Multiplicative derivations on rank-$s$ matrices for relatively small $s$

TL;DR

This paper characterizes maps on matrix rings that behave like derivations on rank-$s$ matrices, showing they are essentially derivations on all matrices up to rank $s$, for fixed $n$ and $s$.

Contribution

It establishes that maps satisfying a derivation-like property on rank-$s$ matrices are actually derivations on all matrices of rank up to $s$, extending the understanding of multiplicative derivations.

Findings

Maps satisfying the derivation property on rank-$s$ matrices are derivations on those matrices.

Such maps coincide with derivations on all matrices of rank up to $s$.

The result holds for fixed $n$ and $s$ with $1 \\leq s \\leq \\frac{n}{2}$.

Abstract

Let $n$ and $s$ be fixed integers such that $n\geq 2$ and $1\leq s\leq \frac{n}{2}$. Let $M_n(\mathbb{K})$ be the ring of all $n\times n$ matrices over a field $\mathbb{K}$. If a map $\delta:M_n(\mathbb{K})\rightarrow M_n(\mathbb{K})$ satisfies that $\delta(xy)=\delta(x)y+x\delta(y)$ for any two rank-$s$ matrices $x,y\in M_n(\mathbb{K})$, then there exists a derivation $D$ of $M_n(\mathbb{K})$ such that $\delta(x)=D(x)$ holds for each rank-$k$ matrix $x\in M_n(\mathbb{K})$ with $0\leq k\leq s$.

Equations

$${\lambda }_1{\delta }_1+{\lambda }_2{\delta }_2\in {\mathcal{D}}_{\mathcal{S}}^{\times }(M_n(\mathbb{K})).$$

$$\begin{array}[]{cl}&(\lambda_{1}\delta_{1}+\lambda_{2}\delta_{2})(xy)\\ =&(\lambda_{1}\delta_{1})(xy)+(\lambda_{2}\delta_{2})(xy)=\lambda_{1}\delta_{1}(xy)+\lambda_{2}\delta_{2}(xy)\\ =&\lambda_{1}\delta_{1}(x)y+\lambda_{1}x\delta_{1}(y)+\lambda_{2}\delta_{2}(x)y+\lambda_{2}x\delta_{2}(y)\\ =&(\lambda_{1}\delta_{1}+\lambda_{2}\delta_{2})(x)y+x(\lambda_{1}\delta_{1}+\lambda_{2}\delta_{2})(y)\end{array}$$

$$y=P\left(\sum _{i=1}^k{e}_{ii}\right)Q,$$

$$y_1=P\left(\sum _{i=1}^k{e}_{ii}+\sum _{j=k+1}^s{e}_{jj}\right),y_2=\left(\sum _{i=1}^k{e}_{ii}+\sum _{j=s+1}^{2s-k}{e}_{jj}\right)Q.$$

$$\left\{\begin{array}[]{l}\delta(e)=\delta(e^{2})=\delta(e)e+e\delta(e)~{}~{}\text{and}\\ \delta(f)=\delta(f^{2})=\delta(f)f+f\delta(f),\end{array}\right.$$

$$\left\{\begin{array}[]{l}\delta(e)=eA(I_{n}-e)+(I_{n}-e)Ae~{}~{}\text{and}\\ \delta(f)=fB(I_{n}-f)+(I_{n}-f)Bf\end{array}\right.$$

$$\left\{\begin{array}[]{lcl}\delta(0)&=&\delta(ef)=e\delta(f)+\delta(e)f\\ &=&efB(I_{n}-f)+e(I_{n}-f)Bf+eA(I_{n}-e)f+(I_{n}-e)Aef=eBf+eAf~{}~{}\text{and}\\ \delta(0)&=&\delta(ge)=\delta(g)e+g\delta(e)=\delta(g)e+geA(I_{n}-e)+g(I_{n}-e)Ae=\delta(g)e+gAe,\end{array}\right.$$

$$\delta (0)=eBf+eAf=(eBf+eAf)f=\delta (0)f=(\delta (g)e+gAe)f=0.$$

$$\left\{\begin{array}[]{l}\delta(x\cdot 0)=\delta(0)=0=x\cdot 0+\delta(x)\cdot 0=x\cdot\delta(0)+\delta(x)\cdot 0~{}~{}\text{and}\\ \delta(0\cdot x)=\delta(0)=0=0\cdot x+0\cdot\delta(x)=\delta(0)\cdot x+0\cdot\delta(x).\end{array}\right.$$

$$\left(e_{11}+\sum _{j=2}^s{e}_{i_j,i_j}\right)QR\left(\sum _{i=1}^s{e}_{ii}\right)\in M_n^s(\mathbb{K})$$

$$\{i_2,i_3,\dots ,i_s\}\cap \{k_2,k_2,\dots ,k_s\}=\varphi .$$

$$\begin{array}[]{lcl}\delta(xy)&=&\delta(x_{1}x_{2}y)=\delta(x_{1}(x_{2}y))=\delta(x_{1})x_{2}y+x_{1}\delta(x_{2}y)=\delta(x_{1})x_{2}y+x_{1}\delta(x_{2})y+x_{1}x_{2}\delta(y)\\ &=&\delta(x_{1}x_{2})y+x_{1}x_{2}\delta(y)=\delta(x)y+x\delta(y).\end{array}$$

$$QR\left(\sum_{i=1}^{s}e_{ii}\right)=\left(\begin{array}[]{cc}O&O\\ G&O\end{array}\right),$$

$$G^T{\alpha }_i=0,i=1,2,\dots ,n-1-s.$$

$$x_{2}=\left(\begin{array}[]{cc}1&O\\ O&H\\ O&O\end{array}\right)Q,$$

$$\begin{array}[]{lcl}\delta(xy)&=&\delta(x_{1}x_{2}y)=\delta(x_{1}(x_{2}y))=\delta(x_{1}\cdot 0)=\delta(x_{1})\cdot 0+x_{1}\cdot\delta(0)=\delta(x_{1})x_{2}y+x_{1}\delta(x_{2}y)\\ &=&\delta(x_{1})x_{2}y+x_{1}\delta(x_{2})y+x_{1}x_{2}\delta(y)=\delta(x_{1}x_{2})y+x_{1}x_{2}\delta(y)=\delta(x)y+x\delta(y).\end{array}$$

$$\begin{array}[]{rcl}\delta(xy)&=&\delta(xy_{1}y_{2})=\delta((xy_{1})y_{2})=\delta(xy_{1})y_{2}+xy_{1}\delta(y_{2})\\ &=&\delta(x)y_{1}y_{2}+x\delta(y_{1})y_{2}+xy_{1}\delta(y_{2})=\delta(x)y_{1}y_{2}+x\delta(y_{1}y_{2})=\delta(x)y+x\delta(y),\end{array}$$

$$\delta (e_{ii})=\delta (e_{ii}{e}_{ii})=e_{ii}\delta (e_{ii})+\delta (e_{ii})e_{ii},$$

$$(s,t)\in \left\{(s,t)\in \underline{n}\times \underline{n}|s\ne i\text{and}t\ne i\}\cup \{(i,i)\right\}.$$

$$0=\delta (0)=\delta (e_{ii}{e}_{jj})=e_{ii}\delta (e_{jj})+\delta (e_{ii})e_{jj}=a_{ij}^{(j)}{e}_{ij}+a_{ij}^{(i)}{e}_{ij},$$

$$a_{ij}^{(j)}=-a_{ij}^{(i)}$$

$$B=\sum _{i=1}^n\delta (e_{ii})e_{ii}.$$

$$e_{jj}B=\sum _{k\in \underline{n}-\{j\}}{a}_{jk}^{(k)}{e}_{jk}=-\sum _{k\in \underline{n}-\{j\}}{a}_{jk}^{(j)}{e}_{jk}=-e_{jj}\delta (e_{jj}).$$

$$\delta (e_{ii})=\delta (e_{ii})e_{ii}+e_{ii}\delta (e_{ii})=B{e}_{ii}-e_{ii}B=[B,e_{ii}].$$

$$\begin{array}[]{rcl}\delta(e_{ij})&=&\delta(e_{ii}e_{ij}e_{jj})=\delta(e_{ii})e_{ij}+e_{ii}\delta(e_{ij})e_{jj}+e_{ij}\delta(e_{jj})\\ &=&\delta(e_{ii})e_{ii}e_{ij}+e_{ii}\delta(e_{ij})e_{jj}+e_{ij}e_{jj}\delta(e_{jj})\\ &=&Be_{ii}e_{ij}+\lambda_{ij}e_{ij}-e_{ij}e_{jj}B=[B,e_{ij}]+\lambda_{ij}e_{ij}.\end{array}$$

$$\delta (e_{ij})=[B,e_{ij}]+{\lambda }_{ij}{e}_{ij}.$$

$$\begin{array}[]{rcl}[B,e_{ik}]+\lambda_{ik}e_{ik}&=&\delta(e_{ik})=\delta(e_{ij}e_{jk})=e_{ij}\delta(e_{jk})+\delta(e_{ij})e_{jk}\\ &=&e_{ij}([B,e_{jk}]+\lambda_{jk}e_{jk})+([B,e_{ij}]+\lambda_{ij}e_{ij})e_{jk}\\ &=&(e_{ij}[B,e_{jk}]+[B,e_{ij}]e_{jk})+(\lambda_{jk}e_{ij}e_{jk}+\lambda_{ij}e_{ij}e_{jk})\\ &=&[B,e_{ik}]+(\lambda_{ij}+\lambda_{jk})e_{ik},\end{array}$$

$$A=B+\sum _{j=1}^n{\lambda }_{j1}{e}_{jj}.$$

$$\begin{array}[]{rcl}\delta(e_{st})&=&[B,e_{st}]+\lambda_{st}e_{st}=[B,e_{st}]+(\lambda_{s1}-\lambda_{t1})e_{st}\\ &=&[B,e_{st}]+[\sum_{j=1}^{n}\lambda_{j1}e_{jj},e_{st}]=[A,e_{st}].\end{array}$$

$${\delta }^{\prime }(a{e}_{ij})={\delta }^{\prime }(e_{ii}(a{e}_{ij})e_{jj})={\delta }^{\prime }(e_{ii})(a{e}_{ij})+e_{ii}{\delta }^{\prime }(a{e}_{ij})e_{jj}+(a{e}_{ij}){\delta }^{\prime }(e_{jj})=e_{ii}{\delta }^{\prime }(a{e}_{ij})e_{jj},$$

Full text

1. Introduction

Franca [5] initialed the research on nonadditive subsets of prime rings in the theory of functional identities by describing the commuting additive map on the set of all $n\times n$ invertible matrices or the set of all $n\times n$ singular matrices rather than the ring of all $n\times n$ matrices over fields. This is an extension of the well-known theorem of Brešar (see the original paper [1, Theorem A], or the survey paper [2, Corollary 3.3], or the book [3, Corollary5.28]). Furthermore, in 2013, Franca [6] (also see Xu et al. [16]) extended the discussion to the set of all rank-$s$ matrices over fields for fixed $2\leq s<n$. In 2014, Liu (see [10, 11]) researched centralizing additive maps and strong commutativity preserving maps on the set of all $n\times n$ invertible matrices or the set of all $n\times n$ singular matrices over division rings and obtained nice conclusions, which developed the corresponding results in the theory of functional identities. Recently, Xu et al. [19, 16] proved that a map $g$ from the ring of all $n\times n$ matrices over a field into itself is additive if and only if $g(A+B)=g(A)+g(B)$ for any two rank-$s$ matrices $A,B\in M_{n}(\mathbb{K})$, where $\frac{n}{2}\leq s\leq n$ is fixed. For further references see [18, 9, 13, 12, 7, 20].

On the other hand, a map $f$ from a ring $R$ into itself is called a multiplicative isomorphism if $f$ is bijective and $f(xy)=f(x)f(y)$ for all $x,y\in R$. A map $f$ from a ring $R$ into itself is called a multiplicative derivation if $f(xy)=f(x)y+xf(y)$ for all $x,y\in R$. The question of when a multiplicative isomorphism is additive has been considered by Rickart [15] and Johnson [8]. Martindale [14] improved the main theorem of Rickart [15, Theorem II].

In 1991, Daif [4] considered the similar question of when a multiplicative derivation is additive. He proved that it is true for the ring $R$ with an idempotent element $e\neq 0,1$ satisfying: (1) $xR=0$ implies $x=0$; (2) $eRx=0$ implies $x=0$; (3) $exeR(1-e)=0$ implies $exe=0$. Note that for $n\geq 2$ the ring $M_{n}(R)$ ($T_{n}(R)$, respectively) of all $n\times n$ (upper triangular) matrices over a unital ring $R$ is a special example of the rings Daif stated. So a multiplicative derivation of $M_{n}(R)$ ($T_{n}(R)$, respectively) must be a derivation, where $R$ is a ring with an identity and $n>1$.

In this short note, we consider the multiplicative derivation on the set of all $n\times n$ rank-$s$ matrices over a field $\mathbb{K}$ other than the ring of all $n\times n$ matrices over $\mathbb{K}$ and prove that for the case $1\leq s\leq\frac{n}{2}$ and $n\geq 2$, if a map $\delta:M_{n}(\mathbb{K})\rightarrow M_{n}(\mathbb{K})$ satisfies that $\delta(xy)=\delta(x)y+x\delta(y)$ for any two rank-$s$ matrices $x,y\in M_{n}(\mathbb{K})$, then there exists a derivation $D$ of $M_{n}(\mathbb{K})$ such that $\delta(x)=D(x)$ for each rank-$k$ matrix $x\in M_{n}(\mathbb{K})$ with $k\leq s$. This means that the multiplicative derivation on rank-$s$ matrices over a field is almost a derivation when restricted on the matrices whose rank is not more than $s$ for relative small $s$. As an application, we will show that the multiplicative derivation on some nonadditive subset of the matrix ring $M_{n}(\mathbb{K})$ over a field $\mathbb{K}$ has to be a derivation.

2. Multiplicative derivations on rank-$s$ matrices for relatively small $s$

In this section, unless stated otherwise, we will always assume that both $n$ and $s$ are fixed integers such that $n\geq 2$ and $1\leq s\leq\frac{n}{2}$, and always denote by $\mathbb{K}$ a field, by $M_{n}(\mathbb{K})$ the ring of all $n\times n$ matrices over $\mathbb{K}$, by $GL_{n}(\mathbb{K})$ the set of all $n\times n$ invertible matrices over $\mathbb{K}$. For $0\leq k\leq n$, the symbol $M_{n}^{k}(\mathbb{K})$ ($M_{n}^{\leq k}(\mathbb{K})$ and $M_{n}^{<k}(\mathbb{K})$, respectively) will always denote the set of all matrices whose rank is equal to (not more than and less than, respectively) $k$ in $M_{n}(\mathbb{K})$. A map $\delta:M_{n}(\mathbb{K})\rightarrow M_{n}(\mathbb{K})$ is called a multiplicative derivation on $\mathcal{S}$ a subset of $M_{n}(\mathbb{K})$ if $\delta(xy)=\delta(x)y+x\delta(y)$ for all $x,y\in\mathcal{S}$. Write $\mathcal{D}_{\mathcal{S}}^{\times}(M_{n}(\mathbb{K}))$ for the set of all multiplicative derivations on the subset $\mathcal{S}$ of $M_{n}(\mathbb{K})$. If $\mathcal{S}=M_{n}^{s}(\mathbb{K})$, we also write $\mathcal{D}_{s}^{\times}(M_{n}(\mathbb{K}))$ for $\mathcal{D}_{\mathcal{S}}^{\times}(M_{n}(\mathbb{K}))$ and call a multiplicative derivation on $M_{n}^{s}(\mathbb{K})$ a multiplicative derivation on rank-$s$ matrices. Write $e_{ij}$ for the $n\times n$ matrix with 1 in the position $(i,j)$ and 0 in every other position. The symbol $\sum_{i=a}^{b}e_{ii}$ will denote zero matrix once $a>b$. Denote by $I_{n}$ the $n\times n$ identity matrix, by $\underline{n}$ the set $\{1,2,\ldots,n\}$ and by $\mathbb{K}^{t}$ the set of all $t\times 1$ matrices over $\mathbb{K}$.

Firstly, we note that the set of all multiplicative derivations on a nonempty subset $\mathcal{S}$ of $M_{n}(\mathbb{K})$ is a vector space.

Lemma 2.1**.**

$\mathcal{D}^{\times}_{\mathcal{S}}(M_{n}(\mathbb{K}))$* is a $\mathbb{K}$-vector space.*

Proof.* We only need to show that for any $\delta_{1},\delta_{2}\in\mathcal{D}_{\mathcal{S}}^{\times}(M_{n}(\mathbb{K}))$ and any $\lambda_{1},\lambda_{2}\in\mathbb{K}$*

[TABLE]

In fact, for any $x,y\in\mathcal{S}$,

[TABLE]

which implies that (2.1) holds. $\Box$

The following Remark 2.2 and Corollary 2.3 will be used in the proof of Lemma 2.5, Theorem 2.6 and Corollary 3.1.

Remark 2.2*.*

Let $M_{n}(\mathbb{K})$ be the ring of all $n\times n$ matrices over a field $\mathbb{K}$ where $n\geq 2$. Let $1\leq s\leq n$ and $2s-n\leq k\leq s$ be integers. Then for each $y\in M_{n}^{k}({\mathbb{K}})$, there exist $y_{1},y_{2}\in M_{n}^{s}({\mathbb{K}})$ such that $y=y_{1}y_{2}$.

Proof.* There exist invertible matrices $P,Q\in GL_{n}(\mathbb{K})$ such that*

[TABLE]

where we denote by $\sum_{i=1}^{k}e_{ii}$ the zero matrix in the case of $k=0$. From $s+s-k=2s-k\leq n$ we have the desired matrices

[TABLE]

$\Box$**

Corollary 2.3**.**

Let $M_{n}(\mathbb{K})$ be the ring of all $n\times n$ matrices over a field $\mathbb{K}$ where $n\geq 2$. If $0\leq s\leq\frac{n}{2}$, then for each $y\in M_{n}^{<s}({\mathbb{K}})$, there exist $y_{1},y_{2}\in M_{n}^{s}({\mathbb{K}})$ such that $y=y_{1}y_{2}$.

Proof.* Denote by $k\in\{0,1,\ldots,s-1\}$ the rank of $y$. From $2s\leq n$ we have $s>k\geq 0\geq 2s-n$. Then Remark 2.2 works. $\Box$*

The following Lemma 2.4 shows that $\delta(0)=0$ and gives a kind of special case for Lemma 2.5. Furthermore, Lemma 2.4 will be used in the proof of Lemma 2.5.

Lemma 2.4**.**

$\delta(0)=0$* for $\delta\in\mathcal{D}_{s}^{\times}(M_{n}(\mathbb{K}))$, where $n\geq 2$ and $1\leq s\leq\frac{n}{2}$ are fixed. In particular, for $x,y\in M_{n}({\mathbb{K}})$ such that $0\in\{x,y\}$, $\delta(xy)=\delta(x)y+x\delta(y)$.*

Proof.* Let $e=\sum_{i=1}^{s}e_{ii}$, $f=\sum_{i=s+1}^{2s}e_{ii}$ and $g=\sum_{i=1}^{s}e_{i,i+s}$. Certainly, $e,f,g\in M_{n}^{s}({\mathbb{K}})$. From the property satisfied by $\delta$, we have*

[TABLE]

which means that

[TABLE]

for some $A,B\in M_{n}({\mathbb{K}})$. By the property satisfied by $\delta$, $ef=0$ and $ge=0$, we have

[TABLE]

which implies

[TABLE]

Particularly, for any $x\in M_{n}^{\leq s}({\mathbb{K}})$,

[TABLE]

$\Box$**

The following lemma will be used in the proof of Theorem 2.6

Lemma 2.5**.**

For $\delta\in\mathcal{D}_{s}^{\times}(M_{n}(\mathbb{K}))$ and $x,y\in M_{n}^{\leq s}({\mathbb{K}})$ such that $\{x,y\}\cap M_{n}^{\leq 1}({\mathbb{K}})\neq\phi$, where $n\geq 2$ and $1\leq s\leq\frac{n}{2}$ are fixed, $\delta(xy)=\delta(x)y+x\delta(y)$.

Proof.* By Lemma 2.4, it is enough to consider the case $\{x,y\}\cap M_{n}^{1}({\mathbb{K}})\neq\phi$. We will only prove the case for $x\in M_{n}^{1}({\mathbb{K}})$ and $y\in M_{n}^{\leq s}({\mathbb{K}})$. The proof of the case for $x\in M_{n}^{\leq s}({\mathbb{K}})$ and $y\in M_{n}^{1}({\mathbb{K}})$ is similar and so omitted.*

Step 1.* We will prove that for all $x\in M_{n}^{1}({\mathbb{K}})$ and all $y\in M_{n}^{s}({\mathbb{K}})$, $\delta(xy)=\delta(x)y+x\delta(y)$. There exist $P,Q,R,S\in GL_{n}({\mathbb{K}})$ such that $x=Pe_{11}Q$ and $y=R(\sum_{i=1}^{s}e_{ii})S$.*

Case-I.* $e_{11}QR(\sum_{i=1}^{s}e_{ii})\neq 0$. Then there exist $2\leq i_{2}<i_{3}<\cdots<i_{s}\leq n$ such that*

[TABLE]

since the rank of $QR(\sum_{i=1}^{s}e_{ii})$ is $s$. By $n\geq 2s$, we can choose $2\leq k_{2}<k_{3}<\cdots<k_{s}\leq n$ such that

[TABLE]

Set $x_{1}=P(e_{11}+\sum_{j=2}^{s}e_{k_{j},k_{j}})$ and $x_{2}=(e_{11}+\sum_{j=2}^{s}e_{i_{j},i_{j}})Q$. In this case, $x_{1},x_{2},x_{2}y\in M_{n}^{s}({\mathbb{K}})$ and $x=x_{1}x_{2}$. Hence

[TABLE]

Case-II.* $e_{11}QR(\sum_{i=1}^{s}e_{ii})=0$, which means that*

[TABLE]

where $G$ is an $(n-1)\times s$ matrix. Note that the rank of $G$ is $s$. So there exist linearly independent $\alpha_{1},\alpha_{2},\ldots,\alpha_{n-1-s}\in\mathbb{K}^{n-1}$ such that

[TABLE]

Since $2s\leq n$, we have $s-1\leq n-1-s$. Then $H=(\alpha_{1},\alpha_{2},\ldots,\alpha_{s-1})^{T}$ is a $(s-1)\times(n-1)$ matrix over $\mathbb{K}$. Note that the rank of $H$ is $s-1$ and $HG=0$. Set $x_{1}=P(e_{11}+\sum_{i=s+1}^{2s-1}e_{ii})$ and

[TABLE]

then $x_{1},x_{2}\in M_{n}^{s}({\mathbb{K}})$, $x=x_{1}x_{2}$ and $x_{2}y=0$. Hence by the property satisfied by $\delta$ and Lemma 2.4, we have

[TABLE]

Step 2.* For $x\in M_{n}^{1}({\mathbb{K}})$ and $y\in M_{n}^{<s}({\mathbb{K}})$, by Corollary 2.3, there exist $y_{1},y_{2}\in M_{n}^{s}({\mathbb{K}})$ such that $y=y_{1}y_{2}$. Furthermore, by the property satisfied by $\delta$, Lemma 2.4 and the conclusion of Step 1, keeping $xy_{1}\in M_{n}^{\leq 1}({\mathbb{K}})$ in mind, we have*

[TABLE]

which completes the proof. $\Box$

Theorem 2.6**.**

Let $n$ and $s$ be integers such that $n>1$ and $1\leq s\leq\frac{n}{2}$. Let $M_{n}(\mathbb{K})$ be the ring of all $n\times n$ matrices over a field $\mathbb{K}$. If a map $\delta:M_{n}(\mathbb{K})\rightarrow M_{n}(\mathbb{K})$ satisfies that $\delta(xy)=\delta(x)y+x\delta(y)$ for any two rank-$s$ matrices $x,y$, then there exists a derivation $D$ of $M_{n}(\mathbb{K})$ such that $\delta=D$ on $M_{n}^{\leq s}(\mathbb{K})$.

Proof.* By Lemma 2.5, for $i\in\underline{n}$,*

[TABLE]

which means that $\delta(e_{ii})=(a_{st}^{(i)})$, where $a_{st}^{(i)}=0$ for all

[TABLE]

By Lemmas 2.4 and 2.5, we have that for $i\neq j\in\underline{n}$

[TABLE]

which means that

[TABLE]

for $i\neq j\in\underline{n}$. Set

[TABLE]

It is easy to see that $Be_{jj}=\delta(e_{jj})e_{jj}$ for each $j\in\underline{n}$. On the other hand, by (2.3)

[TABLE]

Furthermore, from Lemma 2.5, we have that for $i\in\underline{n}$,

[TABLE]

For $i,j\in\underline{n}$, denote by $\lambda_{ij}$ the $(i,j)$ entry of $\delta(e_{ij})$. Note that from the expression of $\delta(e_{ii})$ we have that $\lambda_{ii}=0$ for all $i\in\underline{n}$. For $i\neq j\in\underline{n}$, by Lemma 2.5

[TABLE]

Hence for all $i,j\in\underline{n}$,

[TABLE]

For $i,j,k\in\underline{n}$, by Lemma 2.5 and (2.4)

[TABLE]

which implies that $\lambda_{ik}=\lambda_{ij}+\lambda_{jk}$ for all $i,j,k\in\underline{n}$. In particular, $0=\lambda_{ii}=\lambda_{ij}+\lambda_{ji}$ implies that $\lambda_{ij}=-\lambda_{ji}$ for all $i,j\in\underline{n}$. Set

[TABLE]

Then for all $s,t\in\underline{n}$,

[TABLE]

Let $\delta^{\prime}=\delta-ad_{A}$, so $\delta^{\prime}\in\mathcal{D}_{s}^{\times}(M_{n}(\mathbb{K}))$ has the same property with $\delta$ and $\delta^{\prime}(e_{st})=0$ for all $s,t\in\underline{n}$ by Lemma 2.1. For $a\in\mathbb{K}$ and $i,j\in\underline{n}$, by Lemma 2.5,

[TABLE]

which means that there exists a map $\mu_{ij}:\mathbb{K}\rightarrow\mathbb{K}$ such that $\delta^{\prime}(ae_{ij})=\mu_{ij}(a)e_{ij}$. For $a\in\mathbb{K}$ and $i\neq j\in\underline{n}$, by Lemma 2.5 and $\delta^{\prime}(e_{st})=0$ for all $s,t\in\underline{n}$,

[TABLE]

which means that $u_{ii}=u_{ij}=u_{ji}=u_{jj}$ for all $i\neq j\in\underline{n}$. Moreover for all $s,t,s^{\prime},t^{\prime}\in\underline{n}$,

[TABLE]

Denote $u_{st}$ for $s,t\in\underline{n}$ by $\mu$. For $a,b\in\mathbb{K}$,

[TABLE]

which implies that $\mu(ab)=a\mu(b)+\mu(a)b$ for all $a,b\in\mathbb{K}$. In particular, $\mu(1)=0$ from $\mu(1)=\mu(1\cdot 1)=\mu(1)\cdot 1+1\cdot\mu(1)$. Furthermore, for $x=(x_{ij})\in M_{n}^{\leq s}(\mathbb{K})$, by Lemma 2.5 and $\delta^{\prime}(e_{st})=0$ for all $s,t\in\underline{n}$,

[TABLE]

which means that $\delta^{\prime}((x_{ij}))=(\mu(x_{ij}))$ for all $x=(x_{ij})\in M_{n}^{\leq s}(\mathbb{K})$. Hence for all $x=(x_{ij})\in M_{n}^{\leq s}(\mathbb{K})$

[TABLE]

At last, by Lemma 2.5, $\mu(1)=0$ and $\delta^{\prime}((x_{ij}))=(\mu(x_{ij}))$ for all $x=(x_{ij})\in M_{n}^{\leq s}(\mathbb{K})$, we have that

[TABLE]

which implies that $\mu$ is additive and so $\mu$ is a derivation of $\mathbb{K}$, further inducing a derivation $\overline{\mu}$ of $M_{n}(\mathbb{K})$. Note that the restriction of $\overline{\mu}$ on $M_{n}^{\leq s}(\mathbb{K})$ is $\delta^{\prime}$. Hence $D=ad_{A}+\overline{\mu}$ is the desired. $\Box$

3. Application

For $a\in\mathbb{R}$, let $[a]$ be the least integer being not less than $a$. For example, $[-0.75]=0$, $[1.5]=2$ and $[3]=3$. As an application of Theorem 2.6, we will show that the multiplicative derivation on some nonadditive subset of the matrix ring $M_{n}(\mathbb{K})$ over a field $\mathbb{K}$ has to be a derivation.

Corollary 3.1**.**

Let $n>1$ be an integer. Let $M_{n}(\mathbb{K})$ be the ring of all $n\times n$ matrices over a field $\mathbb{K}$. If a map $\delta:M_{n}(\mathbb{K})\rightarrow M_{n}(\mathbb{K})$ satisfies that $\delta(xy)=\delta(x)y+x\delta(y)$ for any

[TABLE]

then $\delta$ is a derivation of $M_{n}(\mathbb{K})$.

Proof.* From $[\log_{2}(0.5n+1)]\geq\log_{2}(0.5n+1)$, we obtain*

[TABLE]

which gives that there exists a derivation $D$ of $M_{n}(\mathbb{K})$ such that $\delta=D$ on $M_{n}^{\leq s}(\mathbb{K})$ by Theorem 2.6. Let $\delta^{\prime}=\delta-D$, then $\delta^{\prime}(x)=0$ for all $x\in M_{n}^{\leq s}(\mathbb{K})$ and $\delta^{\prime}$ has the same property as $\delta$. Obviously,

[TABLE]

and

[TABLE]

for all $x\in M_{n}(\mathbb{K})$. For any $x\in M_{n}^{1}(\mathbb{K})$ and any $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$, there exists $x_{1},x_{2}\in M_{n}^{s}(\mathbb{K})$ such that $x=x_{1}x_{2}$. Then by $x_{1},x_{2},x_{2}y\in M_{n}^{\leq s}(\mathbb{K})$ and (3.1), we have

[TABLE]

for all $x\in M_{n}^{1}(\mathbb{K})$ and all $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$. Similarly, we have

[TABLE]

for all $x\in M_{n}^{1}(\mathbb{K})$ and all $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$. Hence for all $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$, by (3.1), (3.2) and (3.3), we have that for all $i,j\in\underline{n}$

[TABLE]

which means that $\delta^{\prime}(y)=0$ for all $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$.

Now we only need to show that $\delta^{\prime}(z)=0$ for any rank-$k$* matrix $z$, where $n+1-2^{i+1}<k<n+1-2^{i}$ and $1\leq i<[\log_{2}(0.5n+1)]$. It is easy to see that*

[TABLE]

Then by Remark 2.2, there exist $z_{1},z_{2}\in M_{n}^{n+1-2^{i}}(\mathbb{K})$ such that $z=z_{1}z_{2}$, which implies

[TABLE]

since $\delta^{\prime}(y)=0$ for all $y\in\cup_{i=0}^{[\log_{2}(0.5n+1)]}M_{n}^{n+1-2^{i}}(\mathbb{K})$.

In conclusion, $\delta^{\prime}=0$, which means that $\delta=D$ is a derivation of $M_{n}(\mathbb{K})$. $\Box$