Jordan Derivations of Special Subrings of Matrix Rings

TL;DR

This paper characterizes all Jordan derivations of a specific class of matrix rings over a 2-torsion free ring, showing they decompose into derivations and extremal Jordan derivations.

Contribution

It provides a complete description of Jordan derivations on upper triangular matrix rings with entries in an ideal, extending understanding of their structure.

Findings

Every Jordan derivation decomposes into a derivation plus an extremal Jordan derivation.

Explicit forms of Jordan derivations are characterized.

The structure holds for rings with no 2-torsion and identity.

Abstract

Let $K$ be a 2-torsion free ring with identity and $R_{n}(K,J)$ be 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 $K.$ We describe all Jordan derivations of the matrix ring $R_{n}(K,J)$ in this paper. The main result states that every Jordan derivation $\Delta $ of $R_{n}(K,J)$ is of the form $\Delta =D+\Omega $ where $D$ is a derivation of $R_{n}(K,J)$ and $\Omega $ is an extremal Jordan derivation of $R_{n}(K,J).$

Equations

$$\alpha$$

$$\beta$$

$$\gamma$$

$$y{e}_{1,3}$$

$$y{e}_{1,2}$$

$$y{e}_{2,3}$$

$$y{e}_{1,3}$$

$$y{e}_{i,i}$$

$$y{e}_{2,3}$$

$$y{e}_{1,2}$$

$$[a_{i,j}]\to \left[{\varsigma }_n(a_{1,n})+\sum _{i=1}^{n-1}{\varsigma }_i(a_{i+1,i})\right]e_{n,1}\text{ (3)}$$

$$\bar{\pi }(A)=\sum _{i,j=1}^n\pi (a_{i,j})e_{i,j}\text{ , }A=[a_{i,j}]ϵR$$

$$y{e}_{1,n}$$

$$y{e}_{i,n}$$

$$y{e}_{1,j}$$

$$\Delta (x{e}_{i+1,i})=\sum {\Delta }_{i+1,t}^{i+1,i}(x)e_{i+1,t}+\sum _{s\ne i+1}{\Delta }_{s,i}^{i+1,i}(x)e_{s,i}+{\Delta }_{n,1}^{i+1,i}(x)e_{n,1}\text{ (4)}$$

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

$$0$$

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

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

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

$$\Delta (e_{i,j})$$

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

$$\Delta (x_{n,j}{e}_{n,j})$$

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

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

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

$$\Delta (x_{i,j}{e}_{i,j})$$

$$\Delta \left(e_{i+1,i}\circ e_{i+1,i}\right)$$

$$\Delta (e_{i+1,i}\circ e_{j+1,j})$$

$$\Delta (e_{i+1,i}\circ e_{i+2,i})$$

$$\Delta (e_{i+1,i}\circ e_{i+1,i})$$

$$\Delta (e_{n,n-1}\circ e_{n-1,n-2})$$

$$\Delta (e_{2,1}\circ e_{2,1})$$

$$\Delta (x_{i,1}{e}_{i,1}\circ e_{k,m})$$

$$\Delta (x_{n,i}{e}_{n,i}\circ e_{k,m})$$

$$\Delta (y_{1,i}{e}_{1,i}\circ e_{k,m})$$

$$\Delta (y_{i,n}{e}_{i,n}\circ e_{k,m})$$

$$\Delta (y_{1,n}{e}_{1,n}\circ e_{k,m})$$

$$\Delta (x_{i,j}{e}_{i,j}\circ e_{k,m})$$

Full text

Abstract

Let $K$ be a 2-torsion free ring with identity and $R_{n}(K,J)$ be 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 $K.$ We describe all Jordan derivations of the matrix ring $R_{n}(K,J)$ in this paper. The main result states that every Jordan derivation $\Delta$ of $R_{n}(K,J)$ is of the form $\Delta=D+\Omega$ where $D$ is a derivation of $R_{n}(K,J)$ and $\Omega$ is an extremal Jordan derivation of $R_{n}(K,J).$

1. Introduction

Let $K$ be an associative ring with identity and let $M_{n}(K)$ be the ring of all $n\times n$ matrices over $K.$ Jordan multiplication is defined by $x\circ y=xy+yx$ for any $x,y\in M_{n}(K).$ If an additive map $\Delta$ of $M_{n}(K)$ satisfies $\Delta(x\circ y)=\Delta(x)\circ y+x\circ\Delta(y)=\Delta(x)y+y\Delta(x)+x\Delta(y)+\Delta(y)x$ for all $x,y\in M_{n}(K)$ , then $\Delta$ is called a Jordan derivation of $M_{n}(K).$ Derivations of the ring $M_{n}(K)$ give trivial Jordan derivations. However, there are proper Jordan derivations which are not ring derivations for some matrix rings (see [3], [6], [8]).

In 1957, Herstein ([4]) proved that every Jordan derivation of a prime ring of characteristic not 2 is a derivation. This result was extended for semi-prime rings (see [2]) and for certain triangular matrix algebras and rings with nonzero nilpotent ideals (see [1], [11], [12]).

Let $NT_{n}(K)$ be the ring of all $n\times n$ matrices over $K$ which are all zeros on and above the main diagonal. Derivations and Jordan derivations of the nilpotent ring $NT_{n}(K)$ were described in [10] and [8].

Let $M_{n}(J)$ be the ring of all $n\times n$ matrices over an ideal $J$ of $K$ and $R=R_{n}(K,J)=NT_{n}(K)+M_{n}(J).$ All derivations of the matrix ring $R_{n}(K,J)$ were studied in [9].

Let $e_{i,j}$ denote the matrix with $1$ in the position $(i,j)$ and [math] in every other position. The ring $R_{n}(K,J)$ is generated by the sets $Ke_{i+1,i}$ ($i=1,2,...,n-1$) and $Je_{1,n}.$

The set of ideals $I=\{I_{i,j}:1\leq i,j\leq n\}$ of the ring $K$ are called carpet if $I_{ij}I_{jk}\subseteq I_{ik}$ for any $i,j,k$ (see [5]). By using carpet ideals, we can easily compute powers of the ring $R.$ For any matrix $[x_{i,j}]$ of the ring $R=R_{n}(K,J),$ the (i,j) entry $x_{i,j}$ is an element of $I_{i,j}$ where $I_{i,j}=K$ for $i>j$ and $I_{i,j}=J$  for $i\leq j.$

Let $Ann_{K}J=\{x\in K|\hskip 2.84544ptxJ=Jx=0\}.$ Then $AnnR=(Ann_{K}J)e_{n,1}$ $($see [7]$).$ In this paper we determine the structure of Jordan derivations on $R=R_{n}(K,J).$

2. Construction of Standard Jordan Derivations of $R$

(A1) Choose arbitrary additive group homomorphisms $\alpha,\beta,\gamma:J\rightarrow Ann_{K}J$ satisfying $\alpha(J^{2})=0,$ $\beta(J^{2})=0$ $,\gamma(J^{2})=0.$ Consider the following map of the set of all elementary matrices

\left.\begin{array}[]{c}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}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ ye_{2,n-1}\rightarrow\beta(y)e_{n,1}\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ ye_{2,n}\rightarrow\beta(y)e_{n-1,1}+\gamma(y)e_{n,1}\text{ ,\ \ }y\epsilon J\text{ \ \ \ \ \ \ \ \ }\end{array}\right\}      (1)

We assume that images of the remaining elementary matrices from $R$ are zeros.

If map (1) determines a Jordan derivation of the ring $R,$ then the relations $ye_{1,n}\circ xe_{n,n-1}=yxe_{1,n-1},$ $xe_{2,1}\circ ye_{1,n-1}=xye_{2,n-1},$ $xe_{2,1}\circ ye_{1,n}=xye_{2,n}$ give

\left.\begin{array}[]{c}\alpha(yx)=x\alpha(y)\\ \beta(yx)=x\beta(y)\\ \beta(xy)=\beta(y)x\\ \gamma(xy)=\gamma(y)x\end{array}\right\}                                                 (2)

for all $y\epsilon J$ and $x\epsilon K.$ One can easily check that if $\alpha,\beta,\gamma$ satisfy (2), then (1) determines a proper Jordan derivation of the ring $R$ but not a derivation unless $\alpha,\beta,\gamma$ are all zero maps. This Jordan derivation is called an ’extremal Jordan derivation’.

Example: Let $K_{1}$ be a commutative ring with identity and let $J_{1}$ be an ideal of $K_{1}$ which is nilpotent of class two. Let $K$ be a direct product ($K_{1},K_{1}$) of two copies of the ring $K_{1}$ and let $J=(J_{1},J_{1})$ be an ideal of $K.$ Then the maps

[TABLE]

[TABLE]

[TABLE]

satisfy all the conditions of (2). Moreover, $\alpha(J^{2})=\beta(J^{2})=\gamma(J^{2})=0.$

(A2) For arbitrary additive group homomorphisms $\alpha_{1},\alpha_{2}:J\rightarrow Ann_{K}J$ satisfying $\alpha_{1}(J^{2})=0,$ $\alpha_{2}(J^{2})=0,$ the following map of the set $R_{3}(K,J)$ determines a Jordan derivation of $R_{3}(K,J)$ under the conditions $\alpha_{1}(xy)=\alpha_{1}(y)x,$ $\alpha_{2}(yx)=x\alpha_{2}(y)$ for $x\epsilon K$ and $y\epsilon J.$ We assume that images of the elementary matrices except $ye_{1,3},$ $ye_{1,2}$ and $ye_{2,3}$ are zeros.

[TABLE]

(A3) If $\delta_{i}:J\rightarrow J,$ $\beta_{i}:J\rightarrow K,$ $\theta:J\rightarrow K$ and $\gamma:J\rightarrow K$ $\ (i=1,2,3)$ are additive mappings satisfying $\delta_{2}(J^{2})=0,$ $\beta_{2}(J)\subseteq Ann_{K}J,$ $\delta_{1}(yz)=z\beta_{1}(y)+y\delta_{1}(z)+\delta_{1}(z)y,$ $y\beta_{3}(z)=\delta_{1}(yz),$ $\delta_{1}(yz)=y\theta(z),$ $\delta_{2}(y)z+z\delta_{2}(y)=0,$ $z\gamma(y)+\theta(z)y=0,$ $\delta_{3}(yz)=\beta_{1}(y)z,$ $\delta_{3}(yz)=\delta_{3}(y)z+z\delta_{3}(y)+\beta_{3}(z)y,$ $\gamma(y)z=\delta_{3}(yz),$ $z\delta_{1}(y)+\delta_{3}(y)z=\beta_{1}(yz)+\beta_{3}(yz),$ $\gamma(y)x=\beta_{1}(yx)+\beta_{2}(xy),$ $\beta_{1}(yz+zy)=\beta_{1}(y)z+\beta_{1}(z)y,$ $z\beta_{1}(y)+\beta_{3}(z)y=0,$ $x\theta(y)=\beta_{2}(yx)+\beta_{3}(xy),$ $\theta(yz)=y\theta(z)+z\beta_{2}(y),$ $\theta(y)z+y\beta_{1}(z)=0,$ $\theta(xy)=x\delta_{1}(y)+\delta_{2}(y)x,$ $\theta(yz)=y\beta_{3}(z),$ $z\gamma(y)+\beta_{3}(z)y=0,$ $\gamma(yz)=\beta_{1}(y)z,$ $\gamma(yz)=\gamma(y)z,$ $\delta_{3}(y)x+x\delta_{2}(y)=\gamma(yx),$ $z\gamma(y)+\delta_{1}(z)y+y\delta_{2}(z)=0,$ $\delta_{1}(y)z+z\delta_{3}(y)+\delta_{1}(z)y+y\delta_{3}(z)=0,$ $\delta_{2}(y)z+z\delta_{3}(y)+\theta(z)y=0,$ then the following map is a Jordan derivation of $R_{3}(K,J)$ where $x\epsilon K,$ $y,z\epsilon J.$ We assume that images of the elementary matrices except $ye_{1,3},$ $ye_{1,2}$, $ye_{2,3}$ and $ye_{i,i}$ $(i=1,2,3)$ are zeros.

[TABLE]

We now describe Jordan derivations which are also derivations of the ring $R$ defined in [9].

(B1) Inner Derivations: Let $A\epsilon R$. Then the derivation of $R$ given by $X\rightarrow AX-XA,$ $X=[x_{i,j}]\epsilon R$ is called ’inner derivation’ induced by the matrix $A$.

(B2) Diagonal Derivations: Let $D=\sum\limits_{i=1}^{n}d_{i}e_{i,i}$ ($d_{i}\epsilon K$). Then the derivation of $R$ given by $X\rightarrow DX-XD,$ $X=[x_{i,j}]\epsilon R$ is called ’diagonal derivation’ induced by the matrix $D.$

(B3) Annihilator Derivations: An arbitrary ’annihilator derivation’ of the ring $R$ is of the form

[TABLE]

where $\varsigma_{n}:J\rightarrow Ann_{K}J,$ $\varsigma_{n}(J^{2})=0$ and $\varsigma_{i}:K\rightarrow Ann_{K}(J),$ $\varsigma_{i}(J)=0$ are additive maps for $i=1,2,...,n-1.$

(B4) Ring Derivations: If $\pi$ is a derivation of the coefficient ring $K$ and the restriction of $\pi$ over the ideal $J$ is also a derivation of $J,$ then $\pi$ induces a derivation of the ring $R$ by the rule

[TABLE]

(B5) Almost Annihilator derivation: If additive maps $\alpha,\beta$:$J\rightarrow J,$ $\gamma:J\rightarrow K$ satisfy the relations $\alpha(xy)=x\alpha(y),$ $\ \beta(yx)=\beta(y)x,$ $\ \ \ \ \gamma(y)z=y\gamma(z)=\gamma(yz)=0$ and $\alpha(y)z+y\beta(z)=0,$ then the following map of the set $R$

[TABLE]

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

3. Jordan Derivations of $R_{n}(K,J)$

Throughout this section, $K$ will be a 2-torsion free ring with identity.

Theorem 3.1. Let $K$ be a 2-torsion free ring with identity, $J$ be an ideal of $K$ and $R=R_{n}(K,J).$ If $n\geq 4$ then every Jordan derivation of $R$ is of the form $\Delta=D+\Omega$ where $D$ is a derivation of $R_{n}(K,J)$ and $\Omega$ is an extremal Jordan derivation of $R_{n}(K,J).$ Moreover, $D$ is the sum of certain diagonal, inner, annihilator, ring and almost annihilator derivations.

For an arbitrary Jordan derivation $\Delta$ of the ring $R,$ we write the image of an element $x_{i,j}e_{i,j}$ $\left(1\leq i,j\leq n,\text{ }x_{i,j}\epsilon I_{i,j}\right)$ of the form $\Delta(x_{i,j}e_{i,j})=\sum\limits_{s,t=1}^{n}\Delta_{s,t}^{i,j}(x_{i,j})e_{s,t}$ where $\Delta_{s,t}^{i,j}$ are additive mappings from $I_{i,j}$ to $I_{s,t}$ .

Lemma 3.2. Let $\Delta$ be an arbitrary Jordan derivation of R for $n\geq 4.$ Then for $1<i<n-1$ and $x\epsilon K,$ $y\epsilon J$

[TABLE]

and

[TABLE]

Proof Let us fix $i,j$ and choose $k,m$ such that $k>m.$ If $k\neq j$ and $m\neq i$ then $x_{i,j}e_{i,j}\circ y_{k,m}e_{k,m}=0$. By differentiating $x_{i,j}e_{i,j}\circ y_{k,m}e_{k,m}=0,$ we get

[TABLE]

Putting $y_{k,m}=1,$ the matrix on the right has zeros except $i-th$ , $k-th$  rows and $j-th$ , $m-th$ columns. Hence $\Delta_{s,k}^{i,j}=0$ for $m\neq j,$ $s\neq i,$ $s\neq k$ and $\Delta_{m,t}^{i,j}=0$ for $i\neq k,$ $t\neq m,$ $t\neq j.$ On the other hand, for $k>s>m$ and $k\neq i,j$ , $s\neq i,j$ , $m\neq i,j$ , the (k,m), (s,m) and (k,s) coefficients of the equations $\Delta(x_{i,j}e_{i,j}\circ e_{k,m})=0,$ $\Delta(x_{i,j}e_{i,j}\circ e_{s,m})=0$ and $\Delta(x_{i,j}e_{i,j}\circ e_{k,s})=0$ are $\Delta_{k,k}^{i,j}(x_{i,j})+\Delta_{m,m}^{i,j}(x_{i,j})=0...(\ast),$ $\Delta_{s,s}^{i,j}(x_{i,j})+\Delta_{m,m}^{i,j}(x_{i,j})=0...(\ast\ast)$ and $\Delta_{k,k}^{i,j}(x_{i,j})+\Delta_{s,s}^{i,j}(x_{i,j})=0...(\ast\ast\ast),$ respectively. Comparing $(\ast),(\ast\ast)$ and $(\ast\ast\ast),$ we get $\Delta_{k,k}^{i,j}(x_{i,j})=\Delta_{s,s}^{i,j}(x_{i,j})=\Delta_{m,m}^{i,j}(x_{i,j}).$ By using $(\ast\ast\ast),$ $2\Delta_{k,k}^{i,j}(x_{i,j})=2\Delta_{s,s}^{i,j}(x_{i,j})=0=2\Delta_{m,m}^{i,j}(x_{i,j}).$ Since K is 2-torsion free, $\Delta_{k,k}^{i,j}(x_{i,j})=\Delta_{s,s}^{i,j}(x_{i,j})=\Delta_{m,m}^{i,j}(x_{i,j})=0.$ Therefore, the image of $xe_{i+1,i}$ ($x\epsilon K$) under $\Delta$ is the matrix with zeros outside $i+1-th$ row, $i-th$ column and $(n,1)$ position. For $1<i<n-1,$ $\Delta(xe_{i+1,i})$ has the form (4). In particular, $\Delta(xe_{2,1})=\sum\Delta_{2,t}^{2,1}(x)e_{2,t}+\sum\limits_{s\neq 2}\Delta_{s,1}^{2,1}(x)e_{s,1}+\Delta_{n,2}^{2,1}(x)e_{n,2}+\Delta_{n,3}^{2,1}(x)e_{n,3}$ for $i=1$ and $\Delta(xe_{n,n-1})=\sum\Delta_{n,t}^{n,n-1}(x)e_{n,t}+\sum\limits_{s\neq n}\Delta_{s,n-1}^{n,n-1}(x)e_{s,n-1}+\Delta_{n-1,1}^{n,n-1}(x)e_{n-1,1}+\Delta_{n-2,1}^{n,n-1}(x)e_{n-2,1}$ for $i=n-1.$ By the same relation $\Delta_{n-1,1}^{1,n}(y)\neq 0$ since $k\neq n,$ $k\neq 1,$ $m\neq n-1$ and $\Delta_{n-1,2}^{1,n}(y)\neq 0$ as$\ m\neq n-1,$ $k\neq 2$ for $y\epsilon J$. Similarly, $\Delta_{n,1}^{1,n}(y)\neq 0$ since $k\neq 1,$ $m\neq n$ and $\Delta_{n,2}^{1,n}(y)\neq 0$ as $m\neq n,$ $k\neq 2$ for $y\epsilon J$. Thus we get (5).

Lemma 3.3. Let $\Delta:R\rightarrow R$ be a Jordan derivation. Then there exists a diagonal derivation $D$ of $R$ such that $(i+1,i)-th$  coefficient of the matrix $\left(\Delta-\mathfrak{D}\right)(e_{i+1,i})$ is equal to zero for all $i.$

Proof Let $D=\sum\limits_{i=1}^{n}d_{i}e_{i,i}$ where $d_{1}=0,$ $d_{i+1}=\sum\limits_{k=1}^{i}a_{k}$ and $a_{k}=\Delta_{k+1,k}^{k+1,k}(1).$ Then there exists a diagonal derivation $D:X\rightarrow DX-XD$ by the diagonal matrix $D$ such that $D(e_{i+1,i})=De_{i+1,i}-e_{i+1,i}D=\Delta_{i+1,i}^{i+1,i}(1)e_{i+1,i}$ . Since $(i+1,i)-th$ coefficient of the matrix $\Delta$ is equal to $\Delta_{i+1,i}^{i+1,i}(1),$ the proof is completed.

Lemma 3.4. Let $\Delta:R\rightarrow R$ be a Jordan derivation such that $\left(i+1,i\right)-th$ coefficient of matrices $\Delta(e_{i+1,i})$ are zero for all $1\leq i<n.$ Then there exists an inner derivation $I$ such that each matrix $\left(\Delta-\mathfrak{I}\right)(e_{i+1,i})$ has zero $i-th$ column and $(i+1,1)$ entries.

Proof Define a matrix $A=[A_{i,j}]$ such that $A_{u,u}=0$ ($1\leq u\leq n),$ $A_{u,i+1}=\Delta_{u,i}^{i+1,i}(1)$ ($u\neq i+1,$ $i\neq n$) and $A_{j,1}=0$ ($1\leq j\leq n$). Consider the action of $I_{A}$ on the matrices $e_{i+1,i}$ . Then $I_{A}(e_{i+1,i})=Ae_{i+1,i}-e_{i+1,i}A=\sum\limits_{\begin{subarray}{c}k=1\\ k\neq i+1\end{subarray}}^{n}\Delta_{k,i}^{i+1,i}(1)e_{k,i}+\sum\limits_{\begin{subarray}{c}m=1\\ m\neq i-1\end{subarray}}^{n-1}-\Delta_{i,m}^{m+1,m}(1)e_{i+1,m+1}$ . Therefore, i-th column of each matrix $(\Delta-I_{A})(e_{i+1,i})$ is equal to zero. Now define a matrix $B=[B_{i,j}]$ such that $B=-b_{3}e_{2,1}-b_{4}e_{3,1}-...-b_{n}e_{n-1,1}$ where $b_{i+1}=\Delta_{i+1,1}^{i+1,i}(1)\ (1<i<n)$. Denote by $I_{B}$ the inner derivation induced by the matrix $B.$ It can be easily seen that $I_{B}(e_{i+1,i})=Be_{i+1,i}-e_{i+1,i}B=\Delta_{i+1,1}^{i+1,i}(1)e_{i+1,1}$ ($i=2,...,n-1$). Hence i-th columns and (i+1,1) entries of the matrices ($\Delta-I$)($e_{i+1,i}$) are zeros with $I=I_{A}+I_{B}.$

Lemma 3.5. Let $\Delta:R\rightarrow R$ be a Jordan derivation such that $i-th$ columns and $(i+1,1)$ entries of the matrices $\Delta(e_{i+1,i})$ are all zeros. Then the following equalities are obtained for $x_{k,m}\epsilon I_{k,m},$ $x\epsilon K$ and $y\epsilon J$ ;

[TABLE]

Proof The Equations (1)-(10) can easily be obtained by the relations below when $K$ is 2-torsion free;

[TABLE]

Lemma 3.6. Let $\Delta$ be a Jordan derivation of R satisfying the conditions (1)-(10) in Lemma 3.5. Then there exists an annihilator derivation $\Upsilon$ such that $(\Delta-\Upsilon)$($e_{i+1,i}$) is equal to zero for all $i$ .

Proof Let $x\epsilon K,$ $y,z\epsilon J$ be arbitrary elements. For $i\neq 1,n$ , $\Delta(e_{n,i}\circ ye_{1,n})=\Delta(ye_{1,i})$. This implies that $\Delta_{1,1}^{1,i}=0$ . If we combine $\Delta(e_{n,1}\circ ye_{1,n-1})=\Delta(ye_{n,n-1})$ and $e_{n,1}\circ\Delta(ye_{1,n-1})=\Delta(ye_{n,n-1})$ , we obtain $\Delta_{11}^{1,n-1}(y)=\Delta_{n,1}^{n,n-1}(y)=0$ . Let us say $\varsigma_{i}=\Delta_{n,1}^{i+1,i}$ . Then $\varsigma_{n-1}(J)=0.$ If $k>2$ , $\Delta(e_{2,1}\circ ye_{1,k})=\Delta(ye_{2,k})$ follows that $\Delta_{n,k}^{2,k}=0$. Again for $k>2$ , $\Delta(ye_{2,k}\circ e_{k,1})=\Delta(ye_{2,1})$ and $\Delta(ye_{2,k})\circ e_{k,1}=\Delta(ye_{2,1})$ implies that $\Delta_{n,k}^{2,k}(y)=\Delta_{n,1}^{2,1}(y)=0=\varsigma_{1}(y)$. For $i\neq 1,n-1$ , $\Delta(e_{n,i}\circ ye_{i+1,n})=\Delta(ye_{i+1,i})$ and $e_{n,i}\circ\Delta(ye_{i+1,n})=\Delta(ye_{i+1,i})$ follows $\Delta_{i,1}^{i+1,n}(y)=\Delta_{n,1}^{i+1,i}(y)=0=\varsigma_{i}(y)$. So $\varsigma_{i}(J)$ is zero ($i<n$). For $2\leq i\leq n-2$ , we get $\varsigma_{i}=\Delta_{n,1}^{i+1,i}:K\rightarrow Ann_{K}J$  by using the relations $\Delta(xe_{i+1,i}\circ ye_{1,k})=0$ ($k\neq i+1,n$) and $\Delta(xe_{i+1,i}\circ ye_{k,n})=0$ ($k\neq 1,i$) . By $\Delta(xe_{2,1}\circ ye_{n-1,n})=0$ and $\Delta(xe_{2,1}\circ ye_{1,k})=\Delta(xye_{2,k})$  ($k\neq 2,n$), we have $y\Delta_{n,1}^{2,1}(x)=0$ , $\Delta_{n,1}^{2,1}(x)y=0$ and $\varsigma_{1}=\Delta_{n,1}^{2,1}:K\rightarrow Ann_{K}J$. Similarly, by $\Delta(xe_{n,n-1}\circ ye_{1,2})=0$ and $\Delta(xe_{n,n-1}\circ ye_{1,n})=\Delta(yxe_{1,n-1})$ we obtain $\Delta_{n,1}^{n,n-1}(x)y=0$ , $y\Delta_{n,1}^{n,n-1}(x)=0$ and $\varsigma_{n-1}=\Delta_{n,1}^{n,n-1}:K\rightarrow Ann_{K}J$. Now consider the product $\Delta(ye_{1,k}\circ ze_{k,n})=\Delta(yze_{1,n}).$ Hence $\Delta_{n,1}^{1,n}(yz)=0.$ Say $\varsigma_{n}=\Delta_{n,1}^{1,n}$ , then $\varsigma_{n}(J^{2})=0$. By $\Delta(ye_{1,1}\circ ze_{1,n})=\Delta(yze_{1,n})$ and $\Delta(ye_{1,n}\circ ze_{n,n})=\Delta(yze_{1,n})$ , we have $0=\Delta_{n,1}^{1,n}(z)y=z\Delta_{n,1}^{1,n}(y)$ and $\varsigma_{n}:J\rightarrow Ann_{K}J$ . Thus $\Upsilon:[x_{i,j}]\rightarrow\left(\varsigma_{n}(x_{1,n})+\sum\limits_{i=1}^{n-1}\varsigma_{i}(x_{i+1,i})\right)e_{n,1}$ is an annihilator derivation and $(\Delta-\Upsilon)(e_{i+1,i})=0$  for all $i.$ Say $\Theta=\Delta-\Upsilon.$ Hence (n,1)-coefficients of $\Theta(xe_{i+1,i})$ and $\Theta(ye_{1,n})$ are equal to zero.

Lemma 3.7. Let $\Theta=\Delta-\Upsilon$ be a Jordan derivation of the ring R as in Lemma 3.6. Then there exists a ring derivation $\bar{\theta}$ of $R$ such that $(i,j)$ coefficient of $\Theta(x_{i,j}e_{i,j})$ is equal to zero.

Proof Let $x,x_{1},x_{2}\epsilon K$ and $y\epsilon J$ be arbitrary elements. By using the relation $x_{1}e_{i,j}\circ x_{2}e_{j,k}=x_{1}x_{2}e_{i,k}$ for $i>j>k$ we have $\Theta_{i,j}^{i,j}(x_{1})x_{2}+x_{1}\Theta_{i,j}^{i,j}(x_{2})=\Theta_{i,j}^{i,j}(x_{1}x_{2})$. If we say $\Theta_{i,j}^{i,j}=\theta$ , $\theta$ is a ring derivation of K. Similarly, $\Theta_{i,j}^{i,j}(y)=\Theta_{i,i-1}^{i,i-1}(y)$ by $ye_{i,j}\circ e_{j,i-1}=ye_{i,i-1}$ for $i\leq j$. This implies that $\theta$ is a derivation of J as well. So $\bar{\theta}:[x_{i,j}]\rightarrow\sum\limits_{i,j}\theta(x_{i,j})e_{i,j}$ is a ring derivation of R.

Let us say $\Xi=\Theta-\bar{\theta}$. Thus (i,j)-coefficients of the matrices $\Xi(x_{i,j}e_{i,j})$ are equal to zero.

Lemma 3.8. Let $\Xi$ be a Jordan derivation of R as in Lemma 3.7. Then $\Xi(xe_{i,j})=0$ for all $i>j$ , $\Xi(ye_{i,i})=\Xi_{n,1}^{i,i}(y)e_{n,1}$ for all $i,$ $\Xi(ye_{1,j})=\Xi_{n,j}^{1,j}(y)e_{n,j}$ for $1<j<n-1$ , $\Xi(ye_{i,n})=\Xi_{i,1}^{i,n}(y)e_{i,1}$ for $i\neq 1,2$ where $x\epsilon K,$ $y\epsilon J$ are arbitrary elements.

Proof Let $x,x_{1},x_{2}\epsilon K$ and $y\epsilon J$ be arbitrary elements. For $1<i<n$, by combining $\Xi(x_{1}e_{i+1,i}\circ x_{2}e_{i,i-1})=\Xi(x_{1}x_{2}e_{i+1,i-1})$ and \Xi(x_{1}e_{i+1,i})\circ x_{2}e_{i,i-1}+x_{1}e_{i+1,i}\Xi(x_{2}e_{i,i-1})=\Xi(x_{1}x_{2}e_{i+1,i-1})\it can be easily seen that $\Xi_{n,i-1}^{i+1,i-1}=0=\Xi_{i+1,1}^{i+1,i-1}.$ Then $\Xi(x_{1}x_{2}e_{i+1,i-1})=0$ . By using $\Xi(x_{1}e_{i+1,i}\circ x_{2}e_{i,i-1})=\Xi(x_{1}x_{2}e_{i+1,i-1})=0$ , we get $\Xi_{n,i}^{i+1,i}=0$, $\Xi_{i,1}^{i,i-1}=0$. Thus $\Xi(xe_{i,j})=0$ for $i>j$. To obtain $\Xi(ye_{i,i})=\Xi_{n,1}^{i,i}(y)e_{n,1}$ , we use the following relations; $\Xi(xe_{i+2,i+1}\circ ye_{i+1,i+1})=\Xi(xye_{i+2,i+1})=0$ ($1\leq i<n-1$), $\Xi(xe_{2,1}\circ ye_{1,1})=\Xi(xye_{2,1})=0$ and $\Xi(xe_{i+1,i}\circ ye_{i+1,i+1})=\Xi(yxe_{i+1,i})=0$ ($1\leq i<n$). By $\Xi(xe_{i,1}\circ ye_{1,j})=\Xi(xye_{i,j})=0$ for $i>j$ and $\Xi(xe_{n,j}\circ ye_{1,n})=\Xi(yxe_{1,j})$ for $1<j<n-1$ , we have $\Xi(ye_{1,j})=\Xi_{n,j}^{1,j}(y)e_{n,j}$ for $1<j<n-1$. Finally by $\Xi(xe_{i,1}\circ ye_{1,n})=\Xi(xye_{i,n})$ for $1<i<n$ we have $\Xi(ye_{i,n})=\Xi_{i,1}^{i,n}(y)e_{i,1}$.

Lemma 3.9. Let $\Xi$ be a Jordan derivation of R as in Lemma 3.8. Then there exists an almost annihilator derivation $\Gamma$ of R such that $\Xi-\Gamma$ is an extremal Jordan derivation of R which is defined in Section 2.

Proof Let $\bar{\alpha}=\Xi_{1,1}^{1,n},$ $\bar{\beta}=\Xi_{n,n}^{1,n},$ $x\epsilon K$ and $y,z\epsilon J$. By using the corresponding appropriate relations and considering K as 2-torsion free, the additive map $\Gamma$ on $R$ defined by $\Gamma(ye_{1,n})=\bar{\alpha}(y)e_{1,1}+\bar{\beta}(y)e_{n,n}$ , $\Gamma(ye_{i,n})=\bar{\alpha}(y)e_{i,1}$ ($1<i\leq n$), $\Gamma(ye_{1,j})=\bar{\beta}(y)e_{n,j}$ $(1\leq j<n)$, $\Gamma(x_{i,j}e_{i,j})=0$ $(1<i$ and $j<n)$ is obviously an almost annihilator derivation of R.

Consider the relations $\Xi(xe_{n-1,j}\circ ye_{1,n-1})=\Xi(yxe_{1,j})$  for $1<j<n-1$ and $\Xi(xe_{i,1}\circ ye_{1,j})=\Xi(xye_{i,j})$, $\Xi(ye_{1,j})\circ xe_{i,1}=\Xi(xye_{i,j})$ for $1<i<j<n$ where $(i,j)\neq(2,n-1),$ then we obtain $\Xi(ye_{1,j})=0$ for $j\neq n,n-1$ and $\Xi(ye_{i,j})=0$  for $1<i<j<n$ respectively.

Finally, if we say $\Pi=\Xi-\Gamma$ then$\ \Pi(ye_{1,n})=\Pi_{n-1,1}^{1,n}(y)e_{n-1,1}+\Pi_{n-1,2}^{1,n}(y)e_{n-1,2}+\Pi_{n,2}^{1,n}(y)e_{n,2}$ , $\Pi(ye_{1,n-1})=\Pi_{n,1}^{1,n-1}(y)e_{n,1}+\Pi_{n,2}^{1,n-1}(y)e_{n,2}$, $\Pi(ye_{2,n-1})=\Pi_{n,1}^{2,n-1}(y)e_{n,1}$ , $\Pi(ye_{2,n})=\Pi_{n-1,1}^{2,n}(y)e_{n-1,1}+\Pi_{n,1}^{2,n}(y)e_{n,1}$ and $\Pi(x_{i,j}e_{i,j})=0$ for $(i,j)\neq(1,n),(2,n),(1,n-1),(2,n-1)$. By using relations $ye_{1,n}\circ xe_{n,n-1}=yxe_{1,n-1}$ , $xe_{2,1}\circ ye_{1,n-1}=xye_{2,n-1}$ , $xe_{2,1}\circ ye_{1,n}=xye_{2,n}$ , we get $\alpha=\Pi_{n-1,1}^{1,n}=\Pi_{n,1}^{1,n-1}$ , $\beta=\Pi_{n-1,2}^{1,n}=\Pi_{n,2}^{1,n-1}=\Pi_{n,1}^{2,n-1}=\Pi_{n-1,1}^{2,n}$ , $\gamma=\Pi_{n,2}^{1,n}=\Pi_{n,1}^{2,n}$ and $\alpha(yx)=x\alpha(y),$ $\beta(yx)=x\beta(y),$ $\beta(xy)=\beta(y)x,$ $\gamma(xy)=\gamma(y)x$. Then for $n>4,$ the relations $0=\Pi(ye_{1,3}\circ ze_{3,n})=\Pi(yze_{1,n})$, $0=\Pi(yze_{1,n})=\Pi(ye_{1,n-1}\circ ze_{n-1,n})=\Pi(ye_{1,1}\circ ze_{1,n})$, $0=\Pi(yze_{1,n})=\Pi(ye_{1,2}\circ ze_{2,n})=\Pi(ye_{1,n}\circ ze_{n-1,n-1})$ and $0=\Pi(ye_{1,n}\circ ze_{3,n})$ gives all conditions of extremal Jordan derivation defined in A1.

If $n=4$ , we use the following relations to show all conditions of extremal Jordan derivation of $R_{4}(K,J)$ are satisfied;

By $ye_{1,3}\circ ze_{3,4}=yze_{1,4}$ , $ye_{1,2}\circ ze_{2,4}=yze_{1,4}$ , $ye_{1,1}\circ ze_{1,4}=yze_{1,4}$ we get $\alpha(J^{2})=0,$ $\beta(J^{2})=0,$ $\gamma(J^{2}),$ respectively and by $ye_{1,1}\circ ze_{1,4}=ye_{1,2}\circ ze_{2,4}=ye_{1,3}\circ ze_{3,4}=yze_{1,4}$ , $ye_{1,4}\circ ze_{3,4}=0$ .

Now Theorem 3.1 follows easily by the Lemmas 3.2 - 3.9.

For n=3, after applying Lemmas 3.2-3.7, it is obtained that $\Xi$ is equal to the sum of the Jordan derivations A2 and A3 described in Section 2.