DOI 10.1007/s10589-009-9301-2
The penalized Fischer-Burmeister SOC complementarity function
Shaohua Pan·Jein-Shan Chen·Sangho Kum· Yongdo Lim
Received: 23 May 2008 / Published online: 20 November 2009
© Springer Science+Business Media, LLC 2009
Abstract In this paper, we study the properties of the penalized Fischer-Burmeister (FB) second-order cone (SOC) complementarity function. We show that the func- tion possesses similar desirable properties of the FB SOC complementarity function for local convergence; for example, with the function the second-order cone comple- mentarity problem (SOCCP) can be reformulated as a (strongly) semismooth system of equations, and the corresponding nonsmooth Newton method has local quadratic convergence without strict complementarity of solutions. In addition, the penalized FB merit function has bounded level sets under a rather weak condition which can be satisfied by strictly feasible monotone SOCCPs or SOCCPs with the CartesianR01- property, although it is not continuously differentiable. Numerical results are included to illustrate the theoretical considerations.
Work of S. Pan is supported by National Young Natural Science Foundation (No. 10901058) and Guangdong Natural Science Foundation (No. 9251802902000001).
J.-S. Chen is a member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is partially supported by National Science Council of Taiwan.
S. Pan
School of Mathematical Sciences, South China University of Technology, Guangzhou 510640, China e-mail:[email protected]
J.-S. Chen (
)Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan e-mail:[email protected]
S. Kum
Department of Mathematics Education, Chungbuk National University, Cheongju 361-763, Korea Y. Lim
Department of Mathematics, Kyungpook National University, Taegu 702-701, Korea
Keywords Second-order cone complementarity problem·Penalized
Fischer-Burmeister function·Nonsmooth Newton method·B-subdifferential· Coerciveness
1 Introduction
LetF andGbe continuously differentiable mappings fromRn toRn. We consider the second-order cone complementarity problem (SOCCP) which is to find aζ ∈Rn such that
F (ζ )∈K, G(ζ )∈K, F (ζ ), G(ζ ) =0, (1) where·,·is the Euclidean inner product, andK=Kn1 ×Kn2× · · · ×Knr is the Cartesian product of second-order cones (SOCs) withr, n1, . . . , nr ≥1,n1+ · · · + nr=n, and
Kni:=
(xi1, xi2)∈R×Rni−1|xi1≥ xi2 .
In the rest of this paper, corresponding to the Cartesian structure ofK, we write F=(F1, . . . , Fr)andG=(G1, . . . , Gr)withFi, Gi:Rn→Rni fori=1,2, . . . , r.
An important special case of the SOCCP corresponds toG(ζ )≡ζ. Then (1) re- duces to
F (ζ )∈K, ζ ∈K, F (ζ ), ζ =0, (2) which is a natural extension of the nonlinear complementarity problem (NCP) over the nonnegative orthant coneK=Rn+; see [9,11]. Another important special case corresponds to the optimality conditions of the second-order cone program (SOCP):
min g(x)
s.t. Ax=b, x∈K, (3)
whereg:Rn→Ris a twice continuously differentiable function,A∈Rm×nhas full row rank andb∈Rm. From [6], the KKT conditions of (3) can be reformulated as (1) with
F (ζ ):= ¯x+(I−AT(AAT)−1A)ζ, G(ζ ):= ∇g(F (ζ ))−AT(AAT)−1Aζ, (4) wherex¯ ∈Rn satisfies Ax¯=b. The SOCP (3) has numerous applications arising from the fields of engineering design, finance, robust optimization, combinatorial op- timization, and includes as special cases convex quadratically constrained quadratic programs; see [1,18].
In the past ten years, there have proposed various methods for SOCPs and SOC- CPs. They include the interior-point methods [2,19,27,28], the smoothing Newton methods [7,13, 14], the merit function method [6], and the semismooth Newton method [16]. Among others, SOC complementarity functions play a key role in the
last three ones. Specifically, we call φ:Rn×Rn→Rn an SOC complementarity function associated withKnif
φ (x, y)=0 ⇐⇒ x∈Kn, y∈Kn, x, y =0. (5) Clearly, whenn=1, an SOC complementarity function reduces to an NCP function.
A popular choice ofφis the vector-valued FB functionφFB:Rn×Rn→Rngiven by
φFB(x, y) := (x+y)−(x2+y2)1/2, (6) wherex2=x◦xdenotes the Jordan product ofxand itself,x1/2is a vector such that (x1/2)2=x for anyx∈Kn, andx+y means the usual componentwise addition of vectors. This function was shown to be strongly semismooth in [4,26] via different ways, and its squared norm ψFB = 12φFB2 induces a continuously differentiable merit function [6]. Recently, we analyze that, to guarantee the boundedness of the level sets of the function
FB(ζ ) :=
r i=1
ψFB(Fi(ζ ), Gi(ζ )),
it requires thatF andG at least have the uniform CartesianP-property; see [21]
for details. This means thatφFB has a limitation in handling monotone SOCCPs. In fact, observing the disadvantage of the FB merit functionFB, Chen [5] extended two classes of regularized merit functions for the NCPs to deal with the monotone SOCCPs. But, those functions can not be used to design fast Newton-type methods.
We notice that in the setting of NCPs, the penalized FB function was proposed in [3]
to overcome such shortcoming of the FB function. Thus, it is natural to ask whether the extension of the penalized FB function to the SOCs can become an effective tool in the solution of monotone SOCCPs or not. The main contribution of this paper is to offer partial answers to this question.
The vector-valued penalized FB functionφρ:Rn×Rn→Rnis defined as follows φρ(x, y):=ρφFB(x, y)+(1−ρ)
(x)+◦(y)+
, (7)
where ρ ∈ (0,1) is an arbitrary but fixed parameter, and (·)+ means the mini- mum Euclidean distance projection onKn. We show thatφρ has similar favorable properties ofφFB for local convergence. For example,φρ is a strongly semismooth SOC complementarity function, by which the SOCCP (1) can be reformulated as a (strongly) semismooth system
ρ(ζ ):=
⎛
⎜⎝
φρ(F1(ζ ), G1(ζ )) ...
φρ(Fr(ζ ), Gr(ζ ))
⎞
⎟⎠=0, (8)
and consequently one can apply the nonsmooth Newton method in [24,25], i.e., ζk+1:=ζk−Wk−1ρ(ζk), Wk∈∂Bρ(ζk), k=0,1,2, . . . (9)
to solve (1). Particularly, we establish that the nonsmooth method has the local quadratic convergence without strict complementarity of solutions, although their nondegeneracy is still necessary just like the FB semismooth method [20]. This is an advantage compared to interior-point methods where singular Jacobians will occur if strict complementarity is not satisfied. In addition, we also prove that the penalized FB merit function
ρ(ζ ) :=
r i=1
ψρ(Fi(ζ ), Gi(ζ )) (10) with
ψρ(x, y) := 1
2φρ(x, y)2 (11)
has bounded level sets under a rather weak condition, which can be satisfied by strictly feasible monotone SOCCPs or SOCCPs with the CartesianR01-property. In other words,ρ enjoys some nice features of the merit functions studied by [5] for global convergence. However, unlike its counterpart in the NCP setting,ρ is not smooth even not differentiable.
Throughout this paper,I represents an identity matrix of suitable dimension,Rn denotes the space ofn-dimensional real column vectors, andRn1×· · ·×Rnr is identi- fied withRn1+···+nr. We denote by int(Kn)and bd(Kn)the interior and the boundary of Kn, respectively. For any x ∈Rn, (x)+ and (x)− denote the Euclidean projec- tion ofxontoKnand−Kn, respectively. For a differentiable mappingF,F(x)and
∇F (x)denotes the Jacobian ofF and the transposed Jacobian atx, respectively. For a matrixB∈Rn×n, ifx, Bx ≥0(>0)for all 0=x∈Rn, we say thatBis positive semidefinite (positive definite) and use the symbolB0(0)to mean thatB is symmetric and positive semidefinite (positive definite). For anyx∈Rl withl >1, we writex=(x1, x2)wherex1is the first component ofx, andx2means the vector consisting of the restl−1 components.
2 Preliminaries
This section recalls some background materials and shows thatφρis a strongly semi- smooth SOC complementarity function. We start with the definition of Jordan product [10]:
x◦y:=(x, y, x1y2+y1x2) ∀x, y∈Rn.
The Jordan product, unlike matrix multiplication, is not associative in general. The identity element under this product ise:=(1,0, . . . ,0)T ∈Rn, i.e.,e◦x=xfor any x∈Rn. Let
Lx:=
x1 x2T x2 x1I
,
which can be viewed as a linear mapping fromRn toRnwithLxy=x◦y for any y∈Rn.
From [10] we recall that each x ∈Rn has a spectral factorization associated withKn
x = λ1(x)·u(1)x +λ2(x)·u(2)x ,
whereλi(x) and u(i)x for i=1,2 are the spectral values of x and the associated spectral vectors, respectively, defined by
λi(x):=x1+(−1)ix2, u(i)x :=1 2
1, (−1)ix¯2
,
withx¯2=xx22ifx2=0 and otherwisex¯2being any vector fromRn−1with ¯x2 =1.
If x2=0, the factorization is unique. It is easy to verify that the following rela- tion holds between the spectral factorization ofxand the eigenvalue decomposition ofLx.
Lemma 2.1 For anyx∈Rn,Lx has two single eigenvaluesλ1(x)and λ2(x)with u(1)x andu(2)x being the corresponding eigenvectors, and the remainingn−2 eigen- values are identicallyx1=(λ2(x)+λ1(x))/2 with the corresponding eigenvectors of the form(0,v), where¯ v¯lies in the linear subspace ofRn−1orthogonal tox2.
To show that φρ is an SOC complementarity function, we need the following lemma.
Lemma 2.2 Let φFB and φρ be given by (6) and (7), respectively. Then, for any x, y∈Rn,
[φFB(x, y)]1≥ −2((x)− + (y)−) and φρ(x, y) ≥ρmax{(x)−,(y)−}.
Proof The first inequality follows from the trace inequality of [17, Theorem 3.1], since
2[φFB(x, y)]1=tr(x+y)−tr
(x2+y2)1/2
≥tr(x)+tr(y)−tr(|x|)−tr(|y|)
=2 tr[(x)−] +2 tr[(y)−]
≥ −4(x− + y−).
We next prove the second inequality. Usingx=(x)++(x)−, it follows that φρ(x, y)2=ρ[x+y−(x2+y2)1/2] +(1−ρ)[(x)+◦(y)+]2
=ρ(x)−+ρ[(x)++y−(x2+y2)1/2] +(1−ρ)[(x)+◦(y)+]2
=ρ2(x)−2
+ρ[(x)++y−(x2+y2)1/2] +(1−ρ)[(x)+◦(y)+]2 +2ρ[(x)−]T
ρ
(x)++y−(x2+y2)1/2
+(1−ρ)(x)+◦(y)+
≥ρ2(x)−2+2ρ2[(x)−]T[(x)+] +2ρ2[(x)−]T
y−(x2+y2)1/2 +2ρ(1−ρ)[(x)−]T
(x)+◦(y)+
≥ρ2(x)−2
where the last inequality is since(x)−, y−(x2+y2)1/2∈ −Kn,(x)+, (x)− =0 and
(x◦y)Tz=(y◦z)Tx=(z◦x)Ty for allx, y, z∈Rn.
By the symmetry ofxandy inφρ, similarly, we haveφρ(x, y) ≥ρ(y)−. Proposition 2.1 Letφρandρ be defined as in (7) and (8), respectively. Then, (a) φρ(x, y)=0⇔x∈Kn, y∈Knandx, y =0.
(b) φρ is strongly semismooth.
(c) ρ is semismooth, and strongly semismooth ifF, Gare locally Lipschitz con- tinuous.
Proof (a) The sufficiency is direct by noting that x, y = 0 ⇔ x ◦ y = (x)+◦(y)+=0 andφFB is an SOC complementarity function. Now suppose that φρ(x, y)=0. From the second inequality of Lemma 2.2, we have (x)−=0 and (y)−=0, which implies x, y∈Kn, and hence xTy ≥0. In addition, by the first inequality of Lemma2.2and[φρ(x, y)]1=0,
0=ρ[φFB(x, y)]1+(1−ρ)xTy≥ −2ρ((x)− + (y)−)+(1−ρ)xTy
=(1−ρ)xTy.
The two sides show thatx∈Kn,y∈Knandx, y =0.
(b) Since the FB functionφFBand the projection function(·)+are strongly semi- smooth by [26, Corollary 3.3] and [14, Proposition 4.5], respectively, the result fol- lows from the fact given by [12] that the composite of (strongly) semismooth func- tions is (strongly) semismooth.
(c) The result is immediate by using part (b) and Theorem 19 of [12].
To close this section, we review several concepts that will be used in the sequel.
Definition 2.1 (a) Two mappingsF, G:Rn→Rnare said to have the jointly Carte- sianR01-property if for any sequence{ζk}satisfying
ζk → +∞, [−G(ζk)]+
ζk →0, [−F (ζk)]+
ζk →0, (12)
there exists an indexν∈ {1,2, . . . , r}such that lim inf
k→+∞
Fν(ζk), Gν(ζk) ζk >0.
(b) A mappingF :Rn→Rnis said to have the Cartesian weak coercive property with respect to an element ξ ∈Rn, if there exists an indexν∈ {1,2, . . . , r}such that
lim inf
ζ→∞
ζν−ξν, Fν(ζ ) ζ−ξ >0.
Given a mappingH :Rn→Rm, ifH is locally Lipschitz continuous, then the set
∂BH (z):=
V ∈Rm×n| ∃{zk} ⊆DH:zk→z, H(zk)→V
is nonempty and called the B-subdifferential ofH atz, whereDH ⊆Rn is the set of points at whichHis differentiable. The convex hull∂H (z):=conv∂BH (z)is the generalized Jacobian ofH atzin the sense of Clarke [8]. We assume that the reader is familiar with the concepts of (strongly) semismooth functions, and refer to [24,25]
for details.
3 B-subdifferential
In this section, we present the representation of the elements in the B-subdifferential ofφρat a general point, and then concentrate on the elements of the B-subdifferential ofφρat complementarity pairs(xi∗, y∗i)withi∈ {1,2, . . . , r}, i.e., each pair ofxi∗and yi∗satisfies
xi∗∈Kni, y∗i ∈Kni, xi∗, yi∗ =0. (13) For this purpose, we need the following two lemmas which overestimate the B- subdifferential of the FB functionφFB and the projection function(·)+at a general point, respectively.
Lemma 3.1 [20, Proposition 3.1] For any given(x, y)∈Rn×Rn, each element [U V] ∈∂BφFB(x, y)has the following representation:
(a) Ifx2+y2∈int(Kn), thenφFBis continuously differentiable at(x, y)with U= ∇xφFB(x, y)T =I−L−1
(x2+y2)1/2Lx, V = ∇yφFB(x, y)T =I−L−1
(x2+y2)1/2Ly. (b) Ifx2+y2∈bd(Kn)and(x, y)=(0,0), then
U V
∈
I−H Lx−1 2
1
− ¯w2
uT I−H Ly−1 2
1
− ¯w2
vT
for someu, vsatisfying|u1| ≤ u2 ≤1, |v1| ≤ v2 ≤1, (u1−v1)≤ u2−v2, (u1+v1)≤ u2+v2,
(u1−v1)2+ u2+v22≤2, (u1+v1)2+ u2−v22≤2, (1,w¯T2)u=0, (1,− ¯w2T)u=2u1, (1,w¯2T)v=0,
(1,− ¯wT2)v=2v1
, (14)
where
H= 1
4
x12+y12
1 w¯T2
¯
w2 4I−3w¯2w¯T2
withw¯2= x1x2+y1y2 x1x2+y1y2.
(c) If(x, y)=(0,0), then[U V] ∈ {[I−LgI−Lh]for someg2+h2=e}or [U V] ∈
I−1
2 1
− ¯w2
uT−1
2 1
¯ w2
ξT −
0 0 0 I− ¯w2w¯2T
Ls
I−1 2
1
− ¯w2
vT −1
2 1
¯ w2
ηT −
0 0 0 I− ¯w2w¯2T
Lω
for some ¯w2 =1,someu, vsatisfying(14)with suchw¯2, someξ, η∈Rnsatisfying|ξ1| ≤ ξ2 ≤1,|η1| ≤ η2 ≤1, (ξ1−η1)≤ ξ2−η2, (ξ1+η1)≤ ξ2+η2,
(ξ1−η1)2+ ξ2+η22≤2, (ξ1+η1)2+ ξ2−η22≤2, (1,w¯2T)ξ=2ξ1,
(1,− ¯wT2)ξ=0, (1,w¯2T)η=2η1, (1,− ¯w2T)η=0, ands=σ u+(1−σ )ξ, ω=σ v+(1−σ )η forσ∈ [0,1/2]with 1/2≤ s2+ ω2≤2
. (15)
Furthermore, allU VT +V UT are symmetric and positive semidefinite.
Lemma 3.2 [16] For anyx∈Rn, eachX∈∂B(x)+ has the following representa- tion:
(a) Ifx1= ±x2, then(x)+is continuously differentiable atxwith
X=(x)+=
⎧⎪
⎪⎨
⎪⎪
⎩
0 ifx1<−x2 I ifx1>x2
1 2
1 x¯T
2
¯
x2 X¯ if−x2< x1<x2, where
¯
x2:= x2
x2, X¯ :=
x1 x2+1
I− x1 x2x¯2x¯T2. (b) Ifx2=0 andx1= x2, then
X∈
! I, 1
2
1 x¯2T
¯ x2 X¯
"
, wherex¯2:= x2
x2andX¯ :=2I− ¯x2x¯2T. (c) Ifx2=0 andx1= −x2, then
X∈
! 0, 1
2
1 x¯2T
¯ x2 X¯
"
, wherex¯2:= x2
x2 andX¯ := ¯x2x¯2T. (d) Ifx=0, then eitherX=0 orX=I orXbelongs to the set
! 1 2
1 x¯2T
¯ x2 X¯
####X¯ =(x0+1)I−x0x¯2x¯2T for some|x0| ≤1 and ¯x2 =1
"
.
Proposition 3.1 Letφρ be defined as in (7). Then, for any given(x, y)∈Rn×Rn, each element[S T] ∈∂Bφρ(x, y)has the following representation:
S=ρU+(1−ρ)L(y)+X, T =ρV +(1−ρ)L(x)+Y, where[U V] ∈∂BφFB(x, y),X∈∂B(x)+andY∈∂B(y)+.
Proof From the definition of the B-subdifferential, it is not hard to verify that
∂Bφρ(x, y) ⊆ ρ∂BφFB(x, y)+(1−ρ)∂B[(x)+◦(y)+]. (16) LetF (x, y):=x◦yandG(x, y):=(x)+
(y)+ . Then,(x)+◦(y)+=F (G(x, y)). Using Proposition 7 and Lemma 14 of [22] and noting that∂BG(x, y)=∂B(x)+×∂B(y)+ yields
∂B[(x)+◦(y)+] =J F (G(x, y))∂BG(x, y)=L(y)+∂B(x)+×L(x)+∂B(y)+, where the set on the right hand side denotes the set of all matrices whose first n columns belong toL(y)+∂B(x)+ and lastncolumns belong toL(x)+∂B(y)+. Com- bining the last two equations, we immediately obtain the desired result.
In the rest of this section, we concentrate on the elements of the B-subdifferential ofφρ at all complementarity pairs(x∗i, yi∗)which satisfies (13). As remarked in [1], the index set{1,2, . . . , r}can be partitioned asJI∪JB∪J0∪JB0∪J0B∪J00with
JI:=$
i|xi∗∈int(Kni), yi∗=0% , JB:=$
i|xi∗∈bd+(Kni), yi∗∈bd+(Kni)% , J0:=$
i|xi∗=0, yi∗∈int(Kni)% , JB0:=$
i|xi∗∈bd+(Kni), yi∗=0% , J0B:=$
i|xi∗=0, yi∗∈bd+(Kni)% , J00:=$
i|xi∗=0, yi∗=0%
(17)
where bd+(Kni)denotes the boundary ofKniexcluding the origion. First of all, let us pay attention to the elements of∂Bφρ(xi∗, yi∗)fori∈JI∪JB∪J0. For convenience, the notation “” in the sequel always represents some real number from the interval (0,+∞).
Proposition 3.2 Let[Si Ti] ∈∂Bφρ(xi∗, yi∗)fori=1,2, . . . , r. Then, for i∈JI ∪ JB ∪J0, there exists an orthogonal matrix Qi such thatSi=QiDiQTi and Ti = QiiQTi where,
(a) ifi∈JI, thenDiandi satisfy one of the following cases:
Di=0, i=ρI, Qi=I; (18)
Di=0, i=diag(, , . . . , , ); (19)
Di=0, i is a nonsingular lower triangular matrix. (20) (b) Ifi∈J0, thenDi andi satisfy one of the following cases:
Di=ρI, i=0, Qi=I; (21)
Di=diag(, , . . . , , ), i=0; (22)
Di is a nonsingular lower triangular matrix, i=0. (23) (c) Ifi∈JB, thenDi andi have one of the following representations:
Di =diag(, , . . . , ,0), i=diag(0, , . . . , , ); (24) Di =diag(0, , . . . , , ), i=diag(, , . . . , ,0). (25) Proof For eachi=1, . . . , r, let[Ui Vi] ∈∂BφFB(xi∗, yi∗),X∗i ∈∂B(xi∗)+ andYi∗∈
∂B(yi∗)+.
(a) From Lemma3.1(a) and Lemma3.2(a) and (d), it follows that Ui=0, Vi=I and X∗i =I, Yi∗=0, I orHi:=1
2
1 w¯iT2
¯ wi2 H¯i
,
whereH¯i=(τ+1)I−τw¯i2w¯Ti2for some|τ| ≤1 and ¯wi2 =1. By Proposition3.1, Si=0 and Ti=ρI, ρI+(1−ρ)Lx∗
i orρI+(1−ρ)Lx∗
iHi. In what follows, we proceed the arguments by the possible cases ofTi. Case 1:Ti=ρI. The result is obvious withQi=I,Di=0 andi=ρI. Case 2:Ti=ρI+(1−ρ)Lx∗
i. Under this case, letQi= [qi qˆ1 . . . qˆni−2qi]with qi= 1
√2 1
¯ xi2∗
, qˆj= 0
¯ vj
forj =1, . . . , ni−2, qi= 1
√2 1
− ¯x∗i2
, wherex¯i2∗ =xxi2∗∗
i2andv¯1, . . . ,v¯ni−2are arbitrary unit vectors inRni−1that span the linear subspace{¯v∈Rni−1| ¯vTx¯i2∗ =0}. From Lemma2.1, such orthogonal matrix Qi satisfies
Lx∗
i = Qidiag
λ2(xi∗), xi1∗, . . . , xi1∗, λ1(xi∗)
QTi .
Therefore,Si=QiDiQTi andTi=QiiQTi withDi andi having the expression of (19).
Case 3:Ti=ρI+(1−ρ)Lx∗
iHi. NowTi is similar toρI+(1−ρ)(Lx∗
i)1/2Hi× (Lx∗
i)1/2since Ti =(Lx∗
i)1/2[ρI+(1−ρ)(Lx∗
i)1/2Hi(Lx∗
i)1/2](Lx∗ i)−1/2.
SinceHi0 by Lemma 2.7 of [16], all eigenvalues ofTi are positive. Using Theo- rem 2.3.1 of [15], there exists an orthogonalQi such thatTi =QiiQTi , wherei is lower triangular with diagonal entries being the eigenvalues ofTi. SuchQi clearly satisfies the result of (20).
(b) The desired result is due to part (a) and the symmetry of xi∗ and y∗i in φρ(xi∗, yi∗).
(c) In this case, from Lemma3.2(b) and Proposition3.1, it follows that Si=ρUi+(1−ρ)Ly∗
i or Si=ρUi+(1−ρ)Ly∗
iX∗i, Ti=ρVi+(1−ρ)Lx∗
i or Ti=ρVi+(1−ρ)Lx∗
iYi∗, (26)
where X∗i =1
2
1 (x¯i2∗)T
¯
x∗i2 X¯i∗
withx¯∗i2= xi2∗
xi2∗ andX¯∗i =2I− ¯xi∗2(x¯i2∗)T,
Yi∗=1 2
1 (y¯i2∗)T
¯
y∗i2 Y¯i∗
withy¯∗i2= yi2∗
yi2∗ andY¯i∗=2I− ¯yi2∗(y¯∗i2)T. Sincexi∗ andyi∗ share with the same Jordan frame, without loss of generality we assume
xi∗=σ1qi+σ2qi and yi∗=μ1qi+μ2qi
for someσ1, σ2, μ1, μ2∈R, where the Jordan frame{qi, qi}has the form of qi=1
2 1
¯ qi
and qi=1 2
1
− ¯qi
withq¯i∈Rni−1satisfying ¯qi =1.
Note thatx∗i +yi∗∈int(Kni)holds for this case, which along withxi∗, yi∗∈bd+(Kni) implies
σ1+σ2= |σ1−σ2|, μ1+μ2= |μ1−μ2|, σ1+σ2+μ1+μ2>|σ1−σ2+μ1−μ2|.
From this, we deduce that σ1=0, σ2>0, μ1>0, μ2=0 or σ1 >0, σ2 =0, μ1=0, μ2>0. Let Qi = [√
2qi qˆ1 . . . qˆni−2 √
2qi] with qˆj =0
¯
vj for j = 1, . . . , ni−2, wherev¯1, . . . ,v¯ni−2are arbitrary unit vectors that span the linear sub- space{¯v∈Rni−1| ¯vTq¯i=0}. Clearly, suchQi is an orthogonal matrix. We proceed the arguments by two cases.
Case 1:σ1=0, σ2>0, μ1>0, μ2=0. From the proof of Proposition 4.1(c) of [20], Ui=Qidiag(1, , . . . , ,0)QTi , Vi=Qidiag(0, , . . . , ,1)QTi . By Lemma2.1and the expressions ofxi∗andyi∗, it is not difficult to verify that Lx∗
i =Qidiag
λ1(x∗i), xi1∗, . . . , xi1∗, λ2(xi∗) QTi =Qidiag
0,σ2
2 , . . . ,σ2
2 , σ2
QTi , Ly∗
i =Qidiag
λ2(y∗i), yi1∗, . . . , yi1∗, λ1(yi∗) QTi
=Qidiag
μ1,μ1
2 , . . . ,μ1
2 ,0
QTi .
In addition, by Lemma 2.7 of [16] and the expressions ofxi∗andyi∗, we have X∗i =Qidiag(0,1, . . . ,1,1) QTi , Yi∗=Qidiag(1,1, . . . ,1,0) QTi . Combining the last three equations with (26) yields the desired result in (24).
Case 2:σ1>0, σ2=0, μ1=0, μ2>0. By the proof of Proposition 4.1(c) of [20], Ui=Qidiag(0, , . . . , ,1)QTi , Vi=Qidiag(1, , . . . , ,0)QTi . Also, by Lemma2.1, Lemma 2.7 of [16], and the expressions ofxi∗andyi∗, we verify that
Lx∗
i =Qidiag
λ2(x∗i), xi1∗, . . . , xi1∗, λ1(xi∗) QTi =Qidiag
σ1,σ1
2 , . . . ,σ1
2 ,0
QTi , Ly∗
i =Qidiag
λ1(y∗i), yi1∗, . . . , yi1∗, λ2(yi∗) QTi
=Qidiag
0,μ2
2 , . . . ,μ2 2 ,μ2
2
QTi ,
Xi∗=Qidiag(1,1, . . . ,1,0) QTi , Yi∗=Qidiag(0,1, . . . ,1,1) QTi .
The last two equations and (26) gives the result, withDi,igiven by (25).
