Lie Derivations of a Matrix Ring over an Associative Ring

TL;DR

This paper characterizes Lie derivations of a specific matrix ring structure over a 2-torsion free ring, expanding understanding of derivations in algebraic matrix rings.

Contribution

It provides a detailed description of Lie derivations for matrix rings with entries constrained by an ideal, a novel extension in the study of algebraic derivations.

Findings

Explicit description of Lie derivations for the given matrix ring

Extension of derivation theory to rings with ideal constraints

Enhanced understanding of algebraic structure of matrix rings

Abstract

Let $K$ be a 2-torsion free ring with identity. We give a description of the Lie derivations of $R=R_{n}(K,J)=NT_{n}(K)+M_{n}(J)$, the ring of all $n\times n$ matrices over $K$ such that the entries on and above the main diagonal are elements of an ideal $J$ of $K$.

Equations

$$AnnR=\left\{\alpha \in R:\alpha R=R\alpha =0\right\}$$

$$An{n}_K(J)=\left\{\alpha \in K:\alpha J=J\alpha =0\right\},$$

$$(1){\varsigma }_i(J)={\alpha }_i(J)=0,i=1,2,...,n-1,$$

$$\begin{array}[]{llll}\Omega:&R&\rightarrow&C(R)\\ &[x_{i,j}]&\rightarrow&\begin{array}[]{l}\left(\sum\limits_{i=1}^{n-1}\alpha_{i}(x_{i+1,i})+\alpha_{n}(x_{1,n})\right)E\\ +\left(\sum\limits_{j=1}^{n}u(x_{j,j})+\sum\limits_{i=1}^{n-1}\varsigma_{i}(x_{i+1,i})+\varsigma_{n}(x_{1,n})\right)e_{n,1}\end{array}\end{array}$$

$$\begin{array}[]{l}i)\text{ }\gamma(J)=\theta(J)=0\\ ii)\text{ }\gamma(x_{1})x_{2}=\gamma(x_{2})x_{1}\\ iii)\text{ }x_{1}\theta(x_{2})=x_{2}\theta(x_{1}).\end{array}$$

$$\begin{array}[]{llll}\bar{\Gamma}:&R&\rightarrow&R\\ &ye_{1,n}&\rightarrow&\alpha(y)e_{n-1,1}+\beta(y)e_{n-1,2}+\gamma(y)e_{n,2}\\ &ye_{1,n-1}&\rightarrow&-\alpha(y)e_{n,1}-\beta(y)e_{n,2}\\ &ye_{2,n}&\rightarrow&-\beta(y)e_{n-1,1}-\gamma(y)e_{n,1}\text{ ,\ \ }y\epsilon J\\ &ye_{2,n-1}&\rightarrow&\beta(y)e_{n,1}\\ &x_{i,j}e_{i,j}&\rightarrow&0\text{ \ \ \ }((i,j)\neq(1,n),(1,n-1),(2,n-1),(2,n))\end{array}$$

$$\begin{array}[]{llll}\Lambda:&R_{3}(K,J)&\rightarrow&R_{3}(K,J)\\ &ye_{1,3}&\rightarrow&\alpha(y)e_{2,1}+\gamma(y)e_{3,2}\\ &ye_{1,2}&\rightarrow&-\alpha(y)e_{3,1}\\ &ye_{2,3}&\rightarrow&-\gamma(y)e_{3,1}\\ &x_{i,j}e_{i,j}&\rightarrow&0\text{ \ \ \ }((i,j)\neq(1,3),(1,2),(2,3))\end{array}$$

$$\begin{array}[]{llll}\Gamma:&R&\rightarrow&R\\ &ye_{1,n}&\rightarrow&\alpha(y)e_{1,1}+\beta(y)e_{n,n}\\ &ye_{i,n}&\rightarrow&\alpha(y)e_{i,1}\text{ \ , \ \ }1<i\leq n\\ &ye_{1,j}&\rightarrow&\beta(y)e_{n,j}\text{ \ , \ \ }1\leq j<n\\ &x_{i,j}&\rightarrow&0\text{ \ \ ,\ \ \ }(i\neq 1\text{, }j\neq n)\end{array}$$

$$\Delta (x{e}_{i+1,i})$$

$$\Delta (y{e}_{1,n})$$

$$\text{for}1$$

$${\Delta }_{k,k}^{i,j}$$

$$0$$

$$\Delta (x{e}_{2,1})$$

$$\Delta (x{e}_{n,n-1})$$

$${\sigma }_d(e_{i+1,i})$$

$$\Psi (e_{2,1})$$

$$\Psi (e_{n,n-1})$$

$$\Psi (e_{2,1}\ast e_{i+1,i})$$

$$\Psi (e_{i+1,i}\ast e_{j+1,j})$$

$$\Psi (e_{n,n-1}\ast e_{i+1,i})$$

$$\Psi (e_{i+1,i}\ast e_{i,i-1})$$

$$\Psi (e_{i+1,i-1}\ast e_{i,i-1})$$

$$\Psi (e_{i+1,i-1}\ast e_{i+1,i})$$

$$\Psi (e_{i+1,i})$$

$$\Psi (e_{2,1})$$

$$\Psi (e_{n,n-1})$$

$$\begin{array}[]{llll}\Omega:&R&\rightarrow&R\\ &[a_{i,j}]&\rightarrow&\begin{array}[]{l}\left(\sum_{i=1}^{n-1}\alpha_{i}(a_{i+1,i})+\alpha_{n}(a_{1,n})\right).E\\ +\left(\begin{array}[]{c}\sum_{i=1}^{n}u(a_{i,i})+\sum_{i=1}^{n-1}\Psi_{n,1}^{i+1,i}(a_{i+1,i})\\ +\Psi_{n,1}^{1,n}(a_{1,n})\end{array}\right)e_{n,1}\end{array}\end{array}$$

$$\begin{array}[]{llll}\bar{\Pi}:&R&\rightarrow&R\\ &[a_{i,j}]&\rightarrow&[\Pi(a_{i,j})]\end{array}$$

$$\Psi (e_{2,1})$$

$$\Psi (e_{i+1,i})$$

$$\Psi (e_{n,n-1})$$

$$\begin{array}[]{llll}\bar{\Delta}:&R&\rightarrow&R\\ &[a_{i,j}]&\rightarrow&\Psi_{n,2}^{2,1}(a_{2,1})e_{n,2}+\Psi_{n-1,1}^{n,n-1}(a_{n,n-1})e_{n-1,1}\end{array}$$

$$\Psi (x{e}_{i+1,i})$$

$$\Psi (x{e}_{2,1})$$

$$\Psi (x{e}_{n,n-1})$$

$$\ast$$

$$\Psi (y{e}_{1,n})$$

Full text

1. Introduction

Let $K$ be an associative ring with identity. An additive map $\Delta:K\rightarrow K$ is called a Lie derivation of $K$ if $\Delta(x\ast y)=\Delta(x)\ast y+x\ast\Delta(y)$ for all $x,y\in K$ where $x\ast y=xy-yx.$ Clearly, derivations are basic examples of Lie derivations of $K.$ The usual problem is to determine whether a Lie derivation is a derivation. The problem of describing all Jordan derivations, Lie derivations and Lie automorphisms of a ring or algebra was considered by many authors(see [1], [2], [3], [6], [7], [8]). Derivations and Jordan derivations of the ring $R$ are described in [5] and [9], respectively. In this paper, we continue the investigation of Lie derivations of $R=R_{n}(K,J)=NT_{n}(K)+M_{n}(J)$ for $n\geq 3$ where $NT_{n}(K)$ is the niltriangular $n\times n$ matrix ring and $M_{n}(J)$ is the ring of all $n\times n$ matrices over an ideal $J$ of $K.$ If $[x_{i,j}]$ is an element of $R,$ then $x_{i,j}\in I_{i,j}$ where $I_{i,j}$ is equal to $J$ for $i\leq j$ or to $R$ for $i>j.$

The group of automorphisms of this ring $R$ is given in [4] under certain restrictions on $J.$ Throughout this paper, we consider the square matrices of order $n\geq 3$  and we also assume that the 2-torsion free ring $K$ has identity element. Let $e_{i,j}$ denote the matrix unit with 1 in the (i,j) position and 0 in every other position. The ring $R$ is generated by the sets $Ke_{i+1,i}$ $(i=1,2,...,n-1)$ and $Je_{1,n}.$

Let $E$ denote the identity matrix of order n, $\delta_{i,j}=0$ for $i\neq j$ and $\delta_{i,i}=1.$ Then $e_{i,j}\ast e_{r,s}=\delta_{j,r}e_{i,s}-\delta_{s,i}e_{r,j}.$

2. Basic Derivations and the Main Theorem

Let $R=R_{n}(K,J).$ Then the sets of annihilator of $R$ and of $J$ in $K$ are

[TABLE]

and

[TABLE]

respectively. By Lemma 1.3 in [4], we know that $AnnR$ is equal to $Ann_{K}(J)e_{n,1}.$ Let $C(K)$ be the center of a ring $K.$ Clearly, $C(R)=AnnR+(J\cap C(K))E.$

If we replace the original product by the Lie product $x\ast y=xy-yx$ in any ring, the additive group of that ring becomes a Lie ring. If this ring is of characteristic 2, then the Lie product coincides with the Jordan product $x\circ y=xy+yx.$ So we assume that our ring $K$ is of characteristic not 2.

For any ring $K$ and any element $x$ of this ring, the map $\delta_{x}:K\rightarrow K$ defined by the rule $\delta_{x}(y)=xy-yx$ is a derivation. Such derivations are called inner.

If $\theta:K\rightarrow K$ is a derivation of $K$ satisfying $\theta(J)\subset J,$ then $\bar{\theta}:[a_{i,j}]\rightarrow[\theta(a_{i,j})]$ is a derivation of $R.$ These types of derivations are called ring derivations.

Let $d=\sum\limits_{i=1}^{n}d_{i}e_{i,i}$ $(d_{i}\in K).$ For any diagonal matrix $d$ and $x\in R,$ then the map $\sigma_{d}:x\rightarrow dx-xd$ is a derivation called diagonal derivation induced by the matrix $d.$

From now on

• •

$x_{i,j}$, $y_{i,j}$ will denote arbitrary elements in $I_{i,j}$ for any $i$ and $j,$

• •

$x,$ $x_{1},$ $x_{2}$ will denote arbitrary elements in $K,$

• •

$y$, $z$ will denote arbitrary elements in $J.$

Now we construct several types of Lie derivations on $R$ which will be used in order to generate all Lie derivations.

(A) Let $\varsigma_{n}:J\rightarrow Ann_{K}(J)$,  $\varsigma_{i}:K\rightarrow Ann_{K}(J),\ \ \alpha_{n}:J\rightarrow C(K),$  $\alpha_{i}:K\rightarrow C(K)$ and $u:J\rightarrow Ann_{K}(J)$ be additive maps satisfying

[TABLE]

Then the map

[TABLE]

is a Lie derivation. This Lie derivation is called central since $\Omega=0$ modulo $C(R).$

We now define special Lie derivations of $R;$

(B1) Special Lie derivation of type I:

The map \begin{array}[]{llll}\bar{\Delta}:&xe_{2,1}&\rightarrow&\gamma(x)e_{n,2}\text{ \ \ \ \ \ \ \ }\\ &xe_{n,n-1}&\rightarrow&\theta(x)e_{n-1,1}\text{ \ \ \ \ \ \ \ }\\ &x_{i,j}e_{i,j}&\rightarrow&0\text{ \ \ }((i,j)\neq(2,1),(n,n-1))\end{array}

is a Lie derivation of $R$ if the additive maps $\gamma,\theta:K\rightarrow Ann_{K}(J)$ satisfy the conditions

[TABLE]

(B2) Special Lie derivation of type II:

Let $\alpha,\beta,\gamma:J\rightarrow Ann_{K}(J)$ be additive maps with the conditions $\alpha(J^{2})=\beta(J^{2})=\gamma(J^{2})=0,$ $\alpha(yx)=x\alpha(y),$ $\beta(yx)=x\beta(y),$ $\beta(xy)=\beta(y)x$ and$\ \gamma(xy)=\gamma(y)x.$ Then the map

[TABLE]

is a Lie derivation of $R.$

(B3) Special Lie derivation of type III:

Let $\alpha,\beta,\gamma:J\rightarrow Ann_{K}(J)$ be additive maps satisfying $\alpha(J^{2})=\gamma(J^{2})=0,$ $\alpha(yx)=x\alpha(y)$ and$\ \gamma(xy)=\gamma(y)x.$ Then the map

[TABLE]

is a Lie derivation of $R_{3}(K,J).$

(C) Let $\alpha,\beta:J\rightarrow J$ be additive maps satisfying the conditions $\alpha(xy)=x\alpha(y),$ $\beta(yx)=\beta(y)x$ and $\alpha(y)z+y\beta(z)=0.$ Then the map

[TABLE]

determines a derivation of the ring $R$ called almost annihilator derivation.

Main Theorem: Let $K$ be a 2-torsion free ring with identity and $R=NT_{n}(K)+M_{n}(J)$ for $n>3.$ Then every Lie derivation of $R$ is a sum of diagonal, inner, ring, almost annihilator derivations and central Lie, special Lie derivations defined in (B1) and (B2).

In order to prove our main theorem, we give some technical lemmas. Suppose that $\Delta$ is a Lie derivation of $R.$ For any elementary matrix $x_{i,j}e_{i,j}\in R$ $(x_{i,j}\in I_{i,j},$ $1\leq i,j\leq n)$, we will check the properties of $\Delta(x_{i,j}e_{i,j})=\sum\limits_{s,t=1}^{n}\Delta_{s,t}^{i,j}(x_{i,j})e_{s,t}$ where $\Delta_{s,t}^{i,j}$ are additive maps from $I_{i,j}$ to $I_{s,t}.$

The following three lemmas are based on the results from [9].

Lemma 1 Let $\Delta$ be an arbitrary Lie derivation of $R.$ Then

[TABLE]

for $1\leq i,j\leq n$,  $k\neq i,j$  and  $m\neq i,j.$

Proof The equality $x_{i,j}e_{i,j}\ast y_{k,m}e_{k,m}=0$ for $k\neq j$ and $m\neq i$ is equivalent to

[TABLE]

and (k,m) entry of this equality gives (3) for $k>m,$ $\ k\neq i,j$ and $m\neq i,j$. On the other hand, for $m\neq j,$ $s\neq k,i$ and $y_{k,m}=1,$ we have $\Delta_{s,k}^{i,j}=0$. Similarly, for $k>m,$ $\ k\neq i,$ $\ t\neq m,j$ and $y_{k,m}=1,$ we get$\ \Delta_{m,t}^{i,j}=0.$ Therefore, we obtain (1) and (2). In particular, we get

[TABLE]

Lemma 2 Let $\Delta:R\rightarrow R$ be a Lie derivation. Then there exist a diagonal derivation $\sigma_{d}$ and two inner derivations $I_{A}$, $I_{B}$ such that $(\Delta-\sigma_{d}-I_{A}-I_{B})(e_{i+1,i})$ has nonzero entries only on i+1 row, main diagonal and (n,1) entry.

Proof Let $d_{i+1}=\sum_{k=1}^{i}\Delta_{k+1,k}^{k+1,k}(1)$ and $d=\sum_{i=2}^{n}d_{i}e_{i,i}$ be a diagonal matrix. If $\sigma_{d}$ is the diagonal derivation induced by the matrix $d,$ then

[TABLE]

So the $(i+1,i)$ entry of the matrix $(\Delta-\sigma_{d})(e_{i+1,i})$ is zero. Define a matrix $A=\sum_{i=2}^{n-1}(-\Delta_{i+1,1}^{i+1,i}(1))e_{i,1}$ and a matrix$\ B=[b_{i,j}]$ such that $b_{u,u}=0,$ $b_{j,1}=0$, $b_{v,i+1}=\Delta_{v,i}^{i+1,i}(1)$ for $i<n$, $v\neq i,i+1.$ Let $I_{A}$ and $I_{B}$ be the inner derivations induced by the matrices $A$ and $B,$ respectively. If we use $I_{A}(e_{i+1,i})=Ae_{i+1,i}-e_{i+1,i}A$ and $I_{B}(e_{i+1,i})=Be_{i+1,i}-e_{i+1,i}B,$ we obtain (i+1,1) entry and all entries of $(\Psi-I_{A}-I_{B})(e_{i+1,i})$ on i-th column are zeros. The lemma is proved. In particular, for $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}$, we have

[TABLE]

By using the following relations

[TABLE]

we get

[TABLE]

Lemma 3 There exists a central Lie derivation $\Omega$ of $R$ such that (k,k) entries of the matrix $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega)(xe_{i+1,i})$ and (m,m) entries of $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega)(ye_{1,n})$ are equal to zero for $i<n,$ $\ k\neq i,i+1,$ $\ m\neq 1,n.$

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}$ for brevity. If $1<i<n,$ then the equality $0=\Psi(xe_{i+1,i}\ast ye_{1,n})$ gives $\Psi_{n,1}^{i+1,i}(K)\subset Ann_{K}(J).$ Besides, we have $\Psi_{n,1}^{i+1,i}(J)=0$ by the relations $\Psi(xye_{i+1,i})=\Psi(xe_{i+1,i}\ast ye_{i,i})$ and $\Psi(yxe_{i+1,i})=\Psi(ye_{i+1,i+1}\ast xe_{i+1,i}).$ In addition, we get $\Psi_{n,1}^{2,1}(J)=0$ and$\ \Psi_{n,1}^{2,1}(K)\subset Ann_{K}(J)$ by using the relations $\Psi(xye_{2,1})=\Psi(xe_{2,1}\ast ye_{1,1}),$ $\Psi(yxe_{2,1})=\Psi(ye_{2,2}\ast xe_{2,1}),$ $\Psi(ye_{1,2}\ast xe_{2,1})=\Psi(yxe_{1,1}-xye_{2,2})$ and $\Psi(xe_{2,1}\ast ye_{n,n})=0.$ Moreover, we obtain $\Psi_{n,1}^{n,n-1}(K)\subset Ann_{K}(J)$ and $\Psi_{n,1}^{n,n-1}(J)=0$ by the equalities $\Psi(xye_{n,n-1})=\Psi(xe_{n,n-1}\ast ye_{n-1,n-1}),$ $\Psi(yxe_{n,n-1})=\Psi(ye_{n,n}\ast xe_{n,n-1}),$ $\Psi(ye_{n-1,n}\ast xe_{n,n-1})=\Psi(yxe_{n-1,n-1}-xye_{n,n})$ and $\Psi(xe_{n,n-1}\ast ye_{1,1})=0.$ Furthermore, If we consider the relations $\Psi(ye_{1,2}\ast ze_{2,n})=\Psi(yze_{1,n}),$ $\Psi(ye_{1,n}\ast ze_{2,n})=0$ and $\Psi(ye_{1,n}\ast ze_{1,2})=0,$ we have $\Psi_{n,1}^{1,n}(J^{2})=0$ and $\Psi_{n,1}^{1,n}(J)\subset Ann_{J}(K).$ Now let $\alpha_{i}=\Psi_{k,k}^{i+1,i}$ and $\alpha_{n}=\Psi_{m,m}^{1,n}$ where $m\neq 1,n,$ $\ k\neq i,i+1$ and $i<n.$ We know that $\alpha_{i}=\Psi_{k,k}^{i+1,i}:K\rightarrow C(K)$ $(i<n,$  $k\neq i,i+1)$ and $\alpha_{n}=\Psi_{m,m}^{1,n}:J\rightarrow C(K)$ $(m\neq 1,n)$ by (3) in Lemma 1. $\Psi(xye_{i+1,i})=\Psi(xe_{i+1,i}\ast ye_{i,i})$ gives $\alpha_{i}(J)=0$ and we get $\alpha_{n}(J^{2})=0$ by $\Psi(yze_{1,n})=\Psi(ye_{1,n}\ast ze_{n,n}).$ On the other hand, $\Psi(xye_{i+1,i+1}-yxe_{i,i})=\Psi(xe_{i+1,i}\ast ye_{i,i+1})$ gives $\Psi_{n,1}^{i+1,i+1}=\Psi_{n,1}^{i,i}$ and $\Psi_{n,1}^{i,i}(xy)=\Psi_{n,1}^{i,i}(yx)$ for $1<i<n.$ Say $u=\Psi_{n,1}^{i,i}$ for $1<i<n.$ Also we obtain $u:J\rightarrow Ann_{K}(J)$ by the relation $ye_{2,2}\ast ze_{1,n}=0.$ Then the map

[TABLE]

is the required central Lie derivation. The lemma is proved.

Lemma 4 There exists a ring derivation $\bar{\Pi}$ of $R$ such that (i,j) entry of $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi})(x_{i,j}e_{i,j})$ is zero for all $i,j.$

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega$   for brevity. Then (i,k) entry of the relation $\Psi(x_{i,j}y_{j,k}e_{i,k})=\Psi(x_{i,j}e_{i,j}\ast y_{j,k}e_{j,k})$ gives $\Psi_{i,k}^{i,k}(x_{i,j}y_{j,k})=\Psi_{i,j}^{i,j}(x_{i,j})y_{j,k}+x_{i,j}\Psi_{j,k}^{j,k}(y_{j,k})$ for all $i,j,k$ where $i\neq k.$ This means all $\Psi_{u,v}^{u,v}$ are equal to each other and $\Psi_{u,v}^{u,v}$ is a derivation of both $K$ and $J.$ Let $\Pi=\Psi_{u,v}^{u,v}.$ Then the map

[TABLE]

is the ring derivation we need. The lemma is proved.

Now we have the following equalities for $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}.$

[TABLE]

Lemma 5 There exists a special Lie derivation $\bar{\Delta}$ of $R$ such that (n,2) entry of $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta})(xe_{2,1})$ and (n-1,1) entry of $\ (\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta})(xe_{n,n-1})$ are zeros.

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}$ for brevity. The relations $\Psi(x_{1}e_{2,1}\ast x_{2}e_{2,1})=0$ and $\Psi(x_{1}e_{n,n-1}\ast x_{2}e_{n,n-1})=0$ gives $\Psi_{n,2}^{2,1}(x_{1})x_{2}-\Psi_{n,2}^{2,1}(x_{2})x_{1}=0$ and $x_{1}\Psi_{n-1,1}^{n,n-1}(x_{2})-x_{2}\Psi_{n-1,1}^{n,n-1}(x_{1})=0,$ respectively. Besides, the relations $\Psi(xye_{2,1})=\Psi(xe_{2,1}\ast ye_{1,1})$, $\Psi(yxe_{2,1})=\Psi(ye_{2,2}\ast xe_{2,1})$ and $\Psi(xe_{2,1}\ast ye_{n,n})=0$ gives $\Psi_{n,2}^{2,1}(J)=0$ and$\ \Psi_{n,2}^{2,1}(K)\subset Ann_{K}(J)$. Moreover, we obtain $\Psi_{n-1,1}^{n,n-1}(J)=0$ and $\Psi_{n-1,1}^{n,n-1}(K)\subset Ann_{K}(J)$ by the relations $\Psi(yxe_{n,n-1})=\Psi(ye_{n,n}\ast xe_{n,n-1}),$ $\Psi(xe_{n,n-1}\ast ye_{1,1})=0$ and $\Psi(xye_{n,n-1})=\Psi(xe_{n,n-1}\ast ye_{n-1,n-1}),$ respectively. Then the map

[TABLE]

is a special Lie derivation of type I defined in (B1). The lemma is proved.

Lemma 6 There exists an inner derivation $I_{C}$ such that $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C})(e_{i+1,i})$ has zero (i,i) entry for all $i<n.$

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}$ for brevity and$\ C=\sum_{k=1}^{n-1}\Psi_{k,k}^{k+1,k}(1)e_{k,k+1}.$ Then the inner derivation $I_{C}$ induced by the matrix $C$ is the required inner derivation. Note that the positions with nonzero entries of $I_{C}(e_{i,j})$ and $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta})(e_{i,j})$ are the same for $i>j.$

Now let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}$ for brevity. Then $\Psi(e_{i+1,i})=\Psi_{i+1,i+1}^{i+1,i}(1)e_{i+1,i+1}$ for $i<n.$ But we have $\Psi(e_{i+1,i-1})=0$ and then $\Psi(e_{i+1,i})=0$ by the relations $\Psi(e_{i+1,i}\ast e_{i,i-1})=\Psi(e_{i+1,i-1})$ and $\Psi(e_{i+1,i-1}\ast e_{i,i-1})=0$. It means $\Psi(e_{i,j})=0$ for all $i>j.$

For any numbers $k$ and $m$ such that $k>m,$ $\ k\neq 1,i$, $\ m\neq i+1,n,$ we obtain

[TABLE]

by the relation \Psi(xe_{i+1,i}\ast e_{k,m})=\Psi(xe_{i+1,i})\ast e_{k,m}=0.\Besides, for   $2<i<n-1,$ the relation $\Psi(x_{1}x_{2}e_{i+1,i-1})=\Psi(x_{1}e_{i+1,i}\ast x_{2}e_{i,i-1})$ gives $\Psi_{n,i}^{i+1,i}=0$ and $\Psi_{i,1}^{i,i-1}=0.$ In addition, we get $\Psi_{n,2}^{3,2}=0=\Psi_{n,3}^{2,1}$ and $\Psi_{n-1,1}^{n-1,n-2}=0=\Psi_{n-2,1}^{n,n-1}$ by the equalities $\Psi(x_{1}x_{2}e_{3,1})=\Psi(x_{1}e_{3,2}\ast x_{2}e_{2,1})$ and $\Psi(x_{1}x_{2}e_{n,n-2})=\Psi(x_{1}e_{n,n-1}\ast x_{2}e_{n-1,n-2})$. Finally, we have

[TABLE]

for any $i<n$ and $k>m.$ Also we obtain

[TABLE]

by the relations $\Psi(ye_{1,n-1})=\Psi(ye_{1,n}\ast e_{n,n-1})$ and $\Psi(ye_{2,n})=\Psi(e_{2,1}\ast ye_{1,n}).$

Lemma 7 There exists an almost annihilator derivation $\Gamma$ of $R$ such that (n,i) entry of $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma)(ye_{1,i})$ and (i,1) entry of $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma)(ye_{i,n})$ are zeros for $1\leq i\leq n$.

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}$ for brevity. Then we obtain $\Psi_{n,n}^{1,n}=\Psi_{n,i}^{1,i}$, $\Psi_{n,n}^{1,n}(yx)=\Psi_{n,n}^{1,n}(y)x$ and $\Psi_{1,1}^{1,n}=\Psi_{i,1}^{i,n},$ $\Psi_{1,1}^{1,n}(xy)=x\Psi_{1,1}^{1,n}(y)$ by the relations $\Psi(ye_{1,n}\ast xe_{n,i})=\Psi(yxe_{1,i})$ and $\Psi(xe_{i,1}\ast ye_{1,n})=\Psi(xye_{i,n})$ for $1<i<n.$ Furthermore, we get $\Psi_{n,1}^{1,1}=\Psi_{n,2}^{1,2}$ and $\Psi_{n,1}^{n,n}=\Psi_{n-1,1}^{n-1,n}$ by $\Psi(xe_{2,1}\ast ye_{1,2})=\Psi(xye_{2,2}-yxe_{1,1})$ and $\Psi(xe_{n,n-1}\ast ye_{n-1,n})=\Psi(xye_{n,n}-yxe_{n-1,n-1}).$ Now we have $\Psi_{n,n}^{1,n}=\Psi_{n,i}^{1,i}$ and $\Psi_{1,1}^{1,n}=\Psi_{i,1}^{i,n}$. Let $\lambda=\Psi_{1,1}^{1,n}=\Psi_{i,1}^{i,n}$ and $\mu=\Psi_{n,n}^{1,n}=\Psi_{n,i}^{1,i}.$ Now consider the relation $\Psi(yze_{1,n})=\Psi(ye_{1,2}\ast ze_{2,n}).$ This relation gives $0=z\mu(y)+\lambda(z)y.$ Then the map

[TABLE]

is an almost annihilator derivation of $R$ and this proves the lemma.

Lemma 8 There exists a special Lie derivation $\Theta$ of $R$ such that $\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma-\Theta$ is equal to zero map.

Proof Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma$ for brevity. Then the relations $\Psi(yxe_{1,n-1})=\Psi(ye_{1,n}\ast xe_{n,n-1}),$ $\Psi(xye_{2,n})=\Psi(xe_{2,1}\ast ye_{1,n})$ and $\Psi(yxe_{2,n-1})=\Psi(ye_{2,n}\ast xe_{n,n-1})$ give $\Psi_{n,1}^{1,n-1}(yx)=-x\Psi_{n-1,1}^{1,n}(y),$ $\Psi_{n,2}^{1,n-1}(yx)=-x\Psi_{n-1,2}^{1,n},$ $\Psi_{n-1,1}^{2,n}(xy)=-\Psi_{n-1,2}^{1,n}(y)x,$ $\Psi_{n,1}^{2,n}(xy)=-\Psi_{n,2}^{1,n}(y)x$ and$\ \Psi_{n,1}^{2,n-1}(yx)=-x\Psi_{n-1,1}^{2,n}(y).$ Let $\alpha=\Psi_{n-1,1}^{1,n},$ $\beta=\Psi_{n-1,2}^{1,n}$ and $\gamma=\Psi_{n,2}^{1,n}.$ Then one can see that $\alpha(yx)=x\alpha(y),$ $\beta(yx)=x\beta(y),$ $\beta(xy)=\beta(y)x,$ $\gamma(xy)=\gamma(y)x,$ $\Psi_{n,1}^{1,n-1}=-\alpha,$ $\Psi_{n,2}^{1,n-1}=\Psi_{n-1,1}^{2,n}=-\Psi_{n,1}^{2,n-1}=-\beta,$ $\Psi_{n,1}^{2,n}=-\gamma.$

If $n>4$ and $2<i<n-1,$ then we obtain \alpha(J^{2})=\beta(J^{2})=\gamma(J^{2})=0\and that$\ \alpha(y),\beta(y),\gamma(y)\subset Ann_{K}(J)$ by the relations $\Psi(yze_{1,n})=\Psi(ye_{1,i}\ast ze_{i,n}),$ $\Psi(yze_{1,n})=\Psi(ye_{1,2}\ast ze_{2,n}),$ $\Psi(yze_{1,n})=\Psi(ye_{1,n-1}\ast ze_{n-1,n}),$ $\Psi(yze_{1,n})=\Psi(ye_{1,1}\ast ze_{1,n})$ and $0=\Psi(ye_{1,n}\ast ze_{2,n})$.

If $n=4,$ then the relations $\Psi(yze_{1,4})=\Psi(ye_{1,j}\ast ze_{j,4})$ and $\Psi(ye_{1,4}\ast ze_{2,4})=0$ gives \alpha(J^{2})=\beta(J^{2})=\gamma(J^{2})=0\and$\ \alpha(y),\beta(y),\gamma(y)\subset Ann_{K}(J)$ for $1\leq j\leq 4.$ So $\alpha,\beta$ and $\gamma$ satisfies the conditions given in (B2) and the map

[TABLE]

becomes a special Lie derivation of type II. For the last part, we need to show $(\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma-\Theta)(x_{i,j}e_{i,j})=0$ for all i,j. Let $\Xi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-I_{C}-\Gamma-\Theta$ for brevity. Then $\Xi(ye_{i,j})=0$ for $i<j$ since we have

[TABLE]

Besides, we know that $\Xi(xe_{i,j})=0$ for $i>j$ by $(\ast).$ It suffices to show that $\Xi(ye_{i,i})=0$ for all i. If $i>j$, then $\Xi(ye_{i,i}-ye_{j,j})=\Xi(e_{i,j}\ast ye_{j,i})=0$ as $\Xi(ye_{k,m})=0$ for $k<m.$ This means $\Xi(ye_{i,i})=\Xi(ye_{j,j})$ for all i,j. So it is enough to check $\Xi(ye_{1,1})$ is equal to $0.$ Consider the products $\Xi(ye_{1,1}\ast e_{i,j})=\Xi(ye_{1,1})\ast e_{i,j}=0$ for $i>j.$ These products give that entries on the $i-th$ column and $j-th$ row of $\Xi(ye_{1,1})$ are zeros for $1<j<n$ and $i>2$ which means $\Xi(ye_{1,1})=\Xi_{1,2}^{1,1}(y)e_{1,2}+\Xi_{n,2}^{1,1}(y)e_{n,2}.$ Finally, the relation $0=\Xi(ye_{2,1})=\Xi(e_{2,1}\ast ye_{1,1})=e_{2,1}\ast\Xi(ye_{1,1})$ gives $\Xi_{1,2}^{1,1}=\Xi_{n,2}^{1,1}=0.$ The lemma is proved.

The proof of the main theorem follows by Lemma 1-8 for $n\geq 4$.

Discussion for n=3:

Let $K$ be commutative now. Then Lemma 1-5 applies. Moreover, the relations $yze_{1,3}=ye_{1,1}\ast ze_{1,3}$, $yze_{1,3}=ye_{1,2}\ast ze_{2,3},$  $yze_{1,3}=ye_{1,3}\ast ze_{3,3},$  $xye_{2,3}=xe_{2,1}\ast ye_{1,3},$  $yxe_{1,2}=ye_{1,3}\ast xe_{3,2}$ gives all the conditions of special Lie derivation $\Lambda$ defined in (B3). Let $\Psi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-\Lambda$ for brevity. Then we have $\Psi(xe_{2,1})=\Psi_{1,1}^{2,1}(x)e_{1,1}+\Psi_{2,2}^{2,1}(x)e_{2,2},$  $\Psi(xe_{3,2})=\Psi_{2,2}^{3,2}(x)e_{2,2}+\Psi_{3,3}^{3,2}(x)e_{3,3}$ and$\ \Psi(xe_{3,1})=\Psi_{2,1}^{3,1}(x)e_{2,1}+\Psi_{3,2}^{3,1}(x)e_{3,2}.$  Now Lemma 6 applies and then we find that the images of $xe_{i+1,i}$ under $\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-\Lambda-I_{C}$  are zeros by the equalities $x_{1}x_{2}e_{3,1}=x_{1}e_{3,2}\ast x_{2}e_{2,1},$  $x_{1}e_{2,1}\ast x_{2}e_{3,1}=0$ and $x_{1}e_{3,2}\ast x_{2}e_{3,1}=0.$ After applying Lemma 7, we also obtain the image of $ye_{1,3}$ is zero under $\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-\Lambda-I_{C}-\Gamma.$

Now let $\Xi=\Delta-\sigma_{d}-I_{A}-I_{B}-\Omega-\bar{\Pi}-\bar{\Delta}-\Lambda-I_{C}-\Gamma$ and consider the relations $ye_{1,2}=ye_{1,3}\ast e_{3,2},$  $ye_{2,3}=e_{2,1}\ast ye_{1,3}$. Then we find $\Xi(ye_{i,j})=0$ for $i<j.$ Furthermore, we obtain $\Xi(ye_{k,k})=0$ by $e_{i,j}\ast ye_{k,k}=0.$ So our main theorem holds for n=3 if $K$ is commutative. Note that the ring $K$ is chosen to be commutative to obtain central Lie derivations of $R_{3}(K,J).$