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EXTRACTA

MATHEMATICAE

Volumen 33, Número 2, 2018

instituto de investigación de matemáticas de la

universidad de extremadura

EXTRACTA MATHEMATICAE

Vol. 34, Num. 1 (2019), 77 – 83

doi:10.17398/2605-5686.34.1.77

Available online February 20, 2019

Non-additive Lie centralizer of strictly
upper triangular matrices

Driss Aiat Hadj Ahmed

Centre Régional des Metiers d’Education et de Formation (CRMEF)

Tangier, Morocco

ait hadj@yahoo.com

Received December 31, 2018 Presented by Consuelo Mart́ınez
Accepted February 4, 2019

Abstract: Let F be a field of zero characteristic, let Nn(F) denote the algebra of n×n strictly upper
triangular matrices with entries in F, and let f : Nn(F) → Nn(F) be a non-additive Lie centralizer
of Nn(F), that is, a map satisfying that f([X,Y ]) = [f(X),Y ] for all X,Y ∈ Nn(F). We prove that
f(X) = λX + η (X) where λ ∈ F and η is a map from Nn(F) into its center Z (Nn(F)) satisfying
that η([X,Y ]) = 0 for every X,Y in Nn(F).

Key words: Lie centralizer, strictly upper triangular matrices, commuting map.

AMS Subject Class. (2010): 16S50, 15A27,16U80, 15B99, 47B47, 16R60.

1. Introduction

Consider a ring R. An additive mapping T : R → R is called a left
(respectively right) centralizer if T(ab) = T(a)b (respectively T(ab) = aT(b))
for all a,b ∈ R. The map T is called a centralizer if it is a left and a right
centralizer. The characterization of centralizers on algebras or rings has been
a widely discussed subject in various areas of mathematics.

In [13] Zalar proved the following interesting result: if R is a 2-torsion
free semiprime ring and T is an additive mapping such that T(a2) = T(a)a
(or T(a2) = aT(a)), then T is a centralizer. Vukman [12] considered additive
maps satisfying similar conditions, namely 2T(a2) = T(a)a + aT(a) for any
a ∈ R, and showed that if R is a 2-torsion free semiprime ring then T is also
a centralizer. Since then, the centralizers have been intensively investigated
by many mathematicians (see, e.g., [3, 4, 5, 6, 8]).

Let R be a ring. An additive map f : R → R, is called a Lie centralizer of
R if

f([x,y]) = [f(x),y] for all x,y ∈ R, (1.1)

where [x,y] is the Lie product of x and y.

ISSN: 0213-8743 (print), 2605-5686 (online)

https://doi.org/10.17398/2605-5686.34.1.77
mailto:ait_hadj@yahoo.com
https://www.eweb.unex.es/eweb/extracta/
https://creativecommons.org/licenses/by-nc/3.0/


78 driss aiat hadj ahmed

Recently, Ghomanjani and Bahmani [9] dealt with the structure of Lie
centralizers of trivial extension algebras, whereas Fošner and Jing [7] studied
Lie centralizers of triangular rings.

The inspiration of this paper comes from the articles [1, 5, 7] in which the
authors deal with the Lie centralizer maps of triangular algebras and rings.
In this note we will consider non-additive Lie centralizers on strictly upper
triangular matrices over a field of zero characteristic.

Throughout this article, F is a field of zero characteristic. Let Mn(F)
and Nn(F) denote the algebra of all n × n matrices and the algebra of
all n × n strictly upper triangular matrices over F, respectively. We use
diag(a1,a2, . . . ,an) to represent a diagonal matrix with diagonal (a1,a2, . . . ,an)
where ai ∈ F. The set of all n × n diagonal matrices over F is denoted by

Dn(F). Let In be the identity in Mn(F), J =
n−1∑
i=1

Ei,i+1 and {Eij : 1 ≤ i,j ≤ n}

the canonical basis of Mn(F), where Eij is the matrix with 1 in the (i,j) po-
sition and zeros elsewhere. By CNn(F)(X) we will denote the centralizer of
the element X in the ring Nn(F).

The notation f : Nn(F) → Nn(F) means a non-additive map satisfying
f([X,Y ]) = [f(X),Y ] for all X,Y ∈ Nn(F.

Notice that it is easy to check that Z (Nn(F)) = FE1n.
The main result in this paper is the following:

Theorem 1.1. Let F be a field of zero characteristic. If f : Nn(F) →
Nn(F) is a non-additive Lie centralizer then there exists λ ∈ F and a map
η : Nn(F) → Z (Nn(F)) satisfying η([X,Y ]) = 0 for every X,Y in Nn(F)
such that f(X) = λX + η (X) for all X in Nn(F).

Notice that the converse is trivially true: every map f(X) = λX + η (X)
with η satisfying the condition in Theorem 1.1 is a (non-additive) Lie central-
izer.

2. Proofs

Let’s start with some basic properties of Lie centralizers.

Lemma 2.1. Let f be a non-additive Lie centralizer of Nn(F). Then:

(1) f(0) = 0;

(2) for every X,Y ∈ Nn(F), we have f([X,Y ]) = [X,f(Y )];



non-additive lie centralizer 79

(3) f is a commuting map, i.e., f(X)X = Xf(X) for all X ∈ Nn(F).

Proof. To prove (1) it suffices to notice that

f(0) = f([0, 0]) = [f(0), 0] = 0.

(2) Observe that if f([X,Y ]) = [f(X),Y ], then we have

f(XY −Y X) = f(X)Y −Y f(X).

Interchanging X and Y in the above identity, we have

f(Y X −XY ) = f(Y )X −Xf(Y ).

Replacing X with −X in the above relation, we arrive at f(XY − Y X) =
Xf(Y ) −f(Y )X which can be written as f([X,Y ]) = [X,f(Y )].

From (1) one also gets (3):

[f(X),X] = f([X,X]) = f(0) = 0.

Remark 2.1. Let f be a non-additive Lie centralizer of Nn(F) and X ∈
CNn(F)(Y ). Then f(X) ∈ CNn(F)(Y ). Indeed, if X ∈ CNn(F)(Y ), then
[X,Y ] = 0 and

0 = f(0) = f([X,Y ]) = [f(X),Y ].

Lemma 2.2. Let f be a non-additive Lie centralizer of Nn(F). Then:

(1) f

(
n−1∑
i=1

aiEi,i+1

)
=

n−1∑
i=1

biEi,i+1;

(2) there exists λ ∈F such that f(J) = λJ.

Proof. Let D0 =
n∑
i=1

(n− i) Ei,i.

(1) Consider A ∈ Mn(F). It is well known that [D0,A] = A if and only if

A =
n−1∑
i=1

aiEi,i+1.

Hence, if A =
n−1∑
i=1

aiEi,i+1, we have [D0,A] = A. Thus f ([D0,A]) =

[D0,f (A)] = f (A). Therefore f(A) =
n−1∑
i=1

biEi,i+1.



80 driss aiat hadj ahmed

(2) As in (1), consider A =
n−1∑
i=1

aiEi,i+1 for some ai ∈F. Then [J,A] = 0

if and only if A = aJ for some a ∈F.

Indeed, f(J) =
n−1∑
i=1

aiEi,i+1 by (1). Thus, 0 = f(0) = f([J,J]) = [J,f (J)].

Hence, there exists λ ∈F such that f(J) = λJ.

We will need the following lemma.

Lemma 2.3. (Lemma 2.1, [14]) Suppose that F is an arbitrary field. If
G,H ∈ UTn(F) are such that gi,i+1 = hi,i+1 6= 0 for all 1 ≤ i ≤ n−1, then G
and H are conjugated in UTn(F).

Here UTn(F) is the multiplicative group of n×n upper triangular matrices
with only 1’s in the main diagonal. From the lemma above we obtain the
following corollary.

Corollary 2.1. Let F be a field. For every A =
∑

1≤i<j≤n
aijEij, where

ai,i+1 6= 0 for all 1 ≤ i ≤ n−1, there exists B ∈ Tn(F) such that B−1AB = J
and Tn(F) is the ring of upper triangular matrices.

Proof. Let A be a matrix in Nn(F) of the mentioned form. Then In + A
is a unitriangular matrix. Let’s notice first that there exists B1 ∈ Dn(F)
such that (B−11 AB1)i,i+1 = 1 for all i ∈ N. We can construct B1 ∈ Dn(F)
recursively by:

(B1)11 = 1, (B1)i+1,i+1 = (B1)ii · (Ai,i+1)−1 for i ≥ 1.

Consider the matrix In + B
−1
1 AB ∈ UTn(F). The unitriangular matrices

In + J and In + B
−1
1 AB fulfill the condition in Lemma 2.3. Hence, there

exists B2 ∈ UTn(F) such that

In + J = B
−1
2 (In + B

−1
1 AB1)B2.

Then J = B−12 (B
−1
1 AB1)B2. Taking B = B1B2 ∈ Tn(F), we get J = B

−1AB
as wanted.

Lemma 2.4. Let A =
∑

i<j aijEij be a matrix in Nn(F) with ai,i+1 6= 0
for every i = 1, . . . ,n− 1 . Then there exists λA ∈F such that f(A) = λAA.



non-additive lie centralizer 81

Proof. Since A =
∑

1≤i<j≤n
aijEij, where ai,i+1 6= 0, there exists T ∈ Tn(F)

such that TAT−1 = J by the previous corollary. Define h : Nn(F) → Nn(F)
by h(X) = Tf(T−1XT)T−1. Then h is a non-additive Lie centralizer. Indeed,
for all A,B ∈ Nn(F) we have:

h ([A,B]) = Tf
(
T−1[A,B]T

)
T−1

= Tf
(
T−1 (AB −BA) T

)
T−1

= Tf
(
T−1ATT−1BT −T−1BTT−1AT

)
T−1

= Tf
([
T−1AT,T−1BT

])
T−1

= T
[
f
(
T−1AT

)
,T−1BT

]
T−1

= T
(
f
(
T−1AT

)
T−1BT −T−1BTf

(
T−1AT

))
T−1

= Tf
(
T−1AT

)
T−1B −BTf

(
T−1AT

)
T−1

=
[
Tf
(
T−1AT

)
T−1,B

]
= [h(A),B] .

Hence, h(J) = λAJ by Lemma 2.2. Then

Tf(A)T−1 = Tf(T−1(TAT−1)T)T−1 = h(J) = λAJ = λATAT
−1.

Multiplying the left and right sides by T−1 and T respectively yields
f(A) = λAA.

Now we wish to extend Lemma 2.4 to all elements of Nn(F). In order to
do this, let’s introduce the following set:

S = {B = (bij) ∈ Nn(F) : bi,i+1 6= 0 ∀ i = 1, . . . ,n− 1} .

This set has an important property that is established below.

Lemma 2.5. Let F be a field. Every element of Nn(F) can be written as
a sum of at most two elements of S.

Proof. If ai,i+1 6= 0 for all i = 1, . . . ,n− 1, then A belongs to S, so there
is nothing to prove. If A is not in S, then we can define B1 and B2 as follows:

(B1)ij =

{
ai,i+1 − bi if j = i + 1,
aij if j > i + 1,

(B2)ij =

{
bi if j = i + 1,

0 otherwise,

where bi is an element in F different from ai,i+1. It is easy to see that B1, B2
are in S, and A = B1 + B2, so we wanted.



82 driss aiat hadj ahmed

Lemma 2.6. Let F be a field. For arbitrary elements A,B of Nn(F), there
exists λA,B ∈ F such that

f(A + B) = f(A) + f(B) + λA,BE1n.

Proof. For any A,B,X of Nn(F), we have

[f(A + B),X] = f ([A + B,X])

= [A + B,f(X)]

= [A,f(X)] + [B,f(X)]

= [f(A),X] + [f(B),X]

= [f(A) + f(B),X],

which implies that f(A + B) −f(A) −f(B) ∈Z (Nn(F)). Thus, there exists
λA,B ∈F such that f(A + B) = f(A) + f(B) + λA,BE1n.

Now we can prove the main theorem.

Proof of Theorem 1.1. Let A, B ∈ S be two non-commuting elements.
By Lemma 2.4, f(A) = λAA, f(B) = λBB , λA,λB ∈F.

Since f is a non-additive Lie centralizer, we get,

f ([A,B]) = [f (A) ,B] = λA[A,B]

= [A,f(B)] = λB[A,B].

Then, [A,B] 6= 0 implies that λA = λB.
If A, B ∈S commute, then we take C ∈S that does not commute neither

with A nor with B. As we have just seen, λA = λC and λB = λC. So
λA = λB = λ for arbitrary elements A,B ∈ S. Given X ∈ Nn(F) we know,
by Lemma 2.5, that there exists A,B ∈ S such that X = A + B (we can
assume that X /∈ S). Then f(X)−f(A)−f(B) ∈Z (Nn(F)) by Lemma 2.6.

That is f(X) − λAA − λBB = f(X) − λX ∈ Z (Nn(F)) for λ ∈ F such
that f(A) = λA for each A ∈ S.

We can define η : Nn(F) → Z (Nn(F)) such that η (X) = f(X) − λX,
that is, f(X) = λX + η (X).

Notice that η(A) = 0 for each A ∈ S. Furthermore, if X,Y ∈ Nn(F), then

f ([X,Y ]) = λ [X,Y ] + η ([X,Y ]) = [f (X) ,Y ]

= [λX + η (X) ,Y ] = λ [X,Y ] ,

since η (X) ∈Z (Nn(F)).
Consequently, η ([X,Y ]) = 0 and Theorem 1.1 is proved.



non-additive lie centralizer 83

Acknowledgements

The author would like to thank the referee for providing useful sug-
gestions which served to improve this paper.

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	Introduction
	Proofs