Lie Derivations on Generalized Matrix Algebras by Local Actions

Abstract

Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{B},\mathcal{M},\mathcal{N})$ be a generalized matrix algebra. A linear map $\Delta :\mathcal{G}\to \mathcal{G}$ is called a Lie derivation at $E\in \mathcal{G}$ if $\Delta \left(\right[U,V\left]\right)=[\Delta (U),V]+[U,\Delta (V\left)\right]$ for all pairs $U,V\in \mathcal{G}$ such that $UV=E$. In this paper, we use techniques of matrix decomposition and algebraic identity analysis to fully characterize the general form of Lie derivations at $E=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)$, where $e_0$ is an arbitrary fixed element in $\mathcal{A}$. Our main result establishes a necessary and sufficient condition for a Lie derivation at $E=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)$ to be decomposable into the sum of a derivation of $\mathcal{G}$ and a center-valued linear map. This characterization significantly extends the classical results concerning global Lie derivations and provides a deeper insight into the local Lie-type behavior in operator algebras.

1. Introduction

Let R be a commutative ring with identity, and let $\mathcal{A}$ be an algebra over R; we denote the center of $\mathcal{A}$ by $Z\left(\mathcal{A}\right)$. A linear map $d:\mathcal{A}\to \mathcal{A}$ is termed a derivation if $d\left(xy\right)=d\left(x\right)y+xd\left(y\right)$ for all $x,y\in \mathcal{A}$. For any $x,y\in \mathcal{A}$, the Lie bracket of x and y is defined by $[x,y]=xy-yx$. Correspondingly, a linear map $L:\mathcal{A}\to \mathcal{A}$ is called a Lie derivation if it satisfies the derivation property with respect to this Lie bracket, that is, $L\left(\right[x,y\left]\right)=\left[L\right(x),y]+[x,L(y\left)\right]$ for any $x,y\in \mathcal{A}.$ It is evident that every derivation is a Lie derivation. However, the converse does not generally hold. A canonical method for constructing Lie derivations is to consider maps of the form $L=d+\tau$, where d is a derivation and $\tau :\mathcal{A}\to Z\left(\mathcal{A}\right)$ is a linear map vanishing on all commutators $[x,y]$. Such Lie derivations are said to be of standard form.

The question of standard representation for Lie derivations has been extensively studied in various algebraic contexts [1,2,3,4,5,6,7,8]. Recently, research focus has shifted towards a more refined concept: characterizing Lie derivations by their local actions. Specifically, for a fixed point $E\in \mathcal{A}$, a linear map $\Delta$ is a Lie derivation at E if $\Delta \left(\right[x,y\left]\right)=[\Delta (x),y]+[x,\Delta (y\left)\right]$ for any $x,y\in \mathcal{A}$ with the property that $xy=E$. This local condition is strictly weaker than that of a global Lie derivation.

Significant progress has been made in understanding these local maps. The pioneering work of Lu and Jing [9] established that every Lie derivation at zero point or a fixed nontrivial idempotent P on a standard operator algebra $B\left(X\right)$ is standard. This result was subsequently extended to triangular algebras by Ji and Qi [10]. In a parallel direction, Du and Wang [6] generalized the zero-point case of [10] to generalized matrix algebras, while Yuan [11] later accomplished a similar generalization for the standard idempotent case initially treated in [10].

Despite these advances, the structure of Lie derivations at E on generalized matrix algebras remains largely unexplored. Existing results are confined to specific forms of E—such as the zero element 0 for [6] or the standard idempotent $\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ for [11]—and their analytical methods heavily rely on the particular properties of these elements. In this paper, we address this gap by investigating Lie derivations at $E=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)$ on a generalized matrix algebra $\mathcal{G}$, where $e_0\in \mathcal{A}$ is an arbitrary fixed element (not restricted to 0 or $1_{\mathcal{A}}$). Our main result establishes a necessary and sufficient condition for such a Lie derivation to be of the standard form. This work not only unifies the previous findings for specific points but also significantly expands the theory to a much broader class of local Lie derivations, thereby generalizing the results of [10,11].

2. Preliminaries

This section will provide some definitions and preliminary lemmas.

Definition 1.

Let $\mathcal{A}$ be an algebra over a commutative ring R. An algebraic ideal is a nonempty subset $S\subseteq \mathcal{A}$ that is both a ring-theoretic ideal and an R-submodule of $\mathcal{A}$. If S additionally satisfies $S\subseteq Z\left(\mathcal{A}\right)$, then S is called a central ideal.

Definition 2.

An algebra $\mathcal{X}$ is called 2-torsion fre if for any $x\in \mathcal{X}$, $2x=0$ implies $x=0$.

Definition 3.

Let R be a commutative ring with unity, and let $\mathcal{A}$ and $\mathcal{B}$ be R-algebras. Consider an $(\mathcal{A},\mathcal{B})$-bimodule $\mathcal{M}$ and a $(\mathcal{B},\mathcal{A})$-bimodule $\mathcal{N}$. Suppose there exist bimodule homomorphisms

$$\Phi :\mathcal{M}{\otimes }_{\mathcal{B}}\mathcal{N}\to \mathcal{A}\mathit{and}\Psi :\mathcal{N}{\otimes }_{\mathcal{A}}\mathcal{M}\to \mathcal{B},$$

that satisfy the following compatibility relations for all $m,m^{\prime }\in \mathcal{M}$ and $n,n^{\prime }\in \mathcal{N}$:

$$\begin{array}{cc}\hfill \Phi (m\otimes n)·m^{\prime }& =m·\Psi (n\otimes m^{\prime })\hfill \\ \hfill \Psi (n\otimes m)·n^{\prime }& =n·\Phi (m\otimes n^{\prime }).\hfill \end{array}$$

Under these assumptions, the set of all formal $2\times 2$ matrices

$$\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{B},\mathcal{M},\mathcal{N})=\left\{\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}:a\in \mathcal{A},b\in \mathcal{B},m\in \mathcal{M},n\in \mathcal{N}\right\}$$

forms an R-algebra when equipped with usual matrix operations. Specifically, matrix multiplication is defined by:

$$\left(\begin{array}{cc}{a}_1& m_1\\ n_1& b_1\end{array}\right)\left(\begin{array}{cc}{a}_2& m_2\\ n_2& b_2\end{array}\right)=\left(\begin{array}{cc}{a}_1{a}_2+\Phi (m_1\otimes n_2)& a_1{m}_2+m_1{b}_2\\ b_1{n}_2+n_1{a}_2& \Psi (n_1\otimes m_2)+b_1{b}_2\end{array}\right).$$

This R-algebra $\mathcal{G}$ is referred to as a generalized matrix algebra.

When $\mathcal{N}=0$ (the zero bimodule), this construction simplifies to a triangular algebra, denoted $\mathcal{T}=\mathcal{T}(\mathcal{A},\mathcal{M},\mathcal{B})$. Notable examples of generalized matrix algebras include full matrix algebras $M_n\left(R\right)$ (for $n\ge 2$) and all triangular algebras.

Definition 4.

An $(\mathcal{A},\mathcal{B})$-bimodule $\mathcal{M}$ is called faithful if it satisfies two conditions: (i) for all $a\in \mathcal{A}$, $a\mathcal{M}=0$ implies $a=0$; (ii) for all $b\in \mathcal{B}$, $\mathcal{M}b=0$ implies $b=0$.

We define the natural projection maps ${\pi }_{\mathcal{A}}:\mathcal{G}\to \mathcal{A}$ and ${\pi }_{\mathcal{B}}:\mathcal{G}\to \mathcal{B}$ by

$${\pi }_{\mathcal{A}}\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}=aand{\pi }_{\mathcal{B}}\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}=b,$$

where $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$.

Lemma 1

([12]). Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra. If $\mathcal{M}$ be a faithful $(\mathcal{A},\mathcal{B})-$bimodule, then

$$Z\left(\mathcal{G}\right)=\left.\left\{\left.\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)\right|\right.am=mb,na=bn,\forall m\in \mathcal{M},n\in \mathcal{N}\right\}.$$

Furthermore, there exists a unique algebra isomorphism $\eta :{\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)\to {\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right)$ satisfying $am=m\eta \left(a\right)$ and $na=\eta \left(a\right)n$ for all $m\in \mathcal{M},n\in \mathcal{N}.$

Lemma 2

([13]). Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra, and let $d:\mathcal{G}\to \mathcal{G}$ be a map. Then d is a derivation of $\mathcal{G}$ if and only if, for any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$,

$$d\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{d}_{11}\left(a\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\gamma }_{21}\left(n\right)& d_{22}\left(b\right)+n_{0}m+n{m}_0\end{array}\right),$$

where $m_0\in \mathcal{M}$, $n_0\in \mathcal{N}$, $d_{11}$ is a derivation of $\mathcal{A}$, $d_{22}$ is a derivation of $\mathcal{B}$, and the linear maps ${\mu }_{12}:\mathcal{M}\to \mathcal{M}$, ${\gamma }_{21}:\mathcal{N}\to \mathcal{N}$ satisfy the following conditions:

(i) 

$d_{11}\left(mn\right)={\mu }_{12}\left(m\right)n+m{\gamma }_{21}\left(n\right)$;

(ii) 

$d_{22}\left(nm\right)=n{\mu }_{12}\left(m\right)+{\gamma }_{21}\left(n\right)m$;

(iii) 

${\mu }_{12}\left(am\right)=d_{11}\left(a\right)m+a{\mu }_{12}\left(m\right)$ and ${\mu }_{12}\left(mb\right)={\mu }_{12}\left(m\right)b+m{d}_{22}\left(b\right)$;

(iv) 

${\gamma }_{21}\left(na\right)={\gamma }_{21}\left(n\right)a+n{d}_{11}\left(a\right)$ and ${\gamma }_{21}\left(bn\right)=d_{22}\left(b\right)n+b{\gamma }_{21}\left(n\right)$.

Throughout this paper, all algebras are assumed to be 2-torsion free, and the $(\mathcal{A},\mathcal{B})$-bimodule $\mathcal{M}$ is taken to be faithful. For notation consistency: $\mathbb{Z}$ denotes the ring of integers; for any algebra $\mathcal{X}$, $Z\left(\mathcal{X}\right)$ stands for the center of $\mathcal{X}$; diagonal matrices are denoted by $a\oplus b:=\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)$; and $\Delta \left(\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\right)$ is noted as $\Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)$.

3. Main Results and Proofs

This section characterizes the Lie derivations at $E=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)\in \mathcal{G}$, where $e_0$ is an arbitrary fixed element in $\mathcal{A}$. The proof of our main results (Theorems 1 and 2) is structured into three key stages:

1. Block-wise decomposition of $\Delta$: We first derive the explicit matrix-block structure of $\Delta$, leading to the general form given in Equation (1).

2. Extraction of the inner derivation via $U_0$: Using a suitably chosen element $U_0$, we remove the inner derivation component and verify the compatibility conditions (i)–(iii) stated in Theorem 1.

3. Construction and verification of d and $\tau$: We explicitly construct a derivation d and a center-valued map $\tau$, and rigorously confirm that $\Delta =d+\tau$ under the given hypotheses. 

We now proceed to the detailed proof of Theorem 1.

Theorem 1.

Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra, and let $\Delta :\mathcal{G}\to \mathcal{G}$ be a Lie derivation at E. Assume that, for each $a\in \mathcal{A}$, $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$. Then

$$\Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\xi }_{11}\left(b\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(a\right)+{\xi }_{22}\left(b\right)+n_{0}m+n{m}_0\end{array}\right)$$

(1)

for any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$, where $m_0\in \mathcal{M}$, $n_0\in \mathcal{N}$, and the linear maps ${\kappa }_{11}:\mathcal{A}\to \mathcal{A}$, ${\xi }_{11}:\mathcal{B}\to Z\left(\mathcal{A}\right)$, ${\mu }_{12}:\mathcal{M}\to \mathcal{M}$, ${\lambda }_{21}:\mathcal{N}\to \mathcal{N}$, ${\kappa }_{22}:\mathcal{A}\to Z\left(\mathcal{B}\right)$ and ${\xi }_{22}:\mathcal{B}\to \mathcal{B}$ satisfy the following conditions:

(i) 

${\mu }_{12}\left(am\right)={\kappa }_{11}\left(a\right)m-m{\kappa }_{22}\left(a\right)+a{\mu }_{12}\left(m\right)$, ${\mu }_{12}\left(mb\right)=m{\xi }_{22}\left(b\right)-{\xi }_{11}\left(b\right)m+{\mu }_{12}\left(m\right)b$;

(ii) 

${\lambda }_{21}\left(na\right)=n{\kappa }_{11}\left(a\right)-{\kappa }_{22}\left(a\right)n+{\lambda }_{21}\left(n\right)a$, ${\lambda }_{21}\left(bn\right)={\xi }_{22}\left(b\right)n-n{\xi }_{11}\left(b\right)+b{\lambda }_{21}\left(n\right)$;

(iii) 

${\kappa }_{11}\left(mn\right)-{\xi }_{11}\left(nm\right)={\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right),{\xi }_{22}\left(nm\right)-{\kappa }_{22}\left(mn\right)=n{\mu }_{12}\left(m\right)+{\lambda }_{21}\left(n\right)m$.

Proof of Theorem 1.

Let $\Delta :\mathcal{G}\to \mathcal{G}$ be a Lie derivation at E. Its linearity implies the existence of linear maps ${\kappa }_{ij}:\mathcal{A}\to {\mathcal{G}}_{ij}$, ${\mu }_{ij}:\mathcal{M}\to {\mathcal{G}}_{ij}$, ${\lambda }_{ij}:\mathcal{N}\to {\mathcal{G}}_{ij}$ and ${\xi }_{ij}:\mathcal{B}\to {\mathcal{G}}_{ij}(i,j=1,2)$ such that

$$\begin{array}{c}\hfill \Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\mu }_{11}\left(m\right)+{\lambda }_{11}\left(n\right)+{\xi }_{11}\left(b\right)& {\kappa }_{12}\left(a\right)+{\mu }_{12}\left(m\right)+{\lambda }_{12}\left(n\right)+{\xi }_{12}\left(b\right)\\ {\kappa }_{21}\left(a\right)+{\mu }_{21}\left(m\right)+{\lambda }_{21}\left(n\right)+{\xi }_{21}\left(b\right)& {\kappa }_{22}\left(a\right)+{\mu }_{22}\left(m\right)+{\lambda }_{22}\left(n\right)+{\xi }_{22}\left(b\right)\end{array}\right)\end{array}$$

for any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$, where ${\mathcal{G}}_{11}=\mathcal{A},{\mathcal{G}}_{12}=\mathcal{M},{\mathcal{G}}_{21}=\mathcal{N},{\mathcal{G}}_{22}=\mathcal{B}$.

For any $U,V\in \mathcal{G}$, if $UV=E$, then

$$\Delta \left(\right[U,V\left]\right)=[\Delta (U),V]+[U,\Delta (V\left)\right].$$

(2)

By taking $U=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ in (2), we get

$$\Delta \left(\right[U,V\left]\right)=O$$

and

$$[\Delta \left(U\right),V]+[U,\Delta \left(V\right)]=\left[\Delta \left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right],$$

and so

$$\left[\Delta \left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]=O.$$

(3)

By taking $U=\left(\begin{array}{cc}{e}_0& m\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ in (2),using Equation (3), we deduce that

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}-{\mu }_{11}\left(m\right)& -{\mu }_{12}\left(m\right)\\ -{\mu }_{21}\left(m\right)& -{\mu }_{22}\left(m\right)\end{array}\right)& =\left[\Delta \left(\begin{array}{cc}{e}_0& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0& m\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]\hfill \\ \hfill & =\left[\Delta \left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}m{\kappa }_{21}\left(1_{\mathcal{A}}\right)& -{\mu }_{12}\left(m\right)+m{\kappa }_{22}\left(1_{\mathcal{A}}\right)-{\kappa }_{11}\left(1_{\mathcal{A}}\right)m\\ {\mu }_{21}\left(m\right)& -{\kappa }_{21}\left(1_{\mathcal{A}}\right)m\end{array}\right).\hfill \end{array}$$

Then

$${\mu }_{11}\left(m\right)=-m{\kappa }_{21}\left(1_{\mathcal{A}}\right),{\mu }_{22}\left(m\right)={\kappa }_{21}\left(1_{\mathcal{A}}\right)m$$

and

$$-{\mu }_{21}\left(m\right)={\mu }_{21}\left(m\right),-{\mu }_{12}\left(m\right)=-{\mu }_{12}\left(m\right)+m{\kappa }_{22}\left(1_{\mathcal{A}}\right)-{\kappa }_{11}\left(1_{\mathcal{A}}\right)m.$$

This implies

$$m{\kappa }_{22}\left(1_{\mathcal{A}}\right)={\kappa }_{11}\left(1_{\mathcal{A}}\right)m.$$

(4)

From $-{\mu }_{21}\left(m\right)={\mu }_{21}\left(m\right)$, we have $2{\mu }_{21}\left(m\right)=0$. Since $\mathcal{G}$ is 2-torsion free, then ${\mu }_{21}\left(m\right)=0$. Let ${\kappa }_{21}\left(1_{\mathcal{A}}\right)=n_0$. Then ${\mu }_{11}\left(m\right)=-m{n}_0,{\mu }_{22}\left(m\right)=n_{0}m$, and so

$$\Delta \left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}-m{n}_0& {\mu }_{12}\left(m\right)\\ 0& n_{0}m\end{array}\right)$$

for every $m\in \mathcal{M}$.

By taking $U=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{e}_0& 0\\ n& 0\end{array}\right)$ in (2), using Equation (3), we get

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}-{\lambda }_{11}\left(n\right)& -{\lambda }_{12}\left(n\right)\\ -{\lambda }_{21}\left(n\right)& -{\lambda }_{22}\left(n\right)\end{array}\right)& =\left[\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{e}_0& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{e}_0& 0\\ n& 0\end{array}\right)\right]\hfill \\ \hfill & =\left[\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}{\kappa }_{12}\left(1_{\mathcal{A}}\right)n& {\lambda }_{12}\left(n\right)\\ -{\lambda }_{21}\left(n\right)+{\kappa }_{22}\left(1_{\mathcal{A}}\right)n-n{\kappa }_{11}\left(1_{\mathcal{A}}\right)& -n{\kappa }_{12}\left(1_{\mathcal{A}}\right)\end{array}\right).\hfill \end{array}$$

Then

$${\lambda }_{11}\left(n\right)=-{\kappa }_{12}\left(1_{\mathcal{A}}\right)n,{\lambda }_{12}\left(n\right)=0,{\lambda }_{22}\left(n\right)=n{\kappa }_{12}\left(1_{\mathcal{A}}\right)$$

and

$$-{\lambda }_{12}\left(n\right)={\lambda }_{12}\left(n\right),-{\lambda }_{21}\left(n\right)=-{\lambda }_{21}\left(n\right)+{\kappa }_{22}\left(1_{\mathcal{A}}\right)n-n{\kappa }_{11}\left(1_{\mathcal{A}}\right).$$

This implies

$${\kappa }_{22}\left(1_{\mathcal{A}}\right)n=n{\kappa }_{11}\left(1_{\mathcal{A}}\right).$$

(5)

From $-{\lambda }_{12}\left(n\right)={\lambda }_{12}\left(n\right)$, we have $2{\lambda }_{12}\left(n\right)=0$. Since $\mathcal{G}$ is 2-torsion free, then ${\lambda }_{12}\left(n\right)=0$. Let ${\kappa }_{12}\left(1_{\mathcal{A}}\right)=m_0$. Then ${\lambda }_{11}\left(n\right)=-m_{0}n,{\lambda }_{22}\left(n\right)=n{m}_0$, and so

$$\Delta \left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)=\left(\begin{array}{cc}-m_{0}n& 0\\ {\lambda }_{21}\left(n\right)& n{m}_0\end{array}\right)$$

for all $n\in \mathcal{N}$.

By using Lemma 1, we may deduce from (4) and (5) that ${\kappa }_{11}\left(1_{\mathcal{A}}\right)\oplus {\kappa }_{22}\left(1_{\mathcal{A}}\right)\in Z\left(\mathcal{G}\right)$.

For any invertible element $a\in \mathcal{A}$, we put $U=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)$ in (2), we have

$$\Delta \left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right]=\left[\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right].$$

Specifically, let $b=0$, we have

$$\Delta \left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right]=\left[\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right].$$

Since

$$\Delta \left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right]=\Delta \left(\begin{array}{cc}{e}_0-a^{-1}{e}_{0}a& 0\\ 0& 0\end{array}\right)=\Delta \left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right],$$

we have

$$\begin{array}{cc}\hfill & \left[\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& b\end{array}\right)\right]\hfill \\ \hfill =& \left[\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right],\hfill \end{array}$$

and so

$$\left[\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\Delta \left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right)\right]=0.$$

(6)

By using $\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)& {\kappa }_{12}\left(a\right)\\ {\kappa }_{21}\left(a\right)& {\kappa }_{22}\left(a\right)\end{array}\right)$ and $\Delta \left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{\xi }_{11}\left(b\right)& {\xi }_{12}\left(b\right)\\ {\xi }_{21}\left(b\right)& {\xi }_{22}\left(b\right)\end{array}\right)$ in (6), we have

$$\left(\begin{array}{cc}a{\xi }_{11}\left(b\right)-{\xi }_{11}\left(b\right)a& {\kappa }_{12}\left(a\right)b+a{\xi }_{12}\left(b\right)\\ -b{\kappa }_{21}\left(a\right)-{\xi }_{21}\left(b\right)a& {\kappa }_{22}\left(a\right)b-b{\kappa }_{22}\left(a\right)\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$$

Then

$$\begin{array}{cc}\hfill & a{\xi }_{11}\left(b\right)-{\xi }_{11}\left(b\right)a=0,\hfill \\ \hfill & {\kappa }_{12}\left(a\right)b+a{\xi }_{12}\left(b\right)=0,\hfill \\ \hfill & -b{\kappa }_{21}\left(a\right)-{\xi }_{21}\left(b\right)a=0,\hfill \\ \hfill & {\kappa }_{22}\left(a\right)b-b{\kappa }_{22}\left(a\right)=0,\hfill \end{array}$$

and so

$$\begin{array}{cc}\hfill & a{\xi }_{11}\left(b\right)={\xi }_{11}\left(b\right)a,\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & a{\xi }_{12}\left(b\right)=-{\kappa }_{12}\left(a\right)b,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & {\xi }_{21}\left(b\right)a=-b{\kappa }_{21}\left(a\right),\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & {\kappa }_{22}\left(a\right)b=b{\kappa }_{22}\left(a\right).\hfill \end{array}$$

(10)

By taking $a=1_{\mathcal{A}}$ in (8) and (9), we deduce that

$${\xi }_{12}\left(b\right)=-{\kappa }_{12}\left(1_{\mathcal{A}}\right)b=-m_{0}b,{\xi }_{21}\left(b\right)=-b{\kappa }_{21}\left(1_{\mathcal{A}}\right)=-b{n}_0$$

for any $b\in \mathcal{B}$. This implies ${\xi }_{12}\left(1_{\mathcal{B}}\right)=-m_0,{\xi }_{21}\left(1_{\mathcal{B}}\right)=-n_0$. Thus

$$\Delta \left(\begin{array}{cc}0& 0\\ 0& b\end{array}\right)=\left(\begin{array}{cc}{\xi }_{11}\left(b\right)& -m_{0}b\\ -b{n}_0& {\xi }_{22}\left(b\right)\end{array}\right)$$

for all $b\in \mathcal{B}$.

Using a similar method, by putting $b=1_{\mathcal{B}}$ in (8) and (9), one can obtain

$${\kappa }_{12}\left(a\right)=-a{\xi }_{12}\left(1_{\mathcal{B}}\right)=a{m}_0,{\kappa }_{21}\left(a\right)=-{\xi }_{21}\left(1_{\mathcal{B}}\right)a=n_{0}a$$

(11)

for any invertible element $a\in \mathcal{A}$.

If a is non-invertible in $\mathcal{A}$, by assumption of Theorem 1, then $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$.

By (7), $k{I}_{\mathcal{A}}{\xi }_{11}\left(b\right)={\xi }_{11}\left(b\right)\left(k{I}_{\mathcal{A}}\right),(k{I}_{\mathcal{A}}-a){\xi }_{11}\left(b\right)={\xi }_{11}\left(b\right)(k{I}_{\mathcal{A}}-a)$. Then

$$a{\xi }_{11}\left(b\right)=k{I}_A{\xi }_{11}\left(b\right)-(k{I}_{\mathcal{A}}-a){\xi }_{11}\left(b\right)={\xi }_{11}\left(b\right)\left(k{I}_{\mathcal{A}}\right)-{\xi }_{11}\left(b\right)(k{I}_{\mathcal{A}}-a)={\xi }_{11}\left(b\right)a.$$

Hence $a{\xi }_{11}\left(b\right)={\xi }_{11}\left(b\right)a$ for any $a\in \mathcal{A}$. This implies ${\xi }_{11}\left(b\right)\in Z\left(\mathcal{A}\right)$ for any $b\in \mathcal{B}$.

From (10), we have ${\kappa }_{22}(k{I}_{\mathcal{A}}-a)b=b{\kappa }_{22}(I_{\mathcal{A}}-a),{\kappa }_{22}\left(I_{\mathcal{A}}\right)b=b{\kappa }_{22}\left(k{I}_{\mathcal{A}}\right)$. By using the linearity of ${\kappa }_{22}$, we get

$${\kappa }_{22}\left(a\right)b=k{\kappa }_{22}\left(I_{\mathcal{A}}\right)b-{\kappa }_{22}(k{I}_{\mathcal{A}}-a)b=b{\kappa }_{22}\left(k{I}_{\mathcal{A}}\right)-b{\kappa }_{22}(k{I}_{\mathcal{A}}-a)=b{\kappa }_{22}\left(a\right).$$

Then ${\kappa }_{22}\left(a\right)b=b{\kappa }_{22}\left(a\right)$ for any $a\in \mathcal{A}$. Therefore, ${\kappa }_{22}\left(a\right)\in Z\left(\mathcal{B}\right)$ for any $a\in \mathcal{A}$.

Similarly, from (11), one can obtain ${\kappa }_{12}\left(a\right)=a{m}_0,{\kappa }_{21}\left(a\right)=n_{0}a$ for any $a\in \mathcal{A}$. Thus

$$\Delta \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)& a{m}_0\\ n_{0}a& {\kappa }_{22}\left(a\right)\end{array}\right)$$

for any $a\in \mathcal{A}$.

In conclusion,

$$\Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\xi }_{11}\left(b\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(a\right)+{\xi }_{22}\left(b\right)+n{m}_0+n_{0}m\end{array}\right)$$

for any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$. This proves (1).

Let $U_0=\left(\begin{array}{cc}0& -m_0\\ n_0& 0\end{array}\right)$. Define a map $L:\mathcal{G}\to \mathcal{G}$ as $L\left(U\right)=\Delta \left(U\right)-[U_0,U]$. Then

$$L\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\xi }_{11}\left(b\right)& {\mu }_{12}\left(m\right)\\ {\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(a\right)+{\xi }_{22}\left(b\right)\end{array}\right)$$

for any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$. By applying $L\left(U\right)=\Delta \left(U\right)-[U_0,U]$, it is straightforward to verify that L is a linear map and

$$L\left(\right[U,V\left]\right)=\left[L\right(U),V]+[U,L(V\left)\right]$$

(12)

for all $U,V\in \mathcal{G}$ with $UV=E$. The following proves that the conditions (i)–(iii) are also valid.

For any invertible element $a\in \mathcal{A}$, we put $U=\left(\begin{array}{cc}{e}_0{a}^{-1}& m\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$ in (12), we have

$$L\left(\begin{array}{cc}{e}_0-a{e}_0{a}^{-1}& -am\\ 0& 0\end{array}\right)=\left[L\left(\begin{array}{cc}{e}_0{a}^{-1}& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0{a}^{-1}& m\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right].$$

(13)

Specifically, let $m=0$, we have

$$L\left(\begin{array}{cc}{e}_0-a{e}_0{a}^{-1}& 0\\ 0& 0\end{array}\right)=\left[L\left(\begin{array}{cc}{e}_0{a}^{-1}& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0{a}^{-1}& 0\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right].$$

(14)

Combining (13) with (14), we obtain

$$L\left(\begin{array}{cc}0& -am\\ 0& 0\end{array}\right)=\left[L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\right].$$

By using $L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)& 0\\ 0& {\kappa }_{22}\left(a\right)\end{array}\right)$ and $L\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& {\mu }_{12}\left(m\right)\\ 0& 0\end{array}\right)$ in the above equation, we have

$$\left(\begin{array}{cc}0& -{\mu }_{12}\left(am\right)\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& -a{\mu }_{12}\left(m\right)+m{\kappa }_{22}\left(a\right)-{\kappa }_{11}\left(a\right)m\\ 0& 0\end{array}\right).$$

Then ${\mu }_{12}\left(am\right)={\kappa }_{11}\left(a\right)m-m{\kappa }_{22}\left(a\right)+a{\mu }_{12}\left(m\right)$ for any invertible element $a\in \mathcal{A}$.

If a is non-invertible in $\mathcal{A}$, by assumption of Theorem 1, then $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$, and so

$${\mu }_{12}\left((k{I}_{\mathcal{A}}-a)m\right)={\kappa }_{11}(k{I}_{\mathcal{A}}-a)m-m{\kappa }_{22}(k{I}_{\mathcal{A}}-a)+(k{I}_{\mathcal{A}}-a){\mu }_{12}\left(m\right).$$

By using the linearity of ${\mu }_{12},{\kappa }_{11}$ and ${\kappa }_{22}$, we get

$$\begin{array}{cc}\hfill {\mu }_{12}\left(am\right)& ={\mu }_{12}(km-(k{I}_{\mathcal{A}}-a)m)=k{\mu }_{12}\left(m\right)-{\mu }_{12}\left((k{I}_{\mathcal{A}}-a)m\right)\hfill \\ \hfill & =k{\mu }_{12}\left(m\right)-({\kappa }_{11}(k{I}_{\mathcal{A}}-a)m-m{\kappa }_{22}(k{I}_{\mathcal{A}}-a)+(k{I}_{\mathcal{A}}-a){\mu }_{12}\left(m\right))\hfill \\ \hfill & =-{\kappa }_{11}(k{I}_{\mathcal{A}}-a)m+m{\kappa }_{22}(k{I}_{\mathcal{A}}-a)+a{\mu }_{12}\left(m\right)\hfill \\ \hfill & =-k({\kappa }_{11}\left(I_{\mathcal{A}}\right)m-m{\kappa }_{22}\left(I_{\mathcal{A}}\right))+{\kappa }_{11}\left(a\right)m+m{\kappa }_{22}\left(a\right)+a{\mu }_{12}\left(m\right)\hfill \\ \hfill & ={\kappa }_{11}\left(a\right)m+m{\kappa }_{22}\left(a\right)+a{\mu }_{12}\left(m\right)\left(\mathrm{by}\left(4\right)\right).\hfill \end{array}$$

Then

$${\mu }_{12}\left(am\right)={\kappa }_{11}\left(a\right)m-m{\kappa }_{22}\left(a\right)+a{\mu }_{12}\left(m\right)$$

(15)

for all $a\in \mathcal{A}$.

For any $b\in \mathcal{B}$, we put $U=\left(\begin{array}{cc}{1}_{\mathcal{A}}& m\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{e}_0& -mb\\ 0& b\end{array}\right)$ in (12), and by using the fact ${\kappa }_{11}\left(1_{\mathcal{A}}\right)\oplus {\kappa }_{22}\left(1_{\mathcal{A}}\right)\in Z\left(\mathcal{G}\right)$, we have

$$\begin{array}{cc}\hfill & \left(\begin{array}{cc}0& -{\mu }_{12}\left(e_{0}m\right)\\ 0& 0\end{array}\right)=L\left(\begin{array}{cc}0& -e_{0}m\\ 0& 0\end{array}\right)=L\left([U,V]\right)=[L\left(U\right),V]+[U,L\left(V\right)]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}{\kappa }_{11}\left(1_{\mathcal{A}}\right)& {\mu }_{12}\left(m\right)\\ 0& {\kappa }_{22}\left(1_{\mathcal{A}}\right)\end{array}\right),\left(\begin{array}{cc}{e}_0& -mb\\ 0& b\end{array}\right)\right]+\left[\left(\begin{array}{cc}{1}_{\mathcal{A}}& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)+{\xi }_{11}\left(b\right)& -{\mu }_{12}\left(mb\right)\\ 0& {\kappa }_{22}\left(e_0\right)+{\xi }_{22}\left(b\right)\end{array}\right)\right]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}0& {\mu }_{12}\left(m\right)\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{e}_0& -mb\\ 0& b\end{array}\right)\right]+\left[\left(\begin{array}{cc}{1}_{\mathcal{A}}& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)+{\xi }_{11}\left(b\right)& -{\mu }_{12}\left(mb\right)\\ 0& {\kappa }_{22}\left(e_0\right)+{\xi }_{22}\left(b\right)\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}0& {\mu }_{12}\left(m\right)b-e_0{\mu }_{12}\left(m\right)-{\mu }_{12}\left(mb\right)+m{\kappa }_{22}\left(e_0\right)+m{\xi }_{22}\left(b\right)-{\kappa }_{11}\left(e_0\right)m-{\xi }_{11}\left(b\right)m\\ 0& 0\end{array}\right),\hfill \end{array}$$

and so

$$-{\mu }_{12}\left(e_{0}m\right)={\mu }_{12}\left(m\right)b-e_0{\mu }_{12}\left(m\right)-{\mu }_{12}\left(mb\right)+m{\kappa }_{22}\left(e_0\right)+m{\xi }_{22}\left(b\right)-{\kappa }_{11}\left(e_0\right)m-{\xi }_{11}\left(b\right)m.$$

From (15), we have ${\mu }_{12}\left(e_{0}m\right)={\kappa }_{11}\left(e_0\right)m-m{\kappa }_{22}\left(e_0\right)+e_0{\mu }_{12}\left(m\right)$, thus

$${\mu }_{12}\left(mb\right)=m{\xi }_{22}\left(b\right)-{\xi }_{11}\left(b\right)m+{\mu }_{12}\left(m\right)b$$

for any $b\in \mathcal{B}$. This proves (i).

For any invertible element $a\in \mathcal{A}$, we put $U=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ n& 0\end{array}\right)$ in (12), we have

$$L\left(\begin{array}{cc}{e}_0-a^{-1}{e}_{0}a& 0\\ -na& 0\end{array}\right)=\left[L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ n& 0\end{array}\right)\right].$$

(16)

Specifically, let $n=0$, we have

$$L\left(\begin{array}{cc}{e}_0-a^{-1}{e}_{0}a& 0\\ 0& 0\end{array}\right)=\left[L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}{a}^{-1}{e}_0& 0\\ 0& 0\end{array}\right)\right].$$

(17)

Combining (16) with (17), we obtain

$$L\left(\begin{array}{cc}0& 0\\ -na& 0\end{array}\right)=\left[L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),L\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\right].$$

By using $L\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)& 0\\ 0& {\kappa }_{22}\left(a\right)\end{array}\right)$ and $L\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ {\lambda }_{21}\left(n\right)& 0\end{array}\right)$ in the above equation, we have

$$\left(\begin{array}{cc}0& 0\\ -{\lambda }_{21}\left(na\right)& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ {\kappa }_{22}\left(a\right)n-n{\kappa }_{11}\left(a\right)-{\lambda }_{21}\left(n\right)a& 0\end{array}\right).$$

Then ${\lambda }_{21}\left(na\right)=n{\kappa }_{11}\left(a\right)-{\kappa }_{22}\left(a\right)n+{\lambda }_{21}\left(n\right)a$ for any invertible element $a\in \mathcal{A}$.

If a is non-invertible in $\mathcal{A}$, by assumption of Theorem 1, then $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$, and so

$${\lambda }_{21}\left(n(k{I}_{\mathcal{A}}-a)\right)=n{\kappa }_{11}(k{I}_{\mathcal{A}}-a)-{\kappa }_{22}(k{I}_{\mathcal{A}}-a)n+{\lambda }_{21}\left(n\right)(k{I}_{\mathcal{A}}-a).$$

By using the linearity of ${\lambda }_{21},{\kappa }_{11}$ and ${\kappa }_{22}$, we get

$$\begin{array}{cc}\hfill {\lambda }_{21}\left(na\right)& ={\lambda }_{21}(kn-\left(n(k{I}_{\mathcal{A}}-a)\right))=k{\lambda }_{21}\left(n\right)-{\lambda }_{21}\left(n(k{I}_{\mathcal{A}}-a)\right)\hfill \\ \hfill & =k{\lambda }_{21}\left(n\right)-(n{\kappa }_{11}(k{I}_{\mathcal{A}}-a)-{\kappa }_{22}(k{I}_{\mathcal{A}}-a)n+{\lambda }_{21}\left(n\right)(k{I}_{\mathcal{A}}-a))\hfill \\ \hfill & =-n{\kappa }_{11}(k{I}_{\mathcal{A}}-a)+{\kappa }_{22}(k{I}_{\mathcal{A}}-a)n+{\lambda }_{21}\left(na\right)\hfill \\ \hfill & =-k(n{\kappa }_{11}\left(I_{\mathcal{A}}\right)-{\kappa }_{22}\left(I_{\mathcal{A}}\right)n)+n{\kappa }_{11}\left(a\right)-{\kappa }_{22}\left(a\right)n+{\lambda }_{21}\left(na\right)\hfill \\ \hfill & =n{\kappa }_{11}\left(a\right)-{\kappa }_{22}\left(a\right)n+{\lambda }_{21}\left(na\right)\left(\mathrm{by}\left(5\right)\right).\hfill \end{array}$$

Then

$${\lambda }_{21}\left(na\right)=n{\kappa }_{11}\left(a\right)-{\kappa }_{22}\left(a\right)n+{\lambda }_{21}\left(n\right)a$$

(18)

for any $a\in \mathcal{A}$.

For any $b\in \mathcal{B}$, we put $U=\left(\begin{array}{cc}{e}_0& 0\\ -bn& b\end{array}\right)$ and $V=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)$ in (12), and by using the fact ${\kappa }_{11}\left(1_{\mathcal{A}}\right)\oplus {\kappa }_{22}\left(1_{\mathcal{A}}\right)\in Z\left(\mathcal{G}\right)$, we have

$$\begin{array}{cc}\hfill & \left(\begin{array}{cc}0& 0\\ -{\lambda }_{21}\left(n{e}_0\right)& 0\end{array}\right)=L\left(\begin{array}{cc}0& 0\\ -n{e}_0& 0\end{array}\right)=L\left([U,V]\right)=[L\left(U\right),V]+[U,L\left(V\right)]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)+{\xi }_{11}\left(b\right)& 0\\ -{\lambda }_{21}\left(bn\right)& {\kappa }_{22}\left(e_0\right)+{\xi }_{22}\left(b\right)\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0& 0\\ -bn& b\end{array}\right),\left(\begin{array}{cc}{\kappa }_{11}\left(1_{\mathcal{A}}\right)& 0\\ {\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(1_{\mathcal{A}}\right)\end{array}\right)\right]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)+{\xi }_{11}\left(b\right)& 0\\ -{\lambda }_{21}\left(bn\right)& {\kappa }_{22}\left(e_0\right)+{\xi }_{22}\left(b\right)\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0& 0\\ -bn& b\end{array}\right),\left(\begin{array}{cc}0& 0\\ {\lambda }_{21}\left(n\right)& 0\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}0& 0\\ -{\lambda }_{21}\left(bn\right)+{\kappa }_{22}\left(e_0\right)n+{\xi }_{22}\left(b\right)n-n{\kappa }_{11}\left(e_0\right)-n{\xi }_{11}\left(b\right)+b{\lambda }_{21}\left(n\right)-{\lambda }_{21}\left(n\right)e_0& 0\end{array}\right),\hfill \end{array}$$

and so

$$-{\lambda }_{21}\left(n{e}_0\right)=-{\lambda }_{21}\left(bn\right)+{\kappa }_{22}\left(e_0\right)n+{\xi }_{22}\left(b\right)n-n{\kappa }_{11}\left(e_0\right)-n{\xi }_{11}\left(b\right)+b{\lambda }_{21}\left(n\right)-{\lambda }_{21}\left(n\right)e_0.$$

From (18), we have ${\lambda }_{21}\left(n{e}_0\right)=n{\kappa }_{11}\left(e_0\right)-{\kappa }_{22}\left(e_0\right)n+{\lambda }_{21}\left(n\right)e_0$, thus

$${\lambda }_{21}\left(bn\right)={\xi }_{22}\left(b\right)n-n{\xi }_{11}\left(b\right)+b{\lambda }_{21}\left(n\right)$$

for any $b\in \mathcal{B}$. This proves (ii).

For any $m\in \mathcal{M},n\in \mathcal{N}$, we put $U=\left(\begin{array}{cc}{e}_0-mn& m\\ 0& 0\end{array}\right)$ and $V=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)$ in (12). Then

$$L\left([U,V]\right)=L\left(\begin{array}{cc}mn& -m\\ n\left(mn\right)-n{e}_0& -nm\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(mn\right)-{\xi }_{11}\left(nm\right)& -{\mu }_{12}\left(m\right)\\ {\lambda }_{21}\left(nmn\right)-{\lambda }_{21}\left(n{e}_0\right)& {\kappa }_{22}\left(mn\right)-{\xi }_{22}\left(nm\right)\end{array}\right).$$

By using the fact ${\kappa }_{11}\left(1_{\mathcal{A}}\right)\oplus {\kappa }_{22}\left(1_{\mathcal{A}}\right)\in Z\left(\mathcal{G}\right)$, we have

$$\begin{array}{cc}\hfill & \left[L\right(U),V]+[U,L(V\left)\right]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)-{\kappa }_{11}\left(mn\right)& {\mu }_2\left(m\right)\\ 0& {\kappa }_{22}\left(e_0\right)-{\kappa }_{22}\left(mn\right)\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)\right]\hfill \\ \hfill & +\left[\left(\begin{array}{cc}{e}_0-mn& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}{\kappa }_{11}\left(1_{\mathcal{A}}\right)& 0\\ {\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(1_{\mathcal{A}}\right)\end{array}\right)\right]\hfill \\ \hfill & =\left[\left(\begin{array}{cc}{\kappa }_{11}\left(e_0\right)-{\kappa }_{11}\left(mn\right)& {\mu }_{12}\left(m\right)\\ 0& {\kappa }_{22}\left(e_0\right)-{\kappa }_{22}\left(mn\right)\end{array}\right),\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ n& 0\end{array}\right)\right]+\left[\left(\begin{array}{cc}{e}_0-mn& m\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 0\\ {\lambda }_{21}\left(n\right)& 0\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}{\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right)& -{\mu }_{12}\left(m\right)\\ w(m,n)& -n{\mu }_{12}\left(m\right)-{\lambda }_{21}\left(n\right)m\end{array}\right),\hfill \end{array}$$

where $w(m,n)={\kappa }_{22}\left(e_0\right)n-{\kappa }_{22}\left(mn\right)n-n{\kappa }_{11}\left(e_0\right)+n{\kappa }_{11}\left(mn\right)-{\lambda }_{21}\left(n\right)e_0+{\lambda }_{21}\left(n\right)mn$, and so

$${\kappa }_{11}\left(mn\right)-{\xi }_{11}\left(nm\right)={\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right),{\xi }_{22}\left(nm\right)-{\kappa }_{22}\left(mn\right)=n{\mu }_{12}\left(m\right)+{\lambda }_{21}\left(n\right)m.$$

This proves (iii). □

Remark 1.

The assumption that for every $a\in \mathcal{A}$ there exists an integer $k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$ is a standard and effective tool in the theory of local mappings. It enables the extension of identities from invertible elements to the whole algebra, serving as a unifying technical condition. This hypothesis is naturally fulfilled in several central classes of algebras, such as finite-dimensional algebras over an infinite field $\mathbb{F}$, unital complex Banach algebras, and full matrix algebras over an infinite field $\mathbb{F}$. Thus, Theorem 1 applies to a wide range of operator-algebraic settings encountered in practice.

Theorem 2.

Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra. Assume that

(i) 

${\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{A}\right),{\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{B}\right)$;

(ii) 

for each $a\in \mathcal{A}$, there $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$.

Let $\Delta :\mathcal{G}\to \mathcal{G}$ be a Lie derivation at E. Then $\Delta =d+\tau$, where d is a derivation on $\mathcal{G}$ and $\tau :\mathcal{G}\to Z\left(\mathcal{G}\right)$ is a $R-$linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$, if and only if ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$ for any $m\in \mathcal{M},n\in \mathcal{N}$, where ${\kappa }_{22}$ and ${\xi }_{11}$ are as in Theorem 1.

Proof of Theorem 2.

Set $\Delta :\mathcal{G}\to \mathcal{G}$ be a Lie derivation at E. For all $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$, by Theorem 1, we get

$$\Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\xi }_{11}\left(b\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(a\right)+{\xi }_{22}\left(b\right)+n_{0}m+n{m}_0\end{array}\right)$$

where $m_0\in \mathcal{M}$, $n_0\in \mathcal{N}$, and the linear maps ${\kappa }_{11}:\mathcal{A}\to \mathcal{A}$, ${\xi }_{11}:\mathcal{B}\to Z\left(\mathcal{A}\right)$, ${\mu }_{12}:\mathcal{M}\to \mathcal{M}$, ${\lambda }_{21}:\mathcal{N}\to \mathcal{N}$, ${\kappa }_{22}:\mathcal{A}\to Z\left(\mathcal{B}\right)$ and ${\xi }_{22}:\mathcal{B}\to \mathcal{B}$ satisfy the conditions (i)–(iii) in Theorem 1.

Let us assume that $\Delta =d+\tau$, where d is a derivation on $\mathcal{G}$ and $\tau :\mathcal{G}\to Z\left(\mathcal{G}\right)$ is a linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$. The following will prove that ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$ for all $m\in \mathcal{M},n\in \mathcal{N}$.

By Lemma 2, we may assume that

$$d\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{d}_{11}\left(1_{\mathcal{A}}\right)& {m^{\prime }}_0\\ {n^{\prime }}_0& 0\end{array}\right),d\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}-m{n^{\prime }}_0& {{\mu }^{\prime }}_{12}\left(m\right)\\ 0& {n^{\prime }}_{0}m\end{array}\right)$$

for all $m\in \mathcal{M}$, where ${m^{\prime }}_0\in \mathcal{M},{n^{\prime }}_0\in \mathcal{N}$, $d_{11}$ is a derivation of $\mathcal{A}$, ${{\mu }^{\prime }}_{12}$ is a linear map on $\mathcal{M}$.

Since

$$\begin{array}{cc}\hfill \tau \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)& =\Delta \left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)-d\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(1_{\mathcal{A}}\right)& m_0\\ n_0& {\kappa }_{22}\left(1_{\mathcal{A}}\right)\end{array}\right)-\left(\begin{array}{cc}{d}_{11}\left(1_{\mathcal{A}}\right)& {m^{\prime }}_0\\ {n^{\prime }}_0& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\kappa }_{11}\left(1_{\mathcal{A}}\right)-d_{11}\left(1_{\mathcal{A}}\right)& m_0-{m^{\prime }}_0\\ n_0-{n^{\prime }}_0& {\kappa }_{22}\left(1_{\mathcal{A}}\right)\end{array}\right)\in Z\left(\mathcal{G}\right),\hfill \end{array}$$

by Lemma 1, we have ${m^{\prime }}_0=m_0,{n^{\prime }}_0=n_0$.

For all $m\in \mathcal{M}$,

$$\begin{array}{cc}\hfill \tau \left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)& =\Delta \left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)-d\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}-m{n}_0& {\mu }_{12}\left(m\right)\\ 0& n_{0}m\end{array}\right)-\left(\begin{array}{cc}-m{n^{\prime }}_0& {{\mu }^{\prime }}_{12}\left(m\right)\\ 0& {n^{\prime }}_{0}m\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}0& {\mu }_{12}\left(m\right)-{{\mu }^{\prime }}_{12}\left(m\right)\\ 0& 0\end{array}\right)\in Z\left(\mathcal{G}\right).\hfill \end{array}$$

This implies ${{\mu }^{\prime }}_{12}\left(m\right)={\mu }_{12}\left(m\right)$. Thus $d\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}-m{n}_0& {\mu }_{12}\left(m\right)\\ 0& n_{0}m\end{array}\right)$ for any $m\in \mathcal{M}$.

Likewise, we can show that $d\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)=\left(\begin{array}{cc}-m_{0}n& 0\\ {\lambda }_{21}\left(n\right)& n{m}_0\end{array}\right)$ for all $n\in \mathcal{N}$.

For all $m\in \mathcal{M}$, $n\in \mathcal{N}$,

$$\begin{array}{cc}\hfill d\left(\begin{array}{cc}mn& 0\\ 0& 0\end{array}\right)& =d\left(\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\right)=d\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)·\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)+\left(\begin{array}{cc}0& m\\ 0& 0\end{array}\right)·d\left(\begin{array}{cc}0& 0\\ n& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right)& m\left(n{m}_0\right)\\ \left(n_{0}m\right)n& 0\end{array}\right).\hfill \end{array}$$

By using $\Delta \left(\begin{array}{cc}mn& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(mn\right)& \left(mn\right)m_0\\ n_0\left(mn\right)& {\kappa }_{22}\left(mn\right)\end{array}\right)$ and the condition (iii) in Theorem 1, we have

$$\begin{array}{cc}\hfill \tau \left(\begin{array}{cc}mn& 0\\ 0& 0\end{array}\right)& =\Delta \left(\begin{array}{cc}mn& 0\\ 0& 0\end{array}\right)-d\left(\begin{array}{cc}mn& 0\\ 0& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\kappa }_{11}\left(mn\right)& \left(mn\right)m_0\\ n_0\left(mn\right)& {\kappa }_{22}\left(mn\right)\end{array}\right)-\left(\begin{array}{cc}{\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right)& m\left(n{m}_0\right)\\ \left(n_{0}m\right)n& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\kappa }_{11}\left(mn\right)-{\mu }_{12}\left(m\right)n-m{\lambda }_{21}\left(n\right)& 0\\ 0& {\kappa }_{22}\left(mn\right)\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\xi }_{11}\left(nm\right)& 0\\ 0& {\kappa }_{22}\left(mn\right)\end{array}\right)\in Z\left(\mathcal{G}\right).\hfill \end{array}$$

Then ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$.

Conversely, assume that for all $m\in \mathcal{M},n\in \mathcal{N}$, ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$.

Define a map $d:\mathcal{G}\to \mathcal{G}$ as

$$d\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& {\xi }_{22}\left(b\right)-\eta \left({\xi }_{11}\left(b\right)\right)+n_{0}m+n{m}_0\end{array}\right)$$

and a map $\tau :\mathcal{G}\to \mathcal{G}$ as

$$\tau \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)+{\xi }_{11}\left(b\right)& 0\\ 0& {\kappa }_{22}\left(a\right)+\eta \left({\xi }_{11}\left(b\right)\right)\end{array}\right).$$

Clearly, d, $\tau$ are linear maps, and $\Delta =d+\tau$.

Let us first prove that d is a derivation on $\mathcal{G}$.

Since ${\kappa }_{22}\left(a\right)\in Z\left(\mathcal{B}\right)={\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right),{\xi }_{11}\left(b\right)\in Z\left(\mathcal{A}\right)={\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)$, by Lemma 1, there exists a unique algebra isomorphism $\eta :Z\left(\mathcal{A}\right)\to Z\left(\mathcal{B}\right)$ such that

$$\begin{array}{cc}\hfill & {\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)m=m{\kappa }_{22}\left(a\right),n{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)={\kappa }_{22}\left(a\right)n,\hfill \\ \hfill & {\xi }_{11}\left(b\right)m=m\eta \left({\xi }_{11}\left(b\right)\right),n{\xi }_{11}\left(b\right)=\eta \left({\xi }_{11}\left(b\right)\right)n\hfill \end{array}$$

(19)

for any $m\in \mathcal{M},n\in \mathcal{N}$.

Let us define $d_{11}:\mathcal{A}\to \mathcal{A}$, $d_{22}:\mathcal{B}\to \mathcal{B}$ as

$$d_{11}\left(a\right)={\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right),d_{22}\left(b\right)={\xi }_{22}\left(b\right)-\eta \left({\xi }_{11}\left(b\right)\right).$$

Clearly, $d_{11},d_{22}$ are linear maps and

$$d\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{d}_{11}\left(a\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& d_{22}\left(b\right)+n_{0}m+n{m}_0\end{array}\right).$$

According to conditions (i)–(ii) in Theorem 1 and (19), we can deduce that

$$\begin{array}{cc}\hfill & {\mu }_{12}\left(am\right)=({\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right))m+a{\mu }_{12}\left(m\right)=d_{11}\left(a\right)m+a{\mu }_{12}\left(m\right),\hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & {\mu }_{12}\left(mb\right)={\mu }_{12}\left(m\right)b+m({\xi }_{22}\left(b\right)-\eta \left({\xi }_{11}\left(b\right)\right))={\mu }_{12}\left(m\right)b+m{d}_{22}\left(b\right),\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & {\lambda }_{21}\left(na\right)={\lambda }_{21}\left(n\right)a+n({\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right))={\lambda }_{21}\left(n\right)a+n{d}_{11}\left(a\right),\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & {\lambda }_{21}\left(bn\right)=({\xi }_{22}\left(b\right)-\eta \left({\xi }_{11}\left(b\right)\right))n+b{\lambda }_{21}\left(n\right)=d_{22}\left(b\right)n+b{\lambda }_{21}\left(n\right)\hfill \end{array}$$

(23)

for any $a\in \mathcal{A}$, $b\in \mathcal{B}$, $m\in \mathcal{M}$, $n\in \mathcal{N}$.

From (20), we get

$${\mu }_{12}\left(a{a}^{\prime }m\right)=d_{11}\left(a{a}^{\prime }\right)m+a{a}^{\prime }{\mu }_{12}\left(m\right)$$

and

$$\begin{array}{cc}\hfill {\mu }_{12}\left(a{a}^{\prime }m\right)& ={\mu }_{12}\left(a\left(a^{\prime }m\right)\right)=d_{11}\left(a\right)a^{\prime }m+a{\mu }_{12}\left(a^{\prime }m\right)=d_{11}\left(a\right)a^{\prime }m+a(d_{11}\left(a^{\prime }\right)m+a^{\prime }{\mu }_{12}\left(m\right))\hfill \\ \hfill & =d_{11}\left(a\right)a^{\prime }m+a{d}_{11}\left(a^{\prime }\right)m+a{a}^{\prime }{\mu }_{12}\left(m\right)\hfill \end{array}$$

for any $a,a^{\prime }\in \mathcal{A}$, $m\in \mathcal{M}$, and so $d_{11}\left(a{a}^{\prime }\right)m=(d_{11}\left(a\right)a^{\prime }+a{d}_{11}\left(a^{\prime }\right))m$.

Because $\mathcal{M}$ is a faithful $(\mathcal{A},\mathcal{B})-$bimodule, one can obtain $d_{11}\left(a{a}^{\prime }\right)=d_{11}\left(a\right)a^{\prime }+a{d}_{11}\left(a^{\prime }\right)$. Therefore, $d_{11}$ is a derivation on $\mathcal{A}$.

Similarly, from (21), we can obtain

$${\mu }_{12}\left(mb{b}^{\prime }\right)={\mu }_{12}\left(m\right)b{b}^{\prime }+m{d}_{22}\left(b{b}^{\prime }\right)$$

and

$$\begin{array}{cc}\hfill {\mu }_{12}\left(mb{b}^{\prime }\right)& ={\mu }_{12}\left(\left(mb\right)b^{\prime }\right)={\mu }_{12}\left(mb\right)b^{\prime }+mb{d}_{22}\left(b^{\prime }\right)=({\mu }_{12}\left(m\right)b+m{d}_{22}\left(b\right))b^{\prime }+mb{d}_{22}\left(b^{\prime }\right)\hfill \\ \hfill & ={\mu }_{12}\left(m\right)b{b}^{\prime }+m{d}_{22}\left(b\right)b^{\prime }+mb{d}_{22}\left(b^{\prime }\right)\hfill \end{array}$$

for any $m\in \mathcal{M}$, $b,b^{\prime }\in \mathcal{B}$, and so $m{d}_{22}\left(b{b}^{\prime }\right)=m(d_{22}\left(b\right)b^{\prime }+b{d}_{22}\left(b^{\prime }\right))$.

By the faithfulness of $\mathcal{M}$, we get $d_{22}\left(b{b}^{\prime }\right)=d_{22}\left(b\right)b^{\prime }+b{d}_{22}\left(b^{\prime }\right)$. Therefore, $d_{22}$ is a derivation on $\mathcal{B}$.

Since ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$ for any $m\in \mathcal{M},n\in \mathcal{N}$, by Lemma 1, one can obtain

$$\eta \left({\xi }_{11}\left(nm\right)\right)={\kappa }_{22}\left(mn\right),{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right)={\xi }_{11}\left(nm\right).$$

According to condition (iii) in Theorem 1, we have

$$\begin{array}{cc}\hfill & d_{11}\left(mn\right)={\kappa }_{11}\left(mn\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right)={\kappa }_{11}\left(mn\right)-{\xi }_{11}\left(nm\right)={\mu }_{12}\left(m\right)n+m{\lambda }_{21}\left(n\right),\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill & d_{22}\left(nm\right)={\xi }_{22}\left(nm\right)-\eta \left({\xi }_{11}\left(nm\right)\right)={\xi }_{22}\left(nm\right)-{\kappa }_{22}\left(mn\right)=n{\mu }_{12}\left(m\right)+{\lambda }_{21}\left(n\right)m.\hfill \end{array}$$

(25)

According to the relations (20)-(25) and by Lemma 2, we can prove that d is a derivation of $\mathcal{G}$.

In the following, we will prove that $\tau \left(U\right)\in Z\left(\mathcal{G}\right)$ and $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$.

Since ${\kappa }_{22}\left(a\right)\in Z\left(\mathcal{B}\right)={\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right),{\xi }_{11}\left(b\right)\in Z\left(\mathcal{A}\right)={\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)$, we have

$$\begin{array}{cc}\hfill \tau \left(U\right)& =\left(\begin{array}{cc}{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)+{\xi }_{11}\left(b\right)& 0\\ 0& {\kappa }_{22}\left(a\right)+\eta \left({\xi }_{11}\left(b\right)\right)\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)& 0\\ 0& {\kappa }_{22}\left(a\right)\end{array}\right)+\left(\begin{array}{cc}{\xi }_{11}\left(b\right)& 0\\ 0& \eta \left({\xi }_{11}\left(b\right)\right)\end{array}\right)\in Z\left(\mathcal{G}\right)\hfill \end{array}$$

for any $U=\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$.

Since d is a derivation on $\mathcal{G}$, we have d is a Lie derivation at E. This implies that $\tau =\Delta -d$ is a Lie derivation at E. Therefore,

$$\tau \left(\right[U,V\left]\right)=\left[\tau \right(U),V]+[U,\tau (V\left)\right]=0$$

for any $U,V\in \mathcal{G}$ with $UV=E$. □

In Theorem 2, if $\mathcal{N}=0$, then ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)={\xi }_{11}\left(0\right)\oplus {\kappa }_{22}\left(0\right)=0\in Z\left(\mathcal{G}\right).$ By Theorem 2, one can obtain

Corollary 1.

Let $\mathcal{T}=\mathcal{T}(\mathcal{A},\mathcal{M},\mathcal{B})$ be a triangular algebra. Assume that

(i) 

${\pi }_{\mathcal{A}}\left(Z\left(\mathcal{T}\right)\right)=Z\left(\mathcal{A}\right),{\pi }_{\mathcal{B}}\left(Z\left(\mathcal{T}\right)\right)=Z\left(\mathcal{B}\right)$;

(ii) 

for each $a\in \mathcal{A}$, $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$.

Let $\Delta :\mathcal{T}\to \mathcal{T}$ be a Lie derivation at E. Then $\Delta =d+\tau$, where d is a derivation on $\mathcal{T}$, $\tau :\mathcal{T}\to Z\left(\mathcal{T}\right)$ is a linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$.

Remark 2.

Corollary 1 generalizes Theorem 2.2 in [10].

Theorem 3.

Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra. Assume the following:

(i) 

${\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{A}\right),{\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{B}\right)$;

(ii) 

For each $a\in \mathcal{A}$, there $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$;

(iii) 

Either $\mathcal{A}$ or $\mathcal{B}$ does not contain non-zero central ideals.

If $\Delta :\mathcal{G}\to \mathcal{G}$ is a Lie derivation at E, then $\Delta =d+\tau$, where d is a derivation on $\mathcal{G}$, $\tau :\mathcal{G}\to Z\left(\mathcal{G}\right)$ is a linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$.

Proof of Theorem 3.

According to Theorem 2, it is only necessary to prove that ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$ for any $m\in \mathcal{M},n\in \mathcal{N}$.

Let $\Delta :\mathcal{G}\to \mathcal{G}$ be a Lie derivation at E. For any $\left(\begin{array}{cc}a& m\\ n& b\end{array}\right)\in \mathcal{G}$, by Theorem 1, we have

$$\Delta \left(\begin{array}{cc}a& m\\ n& b\end{array}\right)=\left(\begin{array}{cc}{\kappa }_{11}\left(a\right)+{\xi }_{11}\left(b\right)-m{n}_0-m_{0}n& a{m}_0-m_{0}b+{\mu }_{12}\left(m\right)\\ n_{0}a-b{n}_0+{\lambda }_{21}\left(n\right)& {\kappa }_{22}\left(a\right)+{\xi }_{22}\left(b\right)+n_{0}m+n{m}_0\end{array}\right),$$

where $m_0\in \mathcal{M}$, $n_0\in \mathcal{N}$, and the linear maps ${\kappa }_{11}:\mathcal{A}\to \mathcal{A}$, ${\xi }_{11}:\mathcal{B}\to Z\left(\mathcal{A}\right)$, ${\mu }_{12}:\mathcal{M}\to \mathcal{M}$, ${\lambda }_{21}:\mathcal{N}\to \mathcal{N}$, ${\kappa }_{22}:\mathcal{A}\to Z\left(\mathcal{B}\right)$ and ${\xi }_{22}:\mathcal{B}\to \mathcal{B}$ satisfy the conditions (i)–(iii) in Theorem 1.

By Lemma 1, there exists a unique algebra isomorphism $\eta :Z\left(\mathcal{A}\right)\to Z\left(\mathcal{B}\right)$ such that

$$\begin{array}{c}\hfill {\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)m=m{\kappa }_{22}\left(a\right)\end{array}$$

(26)

for any $m\in \mathcal{M},n\in \mathcal{N}$.

Let $d_{11}\left(a\right)={\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right)$. According to condition (i) in Theorem 1 and (26), we get

$$\begin{array}{c}\hfill {\mu }_{12}\left(am\right)=({\kappa }_{11}\left(a\right)-{\eta }^{-1}\left({\kappa }_{22}\left(a\right)\right))m+a{\mu }_{12}\left(m\right)=d_{11}\left(a\right)m+a{\mu }_{12}\left(m\right)\end{array}$$

(27)

for any $a\in \mathcal{A}$, $m\in \mathcal{M}$.

By using (27), we obtain

$${\mu }_{12}\left(a{a}^{\prime }m\right)=d_{11}\left(a{a}^{\prime }\right)m+a{a}^{\prime }{\mu }_{12}\left(m\right)$$

and

$$\begin{array}{cc}\hfill {\mu }_{12}\left(a{a}^{\prime }m\right)={\mu }_{12}\left(a\left(a^{\prime }m\right)\right)& =d_{11}\left(a\right)a^{\prime }m+a{\mu }_{12}\left(a^{\prime }m\right)=d_{11}\left(a\right)a^{\prime }m+a(d_{11}\left(a^{\prime }\right)m+a^{\prime }{\mu }_{12}\left(m\right))\hfill \\ \hfill & =d_{11}\left(a\right)a^{\prime }m+a{d}_{11}\left(a^{\prime }\right)m+a{a}^{\prime }{\mu }_{12}\left(m\right)\hfill \end{array}$$

for any $a,a^{\prime }\in \mathcal{A}$, $m\in \mathcal{M}$. Thus, $d_{11}\left(a{a}^{\prime }\right)m=(d_{11}\left(a\right)a^{\prime }+a{d}_{11}\left(a^{\prime }\right))m$.

Because $\mathcal{M}$ is a faithful $(\mathcal{A},\mathcal{B})-$bimodule, we may deduce that

$$\begin{array}{c}\hfill d_{11}\left(a{a}^{\prime }\right)=d_{11}\left(a\right)a^{\prime }+a{d}_{11}\left(a^{\prime }\right).\end{array}$$

(28)

Without losing generality, let us assume that $\mathcal{A}$ does not contain non-zero central ideals.

For any $a\in \mathcal{A}$, $m\in \mathcal{M}$, $n\in \mathcal{N}$,

$$\begin{array}{cc}\hfill & a({\xi }_{11}\left(nm\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right))\hfill \\ \hfill =& a({\kappa }_{11}\left(mn\right)-{\mu }_{12}\left(m\right)n-m{\lambda }_{21}\left(n\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right))\left(\mathrm{by}\left(\mathrm{iii}\right)\mathrm{in}\mathrm{Theorem}1\right)\hfill \\ \hfill =& a({\kappa }_{11}\left(mn\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right))-a{\mu }_{12}\left(m\right)n-am{\lambda }_{21}\left(n\right)\hfill \\ \hfill =& a{d}_{11}\left(mn\right)-a{\mu }_{12}\left(m\right)n-am{\lambda }_{21}\left(n\right)\hfill \\ \hfill =& d_{11}\left(amn\right)-d_{11}\left(a\right)mn-a{\mu }_{12}\left(m\right)n-am{\lambda }_{21}\left(n\right)(\mathrm{by}\left(28\right))\hfill \\ \hfill =& d_{11}\left(amn\right)-{\mu }_{12}\left(am\right)n-am{\lambda }_{21}\left(n\right)\left(\mathrm{by}\right(27\left)\right)\hfill \\ \hfill =& {\kappa }_{11}\left(amn\right)-{\eta }^{-1}\left({\kappa }_{22}\left(amn\right)\right)-{\mu }_{12}\left(am\right)n-am{\lambda }_{21}\left(n\right)\hfill \\ \hfill =& {\xi }_{11}\left(nam\right)-{\eta }^{-1}\left({\kappa }_{22}\left(amn\right)\right)\left(\mathrm{by}\left(\mathrm{iii}\right)\mathrm{in}\mathrm{Theorem}1\right).\hfill \end{array}$$

Because ${\kappa }_{22}\left(amn\right)\in Z\left(\mathcal{B}\right)$, we have ${\eta }^{-1}\left({\kappa }_{22}\left(amn\right)\right)\in Z\left(\mathcal{A}\right)$. By using the fact ${\xi }_{11}\left(nam\right)\in Z\left(\mathcal{A}\right)$, we have

$${\xi }_{11}\left(nam\right)-{\eta }^{-1}\left({\kappa }_{22}\left(amn\right)\right)\in Z\left(\mathcal{A}\right).$$

Then $a({\xi }_{11}\left(nm\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right))\in Z\left(\mathcal{A}\right)$, and so $\mathcal{A}({\xi }_{11}\left(nm\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right))$ is a central ideal of $\mathcal{A}$. Since $\mathcal{A}$ does not contain any nonzero central ideal, we have ${\xi }_{11}\left(nm\right)-{\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right)=0$, that is ${\xi }_{11}\left(nm\right)={\eta }^{-1}\left({\kappa }_{22}\left(mn\right)\right)$. By Lemma 1, we have ${\xi }_{11}\left(nm\right)\oplus {\kappa }_{22}\left(mn\right)\in Z\left(\mathcal{G}\right)$. □

Taking $E=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$ in Theorem 3, one can obtain

Corollary 2

([11]). Let $\mathcal{G}=\mathcal{G}(\mathcal{A},\mathcal{M},\mathcal{N},\mathcal{B})$ be a generalized matrix algebra. Assume that

(i) 

${\pi }_{\mathcal{A}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{A}\right),{\pi }_{\mathcal{B}}\left(Z\left(\mathcal{G}\right)\right)=Z\left(\mathcal{B}\right)$;

(ii) 

for each $a\in \mathcal{A}$, $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$;

(iii) 

either $\mathcal{A}$ or $\mathcal{B}$ does not contain nonzero central ideals,

If $\Delta :\mathcal{G}\to \mathcal{G}$ is a Lie derivation at $\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$, then $\Delta =d+\tau$, where d is a derivation on $\mathcal{G}$, $\tau :\mathcal{G}\to Z\left(\mathcal{G}\right)$ is a linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=\left(\begin{array}{cc}{1}_{\mathcal{A}}& 0\\ 0& 0\end{array}\right)$.

Let $\mathcal{A}$ be an algebra with unity and $M_n\left(\mathcal{A}\right)$ the algebra of all $n\times n$ matrices over $\mathcal{A}$. Then $M_n\left(\mathcal{A}\right)$$(n\ge 2)$ can be written as a generalized matrix algebra in the form: $M_n\left(\mathcal{A}\right)\cong \left(\begin{array}{cc}\mathcal{A}& M_{1\times (n-1)}\left(\mathcal{A}\right)\\ M_{(n-1)\times 1}\left(\mathcal{A}\right)& M_{(n-1)\times (n-1)}\left(\mathcal{A}\right)\end{array}\right)$. It’s easy to verified that $Z\left(M_n\left(\mathcal{A}\right)\right)=Z\left(\mathcal{A}\right)·I_n$ and $M_n\left(\mathcal{A}\right)(n\ge 2)$ does not contain any nonzero central ideal. By Theorem 3, we have

Corollary 3.

Let $\mathcal{A}$ be an algebra with unity and $M_n\left(\mathcal{A}\right)$$(n\ge 3)$ the algebra of all $n\times n$ matrices over $\mathcal{A}$. Assume that for each $a\in \mathcal{A}$, $\exists k\in \mathbb{Z}$ such that $k{I}_{\mathcal{A}}-a$ is invertible in $\mathcal{A}$. Then each Lie derivation Δ at E has the standard form, that is, $\Delta =d+\tau$, where d is a derivation of $M_n\left(\mathcal{A}\right)$, $\tau :M_n\left(\mathcal{A}\right)\to Z\left(\mathcal{A}\right)·I_n$ is a linear map satisfying $\tau \left(\right[U,V\left]\right)=0$ with $UV=E$.

4. Conclusions

In this paper, we investigate the structure of Lie derivations at a fixed point $E=\left(\begin{array}{cc}{e}_0& 0\\ 0& 0\end{array}\right)$ on a generalized matrix algebra $\mathcal{G}$. Our main result provides a necessary and sufficient condition for such a Lie derivation to be standard, i.e., expressible as the sum of a derivation and a center-valued linear map. This finding not only clarifies the structural decomposition of these localized mappings but also generalizes several classical results in the literature. As direct applications, we have derived characterizations for the cases of triangular algebras and full matrix algebras, demonstrating the broad utility of our main theorem. Future work will focus on extending these results to more general elements.