GENERALIZED TRIANGULAR MATRIX ALGEBRA

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GENERALIZED TRIANGULAR MATRIX ALGEBRA

AIAT HADJ AHMED DRISS AND BEN YAKOUB L’MOUFADAL Received 25 July 2004 and in revised form 14 February 2005

We investigate Jordan automorphisms and Jordan derivations of a class of algebras called generalized triangular matrix algebras. We prove that any Jordan automorphism on such an algebra is either an automorphism or an antiautomorphism and any Jordan derivation on such an algebra is a derivation.

1. Introduction

Throughout this paper, letRbe a 2-torsion-free commutative ring with identity 1. Con- sider an associative algebraAoverR, thenAcan be viewed as a Jordan algebra with the usual productx◦y=(1/2)(xy+yx). AnR-linear mapδ:A→Ais called a derivation (resp., Jordan derivation) ofAif

δ(ab)=δ(a)b+aδ(b), ∀a,b∈A resp.,δa²=δ(a)a+aδ(a)∀a∈A. (1.1) AnR-linear mapθ:A→Ais said to be a Jordan homomorphism ofAif

θ(a◦b)=θ(a)◦θ(b), ∀a,b∈A, or, equivalently,θa²=

θ(a)², ∀a∈A.

(1.2) Derivations, Jordan derivations, as well as automorphisms and Jordan automorphisms of the algebra of triangular matrices and some class of their subalgebras have been the object of active research for a long time [1,2,5,6,9,10].

A well-know result of Herstein [11] states that every Jordan isomorphism on a prime ring of characteristic diﬀerent from 2 is either an isomorphism or an anti-isomorphism.

We remark that the situation where the rings are semiprime rings does not hold. In the same time, he showed that every Jordan derivation on a prime ring of characteristic dif- ferent from 2 is a derivation [12]. A brief proof of this result can be found in [4]. This result is extended by [3,8] to the semiprime case.

International Journal of Mathematics and Mathematical Sciences 2005:13 (2005) 2125–2132 DOI:10.1155/IJMMS.2005.2125

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Let nowᐁbe the algebra of the form ᐁ=

A M

B

, (1.3)

whereAandBare unitalR-algebras andM is an (A,B)-bimodule. This algebraᐁ, en- dowed with the usual formal matrix addition and multiplication, will be called a generalized triangular matrix algebra. Many widely studied algebras, including upper-triangular matrix algebras, block-triangular matrix algebras, nest algebras, semi-nest algebras, and triangular Banach algebras, may be viewed as triangular algebras.

Khazal et al. [13] discuss the automorphism group ofᐁso thatAandBhave only trivial idempotents. Cheung [7] gives suﬃcient conditions under which every Lie derivation is a sum of derivation onᐁand a mapping fromᐁto its center.

In this paper, we consider linear operators on a class of algebras of the formᐁ; specifi- cally, Jordan derivations and Jordan automorphisms.Mis assumed to be faithful as a left A-module as well as a rightB-module. We will prove that if bothAandBhave only trivial idempotents, any Jordan automorphism of the ringᐁis either an automorphism or an antiautomorphism, and we will prove that any Jordan derivation of such an algebraᐁis a derivation ofᐁ.

2. The Jordan automorphism of generalized triangular matrix algebra In this section, we suppose thatᐁis the algebra of the form

ᐁ=

A M

B

, (2.1)

whereAandB are unitalR-algebras andM is an (A,B)-bimodule, bothAandBhave only trivial idempotents.

This section is devoted to prove the following result.

Theorem2.1. IfM is faithful as a leftA-module as well as a rightB-module and if both AandBhave only trivial idempotents, then any Jordan automorphismθofᐁis either an automorphism or an antiautomorphism.

First, we start by recalling the next statement concerning the set of all idempotents ofᐁ.

Lemma2.2 [13]. The idempotents ofᐁare the elements of the forms(^{0 0}₀),(^{1 0}₁),(¹^x₀), and (⁰^x₁), for anyx∈M.

We now introduce the notations Ex=(¹^x₀) , Fx=(⁰^x₁), and X=(⁰^x₀) , for some x∈M.

Then it is easy to check the following relations:

(i)ExEy=Ey,FxFy=Fx,FxEy=0,ExFy=(⁰^x+y₀ );

(ii)aE0=E0◦(aE0),bF0=F0◦(bF0),X=(2E0)◦X,X=(2F0)◦X;

(iii) 2(aE0)◦X=(⁰^ax₀), 2(bF0)◦X=(⁰^xb₀).

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On the other hand, ifθ is a Jordan automorphism of ᐁ, then either θ(E0)=Eu or θ(E0)=Fufor someu∈M, sinceE0is an idempotent.

The proof ofTheorem 2.1is an immediate consequence of the following two lemmas.

Lemma2.3. Assume thatθ(E0)=Eufor someu∈M, thenθis an automorphism ofᐁ.

Proof. Sinceθ(E0)=Eu, we have necessarilyθ(F0)=Fvfor somev∈M. Indeed, ifθ(F0)

=Evfor somev∈M, we obtain the contradictionθ(E0◦F0)=0. Now, by the relation θ(E0◦F0)=θ(E0)◦θ(F0), we haveu+v=0, henceθ(F0)=F−u. We observe thatW= (¹⁻₁^u) is invertible of inverse (¹^u₁). So, one may consider the inner automorphismσW

of ᐁdefined byσW(Y)=WYW⁻¹. It is not diﬃcult to see thatθ(E0)=σW(E0) and θ(F0)=σW(F0), which furnishesθ1(E0)=E0andθ1(F0)=F0, whereθ1=σW⁻¹◦θis also a Jordan automorphism ofᐁ.

By applyingθ1 toaE0=E0◦(aE0) andbF0=F0◦(bF0) fora∈Aandb∈B, we get thatθ1(aE0)=ϕA(a)E0andθ1(bF0)=ϕB(b)F0, whereϕA:A→AandϕB:B→Bare additive and bijective maps.

Now if we applyθ1 to (a²E0)=(aE0)² and (b²F0)=(bF0)² fora∈Aandb∈B, we haveϕA(a²)=(ϕA(a))² andϕB(b²)=(ϕB(b))², that isϕA andϕB are Jordan automorphisms ofAandB, respectively.

Applyingθ1toX=2E0◦Xyieldsθ1(X)=(⁰ ^f^(x)₀ ). It follows that θ1

a x

b

=

ϕA(a) f(x) ϕB(b)

. (2.2)

Applying againθ1to (⁰^ax₀)=2(aE0)◦Xand (⁰^xb₀)=2(bF0)◦Xfora∈A, forx∈Mand b∈B, we obtain f(ax)=ϕA(a)f(x) and f(xb)= f(x)ϕB(b).

Sinceθ=σW◦θ1, we deduce that θ

a x

b

=

ϕA(a) f(x) +ϕA(a)u−uϕB(b) ϕB(b)

, (2.3)

whereᐁ∈M, andϕA:A→A,ϕB:B→B, f :M→Mare maps satisfying that (i) ϕAis a Jordan automorphism ofA,f(ax)=ϕA(a)f(x),

(ii)ϕBis a Jordan automorphism ofB,f(xb)= f(x)ϕB(b).

As a consequence, the following two identities are valid for alla1,a2∈Aandx∈M: fa1a2x=ϕAa1a2f(x), fa1a2x=ϕAa1fa2x=ϕAa1ϕAa2f(x).

(2.4) Thus,

ϕA a1a2

f(x)=ϕA a1

ϕA a2

f(x). (2.5)

SinceMis faithful, we haveϕA(a1a2)=ϕA(a1)ϕA(a2), which means thatϕAis an automorphism ofA. Similarly,ϕBis an automorphism ofB. Finally, in view of these arguments, one can easily check thatθis an automorphism ofᐁ, which concludes the proof

of the lemma.

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Lemma2.4. Assume thatθ(E0)=Fufor someu∈M, thenθis an antiautomorphism ofᐁ.

Proof. By hypothesis, we haveθ(F0)=Fv for somev∈M. Hence, it follows from the equalityθ(E0◦F0)=θ(E0)◦θ(F0) thatu+v=0, soθ(F0)=E−u.

ApplyingθtoaE0=E0◦(aE0) andbF0=F0◦(bF0) fora∈Aandb∈B, we have θaE0

=

0 uϕ1(a) ϕ1(a)

, θbF0

=

ϕ2(b) −ϕ2(b)u 0

, (2.6)

whereϕ1:A→Bandϕ2:B→Aare obviously additive and bijective maps. In addition, by application ofθto (a²E0)=(aE0)²and (b²F0)=(bF0)²fora∈Aandb∈B, we can show by simple calculus thatϕ1andϕ2are Jordan isomorphisms.

Applying nowθtoX=2E0◦Xgivesθ(X)=(⁰ ^f^(x)₀ ).

Therefore, θ

a x

b

=

ϕ2(b) f(x) +uϕ1(a)−ϕ2(b)u ϕ1(a)

. (2.7)

Applyingθto 2(aE0)◦X=(⁰^ax₀) and 2(bF0)◦X=(⁰^xb₀) fora∈A,x∈M, andb∈M, we have

f(ax)=f(x)ϕ1(a), f(xb)=ϕ2(b)f(x). (2.8) It follows that

θ

a x

b

=

ϕ2(b) f(x) +uϕ1(a)−ϕ2(b)u ϕ1(a)

, (2.9)

whereu∈Mandϕ2:B→A,ϕ1:A→B, f :M→Mare maps satisfying that (i)ϕAis an Jordan isomorphism onAintoBand f(ax)=f(x)ϕ1(a), (ii)ϕBis an Jordan isomorphism onBintoAand f(xb)=ϕ2(b)f(x).

Hence, we have the following two identities:

fa1a2

x=f(x)ϕ1

a1a2

, fa1

a2x= fa2xϕ1

a1

=f(x)ϕ1

a2

ϕ1

a1

(2.10) for anya1,a2∈Aandx∈M. Consequently,

f(x)ϕ1

a1a2

=f(x)ϕ1

a2

ϕ1

a1

, (2.11)

which shows that ϕ1 is an anti-isomorphism fromA ontoB, sinceM is faithful. It is proved analogously thatϕ2is an anti- isomorphism fromBontoA. Finally, the preceding arguments allows us to get by simple calculus thatθis an antiautomorphism ofᐁ. This

completes the proof of the lemma.

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3. The Jordan derivations of generalized triangular matrix algebra In this section, we suppose thatᐁis the algebra of the form

ᐁ=

A M

B

, (3.1)

whereAand Bare unitalR-algebras andM is an (A,B)-bimodule. The first principal result of this paper is the following.

Theorem3.1. IfM is faithful as a leftA-module as well as a rightB-module, then any Jordan derivation ofᐁis an ordinary derivation.

Before proving this theorem, we need to describe all Jordan derivations ofᐁ. Lemma3.2. Every Jordan derivation∂ofᐁis of the from

∂

a x

b

=

gA(a) au−ub+f(x) gB(b)

, (3.2)

whereu∈MandgA:A→A,gB:B→B,f :M→Mare linear maps satisfying that (i)gAis a Jordan derivation ofAandf(ax)=gA(a)x+a f(x),

(ii)gBis a Jordan derivation ofBand f(xb)=xgB(b) +f(x)b.

Proof. Write

∂

a x

b

=

gA(a) +hB(b) +kA(x) fA(a) +fB(b) +f(x) hA(a) +gB(b) +kB(x)

, (3.3)

wheregA:A→A,hB:B→A,kA:M→A, fA:A→M, fB:B→M, f :M→M,hA:A→ B,gB:B→B, andkB:M→Bare clearly linear maps.

TakeX=(^{1 0}₀) andY=(⁰^x₀) in the equation

∂(X◦Y)=∂(X)◦Y+X◦∂(Y). (3.4)

We have∂(X◦Y)=(1/2)^k^A^(x)_k^f_B^(x)_(x), while 2∂(X)◦Y+X◦∂(Y)=

2kA(x) gA(1)x+xhA(1) +f(x) 0

. (3.5)

Hence,kA(x)=0 andkB(x)=0, which allow us to have

∂

a x

b

=

gA(a) +hB(b) fA(a) +fB(b) +f(x) hA(a) +gB(b)

. (3.6)

Putting now X=(^a⁰₀) and Y =(^a⁰₀), we obtain ∂(X◦Y)=_g_A_(a_◦_a) fA(a◦a) hA(a◦a)

and

∂(X)◦Y+X◦∂(Y)=(^gÂ^(a)^◦â^+a^◦^gÂ^(a⁾â^fÂ^{(a)+a f}₀ Â^(a⁾).

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Then,gA(a◦a)=gA(a)◦a+a◦gA(a), that is, gA is a Jordan derivation ofA, and hA(a◦a)=0. Replacingaby 1 in the last relation yieldshA(a)=0. Therefore,

∂

a x

b

=

gA(a) +hB(b) fA(a) +fB(b) +f(x) gB(b)

. (3.7)

Let nowX=(^{0 0}_b) andY=(^{0 0}_b). Then∂(X◦Y)=_h_B_(b_◦_b) fB(b◦b) gB(b◦b)

and

∂(X)◦Y+X◦∂(Y)=

0 fB(b)b+ fB(b)b gB(b)◦b+b◦gB(b)

, (3.8)

showing that

gB(b◦b)=gB(b)◦b+b◦gB(b), (3.9) which means thatgBis a Jordan derivation ofB, andhB(bb+bb)=0. Substituting now b=1 in the latter identity implieshB(b)=0. Consequently,∂(^{a x}b)=(^g^A^(a) ^f^A^(a)+f_g_B^B_(b)^(b)+f^(x)).

We continue with the same method by takingX=(^a_b⁰) andY =(^{1 0}₀). It furnishes that∂(X◦Y)=_g

A(2a) fA(2a) 0

and

∂(X)◦Y+X◦∂(Y)=

2gA(a) +gA(1)a+agA(1) fA(a) +fB(b) +fA(1)b+a fA(1) 0

=

2gA(a) fA(a) +fB(b) +fA(1)b+a fA(1) 0

,

(3.10) sincegA(1)=0.

Hence fA(2a)= fA(a) + fB(b) + fA(1)b+a fA(1), which implies clearly that fA(a)= a fA(1) and fB(b) +fA(1)b=0. So, fA(a)=auand fB(b)= −ub, whereu= fA(1).

It follows that

∂

a x

b

=

. (3.11)

Now, if we takeX=(^a⁰₀) andY=(¹^x₀), we find that

∂(X◦Y)=

gA(2a) 2au+f(ax) 0

,

∂(X)◦Y+X◦∂(Y)=

2gA(a) gA(a)x+ 2au+a f(x) 0

,

(3.12)

deducing the identityf(ax)=gA(a)x+a f(x).

Finally, puttingX=(^{0 0}_b) andY=(⁰^x₀) gives∂(XY+YX)=(⁰ ^f^(xb)₀ ) and

∂(X)◦Y+X◦∂(Y)=

0 xgB(b) +f(x)b 0

. (3.13)

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Hence,f(xb)=xgB(b) +f(x)b, which ends the proof of the lemma.

Now we are ready to establish our first principal theorem.

Proof ofTheorem 3.1. Let∂be a Jordan derivation ofᐁ, we have

∂

a x

b

=

, (3.14)

whereu∈M,gA:A→A,gB:B→B, and f :M→Mare linear maps satisfying that (i)gAis a Jordan derivation ofA,f(ax)=gA(a)x+a f(x),

(ii)gBis a Jordan derivation ofB, f(xb)=xgB(b) + f(x)b.

Hence, we have the following two identities:

f(aa)x=gA(aa)x+aaf(x),

fa(ax)=gA(a)ax+a f(ax)=gA(a)ax+agA(a)x+aaf(x). (3.15) As a consequence, we get that

gA(aa)x=gA(a)ax+agA(a)x. (3.16) SinceM is faithful,gA(aa)=gA(a)a+agA(a) andgAis a derivation ofA. A similar argument shows thatgBis a derivation ofB.

Finally, one can now easily check that∂is a derivation ofᐁ. Indeed, letX=(^{a x}_b) and Y=(^a^x_b) be arbitrary elements inᐁ. By straightforward computations, we have

∂(XY)=

gA(aa) aau−ubb+f(ax+xb) gB(bb)

,

∂(X)Y+X∂(Y)=

gA(a)a+agA(a) aau−ubb+gA(a)x+a f(x)+xgB(b)+f(x)b gB(b)b+bgB(b)

. (3.17) This shows that∂(XY)=∂(X)Y+X∂(Y) and concludes the proof of the theorem.

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[7] W. Cheung,Lie derivations of triangular algebras, Linear Multilinear Algebra51(2003), no. 3, 299–310.

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Aiat Hadj Ahmed Driss: Département de Mathématiques et Informatique, Faculté des Sciences, Université Abdelmalek Essaâdi, B.P. 2121, 93002 Tétouan, Maroc

E-mail address:ait [email protected]

Ben Yakoub l’Moufadal: Département de Mathématiques et Informatique, Faculté des Sciences, Université Abdelmalek Essaâdi, B.P. 2121, 93002 Tétouan, Maroc

E-mail address:[email protected]

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Call for Papers

Thinking about nonlinearity in engineering areas, up to the 70s, was focused on intentionally built nonlinear parts in order to improve the operational characteristics of a device or system. Keying, saturation, hysteretic phenomena, and dead zones were added to existing devices increasing their behavior diversity and precision. In this context, an intrinsic nonlinearity was treated just as a linear approximation, around equilibrium points.

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