# The Characterization of Generalized Jordan Centralizers on Triangular Algebras

Authors: Quanyuan Chen; Xiaochun Fang; Changjing Li
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, it is shown that ifT=Tri(A,M,B) is a triangular algebra and ϕ is an additive operator on T such that (m+n+k+l)ϕ(T2)-(mϕ(T)T+nTϕ(T)+kϕ(I)T2+lT2ϕ(I))∈FI for any T∈T, then ϕ is a centralizer. It follows that an (m,n)- Jordan centralizer on a triangular algebra is a centralizer.


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## Body

## 1. Introduction and Preliminaries

The triangular algebras are firstly introduced in [1]. Let A and B be two unital algebras over a commutative ring R, where R is a unital commutative ring, and let M be a unital (A,B)- bimodule. Suppose that M is a faithful (A,B)- bimodule; that is, M is an (A,B)- bimodule satisfying A∈A: AM=0 and implies that A=0, and B∈B: MB=0 implies that B=0. Consider the set(1)T=TriA,M,B=AMB:A∈A,M∈M,B∈B.The usual matrix addition and matrix multiplication on Tri(A,M,B) are defined as follows: for any A1,A2∈A,M1,M2∈M,B1,B2∈B,(2)A1M1B1+A2M2B2=A1+A2M1+M2B1+B2,A1M1B1A2M2B2=A1A2A1M2+M1B2B1B2.It is easy to see that T=Tri(A,M,B) is an algebra under the usual matrix addition and matrix multiplication and is called a triangular algebra. We denote AMB by Tri(A,M,B) for simplicity.LetR be a ring or an algebra and ϕ be an additive mapping on R. If ϕ(AB)=ϕ(A)B (resp., ϕ(AB)=Aϕ(B)) for any A,B∈R, then ϕ is called a left centralizer (resp., a right centralizer). A centralizer of R is an additive mapping which is both a left centralizer and a right centralizer. An additive mapping ϕ:R→R is called a left (resp., right) Jordan centralizer, if ϕ(A2)=ϕ(A)A (resp., ϕ(A2)=Aϕ(A)) for any A∈R. A Jordan centralizer of R is an additive mapping which is both a left Jordan and a right Jordan centralizer. An (m,n)- Jordan centralizer is defined in ([2]) as follows: let R be a ring or an algebra and ϕ be an additive mapping on R. If there are m,n∈N:m+n≠0 such that (m+n)ϕ(A2)=mϕ(A)A+nAϕ(A) for any A∈R, then ϕ is called an (m,n)- Jordan centralizer.The characterization of centralizers on algebras or rings is a subject in various areas. Bresar and Zalar ([3]) proved that if R is a prime ring and ϕ is an additive mapping on R such that ϕ(A2)=ϕ(A)A (resp., ϕ(A2)=Aϕ(A)) for any A∈R, then ϕ is a left (resp., a right) centralizer. Zalar ([4]) generalized the result to 2-torsion free semiprime rings as follows: if R is a 2-torsion free semiprime ring and ϕ is an additive mapping on R such that ϕ(A2)=ϕ(A)A (resp., ϕ(A2)=Aϕ(A)) for any A∈R, then ϕ is a left (resp., a right) centralizer. Vukman ([5]) proved that if R is a 2-torsion free semiprime ring and ϕ is an additive mapping on R such that 2ϕ(A2)=ϕ(A)A+Aϕ(A) for any A∈R, then ϕ is a centralizer. Benkovic and Eremita ([6]) proved that if R is a prime ring with Ch(R)=0 or Ch(R)≥n, where n is a fixed positive integer such that n≥2, and ϕ is an additive mapping on R such that ϕ(An)=ϕ(A)An-1 for any A∈R, then ϕ is a centralizer. Vukman and Kosi-Ulbl ([7]) proved that if X is a Banach space over the field F, and A is a standard subalgebra of B(X) and ϕ:A→B(X) is an additive mapping such that ϕ(Am+n+1)=Amϕ(A)An for any A∈A, where m,n∈Z+, then ϕ is a centralizer. Qi et al. ([8]) proved that if A is a standard subalgebra of B(X) with the identity I and ϕ:A→B(X) is an additive mapping such that ϕ(Am+n+1)-Amϕ(A)An∈FI for any A∈A, where X is a Banach space over the field F and m,n∈Z+, then ϕ is a centralizer. Yang and Zhang ([9]) proved that, if ϕ:τ(N)→τ(N) is an additive mapping on a nest algebra τ(N), such that (m+n)ϕ(Ap+1)=mϕ(A)Ap+nApϕ(A) or ϕ(Am+n+1)=Amϕ(A)An for any A∈τ(N), where N is a nontrivial nest on H, τ(N) is the corresponding nest algebra, and m,n,p∈Z+, then ϕ is a centralizer. J. Vukman ([2]) proved that an (m,n)- Jordan centralizer on a prime ring with Ch(R)≠6mn(m+n) is a centralizer. Li et al. ([10]) proved that a Jordan centralizer on a CSL subalgebra of a von Neumann algebra is a centralizer. Chen et al. ([11]) proved that an additive mapping ϕ on a CSL subalgebra of a von Neumann algebra such that (m+n+k+l)ϕ(A2)-(mϕ(A)A+nAϕ(A)+kϕ(I)A2+lA2ϕ(I))∈FI for any A∈A is a centralizer. Lately several authors investigated (m,n)- Jordan centralizers or their related mappings on rings and algebras. Some of the results can be found in ([5, 12–20]).Motivated by these results, we are concerned with an additive mappingϕ such that (m+n+k+l)ϕ(A2)-(mϕ(A)A+nAϕ(A)+kϕ(I)A2+lA2ϕ(I))∈FI for any A∈A. It is shown that if T=Tri(A,M,B) is a triangular algebra and ϕ is an additive operator on T such that (m+n+k+l)ϕ(T2)-(mϕ(T)T+nTϕ(T)+kϕ(I)T2+lT2ϕ(I))∈FI for any T∈T, then ϕ is a centralizer. And it follows that an (m,n)- Jordan centralizer on a triangular algebra is a centralizer. As applications, (m,n)- Jordan centralizers on (block) upper triangular matrix algebras and nest algebras are centralizers.


## 2. Generalized Jordan Centralizers on Triangular Algebras

IfR is a unital commutative ring, A and B are unital algebras over R with units 1A and 1B, respectively, and M is a unital, faithful (A,B)- bimodule, then(3)T=TriA,M,B=AMB:A∈A,M∈M,B∈Bis a triangular algebra.LetTri(A,M,B) be a triangular algebra. And let P1=1A00, P2=001B, where 1A and 1B are the identity elements of algebras A and B, respectively. And let Tij=PiTPj (1≤i≤j≤2). Then T=T11⊕T12⊕T22 and P1P2=P2P1=0.The main result is as follows.Theorem 1.
LetT=Tri(A,M,B) be a triangular algebra and ϕ:T→T be an additive mapping. If there exist numbers m,n,k,l≥0:mn≠0 such that(4)m+n+k+lϕT2-mϕTT+nTϕT+kϕIT2+lT2ϕI∈FIfor any T=AMB∈T, then ϕ is a centralizer; that is, ϕ(I)T=Tϕ(I) and ϕ(T)=ϕ(I)T=Tϕ(I) for any T=AMB∈T, where I=1A01B.The proof of Theorem1 can be divided into several lemmas.Lemma 2.
Suppose thatT is a unital algebra, and ϕ is an additive mapping on T such that(5)m+n+k+lϕT2-mϕTT+nTϕT+kϕIT2+lT2ϕI∈FIThat is, for any T∈T, there is μT∈F (depending on T) such that(6)m+n+k+lϕT2=mϕTT+nTϕT+kϕIT2+lT2ϕI+μTI,where m>0,n>0,k≥0,l≥0. Then(1)
(7)m+n+k+lϕTS+ST=mϕTS+nTϕS+mϕST+nSϕT+kϕITS+kϕIST+lTSϕI+lSTϕI+μT+S-μT-μSI;(2)
(8)m+n+2k+2lϕT=m+2kϕIT+n+2lTϕI+μT+I-μTIfor any T,S∈T.Proof.
For anyT,S∈T,(9)m+n+k+lϕT+S2=mϕT+ST+S+nT+SϕT+S+kϕIT+S2+lT+S2ϕI+μT+SI=mϕTT+mϕTS+mϕST+mϕSS+nTϕT+nTϕS+nSϕT+nSϕS+kϕIT2+kϕIST+kϕITS+kϕIS2+lT2ϕI+lTSϕI+lSTϕI+lS2ϕI+μT+SI.On the other hand,(10)m+n+k+lϕT+S2=m+n+k+lϕT2+TS+ST+S2=mϕTT+nTϕT+kϕIT2+lT2ϕI+nSϕS+mϕSS+kϕIS2+lS2ϕI+m+n+k+lϕTS+ST+μTI+μSI.Comparing the above two equalities, we obtain that(11)m+n+k+lϕTS+ST=mϕTS+nTϕS+mϕST+nSϕT+kϕITS+kϕIST+lTSϕI+lSTϕI+μT+S-μT-μSI.Putting in (11) S=I, it follows from μI=0 that(12)m+n+2k+2lϕT=m+2kϕIT+n+2lTϕI+μT+I-μTI.Lemma 3.
LetT be a unital algebra and ϕ be an additive mapping on T satisfying (4). If T0∈T with T0∉FI such that T0ϕ(I)=ϕ(I)T0, then μT0+I-μT0=0 and ϕ(T0)=T0ϕ(I)=ϕ(I)T0.Proof.
SinceT0ϕ(I)=ϕ(I)T0, T02ϕ(I)=ϕ(I)T02=T0ϕ(I)T0. By (12), we have that(13)ϕT0=ϕIT0+1m+n+2k+2lμT0+I-μT0Iand(14)ϕT02=ϕIT02+1m+n+2k+2lμT02+I-μT02I.Hence(15)m+n+k+lϕT02=m+n+k+lϕIT02+m+n+k+lm+n+2k+2lμT02+I-μT02I.On the other hand,(16)m+n+k+lϕT02=mϕT0T0+nT0ϕT0+kϕIT02+lT02ϕI+μT0I=mϕIT0+1m+n+2k+2lμT0+I-μT0IT0+nT0ϕIT0+1m+n+2k+2lμT0+I-μT0I+kϕIT02+lT02ϕI+μT0I.Comparing the two equalities, we have that m+n/m+n+2k+2l(μT0+I-μT0)T0∈FI. Since T0∉FI and m+n>0, μT0+I-μT0=0 and ϕ(T0)=T0ϕ(I)=ϕ(I)T0.Lemma 4.
LetT be a unital algebra and ϕ be an additive mapping on T satisfying (4). If P∈T with P2=P, then ϕ(P)=Pϕ(I)=ϕ(I)P=ϕ(P)P=Pϕ(P).Proof.
IfP=0 or P=I, the result is trivial.
LetP be a nontrivial idempotent; that is, P≠0 and P≠I. By (4),(17)m+n+k+lϕP=mϕPP+nPϕP+kϕIP+lPϕI+μPI.By (12),(18)m+n+2k+2lϕP=m+2kϕIP+n+2lPϕI+μP+I-μPI.Multiplying (17) by P from the left and the right sides gives(19)k+lPϕPP=k+lPϕIP+μPP.Multiplying (18) by P from the left and the right sides, we have that(20)PϕPP=PϕIP+1m+n+2k+2lμP+I-μPP.By comparing the above two equalities,(21)m+n+2k+2lμP=k+lμP+I-μP.Multiplying (17) by P from the left side gives that(22)m+n+k+lPϕP=mPϕPP+nPϕP+kPϕIP+lPϕI+μPP.That is,(23)m+k+lPϕP=mPϕPP+kPϕIP+lPϕI+μPP.It follows from (20) that(24)m+k+lPϕP=m+kPϕIP+lPϕI+mm+n+2k+2lμP+I-μP+μPP.Thus24′m+n+2k+2lm+k+lPϕP=m+km+n+2k+2lPϕIP+lm+n+2k+2lPϕI+mμP+I-μP+m+n+2k+2lμPP.Multiplying (18) by P from the left side yields that(25)m+n+2k+2lPϕP=m+2kPϕIP+n+2lPϕI+μP+I-μPP,and comparing 24′ and (25), we obtain that(26)m+k+lm+2kPϕIP+n+2lm+k+lPϕI+m+k+lμP+I-μPP=m+km+n+2k+2lPϕIP+lm+n+2k+2lPϕI+mμP+I-μP+m+n+2k+2lμPP.It follows from (21) that (m+k+l)(μP+I-μP)P=(m(μP+I-μP)+(m+n+2k+2l)μP)P and(27)PϕI=PϕIP.It follows from (24) that(28)PϕP=PϕIP+1m+n+2k+2lμP+I-μPP.Similarly,(29)ϕIP=PϕIPand(30)ϕPP=PϕIP+1m+n+2k+2lμP+I-μPP.Equations (27) and (29) yield that ϕ(I)P=Pϕ(I). And (31)ϕPP=PϕP=PϕPPby (28) and (30). By Lemma 3 and ϕ(I)P=Pϕ(I), it follows that ϕ(P)=ϕ(I)P=Pϕ(I) and μP+I-μP=0. And by (21), μP=0 and μP+I=μP=0. Identity (17) yields that (32)ϕP=ϕIP=PϕI=PϕIP=PϕPP=ϕPP=PϕP.Lemma 5.
LetT be a unital algebra and ϕ be an additive mapping on T satisfying (4). If T,P∈A with P2=P, then(1)
ϕ(TP)=ϕ(T)P+μ(TP)I-μ(T)P,(2)
ϕ(PT)=Pϕ(T)+μ(PT)I-μ(T)P, where μ(T)=1/m+n+2k+2l(μT+I-μT).Proof.
By (12), (33)ϕTP=m+2km+n+2k+2lϕITP+n+2lm+n+2k+2lTPϕI+1m+n+2k+2lμTP+I-μTPI=m+2km+n+2k+2lϕIT+n+2lm+n+2k+2lTϕIP+1m+n+2k+2lμTP+I-μTPI=ϕTP+μTPI-μTP.Similarly, ϕ(PT)=Pϕ(T)+μ(PT)I-μ(T)P.Lemma 6.
LetT be a unital algebra and ϕ be an additive mapping on T satisfying (4). If T,P∈T with P2=P, then ϕ(PTP)=ϕ(PTP)P=Pϕ(PTP)=Pϕ(PTP)P.Proof.
IfP=0 or P=I, the result is trivial.
LetP be a nontrivial idempotent; that is, P≠0 and P≠I. It follows from Lemma 5 that (34)ϕPTP=ϕPTPP=ϕPTPP+μPTPI-μPTPP,ϕPTP=ϕPPTP=PϕPTP+μPTPI-μPTPP.Comparing the above two equalities, we have that (35)PϕPTP=ϕPTPP.It follows from Lemma 2(2) that (36)m+n+2k+2lϕPTP=m+2kϕIPTP+n+2lPTPϕI+μPTP+I-μPTPI.By Lemma 4, we have that ϕ(I)PTP=ϕ(P)PTP, PTPϕ(I)=PTPϕ(P), and μP=0. Putting PTP for T and P for S in (11), we have that (37)2m+n+k+lϕPTP=m+n+k+lϕPTPP+PPTP=mϕPTPP+nPϕPTP+nPTPϕP+mϕPPTP+2kϕIPTP+2lPTPϕI+μPTP+P-μPTP-μPI=m+nPϕPTP+m+2kϕIPTP+n+2lPTPϕI+μPTP+P-μPTPI.By (36) with (37), (38)2m+n+k+lm+n+2k+2lϕPTP=m+nPm+2kϕIPTP+n+2lPTPϕI+μPTP+I-μPTPI+m+2km+n+2k+2lϕIPTP+n+2lm+n+2k+2lPTPϕI+m+n+2k+2lμPTP+P-μPTPI=2m+2km+n+k+lϕIPTP+2n+2lm+n+k+lPTPϕI+m+n+2k+2lμPTP+P-μPTPI+m+nμPTP+I-μPTPP.It follows from (36) that (39)2m+n+k+lm+n+2k+2lϕPTP=2m+2km+n+k+lϕIPTP+2n+2lm+n+k+lPTPϕI+2m+n+k+lμPTP+I-μPTPI.Comparing (38) and (39), we have that (40)m+n+2k+2lμPTP+P-μPTPI+m+nμPTP+I-μPTPP=2m+n+k+lμPTP+I-μPTPI.Multiplying (36) by P from the left and the right sides gives that (41)m+n+2k+2lPϕPTPP=m+2kϕIPTP+n+2lPTPϕI+μPTP+I-μPTPP.Multiplying (37) by P from both sides yields that (42)2m+n+k+lPϕPTPP=m+nPϕPTPP+m+2kϕIPTP+n+2lPTPϕI+μPTP+P-μPTPP.It follows that (43)m+n+2k+2lPϕPTPP=m+2kϕIPTP+n+2lPTPϕI+μPTP+P-μPTPP.Comparing (41) and (43), we have that (44)μPTP+P-μPTP=μPTP+I-μPTP.It follows from the above equality and (40) that (m+n)(μPTP+P-μPTP)=0. Since m+n>0, μPTP+P-μPTP=0 and (45)μPTP+P-μPTP=μPTP+I-μPTP=0.By (37) and (45), (46)2m+n+k+lϕPTP=m+nPϕPTP+m+2kϕIPTP+n+2lPTPϕI.By (36) and (45), (47)m+n+2k+2lϕPTP=m+2kϕIPTP+n+2lPTPϕI.Comparing it with (46), we have that (m+n)ϕ(PTP)=(m+n)Pϕ(PTP) and (48)ϕPTP=PϕPTP=ϕPTPP=PϕPTPP.Lemma 7.
LetT=Tri(A,M,B) be a triangular algebra. Suppose that Tij∈Tij(1≤i≤j≤2).(1)
IfT11T12=0, then T11=0;(2)
IfT12T22=0, then T22=0.Proof.
(1) IfT11T12=0 and T11=1A00A0M0B01A00, then A0001A00AMB001B=0 for any A∈A,M∈M,B∈B. So A0M=0 for any M∈M. It follows from the definition of a triangular algebra that A0=0. And T11=1A00A0M0B01A00=0.
(2) It is similar to the proof of (1).Lemma 8.
LetT=Tri(A,M,B) be a triangular algebra, and ϕ:T→T be an additive mapping satisfying (4). Then ϕ(Tij)⊆Tij(1≤i≤j≤2).Proof.
It follows thatϕ(T11)⊆T11 and ϕ(T22)⊆T22 from Lemma 6.
For anyT∈T, by Lemma 5, we have that (49)ϕP1TP2=ϕP1TP2P2=ϕP1TP2P2+μP1TP2PI-μP1TP2P2=ϕP1TP2P2+μP1TP2PP1and (50)ϕP1TP2=ϕP1P1TP2=P1ϕP1TP2+μP1TP2PI-μP1TP2P1=P1ϕP1TP2+μP1TP2PP2.It follows that (51)ϕP1TP2P1=μP1TP2PP1and(52)P2ϕP1TP2=μP1TP2PP2.By (12), we have that (53)m+n+2k+2lϕP1TP2=m+2kϕIP1TP2+n+2lP1TP2ϕI+μP1TP2+I-μP1TP2I.Multiplying the above equality by P1 from the right side gives that (54)m+n+2k+2lϕP1TP2P1=μP1TP2+I-μP1TP2P1=μP1TP2PP1.It follows that (55)m+n+2k+2lϕP1TP2P1=μP1TP2PP1=ϕP1TP2P1and (56)ϕP1TP2P1=0.Similarly, we have that (57)P2ϕP1TP2=0.It follows that (58)ϕP1TP2=P1ϕP1TP2=ϕP1TP2P2and ϕ(T12)⊆T12.Lemma 9.
LetT=Tri(A,M,B) be a triangular algebra, and ϕ:T→T be an additive mapping satisfying (4). Let Tij,Sij∈Tij(1≤i≤j≤2). Then(1)
ϕ(T11S12)=ϕ(T11)S12=T11ϕ(S12);(2)
ϕ(T12S22)=ϕ(T12)S22=T12ϕ(S22);(3)
ϕ(T11S11)=ϕ(T11)S11=T11ϕ(S11);(4)
ϕ(T22S22)=ϕ(T22)S22=T22ϕ(S22).Proof.
(1) By (11), (59)m+n+k+lϕT11S12+S12T11=mϕT11S12+nT11ϕS12+mϕS12T11+nS12ϕT11+kϕIT11S12+kϕIS12T11+lT11S12ϕI+lS12T11ϕI+μT11+S12-μT11-μS12I.Since S12T11=0, ϕ(S12)T11∈T12T11=0, and S12ϕ(T11)∈S12T11=0, it follows that (60)m+n+k+lϕT11S12=mϕT11S12+nT11ϕS12+kϕIT11S12+lT11S12ϕI+μT11+S12-μT11-μS12I.Multiplying (60) by P1 from the right side, using the fact that ϕ(S12),ϕ(T11S12)∈T12, yields that (μT11+S12-μT11-μS12)P1=0 and μT11+S12-μT11-μS12=0. Since S12∈T12, it follows from Lemma 7(2) that (61)ϕS12=ϕIS12=S12ϕI.Since T11S12∈T12, (62)ϕT11S12=ϕIT11S12=T11S12ϕI=T11ϕS12.Combining it with (60), we have that mϕ(T11S12)=mϕ(T11)S12. Since m≠0, (63)ϕT11S12=T11S12ϕI=ϕT11S12=T11ϕS12=ϕIT11S12.(2) By (11), (64)m+n+k+lϕT12S22+S22T12=mϕT12S22+nT12ϕS22+mϕS22T12+nS22ϕT12+kϕIT12S22+kϕIS22T12+lT12S22ϕI+lS22T12ϕI+μT12+S22-μT12-μS22I.Using the fact that ϕ(S22)T12,S22ϕ(T12)∈T22T12=0 yields that (65)m+n+k+lϕT12S22=mϕT12S22+nT12ϕS22+kϕIT12S22+lT12S22ϕI+μT12+S22-μT12-μS22I.Multiplying (65) by P1 from the right side, we have that (μT12+S22-μT12-μS22)P1=0 and μT12+S22-μT12-μS22=0. It follows from Lemma 7(2) that ϕ(T12)=ϕ(I)T12=T12ϕ(I) and (66)ϕT12S22=T12S22ϕI=ϕIT12S22=ϕT12S22.Combining it with (65), we have that nϕ(T12S22)=nT12ϕ(S22). Since n≠0, (67)ϕT12S22=ϕT12S22=T12ϕS22=ϕIT12S22=T12S22ϕI.
(3) By (1), for anyR12∈T12, (68)ϕT11S11R12=ϕT11S11R12=ϕT11S11R12,and (69)ϕT11S11R12=ϕT11S11R12=T11ϕS11R12=T11ϕS11R12.ϕ(T11S11)=ϕ(T11)S11=T11ϕ(S11) by Lemma 7(1).
(4) By (2), for anyR12∈T12, (70)R12ϕT22S22=ϕR12T22S22=R12T22ϕS22,and(71)R12ϕT22S22=ϕR12T22S22=ϕR12T22S22=R12ϕT22S22.ϕ(T22S22)=ϕ(T22)S22=T22ϕ(S22) by Lemma 7(2).Proof of Theorem1.
LetT,S∈T. Then T=T11+T12+T22 and S=S11+S12+S22. By Lemmas 8 and 9, we have that (72)ϕTS=ϕT11S11+T11S12+T12S22+T22S22=ϕT11S11+ϕT11S12+ϕT12S22+ϕT22S22=ϕT11S11+ϕT11S12+ϕT12S22+ϕT22S22=T11ϕS11+T11ϕS12+T12ϕS22+T22ϕS22=ϕT11+T12+T22S11+S12+S22=T11+T12+T22ϕS11+S12+S22=ϕTS=TϕS.So that ϕ is a centralizer.Corollary 10.
LetT=Tri(A,M,B) be a triangular algebra, and ϕ:T→T be an additive mapping. If there are positive numbers m,n, such that (m+n)ϕ(T2)=mϕ(T)T+nTϕ(T) for any T∈T, then ϕ is a centralizer.Theorem 11.
LetT=Tri(A,M,B) be a triangular algebra, and ϕ:T→T be an additive mapping. If there are positive numbers m,n and a positive integer p, such that (73)m+nϕTp+1=mϕTTp+nTpϕTfor any T∈T, then ϕ is a centralizer.Proof.
PuttingT+xI for T in (73), where x is any rational number, we get that (m+n)ϕ(T2)=mϕ(T)T+nTϕ(T) for any T∈T, so that ϕ is a centralizer by Corollary 10.Theorem 12.
LetT=Tri(A,M,B) be a triangular algebra and ϕ:T→T be an additive mapping. If there are positive integers m,n, such that (74)ϕTm+n+1=TmϕTTnfor any T∈T, then ϕ is a centralizer.Proof.
PuttingT+xI for T in (74), where x is any rational number, we can also get that (m+n)ϕ(T2)=mϕ(T)T+nTϕ(T) for any T∈T, so that ϕ is a centralizer by Corollary 10.


## 3. Applications

Upper triangular matrix algebras, block upper triangular matrix algebras, and nest algebras over a Hilbert spaceH are important examples of triangular algebras.Upper triangular matrix algebras. Let Mk×m(R) be the set of all k×m matrices and Tk(R) be the algebra of all k×k upper triangular matrices over a ring R. For k≥2 and each 1≤l≤k-1, the algebra Tk(R) can be represented as a triangular algebra of the form Tk(R)=Tri(Tl(R),Ml×(k-l)(R),Tk-l(R)).Block upper triangular matrix algebras. Let N be the set of all positive integers and let k∈N. For every positive integer s, s≤k, we denote by l¯=(l1,l2,…,ls)∈Ns an ordered s-vector of positive integers such that k=l1+l2+…+ls. The block upper triangular matrix algebra Bk(R) with corresponding vector l¯ is a subalgebra of Tk(R) which contains all block upper triangular matrices, where diagonal blocks have sizes l1,l2,…,ls, respectively. Note that Mk(R) and Tk(R) are two special cases of block upper triangular matrix algebras. If we have that k≥2 and Bk(R)≠Mk(R), then Bk(R) is a triangular algebra and can be represented as Bk(R)=Tri(Bl(R),Ml×(k-l)(R),Bk-l(R)), where 1≤l≤k-1 and Bl(R) and Bk-l(R) are block upper triangular algebras with suitable vectors l1¯∈Nl,l2¯∈Nk-l.Corollary 13.
Every(m,n)-Jordan centralizer of a block upper triangular matrix algebra Bk(C)(k≥2) is a centralizer, where C is the complex field.Corollary 14.
LetBk(C) be a block upper triangular matrix algebra and ϕ:Bk(C)→Bk(C) be an additive mapping on Bk(C)(k≥2). If there are positive numbers m,n and a positive integer p, such that (75)m+nϕTp+1=mϕTTp+nTpϕTfor any T∈Bk(C), then ϕ is a centralizer.Corollary 15.
LetBk(C) be a block upper triangular matrix algebra and ϕ:Bk(C)→Bk(C) be an additive mapping on Bk(C)(k≥2). If there are positive numbers m,n and a positive integer p, such that (76)ϕTm+n+1=TmϕTTnfor any T∈Bk(C), then ϕ is a centralizer.Nest algebras. A nest is a chain N of closed subspaces of a real or complex Hilbert space H containing 0 and H, which is closed under arbitrary intersections and closed liner span. The nest algebra associated with N is the algebra (77)TN=T:T∈BHandTN⊆NforallN∈N.A nest N is called trivial if N=0,H. A nontrivial nest algebra over a Hilbert space is a triangular algebra.In fact, by Peirce decomposition, a nontrivial nest algebra is a triangular algebra. Suppose thatA=T(N) is a nontrivial nest algebra, and N is the corresponding nest. Choose any P1∈N such that P1≠0, P1≠I. Set P2=I-P1, A11=P1AP1, A12=P1AP2, and A22=P2AP2. Then A=A11⊕A12⊕A22; that is, (78)A=A11A12A22.Since A12 is a faithful (A11,A22)- bimodule, the nest algebra A is a triangular algebra.Corollary 16.
LetT(N) be a nontrivial nest algebra over a Hilbert space. Then every (m,n)-Jordan centralizer on the nest algebra T(N) is a centralizer.Corollary 17 (see [9]).
LetT(N) be a nontrivial nest algebra over a Hilbert space and let ϕ:T(N)→T(N) be an additive mapping on T(N). If there are positive numbers m,n and a positive integer p, such that (79)m+nϕTp+1=mϕTTp+nTpϕTfor any T∈T(N), then ϕ is a centralizer.Corollary 18 (see [9]).
LetT(N) be a nontrivial nest algebra over a Hilbert space and let ϕ:T(N)→T(N) be an additive mapping on T(N). If there are positive numbers m,n and a positive integer p, such that (80)ϕTm+n+1=TmϕTTnfor any T∈T(N), then ϕ is a centralizer.



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*Source: 6037615-2018-09-16.xml*