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# Linear K-Power Preservers and Trace of Power-Product Preservers

Written By

Huajun Huang and Ming-Cheng Tsai

Reviewed: February 15th, 2022 Published: April 17th, 2022

DOI: 10.5772/intechopen.103713

From the Edited Volume

## Matrix Theory - Classics and Advances [Working Title]

Dr. Mykhaylo I. Andriychuk

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## Abstract

Let V be the set of n×n complex or real general matrices, Hermitian matrices, symmetric matrices, positive definite (resp. semi-definite) matrices, diagonal matrices, or upper triangular matrices. Fix k∈Z\\01. We characterize linear maps ψ:V→V that satisfy ψAk=ψAk on an open neighborhood S of In in V. The k-power preservers are necessarily k-potent preservers, and k=2 corresponds to Jordan homomorphisms. Applying the results, we characterize maps ϕ,ψ:V→V that satisfy “trϕAψBk=trABk for all A∈V, B∈S, and ψ is linear” or “trϕAψBk=trABk for all A,B∈S and both ϕ and ψ are linear.” The characterizations systematically extend existing results in literature, and they have many applications in areas like quantum information theory. Some structural theorems and power series over matrices are widely used in our characterizations.

### Keywords

• k-power
• power preserver
• trace preserver
• power series of matrices

## 1. Introduction

Preserver problem is one of the most active research areas in matrix theory (e.g. [1, 2, 3, 4]). Researchers would like to characterize the maps on a given space of matrices preserving certain subsets, functions or relations. One of the preserver problems concerns maps ψon some sets Vof matrices which preserves k-power for a fixed integer k2, that is, ψAk=ψAkfor any AV(e.g. [3, 5, 6]). The k-power preservers form a special class of polynomial preservers. One important reason of this problem lies on the fact that the case k=2corresponds to Jordan homomorphisms. Moreover, every k-power preserver is also a k-potent preserver, that is, Ak=Aimply that ψAk=ψAfor any AV. Some researches on k-potent preservers can be found in [6, 7, 8].

Given a field F, let MnF, SnF, DnF, NnF, and TnFdenote the set of n×ngeneral, symmetric, diagonal, strictly upper triangular, and upper triangular matrices over F, respectively. When Fis the complex field C, we may write Mninstead of MnC, and so on. Let Hn, Pn, and Pn¯denote the set of complex Hermitian, positive definite, and positive semidefinite matrices, and HnR=SnR, PnR, and PnR¯the corresponding set of real matrices, respectively. A matrix space is a subspace of Mm,nFfor certain m,nZ+. Let At(resp. A) denote the transpose (resp. conjugate transpose) of a matrix A.

In 1951, Kadison [9] showed that a Jordan -isomorphism on Mn, namely, a bijective linear map with ψA2=ψA2and ψA=ψAfor all AMn, is the direct sum of a -isomorphism and a -anti-isomorphism. Hence ψA=UAUfor all AMnor ψA=UATUfor all AMnby [[3], Theorem A.8]. Let k2be a fixed integer. In 1992, Chan and Lim ([5]) determined a nonzero linear operator ψ:MnFMnF(resp. ψ:SnFSnF) such that ψAk=ψAkfor all AMnF(resp. SnF) (See Theorems 3.1 and 5.1). In 1998, Brešar, Martindale, and Miers considered additive maps of general prime rings to solve an analogous problem by using the deep algebraic techniques ([10]). Monlár [[3], P6] described a particular case of their result which extends Theorem 3.1 to surjective linear operators on BH. In 2004, Cao and Zhang determined additive k-power preserver on MnFand SnF([11]). They also characterized injective additive k-power preserver on TnF([12] or [[6], Theorem 6.5.2]), which leads to injective linear k-power preserver on TnF(see Theorem 8.1). In 2006, Cao and Zhang also characterized linear k-power preservers from MnFto MmFand from SnFto MmF(resp. SmF) [8].

In this article, given an integer kZ\01, we show that a unital linear map ψ:VWbetween matrix spaces preserving k-powers on a neighborhood of identity must preserve all integer powers (Theorem 2.1). Then we characterize, for F=Cand R, linear operators on sets V=MnF, Hn, SnF, Pn, PnR, DnF, and TnFthat satisfy ψAk=ψAkon an open neighborhood of Inin V. In the following descriptions, PMnFis invertible, UMnFis unitary, OMnFis orthogonal, and λFsatisfies that λk1=1.

1. V=MnF(Theorem 3.4): ψA=λPAP1or ψA=λPAtP1.

2. V=Hn(Theorem 4.1): When kis even, ψA=UAUor ψA=UAtU. When kis odd, ψA=±UAUor ψA=±UAtU.

3. V=SnF(Theorem 5.2): ψA=λOAOt.

4. V=Pnor PnR(Theorem 6.1): ψA=UAUor ψA=UAtU.

5. V=DnF(Theorem 7.1): ψA=ψIndiagfp1AfpnA, in which ψInk=ψIn, p:1n01nis a function, and fi:DnFFi=01nsatisfy that, for A=diaga1an, f0A=0and fiA=aifor i=1,,n.

6. V=TnF(Theorem 8.4 for n3): ψA=λPAP1or ψA=λPAP1, in which PTnFand A=an+1j,n+1iif A=aij.

Our results on MnFand SnFextend Chan and Lim’s results in Theorems 3.1 and 5.1, and result on TnFextend Cao and Zhang’s linear version result in [12].

Another topic is the study of a linear map ϕfrom a matrix set Sto another matrix set Tpreserving trace equation. In 1931, Wigner’s unitary-antiunitary theorem [[3], p. 12] says that if ϕis a bijective map defined on the set of all rank one projections on a Hilbert space Hsatisfying

trϕAϕB=trAB,E1

then there is an either unitary or antiunitary operator Uon Hsuch that ϕP=UPUor ϕP=UPtUfor all rank one projections P. In 1963, Uhlhorn generalized Wigner’s theorem to show that the same conclusion holds if the equality trϕPϕQ=trPQis replaced by trϕPϕQ=0trPQ=0(see [13]).

In 2002, Molnár (in the proof of [[14], Theorem 1]) showed that maps ϕon the space of all bounded linear operators on a Banach space BXsatisfying (1) for ABX, rank one operator BBXare linear. In 2012, Li, Plevnik, and Šemrl [15] characterized bijective maps ϕ:SSsatisfying trϕAϕB=ctrAB=cfor a given real number c, where Sis Hn, SnR, or the set of rank one projections.

In [[16], Lemma 3.6], Huang et al. showed that the following statements are equivalent for a unital map ϕon Pn:

1. trϕAϕB=trABfor A,BPn;

2. trϕAϕB1=trAB1for A,BPn;

3. ϕA=UAUor UAtUfor a unitary matrix U.

The authors also determined the cases if ϕis not assuming unital, the set Pnis replaced by another set like Mn, Sn, Tn, or Dn. In [[17], Theorem 3.8], Leung, Ng, and Wong considered the relation (1) on infinite dimensional space.

Let Sdenote the subspace spanned by a subset Sof a vector space. Recently, Huang and Tsai studied two maps preserving trace of product [18]. Suppose two maps ϕ:V1W1and ψ:V2W2between subsets of matrix spaces over a field Funder some conditions satisfy

trϕAψB=trABE2

for all AV1, BV2. The authors showed that these two maps can be extended to bijective linear maps ϕ˜:V1W1and ψ˜:V2W2that satisfy trϕ˜Aψ˜B=trABfor all AV1, BV2(see Theorem 2.2). Hence when a matrix space Vis closed under conjugate transpose, every linear bijection ϕ:VVcorresponds to a unique linear bijection ψ:VVthat makes (2) hold (see Corollary 2.3). Therefore, each of ϕand ψhas no specific form.

One natural question is to ask when the following equality holds for a fixed kZ\01:

trϕAψBk=trABk.E3

The second major work of this paper is to use our descriptions of linear k-power preservers on an open neighborhood Sof Inin Vto characterize maps ϕ,ψ:VVunder one of the assumptions:

1. equality (3) holds for all AV, BS, and ψis linear, or

2. equality (3) holds for all A,BSand both ϕand ψare linear,

for the sets V=Mn, Hn, Pn, Sn, Dn, Tn, and their real counterparts. These results, together with Theorem 2.2 and the characterizations of maps ϕ1,,ϕm:VV(m3) that satisfy trϕ1A1ϕmAm=trA1Amin [18], make a comprehensive picture of the preservers of trace of matrix products in the related matrix spaces and sets.

In the following characterizations, F=Cor R, P,QMnFare invertible, UMnFis unitary, OMnFis orthogonal, and cF\0.

1. V=MnF(Theorem 3.5):

1. When k=1, ϕA=PAQand ψB=PBQ, or ϕA=PAtQand ψB=PBtQ.

2. When kZ\1,0,1, ϕA=ckPAP1and ψB=cPBP1, or ϕA=ckPAtP1and ψB=cPBtP1.

2. V=Hn(Theorem 4.2):

1. When k=1, ϕA=cPAPand ψB=cPBP, or ϕA=cPAtPand ψB=cPBtP, for c11.

2. When kZ\1,0,1, ϕA=ckUAUand ψB=cUBU, or ϕA=ckUAtUand ψB=cUBtU, for cR\0.

3. V=SnF(Theorem 5.3):

1. When k=1, ϕA=cPAPtand ψB=cPBPt.

2. When kZ\1,0,1, ϕA=ckOAOtand ψB=cOBOt.

4. V=Pnand PnR(Theorem 6.4): ϕA=ckUAUand ψB=cUBU, or ϕA=ckUAtUand ψB=cUBtU, in which cR+. Characterizations under some other assumptions are also given as special cases of Theorem 6.2 (Huang, Tsai [18]).

5. V=DnF(Theorem 7.2): ϕA=PCkAP1,ψB=PCBP1where Pis a permutation matrix and C=DnFis diagonal and invertible.

6. V=TnF(Theorem 8.5): ϕand ψsend NnFto NnF, DϕDnFand DψDnFare characterized by Theorem 7.2, and Dϕ=DϕD. Here Ddenotes the map that sends ATnFto the diagonal matrix with the same diagonal as A.

The sets Mn, Hn, Pn, Sn,Dn, and their real counterparts are closed under conjugate transpose. In these sets, trAB=ABfor the standard inner product. Our trace of product preservers can also be interpreted as inner product preservers, which have wide applications in research areas like quantum information theory.

## 2. Preliminary

### 2.1 Linear operators preserving powers

We show below that: given kZ\01, a unital linear map ψ:VWbetween matrix spaces preserving k-powers on a neighborhood of identity in Vmust preserve all integer powers. Let Z+(resp. Z) denote the set of all positive (resp. negative) integers.

Theorem 2.1. LetF=CorR. LetVMpFandWMqFbe matrix spaces. FixkZ\01.

1. Suppose the identity matrixIpVandAkVfor all matricesAin an open neighborhoodSVofIpinVconsisting of invertible matrices. Then

AB+BA:ABVV,E4
A1:AVis invertibleV.E5

In particular,

Ar:AVV,rZ+,andE6
Ar:AVis invertibleV,rZ.E7

1. 2. SupposeIpV, IqW, andAkVfor all matricesAin an open neighborhoodSVofIpinVconsisting of invertible matrices. Supposeψ:VWis a linear map that satisfies the following conditions:

ψIp=Iq,E8
ψAk=ψAk,ASV.E9

Then

ψAB+BA=ψAψB+ψBψA,A,BV,E10
ψA1=ψA1,invertibleAV.E11

In particular,

ψAr=ψAr,AV,rZ+,andE12
ψAr=ψAr,invertibleAV,rZ.E13

Proof.We prove the complex case. The real case is done similarly.

1. 1. For each AV\0, there is ϵ>0such that Ip+xASVfor all xCwith x<minϵ1A. Thus

Ip+xAk=Ip+xkA+x2kk12A2+V.E14

The second derivative

d2dx2Ip+xAkx=0=kk1A2V.E15

Since k01, we have A2Vfor all AV. Therefore, for A,BV,

AB+BA=A+B2A2B2V.E16

In particular, AVimplies that ArVfor all rZ+.

Cayley-Hamilton theorem implies that every invertible matrix Asatisfies that A1=fAfor a certain polynomial fxFx. Therefore, A1V, so that ArVfor all rZ.

1. 2. Now suppose (8) and (9) hold. The proof is proceeded similarly to the proof of part (1). For every AV, there is ϵ>0such that for all xCwith x<minϵ1A1ψA,

ψIp+xAk=Iq+xkψA+x2kk12ψA2+W,E17
ψIp+xAk=Iq+xkψA+x2kk12ψA2+W.E18

since (17) and (18) equal, we have

ψA2=ψA2,AV.E19

Therefore, for A,BV,

ψA+B2=ψA+B2E20

We get (10): ψAB+BA=ψAψB+ψBψA. In particular ψAr=ψArfor all AVand rZ+.

Every invertible AVcan be expressed as A1=fAfor a certain polynomial fxFx. Then ψA1=ψfA=fψAis commuting with ψA. Hence

2ψA1ψA=ψA1ψA+ψAψA1=ψA1A+AA1=2Iq.E21

We get ψA1=ψA1. Therefore, ψAr=ψArfor all rZ.

Theorem 2.1 is powerful in exploring k-power preservers in matrix spaces. Note that every k-power preserver is a k-potent preserver. Theorem 2.1 can also be used to investigate k-potent preservers in matrix spaces.

### 2.2 Two maps preserving trace of product

We recall two results about two maps preserving trace of product in [18]. They are handy in proving linear bijectivity of maps preserving trace of products. Recall that if Sis a subset of a vector space, then Sdenotes the subspace spanned by S.

Theorem 2.2 (Huang, Tsai [18]). Letϕ:V1W1andψ:V2W2be two maps between subsets of matrix spaces over a fieldFsuch that:

dimV1=dimV2maxdimW1dimW2.

1. ABare well-defined square matrices forABV1×V2W1×W2.

2. IfAV1satisfies thattrAB=0for allBV2, thenA=0.

3. ϕandψsatisfy that

trϕAψB=trAB,AV1,BV2.E22

Then dimV1=dimV2=dimW1=dimW2and ϕand ψcan be extended to bijective linear map ϕ˜:V1W1and ψ˜:V2W2, respectively, such that

trϕ˜Aψ˜B=trAB,AV1,BV2.E23

A subset Vof Mnis closed under conjugate transpose if A:AVV. A real or complex matrix space Vis closed under conjugate transpose if and only if Vequals the direct sum of its subspace of Hermitian matrices and its subspace of skew-Hermitian matrices.

Corollary 2.3 (Huang, Tsai [18]). LetVbe a subset ofMnclosed under conjugate transpose. Suppose two mapsϕ,ψ:VVsatisfy that

trϕAψB=trAB,A,BV.E24

Thenϕandψcan be extended to linear bijections onV. Moreover, whenVis a vector space, every linear bijectionϕ:VVcorresponds to a unique linear bijectionψ:VVsuch that(24) holds. Explicitly, given an orthonormal basisA1AofVwith respect to the inner productAB=trAB, ψis defined byψAi=Biin whichB1Bis a basis ofVwithtrϕAiBj=δi,jfor alli,j1.

Corollary 2.3 shows that when a matrix space Vis closed under conjugate transpose, every linear bijection ϕ:VVcorresponds to a unique linear bijection ψ:VVthat makes (24) hold. The next natural thing is to determine ϕand ψthat satisfy trϕAψBk=trABkfor a fixed kZ\01..

From now on, we focus on the fields F=Cor R.

## 3. k-power linear preservers and trace of power-product preservers on Mnand MnR

### 3.1 k-power preservers on Mnand MnR

Chan and Lim described the linear k-power preservers on Mnand MnRfor k2in [7, Theorem 1] as follows.

Theorem 3.1. (Chan, Lim[5]) Let an integerk2. LetFbe a field withcharF=0orcharF>k. Suppose thatψ:MnFMnFis a nonzero linear operator such thatψAk=ψAkfor allAMnF. Then there existλFwithλk1=1and an invertible matrixPMnFsuch that

ψA=λPAP1,AMnF,orE25
ψA=λPAtP1,AMnF.E26

(25) and (26) need not hold if ψis zero or is a map on a subspace of MnF. The following are two examples. Another example can be found in maps on DnF(Theorem 7.1).

Example 3.2. The zero mapψA0clearly satisfiesψAk=ψAkfor allAMnbut they are not of the form(25) or(26).

Example 3.3. Letn=k+m, k,m2, and consider the operatorψon the subspaceW=MkMmofMndefined byψAB=ABtforAMkandBMm.ThenψAk=ψAkfor allAWandkZ+, butψis not of the form(25) or(26).

We now generalize Theorem 3.1 to include negative integers kand to assume the k-power preserving condition ψAk=ψAkonly on matrices nearby the identity.

Theorem 3.4. LetF=CorR. Let an integerkZ\01. Suppose thatψ:MnFMnFis a nonzero linear map such thatψAk=ψAkfor allAin an open neighborhood ofInconsisting of invertible matrices. Then there existλFwithλk1=1and an invertible matrixPMnFsuch that

ψA=λPAP1,AMnF,orE27
ψA=λPAtP1,AMnF.E28

Proof. We prove for the case F=C. The case F=Rcan be done similarly. Obviously, ψIn=ψInk=ψInk.

1. First suppose k2. For each AMn, there exists ϵ>0such that for all xCwith x<ϵ, the following two power series converge and equal:

ψIn+xAk=ψIn+xi=0k1ψIniψAψInk1i+x2i=0k2j=0k2iψIniψAψInjψAψInk2ij+E29
ψIn+xAk=ψIn+xkψA+x2kk12ψA2+E30

Equating degree one terms above, we get

A=i=0k1ψIniψAψInk1i.E31

Applying (31), we have

InψAAψIn=ψInkψAψAψInk=ψInψAψAψIn.E32

Hence ψInψA=ψAψInfor AMn, that is, ψIncommutes with the range of ψ.

Now equating degree two terms of (29) and (30) and taking into account that k01, we have

ψInk2ψA2=ψA2.E33

Define ψ1A=ψInk2ψAfor AMn. Then ψ1A2=ψ1A2for all AMn. (31) and the assumption that ψis nonzero imply that ψIn0. So ψ1InψIn=ψInk=ψIn0. Thus ψ1In0and ψ1is nonzero. By Theorem 3.1, there exists an invertible PMnsuch that ψ1A=PAP1for AMnor ψ1A=PAtP1for AMn. Moreover, ψIncommutes with all ψ1A, so that ψIn=λInfor a λC. By In=ψ1In=ψInk1, we get λk1=1. Therefore, ψA=λψ1A. We get (27) and (28).

1. Next Suppose k<0. For every AMn, the power series expansions of ψIn+xAkand ψIn+xAk1are equal when xis sufficiently small:

ψIn+xAk=ψIn1+xi=0k1ψIniψAψInk1i+E34
ψIn+xAk1=ψIn1xkψIn1ψAψIn1+E35

Equating degree one terms of (34) and (35), we get

In1ψAψIn1=i=0k1ψIniψAψInk1i.E36

Therefore,

kψAψIn1ψIn1ψA=ψIni=0k1ψIniψAψInk1ii=0k1ψIniψAψInk1iψIn=ψInkψAψAψInk=ψIn1ψAψAψIn1.E37

We get ψIn1ψA=ψAψIn1for AMn. So ψIn1and ψIncommute with the range of ψ. The following power series are equal for every AMnwhen xis sufficiently small:

ψIn+xAk=ψIn+xkψInk1ψA+x2kk12ψInk2ψA2+E38
ψIn+xAk=ψIn+xkψA+x2kk12ψA2+E39

Equating degree two terms of (38) and (39), we get ψInk2ψA2=ψA2.Let ψ1AψInk2ψA=ψIn1ψA. Then ψ1A2=ψ1A2and ψ1is nonzero. Using Theorem 3.1, we can get (27) and (39).

### 3.2 Trace of power-product preserers on Mnand MnR

Corollary 2.3 shows that every linear bijection ϕ:MnFMnFcorresponds to another linear bijection ψ:MnFMnFsuch that trϕAψB=trABfor all A,BMnF. When m3, maps ϕ1,,ϕmon MnFthat satisfy trϕ1A1ϕmAm=trA1Amfor A1,,AmMnFare determined in [18].

If two maps on MnFsatisfy the following trace condition about k-powers, then they have specific forms.

Theorem 3.5. LetF=CorR. LetkZ\01. LetSbe an open neighborhood ofInconsisting of invertible matrices. Then two mapsϕ,ψ:MnFMnFsatisfy that

trϕAψBk=trABk,E40

1. for allAMnF,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only if ϕand ψtake the following forms:

1. a. Whenk=1, there exist invertible matricesP,QMnFsuch that

ϕA=PAQψB=PBQorϕA=PAtQψB=PBtQA,BMnF.E41

1. b. WhenkZ\1,0,1, there existcF\0and an invertible matrixPMnFsuch that

ϕA=ckPAP1ψB=cPBP1orϕA=ckPAtP1ψB=cPBtP1A,BMnF.E42

Proof.We prove the case F=C; the case F=Rcan be done similarly.

Suppose assumption (2) holds. Then for every AMnF, there exists cF\0such that IncAS, so that for all BS:

trBk=trϕIncA+AψBk=trIncABk+ctrϕAψBk.E43

Thus trϕAψBk=trABkfor AMnFand BS, which leads to assumption (1).

Now we prove the theorem under assumption (1), that is, (40) holds for all AMnFand BS, and ψis linear. Only the necessary part is needed to prove.

Let S=BPn¯:B1/kS, which is an open neighborhood of Inin Pn¯. Define ψ˜:SMnsuch that ψ˜B=ψB1/kk. Then (40) implies that

trϕAψ˜B=trϕAψB1/kk=trAB,AMn,BS.E44

The complex span of Sis Mn. By Theorem 2.2, ϕis bijective linear, and ψ˜can be extended to a linear bijection on Mn.

The linearity of ψand (40) imply that for every BMn, there exists ϵ>0such that In+xBSand the power series of In+xBkconverges whenever x<ϵ. Then

trϕAψIn+Bk=trAIn+xBk,AMn,x<ϵ.E45

1. 1. First suppose k2. Equating degree one terms and degree k1terms on both sides of (45) respectively, we get the following identities for A,BMn:

trϕAi=0k1ψInk1iψBψIni=trkAB,E46
trϕAi=0k1ψBiψInψBk1i=trkABk1.E47

Let Ci:i=1n2be a basis of projection matrices (i.e. Ci2=Ci) in Mn. For example, we may choose the following basis of rank 1 projections:

Eii:1in12Eii+Ejj+δEij+δ¯Eji:1i<jnδ{1i}.E48

By (40) and (47), for AMnand i=1,,n2,

trAψCik=trkACi=trϕAj=0k1ψCijψInψCik1j.E49

By the bijectivity of ϕ,

Cik=j=0k1ψCijψIψCik1j.E50

Therefore, for i=1,,n2,

0=ψCij=0k1ψCijψInψCik1jj=0k1ψCijψInψCik1jψCi=ψCikψInψInψCik.E51

Since

trCik=trϕ1ACik=trϕ1ACi,AMn,i=1,,n2,E52

the only matrix AMnsuch that trCik=0for all i1n2is the zero matrix. So ψCik:i=1n2is a basis of Mn. (51) implies that ψIn=cInfor certain cC\0.

(46) shows that

ck1trϕAψB=trAB,A,BMn.E53

Therefore,

ck1trϕAψBk=trABk=trϕAψBk,AMn,BS.E54

The bijectivity of ϕshows that ck1ψBk=ψBkfor BS, that is,

c1ψBk=c1ψBk,BS.E55

Notice that c1ψIn=In. By Theorem 3.4, there is an invertible PMnsuch that ψis of the form ψB=cPBP1or ψB=cPBtP1for BMn. Consequently, we get (42).

1. 2. Now suppose k<0. Then ψInis invertible. For every BMnand sufficiently small x, we have the power series expansion:

ψIn+Bk=In+In1ψB1ψIn1k=InIn1ψB+x2ψIn1ψBψIn1ψB+ψIn1k=ψInkxi=1kψIniψBψInk1+i+x2i=1kj=1k+1iψIniψBψInjψBψInk2+i+j+E56

Equating degree one terms and degree two terms of (45) respectively and using (56), we get the following identities for A,BMn:

trϕAi=1kψIniψBψInk1+i=trkAB,E57
trϕAi=1kj=1k+1iψIniψBψInjψBψInk2+i+j=trkk12AB2.E58

(57) and (40) imply that

i=1kψIniψBkψInk1+i=kψBk,BS.E59

Let FrBdenote the degree rcoefficient in the power series of ψIn+Bk. Then (57) and (58) show that:

k12F1B2=F2B,BMn.E60

Denote ψ1BψBψIn1. We discuss the cases k=1and k1.

1. When k=1, (60) leads to

ψB2=ψBψIn1ψB,BMn.E61

So ψ1B2=ψ1B2for BMn. Note that ψ1In=In. By Theorem 3.4, there exists an invertible PMnsuch that ψ1B=PBP1or ψ1B=PBtP1for BMn. Let QP1ψIn. Then Qis invertible, and ψB=PBQor ψB=PBtQfor BMn. Using (40), we get (41).

1. b. Suppose the integer k<1. Then (60) implies that

k12ψIn1F1B2F2B2ψIn1=ψIn1F2BF2BψIn1,E62

which gives

1k2ψInkψB2ψB2ψInk=ψInkψBψIn1ψBψBψIn1ψBψInk.E63

In other words, for BMn:

ψInk1k2ψ1B2ψ1B2=1k2ψ1B2ψ1B2ψInk.E64

Let B=In+xEfor an arbitrary matrix EMn. Then (64) becomes

ψInkx1kψ1E+x21k2ψ1E2ψ1E2=x1kψ1E+x21k2ψ1E2ψ1E2ψInk.E65

The equality on degree one terms shows that ψInkcommutes with all ψ1E. Hence ψInkcommutes with the range of ψ. (??) can be rewritten as

tri=1kψInk1+iϕAψIniψB=trkAB,A,BMn.E66

By Theorem 2.2, ψis a linear bijection and its range is Mn. So ψInk=μInfor certain μC.

Now by (59), for BS:

kψInψBkkψBkψIn=ψIni=1kψIniψBkψInk1+ii=1kψIniψBkψInk1+iψIn=ψBkψInkψInkψBk=0E67

So ψIncommutes with ψBkfor BS. In particular, ψIncommutes with ψ˜B=ψB1/kkfor BS. The complex span of Sis Mn, and ψ˜can be extended to a linear bijection on Mn. Hence ψIn=cInfor certain cC\0. By (59), we get ψ1Bk=ψ1Bkfor BS. Note that ψ1In=In. By Theorem 3.4, there is an invertible PMnsuch that ψ1B=PBP1or ψ1B=PBtP1. Then ψB=cPBP1or ψB=cPBtP1. Using (40), we get (42).

Remark 3.6 The following modifications could be applied to the proof of Theorem 3.5 forF=R:

1. LetSbe the collection of matricesA=QDQ1, in whichDis nonnegative diagonal andQMnRis invertible, such thatA1/k=QD1/kQ1S.

2. We may choose the following basis of rank 1 projections ofMnRto substitute(48):

Eii:1in12Eii+Ejj+Eij+Eji:1i<jnω1Eii+ω2Ejj+EijEji:1i<jn,E68

in which ω1,ω2are distinct roots of x2x1=0..

The arguments in the above proof will be applied analogously to maps on the other sets discussed in this paper.

## 4. k-power linear preservers and trace of power-product preservers on Hn

### 4.1 k-power linear preservers on Hn

We give a result that determine linear operators on Hnthat satisfy ψAk=ψAkon a neighborhood of Inin Hnfor certain kZ\01

Theorem 4.1. FixkZ\01. A nonzero linear mapψ:HnHnsatisfies that

ψAk=ψAkE69

on an open neighborhood of Inconsisting of invertible matrices if and only if ψis of the following forms for certain unitary matrix UMn:

1. Whenkis even,

ψA=UAU,AHn;orψA=UAtU,AHn.E70

2. Whenkis odd,

ψA=±UAU,AHn;orψA=±UAtU,AHn.E71

Proof.It suffices to prove the necessary part. Suppose (69) holds on an open neighborhood Sof Inin Hn.

1. 1. First assume k2. Replacing Mnby Hnin part (1) of the proof of Theorem 3.4 up to (33), we can prove that ψIncommutes with the range of ψ, and ψ1AψInk2ψAis a nonzero linear map that satisfies ψ1A2=ψ1A2for AHn.

Every matrix in Mncan be uniquely expressed as A+iBfor A,BHn. Extend ψ1to a map ψ˜:MnMnsuch that

ψ˜A+iB=ψ1A+iψ1B,A,BHn.E72

It is straightforward to check that ψ˜is a complex linear bijection. Moreover, for A,BHn,

ψ1AB+BA=ψ1A+B2ψ1A2ψ1B2=ψ1A+B2ψ1A2ψ1B2=ψ1Aψ1B+ψ1Bψ1A.

It implies that

ψ˜A+iB2=ψ˜A+iB2,A,BHn.

By Theorem 3.1, there is an invertible matrix UMnsuch that

1. ψ˜A=UAU1for all AMn, or

2. ψ˜A=UAtU1for all AMn.

First suppose ψ˜A=UAU1. The restriction of ψ˜on Hnis ψ1:HnHn. Hence for AHn, we have UAU1=UAU1=UAU; then UUA=AUUfor all AHn, which shows that UU=cInfor certain cR+.By adjusting a scalar if necessary, we may assume that Uis unitary. So ψInk2ψA=UAU. Then ψInk1=In, so that ψIn=Inwhen kis even and ψInInInwhen kis odd. Thus ψA=UAUwhen kis even and ψA=±UAUwhen kis odd. Similarly for the case ψ˜A=UAtU1. Therefore, (70) or (71) holds.

1. 2. Now assume that k<0. Replacing Mnby Hnin part (2) of the proof of Theorem 3.4, we can show that ψIncommutes with the range of ψ, and furthermore the nonzero linear map ψ1AψIn1ψAsatisfies that ψ1A2=ψ1A2. By arguments in the preceding paragraphs, we get (70) or (71).

### 4.2 Trace of power-product preservers on Hn

By Corollary 2.3, every linear bijection ϕ:HnHncorresponds to another linear bijection ψ:HnHnsuch that trϕAψB=trABfor all A,BHn. When m3, linear maps ϕ1,,ϕm:HnHnthat satisfy trϕ1A1ϕmAm=trA1Amare characterized in [18].

Theorem 4.2. LetkZ\01. LetSbe an open neighborhood ofIninHnconsisting of invertible Hermitian matrices. Then two mapsϕ,ψ:HnHnsatisfy that

trϕAψBk=trABk,E73

1. for allAHn,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only if ϕand ψtake the following forms:

1. Whenk=1, there exist an invertible matrixPMnandc11such that

ϕA=cPAPψB=cPBPA,BHn;orϕA=cPAtPψB=cPBtPA,BHn.E74

1. b. WhenkZ\1,0,1, there exist a unitary matrixUMnandcR\0such that

ϕA=ckUAUψB=cUBUA,BHn;orϕA=ckUAtUψB=cUBtUA,BHn.E75

Proof.Assumption (2) leads to assumption (1) (cf. the proof of Theorem 3.5). We prove the theorem under assumption (1). It suffices to prove the necessary part.

1. When k2, in the part (1) of proof of Theorem 3.5, through replacing Mnby Hn, complex numbers by real numbers, and Theorem 3.1 or Theorem 3.4 by Theorem 4.1, we can prove that ϕψhas the forms in (75).

2. When k<0, in the part (2) of proof of Theorem 3.5, through replacing Mnby Hnand complex numbers by real numbers, we can get the corresponding equalities of (56) (60) on Hn. The case k<1can be proved completely analogously with the help of Theorem 4.1.

For the case k=1, the equality corresponding to (60) can be simplified as

ψB2=ψBψIn1ψB,BHn.E76

Let ψ1BψIn1ψB. Then ψ1:HnMnis a nonzero real linear map that satisfies ψ1B2=ψ1B2for BHn. Extend ψ1to a complex linear map ψ˜:MnMnsuch that

ψ˜A+iBψ1A+iψ1B,A,BHn.E77

Similarly to the arguments in part (1) of the proof of Theorem 4.1, we have ψ˜A+iB2=ψ˜A+iB2for all A,BHn. Using Theorem 3.4 and the fact that ψ˜In=ψ1In=In, we can prove that there is an invertible PMnsuch that for all BHn, either ψ1B=P1BPor ψ1B=P1BtP. So

ψB=ψInP1BP,BHn;orE78
ψB=ψInP1BtP,BHn.E79

If ψB=ψInP1BPfor BHn, then ψInP1BP=ψInP1BP=PBPψIn,which gives

PψInP1B=BPψInP1,BHn.E80

Hence PψInP1=cInfor certain cR\0.We have ψIn=cPPso that ψB=cPBPfor BHn. Similarly for the case ψB=ψInP1BtP. Adjusting cand Pby scalar factors simultaneously, we may assume that c11. It implies (74).

Remark 4.3. Theorem 4.2 does not hold ifψis not assumed to be linear. Letkbe a positive even integer. Letψ˜:HnHnbe any bijective linear map such thatψ˜Pn¯Pn¯. For example,ψ˜may be a completely positive map of the formψ˜B=i=1rNiBNiforr2, N1,,NrMnlinearly independent, and at least one ofN1,,Nris invertible. By Corollary 2.3, there is a linear bijectionϕ:HnHnsuch thattrϕAψ˜B=trABfor allA,BHn. Letψ:HnHnbe defined byψB=ψ˜Bk1/k.Then

trϕAψBk=trϕAψ˜Bk=trABk,A,BHn.

Obviously, ψmay be non-linear, and the choices of pairs ϕψare much more than those in (74) and (75).

## 5. k-power linear preservers and trace of power-product preservers on Snand SnR

### 5.1 k-power linear preservers on Snand SnR

Chan and Lim described the linear k-power preservers on SnFfor k2in [7, Theorem 2] as follows.

Theorem 5.1. (Chan, Lim[5]) Let an integerk2. LetFbe an algebraic closed field withcharF=0orcharF>k. Suppose thatψ:SnFSnFis a nonzero linear operator such thatψAk=ψAkfor allASnF. Then there existλFwithλk1=1and an orthogonal matrixOMnFsuch that

ψA=λOAOt,ASn.E81

We generalize Theorem 5.1 to include the case SnR, to include negative integers k, and to assume the k-power preserving condition only on matrices nearby the identity.

Theorem 5.2. LetkZ\01. LetF=CorR. Suppose thatψ:SnFSnFis a nonzero linear map such thatψAk=ψAkfor allAin an open neighborhood ofIninSnFconsisting of invertible matrices. Then there existλFwithλk1=1and an orthogonal matrixOMnFsuch that

ψA=λOAOt,ASnF.E82

Proof.It suffices to prove the necessary part. In both k2and k<0cases, using analogous arguments as parts (1) and (2) of the proof of Theorem 3.4, we get that ψIncommutes with the range of ψ, and the nonzero map ψ1AψInk2ψAsatisfies that ψ1A2=ψ1A2for ASnF. Then

ψ1Aψ1B+ψ1Bψ1A=ψ1A+B2ψ1A2ψ1B2=ψ1A+B2ψ1A2ψ1B2=ψ1AB+BA.E83

In particular, ψ1Aψ1Ar+ψ1Arψ1A=2ψ1Ar+1for rZ+. Using induction, we get ψ1A=ψ1Afor all ASnFand Z+. By [26, Corollary 6.5.4], there is an orthogonal matrix OMnFsuch that ψ1A=OAOt. Since ψIncommutes with the range of ψ1, we have ψIn=λInfor certain λFin which λk1=1. So ψA=λOAOtas in (82).

Obviously, in F=Rcase, (82) has λ=1when kis even and λ11when kis odd.

### 5.2 Trace of power-product preservers on Snand SnR

Corollary 2.3 shows that every linear bijection ϕ:SnFSnFcorresponds to another linear bijection ψ:SnFSnFsuch that trϕAψB=trABfor all A,BSnF. When m3, maps ϕ1,,ϕm:SnFSnFthat satisfy trϕ1A1ϕmAm=trA1Amare determined in [18].

We characterize the trace of power-product preserver for SnFhere.

Theorem 5.3. LetF=CorR. LetkZ\01. LetSbe an open neighborhood ofIninSnFconsisting of invertible matrices. Then two mapsϕ,ψ:SnFSnFsatisfy that

trϕAψBk=trABk,E84

1. for allASnF,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only if ϕand ψtake the following forms:

1. Whenk=1, there exist an invertible matrixPMnFandcF\0such that

ϕA=cPAPt,ψB=cPBPt,A,BSnF.E85

We may choose c=1for F=Cand c11for F=R.

1. b. WhenkZ\1,0,1, there existcF\0and an orthogonal matrixOMnFsuch that

ϕA=ckOAOt,ψB=cOBOt,A,BSnF.E86

Proof.Assumption (2) leads to assumption (1) (cf. the proof of Theorem 3.5). We prove the theorem under assumption (1). It suffices to prove the necessary part.

Obviously, SnHn=SnRand SnPn=PnR. Let SBPnR:B1/kS, which is an open neighborhood of Inin PnRand whose real (resp. complex) span is SnR(resp. Sn). Using an analogous argument of the proof of Theorem 3.5, and replacing Mnby SnF, replacing the basis (48) of Mnby the following basis of rank 1 projections in SnF:

Eii:1in12Eii+Ejj+Eij+Eji:1i<jn,E87

and replacing the usage of Theorem 3.4 by that of Theorem 5.2, we can prove the case k2, and for k<0, we can get the corresponding equalities up to (60).

Define a linear map ψ1:SnFMnFby ψ1BψBψIn1.

When k=1, we get the corresponding equality of (61), so that ψ1B2=ψ1B2for BSnF. Similar to the proof of Theorem 5.2, we get ψ1Br=ψ1Brfor all rZ+. By [26, Theorem 6.5.3], there is an invertible matrix PMnFsuch that ψ1B=PBP1, so that ψB=PBP1ψInfor BSnF. Since ψB=ψBt, we get

P1ψInPtB=BP1ψInPt,BSnF.E88

Therefore, P1ψInPt=cInfor certain cF\0, so that ψB=cPBPtfor all BSnF. Consequently, we get (85). The remaining claims are obvious.

When k<1, using analogous argument as in the proof of k<1case of Theorem 3.5 and applying Theorem 5.2, we can get (86).

## 6. k-power linear preservers and trace of power-product preservers on Pnand PnR

In this section, we will determine k-power linear preservers and trace of power-product preservers on maps PnPn¯(resp. PnRPnR¯). Properties of such maps can be applied to maps PnPnand Pn¯Pn¯(resp. PnRPnRand PnR¯PnR¯).

### 6.1 k-power linear preservers on Pnand PnR

Theorem 6.1. FixkZ\01. A nonzero linear mapψ:PnPn¯(resp.ψ:PnRPnR¯) satisfies that

ψAk=ψAkE89

on an open neighborhood of Inin Pn(resp. PnR) if and only if there is a unitary (resp. real orthogonal) matrix UMnsuch that

ψA=UAU,APn;orψA=UAtU,APn.E90

Proof.We prove the case ψ:PnPn¯. The sufficient part is obvious. About the necessary part, the nonzero linear map ψ:PnPn¯can be easily extended to a linear map ψ˜:HnHnthat satisfies ψ˜Ak=ψ˜Akon an open neighborhood of In. By Theorem 4.1, we immediately get (90).

The case ψ:PnRPnR¯can be similarly proved using Theorem 5.2.

### 6.2 Trace of powered product preservers on Pnand PnR

Now consider the maps PnPn¯(resp. PnRPnR¯) that preserve trace of powered products. Unlike Mnand Hn, the set Pn(resp. PnR) is not a vector space. The trace of powered product preservers of two maps have the following forms.

Theorem 6.2 (Huang, Tsai [18]). Leta,b,c,dR\0. Two mapsϕ,ψ:PnPn¯satisfy

trϕAaψBb=trAcBd,A,BPn,E91

if and only if there exists an invertible PMnsuch that

ϕA=PAcP1/aψB=P1BdP1/borϕA=PAtcP1/aψB=P1BtdP1/bA,BPn.E92

Theorem 6.3 (Huang, Tsai [18]). Given an integerm3and real numbersα1,,αm,β1,,βmR\0, mapsϕi:PnPn¯(i=1,,m) satisfy that

trϕ1A1α1ϕmAmαm=trA1β1Amβ1,A1,,AmPn,E93

if and only if they have the following forms for certain c1,,cmR+with c1cm=1:

1. Whenmis odd, there exists a unitary matrixUMnsuch that fori=1,,m:

ϕiA=ci1/αiUAβi/αiU,APn.E94

2. Whenmis even, there exists an invertibleMMnsuch that fori=1,,m:

ϕiA=ci1/αiMAβiM1/αi,iisodd,ci1/αiM1AβiM1/αi,iis even,APn.E95

Both Theorems 6.2 and 6.3 can be analogously extended to maps PnRPnR¯without difficulties.

Theorem 6.2 determines maps ϕ,ψ:PnPn¯that satisfy (91) throughout their domain. If we only assume the equality (91) for ABin certain subset of Pn×Pnand assume certain linearity of ϕand ψ, then ϕand ψmay have slightly different forms. We determine the case a=c=1and b=d=kZ\0here.

Theorem 6.4. LetkZ\0. LetSbe an open neighborhood ofIninPn. Two mapsϕ,ψ:PnPn¯satisfy

trϕAψBk=trABk,E96

3. for allA,BPn, or

4. for allAS,BPn, andϕis linear,

if and only if there exists an invertible PMnsuch that

ϕA=PAPψB=P1BkP1/korϕA=PAtPψB=P1BtkP1/kA,BPn.E97

The mapsϕandψsatisfy(96)

1. for allAPn,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only if whenk11, ϕandψtake the form(97), and whenkZ\1,0,1, there exist a unitary matrixUMnandcR+such that

ϕA=ckUAUψB=cUBUorϕA=ckUAtUψB=cUBtUA,BPn.E98

Proof.It suffices to prove the necessary part.

The case of assumption (1) has been proved by Theorem 6.2.

Similar to the proof of Theorem 3.5, assumption (2) implies assumption (1); assumption (4) implies assumption (3). It remains to prove the case with assumption (3).

When k=1, assumption (3) is analogous to assumption (2), and we get (97).

Suppose kZ\10. Let ψ1:PnPn¯be defined by ψ1BψB1/kk. Let S1BPn:B1/kS.Then (97) with assumption (2) becomes

trϕAψ1B=trAB,APn,BS1.E99

Let ψ˜:HnHnbe the linear extension of ψ. By Theorem 2.2, ϕcan be extended to a linear bijection ϕ˜:HnHnsuch that

trϕ˜Aψ˜Bk=trϕ˜Aψ1Bk=trABk,AHn,BS.E100

By Theorem 4.2 and taking into account the ranges of ϕand ψ, we see that when k=1, ϕand ψtake the form of (97), and when kZ\1,0,1, ϕand ψtake the form of (98).

Theorem 6.4 has counterpart results for ϕ,ψ:PnRPnR¯and the proof is analogous using Theorem 5.3 instead of Theorem 4.2.

## 7. k-power linear preservers and trace of power-product preservers on Dnand DnR

Let F=Cor R. Define the function diag:FnDnFto be the linear bijection that sends each c1cntto the diagonal matrix with c1,,cn(in order) as the diagonal entries. Define diag1:DnFFnthe inverse map of diag.

With the settings, every linear map ψ:DnFDnFuniquely corresponds to a matrix LψMnFsuch that

### 7.1 k-power linear preservers on Dnand DnR

We define the linear functionals fi:DnFFi=01n, such that for each A=diaga1anDnF,

f0A=0;fiA=ai,i=1,,n.E102

Theorem 7.1. LetF=CorR. LetkZ\01. LetSbe an open neighborhood ofIninDnF. A linear mapψ:DnFDnFsatisfies that

ψAk=ψAk,AS,E103

if and only if

in whichψInk=ψInandp:1n01nis a function such thatpi0whenk<0fori=1,,n. In particular, a linear bijectionψ:DnFDnFsatisfies(103) if and only if there is a diagonal matrixCMnFwithCk1=Inand a permutation matrixPMnFsuch that

Proof.For every A=diaga1anDnF, when xFis sufficiently close to 0, we have In+xASand the power series of In+xAkconverges, so that ψIn+xAk=ψIn+xAk.

ψIn+xAk=ψIn+xkψA+x2kk12ψA2+E106
ψIn+xAk=ψIn+xkψInk1ψA+x2kk12ψInk2ψA2+E107

ψA=ψInk1ψA,E108
ψA2=ψInk2ψA2.E109

The linear map ψ1AψInk2ψAsatisfies that

By (101), let Lψ1=ijMnFsuch that diag1ψ1A=Lψ1diag1Afor ADnF. Then (110) implies that for all A=diaga1anDnF:

j=1nijaj2=j=1nijaj2,i=1,2,,n.E111

Therefore, each row of Lψ1has at most one nonzero entry and each nonzero entry must be 1. We get

ψ1A=diagfp1AfpnAE112

in which p:1n01nis a function. Suppose ψIn=diagλ1λn. Then (108) implies that ψA=ψInψ1Ahas the form (104). Obviously, ψInk=ψInand when k<0, each pi0for i=1,,n. Moreover, when ψis a linear bijection, (112) shows that ψ1A=PAP1for a permutation matrix P. (105) can be easily derived.

### 7.2 Trace of power-product preservers on Dnand DnR

In [18], we show that two maps ϕ,ψ:DnFDnFsatisfy trϕAψB=trABfor A,BDnFif and only if there exists an invertible NMnFsuch that

ϕA=diagNdiag1A,ψB=diagNtdiag1B,A,BDnF.E113

When m3, the maps ϕ1,,ϕm:DnFDnFsatisfying trϕ1A1ϕmAm=trA1Amfor A1,,AmDnFare also determined in [18].

Next we consider the trace of power-product preserver on DnF.

Theorem 7.2. LetF=CorR. LetkZ\01. LetSbe an open neighborhood ofIninDnF. Two mapsϕ,ψ:DnFDnFsatisfy that

trϕAψBk=trABk,E114

1. for allADnF,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only if there exist an invertible diagonal matrix CDnFand a permutation matrix PMnFsuch that

ϕA=PCkAP1,ψB=PCBP1,A,BDnF.E115

Proof.Assumption (2) leads to assumption (1) (cf. the proof of Theorem 3.5). We prove the theorem under assumption (1).

For every BDnF, In+xBSand the power series of In+xBkconverges when xFis sufficiently close to 0, so that

trϕAψIn+xBk=trAIn+xBkE116

Comparing degree one terms and degree two terms in the power series of the above equality, respectively, we get the following equalities for A,BDnF:

trϕAψBψInk1=trAB,E117
trϕAψB2ψInk2=trAB2.E118

Applying Theorem 2.2 to (117), ψInis invertible and both ϕand ψare linear bijections. (117) and (119) imply that ψB2ψInk1=ψB2ψInk2. Let ψ1BψBψIn1. Then ψ1B2=ψ1B2for BDnF. By Theorem 7.1 and ψ1In=In, there exists a permutation matrix PMnFsuch that ψ1B=PBP1for BDnF. So ψB=ψInPBP1=PCBP1for CP1ψInPDnF. Then (114) implies (115).

## 8. k-power injective linear preservers and trace of power-product preservers on Tnand TnR

### 8.1 k-power preservers on TnF

The characterization of injective linear k-power preserver on TnFcan be derived from Cao and Zhang’s characterization of injective additive k-power preserver on TnF([12] or [[6], Theorem 6.5.2]).

Theorem 8.1 (Cao and Zhang [12]). Letk2andn3. LetFbe a field withcharF=0orcharF>k. Thenψ:TnFTnFis an injective linear map such thatψAk=ψAkfor allATnFif and only if there exists ak1th root of unityλand an invertible matrixPTnFsuch that

ψA=λPAP1,ATnF,orE119
ψA=λPAP1,ATnF,E120

where A=an+1j,n+1iif A=aij.

Example 8.2. Whenn=2, the injective linear maps that satisfyψAk=ψAkforAT2FsendA=a11a120a22to the followingψA:

λa11ca120a22,λa22ca120a11,E121

in which λk1=1and cF\0..

Example 8.3. Theorem 8.1 does not hold ifψis not assumed to be injective. Letn=3and supposeψ:T3FT3Fis a linear map that sendsA=aij3×3T3Fto one of the followingψA(c,dF):

a11ca1200a22da2300a33,a33000a11000a22,a220ca1200000a11.E122

Then each ψsatisfies that ψAk=ψAkfor every positive integer kbut it is not of the forms in Theorem 8.1.

We extend Theorem 8.1 to the following result that includes negative k-powers and that only assumes k-power preserving in a neighborhood of In.

Theorem 8.4. LetF=CorR. Let integersk0,1andn3. Suppose thatψ:TnFTnFis an injective linear map such thatψAk=ψAkfor allAin an open neighborhood ofIninTnFconsisting of invertible matrices. Then there existλFwithλk1=1and an invertible matrixPTnFsuch that

ψA=λPAP1,ATnF,orE123
ψA=λPAP1,ATnF.E124

where A=an+1j,n+1i=JnAtJnif A=aij, Jnis the anti-diagonal identity.

Proof.Obviously ψis a linear bijection. Follow the same process in the proof of Theorem 3.4. In both k2and k<0cases we have ψIncommutes with the range of ψ, so that ψIn=λInfor λFand λk1=1.Moreover, let ψ1AψIn1ψA, then ψ1is injective linear and ψ1A2=ψ1A2for ATnF.Theorem 8.1 shows that ψ1A=PAP1or ψ1A=PAP1for certain invertible PTnF.It leads to (123) and (124).

### 8.2 Trace of power-product preservers on Tnand TnR

Theorem 2.2 or Corollary 2.3 does not work for maps on TnF. However, the following trace preserving result can be easily derived from Theorem 7.2. We have TnF=DnFNnF. Let DAdenote the diagonal matrix that takes the diagonal of ATnF.

Theorem 8.5. LetF=CorR. LetkZ\01. LetSbe an open neighborhood ofIninTnFconsisting of invertible matrices. Then two mapsϕ,ψ:TnFTnFsatisfy that

trϕAψBk=trABk,E125

1. for allATnF,BS, andψis linear, or

2. for allA,BSand bothϕandψare linear,

if and only ifϕandψsendNnFtoNnF, DϕDnFandDψDnFare linear bijections characterized by(115) in Theorem 7.2, andDϕ=DϕD.

Proof.The sufficient part is easy to verify. We prove the necessary part here. Let ϕDϕDnFand ψDψDnF. Then ϕ,ψ:DnFDnFsatisfy trϕAψBk=trABkfor A,BDnF. So they are characterized by (115). The bijectivity of ϕand ψimplies that ϕand ψmust send NnFto NnFin order to satisfy (125). Moreover, ϕshould send matrices with same diagonal to matrices with same diagonal, which implies that Dϕ=DϕD.

## 9. Conclusion

We characterize linear maps ψ:VVthat satisfy ψAk=ψAkon an open neighborhood Sof Inin V, where kZ\01and Vis the set of n×ngeneral matrices, Hermitian matrices, symmetric matrices, positive definite (resp. semi-definite) matrices, diagonal matrices, or upper triangular matrices, over the complex or real field. The characterizations extend the existing results of linear k-power preservers on the spaces of general matrices, symmetric matrices, and upper triangular matrices.

Applying the above results, we determine the maps ϕ,ψ:VVon the preceding sets Vthat satisfy trϕAψBk=trABk

1. for all AV, BS, and ψis linear, or

2. for all A,BSand both ϕand ψare linear.

These results, together with Theorem 2.2 about maps satisfying trϕAψB=trABand the characterizations of maps ϕ1,,ϕm:VV(m3) satisfying trϕ1A1ϕmAm=trA1Amin [18], make a comprehensive picture of the preservers of trace of matrix products in the related matrix spaces and sets. Our results can be interpreted as inner product preservers when Vis close under conjugate transpose, in which wide applications are found.

There are a few prospective directions to further the researches.

First, for a polynomial or an analytic function fxand a matrix set V, we can consider “local” linear f-preservers, that is, linear operators ψ:VVthat satisfy ψfA=fψAon an open subset Sof V. A linear f-preserver ψon Salso preserves matrices annihilated by fon S, that is, fA=0(AS) implies fψA=0. When S=Vis Mn, BH, or some operator algebras, extensive studies have been done on operators preserving elements annihilated by a polynomial f; for examples, the results on Mnby R. Howard in [19], by P. Botta, S. Pierce, and W. Watkins in [20], and by C.-K. Li and S. Pierce in [21], on BHby P. Šemrl [22], on linear maps ψ:BHBKby Z. Bai and J. Hou in [23], and on some operator algebras by J. Hou and S. Hou in [24]. We may further explore linear f-preservers for a multivariable function fx1xr, that is, operator ψsatisfying ψfA1Ar=fψA1ψAr. The corresponding annihilator preserver problem has been studied in some special cases, for example, on Mnfor homogeneous multilinear polynomials by A. E. Guterman and B. Kuzma in [25].

Second, it is interesting to further investigate maps ϕ,ψ:VVthat satisfy trfϕAgψB=trfAgBfor some polynomials or analytic functions fxand gx. This is equivalent to the inner product preserver problem fϕAgψB=fAgBwhen Vis close under conjugate transpose. More generally, given a multivariable function hx1xm, we can ask what combinations of linear operators ϕ1,,ϕm:VVsatisfy that tr(hϕ1A1ϕmAm=trhA1Am. The research on this area seems pretty new. No much has been discovered by the authors.

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Written By

Huajun Huang and Ming-Cheng Tsai

Reviewed: February 15th, 2022 Published: April 17th, 2022