vol22_pp225-251. 238KB Jun 04 2011 12:06:12 AM

PSEUDOMONOTONICITY AND RELATED PROPERTIES IN EUCLIDEAN JORDAN ALGEBRAS∗

JIYUAN TAO†

Abstract. In this paper, we extend the concept of pseudomonotonicity fromRn

to the setting of Euclidean Jordan algebras. We study interconnections between pseudomonotonicity, monotonicity, and the Z-property.

Key words. Euclidean Jordan algebra, Pseudomonotone, Positive subdefinite (PSBD), Copos-itive, Copositive star, Automorphism, Z-property.

AMS subject classifications. 90C33, 17C20, 17C55.

1. Introduction. Given a convex setKinRn_{, a mapf :K→R}n_{is said to be}

pseudomonotone onKif

x, y_∈K, _hf(x), y₋x_{i ≥}0_{⇒ h}f(y), y₋x_{i ≥}0.

This concept is a generalization of monotonicity defined by_hf(x)₋f(y), x₋y_{i ≥}0 for anyx, y_∈Rn. There is an extensive literature associated with this property covering theory and applications, see e.g., [11], [13], [14].

Gowda (see [4], [5], [6]), Crouzeix et al. (see [1]), and Hassouni et al. (see [12]) studied pseudomonotone matrices on Rn

+ and investigated some properties of the

linear complementarity under the condition of pseudomonotonicity.

Motivated by their results, in this paper, we extend the concept of pseudomono-tonicity from Rn _{to the setting of Euclidean Jordan algebras. Specifically, we give}

a characterization of pseudomonotonicity for a linear transformation and a matrix-induced transformation defined on a Euclidean Jordan algebra. We show that pseu-domonotonicity and monotonicity coincide under the condition of the Z-property. Moreover, we present the invariance of pseudomonotonicity under the algebra and cone automorphisms and describe interconnections between pseudomonotonicity of a linear transformation and its principal subtransformations. We note that symmetric cones (see Section 2 for definition) are, in general, nonpolyhedral. Therefore, these generalizations presented in this paper are not routine generalizations.

∗_{Received by the editors on August 8, 2010. Accepted for publication on February 27, 2011.}

Handling Editor: Bit-Shun Tam.

†_{Department of Mathematical Sciences, Loyola University Maryland, Baltimore, MD 21210, USA}

(jtao@loyola.edu).

Here is an outline of the paper. In Section 2, we cover the basic material dealing with Euclidean Jordan algebras and pseudomonotonicity. In Section 3, we present some general results for pseudomonotone transformations. In Section 4, we investi-gate the relation between pseudomonotonicity of a linear transformation and its prin-cipal subtransformations. In Section 5, we study pseudomonotonicity for some special linear transformations. Specifically, we show that pseudomonotone linear transforma-tions are invariant under the algebra and cone automorphisms, we specialize pseu-domonotonicity of a linear transformation with having the Z-property, and we give a characterization of pseudomonotonicity for a matrix-induced transformation.

2. Preliminaries.

2.1. Euclidean Jordan algebras. In this subsection, we briefly recall some concepts, properties, and results from Euclidean Jordan algebras. Most of these can be found in [3], [9].

LetV be a Euclidean Jordan algebra andKbe a symmetric cone inV. A Jordan product is denoted byx_◦yfor any two elementsxandyinV. In addition, an element e_∈V is called theunit element ifx_◦e=xfor allx_∈V. We define

z+:= ΠK(z) and z−:=z+−z,

where ΠK(z) denotes the (orthogonal) projection ofz ontoK.

For an elementz_∈V, we write

z_≥0 (z >0) if and only if z_∈K (z_∈K◦(= interior(K))), andz_≤0 (z <0) when −z_≥0 (−z >0).

An elementc_∈V such thatc2=cis called anidempotentin V; it is aprimitive idempotentif it is nonzero and cannot be written as a sum of two nonzero idempotents. We say that a finite set{e1, e2, . . . , er}of primitive idempotents inV is aJordan frame

ifei◦ej= 0 ifi=6 jand Pr1ei=e.

Givenx_∈V, there exists a Jordan frame_{e1, . . . , er}and real numbersλ1, . . . , λr

such that

x=λ1e1+· · ·+λrer.

(2.1)

The numbersλi are called theeigenvaluesofx, and the representation (2.1) is called

thespectral decomposition (or the spectral expansion)ofx. Given (2.1), we have

x=

r

X

1

λi+ei− r

X

1

λi−ei and h r

X

1

λi+ei, r

X

1

where for a real numberα,α+:= max_{0, α_} andα−:= (α)+ −α. From this we easily verify that x+ ₌ Pr

1λi+ei and x− = Pr1λi−ei, and so

x=x+_−x− _withhx+_{, x}−_{i= 0.}

For an x_∈V, a linear transformation Lx:V →V is defined by Lx(z) = x◦z,

for allz_∈V. We say that two elementsxandy operator commuteifLxLy =LyLx.

It is known thatxandyoperator commute if and only ifxandyhave their spec-tral decompositions with respect to a common Jordan frame (Lemma X.2.2, Faraut and Kor´anyi [3]).

Here are some standard examples.

Example 2.1. Rn _{is a Euclidean Jordan algebra with inner product and Jordan}

product defined respectively byhx, y_i=Pn_i=1xiyi andx◦y=x∗y.HereRn+ is the

corresponding symmetric cone.

Example 2.2. _Sn_{, the set of all} n_×n real symmetric matrices, is a Euclidean Jordan algebra with the inner and Jordan product given by hX, Y_i := trace(XY) andX_◦Y := 1

2(XY +Y X). In this setting, the symmetric coneS+n is the set of all

positive semidefinite matrices in_Sn_{. Also,X andY operator commute if and only if}

XY =Y X.

Example 2.3. _ConsiderRn _{(n >1) where any elementxis written as}

x=

x0

x

withx0∈Randx∈Rn−1. The inner product inRn is the usual inner product. The

Jordan productx_◦y inRn _{is defined by}

x_◦y=

x0 x ◦ y0 y :=

hx, y_i x0y+y0x

.

We shall denote this Euclidean Jordan algebra (Rn_{,◦,h·,·i) by L}n_{. In this algebra,}

the cone of squares, denoted by_Ln

+, is called the Lorentz cone (or the second-order

cone). It is given by_Ln

+={x:||x|| ≤x0}.

The unit element in_Ln _ise=

1 0

. We note the spectral decomposition of any

xwithx₆= 0: x=λ1e1+λ2e2, where λ1:=x0+||x||,λ2:=x0− ||x||, and

e1:= 1

2

"

1

x ||x||

#

, ande2:=1

2

"

1

− x ||x||

#

.

Peirce decomposition. Fix a Jordan frame_{e1, e2, . . . , er} in a Euclidean Jordan

algebraV. Fori, j_{∈ {}1,2, . . . , r_}, define the eigenspaces

Vii:={x∈V : x◦ei=x}=R ei(where R is the set of all real numbers)

and when i₆=j,Vij :={x∈V : x◦ei = 1₂x=x◦ej}. Then, we have the following

theorem.

Theorem 2.4. _{(Theorem IV.2.1, [3]) The spaceV is the orthogonal direct sum}

of the spacesVij (i≤j). Furthermore,

Vij◦Vij ⊂Vii+Vjj

Vij◦Vjk⊂Vik ifi6=k

Vij◦Vkl={0}if {i, j} ∩ {k, l}=∅.

Thus, given any Jordan frame{e1, e2, . . . , er}, we can write any element x∈ V as

x=Pr_i=1xiei+P_i<jxij wherexi ∈Randxij∈Vij.

A Euclidean Jordan algebra is said to be simple if it is not the direct sum of two Euclidean Jordan algebras. It is well known that any Euclidean Jordan algebra is product of simple Euclidean Jordan algebras and every simple Euclidean Jordan algebra is isomorphic to the Jordan spin algebra _Ln _{or to the algebra of all n}

×n real symmetric matrices_Sn _{or ton}

×ncomplex Hermitian matrices_Hn or ton×n

quaternion Hermitian matricesQn or to the algebra of all 3×3 octonion Hermitian

matrices_O3.

2.2. Pseudomonotone and positive subdefiniteness concepts. Through-out this paper, we assume thatV is a Euclidean Jordan algebra with the corresponding symmetric coneK andL:V _→V is a linear transformation.

We say thatLis:

(a) copositive onK ifhL(x), xi ≥0 for allx_∈K; (b) copositive star onK ifLis copositive onK, and

[x_≥0, L(x)_≥0, and _hL(x), x_i= 0]_⇒LT(x)_≤0.

Definition 2.5. _{LetS be a set inV. Lis said to be pseudomonotone onS if}

x_∈S, y_∈S, _hL(x), y−x_{i ≥}0⇒ hL(y), y−x_{i ≥}0.

Definition 2.6. _{LetSbe a set inV. Lis said to be positive subdefinite (PSBD)}

onS if

We note that every monotone linear transformation is vacuously PSBD.

Definition 2.7. _{L is said to be merely positive subdefinite (MPSBD) ifL is}

PSBD but not monotone.

The following definition is a generalization of (Moore-Penrose) pseudo-inverse of a square matrix.

Definition 2.8. _{The pseudo-inverse of}Lis the uniquely defined linear transfor-mationL† _{which satisfies the following conditions:}

LL†L=L, L†LL†=L†, (LL†)T =LL†, and (L†L)T =L†L. We defineLs_=LT_(L+LT₎†_{L. We note thatL}s _{is always self-adjoint.}

Given a self-adjoint linear transformationB onV, its inertia is defined by In(B) := (π(B), ν(B), δ(B)),

where π(B), ν(B), and δ(B) are, respectively, the number of eigenvalues of B with positive, negative, and zero real parts, counting multiplicities. Note that

π(B) +ν(B) +δ(B) = dim(V).

3. General results for pseudomonotone transformations. In this section, we present some general results for pseudomonotone transformations onV.

Specializing Gowda’s result (Proposition 1, [4]) to symmetric cone, we have Lemma 3.1.

Lemma 3.1. _{If Lis pseudomonotone on K, then L is copositive star onK.}

The following lemma is a modification of Theorem 5.2 in [15]. We omit its proof.

Lemma 3.2. _{L is pseudomonotone onK}◦ _{if and only if}

x_∈K◦, v_∈V, _hv, L(x)_i= 0_{⇒ h}v, L(v)_{i ≥}0.

Throughout the paper, pseudomonotone, PSBD, copositive, and copositive star con-cepts are defined relative/onK, in which case, we drop a phrase onK.

Lemma 3.3. _{If Lis pseudomonotone, then L is PSBD.}

Proof. Suppose thatLis pseudomonotone but not PSBD. Then there existsx0,

such that_hL(x0), x0i<0,LT(x0)6≥0, andLT(x0)6≤0. Thus, in view of the spectral

decomposition ofLT_(x

0), there existsw >0 such thathL(w), x0i=hLT(x0), wi= 0.

The following is analogous to Proposition 3.1 in [1].

Theorem 3.4. L is pseudomonotone if and only ifL is PSBD and hL(z), z_i<0, LT(z)_≤0_{⇒ h}z, L(z−)_i<0. (3.1)

Proof. “Only if” part. Suppose that L is pseudomonotone. Then L is PSBD by Lemma 3.3. Now, suppose that hL(z), zi<0 and LT_{(z)≤0.Since hz, L(z}−_)i= hLT_{(z), z}−

i ≤0, we claim that_hz, L(z−₎

i 6= 0. Suppose not, so that

hz, L(z−)i= 0. (3.2)

Then hz+_−z−_{, L(z}−_{)i = 0 ⇒ hz}+_−z−_{, L(z}+_{)i ≥ 0 by pseudomonotonicity of L.}

Hence,

hz, L(z+)i ≥0. (3.3)

Now, from (3.2) and (3.3), we have hz, L(z+ _−z−_{)i ≥ 0, i.e., hL(z), zi ≥ 0, this}

contradicts_hL(z), z_i<0. Thus, the claim is true. Therefore, (3.1) holds. “If” part. Suppose thatLis PSBD and (3.1) holds. The condition

x, y_≥0,_hL(x), y₋x_{i ≥}0_{⇒ h}L(y), y₋x_{i ≥}0 is equivalent to

u_≥0, z_∈V,_hL(z−+u), z_{i ≥}0_{⇒ h}L(z++u), z_{i ≥}0, which is the same as

u_≥0, z∈V,_hu, LT(z)i ≥ −hL(z−), zi ⇒ hu, LT(z)i ≥ −hL(z+), zi. (3.4)

Now, letu_≥0 withhu, LT_{(z)i ≥ −hL(z}−_{), zi. We verify the rightmost inequality in}

(3.4). Without loss of generality, letLT_{(z)6= 0.}

Case 1: LT_(z)

≤0. Then, 0_{≥ h}u, LT_(z)

i ≥ −hL(z−_{), z}

i=_−hz−_{, L}T_(z)

i ≥0 shows that hL(z−_{), zi= 0. By (3.1), hL(z), zi ≥ 0. Now, hL(z), zi ≥ 0 ⇒ −hL(z}−_{), zi ≥} −hL(z+_{), z}

i, thus,_hu, LT_{(z)i ≥ −hL(z}+_{), z} i.

Case 2: LT_(z)

≥0. Then,_hu, LT_(z)

i ≥0_{≥ −h}z+_{, L}T_(z)

i=_−hL(z+_{), z} i.

Case 3: LT_{(z)6≥0 andL}T_{(z)6≤0. ThenhL(z), zi ≥ 0 (otherwise, ifhL(z), zi<0,}

then either LT(z) _≤0 or LT(z)_≥ 0, because L is PSBD). Again, _hL(z), z_{i ≥} 0 _⇒

−hL(z−_{), z}

i ≥ −hL(z+_{), z}

i, thus,_hu, LT_(z)

i ≥ −hL(z+_{), z}

iby (3.4).

The following Lemma is similar to Lemma 3.1 in [1]; we give a proof for com-pleteness.

Lemma 3.5. _{L is pseudomonotone if and only ifL is PSBD and}

hL(z), z_i<0, LT_(z)

≤0_⇒L(z−)₆= 0. (3.5)

Proof. “Only if” part follows from Lemma 3.3 and Theorem 3.4.

“If” part. Assume that (3.5) holds,L is PSBD and not pseudomonotone. Then, by Theorem 3.4, there existszsuch that

hL(z), z_i<0, LT_(z)

≤0 and _hz, L(z−)_{i ≥}0.

From LT_{(z)≤0, it follows thathz, L(z}−_)i=hLT_{(z), z}−_{i ≤0. Hence, for such az,} hz, L(z−₎

i= 0. Now, for av_∈V, there existsδ >0 such that_hL(z+tv), z+tv_i<0,

∀t _∈ [₋δ, δ]. Since L is PSBD, we have either LT(z+tz) _≥0 or LT(z+tz) _≤ 0. If LT_(z+tz)

≥ 0, then 0 _{≤ h}LT_{(z+tv), z}−

i = t_hL(z−_{), v}

i, _∀t _∈ [₋δ, δ]. Hence

hL(z−_{), vi = 0 for all v ∈ V and therefore L(z}−_{) = 0, which is a contradiction.}

Similarly, we can reach a contradiction ifLT_{(z+tz)≤0.Therefore, (3.5) holds.}

Lemma 3.6. _Let Lbe copositive star. Then hL(z), z_i<0, LT_(z)

≤0_⇒L(z−₎

6

= 0. (3.6)

Proof. Suppose thathL(z), zi<0,LT_{(z)≤0 andL(z}−_{) = 0.ThenhL(z}−_{), z}−_i=

0. SinceLis copositive star, we have that LT_(z−₎

≤0. Then

0>_hL(z), zi=hz+₋z−, L(z+)i=hz+, L(z+)i − hLT(z−), z+i ≥0. This is a contradiction. Therefore, (3.6) holds.

Theorem 3.4, Lemmas 3.1, 3.5, and 3.6 result in the following theorem.

Theorem 3.7. _{L is pseudomonotone if and only if L is PSBD and copositive}

star.

The following theorem is a characterization of MPSBD.

Theorem 3.8. _{Suppose that Rank(L)≥ 2. Then L is MPSBD if and only if}

δ(L+LT_{) = 1,L(V) =L}T_{(V) = (L+L}T_{)(V), and}

−Ls _{is copositive.}

The proof of the above result is based on several lemmas.

Lemma 3.9. _{Assume that} C is a closed convex cone and C_∩(₋C) = _{0_}. If

Proof. Suppose that this is not true. Then the set_{ta+(1₋t)b:t_∈[0,1]_{} ⊆}A_∪B, whereA:=C_\{0_}andB:=₋C_\{0_}. SinceA_∩B=_∅=A_∩B, we have a separation of the connected set _{ta+ (1₋t)b:t _∈[0,1]_}, we reach a contradiction. Hence the conclusion.

Corollary 3.10. _{Assume that C is a closed convex cone andC∩(−C) ={0}.}

Ifa+t1b∈C,a+t2b∈ −C for some t1, t2∈R, anda+tb∈C∪ −C for allt∈R, then there existst0∈R such thata+t0b= 0.

The following lemma and its proof are similar to Proposition 2.2 and its proof in [1]; we give a proof for completeness.

Lemma 3.11. _{Suppose that Lis MPSBD. Then}

(i) ν(L+LT_{) = 1;}

(ii) (L+LT_{)z= 0⇒L(z) =L}T_{(z) = 0.} Proof. LetB :=L+LT.

(i)B has at least one negative eigenvalue since Lis not monotone. Assume, for contradiction, that there existλ1≤λ2<0,z1, andz2 such that

B(z1) =λ1z1, B(z2) =λ2z2, ||z1||2=||z2||2= 1, and hz1, z2i= 0.

(3.7)

ThenhB(z1), z1i<0 andhB(z2), z2i<0, and hencehL(z1), z1i<0 andhL(z2), z2i<

0. Since L is MPSBD, without loss of generality, we assume that LT_(z

1) ≤ 0 and

LT(z2)≥0. Now, we definez(t) =tz1+ (1−t)z2fort∈[0,1]. ThenhB(z(t)), z(t)i=

t2_λ

1+(1−t)2λ2<0.HencehL(z(t)), z(t)i<0 for allt∈[0,1]. This implies that either

LT_{(z(t))≤0 orL}T_{(z(t))≥0, therefore,L}T_{(z(t)) =tL}T_(z

1)+(1−t)LT(z2)∈K∪−K.

Since LT_{(z(0)) = L}T_(z

1) ≤ 0 and LT(z(1)) =LT(z2)≥ 0, there exists t0 ∈ (0,1)

such thatLT(z(t0)) = 0 by the above lemma; this is a contradiction. Thus, (i) holds.

(ii) Let z1, λ1 be defined as in (3.7); let z be such that B(z) = 0. Then we

have_hL(z1), z1i=1₂hB(z1), z1i<0, and henceLT(z1)6= 0. Now, fort∈R, we define

w(t) =z1+tz. Then we havehB(w(t)), w(t)i=λ1<0. HencehL(w(t)), w(t)i<0 for

allt_∈R. This implies that eitherLT_{(w(t))6= 0 and L}T_{(w(t))≤0 or L}T_{(w(t))6= 0}

and LT_{(w(t)) ≥ 0. Therefore, for all t ∈ R, 0 6= L}T_{(w(t)) = L}T_(z

1) +tLT(z) ∈

K_{∪ −}K. We claim thatLT_{(z) = 0. First, we show that eitherL}T_{(w(t))∈K for all}

t_∈RorLT_(w(t))

∈ −Kfor allt_∈R. Suppose not. Then there existt1, t2∈Rsuch

that LT_(z

1) +t1LT(z)∈K andLT(z1) +t2LT(z)∈ −K. Thus, there exists t0 ∈R

such thatLT_(z

1) +t0LT(z) = 0 by Corollary 3.10. This is a contradiction. Now, we

consider the following two cases:

Case 1: LT_(z

1) +tLT(z)∈K for all t∈R. Then 1_tLT(z1) +LT(z) = 1_t(LT(z1) +

is closed. Similarly, 1

tLT(z1) +LT(z)∈ −K for a large −t >0. This implies that

LT(z)_{∈ −}Kas t_{→ ∞}sinceK is closed. Thus,LT(z) = 0. Case 2: LT_(z

1) +tLT(z)∈ −K for allt∈R. The proof is similar to Case 1.

Therefore, the claim holds. HenceL(z) = 0 fromB(z) = 0.

Remark 3.12. _{SupposeM is ann×nmatrix. Then the following are equivalent}

(see [1]):

(i) (M+MT_{)z= 0}

⇒M z=MT_{z= 0;} (ii) M(Rn_{)⊆(M +M}T_)(Rn_).

Now, givenL:V _→V linear transformation, by identifying V with someRk _{and L}

with a matrix, we get the following equivalent statements:

(i) (L+LT_{)(z) = 0⇒L(z) =L}T_{(z) = 0;} (ii) L(V)_⊆(L+LT_)(V).

Using the same argument as in the above remark, Lemma 2 in [2] and Proposition 2.3 in [1] reduce to the following two lemmas.

Lemma 3.13. _{Suppose that L(V)⊆(L+L}T_)(V).Then

π(Ls_{) =π(L+L}T_{) +δ(L+L}T₎ −k, ν(Ls_{) =ν(L+L}T_{) +δ(L+L}T₎

−k, δ(Ls) = 2k−δ(L+LT),

wherek is the dimension of the kernel ofL.

Lemma 3.14. _{Suppose that B is self-adjoint linear transformation defined on V}

andν(B) = 1. Then there exists a closed convex cone T such that

T_{∪ −}T =_{z:_hB(z), z_{i ≤}0_} and int(T)_{∪ −}int(T) =_{z:_hB(z), z_i<0_}.

Furthermore,

T△_{∩ −}T△=_{0_} and T△_{∪ −}T△=_{y:_hB†(y), y_{i ≤}0_{} ∩}B(V).

WhereT△ _{is the polar cone ofT defined byT}△_={x⋆_:hx⋆_{, xi ≤0 ∀x∈T}.}

The following lemma gives a characterization of MPSBD in terms of such a closed convex cone as in the above lemma.

Lemma 3.15. _{Given L onV. The following are equivalent:}

(i) L is MPSBD.

(iii) ForT in (ii), eitherL(K)_⊆T△or L(K)_{⊆ −}T△. Proof. LetB =L+LT_.

(ii)_⇒(i): Suppose (ii) holds and let_hL(x), x_i<0. Then_hB(x), x_i<0, and hence x_∈int(T)_{∪ −}int(T) by the above lemma. Thus, either x_∈int(T)_{⊆ {}z:LT_(z)≤0}

or x_{∈ −}int(T)_{⊆ {}z:LT_(z)

≤0_}. Therefore, LT_(x)

≤0 or LT_(x)

≥0. HenceL is MPSBD.

(i)_⇒ (ii): Suppose (i) holds. Then ν(B) = 1 by Lemma 3.11. Hence there exists a closed convex coneT defined in the above lemma. It is enough to show that int(T) ⊆ {z : LT_{(z) ≤0} or int(T) ⊆ {z : L}T_{(z)≥ 0}. Suppose not. Then there}

existu,v in int(T) such thatLT_{(u)6≤0 andL}T_{(v)6≥0. We note thatu6=v sinceL}

is MPSBD. Again, sinceLis MPSBD, we haveLT(u)_≥0 andLT(v)_≤0. Now, we definew(t) =tu+ (1−t)v fort_∈[0,1]. Then w(t)∈int(T) sinceu,v in int(T), and LT_{(w(t)) =tL}T_{(u) + (1−t)L}T_{(v). HencehB(w(t)), w(t)i<0⇒ hL(w(t)), w(t)i<0.}

The last inequality implies that LT_{(w(t))≤ 0 orL}T_{(w(t))≥0 since L is MPSBD.}

Thus,LT(w(t))_∈K_{∪ −}K. Also,LT(w(0)) =LT(v)_≤0 andLT(w(1)) =LT(u)_≥0. Therefore, there exists t0∈(0,1) such that LT(w(t0)) = 0 by Lemma 3.9. This is a

contradiction.

(ii)_⇒(iii): For any y_∈L(K), there exists x_∈K such thaty=L(x). Now, we claim that eithery_∈T△_{ory∈ −T}△_.

Case 1: z_∈T. Then we haveLT_{(z)≤0. Thus,hz, yi=hz, L(x)i=hL}T_{(z), xi ≤0.}

Hencey_∈T△_.

Case 2: z_{∈ −}T, Then we haveLT_{(z)≤0. Thus,hz, yi=hz, L(x)i=hL}T_{(z), xi ≤0.}

Hencey_{∈ −}T△.

Therefore, the claim holds.

(iii)_⇒(ii): SupposeL(K)_⊆T△. ThenL(x)_∈T△for allx_∈K. Hence, for any z_∈T, we have 0_{≥ h}z, L(x)_i=_hLT_{(z), x}

i. This is implies thatLT_(z)

≤0. Therefore, T _{⊆ {}z:LT_{(z)≤0}. Similarly, ifL(K)⊆ −T}△_{, then−T ⊆ {z:L}T_(z)≤0}.

Proof of Theorem 3.8. Suppose thatL is MPSBD. Then ν(L+LT_{) = 1 by}

Lemma 3.11, and hence there exists a closed convex coneT that satisfies the conditions in Lemma 3.14. Also, L(K) ⊆ T△ _{or L(K) ⊆ −T}△ _{by Lemma 3.15. Without}

loss of generality, we assume that L(K) _⊆T△. Then for any x _≥0, _hLs_{(x), xi =} hLT(L+LT)†_{L(x), x}

i = _h(L+LT)†_{L(x), L(x)}

i ≤0. Therefore, ₋Ls is copositive. From Lemma 3.11 and Remark 3.12, we haveL(V)_⊆(L+LT_{)(V). Thus, Rank(L)}

≤

Rank(L+LT_{). Since L}s _{= L}T_(L+LT₎†_{L, we have Rank(L}s_{)≤ Rank(L). Hence}

Rank(Ls_{)≤ Rank(L)≤ Rank(L+L}T_{). Since Rank(L}s_{)= π(L}s_{) +ν(L}s_{) = π(L+}

Rank(L+LT_{), we haveπ(L+L}T_{) + 1 + 2δ(L+L}T_{)−2k≤π(L+L}T_{) +ν(L+L}T_).

This implies thatk_≥δ(L+LT). Since 0_≤ν(Ls_{) =ν(L+L}T_{) +δ(L+L}T₎

−k(see Lemma 3.13), we havek_≤1 +δ(L+LT_{). Sincek is an integer, we have that either}

k=δ(L+LT_{) ork= 1 +δ(L+L}T_).

Case 1: Supposek= 1 +δ(L+LT_{). Thenν(L+L}T_{) = 1 implies thatν(L}s_{) = 0 (see}

Lemma 3.13) and soLs_{is monotone. Since}

−Ls_{is copositive, we have}

hLs_{(x), x} i= 0, for all x _∈ K. We claim that Ls _{= 0. Take any y ∈ V, y = y}+ _−y−_{, since}

y+, y− _∈ K, we have y+ +y− _∈ K. Thus, _hLs_(y+_+y−_{), y}+_+y−

i = 0, which implies that_hLs(y+_{), y}−

i= 0. Now, it is easy to verify that_hLs(y), y_i =_hLs(y+ −

y−_{), y}+ _−y−_{i = 0. Thus, L}s_{(y) = 0 for all y ∈ V. Hence L}s _{= 0. Therefore,}

dim(V) =δ(Ls_{) = 2k−δ(L+L}T_{) = 2k−(k−1) =k+1. Since dim(V) = Rank(L)+k,}

we have Rank(L)= 1. This is a contradiction.

Case 2: Suppose k = δ(L+LT_{). Then from Lemma 3.13, δ(L}s_{) = δ(L+L}T_),

ν(Ls_{) = ν(L+L}T_{), and π(L}s_{) = π(L+L}T_{), i.e., L+L}T _{and L}s _{have the same}

inertia. Therefore, Rank(L+LT_{)= Rank(L}s_).

Subcase 2.1: SupposeL+LT _{is invertible. Then (L+L}T_{)(V) =V. Since Rank(L+}

LT_{)= Rank(L}s_{), we have thatL}s _{is invertible. SinceL}s _=LT_(L+LT₎†_{L, we have}

that Land LT _{are invertible. ThusL(V) = V andL}T_{(V) =V. Therefore,L(V) =}

LT_{(V) = (L+L}T_)(V).

Subcase 2.2: SupposeL+LT _{is not invertible. SinceL(V)⊆(L+L}T_{)(V), we have}

Rank(L)_≤Rank(L+LT_{). Since Rank(L+L}T_{)= Rank(L}s₎

≤Rank(L)_≤Rank(L+LT_),

we have Rank(L+LT_{)= Rank(L). Again, since L(V)}

⊆ (L+LT_{)(V), we have}

L(V) = (L+LT_{)(V). By symmetry, we haveL}T_{(V) = (L+L}T_{)(V). This completes}

the “Only if” part of Theorem 3.3.

Conversely, suppose thatν(L+LT_{) = 1,L(V) =L}T_{(V) = (L+L}T_{)(V), and} −Ls

is copositive. Then there exists a closed convex cone T that satisfies the conditions in Lemma 3.14. We claim that

L(K)_⊆T△_{∪ −}T△=_{y:_h(L+LT)†_{(y), y}

i ≤0_{} ∩}(L+LT)(V). Now, for anyy_∈L(K), there exists x_∈K such thaty=L(x).

Theny=L(x)_∈L(V) = (L+LT)(V). Hence

h(L+LT)†_{(y), y}

i=_h(L+LT)†_{L(x), L(x)}

i=_hLT(L+LT)†_{L(x), x}

i=_hLs(x), x_{i ≤}0. Thus, y _∈T△_{∪ −}T△, and hence L(K) _⊆T△_{∪ −}T△. Therefore, the claim holds. Now, it is enough to show that the condition

is equivalent to the condition (iii) in Lemma 3.15. Suppose that the condition (iii) holds. Then it is obvious that the condition (3.8) holds. Conversely, Suppose that the condition (3.8) holds, but the condition (iii) does not hold.

ThenL(K)_{\ {}0_{} ⊆}T△_{\ {}0_{} ∪ −}T△_{\ {}0_}. It is clear that T△_{\ {}0_{} ∪ −}T△_{\ {}0_} is a separation of the setL(K)_{\ {}0_}. Now, we show thatL(K)_{\ {}0_}is connected set. We note that L(K) is a convex cone, and dimL(K) = dimL(K₋K) = dimL(V) = Rank(L)≥2. We consider the following cases:

Case 1: Suppose that 0 is on the boundary ofL(K). ThenL(K)_{\ {}0_} is connected set.

Case 2: Suppose that 0 is in the relative interior ofL(K). ThenL(K) is a subspace. HenceL(K)_{\ {}0_}is (path) connected because of Rank(L(K))_≥2.

This is a contradiction. Thus, the condition (iii) holds, and hence Lis MPSBD by Lemma 3.15.

Theorem 3.8 immediately yields the following.

Lemma 3.16. _{Suppose that Rank(L)≥2. ThenL is MPSBD if and only ifL}T is MPSBD.

When Rank(L) = 1, the above lemma is not true, as the following example shows.

Example 3.17. _Let

L=

 

1 0 0 2 0 0 0 0 0

 :L3

→ L3_.

Then it is easy to verify thatLis MPSBD butLT _{is not.}

Theorem 3.18. _{Suppose that L is pseudomonotone with Rank(L)≥ 2. Then}

LT _{is pseudomonotone.}

Proof. Assume without loss of generality that L is not monotone. Hence, by Theorem 3.7, L is PSBD and copositive. Thus, LT is copositive and by the above lemma, LT _{is PSBD. Now, suppose that there exists x such that hL}T_{(x), xi < 0,}

L(x)_≤0, andLT_(x−_{) = 0. SinceL is PSBD, by Theorem 3.8, L(V) =L}T_{(V), and}

hence Ker(L)=Ker(LT_{). Thus,L(x}−_{) = 0 and}

0>_hLT(x), xi=hL(x), xi=hL(x+−x−), x+−x−_i =_hL(x+), x+₋x−_i

=_hL(x+), x+_{i − h}x+, LT(x−)_i =_hL(x+_{), x}+

The last inequality comes from copositivity ofL. This is a contradiction. Using Lemma 3.5 shows thatLT is pseudomonotone.

When Rank(L)=1, the above theorem is not true, as the following example shows.

Example 3.19. _Let

L=

0 0 1 1

:R2_→R2.

Then it is easy to verify thatLis pseudomonotone, but LT _{is not.}

In the matrix case, Gowda ([6]) conjectured that pseudomonotonicity ofLimplies pseudomonotonicity ofLT _{whenLis normal (LL}T _=LT_{L). In what follows, we give}

a positive answer in the setting of Euclidean Jordan algebras.

First, we prove the following lemma.

Lemma 3.20. _{Suppose L is pseudomonotone defined by L(x) = ha, xib with}

0₆=a_≥0 and0₆=b_∈V. Then _ha, x_ihb, x_{i ≥}0 for allx_≥0 impliesb_≥0. Proof. SinceL(x) =_ha, x_ib, we haveLT_{(x) =}

hb, x_ia. SinceLis pseudomonotone, L is PSBD and copositive star by Theorem 3.7. Since ha, x_{i ≥} 0 for all x_≥ 0, we consider the following cases:

Case 1: Ifha, x_i>0 for allx_≥0, thenha, x_ihb, x_{i ≥}0 for allx_≥0 implieshb, x_{i ≥}0 for allx_≥0. Hence,b_≥0.

Case 2: If _ha, x0i= 0 for somex0≥0, thenL(x0) = 0. By copositive star property

ofL, LT(x0)≤0, so hb, x0ia≤0. Ifhb, x0i<0, there exist au >0 andǫ >0, such

that hb, x0+ǫui<0. Since x0+ǫu≥0, we haveha, x0+ǫuihb, x0+ǫui ≥0. Since ha, x0+ǫui=ǫha, ui>0, we haveha, x0+ǫuihb, x0+ǫui<0. This is a contradiction.

Thus,_hb, x0i= 0.

Now, combining cases 1 and 2, we get that_ha, x_ihb, x_{i ≥}0 for allx_≥0, which implies

hb, x_{i ≥}0. Hence,b_≥0.

Remark 3.21. _{If 0 6=a≤ 0 in the above lemma, then replacinga by−a and}

repeating the argument in the proof, we haveb_≤0.

Theorem 3.22. _{IfLis a normal and pseudomonotone, thenL}T _is pseudomono-tone.

Proof. When Rank(L)=2, the implication follows from Theorem 3.18.

When Rank(L)=0, it is obvious. We will show that it is true for Rank(L)=1 case. Since Rank(L)=1, there exist nonzeroa, b_∈V, such thatL(x) =_ha, x_ib. Then we have LT_{(x) = hb, xia. Since L is pseudomonotone, L is PSBD and copositive}

ha, x_ihb, x_i < 0, which implies _hL(x), x_i < 0. Then we have that either LT_{(x) =} hb, x_ia _≥ 0 or LT(x) = _hb, x_ia _≤ 0 by PSBD of L. Since _hb, x_{i 6}= 0, we have that either a _≥ 0 or a _≤ 0. Since L is copositive, we have _ha, y_ihb, y_{i ≥} 0 for all y _≥ 0, which implies that either a _≥ 0 and b _≥ 0 or a _≤ 0 and b _≤ 0 by the above lemma and remark. This implies that either L(x) _≥ 0 or L(x) _≤ 0. Therefore,LT is PSBD. Now, suppose that_hLT(x), x_i<0,L(x)_≤0 andLT(x−_{) = 0.}

Then hL(x−_{), L(x}−_{)i = hx}−_{, L}T_L(x−_{)i = hx}−_{, LL}T_(x−_{)i = 0, which implies that}

L(x−) = 0. Then, as in the proof of Theorem 3.18,

0>_hLT(x), xi=hL(x), xi=hL(x+−x−), x+−x−_i =_hL(x+), x+₋x−_i

=_hL(x+), x+_{i − h}x+, LT(x−)_i =_hL(x+), x+_{i ≥}0.

The last inequality comes from copositivity ofL. This is a contradiction. By Lemma 3.5,LT is pseudomonotone.

In the examples below, we describe matrices onL3 _{of rank 1, 2, and 3 that are}

pseudomonotone, but not monotone. The proofs are given in Appendix.

Example 3.23. _Let

L=

 

a 0 0 b 0 0 c 0 0



:_L3_{→ L}3

be such thata_≥√b2_+c2_{. ThenLis pseudomonotone.}

Remark 3.24. _{By puttinga=b= 1 and c= 0, we get that}

L=

 

1 0 0 1 0 0 0 0 0



:_L3_{→ L}3

is pseudomonotone, but not monotone.

Example 3.25. _Let

L=

 

1 c 0 b 0 0 0 0 0



:_L3_{→ L}3

be such that

(c) b <0.

ThenLis pseudomonotone.

Remark 3.26. _{By puttingb=−}1

4 andc= 1

2, we get that

L=

 

1 1₂ 0

−1 4 0 0

0 0 0

 :_L3

→ L3

is pseudomonotone, but not monotone.

Example 3.27. _Let

L=

 

1 c 0 b 0 0 0 0 ǫ

 :L3

→ L3

be such that

(a) 0< b+c <1, (b) 0< c <√2, (c) b <0, (d) 0< ǫ_≤1,

(e) (3b+c)2_−4b2_(2−c2_)<0.

ThenLis pseudomonotone.

Remark 3.28. _{By puttingb=−}1

4,c= 1

2 andǫ= 1, we get that

L=

 

1 1 2 0 −14 0 0

0 0 1

 :L3

→ L3

is pseudomonotone, but not monotone.

Theorem 3.29. _{Suppose thatL is self-adjoint and copositive. Then Lis PSBD}

if and only if Lis monotone.

Proof. As the “If” part is obvious, we prove the “Only if” part. SinceLis self-adjoint, every eigenvalue of Lis real. Assume there exists an eigenvalueλ <0, and nonzerox_∈V such thatL(x) =λx. Then we havehL(x), xi=hλx, x_i=λ_||x_||2_<0;

it follows that LT_{(x) = L(x) = λx ≥ 0 or L}T_{(x) = L(x) = λx ≤ 0 since L is}

Corollary 3.30. _{Suppose that Lis self-adjoint. Then Lis pseudomonotone if}

and only ifL is monotone.

4. Pseudomonotonicity and principal subtransformations. In the matrix case, Mohan, Neogy and Das (see [16]) showed that ifL is PSBD, then its principal subtransformations are also PSBD. In this section, we study the relation between pseudomonotonicity ofL and its principal subtransformations.

First, we recall the notion of “principal subtransformations” of a given linear transformation onV.

Given a Jordan frame{e1, . . . , er}in a Euclidean Jordan algebraV, we define

V(l):=V(e1+e2+· · ·+el,1) :={x∈V :x◦(e1+e2+· · ·+el) =x}

for 1 _≤ l _≤ r. Corresponding to V(l), we consider the orthogonal projection P(l) : V _7→V(l)_{. For a given linear transformationL:V}

7→V, the transformationP(l) ◦L: V(l) _{7→ V}(l) _{is, by definition, a principal subtransformation of L corresponding to} {e1, . . . , el}, and is denoted byL{e1,...,el}(the symbol “◦” means here the composition

rather than Jordan multiplication).

We note that for a given Jordan frame{e1, . . . , er}, we can permute the objects

and select the first l objects (for any 1_≤ l _≤r). Thus, there are 2r_{−1 principal}

subtransformations corresponding to a Jordan frame. Of course, by taking other Jordan frames, we generate other principal subtransformations.

Proposition 4.1. _{Fix a Jordan frame {e1, . . . , er}. Suppose that L is PSBD.}

ThenP(l)_◦Lis PSBD for anyl,1_≤l_≤r.

Proof. Let Lb = P(l)_{◦L : V}(l) _{→ V}(l)_{. Then for x ∈ V}(l)_{, 0 > hx,L(x)b i =} hx, P(l)_◦L(x)_i = _hx, L(x)_i, hence, LT_{(x) ≤ 0 orL}T_{(x) ≥ 0, because L is PSBD.}

Suppose LT(x)_≤ 0. Then for any u _≥0 in V(l), _hu,LbT(x)_i= _hL(u), xb _i = _hP(l)_◦ L(u), x_i=_hL(u), P(l)_(x)

i =_hL(u), x_i=_hu, LT_(x)

i ≤0. Hence, LbT_(x)

≤0 in V(l)_.

Similarly, we show that LbT_{(x)≥ 0 in V}(l) _{when L}T_{(x) ≥0. Therefore, P}(l) ◦L is PSBD for anyl, 1_≤l_≤r.

Lemma 4.2. _{Fix a Jordan frame {e1, . . . , er}. If L is copositive on V, then}

P(l)_◦Lis copositive onV(l) for anyl,1≤l_≤r.

Proof. Suppose x _≥ 0 in V(l). Then _hx, P(l)_◦ L(x)_i = _hP(l)(x), L(x)_i =

hx, L(x)i ≥0.

Theorem 4.3. _{Fix a Jordan frame{e1, . . . , er}. IfLis pseudomonotone and}

Rank(P(l)

◦L)_≥2for1< l_≤r, thenP(l)

3.5, suppose that L is pseudomonotone. Then L is PSBD and copositive star by Theorem 3.7. HenceP(l)_◦Lis PSBD for 1_≤l_≤rby Proposition 4.1 and copositive by the above lemma. Suppose there existsl0(1< l0≤r) such thatLc0:=P(l0)◦Lis not

pseudomonotone. Then there existsx_∈V(l0)_{such that}

hx,Lc0(x)i<0,LbT0(x)≤0 and

b

L0(x−) = 0. SinceLc0is PSBD and Rank(Lc0)≥2, we haveLc0(V(l0)) =LbT0(V(l0)) by

Theorem 3.8; thus Ker(Lc0)=Ker(Lb0T). HenceLbT0(x−) = 0. Repeating the argument

given in the proof of Theorem 3.18 and 3.22, we come to a contradiction, soLb0(x−)6=

0. Thus,P(l)_◦Lis pseudomonotone for anyl, 1< l_≤r.

5. Pseudomonotonicity of some specialized transformations.

5.1. Quadratic representations. We now characterize the pseudomonotonic-ity for quadratic representations.

Given any elementainV, thequadratic representationofais defined by Pa(x) := 2a◦(a◦x)−a2◦x.

RecallPa(K)⊆K, soPa is copositive.

Theorem 5.1. _{Fora∈V, the following are equivalent:}

(a) Pa is monotone on V. (b) Pa is pseudomonotone.

If, in addition,V is simple, then the above conditions are further equivalent to

(c) _±a_∈K.

Proof. SincePa is self-adjoint and copositive, we only need to show (a) is

equiv-alent to (c), when V is simple. For a given a _∈ V, there exits a Jordan frame

{e1, e2, . . . , er} such that

a=a1e1+a2e2+· · ·+arer.

For any x_∈ V, we have the Peirce decomposition of xwith respect to this Jordan frame

x=

r

X

i=1

xiei+

X

i<j

xij

(withxi ∈Randxij∈Vij). It can be easily verified that

Pa(x) = r

X

i=1

ai2xiei+

X

i<j

WhenV is simple, Vij is nonzero for each i≤j (see Corollary IV.2.4 in [3]), so we

have

0_{≤ h}x, Pa(x)i = r

X

i=1

ai2xi2||ei||2+

X

i<j

aiaj||xij||2 (∀x∈V) ⇔aiaj≥0 (i≤j)

⇔ai≥0 orai≤0,∀i.

Hence, whenV is simple,Pa is monotone onV if and only if±a∈K.

Remark 5.2. _{WhenV =S}n_{, for a realn×nmatrixA, thetwo sided} multiplica-tive transformationis defined by

MA(X) :=AXAT.

Clearly,MAis copositive. If we specializePa onSn, thenPA(X) =AXAforA∈ Sn.

Thus forMA, the following are equivalent whenAis a real symmetric square matrix.

(a) Ais either positive semidefinite or negative semidefinite. (b) MA is monotone.

(c) MA is pseudomonotone.

When A is any nonsingular matrix, we claim that MA is PSBD if and only if MA

is monotone. We only need to show thatMA is monotone ifMA is PSBD. Suppose

there exists X _{∈ S}n_{, such that hX, M}

A(X)i < 0. Then MAT(X) = MAT(X) 0

or M_AT(X) 0. Let 0 Y := M

AT(X) = ATXA. Then X = (A−1)TY A−1

0. As MA is copositive, this contradicts hX, MA(X)i < 0. Similarly, we can get

the contradiction whenY 0. Since monotonicity implies pseudomonotonicity and pseudomonotonicity implies PSBD, we haveMAis pseudomonotone if and only ifMA

is monotone when Ais any nonsingular matrix. This need not be true when A is a singular matrix, as the following example shows.

Example 5.3. _Let

A=

0 0 1 1

.

Then, it is easy to verify thatMAis PSBD, but not monotone.

5.2. Automorphisms. In [16], Mohan, Neogy and Das showed that PSBD ma-trices are invariant under principal rearrangement. In this section, we show that PSBD property is invariant for cone invariant transformations and pseudomonotonic-ity is invariant under algebra and cone automorphisms.

A linear transformation Γ :V _→V is said to be acone automorphismif Γ(K) = K. BecauseK◦₆=_∅, such a transformation is necessarily invertible. We denote the set of all automorphisms ofK by Aut(K). It is immediate that Aut(V)_⊆Aut(K).

We use Π(K) to denote the set of all linear transformations onV that leave K invariant, i.e.,L(K)_⊆Kfor anyL_∈Π(K).

We note that Aut(V)⊆Aut(K)⊆Π(K).

To illustrate these concepts, we recall the following examples from [9].

Example 5.4. _{ConsiderV =R}n_{. In this case, the permutation matrices are the}

automorphisms ofRn_{and any automorphism ofR}n

+is a product of a positive definite

diagonal matrix and a permutation matrix.

Example 5.5. _{(see [9]) ConsiderV =S}n_{. In this case, for any Γ ∈ Aut(S}n +),

there exists an invertible matrix Q _∈ Rn×n _{such that Γ(Z) = QZQ}T_{, ∀Z ∈ S}n_.

In particular, for Λ _∈Aut(_Sn_{), there exists a real orthogonal matrix U such that}

Λ(Z) =U ZUT,

∀Z_{∈ S}n_.

Example 5.6. _{(see [9]) Consider V = L}n_{. In this case, an n×n matrix}

A _∈Aut(_Ln

+) if and only if there exists a µ > 0 such that ATJnA = µJn, where

Jn = diag(1,−1, . . . ,−1). In particular, for A ∈Aut(Ln), A =

_{1 0}

0 D

, where

D:Rn−1_→Rn−1_{is an orthogonal matrix.}

Theorem 5.7. _{If L is PSBD, thenP LP}T _{is PSBD for allP ∈Π(K).}

Proof. Suppose hx, P LPT_{(x)i < 0. Then we have hP}T_{(x), LP}T_{(x)i < 0. Let}

y = PT_{(x). Then hy, L(y)i < 0, which implies either L}T_{(y) ≥ 0 or L}T_{(y) ≤ 0,}

becauseLis PSBD. SinceP _∈Π(K), we have eitherP LT_{(y)≥0 orP L}T_{(y)≤0, i.e.,}

P LT_PT_(x)

≥0 orP LT_PT_(x)

≤0. Hence,P LPT _{is PSBD.}

Recall the following proposition from [8].

Proposition 5.8. _{(Proposition 4.1, [8]) Let Γ ∈Aut(K). Then Γ}−1 _{and Γ}T _∈

Aut(K).

Theorem 5.9. _{Let Γ∈Aut(K). ThenLis pseudomonotone if and only ifΓLΓ}T is pseudomonotone.

Proof. To use Theorem 3.7, first, we show thatL is PSBD if and only if ΓLΓT

is PSBD. Suppose that L is PSBD. The implication of that ΓLΓT _{is PSBD follows}

from Theorem 5.7 because Γ_∈Aut(K)_⊆Π(K). The converse follows from the fact that Γ−1_,ΓT _{∈Aut(K)⊆Π(K). Now, we prove thatLis copositive star if and only}

and _hx,ΓLΓT_{(x)i= 0. Then, we have Γ}T_{(x) ≥ 0, because Γ}T _{∈ Aut(K)⊆ Π(K);}

LΓT_(x)

≥ 0, because Γ−1

∈ Aut(K) _⊆ Π(K) and _hΓT_{(x), LΓ}T_(x)

i = 0. Since L is copositive star, we have LT_ΓT_(x)

≤ 0, so ΓLT_ΓT_(x)

≤ 0. Therefore, ΓLΓT _is

copositive star. Again, the converse follows from the fact that Γ−1_,ΓT _∈Aut(K)⊆

Π(K).

We note that the above theorem holds if Γ is replaced by Λ because Aut(V)_⊆ Aut(K).

5.3. Z transformations. We say thatLhas the Z-property on V if x, y_∈K, and_hx, y_i= 0_{⇒ h}L(x), y_{i ≤}0.

Recently, Gowda and Tao (see [10]) introduced and studied the properties of such transformations.

Theorem 5.10. _{Suppose that L has the Z and copositive properties. ThenL is}

monotone.

Proof. Assume that L has the Z and copositive properties. Then LT _{has the}

Z and copositive properties. So A := L+LT _{has the Z and copositive properties.}

Letλbe the minimum eigenvalue ofA. Then a corresponding eigenvectoruis inK (see Theorem 6, [17]). It follows that 0≤ hA(u), ui=λ_||u_||2_{, so λ≥0. Thus A is}

monotone. HenceLis monotone.

This theorem and Lemma 3.1 immediately yield the following result.

Corollary 5.11. _{Suppose thatLhas the Z-property andL is pseudomonotone.}

ThenLis monotone.

Example 5.12. _{WhenV =S}n_{, for a realn×nmatrixA, theLyapunov} trans-formationis defined by

LA(X) :=AX+XAT.

Then LA has the Z-property (see [10]). It is easy to verify that hLA(c), ci ≥0 for

all primitive idempotents c in Sn

+ if and only if A is positive semidefinite. Since hLA(c), ci ≥0 for all primitive idempotentscinS+n if and only ifLAis monotone (see

the proof of Theorem 7.1 in [7]), we have that LA is pseudomonotone if and only if

Ais positive semidefinite by the above theorem.

5.4. Relaxation transformations. In this section, we apply the ideas of the previous sections to study the transformationRA:V →V that arises from a matrix

Suppose we are given a Jordan frame _{e1, . . . , er} in V and A ∈ Rr×r. We define

RA:V →V as follows. For anyx∈V, we write the Peirce decomposition

x=

r

X

1

xiei+

X

i<j

xij.

Then

RA(x) := r

X

1

yiei+

X

i<j

xij,

where

[y1. . . yr]T =A [x1. . . xr]T

.

Our objective in this section is to study some interconnections between the prop-erties ofA and the properties ofRA. Such a study has found to be quite interesting

and useful in the context of matrix based linear transformations on V = _Sn _and

V =_Ln_{. In particular, it will provide examples to study complementarity problems}

onV (see [18]). In what follows, we will study the relationship between a matrix A and a transformationRA, when RA is pseudomonotone.

Through this discussion, we denoteD:=diag(_||e1||2, . . . ,||er||2).

Theorem 5.13. _{A is pseudomonotone on R}r

+ if and only if RD−1_A+I −I ′ _is

pseudomonotone on K, where I is an identity matrix on Rr, and I′ :V _→V is an identity transformation onV.

Proof. “Only if” part. LetB :=D−1A+I. ThenA=D(B₋I). For allx, y_≥0, let

x=

r

X

1

xiei+

X

i<j

xij and y= r

X

1

yiei+

X

i<j

yij.

Thenxi≥0 andyi≥0 for alli= 1, . . . , rand

RB(x) := r

X

1

¯ xiei+

X

i<j

xij,

where

[¯x1. . .x¯r]T =B [x1. . . xr]T.

Then we have

hRB(x), y−xi= r

X

1

¯

xi(yi−xi)||ei||2+

X

i<j

We note that

hx, y₋x_i=

r

X

1

xi(yi−xi)||ei||2+

X

i<j

hxij, yij−xiji.

We define ˆx:= [x1. . . xr]T and ˆy:= [y1. . . yr]T. Then we have

h(RB−I′)(x), y−xi= r

X

1

(¯xi−xi)(yi−xi)||ei||2

=_h(B₋I)ˆx, D(ˆy₋x)ˆ _i =_hD(B₋I)ˆx,(ˆy₋x)ˆ _i =_hAˆx,yˆ₋xˆ_i.

Similarly,h(RB−I′)(y), y−xi=hAˆy,yˆ−xˆi. Assume thath(RB−I′)(x), y−xi ≥0,

which means_hAˆx,yˆ₋xˆ_{i ≥}0.SinceAis pseudomonotone, we have_hAˆy,yˆ₋xˆ_{i ≥}0, and hence_h(RB−I′)(y), y−xi ≥0. Therefore,RB−I′=RD−1A+I−I

′_{is pseudomonotone}

onK.

“If” part. Let ˆx= [x1. . . xr]T and ˆy = [y1. . . yr]T withxi≥0 andyi≥0 for all

1_≤i_≤r. Suppose_hAˆx,yˆ₋xˆ_{i ≥}0. Then we definex=Pr₁xieiandy=Pr1yiei. So

x, y_≥0. Then thatAis pseudomonotone follows from the proof of “Only if” part.

Theorem 5.14. _{If RA is pseudomonotone on K, thenDA is pseudomonotone}

onRr +.

Proof. Let x= [x1. . . xr]T and y = [y1. . . yr]T with xi ≥ 0 and yi ≥ 0 for all

1 _≤i_≤ r. Suppose_h(DA)x, y₋x_{i ≥} 0. Define ¯x=P_i=1r xiei and ¯y =Pri=1yiei.

Thus, ¯x_≥0, ¯y_≥0, and

RA(¯x) := r

X

1

ziei,

where

[z1. . . zr]T =A [x1. . . xr]T

. Therefore,

hRA(¯x),y¯−x¯i= r

X

1

zi(yi−xi)||ei||2=hAx, D(y−x)i=h(DA)x, y−xi ≥0.

The following example shows that the converse of the above theorem may not be true.

Example 5.15. _Consider RA:L3→ L3, where A=

0 0 1 1

.

Fix a Jordan frame{e1, e2}, wheree1=1₂[1 1 0]T ande2= 1₂[1−1 0]T. Then we

have||e1||2=||e2||2= 1₂ andD=1₂I.

It is easy to verify thatDAis pseudomonotone. Now, takex=e1+2e2+[0 0 −1]T

and y = 12e1+e2+ [0 0 −3]T. It is easy to verify that x, y≥0. Then RA(x) =

3e2+ [0 0 −1]T,RA(y) = 13e2+ [0 0 −3]T andy−x= 11e1−e2+ [0 0 −2]T. By

simple calculation, we have_hRA(x), y−xi= 1₂ >0 andhRA(y), y−xi=−1₂ <0.

6. Concluding remarks. In this paper, we have extended the concept of pseu-domonotonicity fromRn _{to the setting of Euclidean Jordan algebras. Some}

intercon-nections between pseudomonotonicity, monotonicity, and the Z-property were stud-ied. In particular, we have given a characterization of pseudomonotonicity for a linear transformation and a matrix-induced transformation defined on a Euclidean Jordan algebra. We have shown that pseudomonotonicity and monotonicity coincide un-der the condition of the Z-property. In addition, we have proved the invariance of pseudomonotonicity under the algebra and cone automorphisms and described inter-connections between pseudomonotonicity of a linear transformation and its principal subtransformations.

7. Appendix. In this section, we use Theorem 3.7 to prove claims made in Examples 3.23, 3.25, and 3.27 from Section 3.

Example 7.1. _Let

L=

 

a 0 0 b 0 0 c 0 0

 :L3

→ L3

be such thata_≥√b2_+c2_{. ThenLis pseudomonotone.}

Proof. First, we show that L is PSBD. Let x = [x0 x1 x2]T. Then, from hL(x), x_i < 0, we have x0(ax0+bx1+cx2) < 0. Thus, we have either x0 > 0

andax0+bx1+cx2<0 orx0<0 and ax0+bx1+cx2>0. Hence, we have either

LT_{(x) = (ax}

0+bx1+cx2)e >0 orLT(x)<0. Therefore,Lis PSBD. Next, we need

to show thatL is copositive star. For x_≥0, we have thatx0≥√x12+x22. Using

the condition ofa_≥√b2_+c2_{, we have}

ax0 ≥

p

x12+x22

p

b2_+c2

= pb2_x

≥ pb2_x

12+c2x22+ 2bcx1x2

= p(bx1+cx2)2=|bx1+cx2| ⇒ax0+bx1+cx2≥0.

SincehL(x), xi=x0(ax0+bx1+cx2), we havehL(x), xi ≥0 forx≥0. Thus,L is

copositive. It can be easily verify that

[x≥0, L(x)≥0, hL(x), xi= 0]⇒LT(x) = 0.

This shows that L is copositive star. Therefore, by Theorem 3.7, L is pseudomo-notone.

Example 7.2. _Let

L=

 

1 c 0 b 0 0 0 0 0

 :_L3

→ L3

be such that

(a) 0_≤b+c_≤1, (b) c >0,

(c) b <0.

ThenLis pseudomonotone.

Proof. Without loss of generality, we assume thatb+c >0. Letx= [x0 x1 x2]T.

Then

hL(x), xi=x0[x0+ (b+c)x1],

L(x) = [x0+cx1 bx0 0]T,

and LT(x) = [x0+bx1 cx0 0]T.

First, we show thatLis PSBD. Suppose that_hL(x), x_i<0. Then eitherx0>0 and

x0+ (b+c)x1<0 orx0<0 andx0+ (b+c)x1>0.

Case 1: x0>0 andx0+ (b+c)x1<0.Thenx0+ (b+c)x1<0⇒x1<−

x0

b+c <0 by the item (a). Thus, x0+bx1 >0 by the item (c). Since x1 <−

x0

b+c, we have x0+bx1 > x0−

bx0

b+c = cx0

b+c ≥cx0 > 0 by the item (b). Therefore, L

T_{(x) >0.}

Hence,Lis PSBD.

Case 2: x0<0 andx0+ (b+c)x1>0. Thenx0+ (b+c)x1>0⇒x1>−

x0

b+c >0 by the item (a). Thus,x0+bx1<0 by the item (c). Now,−(x0+bx1) =−x0−bx1> −x0+

bx0

b+c =− cx0

b+c ≥ −cx0>0. Therefore,L

Now, we need to show thatLis copositive star. Forx_≥0, we have x0≥

p

x12+x22≥

p

x12=|x1| ≥(b+c)|x1| ⇒x0+ (b+c)x1≥0.

Thus, _hL(x), x_{i ≥} 0, which implies that L is copositive. Now, suppose that x_≥0, L(x)_≥0, and_hL(x), x_i=x0[x0+ (b+c)x1] = 0. Then we have that eitherx0= 0 or

x0+ (b+c)x1= 0. Ifx0= 0, thenx= 0, and henceLT(x) = 0. Ifx0+ (b+c)x1= 0

and x0 > 0, then x1 = −

x0

b+c. Thus, x0 +cx1 = bx0

b+c < 0, which contradicts L(x)≥0. This shows thatLis copositive star. Therefore, Lis pseudomonotone by Theorem 3.7.

Example 7.3. _Let

L=

 

1 c 0 b 0 0 0 0 ǫ

 :L3

→ L3

be such that

(a) 0< b+c <1, (b) 0< c <√2, (c) b <0, (d) 0< ǫ_≤1, (e) (3b+c)2

−4b2₍₂

−c2)<0. ThenLis pseudomonotone.

Proof. Letx= [x0 x1 x2]T. Then we havehL(x), xi=x0[x0+ (b+c)x1] +ǫx22,

L(x) = [x0+cx1 bx0 ǫx2]T and LT(x) = [x0+bx1 cx0 ǫx2]T.

First, we show thatLis PSBD. Suppose thathL(x), xi<0. Thenx0[x0+(b+c)x1]< −ǫx22 ≤0. This implies that either x0 >0 and x0+ (b+c)x1 <0, or x0 <0 and

x0+ (b+c)x1>0.

Case 1: x0 > 0 and x0+ (b+c)x1 < 0. Then we have that x0+bx1 >0 by the

proof in the previous example. Sinceǫ2x22 ≤ǫx22 <−x0[x0+ (b+c)x1], we have

ǫ2x22+c2x02<−x0[x0+(b+c)x1]+c2x02. We show that−x0[x0+(b+c)x1]+c2x02≤

(x0+bx1)2. Now,

−x0[x0+ (b+c)x1] +c2x02≤(x0+bx1)2⇔(2−c2)x02+ (3b+c)x0x1+b2x12≥0.

The last inequality holds because of the items (b) and (e). Therefore,ǫ2_x

22+c2x02≤

(x0+bx1)2. Thus,LT(x)>0, and henceLis PSBD.

Case 2: x0<0 andx0+ (b+c)x1>0. Then we have thatx0+bx1<0 by the proof

in the previous example. From Case 1, we know thatǫ2_x

22+c2x02≤[−(x0+bx1)]2.

Now, we show thatLis copositive star. We know thatx0+(b+c)x1≥0 forx≥0

by the proof in the previous example. Thus, _hL(x), x_{i ≥}0, which implies that L is copositive. It is easy to verify that [x_≥0,L(x)_≥0,_hL(x), x_i= 0]_⇒LT_{(x) = 0 by}

the proof in the previous example. This shows thatL is copositive star. Therefore, by Theorem 3.7,Lis pseudomonotone.

Acknowledgment. I am grateful to Professor M.S. Gowda for helpful sugges-tions.

REFERENCES

[1] J.-P. Crouzeix, A. Hassouni, A. Lahlou, and S. Schaible. Positive subdefinite matrices, gener-alized monotonicity and linear complementarity problems. SIAM J. Matrix Anal. Appl., 22:66–85, 2000.

[2] J.-P. Crouzeix and S. Schaible. Generalized monotone affine maps.SIAM J. Matrix Anal. Appl., 17:992–997, 1996.

[3] J. Faraut and A. Kor´anyi.Analysis on Symmetric Cones. Clarendon Press, Oxford, 1994. [4] M.S. Gowda. Pseudomonotone and copositive star matrices.Linear Algebra Appl., 113:107–118,

1989.

[5] M.S. Gowda. Affine pseudomonotone mappings and the linear complementarity problem. SIAM J. Matrix Anal. Appl., 11:373–380, 1990.

[6] M.S. Gowda. On the transpose of a pseudomonotone matrix and the LCP.Linear Algebra Appl., 140:129–137, 1990.

[7] M.S. Gowda and R. Sznajder. Some global uniqueness and solvability results for linear comple-mentarity problems over symmetric cones. SIAM J. Optim., 18:461–481, 2007.

[8] M.S. Gowda and R. Sznajder. Automorphism invariance of P _{and GUS-properties of linear}

transformations on Euclidean Jordan algebras. Math. Oper. Res., 31:109–123, 2006. [9] M.S. Gowda, R. Sznajder, and J. Tao. Some P-properties for linear transformations on Euclidean

Jordan algebras. Linear Algebra Appl., 393:203–232, 2004.

[10] M.S. Gowda and J. Tao. Z-transformations on proper and symmetric cones. Math. Program., Series B, 117:195–221, 2009.

[11] N. Hadjisavvas and S. Schaible. Quasimonotonicity and pseudomonotonicity in variational in-equalities and equilibrium problems. In J.-P. Crouzeix, J.-E. Martinez-Legaz, and M. Volle, editors,Generalized Convexity, Generalized Monotonicity: Recent Results, Nonconvex Op-tim. Appl., 27, Kluwer Acad. Publ., Dordrecht, 257–275, 1998.

[12] A. Hassouni, A. Lahlou, and A. Lamghri. Existence theorems for linear complementarity prob-lems on solid closed convex cones.J. Optim. Theory Appl., 126:225–246, 2005.

[13] S. Karamardian. Complementarity problems over cones with monotone and pseudomonotone maps. J. Optim. Theory Appl., 18:445–454, 1976.

[14] S. Karamardian and S. Schaible. Seven kinds of monotone maps. J. Optim. Theory Appl., 66:37–46, 1990.

[15] S. Karamardian, S. Schaible and J.-P Crouzeix. Characterizations of generalized monotone maps. J. Optim. Theory Appl., 76:399–413, 1993.

[16] S.R. Mohan, S.K. Neogy and A.K. Das. More on positive subdefinite matrices and the linear complementarity problem. Linear Algebra Appl., 338:275–285, 2001.

[18] J. Tao. Positive principal minor property of linear transformations on Euclidean Jordan algebras.