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to appear in Optimization, 2016

Further relationship between second-order cone and positive semidefinite cone

Jinchuan Zhou 1 Department of Mathematics

School of Science

Shandong University of Technology Zibo 255049, P.R. China E-mail: jinchuanzhou@163.com

Jingyong Tang2

College of Mathematics and Information Science Xinyang Normal University

Xinyang 464000, Henan, P.R.China E-mail: jingyongtang@163.com

Jein-Shan Chen 3 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

February 2, 2016 (1st revised on May 2, 2016) (2nd revised on July 21, 2016)

Abstract. It is well known that second-order cone programming can be regarded as a special case of positive semidefinite programming by using the arrow matrix. This paper further studies the relationship between second-order cones and positive semi- definite matrix cones. In particular, we explore the relationship to expressions regarding

1The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026).

2The author’s work is supported by Basic and Frontier Technology Research Project of Henan Province (162300410071).

3Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan.

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distance, projection, tangent cone, normal cone, and the KKT system. Understanding these relationships will help us to see the connection and difference between the SOC and its PSD reformulation more clearly.

Keywords. Positive semidefinite matrix cone, second-order cone, projection, tangent cone, normal cone, KKT system.

AMS subject classifications. 90C25; 90C22.

1 Introduction

The second-order cone (SOC) in IRn, also called the Lorentz cone, is defined as

Kn:=(x1, x2) ∈ IR × IRn−1| x1 ≥ kx2k , (1) where k · k denotes the Euclidean norm. If n = 1, Kn is the set of nonnegative reals IR+. The positive semidefinite matrix cone (PSD cone), denoted by S+n, is the collection of all symmetric positive semidefinite matrices in IRn×n, i.e.,

S+n := X ∈ IRn×n| X ∈ Sn and X  O

:= X ∈ IRn×n| X = XT and vTXv ≥ 0 ∀v ∈ IRn .

It is well known that second-order cone and positive semidefinite matrix cone both belong to the category of symmetric cones [7], which are unified under Euclidean Jordan algebra.

In [1], for each vector x = (x1, x2) ∈ IR × IRn−1, an arrow-shaped matrix Lx (alterna- tively called an arrow matrix and denoted by Arw(x)) is defined as

Lx :=x1 xT2 x2 x1In−1



. (2)

It can be verified that there is a close relationship between the SOC and the PSD cone as below:

x ∈ Kn ⇐⇒ Lx :=x1 xT2 x2 x1In−1



 O. (3)

Hence, a second-order cone program (SOCP) can be recast as a special semidefinite pro- gram (SDP). In light of this, it seems that we just need to focus on SDP. Nevertheless, this reformulation has some disadvantages. For example, the reference [11] indicates that

“Solving SOCPs via SDP is not a good idea, however. Interior-point methods that solve the SOCP directly have a much better worst-case complexity than an SDP method....

The difference between these numbers is significant if the dimensions of the second-order constraints are large.”. This comment mainly concerns the algorithmic aspects; see [1, 11]

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for more information.

In fact, “reformulation” is usually the main idea behind many approaches to studying various optimization problems and it is necessary to discuss the relationship between the primal problem and the transformed problem. For example, for complementarity prob- lems (or variational inequality problems), we can reformulate these problems to work on a minimization optimization problem via merit functions (or gap functions). The proper- ties of merit functions ensure the solution to complementarity problems is the same as the global optimal solution to the minimization problem. Nonetheless, finding a global opti- mal solution is very difficult. Thus, we turn to study the connection between the solution to complementarity problems and the stationary points of the transformed optimization problem. Similarly, for mathematical programming with complementarity constraints (MPCC), the ordinary KKT conditions do not hold, because the standard constraint qualification fails to hold (due to the existence of complementarity constraints). One therefore considers to recast MPCC to other types of optimization problems with differ- ent approaches. These different approaches also ensure the solution set of MPCC is the same to that of the transformed optimization problems. But, the KKT conditions for these transformed optimization problems are different, which are the source of various concepts of stationary points for MPCC, such as S-, M -, C-stationary points.

A similar question arises from SOCP and its SDP-reformulation. In view of the above discussions, it could be interesting to study their relation from theoretical and numerical aspects. As mentioned above, the reference [11] mainly deals with the SOCP and its SDP-reformulation from the perspective of algorithm. The study on the relationship between SOCP and its corresponding SDP from theoretical aspect is rare. Sim and Zhao [13] discuss the relation between SOCP and its SDP counterpart from the perspective of duality theory. There are already some known relations between the SOC and the PSD cone; for instance,

(a) x ∈ int Kn ⇐⇒ Lx ∈ int S+n; (b) x = 0 ⇐⇒ Lx= 0;

(c) x ∈ bd Kn\ {0} ⇐⇒ Lx ∈ bd S+n\ {O}.

Besides the interior, boundary point set, we know that for an optimization problem, some other topological structures, such as tangent cones, normal cones, projections, and KKT systems, play very important roles. One may wonder whether there exist analogous relationship between the SOC and the PSD cone? We will answer it in this paper. In particular, by comparing the expressions of distance, projection, tangent cone, normal cone, and the KKT system between the SOC and the PSD cone, we will know more about the differences between SOCP and its SDP reformulation.

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2 Preliminaries

In this section, we introduce some background materials that will be used in subsequent analysis. In the space of matrices, if we equip it with the trace inner product and the Frobenius norm

hX, Y iF := tr(XTY ), kXkF :=phX, XiF,

then, for any X ∈ Sn, its (repeated) eigenvalues λ1, λ2, · · · , λn are real and it admits a spectral decomposition of the form:

X = P diag[λ1, λ2, · · · , λn] PT (4) for some P ∈ O. Here O denotes the set of orthogonal P ∈ IRn×n, i.e., PT = P−1.

The above factorization (4) is the well-known spectral decomposition (eigenvalue de- composition) in matrix analysis [9]. There is a similar spectral decomposition associ- ated with Kn. To see this, we first introduce the so-called Jordan product. For any x = (x1, x2) ∈ IR × IRn−1 and y = (y1, y2) ∈ IR × IRn−1, their Jordan product [7] is defined by

x ◦ y := (hx, yi, y1x2+ x1y2) .

Since the Jordan product, unlike scalar or matrix multiplication, is not associative, this is a main source on complication in the analysis of second-order cone complementarity problem (SOCCP). The identity element under this product is e := (1, 0, · · · , 0)T ∈ IRn. It can be verified that the arrow matrix Lx is a linear mapping from IRn to IRn given by Lxy = x ◦ y. For each x = (x1, x2) ∈ IR × IRn−1, x admits a spectral decomposition [4, 5, 6, 7] associated with Kn in the form of

x = λ1(x)u(1)x + λ2(x)u(2)x , (5) where λ1(x), λ2(x) and u(1)x , u(2)x are the spectral values and the corresponding spectral vectors of x, respectively, given by

λi(x) := x1+ (−1)ikx2k and u(i)x := 1 2

 1

(−1)i2



, i = 1, 2, (6) with ¯x2 = x2/kx2k if x2 6= 0, and otherwise ¯x2 being any vector in IRn−1 with k¯x2k = 1.

When x2 6= 0, the spectral decomposition is unique. The following lemma states the relation between the spectral decomposition of x and the eigenvalue decomposition of Lx.

Lemma 2.1. Let x = (x1, x2) ∈ IR × IRn−1 have the spectral decomposition given as in (5)-(6). Then, Lx has the eigenvalue decomposition:

Lx = P diag [λ1(x), λ2(x), x1, · · · , x1] PT

(5)

where

P = h√

2u(1)x

2u(2)x u(3)x · · · u(n)x i

∈ IRn×n

is an orthogonal matrix, and u(i)x for i = 3, · · · , n have the form of (0, ¯ui) with ¯u3, . . . , ¯un being any unit vectors in IRn−1 that span the linear subspace orthogonal to x2.

Proof. Please refer to [5, 6, 8]. 2

From Lemma 2.1, it is not hard to calculate the inverse of Lx whenever it exists:

L−1x = 1 det(x)

x1 −xT2

−x2 det(x) x1 I + 1

x1x2xT2

 (7)

where det(x) := x21− kx2k2 denotes the determinant of x.

Throughout the whole paper, we use ΠC(·) to denote the projection mapping onto a closed and convex set C. In addition, for α ∈ IR, (α)+ := max{α, 0} and (α) :=

min{α, 0}. Given a nonempty subset A in IRn, we define AAT := {uuT| u ∈ A} and LA := {Lu| u ∈ A} respectively. We denote Λn the set of all arrow-shape matrices and Λn+ the set of all positive semidefinite arrow matrices, i.e.,

Λn := {Ly ∈ IRn×n| y ∈ IRn} and Λn+ := {Ly  O | y ∈ IRn}.

Lemma 2.2. Let x = (x1, x2) ∈ IR × IRn−1 have the spectral decomposition given as in (5)-(6). Then, the following hold:

(a) ΠKn(x) = (x1− kx2k)+u(1)x + (x1+ kx2k)+u(2)x ,

(b) ΠSn+(Lx) = P

(x1− kx2k)+ 0 0

0 (x1+ kx2k)+ 0

0 0 (x1)+In−2

PT where P is an orthog- onal matrix of Lx.

Proof. Please see [8, 15] for a proof. 2

3 Relation on Distance and Projection

In this section, we show the relation on distance and projection associated with the SOC and the PSD cone. We begin with some explanation for why we need to do so. First, let us consider the projection of x over Kn. In light of the relationship (3) between the SOC and the PSD cone, one may ask “Can we obtain the expression of projection ΠKn(x) by using ΠS+n(Lx), the projection of Lx over S+n?”. In other words,

Is ΠKn(x) = L−1

ΠS+n(Lx)

or ΠS+n(Lx) = L (ΠKn(x)) right ? (8)

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Here the operator L, defined as L(x) := Lx, is a single-point mapping between IRn and Sn, and L−1 is the inverse mapping of L, which can be achieved as in (7). To see this, take x = (1, 2, 0) ∈ IR3; then applying Lemma 2.1 yields

Lx =

1 2

1 2 0

1

2

1 2 0

0 0 1

−1 0 0 0 3 0 0 0 1

1 21

2 0

1 2

1 2 0

0 0 1

. Hence, by Lemma 2.2, we have

ΠS3

+(Lx) =

1 2

1 2 0

12 12 0

0 0 1

0 0 0 0 3 0 0 0 1

1

212 0

1 2

1 2 0

0 0 1

=

3 2

3 2 0

3 2

3 2 0 0 0 1

,

which is not a form of the arrow matrix as shown in (2), because the diagonal entries are not equal. This means that we cannot seek a vector y such that Ly = ΠS+n(Lx). Note that

ΠKn(x) = (1 + 2)1 2

 1 1 0

=

3 23 2

0

 which gives

L (ΠKn(x)) =

3 2

3 2 0

3 2

3 2 0 0 0 32

. Hence ΠKn(x) 6= L−1Sn

+(Lx)) and ΠSn

+(Lx) 6= L(ΠKn(x)). The distances dist(x, Kn) and dist(Lx, S+3) are also different, since

dist(x, Kn) = kx − ΠKn(x)k =

12

1 2

0

=

√2 2 and

dist(Lx, S+n) = kLx− ΠS+n(Lx)k =

12 12 0

1

212 0

0 0 0

= 1.

The failure of the above approach comes from the fact that the PSD cone is much larger, i.e., there exists a positive semi-definite matrix that is not arrow-shape. Conse- quently, we may ask whether (8) holds if we restrict the positive semi-definite matrices to arrow-shape matrices. Still for x = (1, 2, 0), by the expression given as in Theorem 3.1 below, we know that

ΠΛn+(Lx) =

7 5

7 5 0

7 5

7 5 0 0 0 75

(7)

which implies L−1Λn

+(Lx)) = (75,75, 0). To sum up, ΠKn(x) 6= L−1Λn

+(Lx)) and ΠΛn

+(Lx) 6= L(ΠKn(x)). All the above observations and discussions lead us to further explore some relationship, other than (3), between the SOC and the PSD cone.

Lemma 3.1. The problem of finding the projection of Lx onto Λn+: min kLx− LykF

s.t. Ly ∈ Λn+ (9)

is equivalent to the following optimization problem:

min kLx−ykF

s.t. y ∈ Kn. (10)

Precisely, Ly is an optimal solution to (9) if and only if y is an optimal solution to (10).

Proof. The result follows from the facts that Lx−Ly = Lx−y and Ly ∈ Λn+ ⇐⇒ y ∈ Kn. 2

The result of Lemma 3.1 will help us to find the expressions of the distance and projection of x onto Kn, Lx to S+n and Λn+. In particular, the distance of x onto Kn and Lx to S+n can be obtained by using their expression of the projection given in Lemma 2.2.

Theorem 3.1. Let x = (x1, x2) ∈ IR × IRn−1 have the spectral decomposition given as in (5)-(6). Then, the following holds:

(a) dist(x, Kn) = q1

2(x1− kx2k)2+ 12(x1+ kx2k)2;

(b) dist(Lx, S+n) = p(x1− kx2k)2+ (x1+ kx2k)2+ (n − 2)(x1)2;

(c) ΠΛn+(Lx) =





Lx if x1 ≥ kx2k, O if x1 ≤ −n2kx2k,

1

1+2n x1+ n2kx2k 1 x¯T2

¯

x2 In−1



if −n2kx2k < x1 < kx2k,

(d) dist(Lx, Λn+) = q 2n

n+2(x1− kx2k)2+n+2n2 x1+n2kx2k2

.

Proof. (a) From Lemma 2.2, we know that x = (x1 − kx2k)u(1)x + (x1+ kx2k)u(2)x and ΠKn(x) = (x1− kx2k)+u(1)x + (x1+ kx2k)+u(2)x . Thus, it is clear to see that

dist(x, Kn) = kx − ΠKn(x)k

=

(x1− kx2k)u(1)x + (x1+ kx2k)u(2)x

(8)

= r1

2(x1− kx2k)2+1

2(x1+ kx2k)2 where the last step is derived from ku(i)x k =√

2/2 for i = 1, 2 and hu(1)x , u(2)x i = 0.

(b) By Lemma 2.1 and Lemma 2.2(b),

Lx= P

x1− kx2k 0 0

0 x1+ kx2k 0

0 0 x1In−2

PT and

ΠS+n(Lx) = P

(x1− kx2k)+ 0 0

0 (x1+ kx2k)+ 0

0 0 (x1)+In−2

PT. Combining the above yields

dist(Lx, S+n) =

(x1− kx2k) 0 0

0 (x1+ kx2k) 0

0 0 (x1)In−2

= q

(x1− kx2k)2+ (x1+ kx2k)2+ (n − 2)(x1)2. (c) To find ΠΛn

+(Lx), we need to solve the optimization problem (9). From Lemma 3.1, it is equivalent to look into problem (10). Thus, we first compute

kLx−ykF

= p

(x1− y1− kx2− y2k)2+ (x1− y1+ kx2− y2k)2+ (n − 2)(x1− y1)2

= p

n(x1− y1)2+ 2kx2− y2k2

= √

n r

(x1− y1)2 + 2

nkx2− y2k2

= √

n v u u

t(x1− y1)2 +

r2 nx2

r2 ny2

2

. Now, we denote

y0 := y1, r2

ny2

!

= (y1, γy2) = Γy where γ :=

r2

n and Γ :=1 0 0 γI

 .

Then, y1 ≥ ky2k if and only if y011γky20k; that is, y ∈ Kn if and only if y0 ∈ Lθ with cot θ = γ1, where Lθ := {x = (x1, x2) ∈ IR × IRn−1|x1 ≥ kx2k cot θ}; see [16]. We therefore conclude that the problem (10) is indeed equivalent to the following optimization problem:

min r

(x1− y10)2+

q2

nx2− y20

2

s.t. y0 ∈ Lθ.

(11)

(9)

The optimal solution to the problem (11) is ΠLθ(x0), the projection of x0 := (x1, γx2) = Γx onto Lθ, which according to [16, Theorems 3.1 and 3.2] is expressed by

ΠLθ(x0)

= 1

1 + cot2θ(x01− kx02k cot θ)+

 1

−¯x02cot θ



+ 1

1 + tan2θ(x01+ kx02k tan θ)+

 1

¯ x02tan θ



= γ2

1 + γ2(x1− kx2k)+

 1

1γ2



+ 1

1 + γ2(x1+ γ2kx2k)+

 1 γ ¯x2

 . Hence the optimal solution to (10) is

y = Γ−1y0 = Γ−1ΠLθ(x0) = Γ−1ΠLθ(Γx)

=

" γ2

1+γ2(x1− kx2k)++ 1+γ1 2(x1+ γ2kx2k)+

−1+γ1 2(x1− kx2k)++ 1+γ1 2(x1+ γ2kx2k)+

¯ x2

#

(12)

=





x if x1 ≥ kx2k, 0 if x1 ≤ −2nkx2k,

1

1+γ2 (x1 + γ2kx2k) 1

¯ x2



if −2nkx2k < x1 < kx2k.

By Lemma 3.1, the optimal solution to (9) is Ly, i.e,

Ly = ΠΛn

+(Lx) =





Lx if x1 ≥ kx2k, O if x1 ≤ −n2kx2k,

1

1+n2 x1+n2kx2k 1 x¯T2

¯

x2 In−1



if −n2kx2k < x1 < kx2k.

(d) In view of the expression (12), we can compute the distance dist(Lx, Λn+) as follows.

dist(Lx, Λn+) = kLx− LykF = kLx−ykF

= n



x1− γ2

1 + γ2(x1− kx2k)+− 1

1 + γ2 x1+ γ2kx2k

+

2

+2



kx2k + 1

1 + γ2(x1− kx2k)+− 1

1 + γ2 x1+ γ2kx2k

+

2!12

= n



x1− 2

n + 2(x1− kx2k)+− n

n + 2 x1+ 2 nkx2k

+

2

+2



kx2k + n

n + 2(x1− kx2k)+− n

n + 2 x1+ 2 nkx2k

+

2!12

= n

 2

n + 2(x1− kx2k)+ n

n + 2 x1+ 2 nkx2k

2

(10)

+2



− n

n + 2(x1− kx2k)+ n

n + 2 x1+ 2 nkx2k

2!12

= s

2n

n + 2(x1− kx2k)2+ n2 n + 2



x1 + 2 nkx2k

2

, where the third equation comes from the facts that

x1 = 2

n + 2(x1 − kx2k) + n n + 2



x1+ 2 nkx2k



and

kx2k = − n

n + 2(x1 − kx2k) + n n + 2



x1+ 2 nkx2k

 . 2

Theorem 3.2. For any x = (x1, x2) ∈ IR × IRn−1,

dist(x, Kn) ≤ dist(Lx, S+n) ≤ dist(Lx, Λn+).

In particular, for n = 2, dist(x, K2) =

√2

2 dist(Lx, S+2) and dist(Lx, S+2) = dist(Lx, Λ2+).

Proof. The first inequality follows from the formula of distance given as in Theorem 3.1; the second inequality comes from the fact that Λn+ is a subset of S+n, i.e., Λn+ ⊂ S+n. For n = 2, by part(d) of Theorem 3.1, we have

dist(Lx, Λ2+) = q

(x1− kx2k)2+ (x1+ kx2k)2. Combining this and Theorem 3.1(a)-(b) yields dist(x, K2) =

2

2 dist(Lx, S+2) and dist(Lx, S+2) = dist(Lx, Λ2+). 2

Note that Λ2+ is strictly included in S+2, i.e., Λ2+ $ S+2, because in the arrow matrix, the diagonal element is the same, but positive semidefinite matrix does not impose this requirement. Thus, dist(Lx, Λ2+) ≤ dist(Lx, S+2). In Theorem 3.2, we further show that the equality holds.

In view of Theorem 3.2, a natural question arises here: are these distances equivalent?

Recall that for two functions g, h : IRn → IR, we say that they are equivalent if there exist τ1, τ2 > 0 such that

τ1g(x) ≤ h(x) ≤ τ2g(x), ∀x ∈ IRn.

For instance, 1-norm and 2-norm are equivalent in this sense. To answer this question, we need the following lemma.

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Lemma 3.2. For a, b ∈ IR, the following inequality holds:

 a + b 2

2

≤ 1

2 a2+ b2.

Proof. We assume without loss of generality that a ≤ b. Then, we consider the following four cases to proceed the proof.

Case 1: For a ≥ 0 and b ≥ 0, we have

 a + b 2

2

= 0 = 1

2 a2+ b2.

Case 2: For a ≤ 0 and b ≤ 0, we have

 a + b 2

2

= a + b 2

2

≤ a2+ b2

2 = 1

2 a2+ b2.

Case 3: For a ≤ 0, b ≥ 0, and a ≤ −b, there implies (a + b)/2 ≤ 0. Then, we have

 a + b 2

2

= a + b 2

2

= a2+ b2+ 2ab

4 ≤ a2+ b2

4 ≤ 1

2a2 = 1

2 a2+ b2,

where the first inequality comes from the fact that ab ≤ 0 and the second inequality follows from the fact that a2 ≥ b2 due to a ≤ −b ≤ 0.

Case 4: For a ≤ 0, b ≥ 0, and a ≥ −b, we have

 a + b 2

2

= 0 ≤ 1

2a2 = 1

2 a2+ b2.

2

Theorem 3.3. The distances dist(x, Kn), dist(Lx, S+n), and dist(Lx, Λn+) are all equiva- lent in the sense of

dist(x, Kn) ≤ dist(Lx, S+n) ≤√

n dist(x, Kn) (13)

and

dist(Lx, S+n) ≤ dist(Lx, Λn+) ≤

r 2n

n + 2dist(Lx, S+n). (14) Proof. (i) The key part to prove inequality (13) is to look into dist2(Lx, S+n), which are computed as below:

dist2(Lx, S+n)

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= (x1− kx2k)2+ (x1+ kx2k)2+ (n − 2)(x1)2

= (x1− kx2k)2+ (x1+ kx2k)2+ (n − 2) (x1− kx2k) + (x1+ kx2k) 2

2

≤ (x1− kx2k)2+ (x1+ kx2k)2+n − 2 2



(x1 − kx2k)2+ (x1+ kx2k)2

= n 1

2(x1 − kx2k)2+1

2(x1+ kx2k)2



= n dist2(x, Kn),

where the inequality is due to Lemma 3.2. Hence, we achieve dist(x, Kn) ≤ dist(Lx, S+n) ≤√

n dist(x, Kn),

which indicates that the distance between x to Kn and Lx to S+n is equivalent.

(ii) It remains to show the equivalence between dist(Lx, S+n) and dist(Lx, Λn+). To proceed, we need to consider the following cases.

Case 1: For x1 ≥ kx2k, dist(Lx, S+n) = 0 = dist(Lx, Λn+).

Case 2: For x1 ≤ −kx2k, dist(Lx, Λn+) = pnx21+ 2kx2k2 = dist(Lx, S+n).

Case 3: For 0 ≤ x1 ≤ kx2k, dist(Lx, Λn+) = q 2n

n+2|x1− kx2k| and dist(Lx, S+n) = |x1 − kx2k|.

Case 4: For −n2kx2k ≤ x1 ≤ 0, dist2(Lx, Λn+) = n+22n (x1 − kx2k)2 and dist2(Lx, S+n) = (x1− kx2k)2+ (n − 2)x21. Then,

2n

n + 2dist2(Lx, S+n) = 2n

n + 2 x1− kx2k2

+ 2n

n + 2(n − 2)x21 ≥ dist2(Lx, Λn+).

Case 5: For −kx2k ≤ x1 ≤ −2nkx2k,

dist2(Lx, Λn+) = nx21+ 2kx2k2 and dist2(Lx, S+n) = (x1− kx2k)2+ (n − 2)x21. Note that

dist(Lx, Λn+) ≤

r 2n

n + 2dist(Lx, S+n)

⇐⇒ nx21 + 2kx2k2 ≤ 2n n + 2



(x1− kx2k)2+ (n − 2)x21



⇐⇒ 4kx2knx1+ kx2k ≤ n(n − 4)x21. Since x1 ≤ −2nkx2k, it implies that

4kx2knx1+ kx2k ≤ −4kx2k2 ≤ 4n − 4

n kx2k2 = n(n − 4)



−2 nkx2k

2

≤ n(n − 4)x21,

(13)

where the second inequality is due to the fact n−4n ≥ −1 for all n ≥ 2. Hence,

dist(Lx, Λn+) ≤

r 2n

n + 2dist(Lx, S+n), which is the desired result. 2

The following example demonstrates that the inequalities (13) and (14) in Theorem 3.3 may be strict.

Example 3.1. Consider x = (−1, 2, 0, . . . , 0

| {z }

n−2

) with n ≥ 4. Then,

dist(x, Kn) < dist(Lx, S+n) < √

n dist(x, Kn) and

dist(Lx, S+n) < dist(Lx, Λn+) <

r 2n

n + 2dist(Lx, S+n).

To see this, from Theorem 3.1, we know that

dist(x, Kn) = r9

2, dist(Lx, S+n) = √

n + 7, dist(Lx, Λn+) =√

n + 8. (15) Note that for n ≥ 4, we have

r9 2 <√

n + 7 <

r9n 2 , and

√n + 7 <√

n + 8 <

r 2n n + 2

√n + 7,

which says

dist(x, Kn) < dist(Lx, S+n) <√

n dist(x, Kn), and

dist(Lx, S+n) < dist(Lx, Λn+) <

r 2n

n + 2dist(Lx, S+n).

From this example, we see that the distance related to second-order cone is indepen- dent of n; nonetheless, if we treat it as semi-definite matrix, the distance is dependent on n; see (15).

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4 Relation on Tangent Cone

As shown earlier, all the distances introduced in Section 2 are equivalent. This allows us to study the relation on tangent cone, because the tangent cone can be achieved by distance function [3]. More specifically, for a convex set C, there is

TC(x) := {h | dist(x + th, C) = o(t), t ≥ 0}.

In light of this, this section is devoted to exploring relation on tangent cones.

Theorem 4.1. Let x = (x1, x2) ∈ IR × IRn−1 belong to Kn, i.e., x ∈ Kn. Then

(a) TKn(x) =

Kn if x = 0,

IRn if x ∈ int Kn,

(d1, d2) ∈ IRn

dT2x2− x1d1 ≤ 0

if x ∈ bd Kn\{0}.

(b) TS+n(Lx) =





S+n if x = 0,

Sn if x ∈ int Kn,

n

H ∈ Sn| (u(1)x )THu(1)x ≥ 0o

if x ∈ bd Kn\{0}.

(c) TΛn

+(Lx) = {Lh| h ∈ TKn(x)} = TS+n(Lx) ∩ Λn.

Proof. The formulae of TKn(x) and TS+(Lx) follows from the results given in [2, 14]. To verify part(c), we know that

TΛn

+(Lx) =H ∈ Sn| Lx+ tnHn∈ Λn+, tn → 0+, Hn → H .

Due to tnHn ∈ Λn+− Lx, Hn is also an arrow matrix. This means Hn = Lhn for some hn ∈ IRn. In addition, Hn → H implies H = Lh for some h with hn → h. Thus, we obtain that Lx+ tnHn= Lx+tnhn ∈ Λn+ which is equivalent to saying x + tnhn ∈ Kn, i.e., h ∈ TKn(x). Moreover, since Λn+ = S+n ∩ Λn and S+n, Λn cannot be separated, it yields

TΛn

+(Lx) = TSn+(Lx) ∩ TΛn(Lx) = TSn+(Lx) ∩ Λn

by [12, Theorem 6.42], where the last step comes from the fact that Λn is a subspace.

2

The relation between TKn(x) and TSn+(Lx) can be also characterized by using their expression.

Theorem 4.2. Let x = (x1, x2) ∈ IR × IRn−1 belong to Kn, i.e., x ∈ Kn. Then,

LTKn(x) = TS+n(Lx) ∩ Λn. (16)

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Proof. We proceed the proof by discussing the following three cases.

Case 1: For x ∈ intKn, we have Lx ∈ intS+n. Thus, TKn(x) = IRn and TS+n(Lx) = Sn. This implies

LTKn(x) = LIRn = Λn = Sn∩ Λn = TS+n(Lx) ∩ Λn.

Case 2: For x = 0, we have TKn(x) = Kn and TS+n(Lx) = S+n. Since y ∈ Kn if and only if Ly ∈ S+n,

LTKn(x)= LKn = Λn+= S+n ∩ Λn= TSn

+(Lx) ∩ Λn. Case 3: For x ∈ bd Kn\{0}, take d ∈ TKn(x). Then,

(u(1)x )TLdu(1)x = 1

4 1 − ¯xT2d1 dT2 d2 d1I

  1

−¯x2



= 1

2(d1− dT22) ≥ 0,

where the inequality comes from d ∈ TKn(x). Hence, Ld∈ TS+n(Lx) by Theorem 4.1, i.e., LTKn(x) ⊂ TS+n(Lx) ∩ Λn. The converse inclusion can be proved by a similar argument.

2

The restriction to Λn in (16) is required, which is illustrated by the following example.

Taking x = (1, 1) ∈ IR2, we have

TK2(x) = {d = (d1, d2) ∈ IR2| − d1+ d2 ≤ 0}

and TS2

+(Lx) = H ∈ S2| (u(1)x )THu(1)x ≥ 0 = H ∈ S2| H11− 2H12+ H22≥ 0 . Hence, LTKn(x) does not equal TS+n(Lx).

5 Relation on Normal Cone

In this section, we continue to explore relation on normal cone between the SOC and its PSD reformulation. To this end, we first write out the expressions of NKn(x), NS+n(Lx), and NΛn

+(Lx), respectively.

Theorem 5.1. Let x = (x1, x2) ∈ IR × IRn−1 belong to Kn, i.e., x ∈ Kn. Then

(a) NKn(x) =

−Kn if x = 0, {0} if x ∈ int Kn, IR+(−x1, x2) if x ∈ bd Kn\{0}.

(b) NSn

+(Lx) =





−S+n if x = 0,

{O} if x ∈ int Kn,

 α

 1 −¯xT2

−¯x22T2



α ≤ 0



if x ∈ bd Kn\{0}.

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(c) NΛn

+(Lx) = NSn

+(Lx) + (Λn), where

n)= {H ∈ Sn| tr(H) = 0, H1,i = 0, i = 2, · · · , n} .

Proof. Part (a) and (b) follow from [2] and [14]. For Part (c), since Λn+ = S+n∩ Λn, it follows from [12, Theorem 6.42] that

NΛn+(Lx) = NS+n(Lx) + NΛn(Lx).

Because Λn is a subspace, we know that NΛn(Lx) = (Λn), where

n) = {H ∈ Sn| hH, Lyi = 0, ∀y ∈ IRn} = {H ∈ Sn| tr(H) = 0, H1,i = 0, i = 2, · · · , n} . 2

The relation between NΛn

+(Lx) and NSn+(Lx) is already described in Theorem 5.1.

Next, we further describe the relation between NKn(x) and NSn+(Lx).

Theorem 5.2. Let x = (x1, x2) ∈ IR × IRn−1 belong to Kn, i.e., x ∈ Kn. Then, for x ∈ int Kn and x ∈ bd Kn\{0},

NSn

+(Lx) = −NKn(x)NKn(x)T.

Proof. Case 1: For x ∈ int Kn, NKn(x) = {0} and NS+n(Lx) = {O}. The desired result holds in this case.

Case 2: For x ∈ bd Kn\{0}, it follows from Theorem 5.1 that NS+n(Lx) =

 α

 1 −¯xT2

−¯x22T2



α ≤ 0



=

 α

 1

−¯x2



1, −¯xT2

α ≤ 0



. (17)

Since NKn(x) = {y| y = β ˆx, β ≤ 0} with ˆx := (x1, −x2),

−NKn(x)NKn(x)T = {−β2xˆˆxT| β ≤ 0} =



−(βx1)2

 1

−¯x2



1, −¯xT2

β ≤ 0



. (18) Compared with (17) and (18) yields the desired result. 2

From Theorem 5.1(c), we know that NΛn

+(Lx) ⊃ NS+n(Lx) since O ∈ (Λn). For x = 0, NKn(x) = −Kn and NS+n(Lx) = −S+n. In this case, NSn+(Lx) and −NKn(x)NKn(x)T do not coincide, i.e., Theorem 5.2 fails when x = 0. Below, we give the algebraic expressions for NΛn

+(Lx) and NS+n(Lx) as n = 2, from which we can see the difference between them more clearly.

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Theorem 5.3. For n = 2, the explicit expressions of NS2

+(Lx) and NΛ2

+(Lx) are as below:

NS2

+(Lx) =





















 a b b c



a ≤ 0, c ≤ 0, ac ≥ b2



if x = 0,

{O} if x ∈ int K2,



α 1 −1

−1 1



α ≤ 0



if x ∈ bd K2\{0}, x2 > 0,



α1 1 1 1



α ≤ 0



if x ∈ bd K2\{0}, x2 < 0.

and

NΛ2

+(Lx) =

























 a b b c



a + c ≤ −2|b|



if x = 0,

 a b b c



a + c = 0, b = 0



if x ∈ int K2,

 a b b c



a + c + 2b = 0, b ≥ 0



if x ∈ bd K2\{0}, x2 > 0,

 a b b c



a + c − 2b = 0, b ≤ 0



if x ∈ bd K2\{0}, x2 < 0.

Proof. First, we claim that

2+) = a b b c



a + c ≤ −2|b|

 . In fact

a b b c



∈ (Λ2+) ⇐⇒ a b b c



,x1 x2 x2 x1



≤ 0, ∀x1 ≥ |x2|,

⇐⇒ (a + c)x1+ 2bx2 ≤ 0, ∀x1 ≥ |x2|. (19) If we plug in x1 = |x2| + τ with τ ≥ 0, then (19) can be rewritten as

(a + c)|x2| + 2bx2+ (a + c)τ ≤ 0, ∀x2 ∈ IR and τ ≥ 0, i.e.,

(a + c + 2b)x2+ (a + c)τ ≤ 0, ∀x2 ≥ 0 and τ ≥ 0 (20) and

(−a − c + 2b)x2+ (a + c)τ ≤ 0, ∀x2 ≤ 0 and τ ≥ 0. (21) With the arbitrariness of τ ≥ 0, we have a + c ≤ 0. Likewise, we have a + c + 2b ≤ 0 by (20) and −a − c + 2b ≥ 0 by (21). Thus, a + c ≤ −2b and a + c ≤ 2b. In other words, we conclude that the inequality (19) implies

a + c ≤ min{−2b, 2b} = −2|b|.

參考文獻

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