (or the spectral expansion) of x. Given (2.), the trace of x is defined by tr(x) = λ + + λ r. It is well known that x, y tr := tr(x y) defines another inner product compatible with the Jordan product. Also, when V is simple, there exists a positive number α such that, x, y = α tr(x y) for all x and y (see, Prop. III.4., [7]). More generally, if V is a product of simple algebras V i (i =, 2,...,N) (where the Jordan product is computed componentwise and the inner product is the sum of the inner products in each V i ), then for two objects x = (x, x 2,...,x N ) and y = (y, y 2,..., y N ) in V, N N N x, y = x i, y i = α i tr(x i y i ) = α i x i, y i tr. (2.2) We also note that x, y tr = N x i, y i tr. (2.3) It follows from these that x 0, y 0, x, y = 0 x i 0, y i 0, and x i, y i = 0 i. Let {e, e 2,...,e r } be a fixed Jordan frame in V. For i, j {, 2,...,r}, consider the eigenspaces V ii := {x V : x e i = x} = Re i and when i j, Then we have the following V ij := {x V : x e i = 2 x = x e j}. Theorem 2.2 (Peirce decomposition; Theorem IV.2., [7]) The space V is the orthogonal direct sum of spaces V ij (i j). Furthermore, V ij V ij V ii + V jj, V ij V jk V ik if i k, and V ij V kl = {0} if {i, j} {k, l} =. Thus, given any Jordan frame {e, e 2,..., e r }, we can write any element x V as x = x i e i + i= i<j where x i R and x ij V ij. We refer to this as the Peirce decomposition of x with respect to the given Jordan frame. We will also use the following notation: For k r, V (e + e 2 + + e k, ) := V ij. x ij i j k From now on, we use a boldface letter to denote either a class of linear transformations or a property satisfied by a linear transformation; for example, L P if and only if L has the P-property.

4 Gowda, Tao, Ravindran: Complementarity properties 3. Z-transformations and the P-property Recall that a linear transformation on a Euclidean Jordan algebra V is a Z-transformation if it satisfies condition (.3). This condition can also be described in other equivalent ways: Theorem 3. The following are equivalent: (i) L has the Z-property. (ii) For every Jordan frame {e, e 2,..., e r } in V, it holds that L(e i ), e j 0 for all i j. (iii) e tl (K) K for all t 0 in R. (iv) Any trajectory of the differential equation ẋ + L(x) = 0 which starts in K stays in K. (v) There exists α n in R and linear transformations S n on V such that S n (K) K and L = lim n (α ni S n ). In the above result, the equivalence of (i) and (ii) is given in [5] and the equivalence of (i), (iii), and (iv) can be found in [22], [5]. Our primary example of a Z-transformation is L = I S, where S is a linear transformation satisfying S(K) K. If L and L are both Z-transformations, then L is said to be Lyapunov-like; an example is the Lyapunov transformation L A on Herm(R n n ). In Section 4, we characterize such transformations. The result below shows that the Z-property is independent of the inner product. Theorem 3.2 Suppose L has the Z-property on the Euclidean Jordan algebra V which carries the inner product,. Then L has the Z-property with respect to the canonical inner product, tr and conversely. Proof. The result is obvious when V is simple, as any inner product on a simple algebra is a positive multiple of the trace inner product. We assume that V is a product of simple algebras; for simplicity, let V = V V 2, where V and V 2 are simple. Assume that L has the Z-property with respect to, and consider x = (x, x 2 ) 0, y = (y, y 2 ) 0 with 0 = x, y tr = x, y tr + x 2, y 2 tr. This implies that x i, y i tr = 0 for i =, 2 and hence (via (2.2)), 0 x (y, 0) 0 with respect to,. Hence L(x), (y, 0) 0, which yields L(x), y tr 0. A similar inequality ensues when we replace the subscript by 2. By adding these inequalities and using (2.3), we get L(x), y tr 0. This proves that L has the Z-property with respect to the trace inner product. The converse statement is proved in a similar way. As mentioned in the Introduction, Z P Z S. The reverse inclusion holds in the algebras R n and L n, and for any Lyapunov transformation L A on Herm(R n n ). We address the general situation here. In the result below, e denotes the unit element in V and V carries the trace inner product, in which case, the norm of any primitive idempotent is one. Theorem 3.3 If L Z and L(e) > 0, then L P. More generally, if L Z S, then there is a c > 0 such that P c LP c P. Proof. First suppose that L Z and L(e) > 0. Let x V be such that x and L(x) operator commute with x L(x) 0.

Gowda, Tao, Ravindran: Complementarity properties 5 Then for some Jordan frame {e, e 2,..., e r } in V we have the spectral decompositions, x = x i e i and L(x) = y i e i, where x i y i 0 for all i =, 2,...,r. Hence, x i L(e i ) = L(x) = which leads to y i e i, x i L(e i ), e j = y j j =, 2,..., r, (3.) as e i s are orthogonal and e j = for all j. Now, in view of Theorem 3.(ii), the matrix A := [a ij ] with a ij = L(e i ), e j is a Z-matrix and A T p = q, where p (q) is the column vector in R r whose components are x i (respectively, y i ). As x i y i 0 for all i =, 2,...,r, we see that p A T p 0, where denotes the Hadamard (or the componentwise) product. Now the assumption L(e) > 0 implies that L(e i ), e j = L(e), e j > 0 j =, 2,...,r. This leads to i= A T > 0 (in R r ), where is the vector in R r with all components equal to one. This means that the Z-matrix A T is also an S-matrix. It follows that, see Theorem 3..0 in [2], A T is a P-matrix and hence p A T p 0 p = 0; This proves that x i = 0 for all i. Hence L has the P-property. Now for the general case, assume that L Z and that there is a d > 0 such that L(d) > 0. Put c := d > 0 and consider the quadratic representation P c. Then P c is self-adjoint, invertible, (P c ) = P c, P c (K) = K, and P c (e) = c 2 = d. It is easily seen that L := Pc LP c is a Z-transformation with L(e) = (P c ) (L(d)) > 0. By the first part of the result, L has the P-property. This completes the proof. Remark. It is not known if in the above result, Pc LP c P can be replaced by L P proving the result Z S = Z P. 4. Lyapunov-like transformations Recall that a linear transformation L on V is Lyapunovlike if both L and L are Z-transformations, i.e., if 0 x y 0 L(x), y = 0. In view of Theorem 3., this condition is equivalent to: or For any Jordan frame {e, e 2,..., e r }, L(e i ), e j = 0 i j e tl (K) K t R. Note that the latter condition can be written as e tl (K) = K for all t R, or e tl Aut(K) t R, (4.) where Aut(K) is the set of all (invertible) linear transformations which map K onto K. Since Aut(K) can be regarded as a matrix group, the above condition immediately says (see Section 7.6 in []) that

6 Gowda, Tao, Ravindran: Complementarity properties L is a Lyapunov-like transformation if and only if it belongs to the Lie algebra of Aut(K). As we shall see below, the above result allows us to characterize Lyapunov-like transformations. We begin with our first result that a Lyapunov-like transformation on a product algebra is a product of Lyapunov-like transformations. Theorem 4. Suppose that V is the product of algebras V k, k =, 2,..., N, and L : V V is Lyapunov-like. Then, identifying V k with {0} V k {0}, we have L(V k ) V k and L restricted to any V k is Lyapunov-like. Proof. For simplicity, let V = V V 2. Consider any 0 x V. Then for any 0 y 2 V 2, 0 (x, 0) (0, y 2 ) 0 in V. By the Lyapunov-like property, L(x, 0), (0, y 2 ) = 0. As y 2 is arbitrary, it follows that the second component of L(x, 0) is zero. This proves that L(x, 0) V {0}. By linearity and our identification, L(V ) V. Similarly, L(V 2 ) V 2. The Lyapunov-like property of L on each V k is easily verified. Remark. If, in the above theorem, we let L k denote the restriction of L to V k, then L(x, x 2,...,x N ) = (L (x ), L 2 (x 2 ),..., L N (x N )), so we may regard the Lyapunov-like transformation L as a product of Lyapunov-like transformations L k. Conversely, if each L k is Lyapunov-like on V k, then L defined above is Lyapunov-like on V. We now relate Lyapunov-like transformations with another class of transformations known as derivations. A linear transformation D on V is said to be a derivation if for all x, y V, D(x y) = D(x) y + x D(y). It is well-known, see Page 36, [7], that D is a derivation on V if and only if e td Aut(V ) t R, where Aut(V ) is the group of all algebra automorphisms of V (these are invertible linear transformations on V that preserve the Jordan product). This means that Der(V ), the set of all derivations, is the Lie algebra corresponding to the group Aut(V ). Recalling that for an element a V, the corresponding Lyapunov transformation L a is defined by L a (x) = a x, any finite linear combination of commutators of the form [L a, L b ] := L a L b L b L a is a derivation, called the inner derivation. It is known that on a Euclidean Jordan algebra, every derivation is inner, see, Prop. VI..2, [7] or Theorem 8, [6]. Here is our characterization result. Theorem 4.2 A linear transformation L on a Euclidean Jordan algebra V is Lyapunov-like if and only if it is of the form L = L a + D, where a V and D is a (inner) derivation. In this situation, L a is the symmetric part of L and D is the skew-symmetric part of L. This result has already been observed in Prop. VIII.2.6, [7], perhaps, in a slightly different form. Here, we sketch its proof based on Lie algebraic ideas. A direct and elementary proof is presented in the Appendix.

Gowda, Tao, Ravindran: Complementarity properties 7 If D is a derivation, we have e td Aut(V ) Aut(K) for all t R and so D is Lyapunov-like. It turns out that every derivation D is also skew-symmetric (that is, D T + D = 0) or equivalently, D(e) = 0. Conversely, suppose L is Lyapunov-like and skew-symmetric. From L + L T = 0, we see that L and L T commute and so I = e t(l+lt) = e tl e tlt = e tl (e tl ) T t R. This says that e tl is an orthogonal transformation on V for all t R. Writing Orth(V ) for the group of all orthogonal transformations on V, we have e tl Orth(V ) Aut(K) t R. Now suppose that V is simple. Then, Orth(V ) Aut(K) = Aut(V ), see Pages 56 and 57 in [7]. Thus, e tl Aut(V ) t R. This means that L Der(V ), see Page 36 in [7]. Now when V is not simple, we can write V as a product of simple algebras V k, k =, 2,..., N. Then L can be written as a product of Lyapunov-like transformations L k on V k. It is easy to verify that each L k is skew-symmetric on V k. It follows from our previous discussion that each L k Der(V k ). Since the Jordan product in V is defined in a componentwise manner, we see that L Der(V ). So, every Lyapunov-like and skew-symmetric transformation is a derivation. Now suppose that L is any Lyapunov-like transformation and let a = L(e). Then L L a is Lyapunov-like and moreover, (L L a )(e) = 0. This implies that L L a is Lyapunov-like and skew-symmetric, and hence a derivation, say, D. This gives L = L a + D. We also note that the symmetric part of L, namely, (L+LT ) 2 is L a and the skew-symmetric part, namely, (L L T ) 2 is D. Remarks. It follows from the above theorem that if L is symmetric and Lyapunov-like, it must be of the form L a for some a V. 4. Lyapunov-like transformations on matrix algebras Let F denote the set of all real numbers/complex numbers/quaternions/octonions. Given any element p F, we write Re(p) for its real part and p for its conjugate. We note that quaternions are noncommutative but associative, while octonions are noncommutative and nonassociative. Still, for any three elements a, b and c in F, we have [4], Re(a) = Re(a), Re(ab) = Re(ba), and Re [a(bc)] = Re [(ab)c]. (4.2) For any A F n n, let tr(a) denote the sum of the diagonal elements of A. Then, for any three matrices A, B and C in F n n, see Prop. V.2., [7], Re tr(a) = Re tr(a ), Re tr(ab) = Re tr(ba), and Re tr (A(BC)) = Re tr ((AB)C), (4.3) where A is the conjugate transpose of A. Now, given A F n n, we define the Lyapunov transformation L A on Herm(F n n ) by The following extends a result of Damm [3]. L A (X) = AX + XA. Theorem 4.3 Let F denote real numbers, complex numbers, or quaternions. A linear transformation L on the Euclidean Jordan algebra Herm(F n n ) is Lyapunov-like if and only if there exists an A F n n such that L = L A.

8 Gowda, Tao, Ravindran: Complementarity properties Proof. Suppose that A F n n and consider X, Y Herm(F n n ) such that X 0, Y 0, and X, Y = 0. Then XY = Y X = 0. (This is well known for F = R or C; see Remark 3 in [7] for F = H.) Now, relying on the associativity in F, and using (4.3), L A (X), Y = Re tr(l A (X)Y ) = Re tr(axy + XA Y ) = Re tr(xa Y ) = Re tr(a Y X) = 0. This proves the Lyapunov-like property of L A on Herm(F n n ). Now for the converse. Suppose L is Lyapunov-like on Herm(F n n ). By the previous theorem, L = L A +D, where A Herm(F n n ) and D is a derivation. As D is inner, we can write m D = α i [L Ai, L Bi ], where α i are real numbers and A i, B i Herm(F n n ), i =, 2,..., m. Using the associativity of the ordinary matrix product of matrices in F n n, [L Ai, L Bi ] = L [Ai,B i]. It follows that D = L B, where B := m α i[a i, B i ] F n n. Hence, L = L A + L B = L A+B = L C, where C := A + B F n n. This completes the proof. We next show that a result of the previous type is not valid for matrices over octonions. Theorem 4.4 There exists A O 3 3 such that L A is not Lyapunov-like on Herm(O 3 3 ). Proof. By Remark 3 in [7], there exists a Jordan frame {E, E 2, E 3 } in Herm(O 3 3 ) such that E E 2 0. Now, E E 2 = 0 E E 2 + E 2 E = 0. Let p a b E E 2 := α q c. β γ r Then p ᾱ β E 2 E = (E E 2 ) = ā q γ. b c r From E E 2 + E 2 E = 0, we get Re(p) = Re(q) = Re(r) = 0, a + ᾱ = 0, b + β = 0, and c + γ = 0. We will construct an octonion matrix a a 2 a 3 A := a 2 a 22 a 23 a 3 a 32 a 33 such that L A (E ), E 2 0. As E E 2 0, some row of E E 2 is nonzero. Without loss of generality, assume that the first row [p a b] is nonzero; In this case, we take a ij = 0 for i {2, 3} and j {, 2, 3}. Then, using (4.3), and so Re tr((ae )E 2 ) = Re tr((a(e E 2 )) = Re tr((ae ) E 2 ) L A (E ), E 2 = Re tr(l A (E )E 2 ) = Re tr((ae )E 2 + (AE ) E 2 ) = 2 Re(a p + a 2 α + a 3 β). As [ p a b ] is nonzero, the vector [ p α β ] is also nonzero. Now, if p 0, we can take a = p, a 2 = a 3 = 0. Then L A (E ), E 2 = 2. A similar construction can be made if α or β is nonzero. Thus, A can be constructed so that L A (E ), E 2 0. This means that L A is not Lyapunov-like.

Gowda, Tao, Ravindran: Complementarity properties 9 4.2 Lyapunov-like transformations on L n In this section, we give two characterizations of Lyapunov-like transformations on L n. Our second result, Theorem 4.6, can also be deduced from Theorem 4.2. Recalling that the underlying space of L n is R n (n > ), we fix a matrix A R n n and let J R n n be defined by J := diag(,,..., ). We recall the following from [5]: Lemma 4. The matrix A has the Z-property on L n if and only if there exists γ R such that γj (JA + A T J) is positive semidefinite on L n. As a consequence, we prove Theorem 4.5 The matrix A is Lyapunov-like on L n if and only if there exists β R such that βj + (JA + A T J) = 0. Proof. Suppose A is Lyapunov-like, in which case, A and A have the Z-property. Then by Lemma 4., there exist α and β such that αj (JA+A T J) and βj +(JA+A T J) are positive semidefinite. Therefore, αj (JA+A T J)+βJ +(JA+A T J) = (α+β)j is positive semidefinite. Thus, we have α = β. Now, βj (JA + A T J) = (βj + (JA + A T J)) is positive semidefinite and symmetric, hence βj + (JA + A T J) = 0. The if part is obvious because of the previous lemma. Here is another characterization of Lyapunov-like transformations on L n. Theorem 4.6 A matrix A on L n is a Lyapunov-like if and only if it is of the form [ a b T A = b D where a R, D R (n ) (n ), with D + D T = 2aI. ], (4.4) Proof. To see the if part, take x and y in L n such that 0 x y 0. Assuming that x and y are nonzero, we may write x = [ u ] T, y = [ u] T, where u = (see e.g., [2]). Then Ax, y = a u T Du = 0, where the last equality comes from D + D T = 2aI. This proves the Lyapunov-like property of A. Now for the only if part. Suppose the matrix A is Lyapunov-like and given by [ ] a b T A =. c D Putting x = [ u ] T, y = [ u ] T with u =, we see that 0 x y 0. Since A is Lyapunov-like, we have Ax, y = 0 and so Replacing u by u in (4.5), we have a + (b c) T u u T Du = 0. (4.5) a (b c) T u u T Du = 0. (4.6) The above two equations lead to (b c) T u = 0 for all u with u =. Thus, b = c. Now from the previous result, we have, βj + (JA+A T J) = 0 (for some β). This leads to β = 2a and D + D T = βi. Therefore, D + D T = 2aI. This completes the proof.

0 Gowda, Tao, Ravindran: Complementarity properties 5. The P-property of Lyapunov-like transformations Throughout this section, (in view of Theorem 3.2), we assume that the inner product is given by the trace inner product. This simplifies our proofs somewhat as c = for any primitive idempotent with respect to the trace inner product. Theorem 5. Let L be Lyapunov-like on V and L S. Then L P. Before giving a proof of this result, we prove several lemmas. In these lemmas, we fix a Jordan frame {e, e 2,...,e r } in V and consider the corresponding Peirce decomposition V = i j V ij given in Theorem 2.2. Throughout, we assume that L is Lyapunov-like. Lemma 5. For i j, let x ij V ij with x ij =. Then (a) x := e i + e j + 2x ij 0, (b) y := e i + e j 2x ij 0, and (c) x, y = 0. Proof. (a) In V, consider any d = d i e i + d ij 0. We note that 2d i d j d ij 2 for all i j, see, Exercise 7, Page 80, in [7]. Then, x, d = d i + d j + 2 x ij, d ij d i + d j 2 x ij d ij = d i + d j 2 d ij d i + d j 2 d i dj = ( d i d j ) 2 0. As K is self-dual, x 0. The proof of (b) is similar, and (c) is obvious. Lemma 5.2 The following hold: (a) i j, x ij V ij, x ij = L(x ij ), x ij = 2 ( L(e i), e i + L(e j ), e j ). (b) L(e i ), e i 0 i =, 2,...,r Tr(L) 0, where Tr(L) denotes the trace of the linear transformation L on V. Proof. (a) For a given x ij, let x and y be as in the previous lemma. Since L is Lyapunov-like, we have L(x), y = 0 = L(y), x. Now, L(x), y = 0 L(e i ), e i + L(e j ), e j 2 L(x ij ), x ij = 2 L(x ij ), e i + e j + 2 L(e i ) + L(e j ), x ij. (5.) L(y), x = 0 L(e i ), e i + L(e j ), e j 2 L(x ij ), x ij Adding these equations, we get the desired result. = 2 L(x ij ), e i + e j 2 L(e i ) + L(e j ), x ij. (5.2) (b) Now suppose L(e i ), e i 0 for all i =, 2,..., r. Recall that the trace of the linear transformation L on the (real) Hilbert space V is the sum of numbers of the form L(u), u, as u varies over any orthonormal basis in V. (This can be seen by writing the matrix representation of L with respect to an orthonormal basis and adding the diagonal elements of that matrix.) As V is an orthogonal direct sum of spaces V ij for i j, the union of orthonormal bases in various V ij s is an orthonormal basis in V. By our assumption and Item (a), the sum of numbers of the form L(u), u as u varies over an orthonormal basis in any V ij is nonpositive. Thus, the sum of these, namely, Tr(L), is also nonpositive. This completes the proof.

Gowda, Tao, Ravindran: Complementarity properties Lemma 5.3 Suppose i, k, and l are distinct and x kl V kl. Then L(e i ), x kl = 0 and L(x kl ), e i = 0. Proof. Without loss of generality, we assume that x kl =. Then e k + e l 2x kl 0 by Lemma 5., and e i, e k + e l 2x kl = 0 for i {k, l}. Since L is a Lyapunov-like transformation, we have L(e i ), e k + e l 2x kl = 0 L(e i ), x kl = 0. Similarly, L(e k + e l 2x kl ), e i = 0 L(x kl ), e i = 0. Lemma 5.4 Suppose i, j, k, and l are all distinct, x ij V ij, and x kl V kl. Then L(x ij ), x kl = 0. Proof. Without loss of generality, we assume that x ij = x kl =. Now, put u = e i +e j 2ε x ij and v = e k + e l 2δ x kl, where ε, δ {, }. Then 0 u v 0. Since L is Lyapunov-like, we have L(u), v = 0 and so 2δ L(e i + e j ), x kl + 2εδ L(x ij ), x kl 2ε L(x ij ), e k + e l = 0. (5.3) In (5.3), we replace ε by ε and δ by δ to get 2δ L(ei + e j ), x kl + 2εδ L(x ij ), x kl + 2ε L(x ij ), e k + e l = 0. (5.4) Now adding these two equations, we get L(x ij ), x kl = 0. Lemma 5.5 Corresponding to a Jordan frame {e, e 2,...,e r }, suppose there exists x = k x ie i such that x i 0 for all i =, 2,...,k and y = L(x) = k y ie i. Then L(e i ), z ij = 0, i {, 2,..., k}, j {k +, k + 2,...,r}, z ij V ij. Proof. Since the spaces V ij are mutually orthogonal and z ij V ij, we have L(x), z ij = y, z ij = 0. On expanding, we get x L(e ), z ij + x 2 L(e 2 ), z ij + + x i L(e i ), z ij + x i+ L(e i+ ), z ij + + x k L(e k ), z ij = 0. In the above equality, fix i {, 2,..., k}. By Lemma 5.3, L(e l ), z ij = 0 for all l i. For l = i, x i 0 and so L(e i ), z ij = 0. In our next lemma, we assume that V is simple and of rank 3. As V carries the trace inner product, we have, see Prop. IV..4 and Lemma IV.2.2 in [7], (x ij ) 2 = x ij 2 (e i +e j ), x ij (x ij y jl ) = 2 8 x ij 2 y jl and x ij y jl 2 = 8 x ij 2 y jl 2 (5.5) for all x ij V ij, y jl V jl, with i, j, and l distinct. Lemma 5.6 Suppose that the conditions of the previous lemma are in place. Assume that V is simple and rank (V ) 3. Then L(x ij ), y il = 0, for all i j {, 2,..., k}, l {k +, k + 2,...,r}, x ij V ij, and y il V il. Proof. Fix i j {, 2,..., k} and l {k +, k + 2,...,r}. Assume without loss of generality, x ij = = y il. Let x = e i + e j + 2ε x ij and y = αe i + αe j + e l 2ε α x ij + δ y il 2 2ε δ x ij y il, where ε, δ {, } and α R. Then, using the identities in (5.5), relations of the form x ij e i = 2 x ij and Theorem 2.2, we get 0 y 2 = y y = (2α 2 + 2 )e i + (2α 2 + 2 )e j + 2e l 2ε(2α 2 + 2 )x ij + δ(2α + )y il 2 2εδ(2α + )x ij y il.

2 Gowda, Tao, Ravindran: Complementarity properties It is easy to verify that x, y 2 = 0. Since L is Lyapunov-like, we have L(x), y 2 = 0. This gives, after expanding and putting a = L(e i ), e i and b = L(e j ), e j, 0 = a(2α 2 + 2 ) + b(2α2 + 2 ) 2ε(2α 2 + 2 ) L(e i) + L(e j ), x ij + δ(2α + ) L(e i ) + L(e j ), y il 2 2εδ(2α + ) L(e i ) + L(e j ), x ij y il + (2α 2 + 2 ) 2ε L(x ij ), e i + e j + 2 2ε L(x ij ), e l (4α 2 + ) L(x ij ), x ij (5.6) + 2εδ(2α + ) L(x ij ), y il 4δ(2α + ) L(x ij ), x ij y il. Replacing ε by ε and δ by δ in (5.6), we have 0 = a(2α 2 + 2 ) + b(2α2 + 2 ) + 2ε(2α 2 + 2 ) L(e i) + L(e j ), x ij δ(2α + ) L(e i ) + L(e j ), y il 2 2εδ(2α + ) L(e i ) + L(e j ), x ij y il (2α 2 + 2 ) 2ε L(x ij ), e i + e j 2 2ε L(x ij ), e l (4α 2 + ) L(x ij ), x ij (5.7) + 2εδ(2α + ) L(x ij ), y il + 4δ(2α + ) L(x ij ), x ij y il. Since V ij V il V jl, L(e i ), x ij y il = 0 by Lemma 5.3 and L(e j ), x ij y il = 0 by Lemma 5.5. Adding equations (5.6) and (5.7), we get a(2α 2 + 2 ) + b(2α2 + 2 ) (4α2 + ) L(x ij ), x ij + 2εδ(2α + ) L(x ij ), y il = 0. (5.8) Now replacing α by α and δ by δ in (5.8), we have a(2α 2 + 2 ) + b(2α2 + 2 ) (4α2 + ) L(x ij ), x ij 2εδ( 2α + ) L(x ij ), y il = 0. (5.9) Subtracting equation (5.9) from (5.8), we get L(x ij ), y il = 0. Our previous lemmas lead to the following Proposition 5. Suppose V is simple with rank r and L is Lyapunov-like on V. Further suppose that corresponding to a Jordan frame {e, e 2,...,e r }, there exists x = k x ie i such that x i 0 for all i =, 2,...,k, and y = L(x) = k y i e i. Let W := V (e + e 2 +... + e k, ). Then L(W) W. Proof The result is obvious if k = or k = r. We assume that < k < r. This implies that r 3 and the previous Lemma is applicable. Consider the Peirce decomposition of any x W with respect to the Jordan frame {e, e 2,..., e k }: x = k x ie i + i<j k x ij. We write the Peirce decomposition of z := L(x) with respect to the Jordan frame {e, e 2,..., e r } in V : k x i L(e i )+ k L(x ij ) = ( z i e i + z ij )+( z il )+( z i e i + z il ). i<j k i<j k i k<k+ l r k+ k+ i<l r Now, taking the inner product of the term on the left-hand side of the above expression with z il for i k < k + l r or with z i e i for i k +, or with z il for k + i < l r, and using the previous lemmas, we deduce that the last two (grouped) terms on the right-hand side of the above expression are both zero. Thus, k L(x) = z i e i + z ij W. This completes the proof. i<j k Proof of Theorem 5.. We are given that L is Lyapunov-like on V and L S. To start with suppose that V is simple. Assume, if possible, there is a nonzero x such that

Gowda, Tao, Ravindran: Complementarity properties 3 x and y := L(x) operator commute, and x y 0. Then for some Jordan frame {e, e 2,..., e r } in V, x = x i e i and L(x) = y i e i, where x i y i 0 for all i =, 2,..., r. As x 0, we may assume that x i 0 for i k and x i = 0 for i > k, where k r. Then k x i L(e i ) = L(x) = y i e i, which leads to k x i L(e i ), e j = y j j =, 2,...,r, (5.0) as e j = for all j. Since L is Lyapunov-like, these yield y j = 0 for j k + so that Moreover, L(x) = L( k x i e i ) = k y i e i. L(e i ), e i = y i x i 0 i =, 2,...,k. (5.) Let W := V (e + e 2 + + e k, ). Then by Proposition 5., L(W) W. Denote the restriction of L to W by L. Then L is Lyapunov-like on the (subalgebra) W. Also, the matrix representation of L on V = W W is of the form [ ] A B, 0 C where A is the matrix representation of L on W. Since L is Lyapunov-like and hence a Z- transformation, the S-property is equivalent to the positive stable property, see Theorems 6 and 7 in [5]. This implies, from the above matrix representation, that L also has the positive stable property and so its trace must be positive. However, (5.) implies that L (e i ), e i 0 for all i =, 2,...,k and by Lemma 5.2, applied to L and W, Tr(L ) 0. This is a contradiction. Hence no x exists, completing the proof of the theorem when V is simple. Now for the general case. Suppose that V is a product of simple algebras V k, k =, 2,...,N. By Theorem 4., we may write L as a product of linear transformations L k, where L k is Lyapunov-like on V k. Also, as L S, there is a 0 < d = (d, d 2,..., d n ) such that 0 < L(d) = (L (d ),...,L N (d N )). This means that L k S on V k. Now, by the first part of the proof, each L k has the P-property on V k. Since the P-property depends only on the Jordan product, it can be described in a componentwise manner. This means that the product of L k s, that is, L, has the P-property on V. This completes the proof of the theorem. Remarks. For any a V, L a is Lyapunov-like. In this case, the symmetric transformation L a has the P-property if and only if L a is strictly monotone. This can happen if and only if a > 0. A derivation D, while Lyapunov-like, can never be invertible because D(e) = 0. Hence it can never have any of the complementarity properties S, Q, or P. 6. Appendix In this Appendix, we present a proof of Theorem 4.2 without relying on Lie algebra techniques. We begin with a proposition. Proposition 6. The following are equivalent:

4 Gowda, Tao, Ravindran: Complementarity properties (i) L is a derivation. (ii) L is Lyapunov-like and L(e) = 0. (iii) L is Lyapunov-like and skew-symmetric. Proof. (i) (ii) : Let L be a derivation, {e, e 2,...,e r } be a Jordan frame in V, and i j. Using e i e j = 0 and writing 0 = L(e i e j ) = L(e i ) e j +e i L(e j ), we get, upon taking the inner product with e j, 0 = L(e i ) e j + e i L(e j ), e j = L(e i ), e j e j + L(e j ), e i e j = L(e i ), e j, where we used the properties e j e j = e j and x y, z = x, y z in V. This proves that for all i j, L(e i ), e j = 0, that is, L is Lyapunov-like. For any i, e i e i = e i. Applying L both sides, we get 2 L(e i ) e i = L(e i ) and 2 L(e i ) e i, e i = L(e i ), e i. This implies that L(e i ), e i = 0, i =, 2,..., r. So we have proved that when L is a derivation, for any Jordan frame {e, e 2,..., e r }, L(e i ), e j = 0 i, j =, 2,...,r. Fixing j and summing over i, we get L(e), e j = 0 for all j. As the Jordan frame is arbitrary, writing the spectral decomposition of any x as x = r x je j, we get L(e), x = 0. As x is arbitrary, this gives L(e) = 0. Thus we have proved that L is Lyapunov-like and L(e) = 0. (ii) (iii) : Let {e, e 2,...,e r } be any Jordan frame. Then L(e i ), e j = 0, i j. The condition L(e) = 0 implies that r L(e i) = 0 and hence L(e i ), e j = 0 even when i = j. Now for any x V, we have the spectral expansion x = r x ie i for some Jordan frame {e, e 2,..., e r } and eigenvalues x, x 2,..., x r. Then L(x), x = x i x j L(e i ), e j = 0. i,j This implies that L + L T = 0, i.e., L is skew-symmetric. (iii) (ii) : Assume that L is Lyapunov-like and skew-symmetric. Then for any Jordan frame {e, e 2,..., e r }, we have L(e i ), e j = 0 for all i and j. As in the last part of the proof of (i) (ii), we get L(e) = 0. (iii) (i) : Let D = L be Lyapunov-like and skew-symmetric. We first observe that D is a derivation provided D(x 2 ) = 2x D(x) x V. (6.) For, if this condition holds, then replacing x by x + λy and comparing coefficients of λ, we get D(x y) = x D(y) + y D(x). Now to prove (6.), take any x V and consider its spectral decomposition: x = x i e i. Then x 2 = x 2 i e i, and hence D(x 2 ) = x 2 i D(e i). Now, suppose that for all i and k l, Then 2e i D(e i ) = D(e i ) and e k D(e l ) + e l D(e k ) = 0. (6.2) 2x D(x) = 2( x i e i ) ( x j D(e j )) = 2( i = 2( i x 2 ie i D(e i )) + 2 i<j x 2 ie i D(e i )) x i x j (e i D(e j ) + e j D(e i )) = i x 2 i D(e i) = D(x 2 ).

Gowda, Tao, Ravindran: Complementarity properties 5 We now prove (6.2). Fix an index k, k r. We write the Peirce decomposition of D(e k ) as D(e k ) = x i e i + i<j x ij. Since D is skew symmetric, D(e i ), e i = 0 for i =,...,r. Since D is a Lyapunov-like, D(e k ), e j = 0 for k j. Thus, x i = 0 for i =,...,r. Hence D(e k ) = i<j x ij. Now, e k, e i + e j + 2x ij = 0 for k i, j and e i + e j + 2x ij 0 for i j (see Lemma 5.). Then D(e k ), e i + e j + 2x ij = 0. This implies D(e k ), x ij = 0. Thus, k D(e k ) = x ik + i= j=k+ Multiplying both sides of this equality by e k and using x ik e k = 2 x ik, etc., we get e k D(e k ) = 2 D(e k). This proves the first part of (6.2). Now for the second part. Based on the discussion above, we write the Peirce decomposition of D(e k ) for k =, 2,..., r: x kj. D(e ) = x 2 + x 3 + + x r D(e 2 ) = y 2 + y 23 + + y 2r D(e 3 ) = z 3 + z 23 + + z 3r. =. k D(e k ) = p ik + i=... =. l D(e l ) = q il +. = i= Now, taking sum of these equations, we have. j=k+ j=l+ p kj q lj D(e r ) = w r + w 2r + + w r r. D(e) = (x 2 + y 2 ) + (x 3 + z 3 ) + + (p lk + q lk ) +. Since the grouped terms in D(e) belong to orthogonal Peirce spaces and D(e) = 0, we have p lk +q lk = 0. Since e k D(e l ) = 2 q lk and e l D(e k ) = 2 p lk, we have e k D(e l ) + e l D(e k ) = 2 (p lk + q lk ) = 0. This proves the second part of (6.2) and completes the proof of the proposition. Now to prove Theorem 4.2, let L be Lyapunov-like. Then with a = L(e), D := L L a is Lyapunov-like and D(e) = 0. Hence by the above Proposition, D is a derivation. It follows that L = L a + D completing the proof. Concluding Remarks. In this paper, we discussed the P-property of Z and Lyapunov-like transformations. We proved that for Lyapunov-like transformations, the P-property is equivalent to the S-property. In the case of Z-transformation, we do not have this full equivalence. So the problem of proving the equality Z S = Z P continues to be open. References [] Baker, A. 2002. Matrix Groups, Springer-Verlag, London. [2] Cottle, R.W., J.-S. Pang, R.E. Stone. 992. The Linear Complementarity Problem. Academic Press, Boston. [3] Damm, T. 2004. Positive groups on H n are completely positive. Linear Algebra Appl. 393 27 37.

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