their applications On unitary elements defined on Lorentz cone and Linear Algebra and its Applications

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Linear Algebra and its Applications

www.elsevier.com/locate/laa

On unitary elements defined on Lorentz cone and their applications

Chien-Hao Huang,Jein-Shan Chen

DepartmentofMathematics,NationalTaiwanNormalUniversity,Taipei11677, Taiwan

a r t i c l e i n f o a b s t r a c t

Articlehistory:

Received30August2018 Accepted8December2018 Availableonline11December2018 SubmittedbyP.Semrl

MSC:

90C25 15B99 33E99

Keywords:

Lorentzcone Unitaryelement Triangularinequality Majorization

Inthispaper,weillustrateanewconcept regardingunitary elementsdefinedon Lorentz cone, andestablish somebasic propertiesundertheso-calledunitarytransformationassoci- atedwithLorentzcone.Asanapplicationofunitarytransfor- mation,weachieveaweakerversionofthetriangleinequality andseveral(weak)majorizationsdefinedonLorentzcone.

©2018ElsevierInc.Allrightsreserved.

1. Introduction

Acomplexsquare matrixU iscalledunitary if itsconjugatetransposeU is alsoits inverse,thatis,

UU = U U= I,

* Correspondingauthor.

E-mailaddresses:qqnick0719@yahoo.com.tw(C.-H. Huang),jschen@math.ntnu.edu.tw(J.-S. Chen).

https://doi.org/10.1016/j.laa.2018.12.007

0024-3795/©2018ElsevierInc. Allrightsreserved.

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where I is the identity matrix. In fact, unitary matrices play an important role on manyalgorithms innumericalmatrixanalysisand computingeigenvalues.Theycanbe regarded as a bridge of “changeof basis operators”.No matter from computational or theoretical aspect,it is usually helpfuland convenient to transform agiven matrix by unitary congruence into another matrix with a special form. For example, let f be a real-valued function defined on an interval I ⊆ R. If A is a Hermitian matrix whose eigenvaluesareλj∈ I,wemaychooseaunitarymatrixU suchthatA= U DU,where D is adiagonal matrix.Then, onecandefine theassociated matrix-valued functionby f (A) = U f (D)U, which is heavily used in matrix analysis and designing solutions methods.Formoredetailsaboutusagesandpropertiesofunitarymatrices,pleaserefers to [3,4,10,14,15].

ItisknownthatbothpositivesemidefiniteconeandLorentzconearespecialcasesof symmetriccones[8].ItisinterestingtoknowwhetherthereisasimilarrolelikeU inthe setting ofLorentz cone. In other words, anaturalquestion arises: whatis the concept of unitary looks like in the setting of Lorentz cone? In this paper, we try to answer this question. More specifically,we tryto extend theconcept of unitary to thesetting of Lorentz cone(also called second-ordercone) byobserving therole and propertiesof unitarymatrices.Withtheobservations,weillustratehowwedefinetheunitaryelements associatedwithLorentzcone.Accordingly,theso-calledunitarytransformationisdefined aswell.Then,weestablishsomepropertiesundertheunitarytransformationinSection3.

Moreover,usingtheunitarytransformation,wederiveaweakSOCtriangularinequality, whichisaparallelversiontothematrixcasegivenbyThompsonin[14].Several(weak) majorizations oftheeigenvaluesarededucedandsomeSOCinequalitiesareachievedas well.

2. Preliminary

In this section,we review somebasicconcepts andpropertiesconcerning Jordan al- gebras from the book [8] on symmetric cones and Lorentz cones (second-order cones) [5–7],whichareneededinthesubsequentanalysis.

AEuclideanJordanalgebra isafinitedimensionalinnerproductspace(V,·,·) (V for short) over the field of real numbers R equipped with a bilinear map (x,y) → x◦ y : V× V→ V,whichsatisfiesthefollowingconditions:

(i) x◦ y = y ◦ x forallx,y ∈ V;

(ii) x◦ (x2◦ y)= x2◦ (x◦ y) forallx,y∈ V;

(iii) x◦ y,z=x,y◦ z forallx,y,z∈ V,

where x2:= x◦ x,andx◦ y iscalledtheJordanproduct ofx and y.IfaJordanproduct only satisfiestheconditions(i)and(ii) intheabovedefinition,thealgebraV issaidto be aJordanalgebra.Moreover,ifthereisan(unique)elemente∈ V such thatx◦ e= x forallx∈ V,theelemente iscalledtheJordanidentity inV.NotethataJordanalgebra

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doesnot necessarilyhavean identity element.Throughout this paper, we assumethat V isaEuclidean Jordanalgebrawithanidentityelemente.

In a given Euclidean Jordan algebraV, the set of squares K := {x2 : x ∈ V} is a symmetriccone [8,TheoremIII.2.1].ThismeansthatK isaself-dualclosedconvexcone and,forany twoelements x,y∈ int(K),there existsaninvertible lineartransformation Γ:V→ V suchthatΓ(x)= y andΓ(K)=K.TheLorentzcone,alsocalledsecond-order cone,inRn isanimportantexampleofsymmetriccones, whichisdefinedas follows:

Kn =

x = (x1, x2)∈ R × Rn−1| x1≥ x2 .

For n = 1, Kn denotes the set of nonnegative real numberR+. Since Kn is a pointed closedconvexcone, foranyx,y inRn,wecandefine apartial orderonit:

x Kny ⇐⇒ y − x ∈ Kn; x≺Kny ⇐⇒ y − x ∈ int(Kn).

Note that the relation Kn (or Kn) is only apartial ordering, not a linear ordering inKn.Toseethis,acounterexampleoccurs bytakingx= (1,1) andy = (1,0) inR2.It iscleartoseethatx− y = (0,1)∈ K/ 2,y− x= (0,−1)∈ K/ 2.

Forany x= (x1,x2)∈ R× Rn−1 andy = (y1,y2)∈ R× Rn−1, wedefine theJordan product as

x◦ y =

xTy, y1x2+ x1y2

.

We note that e = (1,0) ∈ R× Rn−1 acts as the Jordan identity. Besides, the Jordan productisnot associative ingeneral.However,itispowerassociative,i.e.,x◦ (x◦ x)= (x◦ x)◦ x forallx∈ Rn.Withoutlossofambiguity, wemaydenotexmfortheproduct of m copiesof x and xm+n = xm◦ xn forany positiveintegers m andn. Here, weset x0= e.Inaddition,Kn isnot closed underJordanproduct.

Givenanyx∈ Kn,itisknownthatthereexistsauniquevectorinKn denotedbyx12 suchthat(x12)2= x12 ◦ x12 = x.Indeed,

x12 = s,x2

2s



, where s =

 1 2

x1+

x21− x22

.

Intheaboveformula, thetermx2/s isdefinedtobethezerovectorifs= 0,i.e.,x= 0.

Forany x∈ Rn, we alwayshave x2 ∈ Kn, i.e., x2 Kn 0.Hence, there exists aunique vector(x2)12 ∈ Kn denotedby|x|. It iseasy toverify that|x|Kn 0 and x2 =|x|2 for anyx∈ Rn.Itis alsoknownthat|x|Knx.Formoredetails,pleasereferto[8,9].

InaEuclideanJordanalgebraV,anelemente(i)∈ V isanidempotent if(e(i))2= e(i), and it is a primitive idempotent if it is nonzero and cannot be written as a sum of two nonzero idempotents. The idempotents e(i) and e(j) are said to be orthogonal if

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e(i) ◦ e(j) = 0. In addition, we say that a finite set {e(1),e(2),· · · ,e(r)} of primitive idempotents inV isaJordan frame if

e(i)◦ e(j)= 0 for i= j, and r i=1

e(i)= e.

Note thate(i),e(j)=e(i)◦ e(j),e= 0 whenever i= j.

Withtheabove,therehavethespectraldecompositionofanelementx inV.

Theorem 2.1.(The Spectral Decomposition Theorem) [8, Theorem III.1.2] Let V be a Euclidean Jordan algebra. Then there is a number r such that, for every x∈ V, there exists aJordanframe{e(1),· · · ,e(r)} andrealnumbers λ1(x),· · · ,λr(x) with

x = λ1(x)e(1)+· · · + λr(x)e(r).

Here, the numbers λi(x) (i = 1,· · · ,r) are called the eigenvalues of x, the expression λ1(x)e(1)+· · · + λr(x)e(r) is called thespectral decomposition of x. Moreover, tr(x):=

r

i=1λi(x) is called the trace of x, and det(x) := λ1(x)λ2(x)· · · λr(x) is called the determinant of x.

In the setting of Lorentz cone in Rn, the vector x = (x1,x2) ∈ R× Rn−1 can be decomposedas

x = λ1(x)u(1)x + λ2(x)u(2)x , (1) where λ1(x),λ2(x) and u(1)x ,u(2)x are the spectral values and the associated spectral vectorsof x,respectively,givenby

λi(x) = x1+ (−1)ix2, (2)

u(i)x =

1 2



1, (−1)i xx22

if x2= 0,

1 2

1, (−1)iv¯

if x2= 0, (3)

fori= 1,2 withv being¯ any vectorinRn−1 satisfying = 1.Ifx2= 0,thedecompo- sition is unique.Accordingly, thedeterminant, thetrace,and theEuclidean norm of x canallberepresentedintermsofλ1(x) andλ2(x):

det(x) = λ1(x)λ2(x) = x21− x22, tr(x) = λ1(x) + λ2(x) = 2x1,

2=1 2

λ1(x)2+ λ2(x)2 .

Forany functionf :R→ R, thefollowing vector-valuedfunctionassociatedwith Kn (n≥ 1) wasconsideredin[5,6]:

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fsoc(x) = f (λ1(x))u(1)x + f (λ2(x))u(2)x , ∀x = (x1, x2)∈ R × Rn−1. (4) If f isdefined onlyon asubset of R,then fsoc is definedon the corresponding subset of Rn. The definition (4) is unambiguous whether x2 = 0 or x2 = 0. The cases of fsoc(x)= x12,x2,exp(x) arediscussedin[8].

In aEuclidean Jordan algebras V, for any x ∈ V, the linear transformation L(x) : V → V is called Lyapunov transformation, which is defined as L(x)y := x◦ y for all y∈ V.Theso-calledquadraticrepresentation P (x) isdefinedby

P (x) := 2L2(x)− L(x2). (5)

For any x ∈ V, the endomorphisms L(x) andP (x) are self-adjoint. For the quadratic representationP (x),ifx isinvertible,thenP (x) isinvertiblewithP (x)−1= P (x−1) and

P (x)K = K and P (x)int(K) = int(K).

Forsubsequentanalysis,welistsomepropertiesofthetraceanddeterminantconcern- ingthemappingP whoseproofscanbefoundinPropositionII.4.3andProposition III.4.2 of[8].

Lemma2.2. LetV beasimple Euclidean Jordanalgebra andx,y,z∈ V.

(a) tr(x◦ (y ◦ z))= tr((x◦ y)◦ z).

(b) det(P (x)y)= det(x)2det(y).

Inaddition,bythedefinitionofP andLemma2.2(a),wealsohave tr(P (x)y) = tr(2x◦ (x ◦ y) − x2◦ y)

= 2tr(x◦ (x ◦ y)) − tr(x2◦ y)

= 2tr((x◦ x) ◦ y) − tr(x2◦ y)

= tr(x2◦ y)

whichis often usedinthe following section. Toclose this section, werecall somebasic propertiesas listedinthefollowingtext.Weomittheproofssincetheycanbefoundin [5,8,9].

Lemma 2.3.For any x,y ∈ Rn with spectral decomposition given as in (1)–(3), the followinghold.

(a) x12 =

λ1(x)u(1)x +

λ2(x)u(2)x wheneverx∈ Kn. (b) |x|=1(x)|u(1)x +2(x)|u(2)x ;

(c) If x Kny, thenλi(x)≤ λi(y) foralli= 1,2.

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3. Unitaryelements

Note that the nonnegative orthant, the cone of positive semidefinite matrices, and the Lorentzcone are specialcasesof symmetric cones. Infact,the spaceSym(n;R) of n× n realsymmetricmatrices withtheJordan productX◦ Y = 12(XY + Y X) andthe bilinear map tr(XY ) formsaEuclidean Jordan algebra.In thiscase, itscorresponding symmetric cone K is exactly the cone of positive definite matrices, and the quadratic representation is

P (X)Y = XY X, ∀ X, Y ∈ Sym(n; R).

As anexampleofEuclidean Jordanalgebra,wenoticethat U◦ U =1

2(UU + U U) = I.

Inaddition,foranymatrixA∈ Rn×n,weshallwriteA= U P forthepolardecomposition of A, where U is aunitaryand P is apositivesemidefinite matrix.We note thatP = (AA)12 is the so-called absolute value of A and is denoted by |A|. Combining with the definitionofabsolute valueassociated withLorentzcone intheprevioussection, it motivates us todefine theunitary element w onLorentz cone,even thoughthere is no concept ofconjugateassociatedwithLorentzconeyet.

Definition 3.1.Let Kn be the Lorentz cone. An element w in Rn is called a unitary element definedonKn ifitsatisfies

w2= w◦ w = e.

After simplecalculation,we obtainthatforw = (w1,w2)∈ R× Rn−1, w2= (w12+w22, 2w1w2) = (1, 0),

whichsaysw1= 0 orw2= 0.Hence,weconcludethatanyunitaryelementw associated with Lorentzconeonlyhasthree types:

w =

⎧⎪

⎪⎩ e;

−e;

(0, ¯w) with ¯w = 1.

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We recall that transformation from one orthonormal basis to another one is ac- complished by unitary matrix. The matrix of unitary transformation relative to an orthonormal basis is also a unitary matrix. In other words, the unitary matrix plays significant importanceon thedecomposition of matrix. Inthe setting of Lorentzcone, let P (·) be finedas in (5), we call P (w) is the unitary transformation associated with

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Lorentzconewheneverw isaunitaryelementdefinedonLorentzcone.Withthisdefini- tion,itisdesiredto seetherole oftheunitarytransformation inthesetting ofLorentz cone.Allthepropositionsaredevotedto answerthis.

Proposition3.2. Foranyunitaryelementw∈ RndefinedonLorentzconegivenasin (6), thereholds |det(w)|= 1.

Proof. Accordingtothespectraldecomposition(1)–(3) andtheform ofunitaryw,itis obviousthespectralvaluesofw iseither1 or−1,whichimplies|det(w)|= 1. 2

We note thatfor any x= (x1,x2) ∈ R× Rn−1,y = (y1,y2)∈ R× Rn−1, if x2 = 0 ory2= 0,thenx,y mayhavethesameJordanframeviaappropriatelychoosinginthe spectraldecomposition.Otherwise,x,y havethesameJordanframeifthereexistsα∈ R such thatx2 = αy2. Furthermore, x,y have the sameordered Jordan frame whenever α > 0 andthereverselyordered Jordanframewheneverα < 0.

Inthe sequel, givenany z = (z1,z2)∈ R× Rn−1, we shalldenote ¯z := zz2

2 for the simplicityofnotation.

Proposition 3.3. For any x,y ∈ Rn, there exists a unitary element w ∈ Rn defined on Lorentz cone such that L(w)y and x havethe same Jordan frame.Moreover, w can be chosensuchthat L(w)y have thedesired orderedJordanframe.

Proof. For any x= (x1,x2)∈ R× Rn−1,y = (y1,y2) ∈ R× Rn−1, it is clearthat the assertionholdsbychoosingw = e wheneverx2= 0 ory2= 0.Supposethatx2= 0 and y2= 0. Foranyw = (0,w) with¯ ¯w= 1, wehave

L(w)y = w◦ y =

yT2w, y¯ 1w¯ .

Accordingtothespectraldecomposition(1)–(3),wecanchooseaunitaryelement

w =

⎧⎪

⎪⎩

(0, ¯x), if y1> 0 (0,−¯x), if y1< 0 (0, ¯w) with ¯w = 1, if y1= 0

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sothatL(w)y havethesameorderedJordanframewithrespecttox.Ontheotherhand, choosingw =−w leadsL(w)y tohavethereverselyorderedJordanframewithrespect tox. 2

Remark3.4.In Proposition 3.3, the Lyapunov transformation L(·) could leadtwo ele- mentsinto the sameJordan frame. However,it doesnotkeep thespectral values after thistransforming.Indeed,from (7),wenoticethatfori= 1,2

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λi(L(w)y) = y2Tw + (¯ −1)iy1w¯ = yT2w + (¯ −1)i|y1| maynotalwayscoincidewithλi(y)= y1+ (−1)iy2,ingeneral.

Whatkindoftransformationwillmakeanytwoelementsx,y∈ Rn tosharethesame Jordan frame and keep the spectral values? The question will be answered gradually from thefollowingpropositions.

Proposition 3.5.Let w ∈ Rn be aunitary element defined on Lorentz cone givenas in (6) and P (w) be the unitary transformation.Then, forany y ∈ Rn, thespectral values of P (w)y coincidewith theonesof y.

Proof. FromLemma2.2,wehave

tr(P (w)y) = tr(w2◦ y) = tr(y), det(P (w)y) = det(w2) det(y) = det(y),

whichimplythatthespectralvaluesofP (w)y andy satisfythesamequadraticequation.

Hence,P (w)y andy havethesamespectralvalues. 2

Proposition 3.6.Let w ∈ Rn be aunitary element defined on Lorentz cone givenas in (6) andP (w) betheunitary transformation.Then, foranyy∈ Rn,thereholds

that is, thenormofy is invariantunder theunitary transformation P (w).

Proof. For any z ∈ Rn, we note that 2 = 121(z)2+ λ2(z)2). Applying Proposi- tion 3.5, we know that the spectral values of P (w)x coincide with the ones of y, and henceweobtain =

InProposition3.5, wealreadyknowthespectral valuesisinvariantunderthetrans- formationP (w) withanyunitaryelementw.ItisnaturaltoaskifP (w) isabletochange theJordanframetoanotheronebysuitablew.Theanswerisaffirmative.Toseethis,we recall atheoremthatFarautandKorányi[8] establishedbythePeircedecomposition.

Theorem 3.7. [8, Theorem IV.2.5] Let V be a simple Euclidean Jordan algebra. If {c(1),. . . ,c(r)} and {d(1),. . . ,d(r)} are two Jordan frames, then there exists an auto- morphismA such that

Ac(j)= d(j) (1≤ j ≤ r), where A= P (w) for somew withw2= e.

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Fig. 1. The geometric view of P (w)y.

In the setting of Lorentz cone, we offer another approach via the geometric view of P (w)y and figure outthe exact form of the suitable unitary element w. Given any y = (y1,y2)∈ R× Rn−1 andunitaryelement w = (0,w),¯ weobservethat

P (w)y = 2w◦ (w ◦ y) − w2◦ y

= 2 (0, ¯w)◦

yT2w, y¯ 1w¯

− e ◦ (y1, y2)

= 2

y1, (y2Tw) ¯¯ w

− (y1, y2) (8)

=

y1, 2(yT2w) ¯¯ w− y2

.

ThegeometricmeaningofP (w)y isdepictedinFig.1.Infact,itnotonlysupportsthe conclusionofProposition3.6,butalsotellsushowtochooseasuitableunitaryelement w definedonLorentzcone.

Proposition 3.8. For any x,y ∈ Rn, there exists a unitary element w ∈ Rn defined on Lorentz cone such that P (w)y and x havethe sameJordan frame. Moreover, w can be chosensuchthat P (w)y havethedesiredordered Jordanframe.

Proof. For any x = (x1,x2) ∈ R× Rn−1 and y = (y1,y2) ∈ R× Rn−1, if x2 = 0 or y2= 0,thenx,y mayhavethesameJordanframewithdesiredorderingviaappropriately choosinginthespectraldecomposition.Thus, weassumethatx2= 0 andy2= 0.

(i)Forx,y withthesameJordanframe,wemaychoosew = e tokeepthesameordered Jordanframe.Ontheother hand,ifwechooseanyw = (0,w) such¯ thatyT2w = 0,¯ then it follows from (8) that P (w)y = (y1,−y2) and henceP (w)y and y have the reversely orderedJordanframe. Inotherwords,P (w) changetheorderofJordanframe.

(ii) For x,y with the different Jordan frame, we choose w =



0,¯x+¯x+¯¯ yy



, and then applying(8) againyields

P (w)y =

y1, 2y2Tx + ¯y)

2x + ¯y)− y2

=

y1, 2y2(¯yTx + 1)¯

2 + 2¯yTx¯ (¯x + ¯y)− y2

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= (y1,y2(¯x + ¯y) − y2)

= (y1,y2¯x).

Hence, P (w)y and x have the same ordered Jordan frame. On the other hand, while choosing w = 

0,−¯x+¯−¯x+¯yy

, it will lead P (w)y and x to have the reversely ordered Jordan frame. 2

Remark 3.9.Proposition 3.8 illustrates that we can change the Jordan frame to the desiredoneviaP (w) withsuitableunitaryelementw.Conversely,foranyJordanframe {e(1),e(2)} andanyunitaryelementw,theset{P (w)e(1),P (w)e(2)} stillformsaJordan frame. Indeed,wenoticethatforany i= 1,2,

tr(P (w)e(i)) = tr(e(i)) = 1, det(P (w)e(i)) = det(e(i)) = 0, whichtellusthatP (w)e(i)isoftheform1

2,12¯v(i)

with¯v(i)= 1 sinceP (w)e(i)∈ Kn. Inaddition,

P (w)e(1)+ P (w)e(2)= P (w)e = w2= e,

whichimplies¯v(1)+ ¯v(2)= 0.Thus,theset{P (w)e(1),P (w)e(2)} actuallyformsaJordan frame. Furthermore, forany pair of unitary elements w,w, there must havea unitary element w suchˆ thatP ( ˆw)= P (w)P (w).

Proposition 3.10.Let w∈ Rn be aunitary element defined on Lorentz cone givenas in (6) andP (w) betheunitary transformation.Then, foranyx∈ Rn,there holds

|P (w)x| = P (w)|x|.

Proof. Denote x= λ1(x)u(1)x + λ2(x)u(2)x by thespectral decomposition (1)–(3). Then, from Propositions3.5–3.8andRemark3.9,thereexists aJordanframe{v(1),v(2)} such that

|P (w)x| =1(x)v(1)+ λ2(x)v(2)

=1(x)|v(1)+2(x)|v(2)

= P (w)|x|.

Thus, thedesiredresultisdeduced. 2

Proposition3.11. Foranyx,y∈ Kn,P (x12)y andP (y12)x havethesamespectralvalues.

Moreover, there existsaunitaryelement w definedonLorentz cone suchthat

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P (x12)y = P (w)(P (y12)x).

Proof. ByLemma2.2, wehave

tr(P (x12)y) = tr(x◦ y) = tr(P (y12)x), det(P (x12)y) = det(x) det(y) = det(P (y12)x),

which says thatthe spectral values of P (x12)y and P (y12)x satisfy the same quadratic equation,and hencetheyhavethesamespectral values.Moreover,thedesiredequality alsoholdsbysimilarargumentsas inProposition3.8. 2

4. Applications

In this section, we provide two applications of unitary elements associated with Lorentz cone. First application is about an extended version of triangular inequality.

Tothisend,webeginwithrecallingthetriangleinequality

|a + b| ≤ |a| + |b|

for any real or complexnumbers a,b. Moreover, inthe Euclidean space Rn or normed linearspaces V ,the triangleinequalityis aproperty aboutdistances, andit iswritten as

foranygivenx,y∈ Rn(orV ).However,theprospectivetriangleinequalityforsymmetric matricesX,Y

|X + Y |  |X| + |Y |

maynotbetrueingeneral.ThenotationX  Y meansY− X isapositivesemidefinite matrix. In fact, Thompson [14] established a weaker version of triangle inequality for twomatrices:forany twomatricesX andY ,there existunitariesV andW suchthat

|X + Y |  V |X|V+ W|Y |W. (9) Recently,Huanget al. [11] alsodiscussthetriangleinequalityonLorentzconeandgive acounterexampletoillustrate theSOCtriangular inequality

|x + y| Kn|x| + |y|

doesnothold.Nevertheless,Huang et al. buildupanother SOCtraceversion oftrian- gular inequality.We noticethat theunitary matrixplays acrucial role inthe proof of

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Thompson’s theorem [14]. Here,we tryto derive aparallel inequalityinthe setting of Lorentz cone by applying the concept of the unitary transformation discussed in Sec- tion3.

Theorem 4.1. For any x,y ∈ Rn, there exist unitary elements w,w ∈ Rn defined on Lorentz cone suchthat

|x + y| KnP (w)|x| + P (w)|y|. (10) Proof. Let x = (x1,x2) ∈ R× Rn−1 and y = (y1,y2) ∈ R× Rn−1. It is clear that inequality (10) holds if x+ y ∈ Kn∪ (−Kn) by choosing w = w = e. Suppose that x+ y /∈ Kn∪ (−Kn),itsaysthatλ1(x+ y)< 0< λ2(x+ y) whichimplies

1(x + y)| = x2+ y2 − x1− y1≤ x2 2 − x1− y1=−λ1(x)− λ1(y),

2(x + y)| = x1+ y1+x2+ y2 ≤ x1+ y1+x2 2 = λ2(x) + λ2(y).

(i) For x,y withthe sameJordan frame, we assumethat x= λ1(x)e(1)+ λ2(x)e(2). Then, wediscusstwo subcases.Ify = λ1(y)e(1)+ λ2(y)e(2),wehave

|x + y| = (λ1(x) + λ1(y))e(1)+ (λ2(x) + λ2(y))e(2)

= 1(x) + λ1(y)|e(1)+2(x) + λ2(y) e(2) Kn (1(x)| + |λ1(y)|)e(1)+ (2(x)| + |λ2(y)|)e(2)

= 1(x)|e(1)+2(x)|e(2)+1(y)|e(1)+2(y)|e(2)

= |x| + |y|.

Similarly,ify = λ1(y)e(2)+ λ2(y)e(1),wecanalsoderive|x+ y| Kn|x|+|y|.Hence,we havethedesiredinequalitybychoosingw = w= e.

(ii)Forx,y withthedifferent Jordanframe,wenotice thatx + ¯¯ y= 0,and

|x + y| = |λ1(x + y)|u(1)x+y+2(x + y)|u(2)x+y

Kn−λ1(x)u(1)x+y− λ1(y)u(1)x+y+ λ2(x)u(2)x+y+ λ2(y)u(2)x+y Kn1(x)|u(1)x+y+2(x)|u(2)x+y+1(y)|u(1)x+y+2(y)|u(2)x+y.

Applying Proposition 3.8, there exist unitary elements w,w defined on Lorentz cone suchthat

P (w)|x| = |λ1(x)|u(1)x+y+2(x)|u(2)x+y, P (w)|y| = |λ1(y)|u(1)x+y+2(y)|u(2)x+y.

Thus, thedesiredinequalityfollowsfrom alltheaboveexpressions. 2

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Anotherapplicationisdevotedtothemajorizationsoftheeigenvalues.Inparticular, wederiveseveralmajorizationsoftheeigenvaluesparalleltothoseformatrixcase.In[1, 2],theauthorsgivemanymatrixversionsofinequalitiesforconvex(orconcave)function.

Inlightoftheseconcepts,weachievetheirparallelinequalitiesinthesettingofLorentz cone. For convenience, we introduce some notations. For any x ∈ Rn, we use λ(x) to meanthevector1(x),λ2(x)). Foranyx,y∈ Rn,we denote

(i) (spectralvalueinequalities)

λ(x) λ(y) ⇐⇒ λi(x)≤ λi(y) (i = 1, 2), (ii) (weakmajorization)

λ(x)≺wλ(y)⇐⇒ λ2(x)≤ λ2(y) and 2 i=1

λi(x)≤ 2 i=1

λi(y),

λ(x)≺w λ(y)⇐⇒ λ1(x)≤ λ1(y) and 2 i=1

λi(x)≤ 2 i=1

λi(y).

Wenotethat

x Kny =⇒ λ(x)  λ(y) =⇒ λ(x) ≺wλ(y) and λ(x)≺wλ(y).

TogetherwithProposition3.8andtheproofofTheorem4.1,wefurtherhavethefollowing implication:

λ(x) λ(y) =⇒ x KnP (w)y for some unitary element w.

Lemma4.2. Supposethatx∈ Rn hasspectralvaluesinI⊆ R.Letf beaconvexfunction onI. Then, foreveryunit vectorv∈ Rn ( = 1),wehave

f (x ◦ v, v) ≤ fsoc(x)◦ v, v.

Inparticular, foranyvector ˜v = (˜v1,˜v2)∈ Kn with ˜v1= 1,there holds

f (x, ˜v) ≤ fsoc(x), ˜v. (11) In addition, if 0 ∈ I and f (0)≤ 0, then for any arbitrary Jordan frame {e(1),e(2)} in Kn andforalli= 1,2, thereholds

f (x, e(i)) ≤ fsoc(x), e(i).

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Proof. First, wewritex= λ1u(1)x + λ2u(2)x .Then, wehave f (x ◦ v, v) = f

(λ1u(1)x + λ2u(2)x )◦ v, v

= f

1u(1)x + λ2u(2)x , v2

= f

λ1u(1)x , v2 + λ2u(2)x , v2 ,

where thesecond equality holdsbythecondition (iii)ofEuclidean Jordan algebra.We note that

u(i)x , v2 ≥ 0 for all i = 1, 2 sinceu(i)x ,v2∈ Kn and

u(2)x , v2 + u(2)x , v2 = e, v2 2= 1.

Bytheconvexityoff ,wehave f (x ◦ v, v) = f

λ1u(1)x , v2 + λ2u(2)x , v2

≤ u(1)x , v2f (λ1) +u(2)x , v2f (λ2)

=f (λ1) u(1)x , v2 + f (λ2) u(2)x , v2

=fsoc(x), v2

=fsoc(x)◦ v, v.

This proves the firstassertion. Inparticular, for any vector ˜v∈ Kn with ˜v1 = 1,there has avectorv ∈ Rn such thatv2= ˜v.Moreover, thecondition ˜v1= 1 implies = 1 bythedefinitionofJordanproduct.Thus,thesecondassertionisobtained.

Iffurther0∈ I andf (0)≤ 0,we obtain

f (t) = f 1

2· 0 +1 2· 2t

1

2f (0) +1

2f (2t)≤ 1 2f (2t) , whichimplies

f (x, e(i)) = f λ1

2 u(1)x , 2e(i) +λ2

2u(2)x , 2e(i)

≤ u(1)x , 2e(i)f λ1

2

+u(2)x , 2e(i)f λ2

2

=u(1)x , e(i)2f λ1

2

+u(2)x , e(i)2f λ2

2

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≤ u(1)x , e(i)f (λ1) +u(2)x , e(i)f (λ2)

=f (λ1) u(1)x , e(i) + f (λ2) u(2)x , e(i)

=fsoc(x), e(i),

foralli= 1,2.Hence,weconcludethethirdassertion. 2

Lemma4.3. Forany x∈ Rn withspectral valuesλ1(x) and λ2(x). Then,we have λ1(x) = minx ◦ v, v and λ2(x) = maxx ◦ v, v,

wheretheminimumandmaximumaretakenoverallchoicesofunitvectorv. Moreover, forany arbitraryJordanframe {e(1),e(2)} inKn,there holds

λ1(x) + λ2(x) = 2 i=1

x ◦√ 2e(i),√

2e(i) = 2 i=1

x, 2e(i)

Proof. Denotex= λ1(x)u(1)x + λ2(x)u(2)x bythe spectral decomposition(1)–(3).Then, foranyunitvectorv∈ Rn,wehave

x ◦ v, v = 

λ1(x)u(1)x + λ2(x)u(2)x

◦ v, v

=1(x)u(1)x , v2 + λ2(x)u(2)x , v2

= λ1(x)u(1)x , v2 + λ2(x)u(2)x , v2.

Fori = 1,2,we notice u(i)x ,v2≥ 0 since u(i)x ,v2 ∈ Kn, and u(2)x ,v2+u(2)x ,v2= 1.

Thistogethertheaboveyields

λ1(x)≤ x ◦ v, v ≤ λ2(x), and the minimum and maximum occur whenever v = ±√

2u(1)x and v = ±√

2u(2)x , re- spectively.

Inaddition,foranyarbitraryJordanframe {e(1),e(2)} inKn,itcanbe verifiedthat 2

i=1

x ◦√ 2e(i),√

2e(i) = 2 i=1

x, 2e(i) = x, 2e = 2x1= λ1(x) + λ2(x).

Hence,theproof iscomplete. 2

We point out that Lemma 4.2–4.3 are the SOC versions of results as in [1, Lemma 2.1–2.2] (alsosee[4]).Weusethesetwolemmas todeduceaseriesofinequalities liketheones in[1] accordingly.

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Theorem 4.4. Let f be a convex real-valued functionon I ⊆ R.Then, forall x,y with spectral valuesinI and0≤ α ≤ 1,there holds

λ(fsoc(αx + (1− α)y)) ≺wλ(αfsoc(x) + (1− α)fsoc(y)).

If further0∈ I and f (0)≤ 0,then

λ(fsoc(P (s)x))≺wλ(P (s)fsoc(x)) forallx∈ Rn withspectral valuesinI and0Kns2 Kne.

Proof. For anyx,y∈ Rn withx= y,wehave

λ1(αx + (1− α)y) = αx1+ (1− α)y1− αx2+ (1− α)y2 ≥ αλ1(x) + (1− α)λ1(y), λ2(αx + (1− α)y) = αx1+ (1− α)y1+αx2+ (1− α)y2 ≤ αλ2(x) + (1− α)λ2(y), whichimplythat

αλ1(x) + (1− α)λ1(y)≤ λ1(αx + (1− α)y) ≤ λ2(αx + (1− α)y) ≤ αλ2(x) + (1− α)λ2(y).

Hence,thespectralvaluesofαx+ (1− α)y arealsoinI.Forthesimplicityofnotation, weletλˆ1,λˆ2bethespectralvaluesofαx+ (1− α)y and{e(1),e(2)} betheJordanframe arrangedsuchthatf (ˆλ1)≤ f(ˆλ2).Then,wehave

λ2(fsoc(αx + (1− α)y)) = f

αx + (1 − α)y, 2e(2)

= f

αx, 2e(2) + (1 − α)y, 2e(2)

 αf

x, 2e(2)

+ (1− α)f

y, 2e(2)



αfsoc(x), 2e(2) + (1 − α)fsoc(y), 2e(2)

=αfsoc(x) + (1− α)fsoc(y), 2e(2)

≤ λ2(αfsoc(x) + (1− α)fsoc(y)),

where thethree inequalitieshold bytheconvexityoff ,inequality(11) andLemma4.3, respectively.Similarly,wehave

2 i=1

λi(fsoc(αx + (1− α)y)) = 2 i=1

f

αx, 2e(i) + (1 − α)y, 2e(i)

2 i=1

 αf

x, 2e(i)

+ (1− α)f

y, 2e(i)

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2 i=1



αfsoc(x), 2e(i) + (1 − α)fsoc(y), 2e(i)

= 2 i=1

αfsoc(x) + (1− α)fsoc(y), 2e(i)

= 2 i=1

λi(fsoc(x) + (1− α)fsoc(y)).

Thisprovesthefirstassertion.

Toprovethesecondassertion,let¯λ1λ2bethespectralvaluesofP (s)x and{d(1),d(2)} be the Jordan frame arranged such that f (¯λ1) ≤ f(¯λ2). Since f (0) ≤ 0, to prove the desiredinequalitywecanassumethatP (s)d(i)= 0 fori= 1,2.Moreover, wenotethat s2 Kne ifand onlyifP (s2) P (e) by[13,Lemma2.3],whichgivesP (s)2 I. Thus, wehave

P (s)d(i)2= d(i)TP (s)TP (s)d(i)≤ d(i)Td(i)= 1 2, whichsays

2P (s)d(i)≤ 1. Inaddition,we note thatd(i)is ontheboundaryof Kn, whichsaysdet(d(i))= 0.Furthermore,it alsoimpliesdet(P (s)d(i))= 0 byLemma2.2.

SincethequadraticrepresentationP (s) isinvariantonKn,weconcludeP (s)d(i)isonthe boundaryofKn as well.Inparticular, thefirst componentofP (s)d(i) is 12P (s)d(i). Ontheotherhand,thequadratic representationP (s) canbeexpressedas

P (s) = 2ssT − det(s)J,

where J :=

1 0T 0 −In−1



with In−1 being theidentity matrix inR(n−1)×(n−1) (see [8, 12]),whichsaysP (s) isasymmetrictransformation.Thus,fori= 1,2,there holds

λi(fsoc(P (s)x)) = f

P (s)x, 2d(i)

= f

x, 2P (s)d(i) . Hence,wehave

λ2(fsoc(P (s)x))

= f √

2P (s)d(2) ·

 x,

√2P (s)d(2) P (s)d(2)



+ (1−√

2P (s)d(2)) · 0

≤√

2P (s)d(2) · f 

x,

√2P (s)d(2) P (s)d(2)



+ (1−√

2P (s)d(2)) · f(0)



≤√

2P (s)d(2) ·

 fsoc(x),

√2P (s)d(2) P (s)d(2)



=fsoc(x), 2P (s)d(2)

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=P (s)fsoc(x), 2d(2)

≤ λ2(P (s)fsoc(x)),

by the convexityof f ,the conditionf (0) ≤ 0,inequality(11) and Lemma4.3, respec- tively. Similarly,wecanverifythat

2 i=1

λi(fsoc(P (s)x))

= 2 i=1

f √

2P (s)d(i) ·

 x,

√2P (s)d(i) P (s)d(i)

 +

1−√

2P (s)d(i)

· 0

2 i=1

√2P (s)d(i) · f 

x,

√2P (s)d(i) P (s)d(i)

 +

1−√

2P (s)d(i)

· f(0)



2 i=1

2P (s)d(i) ·



fsoc(x),

√2P (s)d(i) P (s)d(i)



= 2 i=1

P (s)fsoc(x), 2d(i)

= 2 i=1

λi(P (s)fsoc(x)).

This completestheproof. 2

Remark4.5.AccordingtotheargumentofTheorem4.4,wesimilarlyhavethefollowing weakmajorizationforconcavereal-valuedfunctiong definedonI⊂ R:

λ(αgsoc(x) + (1− α)gsoc(y))≺w λ(gsoc(αx + (1− α)y))

forallx,y withspectralvaluesinI and0≤ α ≤ 1.Iffurther0∈ I andg(0)≥ 0,then λ(P (s)gsoc(x))≺w λ(gsoc(P (s)x))

forallx∈ Rn withspectral valuesinI and0Kn s2 Kn e.

Corollary 4.6.Let f be a nonnegative decreasing convex function on [0,∞). Then, for any x,y ∈ Kn,there holds

λ (fsoc(x + y))≺wλ (fsoc(x) + fsoc(y)) .

Proof. Notethateverynonnegative decreasingfunctionf on[0,∞) satisfies f (2t)≤ 2f(t), ∀t ∈ [0, ∞).

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This furtherimplies thatfsoc(2x)≤ 2fsoc(x) and fsoc(2y) ≤ 2fsoc(y).Using this fact andapplyingTheorem 4.4,wehave

λ (fsoc(x + y))≺wλ

fsoc(2x) + fsoc(2y) 2

wλ (fsoc(x) + fsoc(y)) , whichsaysthedesiredresult. 2

Corollary4.7. Letf beanonnegative convexfunctionon I⊆ R.Then, fsoc(αx + (1 soc(x) + (1− α)fsoc(y)

foranyx,y withspectral valuesinI and0≤ α ≤ 1.If further0∈ I andf (0)= 0, then fsoc(P (s)x) soc(x)

forany x∈ Rn withspectral valuesinI and0Kns2 Kne.

Proof. WedefineafunctionΦ:R2→ [0,∞) by

Φ(a, b) =

1

2(a2+ b2)

12 .

For any z ∈ Rn, we notice that = Φ(λ1(z),λ2(z)). Then, the results follow from Theorem4.4and ProblemII.5.12(iv)[4,page53]. 2

Theorem 4.8.Let f be amonotone convex function on I ⊆ R. Then, forany x,y with spectralvaluesin I and0≤ α ≤ 1,there holds

λ (fsoc(αx + (1− α)y))  λ (αfsoc(x) + (1− α)fsoc(y)) . If further0∈ I and f (0)≤ 0,then

λ (fsoc(P (s)x)) λ (P (s)fsoc(x)) forallx∈ Rn with spectralvaluesinI and0Kns2 Kne.

Proof. Accordingtotheproofof Theorem4.4,itremainstoshow that λ1(fsoc(αx + (1− α)y)) ≤ λ1(αfsoc(x) + (1− α)fsoc(y)).

Iff isincreasing(ordecreasing),then

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