PROBLEMS ASSOCIATED WITH SECOND-ORDER CONE
XINHE MIAO∗, WEI-MING HSU, CHIEU THANH NGUYEN, AND JEIN-SHAN CHEN†
Abstract. In this paper, we study the solvabilities of three optimiza- tion problems associated with second-order cone, including the absolute value equations associated with second-order cone (SOCAVE), eigen- value complementarity problem associated with second-order cone (SOCE- iCP), and quadratic eigenvalue complementarity problem associated with second-order cone (SOCQEiCP). More specifically, we characterize un- der what conditions these optimizations have solution and unique solu- tion, respectively.
1. Introduction
In this paper, we study the solvabilities of three optimization problems associated with second-order cone. The first optimization problem that our target is the so-called absolute value equations associated with second-order cone, abbreviated as SOCAVEs. For SOCAVEs, there have two types of them. The first type is in the form of
(1) Ax − |x| = b.
Another one is a more general SOCAVE, which is in the form of
(2) Ax + B|x| = b,
where A, B ∈ Rn×n, B 6= 0, and b ∈ Rn. Note that, unlike the standard absolute value equation that is presented below, here |x| means the absolute value of x coming from the square root of the Jordan product “◦”, associated with second-order cone (SOC), of x and x, that is, |x| := (x ◦ x)1/2. The second-order cone in Rn (n ≥ 1), also called the Lorentz cone or ice-cream cone, is defined as
Kn:=(x1, x2) ∈ R × Rn−1| kx2k ≤ x1 ,
2010 Mathematics Subject Classification. 26B05, 26B35, 90C33.
Key words and phrases. Solvability, eigenvalue, second-order cone, absolute value equations.
∗The author’s work is supported by National Natural Science Foundation of China (No.
11471241).
†Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan.
1
where k · k denotes the Euclidean norm. If n = 1, then K1 is the set of nonnegative reals R+. In general, a general second-order cone K could be the Cartesian product of SOCs, i.e.,
K := Kn1× · · · × Knr.
For simplicity, we focus on the single second-order cone Kn because all the analysis can be carried over to the setting of Cartesian product. More de- tails about second-order cone, Jordan product, and (·)1/2 will be introduced in Section 2.
Indeed, the SOCAVE (1) (respectively, SOCAVE (2)) is a natural exten- sion of the standard absolute value equation (AVE for short) as bellow:
(3) Ax − |x| = b, (respectively, Ax + B|x| = b)
where |x| denotes the componentwise absolute value of vector x ∈ Rn. It is known that the standard absolute value equation (3) was first introduced by Rohn in [44] and recently has been investigated by many researchers. For standard absolute value equation, there are two main research directions.
One is on the theoretical side in which the corresponding properties of the solution for the AVE (3) are studied, see [21, 25, 28, 29, 32, 35, 42, 44, 52].
The other one focuses on the algorithm for solving the absolute value equa- tion, see [5, 23, 30, 31, 33, 34, 45, 53, 54].
On the theoretical aspect, Mangasarian and Meyer [35] show that the AVE (3) is equivalent to the bilinear program, the generalized LCP (lin- ear complementarity problem), and the standard LCP provided 1 is not an eigenvalue of A. Prokopyev [42] further improves the above equivalence which indicates that the AVE (3) can be equivalently recast as an LCP with- out any assumption on A and B, and also provides a relationship with mixed integer programming. In general, if solvable, the AVE (3) can have either unique solution or multiple (e.g., exponentially many) solutions. Indeed, various sufficient conditions on solvability and non-solvability of the AVE (3) with unique and multiple solutions are discussed in [35, 42]. Moreover, Wu and Guo [52] further study the unique solvability of the AVE (3), and give some new and useful results for the unique solvability of the AVE (3).
Recently, the absolute value equation associated with second-order cone or circular cone are investigated in [22] and [27], respectively. In particular, Hu, Huang and Zhang [22] show that the SOCAVE (2) is equivalent to a class of second-order cone linear complementarity problems, and establish a result regarding the unique solvability of the SOCAVE (2). Along this direction, we further look into the SOCAVEs (1) and (2) in this paper, and achieve some new results about the existence of (unique) solution.
The second optimization problem that we focus on is the so-called second- order cone eigenvalue complementarity problem, SOCEiCP for short. More specifically, given two matrices B, C ∈ Rn×n, the SOCEiCP is to find (x, y, λ) ∈ Rn× Rn× R such that
(4) SOCEiCP(B, C) :
y = λBx − Cx, y Kn 0, x Kn 0, xTy = 0,
aTx = 1,
where a is an arbitrary fixed point with a ∈ int(Kn), and x Kn 0 means that x ∈ Kn, a partial order. The SOCEiCP(B, C) given as in (4) comes naturally from the traditional eigenvalue complementarity problem [43, 47], which seeks to find (λ, x, w) ∈ Rn× Rn× R such that
EiCP(B, C) :
y = λBx − Cx, y ≥ 0, x ≥ 0, xTy = 0, eTx = 1,
where B, C ∈ Rn×n and e = (1, 1, · · · , 1)T ∈ Rn. Usually, the matrix B is assumed to be positive definite. The scalar λ is called a complementary eigenvalue and x is a complementary eigenvector associated to λ for the pair (B, C). The condition xTy = 0 and the nonnegative requirements on x and y imply that either xi = 0 or yi = 0 for 1 ≤ i ≤ n. These two variables are called complementary.
A natural extension of the EiCP goes to the quadratic eigenvalue comple- mentarity problem (QEiCP), whose mathematical format is as below. Given A, B, C ∈ Rn×n, the QEiCP consists of finding (x, y, λ) ∈ Rn× Rn× R such that
QEiCP(A, B, C) :
y = λ2Ax + λBx + Cx, y ≥ 0, x ≥ 0,
xTy = 0, eTx = 1,
where e = (1, 1, · · · , 1)T ∈ Rn. It is clear that when A = 0, the QEiCP(A, B, C) reduces to the EiCP(B,−C). The λ component of a solution to the QEiCP(A, B, C) is called a quadratic complementary eigenvalue for the pair (A, B, C), whereas the x component is called a quadratic complementary eigenvector for the pair (A, B, C).
Following the same idea for creating the SOCEiCP(B, C), the third op- timization problem that we study in this paper is the so-called second- order cone quadratic eigenvalue complementarity problem (SOCQEiCP).
In other words, given matrices A, B, C ∈ Rn×n, the SOCQEiCP seeks to
find (x, y, λ) ∈ Rn× Rn× R such that
(5) SOCQEiCP(A, B, C) :
y = λ2Ax + λBx + Cx, y Kn 0, x Kn 0, xTy = 0,
aTx = 1,
with arbitrary fixed point a ∈ int(Kn). The SOCEiCP (4) and the SOC- QEiCP (5) have been investigated in [2, 3, 19]. The purpose of this paper aims to establish the solvabilities of the SOCEiCP (4) and the SOCQEiCP (5) by reformulating them as second-order cone complementarity problem (SOCCP) and a nonsmooth system of equations (see more details in Section 5).
We point out that the last normalization constraint appeared in the above EiCP, QEiCP, SOCEiCP, and SOCQEiCP has been introduced in order to prevent the x component of a solution to vanish. In other words, “for an arbitrary fixed point a ∈ int(Kn), x ∈ Knsatisfying aTx > 0 is equivalent to x 6= 0”. To see this, we provide some arguments as below. First, it is trivial that aTx > 0 implies that x 6= 0. Now, suppose that x = (x1, x2) ∈ Kn which is nonzero. Then, there must have x1 > 0. Using the definition of
int(Kn) =(x1, x2) ∈ R × Rn−1| kx2k < x1 , we have
aTx = a1x1+ ha2, x2i > |ha2, x2i| + ha2, x2i ≥ 0.
This proves that aTx > 0.
Another thing needs to be pointed out is that the normalization constraint eTx = 1 is good enough for EiCP and QEiCP; moreover, this condition was also used in [2] for SOCEiCP. However, we show that it does not make sense in the settings of SOCEiCP and SOCQEiCP because e /∈ int(Kn).
Indeed, for a counterexample, we consider λ = 1, x =
1
−1
∈ K2, two matrices C =
1 2 2 5
∈ R2×2 and B := I ∈ R2×2. Then, we have λBx − Cx =
1
−1
−
1 2 2 5
1
−1
=
2 2
∈ K2. Hence, xT(λBx − Cx) = 0, but eTx = 0. This is why, in this paper, we require an point a ∈ int(Kn) such that aTx = 1 to serve as the normalization constraint in SOCEiCP and SOCQEiCP.
To close this section, we say a few words about notations. As usual, Rn denotes the space of n-dimensional real column vectors. R+and R++denote the nonnegative and positive reals. For any x, y ∈ Rn, the Euclidean inner product are denoted hx, yi = xTy, and the Euclidean norm kxk are denoted as kxk = phx, xi. Given a matrix A ∈ Rn×n, kAka denotes the arbitrary
matrix norm, for example, kAk1, kAk2 and kAk∞. In addition, ρ(A) means the spectral radius of A, that is, ρ(A) := max{|λ| | λ is eigenvalue of A}, and M (Kn) ⊂ Kn denotes that for any z ∈ Kn, we have M z ∈ Kn. For conve- nience, we say that a pair (x, λ) ∈ Rn× R solves the SOCEiCP(B, C) when the triplet (x, y, λ) with y = λBx − Cx, is a solution to the SOCEiCP(B, C) in the sense defined in (4). Similarly, we say that (x, λ) ∈ Rn× R solves the SOCQEiCP(A, B, C) when the same occurs with the triplet (x, λ), where y = λ2Ax + λBx + Cx.
2. Preliminaries
In this section, we recall some basic concepts and background materials regarding second-order cone and the absolute value of x ∈ Rn, which will be extensively used in the subsequent analysis. More details can be found in [9, 14, 16, 17, 20, 22].
The official definition of second-order cone (SOC) is already defined in Section 1. We begin with introducing the concept of Jordan product. For any two vectors x = (x1, x2) ∈ R × Rn−1 and y = (y1, y2) ∈ R × Rn−1, the Jordan product of x and y associated with Knis given by
x ◦ y :=
xTy y1x2+ x1y2
.
The Jordan product, unlike scalar or matrix multiplication, is not associa- tive, which is a main source of complication in the analysis of optimization problems involved SOC, see [14, 16, 20] and references therein for more de- tails. The identity element under this Jordan product is e = (1, 0, · · · , 0)T ∈ Rn. With these definitions, x2 means the Jordan product of x with itself, i.e., x2 := x ◦ x; while x1/2 with x ∈ Kn denotes the unique vector in Kn such that x1/2◦ x1/2 = x. In light of this, the vector |x| in the SOCAVEs (1) and (2) is computed by
|x| := (x ◦ x)1/2.
However, by the definition of |x|, it is not easy to write out the expression of |x| explicitly. Fortunately, there is another way to reach |x| via spectral decomposition and projection onto second-order cone. We elaborate it as below. For x = (x1, x2) ∈ R × Rn−1, the spectral decomposition of x with respect to SOC is given by
(6) x = λ1(x)u(1)x + λ2(x)u(2)x , where λi(x) = x1+ (−1)ikx2k for i = 1, 2 and
u(i)x =
1 2
1, (−1)i xkxT2
2k
T
if kx2k 6= 0,
1
2 1, (−1)iωTT
if kx2k = 0,
with ω ∈ Rn−1 being any vector satisfying kωk = 1. The two scalars λ1(x) and λ2(x) are called spectral values (or eigenvalues) of x; while the two vectors u(1)x and u(2)x are called the spectral vectors (or eigenvectors) of x.
Moreover, it is obvious that the spectral decomposition of x ∈ Rn is unique if x2 6= 0.
Next, we talk about the projection onto second-order cone. Let x+ be the projection of x onto Kn, while x− be the projection of −x onto its dual cone of Kn. Since second-order cone Kn is self-dual, the dual cone of Kn is itself, i.e., (Kn)∗ = Kn. In fact, the explicit formula of projection of x = (x1, x2) ∈ R × Rn−1 onto Kn is characterized in [14, 16, 17, 20, 18] as below:
x+=
x if x ∈ Kn, 0 if x ∈ −Kn, u otherwise,
where u =
" x1+kx2k
2 x1+kx2k
2
x2
kx2k
# . Similarly, the expression of x− is in the form of
x−=
0 if x ∈ Kn,
−x if x ∈ −Kn, w otherwise,
where w =
"
−x1−kx2 2k
x
1−kx2k 2
x2
kx2k
# . Together with the spectral decomposition (6) of x, it can be verified that x = x+− x− and the expression of x+ and x− have the form:
x+ = (λ1(x))+u(1)x + (λ2(x))+u(2)x , x− = (−λ1(x))+u(1)x + (−λ2(x))+u(2)x , where (α)+= max{0, α} for α ∈ R.
Based on the definitions and expressions of x+ and x−, we introduce another expression of |x| associated with SOC. In fact, the alternative ex- pression is obtained by the so-called SOC-function, which can be found in [10]. For any x ∈ Rn, we define the absolute value |x| of x with respect to SOC as |x| := x++x−. In fact, in the setting of SOC, the form |x| = x++x−
is equivalent to the form |x| = (x ◦ x)1/2. Combining the above expression of x+ and x−, it is easy to see that the expression of the absolute value |x|
is in the form of
|x| = (λ1(x))++ (−λ1(x))+u(1)x +(λ2(x))++ (−λ2(x))+u(2)x
= λ1(x)
u(1)x + λ2(x)
u(2)x .
For the absolute value |x| associated with SOC, Hu, Huang and Zhang [22] have obtained some properties as the following lemmas.
Lemma 2.1. [22, Theorem 2.1] The generalized Jacobian of the absolute value function | · | is given as follows:
(a) Suppose that x2= 0. Then, ∂|x| = {tI | t ∈ sgn(x1)}.
(b) Suppose that x26= 0.
(i) If x1 + kx2k < 0 and x1 − kx2k < 0, then ∂|x| = {∇|x|} =
−1 0T 0 −I
.
(ii) If x1 + kx2k > 0 and x1 − kx2k > 0, then ∂|x| = {∇|x|} =
1 0T 0 I
.
(iii) If x1+ kx2k > 0 and x1− kx2k < 0, then
∂|x| = {∇|x|} =
0 kxxT2
2k x2
kx2k x1
kx2k
I −kxx2xT2
2k2
. (iv) If x1+ kx2k = 0 and x1− kx2k < 0, then
∂|x| =
1 2
t − 1 (t + 1)kxxT2
2k
(t + 1)kxx2
2k −2I + (t + 1)kxx2xT2
2k2
t ∈ sgn(x1+ kx2k)
. (v) If x1+ kx2k > 0 and x1− kx2k = 0, then
∂|x| =
1 2
t + 1 (1 − t)kxxT2
2k
(1 − t)kxx2
2k 2I − (1 − t)kxx2xT2
2k2
t ∈ sgn(x1− kx2k)
,
where the function sgn(·) denotes that sgn(a) =
{1} if a > 0, {t | t ∈ [−1, 1]} if a = 0, {−1} if a < 0.
Lemma 2.2. [22, Theorem 2.2] For any V ∈ ∂|x|, the absolute value of every eigenvalue of V is not greater than 1.
Lemma 2.3. [22, Theorem 2.3] For any V ∈ ∂|x|, we have V x = |x|.
3. Existence of solution to the SOCAVEs
This section is devoted to the existence and nonexistence of solution to SOCAVE (1) and SOCAVE (2).
Theorem 3.1. Let C ∈ Rn×n and b ∈ Rn. (a) If the following system
(7) (C − I)z = b, z ∈ Kn
has a solution, then for any A = ±C the SOCAVE (1) has a solution.
(b) If the following system
(C + B)z = b, z ∈ Kn
has a solution, then for any A = ±C the SOCAVE (2) has a solution.
Proof. (a) Suppose that z := (z1, z2) ∈ R × Rn−1 is a solution to the system (7), i.e.,
(C − I)z = b, z ∈ Kn.
Since z ∈ Kn, it follows that z1 ≥ kz2k. Taking x = ±z, which means x = (±z1, ±z2) ∈ R × Rn−1. Using the definition of |x|, we see that
|x| = λ1(x)
u(1)x + λ2(x)
u(2)x
=
± z1− k ± z2k
1
2
−2kz±z2
2k
+
± z1+ k ± z2k
1
±z22 2kz2k
= z.
Plugging in A = ±C yields that
Ax − |x| = ±Cx − z = (C − I)z = b.
This says that x is a solution to the SOCAVE (1).
(b) The arguments are similar to part (a). 2
Theorem 3.2. Suppose that −b ∈ Kn and A(Kn) ⊆ Kn with ρ(A) < 1.
Then, the SOCAVE (1) has a solution x ∈ Kn.
Proof. We consider the iterative scheme xk+1 = Axk − b with x0 := −b.
Since −b ∈ Kn, it follows that xk ∈ Kn for every k ∈ N. Hence, from the condition ρ(A) < 1, we can conclude that the sequence {xk} converges to a point x∗ such that x∗ = Ax∗− b. Combining with the closeness of Kn, this yields x∗ ∈ Kn, which implies
Ax∗− |x∗| = Ax∗− x∗= b.
Thus, x∗ ∈ Kn is a solution to the SOCAVE (1). 2
Remark 3.1. In fact, if the condition ρ(A) < 1 in Theorem 3.2 is replaced by kAka< 1, where kAka denotes an arbitrary matrix norm, then the result of Theorem 3.2 still holds.
Theorem 3.3. Suppose that 0 6= b ∈ Kn. Then, the following hold.
(a) If the spectral norm kAk < 1 with kAk := p
ρ(AHA), then the SO- CAVE (1) has no solution.
(b) If kAk < 1, B(Kn) ⊂ −Kn and kBxk ≥ kxk for any x ∈ Kn, then the SOCAVE (2) has no solution.
Proof. From Ax − |x| = b and 0 6= b ∈ Kn, it follows that Ax − |x| ∈ Kn. This together with the fact |x| ∈ Kn implies Ax + |x| ∈ Kn. Moreover, by
the self-duality of Kn, we see that
kAxk2− kxk2 = kAxk2− k|x|k2
= hAx + |x|, Ax − |x|i
≥ 0.
Hence, we have
kxk ≤ kAxk ≤ kAkkxk < kxk,
where the last inequality is due to kAk < 1. This is a contradiction. There- fore, the SOCAVE (1) has no solution.
(b) The idea for the proof is similar to part(a), we present it for completeness.
From Ax + B|x| = b and 0 6= b ∈ Kn, we know Ax + B|x| ∈ Kn. Then, it follows from B(Kn) ⊂ −Kn and b ∈ Kn that Ax = b − B|x| ∈ Kn, which says Ax − B|x| ∈ Kn. Moreover, by the self-duality of Kn, we have
kAxk2− kB|x|k2= hAx + B|x|, Ax − B|x|i ≥ 0, which implies
kxk > kAxk ≥ kB|x|k ≥ k|x|k = kxk,
where the first inequality is due to kAk < 1 and the last inequality is due to kBxk ≥ kxk for any x ∈ Kn. This is a contradiction. Hence, the SOCAVE (2) has no solution. 2
4. The unique solvability for the SOCAVEs
In this section, we further investigate the unique solvability of SOCAVE (1) and SOCAVE (2).
Theorem 4.1. (a) If all singular values of A exceed 1, then the SO- CAVE (1) has a unique solution.
(b) If all singular values of A ∈ Rn×n exceed the maximal singular value of B ∈ Rn×n, then the SOCAVE (2) has a unique solution.
Proof. (a) For any V ∈ ∂|x|, by Lemma 2.3, we have |x| = V x, which implies that
Ax − |x| = Ax − V x = (A − V )x,
i.e., the SOCAVE (1) becomes the equation (A − V )x = b. Moreover, by Lemma 2.1, we know that the real matrix V is symmetric. This leads to that the singular values of V are the absolute values of eigenvalue of V . On the other hand, from Lemma 2.2, it follows that all singular values of V are not greater than 1. Combining with the condition that all singular values of A exceed 1, we can assert that the matrix A − V is nonsingular. If not, there exists 0 6= x ∈ Rn such that (A − V )x = 0, i.e., Ax = V x. Hence, we have
kxk2< hAx, Axi = hV x, V xi ≤ kxk2,
which is a contradiction. Thus, the matrix A − V is nonsingular, which says the equation (A − V )x = b has a unique solution. Then, the proof is complete.
(b) The proof is similar to that for part (a), we present it for completeness.
For any V ∈ ∂|x|, by Lemma 2.3 again, we have |x| = V x; and hence Ax + B|x| = (A + BV )x.
Moreover, we also know that all singular values of V are not greater than 1 due to Lemma 2.2. Applying the condition that all singular values of A exceed the maximal singular value of B ∈ Rn×n and [22, Theorem 3.1], we obtain that the matrix A + BV is nonsingular. Thus, the equation (A + BV )x = b has a unique solution, which says the SOCAVE (2) has a unique solution. 2
Remark 4.1. We point out that in [22], Hu, Huang and Zhang have shown that if all singular values of A ∈ Rn×n exceed the maximal singular value of B ∈ Rn×n, the SOCAVE (2) has at least one solution for any b ∈ Rn. In Theorem 4.1(b), we study when the SOCAVE (2) has a unique solution, which is a stronger result than the aforementioned one in [22], although the same condition is used. In other words, under the condition that all singular values of A ∈ Rn×n exceed the maximal singular value of B ∈ Rn×n, it guarantees that the SOCAVE (2) not only has at least one solution, but also has unique solution.
Corollary 4.1. If the matrix A is nonsingular and kA−1k < 1, then the SOCAVE (1) has a unique solution.
Proof. This is an immediate consequence of Theorem 4.1(a), whose proof is similar to that for [35, Proposition 4.1]. Hence, we omit it. 2
Theorem 4.2. (a) If the matrix A = [aij] ∈ Rn×n satisfies
|aii| >√
n +X
j6=i
|aij| ∀i ∈N := {1, 2, · · · , n}, then for any b ∈ Rn the SOCAVE (1) has a unique solution.
(b) If the matrices A = [aij] ∈ Rn×n and B ∈ Rn×n satisfy
|aii| > kBk∞√
n +X
j6=i
|aij| ∀i ∈N := {1, 2, · · · , n}, then for any b ∈ Rn the SOCAVE (2) has a unique solution.
Proof. (a) Again, for any V ∈ ∂|x|, we know that |x| = V x and kV k ≤ 1, which implies that the SOCAVE (1) is equal to the equation (A − V )x = b.
Moreover, by the relationship between the spectral norm and the infinite norm, i.e.,
kV k∞≤√ nkV k, it follows that kV k∞ ≤ √
n. Let [wij] = W := A − V = [aij − vij]. Then, we note that for any i ∈N = {1, 2, · · · , n},
|wii| = |aii− vii| ≥ |aii| − |vii|
> √
n +X
j6=i
|aij| − |vii|
≥ √
n +X
j6=i
|wij| −
n
X
j=1
|vij|
≥ X
j6=i
|wij|,
where the last inequality is due to kV k∞ ≤ √
n. This indicates that the matrix A − V = W is a strictly diagonally dominant by row. Hence, the matrix A − V is nonsingular, which leads to that the equation (A − V )x = b has a unique solution. Thus, the SOCAVE (1) has a unique solution.
(b) The proof is similar to part (a) and we omit it here. 2
Theorem 4.3. If the matrix A ∈ Rn×n can be expressed as
A = αI + M, where M (Kn) ⊆ Kn and α − 1 > ρ(M ), then for any b ∈ Rn, the SOCAVE (1) has a unique solution.
Proof. For any x ∈ Kn and V ∈ ∂|x|, we know that x = |x| = V x and kV k ≤ 1. Note that
Ax − |x| = (αI − V )x + M x = (α − 1)|x| + M x.
This implies that the matrix αI + M − V is a generalized M -matrix with respect to Kn. Hence, we have the matrix αI + M − V is nonsingular. In addition, applying the fact that Ax − |x| = (αI + M − V )x, it yields that the SOCAVE (1) has a unique solution. 2
Lemma 4.1. For any x, y ∈ Rn, let |x|, |y| be the absolute value coming from the square root of x2 and y2 under the Jordan product, respectively.
Then, we have
|x| − |y|
≤ kx − yk.
Proof. First, we note that kx − yk2−
|x| − |y|
2 = hx − y, x − yi − h|x| − |y|, |x| − |y|i
= 2 h|x|, |y|i − hx, yi
= 2 hx++ x−, y++ y−i − hx+− x−, y+− y−i
= 4 hx+, y−i + hx−, y+i
≥ 0.
With this, it is clear to see that
|x| − |y|
≤ kx − yk. Then, the proof is complete. 2
Theorem 4.4. For any β ∈ R, assume that the matrix βI + A is nonsin- gular.
(a) If the matrix A satisfies
(βI + A)−1
< 1
|β| + 1, then the SOCAVE (1) has a unique solution.
(b) If the matrices A and B satisfy (βI + A)−1
< 1
|β| + kBk, then the SOCAVE (2) has a unique solution.
Proof. (a) For the SOCAVE (1), we know that
Ax − |x| = b ⇐⇒ (βI + A)x = βx + |x| + b.
If the matrix βI + A is nonsingular, then we further have
Ax−|x| = b ⇐⇒ (βI+A)x = βx+|x|+b ⇐⇒ x = (βI+A)−1(βx+|x|+b).
In view of this, we consider the following iterative scheme xk+1= (βI + A)−1(βxk+ |xk| + b).
With this, it follows that
xk+1− xk= (βI + A)−1β(xk− xk−1) + (|xk| − |xk−1|).
Hence, we have
xk+1− xk =
(βI + A)−1β(xk− xk−1) + (|xk| − |xk−1|)
≤
(βI + A)−1
|β|kxk− xk−1k + k|xk| − |xk−1|k (8)
≤
(βI + A)−1
(|β| + 1)kxk− xk−1k,
where the last inequality holds due to Lemma 4.1. This together with the assumption that
(βI + A)−1
< |β|+11 yields the sequence {xk} converges to a solution of the SOCAVE (1).
Next, we verify the SOCAVE (1) has a unique solution. If there exist x∗ and ¯x that both satisfy the SOCAVE (1), then as done in (8) we have
kx∗− ¯xk ≤
(βI + A)−1
(|β| + 1)kkx∗− ¯xk.
Since
(βI + A)−1
< |β|+11 , we obtain that x∗ = ¯x. This says that the SOCAVE (1) has a unique solution. Thus, the proof is complete.
(b) The proof is similar to part (a) and we omit it here. 2
5. The solvabilities of SOCEiCP and SOCQEiCP
In this section, we focus on the solvabilities of the other two optimization problems, SOCEiCP(B,C) and SOCQEiCP(A,B,C), which are given as in (4) and (5) respectively. In order to clearly describe our results, we need a few concepts which were introduced in [3, 4].
Definition 5.1. Let Kn be a single second-order cone.
(a) A matrix A ∈ Rn×n is called Kn-regular if xTAx 6= 0 for all nonzero x Kn 0.
(b) A matrix A ∈ Rn×n is called strictly Kn-copositive if xTAx > 0 for all nonzero x Kn0.
(c) A triple (A, B, C) with A, B, C ∈ Rn×n is called Kn-hyperbolic if (xTBx)2≥ 4(xTAx)(xTCx)
for all nonzero x Kn 0.
(d) The class R0(Kn) ⊆ Rn×n consists of those matrices A ∈ Rn×n such that there exists no nonzero x ∈ Kn satisfying Ax ∈ Kn and xTAx = 0.
(e) The class S0(Kn) ⊆ Rn×n consists of those matrices A ∈ Rn×n such that Ax ∈ Kn for at least a nonzero x ∈ Kn.
(f) The class R00(Kn) ⊆ Rn×n consists of those matrices A ∈ Rn×n such that xTAx = 0 for at least a nonzero x ∈ Kn satisfying Ax ∈ Kn. (g) The class S00(Kn) ⊆ Rn×n consists of those matrices A ∈ Rn×n such
that there exists no nonzero x ∈ Kn satisfying Ax ∈ Kn.
In fact, there exist some study in [3, 46, 48], which investigated the eigen- values problems involved with general cones. The solvability results therein automatically include solvabilities of SOCEiCP(B,C) and SOCQEiCP(A,B,C) as special cases. For example, we extract some of them from [3, 46, 48], when the cone reduces to an SOC or is a general cone, and list them as below.
Proposition 5.1. Let Kn be a single second-order cone and consider the SOCEiCP(B,C) given as in (4) and the SOCQEiCP(A,B,C) given as in (5).
(a) If B ∈ Rn×n is strictly Kn-copositive, then SOCEiCP(B, C) has solutions for any C ∈ Rn×n.
(b) If A is Kn-regular and (A, B, C) is Kn-hyperbolic, then SOCQEiCP(A, B, C) has solutions.
(c) The matrix C ∈ R00(Kn) if and only if 0 is a quadratic complementary eigenvalue for SOCQEiCP(A, B, C).
(d) If C ∈ S00(Kn) and A is strictly Kn-copositive, there exist at least one positive and one negative quadratic complementary eigenvalue for SOCQEiCP(A, B, C).
(e) If A ∈ S00(Kn) and C is strictly Kn-copositive, there exist at least one positive and one negative quadratic complementary eigenvalue for SOCQEiCP(A, B, C).
In view of the above existing solvability results in the literature, we aim to seek the solvabilities of SOCEiCP(B,C) and SOCQEiCP(A,B,C) via dif- ferent approach. In this section, we will recast these problems as three reformulations, called Reformulation I, Reformulation II and Refor- mulation III.
The idea of Reformulation I is to recast these problems as a form of second-order cone complementarity problem (SOCCP), which is a natural extension of nonlinear complementarity problem (NCP). To proceed, we first recall the mathematical format of the SOCCP as follows. More details can be found in [6, 7, 8, 9, 12, 13, 14, 16, 36, 37, 38, 39, 40, 41, 50, 51]. Given a continuously differentiable mapping F : Rn → Rn, the SOCCP(F ) is to find x ∈ Rn satisfying
(9) SOCCP(F ) :
x Kn 0, F (x) Kn0, xTF (x) = 0.
It is well know that the KKT conditions of an second-order cone program- ming problem can be rewritten as an SOCCP(F ). We now elaborate how we to recast the SOCEiCP(B, C) as the SOCCP(F ). Suppose that we are given an SOCEiCP(B, C) as in (4), where B, C ∈ Rn×n and the matrix B is assumed to be positive definite. For any x ∈ Rn such that x 6= 0, plug- ging w = λBx − Cx into the complementarity condition xTw = 0 yields λ = xxTTCxBx. Hence, we obtain
w = xTCx
xTBxBx − Cx.
With this, for any x ∈ Rn such that x 6= 0, we define a mapping F : Rn → Rn which is given by
(10) F (x) := xTCx
xTBxBx − Cx.
This mapping F is not good enough to be put into the SOCCP (9) because F (0) is not defined yet. To this end, we show the following lemma to make up the value F (0).
Lemma 5.1. Consider the SOCEiCP(B, C) given as in (4) where B is positive definite. Let F : Rn→ Rnbe defined as in (10) where x 6= 0. Then,
x→0limF (x) = 0.
Proof. Since B is positive definite, from Cholesky factorization, there exists an invertible lower triangle matrix L with positive diagonal entries such that B = LLT. Hence, for x 6= 0, we have
xTBx = xTLLTx = LTxT
LTx and
xTCx = xTLL−1C(LT)−1LTx = LTxT
L−1C(L−1)T
LTx . For convenience, we denote D := L−1C(L−1)T and let M := kDksup =
1≤i,j≤nmax |dij| be the supremum norm of D, where dij means the (i, j)-entry of D. In addition, for x 6= 0, we denote y = (y1, · · · , yn)T := LTx. Then, we obtain
xTCx xTBx
=
yTDy yTy
≤ Pn
i,j=1|dij||yi||yj| Pn
i=1|yi|2 . By Cauchy’s inequality |yi||yj| ≤ y
2 i+y2j
2 , we see that Pn
i,j=1|dij||yi||yj| Pn
i=1|yi|2 ≤ M ·Pn i,j=1
yi2+yj2 2
Pn
i=1y2i = M
2 ·nPn
i=1yi2+ nPn j=1yj2 Pn
i=1yi2 = nM which says
xTCx xTBx
≤ nM.
This further implies that kF (x)k ≤
xTCx xTBx
· kBxk + kCxk ≤ (nM )kBxk + kCxk.
Applying the continuity of linear transformation B and C proves lim
x→0F (x) =
0. 2
Very often, the mapping F in an SOCCP(F ) is required to be differ- entiable. Therefore, in view of Lemma 5.1, it is natural to redefine F (x) as
(11) F (x) =
xTCx
xTBxBx − Cx if x 6= 0,
0 if x = 0.
This enables that the mapping F : Rn → Rn is continuous. Indeed, it is clear that the mapping F : Rn→ Rn is even smooth except for 0. In other
words, F may not be differentiable at 0. To see this, we give an example as below. For n = 2, we take B =
b11 b12 b21 b22
which is positive definite with b12 > 0 and C =
c11 c12
c21 c22
with c226= 0. Because B is positive definite, the entries b11, b22 are positive. If we consider the first term of F (x) as in (10), i.e., xxTTCxBxBx, it can be written out as
c11x21+ (c12+ c21)x1x2+ c22x22 b11x21+ (b12+ b21)x1x2+ b22x22
·
b11x1+ b12x2
b21x1+ b22x2
.
If we denote
f (x) =
f1(x) f2(x)
:= xTCx xTBxBx, using the fact
xlim1→0
c11x21+ (c12+ c21)x1x2+ c22x22 b11x21+ (b12+ b21)x1x2+ b22x22
·
b11+b12x2 x1
= ∞,
we see that ∂f∂x1
1(0) does not exist. This means f is not differentiable at 0, and hence F (x) = f (x) − Cx is not differentiable at 0.
Next, we provide two technical lemmas in order to express the Jacobian matrix of F (x) for x 6= 0.
Lemma 5.2. Suppose that f : Rn → R and gi : Rn → R (1 ≤ i ≤ n)
are real-valued differentiable functions. Denote G(x) =
g1(x) g2(x)
... gn(x)
. Then,
the scalar product function f (x)G(x) =
f (x)g1(x) f (x)g2(x)
... f (x)gn(x)
is a differentiable
function on Rn and its Jacobian matrix ∇(f (x)G(x)) is expressed as
∇ f (x)G(x)
= ∇f (x)(G(x))T + f (x)∇G(x).
Proof. The proof comes from direct computation as below.
∇ f (x)G(x)
=
(∂x∂f
1 · g1+ f · ∂g∂x1
1)(x) (∂x∂f
1 · g2+ f ·∂x∂g2
1)(x) · · · (∂x∂f
1 · gn+ f · ∂g∂xn
1)(x) (∂x∂f
2 · g1+ f · ∂g∂x1
2)(x) (∂x∂f
2 · g2+ f ·∂x∂g2
2)(x) · · · (∂x∂f
2 · gn+ f · ∂g∂xn
2)(x)
... ... . .. ...
(∂x∂f
n · g1+ f · ∂x∂g1
n)(x) (∂x∂f
n · g2+ f ·∂x∂g2
n)(x) · · · (∂x∂f
n · gn+ f · ∂x∂gn
n)(x)
=
∂f
∂x1(x)
∂f
∂x2(x) ...
∂f
∂xn(x)
g1(x) g2(x) · · · gn(x) + f (x)
∂g1
∂x1(x) ∂x∂g2
1(x) · · · ∂g∂xn
1(x)
∂g1
∂x2(x) ∂x∂g2
2(x) · · · ∂g∂xn
2(x) ... ... . .. ...
∂g1
∂xn(x) ∂x∂g2
n(x) · · · ∂x∂gn
n(x)
= ∇f (x)(G(x))T + f (x)∇G(x).
2
Lemma 5.3. Consider the SOCEiCP(B, C) given as in (4) where B is positive definite. Let F : Rn→ Rnbe defined as in (11). Then, F is smooth except for 0 and its Jacobian matrix is expressed as
∇F (x) =(C + CT)xxTB − (B + BT)xxTC xxTBT
(xTBx)2 +xTCx
xTBxBT − CT. Proof. Denote f (x) = xxTTCxBx and g(x) = Bx. Then, F (x) = f (x)g(x) − Cx.
For x 6= 0, we know
∇f (x) = ∇(xTCx) · (xTBx) − (xTCx) · ∇(xTBx) (xTBx)2
= (C + CT)x · (xTBx) − (xTCx) · (B + BT)x (xTBx)2
= (C + CT)xxTB − (B + BT)xxTC x
(xTBx)2 .
Then, this together with Lemma 5.2 lead to the desired result. 2
Now, we sum up the relation between SOCEiCP(B, C) and SOCCP(F ) in the below theorem and we call it Reformulation I for SOCEiCP.
Theorem 5.1 (Reformulation I for SOCEiCP). Consider the SOCEiCP(B, C) given as in (4) where B is positive definite. Let F : Rn → Rn be defined as in (11). Then, the mapping F is smooth except for 0 and its Jacobian matrix is given as in Lemma 5.3. Moreover, the following hold.
(a) If (x∗, λ∗) solves the SOCEiCP(B, C), then x∗solves the SOCCP(F ).
(b) Conversely, if ¯x is a nonzero solution of the SOCCP(F ), then (x∗, λ∗) solves the SOCEiCP(B, C) with λ∗= xx¯¯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 SOCCP(F ) with F given as in (11). Then, we have xx¯¯TTC ¯B ¯xx · B ¯x − C ¯x ∈ Kn, ¯x ∈ Kn, and ¯xT
¯ xTC ¯x
¯
xTB ¯xB ¯x − C ¯x
= 0. Since a ∈ int(Kn) and ¯x ∈ Kn, it yields aT1¯x > 0 by the same arguments as on page 4. From all the above, we conclude that
y∗ := λ∗Bx∗− Cx∗ = aT1¯x
h¯xTC ¯x
¯ xTB ¯x
B ¯x − C ¯xi
∈ Kn, x∗ := aT1x¯x ∈ K¯ n,
aTx∗ = aaTTxx¯¯ = 1, (x∗)Ty∗ = aT1¯x
2h
¯ xT
¯ xTC ¯x
¯
xTB ¯xB ¯x − C ¯xi
= 0.
Thus, (x∗, λ∗) solves SOCEiCP(B, C). 2
Next, we consider the SOCQEiCP(A, B, C) given as in (5), where A, B, C ∈ Rn×nsuch that the matrix A is positive definite (hence A is Kn-regular) and the triplet (A, B, C) is Kn-hyperbolic. For any x ∈ Rnwith x 6= 0, plugging w = λ2Ax + λBx + Cx into the complementarity condition xTw = 0 yields (xTAx)λ2 + (xTBx)λ + (xTCx) = 0. Thus, λ can be obtained by solving this quadratic equation, i.e.,
λ1(x) = −(xTBx) +p
(xTBx)2− 4(xTAx)(xTCx)
2(xTAx) ,
(12)
λ2(x) = −(xTBx) −p
(xTBx)2− 4(xTAx)(xTCx)
2(xTAx) .
(13)
Then, for x 6= 0, we define Fi: Rn→ Rn as
(14) Fi(x) = λ2i(x)Ax + λi(x)Bx + Cx,
where i = 1, 2. In order to guarantee the well-definedness of Fi(0) for i = 1, 2, we need to look into lim
x→0Fi(x).
Lemma 5.4. Consider the SOCQEiCP(A, B, C) given as in (5) where A is positive definite. Let Fi : Rn → Rn be defined as in (14) where x 6= 0.
Then, we have lim
x→0Fi(x) = 0 for i = 1, 2.
Proof. Since A is positive definite, by Cholesky factorization, there exists an invertible lower triangle matrix L with positive diagonal entries such that A = LLT. Using the same techniques in the proof of Lemma 5.1, for x 6= 0, we obtain
xTAx = (LTx)T(LTx),
xTBx = (LTx)T(L−1B(L−1)T)(LTx), xTCx = (LTx)T(L−1C(L−1)T)(LTx).
For convenience, we denote D := L−1B(L−1)T, E := L−1C(L−1)T, M1 :=
kDksup = max
1≤i,j≤n|dij| be the supremum norm of D, and M2 := kEksup =
1≤i,j≤nmax |eij| be the supremum norm of E, where dij is the (i, j)-entry of D and eij is the (i, j)-entry of E. In addition, we also denote y = (y1, · · · , yn)T :=
LTx. Using the same techniques in the proof of Lemma 5.1, we obtain
|xTBx| = |yTDy| ≤ nM1 n
X
i=1
yi2,
|xTCx| = |yTEy| ≤ nM2 n
X
i=1
yi2. Hence, for each i and for x 6= 0, we see that
|λi(x)|
≤ |xTBx| +p|xTBx|2+ 4|xTBx||xTCx|
2|xTAx|
≤
nM1Pn i=1yi2+
q
(nM1Pn
i=1yi2)2+ 4(nM1Pn
i=1yi2)(nM2Pn i=1yi2) 2Pn
i=1y2i
≤ M3 := 1 +√ 5 2
!
n max{M1, M2}.
This yields
kFi(x)k ≤ M32kAxk + M3kBxk + kCxk,
for each i and x 6= 0. Then, by the continuity of linear transformation A, B, and C, the desired result follows. 2
Again, in view of Lemma 5.4, we need to do something to construct a differentiable mapping Fi. In other words, we redefine Fi(x) by
(15) Fi(x) =
λ2i(x)Ax + λi(x)Bx + Cx if x 6= 0,
0 if x = 0.
where λi(x), i = 1, 2 are given as in (12)-(13). From Lemma 5.4, it is clear that the mapping Fi : Rn → Rn is continuous for i = 1, 2. In fact, the mapping Fi : Rn→ Rn is smooth except for 0. To see this fact, we give an example as follows. For n = 2, we take A =
a11 a12 a21 a22
which is positive definite and B =
b11 b12 b21 b22
such that b22 6= 0. Because A is positive definite, the entries a22 are positive. Now for each i = 1, 2, we consider the first two terms of Fi(x) described as in (14), i.e.,
λ2i(x)
a11x1+ a12x2
a21x1+ a22x2
+ λi(x)
b11x1+ b12x2
b21x1+ b22x2
:=
gi1(x) gi2(x)
.
It can be verified that
xlim1→0
gi2(x) x1
= ∞,
which implies that ∂g∂xi2
1(0) does not exist. Therefore, Fi(x) is not differen- tiable at 0.
For x 6= 0, the Jacobian matrix of Fi(x) in (15) is computed as below.
Lemma 5.5. Consider the SOCQEiCP(A, B, C) given as in (5) where A is positive definite. Let Fi : Rn→ Rn be defined as in (15) for i = 1, 2. Then, Fi is smooth except for 0 and its Jacobian matrix is expressed as
∇Fi(x) = ∇λi(x) 2λi(x)xTAT + xTBT + λ2i(x)AT + λi(x)BT + CT,
where
∇λ1(x) = 1
2xTAx(Bx + BTx)
(D(x))−12(xTBx) − 1
− 1
xTAx· (D(x))−12 (Ax + ATx)(xTCx) + (Cx + CTx)(xTAx)
+ 1
2(xTAx)2 h
xTBx −p D(x)i
(Ax + ATx),
∇λ2(x) = − 1
2xTAx(Bx + BTx)((D(x))12(xTBx) + 1)
+ 1
xTAx(D(x))−12 (Ax + ATx)(xTCx) + (Cx + CTx)(xTAx)
+ 1
2(xTAx)2 h
xTBx +p D(x)
i
(Ax + ATx),
and D(x) := (xTBx)2− 4(xTAx)(xTCx).
