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 Tang^{2}

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.

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 IR^{n}, also called the Lorentz cone, is defined as

K^{n}:=(x1, x2) ∈ IR × IR^{n−1}| x1 ≥ kx2k , (1)
where k · k denotes the Euclidean norm. If n = 1, K^{n} 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 IR^{n×n}, i.e.,

S_{+}^{n} := X ∈ IR^{n×n}| X ∈ S^{n} and X O

:= X ∈ IR^{n×n}| X = X^{T} and v^{T}Xv ≥ 0 ∀v ∈ IR^{n} .

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 = (x_{1}, x_{2}) ∈ IR × IR^{n−1}, an arrow-shaped matrix L_{x} (alterna-
tively called an arrow matrix and denoted by Arw(x)) is defined as

L_{x} :=x_{1} x^{T}_{2}
x_{2} x_{1}I_{n−1}

. (2)

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

x ∈ K^{n} ⇐⇒ L_{x} :=x_{1} x^{T}_{2}
x_{2} x_{1}I_{n−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]

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 K^{n} ⇐⇒ L_{x} ∈ int S_{+}^{n};
(b) x = 0 ⇐⇒ L_{x}= 0;

(c) x ∈ bd K^{n}\ {0} ⇐⇒ L_{x} ∈ 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.

### 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 i_{F} := tr(X^{T}Y ), kXk_{F} :=phX, Xi_{F},

then, for any X ∈ S^{n}, its (repeated) eigenvalues λ_{1}, λ_{2}, · · · , λ_{n} are real and it admits a
spectral decomposition of the form:

X = P diag[λ_{1}, λ_{2}, · · · , λ_{n}] P^{T} (4)
for some P ∈ O. Here O denotes the set of orthogonal P ∈ IR^{n×n}, i.e., P^{T} = 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 K^{n}. To see this, we first introduce the so-called Jordan product. For any
x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} and y = (y_{1}, y_{2}) ∈ IR × IR^{n−1}, their Jordan product [7] is
defined by

x ◦ y := (hx, yi, y_{1}x_{2}+ x_{1}y_{2}) .

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} ∈ IR^{n}.
It can be verified that the arrow matrix L_{x} is a linear mapping from IR^{n} to IR^{n} given
by L_{x}y = x ◦ y. For each x = (x_{1}, x_{2}) ∈ IR × IR^{n−1}, x admits a spectral decomposition
[4, 5, 6, 7] associated with K^{n} in the form of

x = λ_{1}(x)u^{(1)}_{x} + λ_{2}(x)u^{(2)}_{x} , (5)
where λ_{1}(x), λ_{2}(x) and u^{(1)}x , u^{(2)}x are the spectral values and the corresponding spectral
vectors of x, respectively, given by

λ_{i}(x) := x_{1}+ (−1)^{i}kx_{2}k and u^{(i)}_{x} := 1
2

1

(−1)^{i}x¯2

, i = 1, 2, (6)
with ¯x_{2} = x_{2}/kx_{2}k if x_{2} 6= 0, and otherwise ¯x_{2} being any vector in IR^{n−1} with k¯x_{2}k = 1.

When x_{2} 6= 0, the spectral decomposition is unique. The following lemma states the
relation between the spectral decomposition of x and the eigenvalue decomposition of
L_{x}.

Lemma 2.1. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} have the spectral decomposition given as in
(5)-(6). Then, L_{x} has the eigenvalue decomposition:

L_{x} = P diag [λ_{1}(x), λ_{2}(x), x_{1}, · · · , x_{1}] P^{T}

where

P = h√

2u^{(1)}_{x} √

2u^{(2)}_{x} u^{(3)}_{x} · · · u^{(n)}_{x} i

∈ IR^{n×n}

is an orthogonal matrix, and u^{(i)}x for i = 3, · · · , n have the form of (0, ¯u_{i}) with ¯u_{3}, . . . , ¯u_{n}
being any unit vectors in IR^{n−1} that span the linear subspace orthogonal to x_{2}.

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

From Lemma 2.1, it is not hard to calculate the inverse of L_{x} whenever it exists:

L^{−1}_{x} = 1
det(x)

x1 −x^{T}_{2}

−x_{2} det(x)
x_{1} I + 1

x_{1}x_{2}x^{T}_{2}

(7)

where det(x) := x^{2}_{1}− kx2k^{2} denotes the determinant of x.

Throughout the whole paper, we use Π_{C}(·) 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 IR^{n}, we define AA^{T} := {uu^{T}| u ∈ A} and
L_{A} := {L_{u}| u ∈ A} respectively. We denote Λ^{n} the set of all arrow-shape matrices and
Λ^{n}_{+} the set of all positive semidefinite arrow matrices, i.e.,

Λ^{n} := {L_{y} ∈ IR^{n×n}| y ∈ IR^{n}} and Λ^{n}_{+} := {L_{y} O | y ∈ IR^{n}}.

Lemma 2.2. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} have the spectral decomposition given as in
(5)-(6). Then, the following hold:

(a) ΠK^{n}(x) = (x_{1}− kx_{2}k)_{+}u^{(1)}x + (x_{1}+ kx_{2}k)_{+}u^{(2)}x ,

(b) ΠS^{n}_{+}(L_{x}) = P

(x_{1}− kx_{2}k)_{+} 0 0

0 (x_{1}+ kx_{2}k)_{+} 0

0 0 (x_{1})_{+}I_{n−2}

P^{T} 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 K^{n}. In light of the relationship (3) between the SOC
and the PSD cone, one may ask “Can we obtain the expression of projection ΠK^{n}(x) by
using ΠS_{+}^{n}(Lx), the projection of Lx over S_{+}^{n}?”. In other words,

Is ΠK^{n}(x) = L^{−1}

ΠS_{+}^{n}(L_{x})

or ΠS_{+}^{n}(L_{x}) = L (ΠK^{n}(x)) right ? (8)

Here the operator L, defined as L(x) := L_{x}, is a single-point mapping between IR^{n} and
S^{n}, and L^{−1} is the inverse mapping of L, which can be achieved as in (7). To see this,
take x = (1, 2, 0) ∈ IR^{3}; then applying Lemma 2.1 yields

L_{x} =

√1 2

√1 2 0

−^{√}^{1}

2

√1 2 0

0 0 1

−1 0 0 0 3 0 0 0 1

√1
2 −^{√}^{1}

2 0

√1 2

√1 2 0

0 0 1

. Hence, by Lemma 2.2, we have

Π_{S}^{3}

+(Lx) =

√1 2

√1 2 0

−^{√}^{1}_{2} ^{√}^{1}_{2} 0

0 0 1

0 0 0 0 3 0 0 0 1

√1

2 −^{√}^{1}_{2} 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 L_{y} = ΠS_{+}^{n}(L_{x}). Note
that

ΠK^{n}(x) = (1 + 2)1
2

1 1 0

=

3 23 2

0

which gives

L (ΠK^{n}(x)) =

3 2

3 2 0

3 2

3
2 0
0 0 ^{3}_{2}

.
Hence Π_{K}^{n}(x) 6= L^{−1}(Π_{S}^{n}

+(L_{x})) and Π_{S}^{n}

+(L_{x}) 6= L(Π_{K}^{n}(x)). The distances dist(x, K^{n})
and dist(L_{x}, S_{+}^{3}) are also different, since

dist(x, K^{n}) = kx − ΠK^{n}(x)k =

−^{1}_{2}

1 2

0

=

√2 2 and

dist(L_{x}, S_{+}^{n}) = kL_{x}− ΠS_{+}^{n}(L_{x})k =

−^{1}_{2} ^{1}_{2} 0

1

2 −^{1}_{2} 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

Π_{Λ}^{n}_{+}(L_{x}) =

7 5

7 5 0

7 5

7
5 0
0 0 ^{7}_{5}

which implies L^{−1}(Π_{Λ}^{n}

+(L_{x})) = (^{7}_{5},^{7}_{5}, 0). To sum up, Π_{K}^{n}(x) 6= L^{−1}(Π_{Λ}^{n}

+(L_{x})) and
Π_{Λ}^{n}

+(L_{x}) 6= L(ΠK^{n}(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 L_{x} onto Λ^{n}_{+}:
min kL_{x}− L_{y}k_{F}

s.t. L_{y} ∈ Λ^{n}_{+} (9)

is equivalent to the following optimization problem:

min kL_{x−y}k_{F}

s.t. y ∈ K^{n}. (10)

Precisely, L_{y} 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 L_{x}−L_{y} = L_{x−y} and L_{y} ∈ Λ^{n}_{+} ⇐⇒ y ∈ K^{n}.
2

The result of Lemma 3.1 will help us to find the expressions of the distance and
projection of x onto K^{n}, L_{x} to S_{+}^{n} and Λ^{n}_{+}. In particular, the distance of x onto K^{n} 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 × IR^{n−1} have the spectral decomposition given as in
(5)-(6). Then, the following holds:

(a) dist(x, K^{n}) =
q1

2(x1− kx2k)^{2}_{−}+ ^{1}_{2}(x1+ kx2k)^{2}_{−};

(b) dist(L_{x}, S_{+}^{n}) = p(x_{1}− kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}+ (n − 2)(x_{1})^{2}_{−};

(c) ΠΛ^{n}_{+}(Lx) =

Lx if x1 ≥ kx2k,
O if x_{1} ≤ −_{n}^{2}kx_{2}k,

1

1+^{2}_{n} x_{1}+ _{n}^{2}kx_{2}k 1 x¯^{T}_{2}

¯

x_{2} I_{n−1}

if −_{n}^{2}kx_{2}k < x_{1} < kx_{2}k,

(d) dist(L_{x}, Λ^{n}_{+}) =
q 2n

n+2(x_{1}− kx_{2}k)^{2}_{−}+_{n+2}^{n}^{2} x_{1}+_{n}^{2}kx_{2}k2

−.

Proof. (a) From Lemma 2.2, we know that x = (x1 − kx2k)u^{(1)}x + (x1+ kx2k)u^{(2)}x and
ΠK^{n}(x) = (x_{1}− kx_{2}k)_{+}u^{(1)}x + (x_{1}+ kx_{2}k)_{+}u^{(2)}x . Thus, it is clear to see that

dist(x, K^{n}) = kx − ΠK^{n}(x)k

=

(x_{1}− kx_{2}k)_{−}u^{(1)}_{x} + (x_{1}+ kx_{2}k)_{−}u^{(2)}_{x}

= r1

2(x_{1}− kx_{2}k)^{2}_{−}+1

2(x_{1}+ kx_{2}k)^{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),

L_{x}= P

x_{1}− kx_{2}k 0 0

0 x_{1}+ kx_{2}k 0

0 0 x_{1}I_{n−2}

P^{T}
and

ΠS_{+}^{n}(L_{x}) = P

(x_{1}− kx_{2}k)_{+} 0 0

0 (x_{1}+ kx_{2}k)_{+} 0

0 0 (x_{1})_{+}I_{n−2}

P^{T}.
Combining the above yields

dist(L_{x}, S_{+}^{n}) =

(x1− kx2k)− 0 0

0 (x_{1}+ kx_{2}k)_{−} 0

0 0 (x_{1})−I_{n−2}

= q

(x_{1}− kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}+ (n − 2)(x_{1})^{2}_{−}.
(c) To find Π_{Λ}^{n}

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

kL_{x−y}k_{F}

= p

(x_{1}− y_{1}− kx_{2}− y_{2}k)^{2}+ (x_{1}− y_{1}+ kx_{2}− y_{2}k)^{2}+ (n − 2)(x_{1}− y_{1})^{2}

= p

n(x1− y1)^{2}+ 2kx2− y2k^{2}

= √

n r

(x_{1}− y_{1})^{2} + 2

nkx_{2}− y_{2}k^{2}

= √

n v u u

t(x1− y1)^{2} +

r2 nx2−

r2 ny2

2

. Now, we denote

y^{0} := y_{1},
r2

ny_{2}

!

= (y_{1}, γy_{2}) = Γy where γ :=

r2

n and Γ :=1 0 0 γI

.

Then, y_{1} ≥ ky_{2}k if and only if y^{0}_{1} ≥ ^{1}_{γ}ky_{2}^{0}k; that is, y ∈ K^{n} if and only if y^{0} ∈ L_{θ} with
cot θ = _{γ}^{1}, where L_{θ} := {x = (x_{1}, x_{2}) ∈ IR × IR^{n−1}|x_{1} ≥ kx_{2}k cot θ}; see [16]. We therefore
conclude that the problem (10) is indeed equivalent to the following optimization problem:

min r

(x1− y_{1}^{0})^{2}+

q2

nx2− y_{2}^{0}

2

s.t. y^{0} ∈ Lθ.

(11)

The optimal solution to the problem (11) is Π_{L}_{θ}(x^{0}), the projection of x^{0} := (x_{1}, γx_{2}) = Γx
onto L_{θ}, which according to [16, Theorems 3.1 and 3.2] is expressed by

ΠL_{θ}(x^{0})

= 1

1 + cot^{2}θ(x^{0}_{1}− kx^{0}_{2}k cot θ)+

1

−¯x^{0}_{2}cot θ

+ 1

1 + tan^{2}θ(x^{0}_{1}+ kx^{0}_{2}k tan θ)+

1

¯
x^{0}_{2}tan θ

= γ^{2}

1 + γ^{2}(x_{1}− kx_{2}k)_{+}

1

−^{1}_{γ}x¯_{2}

+ 1

1 + γ^{2}(x_{1}+ γ^{2}kx_{2}k)_{+}

1
γ ¯x_{2}

. Hence the optimal solution to (10) is

y = Γ^{−1}y^{0} = Γ^{−1}ΠL_{θ}(x^{0}) = Γ^{−1}ΠL_{θ}(Γx)

=

" _{γ}^{2}

1+γ^{2}(x1− kx2k)++ _{1+γ}^{1} 2(x1+ γ^{2}kx2k)+

−_{1+γ}^{1} 2(x_{1}− kx_{2}k)_{+}+ _{1+γ}^{1} 2(x_{1}+ γ^{2}kx_{2}k)_{+}

¯
x_{2}

#

(12)

=

x if x_{1} ≥ kx_{2}k,
0 if x_{1} ≤ −^{2}_{n}kx_{2}k,

1

1+γ^{2} (x_{1} + γ^{2}kx_{2}k) 1

¯
x_{2}

if −^{2}_{n}kx_{2}k < x_{1} < kx_{2}k.

By Lemma 3.1, the optimal solution to (9) is L_{y}, i.e,

L_{y} = Π_{Λ}^{n}

+(L_{x}) =

L_{x} if x_{1} ≥ kx_{2}k,
O if x_{1} ≤ −_{n}^{2}kx_{2}k,

1

1+_{n}^{2} x_{1}+_{n}^{2}kx_{2}k 1 x¯^{T}_{2}

¯

x_{2} I_{n−1}

if −_{n}^{2}kx_{2}k < x_{1} < kx_{2}k.

(d) In view of the expression (12), we can compute the distance dist(Lx, Λ^{n}_{+}) as follows.

dist(L_{x}, Λ^{n}_{+}) = kL_{x}− L_{y}k_{F} = kL_{x−y}k_{F}

= n

x_{1}− γ^{2}

1 + γ^{2}(x_{1}− kx_{2}k)_{+}− 1

1 + γ^{2} x_{1}+ γ^{2}kx_{2}k

+

2

+2

kx_{2}k + 1

1 + γ^{2}(x_{1}− kx_{2}k)_{+}− 1

1 + γ^{2} x_{1}+ γ^{2}kx_{2}k

+

2!^{1}_{2}

= n

x1− 2

n + 2(x1− kx2k)+− n

n + 2 x1+ 2 nkx2k

+

2

+2

kx_{2}k + n

n + 2(x_{1}− kx_{2}k)_{+}− n

n + 2 x_{1}+ 2
nkx_{2}k

+

2!^{1}_{2}

= n

2

n + 2(x_{1}− kx_{2}k)−+ n

n + 2 x_{1}+ 2
nkx_{2}k

−

2

+2

− n

n + 2(x_{1}− kx_{2}k)−+ n

n + 2 x_{1}+ 2
nkx_{2}k

−

2!^{1}_{2}

= s

2n

n + 2(x_{1}− kx_{2}k)^{2}_{−}+ n^{2}
n + 2

x_{1} + 2
nkx_{2}k

2

−

, where the third equation comes from the facts that

x1 = 2

n + 2(x1 − kx2k) + n n + 2

x1+ 2 nkx2k

and

kx_{2}k = − n

n + 2(x_{1} − kx_{2}k) + n
n + 2

x_{1}+ 2
nkx_{2}k

. 2

Theorem 3.2. For any x = (x_{1}, x_{2}) ∈ IR × IR^{n−1},

dist(x, K^{n}) ≤ dist(L_{x}, S_{+}^{n}) ≤ dist(L_{x}, Λ^{n}_{+}).

In particular, for n = 2,
dist(x, K^{2}) =

√2

2 dist(Lx, S_{+}^{2}) and dist(Lx, S_{+}^{2}) = dist(Lx, Λ^{2}_{+}).

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

dist(Lx, Λ^{2}_{+}) =
q

(x1− kx2k)^{2}_{−}+ (x1+ kx2k)^{2}_{−}.
Combining this and Theorem 3.1(a)-(b) yields dist(x, K^{2}) =

√ 2

2 dist(L_{x}, S_{+}^{2}) and dist(L_{x}, S_{+}^{2}) =
dist(L_{x}, Λ^{2}_{+}). 2

Note that Λ^{2}_{+} is strictly included in S_{+}^{2}, i.e., Λ^{2}_{+} $ S+^{2}, because in the arrow matrix,
the diagonal element is the same, but positive semidefinite matrix does not impose this
requirement. Thus, dist(L_{x}, Λ^{2}_{+}) ≤ dist(L_{x}, 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 : IR^{n} → IR, we say that they are equivalent if there
exist τ_{1}, τ_{2} > 0 such that

τ_{1}g(x) ≤ h(x) ≤ τ_{2}g(x), ∀x ∈ IR^{n}.

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

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

a + b 2

2

−

≤ 1

2 a^{2}_{−}+ b^{2}_{−}.

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 a^{2}_{−}+ b^{2}_{−}.

Case 2: For a ≤ 0 and b ≤ 0, we have

a + b 2

2

−

= a + b 2

2

≤ a^{2}+ b^{2}

2 = 1

2 a^{2}_{−}+ b^{2}_{−}.

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

= a^{2}+ b^{2}+ 2ab

4 ≤ a^{2}+ b^{2}

4 ≤ 1

2a^{2} = 1

2 a^{2}_{−}+ b^{2}_{−},

where the first inequality comes from the fact that ab ≤ 0 and the second inequality
follows from the fact that a^{2} ≥ b^{2} due to a ≤ −b ≤ 0.

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

a + b 2

2

−

= 0 ≤ 1

2a^{2} = 1

2 a^{2}_{−}+ b^{2}_{−}.

2

Theorem 3.3. The distances dist(x, K^{n}), dist(L_{x}, S_{+}^{n}), and dist(L_{x}, Λ^{n}_{+}) are all equiva-
lent in the sense of

dist(x, K^{n}) ≤ dist(L_{x}, S_{+}^{n}) ≤√

n dist(x, K^{n}) (13)

and

dist(L_{x}, S_{+}^{n}) ≤ dist(L_{x}, Λ^{n}_{+}) ≤

r 2n

n + 2dist(L_{x}, S_{+}^{n}). (14)
Proof. (i) The key part to prove inequality (13) is to look into dist^{2}(L_{x}, S_{+}^{n}), which are
computed as below:

dist^{2}(L_{x}, S_{+}^{n})

= (x_{1}− kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}+ (n − 2)(x_{1})^{2}_{−}

= (x_{1}− kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}+ (n − 2) (x_{1}− kx_{2}k) + (x_{1}+ kx_{2}k)
2

2

−

≤ (x_{1}− kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}+n − 2
2

(x_{1} − kx_{2}k)^{2}_{−}+ (x_{1}+ kx_{2}k)^{2}_{−}

= n 1

2(x_{1} − kx_{2}k)^{2}_{−}+1

2(x_{1}+ kx_{2}k)^{2}_{−}

= n dist^{2}(x, K^{n}),

where the inequality is due to Lemma 3.2. Hence, we achieve
dist(x, K^{n}) ≤ dist(L_{x}, S_{+}^{n}) ≤√

n dist(x, K^{n}),

which indicates that the distance between x to K^{n} and L_{x} to S_{+}^{n} is equivalent.

(ii) It remains to show the equivalence between dist(L_{x}, S_{+}^{n}) and dist(L_{x}, Λ^{n}_{+}). To proceed,
we need to consider the following cases.

Case 1: For x_{1} ≥ kx_{2}k, dist(L_{x}, S_{+}^{n}) = 0 = dist(L_{x}, Λ^{n}_{+}).

Case 2: For x_{1} ≤ −kx_{2}k, dist(L_{x}, Λ^{n}_{+}) = pnx^{2}_{1}+ 2kx_{2}k^{2} = dist(L_{x}, S_{+}^{n}).

Case 3: For 0 ≤ x1 ≤ kx2k, dist(Lx, Λ^{n}_{+}) =
q 2n

n+2|x1− kx2k| and dist(Lx, S_{+}^{n}) = |x1 −
kx_{2}k|.

Case 4: For −_{n}^{2}kx_{2}k ≤ x_{1} ≤ 0, dist^{2}(L_{x}, Λ^{n}_{+}) = _{n+2}^{2n} (x_{1} − kx_{2}k)^{2} and dist^{2}(L_{x}, S_{+}^{n}) =
(x_{1}− kx_{2}k)^{2}+ (n − 2)x^{2}_{1}. Then,

2n

n + 2dist^{2}(L_{x}, S_{+}^{n}) = 2n

n + 2 x_{1}− kx_{2}k2

+ 2n

n + 2(n − 2)x^{2}_{1} ≥ dist^{2}(L_{x}, Λ^{n}_{+}).

Case 5: For −kx_{2}k ≤ x_{1} ≤ −^{2}_{n}kx_{2}k,

dist^{2}(L_{x}, Λ^{n}_{+}) = nx^{2}_{1}+ 2kx_{2}k^{2} and dist^{2}(L_{x}, S_{+}^{n}) = (x_{1}− kx_{2}k)^{2}+ (n − 2)x^{2}_{1}.
Note that

dist(L_{x}, Λ^{n}_{+}) ≤

r 2n

n + 2dist(L_{x}, S_{+}^{n})

⇐⇒ nx^{2}_{1} + 2kx_{2}k^{2} ≤ 2n
n + 2

(x_{1}− kx_{2}k)^{2}+ (n − 2)x^{2}_{1}

⇐⇒ 4kx2knx1+ kx2k ≤ n(n − 4)x^{2}_{1}.
Since x_{1} ≤ −^{2}_{n}kx_{2}k, it implies that

4kx_{2}knx1+ kx_{2}k ≤ −4kx2k^{2} ≤ 4n − 4

n kx_{2}k^{2} = n(n − 4)

−2
nkx_{2}k

2

≤ n(n − 4)x^{2}_{1},

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

dist(L_{x}, Λ^{n}_{+}) ≤

r 2n

n + 2dist(L_{x}, 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, K^{n}) < dist(L_{x}, S_{+}^{n}) < √

n dist(x, K^{n})
and

dist(Lx, S_{+}^{n}) < dist(Lx, Λ^{n}_{+}) <

r 2n

n + 2dist(Lx, S_{+}^{n}).

To see this, from Theorem 3.1, we know that

dist(x, K^{n}) =
r9

2, dist(L_{x}, S_{+}^{n}) = √

n + 7, dist(L_{x}, Λ^{n}_{+}) =√

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, K^{n}) < dist(L_{x}, S_{+}^{n}) <√

n dist(x, K^{n}),
and

dist(L_{x}, S_{+}^{n}) < dist(L_{x}, Λ^{n}_{+}) <

r 2n

n + 2dist(L_{x}, 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).

### 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

T_{C}(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 = (x_{1}, x_{2}) ∈ IR × IR^{n−1} belong to K^{n}, i.e., x ∈ K^{n}. Then

(a) TK^{n}(x) =

K^{n} if x = 0,

IR^{n} if x ∈ int K^{n},

(d_{1}, d_{2}) ∈ IR^{n}

d^{T}_{2}x_{2}− x_{1}d_{1} ≤ 0

if x ∈ bd K^{n}\{0}.

(b) TS_{+}^{n}(Lx) =

S_{+}^{n} if x = 0,

S^{n} if x ∈ int K^{n},

n

H ∈ S^{n}| (u^{(1)}x )^{T}Hu^{(1)}x ≥ 0o

if x ∈ bd K^{n}\{0}.

(c) T_{Λ}^{n}

+(L_{x}) = {L_{h}| h ∈ TK^{n}(x)} = TS_{+}^{n}(L_{x}) ∩ Λ^{n}.

Proof. The formulae of TK^{n}(x) and TS_{+}(L_{x}) follows from the results given in [2, 14]. To
verify part(c), we know that

T_{Λ}^{n}

+(L_{x}) =H ∈ S^{n}| L_{x}+ t_{n}H_{n}∈ Λ^{n}_{+}, t_{n} → 0^{+}, H_{n} → H .

Due to tnHn ∈ Λ^{n}_{+}− Lx, Hn is also an arrow matrix. This means Hn = Lhn for some
h_{n} ∈ IR^{n}. In addition, H_{n} → H implies H = L_{h} for some h with h_{n} → h. Thus, we
obtain that L_{x}+ t_{n}H_{n}= L_{x+t}_{n}_{h}_{n} ∈ Λ^{n}_{+} which is equivalent to saying x + t_{n}h_{n} ∈ K^{n}, i.e.,
h ∈ TK^{n}(x). Moreover, since Λ^{n}_{+} = S_{+}^{n} ∩ Λ^{n} and S_{+}^{n}, Λ^{n} cannot be separated, it yields

T_{Λ}^{n}

+(L_{x}) = TS^{n}_{+}(L_{x}) ∩ T_{Λ}^{n}(L_{x}) = TS^{n}_{+}(L_{x}) ∩ Λ^{n}

by [12, Theorem 6.42], where the last step comes from the fact that Λ^{n} is a subspace.

2

The relation between TK^{n}(x) and TS^{n}_{+}(L_{x}) can be also characterized by using their
expression.

Theorem 4.2. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} belong to K^{n}, i.e., x ∈ K^{n}. Then,

L_{T}_{Kn}_{(x)} = TS_{+}^{n}(L_{x}) ∩ Λ^{n}. (16)

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

Case 1: For x ∈ intK^{n}, we have Lx ∈ intS_{+}^{n}. Thus, TK^{n}(x) = IR^{n} and TS_{+}^{n}(Lx) = S^{n}.
This implies

L_{T}_{Kn}_{(x)} = L_{IR}^{n} = Λ^{n} = S^{n}∩ Λ^{n} = TS_{+}^{n}(L_{x}) ∩ Λ^{n}.

Case 2: For x = 0, we have TK^{n}(x) = K^{n} and TS_{+}^{n}(L_{x}) = S_{+}^{n}. Since y ∈ K^{n} if and only if
Ly ∈ S_{+}^{n},

L_{T}_{Kn}_{(x)}= L_{K}^{n} = Λ^{n}_{+}= S_{+}^{n} ∩ Λ^{n}= T_{S}^{n}

+(L_{x}) ∩ Λ^{n}.
Case 3: For x ∈ bd K^{n}\{0}, take d ∈ TK^{n}(x). Then,

(u^{(1)}_{x} )^{T}L_{d}u^{(1)}_{x} = 1

4 1 − ¯x^{T}_{2}d_{1} d^{T}_{2}
d_{2} d_{1}I

1

−¯x_{2}

= 1

2(d_{1}− d^{T}_{2}x¯_{2}) ≥ 0,

where the inequality comes from d ∈ TK^{n}(x). Hence, L_{d}∈ TS_{+}^{n}(L_{x}) by Theorem 4.1, i.e.,
L_{T}_{Kn}_{(x)} ⊂ TS_{+}^{n}(L_{x}) ∩ Λ^{n}. The converse inclusion can be proved by a similar argument.

2

The restriction to Λ^{n} in (16) is required, which is illustrated by the following example.

Taking x = (1, 1) ∈ IR^{2}, we have

T_{K}^{2}(x) = {d = (d_{1}, d_{2}) ∈ IR^{2}| − d_{1}+ d_{2} ≤ 0}

and
T_{S}^{2}

+(L_{x}) = H ∈ S^{2}| (u^{(1)}_{x} )^{T}Hu^{(1)}_{x} ≥ 0 = H ∈ S^{2}| H_{11}− 2H_{12}+ H_{22}≥ 0 .
Hence, L_{T}_{Kn}_{(x)} does not equal TS_{+}^{n}(L_{x}).

### 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 NK^{n}(x), NS_{+}^{n}(L_{x}),
and N_{Λ}^{n}

+(L_{x}), respectively.

Theorem 5.1. Let x = (x_{1}, x_{2}) ∈ IR × IR^{n−1} belong to K^{n}, i.e., x ∈ K^{n}. Then

(a) NK^{n}(x) =

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

(b) N_{S}^{n}

+(L_{x}) =

−S_{+}^{n} if x = 0,

{O} if x ∈ int K^{n},

α

1 −¯x^{T}_{2}

−¯x2 x¯2x¯^{T}_{2}

α ≤ 0

if x ∈ bd K^{n}\{0}.

(c) N_{Λ}^{n}

+(L_{x}) = N_{S}^{n}

+(L_{x}) + (Λ^{n})^{⊥}, where

(Λ^{n})^{⊥}= {H ∈ S^{n}| tr(H) = 0, H_{1,i} = 0, i = 2, · · · , n} .

Proof. Part (a) and (b) follow from [2] and [14]. For Part (c), since Λ^{n}_{+} = S_{+}^{n}∩ Λ^{n}, it
follows from [12, Theorem 6.42] that

N_{Λ}^{n}_{+}(L_{x}) = NS_{+}^{n}(L_{x}) + N_{Λ}^{n}(L_{x}).

Because Λ^{n} is a subspace, we know that N_{Λ}^{n}(L_{x}) = (Λ^{n})^{⊥}, where

(Λ^{n})^{⊥} = {H ∈ S^{n}| hH, L_{y}i = 0, ∀y ∈ IR^{n}} = {H ∈ S^{n}| tr(H) = 0, H_{1,i} = 0, i = 2, · · · , n} .
2

The relation between N_{Λ}^{n}

+(L_{x}) and NS^{n}_{+}(L_{x}) is already described in Theorem 5.1.

Next, we further describe the relation between NK^{n}(x) and NS^{n}_{+}(L_{x}).

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

N_{S}^{n}

+(L_{x}) = −N_{K}^{n}(x)N_{K}^{n}(x)^{T}.

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

Case 2: For x ∈ bd K^{n}\{0}, it follows from Theorem 5.1 that
NS_{+}^{n}(L_{x}) =

α

1 −¯x^{T}_{2}

−¯x_{2} x¯_{2}x¯^{T}_{2}

α ≤ 0

=

α

1

−¯x2

1, −¯x^{T}_{2}

α ≤ 0

. (17)

Since NK^{n}(x) = {y| y = β ˆx, β ≤ 0} with ˆx := (x_{1}, −x_{2}),

−N_{K}^{n}(x)N_{K}^{n}(x)^{T} = {−β^{2}xˆˆx^{T}| β ≤ 0} =

−(βx_{1})^{2}

1

−¯x_{2}

1, −¯x^{T}_{2}

β ≤ 0

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

From Theorem 5.1(c), we know that N_{Λ}^{n}

+(L_{x}) ⊃ NS_{+}^{n}(L_{x}) since O ∈ (Λ^{n})^{⊥}. For x = 0,
NK^{n}(x) = −K^{n} and NS_{+}^{n}(L_{x}) = −S_{+}^{n}. In this case, NS^{n}_{+}(L_{x}) and −NK^{n}(x)NK^{n}(x)^{T} do
not coincide, i.e., Theorem 5.2 fails when x = 0. Below, we give the algebraic expressions
for N_{Λ}^{n}

+(L_{x}) and NS_{+}^{n}(L_{x}) as n = 2, from which we can see the difference between them
more clearly.

Theorem 5.3. For n = 2, the explicit expressions of N_{S}^{2}

+(L_{x}) and N_{Λ}^{2}

+(L_{x}) are as below:

N_{S}^{2}

+(L_{x}) =

a b b c

a ≤ 0, c ≤ 0, ac ≥ b^{2}

if x = 0,

{O} if x ∈ int K^{2},

α 1 −1

−1 1

α ≤ 0

if x ∈ bd K^{2}\{0}, x_{2} > 0,

α1 1 1 1

α ≤ 0

if x ∈ bd K^{2}\{0}, x_{2} < 0.

and

N_{Λ}^{2}

+(L_{x}) =

a b b c

a + c ≤ −2|b|

if x = 0,

a b b c

a + c = 0, b = 0

if x ∈ int K^{2},

a b b c

a + c + 2b = 0, b ≥ 0

if x ∈ bd K^{2}\{0}, x_{2} > 0,

a b b c

a + c − 2b = 0, b ≤ 0

if x ∈ bd K^{2}\{0}, x_{2} < 0.

Proof. First, we claim that

(Λ^{2}_{+})^{◦} = a b
b c

a + c ≤ −2|b|

. In fact

a b b c

∈ (Λ^{2}_{+})^{◦} ⇐⇒ a b
b c

,x_{1} x_{2}
x2 x1

≤ 0, ∀x_{1} ≥ |x_{2}|,

⇐⇒ (a + c)x_{1}+ 2bx_{2} ≤ 0, ∀x_{1} ≥ |x_{2}|. (19)
If we plug in x_{1} = |x_{2}| + τ with τ ≥ 0, then (19) can be rewritten as

(a + c)|x_{2}| + 2bx_{2}+ (a + c)τ ≤ 0, ∀x_{2} ∈ 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|.