LNALNA
More specifically, given matrices A, B, C ∈ IRn×n, the QEiCP consists of finding (x, y, λ)∈ IRn× IRn× IR such that
y = λ2Ax + λBx + Cx, x≥ 0, y ≥ 0, xTy = 0, aTx = 1.
In the literature, further extensions of the EiCP and the QEiCP are introduced in [1, 3, 11, 15], which include second-order cone eigenvalue complementarity problem (SOCEiCP) and general closed convex cone eigenvalue complementarity problem.
Their general format are as below:
(1.2)
y = λBx− Cx,
x∈ K, y ∈ K∗, xTy = 0, aTx = 1,
or
y = λ2Ax + λBx + Cx, x∈ K, y ∈ K∗, xTy = 0, aTx = 1,
where K is a closed convex cone, K∗ denotes its dual cone, and a ∈ int(K) is arbitrary fixed point. When K represents the second-order cone (denoted by Kn), they become second-order cone eigenvalue complementarity problem (SOCEiCP) and second-order cone quadratic eigenvalue complementarity problem (SOCQEiCP) [17]. For the concept and related properties of second-order cone, please refer to [2, 6, 7, 8, 9, 12, 13, 14, 26].
Roughly, there are two main research directions regarding the eigenvalue com- plementarity problems. One is on the theoretical side in which their corresponding solution properties are investigated, see [3, 4, 5, 11, 21, 23, 24, 25]. The other one focuses on the numerical algorithm for solving the problems, which include the Lattice projection method, the semismooth Newton methods, the RLT-based branch and bound method (BBRLT) and so on, see [1, 3, 4, 5, 15, 21, 22, 23, 24]
and references therein. As seen in the literature, almost all the attention is paid to the standard eigenvalue complementarity problems and second-order cone eigen- value complementarity problems [17]. However, the study about other special conic eigenvalue complementarity problems is very limited. On the other hand, more and more nonsymmetric cones (for instance, circular cone, p-order cone) appear in plenty of real applications. Hence, in this manuscript, we are therefore motivated to extend the concepts and properties of the EiCP and the QEiCP to the setting of the circular cone Lθ and p-order cone Kp. In other words, the cone K in problem (1.2) is circular coneLθ and p-order cone Kp. For the concepts of circular coneLθ
and p-order cone Kp, we will review them in details in the next section.
In this paper, several results about the solutions of the eigenvalue complementar- ity problem (EiCP) and the quadratic eigenvalue complementarity problems (QE- iCP) are extended to the version of circular conesLθ and p-order conesKn, respec- tively. In the setting of circular cone, the relationship between the solution of the circular cone eigenvalue complementarity problems (CCEiCP) and the solution of the corresponding circular cone complementarity problems (or the solution of the circular cone quadratic eigenvalue complementarity problems (CCQEiCP) and the
solution of the corresponding circular cone complementarity problems) are estab- lished. Besides, for the p-order cone eigenvalue complementarity problems (POCE- iCP) and p-order cone eigenvalue complementarity problems (POCQEiCP), we also sum up the corresponding theoretical results. The solutions of the POCEiCP or the POCQEiCP and the solutions of the corresponding p-order cone complementarity problems correspond to each other. These results build up a theoretical basis for designing numerical solution algorithm or making further study about their corre- sponding eigenvalue complementarity problems in the future.
The remainder of this paper is organized as follows. In section 2, we recall some background materials regarding circular cone and p-order cone. In section 3, we study the properties of the solutions for the eigenvalue complementarity problems and quadratic eigenvalue complementarity problems in the setting of circular cone, respectively. In section 4, based on the cases of circular cone eigenvalue optimiza- tion problems, we study the corresponding properties of the solutions for p-order cone eigenvalue complementarity problems and p-order cone quadratic eigenvalue complementarity problems, respectively.
2. Preliminaries
In this section, we briefly review some basic concepts and background materials about the circular cone Lθ and the p-order coneKp, which will be extensively used in the subsequent sections. More details can be found in [10, 16, 18, 20, 27, 28].
Let Lθ denote the circular cone (see [10, 16, 18, 28]) in IRn, which is defined by Lθ :={x = (x1, x2)∈ IR × IRn−1| ∥x2∥ ≤ x1tan θ}
with ∥ · ∥ denoting the Euclidean norm and θ ∈ (0,π2). It is well known that the dual cone of Lθ can be expressed as
(Lθ)∗ ={x = (x1, x2)∈ IR × IRn−1| ∥x2∥ ≤ x1cot θ} = Lπ2−θ.
Moreover, Zhou and Chen [28] gave the below spectral decomposition of x = (x1, x2)∈ IR × IRn−1 with respect toLθ:
x = µ1(x)vx(1)+ µ2(x)v(2)x , where µ1(x), µ2(x), v(1)x , and vx(2) are expressed as
µ1(x) = x1− ∥x2∥ cot θ, µ2(x) = x1+∥x2∥ tan θ, and
vx(1) = 1 1 + cot2θ
[ 1
− cot θ · w ]
, v(2)x = 1 1 + tan2θ
[ 1
tan θ· w ]
. with w = ∥xx2
2∥ if x2 ̸= 0, or any vector in IRn−1 satisfying ∥w∥ = 1 if x2 = 0.
As for the p-order cone. Let Kp denote the p-order cone (p > 1) (see [19, 20, 27]) in IRn, which is defined by
Kp :={x = (x1, x2)∈ IR × IRn−1| ∥x2∥p≤ x1}
with ∥ · ∥p denoting the p-norm. It is clear to see that when p = 2, K2 is exactly the second-order cone, which confirms that the second-order cone is a special case of p-order cone. It is also well known that the dual cone ofKp can be expressed as
(Kp)∗ ={x = (x1, x2)∈ IR × IRn−1| ∥x2∥q ≤ x1} = Kq,
where q > 1 and satisfies 1p + 1q = 1. Miao, Qi and Chen [20] gave the following spectral decomposition of x = (x1, x2)∈ IR × IRn−1 with respect toKp:
x = α1(x)v(1)x + α2(x)v(2)x , where α1(x), α2(x), v1x, and vx2 are expressed as
α1(x) = (x1+∥x2∥p)
2 , α2(x) = (x1− ∥x2∥p)
2 ,
and
vx(1)= [ 1
w ]
, vx(2)= [ 1
−w ]
. with w = ∥xx2
2∥p if x2 ̸= 0, or any vector in IRn−1 satisfying ∥w∥p = 1 if x2 = 0.
3. Circular cone eigenvalue complementarity problems
Now, we consider the eigenvalue complementarity problems and circular cone quadratic eigenvalue complementarity problems in the setting of the circular cone Lθ. We will sum up the relationship between the solution of circular cone eigen- value complementarity problems (or the solution of circular cone quadratic eigen- value complementarity problems) and the solution of the corresponding circular cone complementarity problems, respectively.
3.1. Circular cone eigenvalue complementarity problems. Consider the cir- cular cone eigenvalue complementarity problems (CCEiCP for short): find (x, y, λ)∈ IRn× IRn× IR such that
(3.1) CCEiCP(B, C) :
y = λBx− Cx,
x∈ Lθ, y∈ (Lθ)∗, xTy = 0, aTx = 1,
where B, C ∈ IRn×n and a is an arbitrary fixed point with a∈ int((Lθ)∗). We first study the solvability of the CCEiCP that will be used in subsequent analysis.
Proposition 1. Suppose that x := (x1, x2)∈ Lθ, y := (y1, y2)∈ (Lθ)∗. Then, the following hold.
(a) xTy≥ 0.
(b) If y ∈ int(Lθ), then xTy > 0 if and only if x̸= 0.
(c) If x̸= 0 and y ̸= 0, then
xTy = 0 ⇐⇒ x1tan θ =∥x2∥ y1cot θ =∥y2∥ and y = α(x1tan2θ,−x2), where α is a positive constant, or x = β(y1cot2θ,−y2) with a positive con- stant β.
Proof. (a) It is obvious by the definition of dual cone.
(b) The result that xTy > 0 implies x ̸= 0 is trivial. We only prove the other direction as follows. Since x ̸= 0, we know that x1 > 0. Otherwise, we have that
∥x2∥ ≤ x1 = 0 implies x = 0. Then, it follows that
xTy = x1y1+ xT2y2≥ x1y1− ∥x2∥∥y2∥ > x1y1− x1tan θ y1cot θ = 0.
(c) The proof of this direction “⇐” is trivial. We only need to prove the direction
“⇒”. By the assumption x ̸= 0, we know that x1 > 0. Similarly, we obtain that y1> 0. Using the condition xTy = 0, the Schwarz’s inequality and the definition of circular cones , we have
(3.2) x1y1=|xT2y2| ≤ ∥x2∥∥y2∥ ≤ (x1tan θ)(y1cot θ) = x1y1. Thus, it follows that
(3.3) (x1tan θ)(y1cot θ) = x1y1 =∥x2∥∥y2∥.
By combining ∥x2∥ ≤ x1tan θ,∥y2∥ ≤ y1cot θ and (3.3), it yields (3.4) x1tan θ =∥x2∥ and y1cot θ =∥y2∥.
Besides, by (3.2), (3.3) and (3.4) again, we have
xT2y2=−x1y1=−∥x2∥∥y2∥.
This implies the equality holds in the Schwarz’s inequality. Hence, there exists a constant k such that
(3.5) y2 = kx2.
Since k∥y2∥2 = xT2y2 =−x1y1 < 0, we have k < 0. Choosing α =−k, it leads to α > 0 and y2 =−αx2. Then, it follows from (3.4) and (3.5) that
y1=∥y2∥ tan θ = ∥ − αx2∥ tan θ = | − α|∥x2∥ tan θ = α∥x2∥ tan θ = αx1tan2θ.
By this, we have y = α(x1tan2θ,−x2). Thus, the proof is complete. 2
In the next theorem, we establish the relation between CCEiCP and the below special circular cone complementarity problem (CCCP):
(3.6) CCCP(F ) : x∈ Lθ, F (x)∈ (Lθ)∗, xTF (x) = 0, where
(3.7) F (x) =
{ xTCx
xTBxBx− Cx if x ̸= 0,
0 if x = 0.
Theorem 3.1. Consider the CCEiCP(B, C) given as in (3.1) where B is positive definite. Let F : IRn→ IRn be defined as in (3.7). Then, the following hold.
(a) If (x∗, λ∗) solves the CCEiCP(B, C), then x∗ solves the CCCP(F ) (3.6).
(b) If ¯x is a nonzero solution of the CCCP(F ) (3.6), then (x∗, λ∗) solves the CCEiCP(B, C) with λ∗ = x¯x¯TTC ¯B ¯xx and x∗= aTx¯x¯.
Proof. Part(a) is trivial and we only need to prove part(b). Suppose that ¯x is a nonzero solution to the CCCP(F ) (3.6). Then, we have
¯
x∈ Lθ, x¯TC ¯x
¯
xTB ¯x · B¯x − C ¯x ∈ (Lθ)∗, and ¯xT
(x¯TC ¯x
¯
xTB ¯xB ¯x− C ¯x )
= 0.
Using a ∈ int((Lθ)∗) and ¯x ∈ Lθ, we have aT1¯x > 0. In view of all the above, we conclude that
x∗ := aT1x¯x¯∈ Lθ,
w∗ := λ∗Bx∗− Cx∗= aT1x¯
[(x¯TC ¯x
¯ xTB ¯x
)
B ¯x− C ¯x]
∈ (Lθ)∗, (x∗)Tw∗ = ( 1
aTx¯
)2[
¯ xT
(¯xTC ¯x
¯
xTB ¯xB ¯x− C ¯x)]
= 0, aTx∗ = aaTTxx¯¯ = 1.
Thus, (x∗, λ∗) is a solution to the CCEiCP(B, C). 2
Remark 3.2. For the CCCP(F ) (3.6), based on Proposition 1, we note that if x is a solution of the CCCP(F ) and x has the spectral decomposition: x = µ1(x)v(1)x + µ2(x)vx(2), F (x) can be written by F (x) = µ1(F (x))vx(2)+ µ2(F (x))v(1)x . Therefore, we only need to check both x ∈ bd(Lθ) and F (x) ∈ bd((Lθ)∗). In addition, we define
H1(x) =
µ1(x) ... µ1(x) µ1(F (x)).
∈ IRn and g1(z) := 1
2∥H1(z)∥2 = 1
2H1(z)TH1(z).
In fact, g is a merit function for the CCCP(F ). Then, we can obtain a solution of the CCEiCP (3.1) by using Newton method for solving the equations H1(x) = 0.
Next, we consider the three special kinds of the circular cone complementarity problems, which have the following forms:
(3.8) CCLCP(−C, 0) : x ∈ Lθ, −Cx ∈ (Lθ)∗, xT(−Cx) = 0;
(3.9) CCCP(G1) :
x∈ Lθ, λ≥ 0,
λBx− Cx ∈ (Lθ)∗, aTx− 1 ≥ 0, xT(λBx− Cx) + λ(aTx− 1) = 0;
and
(3.10) CCCP(G2) :
x∈ Lθ, λ≥ 0,
−λBx − Cx ∈ (Lθ)∗, aTx− 1 ≥ 0, xT(−λBx − Cx) + λ(aTx− 1) = 0.
In fact, the solutions of the CCEiCP (3.1) have a close correspondence with the solutions of the above three kinds of complementarity problems. We shall show the relationship between the CCEiCP and the CCCP in the following theorems.
Theorem 3.3. Let (x∗, λ∗) solves the CCEiCP(B,C) (3.1). Then, the following hold.
(a) If λ∗> 0, then (x∗, λ∗) solves the CCCP(G1) (3.9).
(b) If λ∗< 0, then (x∗,−λ∗) solvesthe CCCP(G2) (3.10).
(c) If λ∗= 0, then x∗ solves the CCLCP(−C,0) (3.8).
Proof. From the assumption that (x∗, λ∗) solves the CCEiCP(B,C) (3.1), we have x∗ ∈ Lθ, λ∗Bx∗− Cx∗∈ (Lθ)∗, (x∗)T(λ∗Bx∗− Cx∗) = 0 and aTx∗ = 1.
To proceed, we discuss three cases of λ∗. (a) If λ∗> 0, then we have
(x∗)T(λ∗Bx∗− Cx∗) + λ∗(aTx∗− 1) = (x∗)T(λ∗Bx∗− Cx∗) = 0, which implies that (x∗, λ∗) solves the CCCP (G1) (3.9).
(b) If λ∗< 0, then we get that −λ∗ > 0 and
(x∗)T(λ∗Bx∗− Cx∗)− λ∗(aTx∗− 1) = (x∗)T(λ∗Bx∗− Cx∗) = 0.
This indicates that (x∗,−λ∗) solves the CCCP(G2) (3.10).
(c) If λ∗ = 0, then it follows that
λ∗Bx∗− Cx∗=−Cx∗ ∈ (Lθ)∗and (x∗)T(−Cx∗) = (x∗)T(λ∗Bx∗− Cx∗) = 0.
This implies that (x∗, 0) solves the CCLCP(−C,0) (3.8).
Based on the above arguments, the proof is complete. 2
Theorem 3.4. (a) If λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G1) (3.9), then (x∗, λ∗) solves the CCEiCP(B,C) (3.1).
(b) If λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G2) (3.10), then (x∗,−λ∗) solves the CCEiCP(B,C) (3.1).
(c) If x∗ solves the CCLCP(−C, 0) (3.8) and x∗ ̸= 0, then (aTx∗x∗, 0) solves the CCEiCP(B, C) (3.1).
Proof. The proof is done by the below three cases..
Case (a): if λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G1) (3.9), then we have
x∗∈ Lθ, λ∗ > 0,
λ∗Bx∗− Cx∗ ∈ (Lθ)∗, aTx∗− 1 ≥ 0, (x∗)T(λ∗Bx∗− Cx∗) + λ∗(aTx∗− 1) = 0.
Combining with the definition of the dual cone again, it gives (x∗)T(λ∗Bx∗− Cx∗) = 0 and λ∗(aTx∗− 1) = 0.
It follows from λ∗ > 0 that aTx∗− 1 = 0. Hence, (x∗, λ∗) solves the CCEiCP(B, C) (3.1).
Case (b): if λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G2) (3.10), we have
x∗ ∈ Lθ, λ∗ > 0,
−λBx∗− Cx∗ ∈ (Lθ)∗, aTx∗− 1 ≥ 0, (x∗)T(λ∗Bx∗− Cx∗) + λ∗(aTx∗− 1) = 0.
Then, it follows that (x∗)T(−λ∗Bx∗−Cx∗) = 0 and aTx∗−1 = 0. Hence, (x∗,−λ∗) solves the CCEiCP(B, C) (3.1).
Case (c): if x∗ is a solution of CCLCP(−C, 0) (3.8) and x∗ ̸= 0, then it is easy to see that (aTx∗x∗, 0) solves the CCEiCP(B, C) (3.1).
Based on the above arguments, the proof is complete. 2
Remark 3.5. If let ϕ1 : IRn×IRn→ IR be a circular cone complementarity function [16], based on Theorem 3.3 and Theorem 3.4, then the CCEiCP(B, C) in (3.1) can be reformulated as the following nonsmooth system of equations
Φ1(z) = Φ1(x, y, λ) :=
ϕ1(x, y) λBx− Cx − y
aTx− 1
= 0.
(3.11)
By this, we can solve the CCEiCP (3.1) via using semismooth Newton method to solve the nonsmooth system of equations (3.11).
Example 1. In the CCEiCP(B, C) (3.1) or the CCCP(G1) (3.9) and the CCCP(G2) (3.10), let θ = π3. Suppose that the matrix B is the identity matrix, i.e., B := I, and the matrix C takes the following matrices, respectively:
C1 :=
[ 1 −1 0 2
]
, C2 :=
[ −1 1 0 −2
]
and C3 :=
[ 0 1
−1 0 ]
.
For the CCEiCP(B, C1), we find that w∗1 := (x∗1= (α, 0)T, λ∗1 = 1) for any α > 0 are the solutions of this problems. At this moment, λ∗1 = 1 > 0, by Theorem 3.3, it follows that w∗1 = (x∗1 = (α, 0)T, λ∗1 = 1) are also the solutions of CCCP(G1) (3.9).
For the CCEiCP(B, C2), we know that w∗2 := (x∗2 = (α, 0)T, λ∗2=−1) for any α > 0 are the solutions of this problems. Here, λ∗2 =−1 < 0, by Theorem 3.3, it follows that (x∗2 = (α, 0)T,−λ∗2 = 1) are the solutions of the CCCP(G2) (3.10).
For the CCEiCP(B, C3), it is easy to see that w∗3 := (x∗3 = (1,−√
3)T, λ∗3 = 0) is a solution of this problems. Now, λ∗3 = 0, by Theorem 3.3 again, we have x∗3 = (1,−√
3)T is a solution of the CCCP(−C3, 0) (3.8).
Conversely, we know that w1∗, w∗2 and x∗3 solve the CCCP(G1) (3.9), the CCCP(G2) (3.10) and the CCCP(−C3, 0) (3.8), respectively. By Theorem 3.4, based on the value of λ∗i for i = 1, 2, 3, it follows that w∗1, (x∗2,−λ∗2) and (aTx∗3x∗3, 0) are the solutions of CCEiCP(B, Ci) for i = 1, 2, 3.
3.2. Circular cone quadratic eigenvalue complementarity problems. In this subsection, we consider the following circular cone quadratic eigenvalue com- plementarity problems (CCQEiCP for short). Given matrices A, B, C ∈ IRn×n, the CCQEiCP seeks to find (x, y, λ) ∈ IRn× IRn× IR such that
(3.12) CCQEiCP(A, B, C) :
y = λ2Ax + λBx + Cx, x∈ Lθ, y∈ (Lθ)∗, xTy = 0, aTx = 1,
where a is an arbitrary fixed point with a ∈ int((Lθ)∗).
Before discussing the properties of the CCQEiCP (3.12), we first introduce the definition of Lθ-hyperbolic that will be used later.
Definition 1. A triple (A, B, C), with A, B, C ∈ IRn×nis called Lθ-hyperbolic if (xTBx)2 ≥ 4(xTAx)(xTCx),
for all nonzero x∈ Lθ.
Again, similar to Theorem 3.1, we build up the relation between the CCQEiCP (3.12) and the CCCP (3.6).
Theorem 3.6. Consider the CCQEiCP(A, B, C) given as in (3.12) where A is positive definite and a triple (A, B, C) is Lθ-hyperbolic. Let Fi : IRn → IRn be defined as follows:
(3.13) Fi(x) =
{ λ2i(x)Ax + λi(x)Bx + Cx if x̸= 0,
0 if x = 0,
where λi(x) = −(x
TBx)+(−1)i+1√
(xTBx)2−4(xTAx)(xTCx)
2(xTAx) for i = 1, 2. Then the follow- ing hold.
(a) If (x∗, λ∗) is a solution to the CCQEiCP(A, B, C), then x∗ solves either the CCCP(F1) or the CCCP(F2).
(b) If ¯x is a nonzero solution to the CCCP(F1) or the CCCP(F2), then (x∗, λ∗) is a solution of the CCQEiCP(A, B, C) with x∗ = aTx¯x¯ and λ∗ = λ1(¯x) or λ2(¯x).
Proof. The result of part(a) is trivial. We only need to prove part(b). Suppose that
¯
x is a nonzero solution to the CCCP(Fi) with Fi as in (3.13) for i = 1, 2. Then, for each i, we know that
¯
x∈ Lθ, λ2i(¯x)A¯x + λi(¯x)B ¯x + C ¯x∈ Lθ and ¯xT(λ2i(¯x)A¯x + λi(¯x)B ¯x + C ¯x) = 0.
Since a∈ int((Lθ)∗) and ¯x∈ Lθ, we have aT1x¯ > 0. From all the above, we conclude that
x∗ := aT1x¯x¯∈ Lθ,
w∗ := (λ∗)2Ax∗+ λ∗Bx∗+ Cx∗ = aT1x¯ (
λ2i(¯x)A¯x + λi(¯x)B ¯x + C ¯x)
∈ (Lθ)∗, (x∗)Tw∗ = ( 1
aTx¯
)2[
¯
xT(λ2i(¯x)A¯x + λi(¯x)B ¯x + C ¯x) ]
= 0, aTx∗ = aaTTxx¯¯ = 1.
Thus, (λ∗, x∗) solves the CCQEiCP(A, B, C) and the proof is complete. 2
Remark 3.7. Similar to Remark 3.2, for the CCCP(Fi), where Fi is defined as in (3.13) (i = 1, 2), we use the same skills to define
Hi1(x) =
µ1(x) ... µ1(x) µ1(Fi(x)).
∈ IRn
with i = 1, 2. Then, we obtain a solution of the CCQEiCP (3.12) via using Newton method for solving the equation Hi1(x) = 0 (i = 1, 2).
Next, we consider two special classes of the circular cone complementarity prob- lems, which have the following forms:
(3.14) CCCP(G4) :
x∈ Lθ, λ≥ 0,
λ2Ax + λBx + Cx∈ (Lθ)∗, aTx− 1 ≥ 0, xT(λ2Ax + λBx + Cx) + λ(aTx− 1) = 0, and
(3.15) CCCP(G5) :
x∈ Lθ, λ≥ 0,
λ2Ax− λBx + Cx ∈ (Lθ)∗, aTx− 1 ≥ 0, xT(λ2Ax− λBx + Cx) + λ(aTx− 1) = 0,
Similar to the cases of the CCEiCP (3.1), We will show the relationship between the CCQEiCP(A, B, C), the CCLCP(C, 0) and the CCCP(Gi) for i = 4, 5.
Theorem 3.8. Let (x∗, λ∗) solves the CCQEiCP(A,B,C) (3.12). Then, the follow- ing hold.
(a) If λ∗ > 0, then (x∗, λ∗) solves the CCCP(G4) (3.14).
(b) If λ∗ < 0, then (x∗,−λ∗) solves the CCCP(G5) (3.15).
(c) If λ∗ = 0, then x∗ solves the CCLCP(C,0) (3.8).
Proof. From the assumption that (x∗, λ∗) solves the CCQEiCP(A,B,C) (3.12), we have
x∗ ∈ Lθ, (λ∗)2Ax∗+ λ∗Bx∗+ Cx∗∈ (Lθ)∗, (x∗)T[
(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗]
= 0 and aTx∗= 1. To proceed, we discuss three cases.
Case (a): if λ∗ > 0, then we have
(x∗)T[(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗] + λ∗(aTx∗− 1) = 0.
This implies that (x∗, λ∗) solves the CCCP(G4) (3.14).
Case (b): if λ∗ < 0, then we have −λ∗> 0 and (x∗)T[
(λ∗)2Ax∗− λ∗Bx∗+ Cx∗]
− λ∗(aTx∗− 1) = 0.
This indicates that (x∗,−λ∗) solves the CCCP(G5) (3.15).
Case (c): if λ∗ = 0, then we have
(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗ = Cx∗∈ Lθ and (x∗)TCx∗ = 0.
This says that (x∗, 0) solves the CCLCP(C,0) (3.8).
Based on the above arguments, we prove the desired result. 2
Theorem 3.9. (a) If λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G4) (3.14), then (x∗, λ∗) solves the CCQEiCP(A,B,C) (3.12).
(b) If (x∗, λ∗) solves the CCCP(G5) (3.15) and λ∗ ̸= 0, then (x∗,−λ∗) solves the CCQEiCP(A,B,C) (3.12).
(c) If x∗ solves CCLCP(−C, 0) and x∗ ̸= 0, then (aTx∗x∗, 0) is a solution to the CCQEiCP(A,B,C) (3.12).
Proof. Again, the proof is done by discussing three cases.
Case (a): if λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G4) (3.14), then we have
x∗∈ Lθ, λ∗ > 0,
(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗ ∈ (Lθ)∗, aTx∗− 1 ≥ 0, (x∗)T[
(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗]
+ λ∗(aTx∗− 1) = 0.
By the definition of the dual cone, this implies (x∗)T[(λ∗)2Ax∗+ λ∗Bx∗+ Cx∗] = 0 and λ∗(aTx∗−1) = 0. In addition, it follows from λ∗> 0 that aTx∗−1 = 0. Hence, we obtain that (x∗, λ∗) solves the CCQEiCP(A,B,C) (3.12).
Case (b): if λ∗ ̸= 0 and (x∗, λ∗) solves the CCCP(G5) (3.15), then we have
x∗∈ Lθ, λ∗> 0,
(−λ∗)2Ax∗+ (−λ∗)Bx∗+ Cx∗ ∈ (Lθ)∗, aTx∗− 1 ≥ 0, (x∗)T[
(−λ∗)2Ax∗+ (−λ∗)Bx∗+ Cx∗]
+ λ∗(aTx∗− 1) = 0.
This says that (x∗)T[(−λ∗)2Ax∗+ (−λ∗)Bx∗+ Cx∗] = 0 and aTx∗− 1 = 0. Hence, we obtain that (x∗,−λ∗) solves the CCQEiCP(A,B,C) (3.12).
Case (c): if x∗ solves the CCLCP(−C, 0) (3.8) and x∗ ̸= 0, then it is easy to see that (aTx∗x∗, 0) solves the CCQEiCP(A,B,C) (3.12).
Based on the above arguments, the proof is complete. 2
Remark 3.10. Similar to Remark 3.5, the CCQEiCP(A, B, C) in (3.12) can be reformulated as the following nonsmooth system of equations
Ψ1(z) = Ψ1(x, y, λ) :=
ϕ1(x, y)
λ2Ax + λBx + Cx− y aTx− 1
= 0, (3.16)
From this, we can solve the CCQEiCP(A, B, C) (3.12) via using semismooth Newton method to solve the nonsmooth system of equations (3.16).
4. p-order cone eigenvalue complementarity problems
In this section, we consider p-order cone eigenvalue complementarity problems and p-order cone quadratic eigenvalue complementarity problems. Similar to the cases of the circular cone, we will show the relationship between the solution of p-order cone eigenvalue complementarity problems (or the solution of p-order cone quadratic eigenvalue complementarity problems) and the solution of the correspond- ing p-order cone complementarity problems, respectively.
4.1. p-order cone eigenvalue complementarity problems. Consider the p- order cone eigenvalue complementarity problems (POCEiCP for short): find (x, y, λ)∈ IRn× IRn× IR such that
(4.1) POCEiCP(B, C) :
y = λBx− Cx,
x∈ Kp, y∈ (Kp)∗, xTy = 0, aTx = 1,
where B, C ∈ IRn×n and a is an arbitrary fixed point with a∈ int((Kp)∗). First, we introduce some notations for the sake of convenience. If x = (x1, x2) ∈ IR × IRn−1 and x2 = (x(1)2 , ..., x(n2 −1))T, then we denote
(|x(1)2 |p,· · · , |x(n2 −1)|p)T
by|x2|p. Similar to the CCEiCP (Proposition 1), by H¨older’s inequality, we achieve the version of p-order cone as the following proposition.
Proposition 2. Suppose that x := (x1, x2)∈ Kp, y := (y1, y2) ∈ (Kp)∗. Then, the following hold.
(a) xTy≥ 0.
(b) If y ∈ int(Kp), then xTy > 0 if and only if x̸= 0.
(c) If x̸= 0 and y ̸= 0, then
xTy = 0 ⇒ x1=∥x2∥p and |y2|q= α|x2|p,
where α is a positive constant. Similarly, if x̸= 0 and y ̸= 0, then xTy = 0 ⇒ y1 =∥y2∥q and |x2|p = β|y2|q,
where β is a positive constant.
Proof. By the definition of dual cones, part(a) is obvious. For part(b), it is also trivial that xTy > 0 implies x ̸= 0. We only prove the other direction as follows.
Since x̸= 0, we know that x1 > 0. Otherwise, we have that∥x2∥p ≤ x1= 0 implies x = 0. Thus, it yields
xTy = x1y1+ xT2y2 ≥ x1y1− ∥x2∥p∥y2∥q> x1y1− x1y1= 0.
Hence, we prove the result of part(b).
(c) “⇐” The proof of this direction is trivial. “⇒” By the assumption x ̸= 0, we know that x1> 0. Similarly, we know that y1 > 0. Using the condition of xTy = 0, the H¨older’s inequality and the definition of circular cones, we have
(4.2) x1y1 =|xT2y2| ≤ ∥x2∥p∥y2∥q≤ x1y1
which leads to
(4.3) x1y1=∥x2∥p∥y2∥q.
From x1> 0, y1 > 0, x1 ≥ ∥x2∥p, y1 ≥ ∥y2∥q and the equality (4.3), it follows that (4.4) x1=∥x2∥p and y1=∥y2∥q.
Moreover, by (4.2), (4.3) and (4.4) again, we have xT2y2=−x1y1=−∥x2∥∥y2∥.
This implies the equality holds in the H¨older’s inequality. Thus, there exists a constant α such that
|y2|q = α|x2|p. Then, the proof is complete. 2
Based on the cases of circular cone, we build up the relation between the POCE- iCP and the below p-order cone complementarity problem (POCCP):
(4.5) POCCP(F ) : x∈ Kp, F (x)∈ (Kp)∗, xTF (x) = 0, where F : IRn→ IRn be defined as in (3.7), i.e.,
F (x) =
{ xTCx
xTBxBx− Cx if x ̸= 0,
0 if x = 0.
Similar to Theorem 3.1, we use the same technique to establish the relation between the POCEiCP and the POCCP.
Theorem 4.1. Consider the POCEiCP(B, C) given as in (4.1) where B is positive definite. Then, the following hold.
(a) If (x∗, λ∗) solves the POCEiCP(B, C), then x∗ solves the POCCP(F ) (4.5).
(b) If ¯x is a nonzero solution to the POCCP(F ) (4.5), then (x∗, λ∗) solves the POCEiCP(B, C) with λ∗ = x¯x¯TTC ¯B ¯xx and x∗= aTx¯x¯.
Proof. The proof is similar to Theorem 3.1. Hence, we omit it. 2
Likewise, we consider the following three special classes of the p-order cone com- plementarity problems:
(4.6) POCLCP(−C, 0) : x ∈ Kp, −Cx ∈ (Kp)∗, xT(−Cx) = 0,
(4.7) POCCP(G1) :
x∈ Kp, λ≥ 0,
λBx− Cx ∈ (Kp)∗, aTx− 1 ≥ 0, xT(λBx− Cx) + λ(aTx− 1) = 0,
and
(4.8) POCCP(G2) :
x∈ Kp, λ≥ 0,
−λBx − Cx ∈ (Kp)∗, aTx− 1 ≥ 0, xT(−λBx − Cx) + λ(aTx− 1) = 0, where the matrixes B, C are given as in the POCEiCP (4.1).
Analogous to Theorem 3.3 and Theorem 3.4, we have the relation between the POCEiCP and the POCCP.
Theorem 4.2. Let (x∗, λ∗) solves thePOCEiCP(B,C) (4.1). Then, the following hold.
(a) If λ∗ > 0, then (x∗, λ∗) solves the POCCP(G1) (4.7).
(b) If λ∗ < 0, then (x∗,−λ∗) solves the POCCP(G2) (4.8).
(c) If λ∗ = 0, then x∗ solves the POCLCP(−C,0)(4.6).
Proof. The proof is similar to Theorem 3.3. Hence, we omit it. 2
Theorem 4.3. (a) If λ∗ ̸= 0 and (x∗, λ∗) solves the POCCP(G1) (4.7), then (x∗, λ∗) solves the POCEiCP(B,C) (4.1).
(b) If λ∗ ̸= 0 and (x∗, λ∗) solves the POCCP(G2) (4.8), then (x∗,−λ∗) solves the POCEiCP(B,C) (4.1).
(c) If x∗ solves the POCLCP(−C, 0)(4.6) and x∗ ̸= 0, then (aTx∗x∗, 0) solves the POCEiCP(B, C) (4.1).
Proof. The proof is similar to Theorem 3.4. Hence, we omit it. 2
Remark 4.4. If let ϕ2: IRn×IRn→ IR be a p-order cone complementarity function, based on Theorem 4.2 and Theorem 4.3, then the POCEiCP(B, C) in (4.1) can be reformulated as the following nonsmooth system of equations:
Φ2(z) = Φ2(x, y, λ) :=
ϕ2(x, y) λBx− Cx − y
aTx− 1
= 0.
(4.9)
Accordingly, we can solve the POCEiCP (4.1) via using semismooth Newton method to solve the nonsmooth system of equations (4.9).
4.2. p-order Cone quadratic eigenvalue complementarity problems. Now, we consider the p-order cone quadratic eigenvalue complementarity problems (POC- QEiCP for short) as follows: given matrices A, B, C ∈ IRn×n, the POCQEiCP seeks to find (x, y, λ)∈ IRn× IRn× IR such that
(4.10) POCQEiCP(A, B, C) :
y = λ2Ax + λBx + Cx, x∈ Kp, y ∈ (Kp)∗, xTy = 0, aTx = 1,
