Transitive Closure Preservers for Matrices over a Class of Semirings

Abstract

We introduce the transitive closure operator $\mathrm{\Gamma }$ on the set of square matrices over a class of semirings, and then define the transitive closure of a matrix. We characterize the surjective linear transformations of matrices preserving (strongly preserving) transitive closures, and show that a surjective linear transformation preserves the transitive closures if and only if it commutes with $\mathrm{\Gamma }$, which is equivalent to preserving the kernel of $\mathrm{\Gamma }$. We also deduce that the strong linear preservers of transitive closure on $2\times 2$ matrices over certain inclines are surjective. In particular, we provide a complete characterization of strong linear preservers of transitive closure on $n\times n$ matrices over the binary Boolean semiring.

1. Introduction

Linear preserver problems (LPPs) concern the characterization of linear transformations on matrix spaces that leave certain functions, subsets, relations etc., invariant. LPPs have been a classical and active topic in matrix algebra and operator theory (see [1,2]). Over the past few years, extensive results have been obtained for matrices over fields or various semirings. For instance, Beasley et al. [3,4] characterized linear preservers of matrix rank. Catalano et al. [5] described linear transformations preserving products equal to a diagonalizable matrix. Chen et al. [6] studied strong commutativity linear preservers of strictly triangular matrix Lie algebras. Costara [7] investigated linear transformations preserving fixed values of products of matrices at fixed vectors. Guterman et al. [8] classified linear isomorphisms preserving Green’s relations for matrices over anti-negative semifields. Kang et al. [9,10] studied linear preservers of regularity. Nikita et al. [11] characterized linear transformations preserving square roots of idempotent and rank-one nilpotent matrices. Pshenitsyna [12] classified linear preservers of invertibility. Deng et al. [13] investigated linear preservers of Kleene stars on matrices over inclines.

Despite these advances, some issues concerning linear preservers on matrices over inclines remain unresolved. Inclines are a special and important class of idempotent semirings that have wide applications in areas such as automata theory, the design of switching circuits, logic of binary relations, Markov chains, information systems, and fuzzy mathematics [14,15]. Transitive closure is a fundamental concept in graph theory and combinatorial matrix theory, which records the existence of paths between vertices in the associated digraph. The transitive closure can be naturally extended to matrices over semirings. While transitive closure linear preservers have been studied on tropical matrices [16], the case of incline matrices has not been systematically investigated. This paper aims to fill this gap.

The main contributions of this paper are as follows. In Section 3, we introduce the transitive closure operator $\mathrm{\Gamma }$ for matrices over commutative inclines without zero divisors, and define the transitive closure of such a matrix. In Section 4, we characterize the surjective linear transformations preserving (or strongly preserving) transitive closures (see Theorem 1). In Section 5, we prove that a surjective linear transformation preserves transitive closures if and only if it commutes with $\mathrm{\Gamma }$, if and only if it preserves the kernel of $\mathrm{\Gamma }$ (see Theorem 4). In Section 6, we show that strong linear preservers of transitive closure on $2\times 2$ matrices over certain inclines are surjective (see Theorems 6 and 7). In particular, we completely characterize strong linear preservers of transitive closure on square matrices over the binary Boolean semiring (see Theorem 8). These results extend the theory of linear preserver problems to incline matrices and enrich the connections between semiring matrix theory and graph-theoretic closure operations.

2. Preliminaries

A semiring is a nonempty set R together with two binary operations + (addition) and · (multiplication) such that $(R,+)$ is a commutative monoid with identity 0, $(R,·)$ is a monoid with identity 1, multiplication distributes over addition from either side, and $r·0=0·r=0$ for all $r\in R$. For simplicity we denote $a·b$ by $ab$.

Let R be a semiring and $a\in R$. a is a left zero divisor (right zero divisor, respectively) if $ab=0$ ($ba=0$, respectively) for some $b\in R\setminus \left\{0\right\}$ and a is a zero divisor if it is both a left and a right zero divisor. The semiring R is commutative if $ab=ba$ for all $a,b\in R$, R is anti-negative if $a+b=0$ implies that $a=b=0$ for all $a,b\in R$, R is idempotent if $a+a=a$ for all $a\in R$ (see [3,17]). An idempotent semiring R is called an incline if $a+ab=a$ and $b+ab=b$ for all $a,b\in R$ (see [14]). It is easily seen that if a semiring R is idempotent, then R is anti-negative and so the incline is anti-negative.

Now define a binary relation ≤ on an idempotent semiring R by

$$a\le b⟺a+b=b.$$

(1)

Then ≤ is a partial order on R (see [18]). Also, ≤ is compatible with + and ·, that is, for any $a,b,x\in R$,

$$a\le b⟹a+x\le b+x,ax\le bx\mathrm{and}xa\le xb.$$

(2)

In the sequel, S always denotes the commutative incline without zero divisors and with multiplicative identity 1, where $1\ne 0$. Clearly,

$$(\forall a\in S)a+1=1+a=1.$$

As defined in (1), ≤ is a partial order on S, and $0\le a\le 1$ for all $a\in S$. Moreover, for any $a_1,\dots ,a_k,b_1,\dots ,b_{k+1}\in S$,

$$b_1{a}_1\cdots b_k{a}_k{b}_{k+1}\le b_1\cdots b_k{b}_{k+1}.$$

(3)

It is well known that the binary Boolean semiring $\mathbb{B}=\left(\right\{0,1\},+,·)$, the fuzzy semiring $\mathcal{F}=\left(\right[0,1],+,·)$ (where $a+b=max\{a,b\}$ and $a·b=min\{a,b\}$), $\mathcal{R}=({\mathbb{R}}^{\le 0}\cup \{-\infty \},+,·)$ (where ${\mathbb{R}}^{\le 0}$ is the set of non-positive real numbers, $a+b=max\{a,b\}$ and $a·b$ is defined as the arithmetic sum of a and b) are examples of such incline S (see [4,9,15,19]).

Let $M_n\left(S\right)$ denote the set of all $n\times n$ matrices over S. For $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_n\left(S\right)$ and $x\in S$, define $A+B$, $AB$, $xA$ by

$$A+B=(a_{ij}+b_{ij}),AB=\left(\sum _{k=1}^n{a}_{ik}{b}_{kj}\right),xA=\left(x{a}_{ij}\right).$$

Notice that $(S,+)$ and $(S,·)$ are both commutative. Then $A+B=B+A$ and $Ax=xA$. With respect to the above addition and multiplication, $M_n\left(S\right)$ forms an idempotent semiring with an additive identity element $Z_n$ (the $n\times n$ matrix with all entries 0) and a multiplicative identity element $I_n$ (the $n\times n$ matrix with all diagonal entries 1 and all off-diagonal ones 0). Also, with respect to the above addition and scalar multiplication, $M_n\left(S\right)$ forms a S-semimodule.

Let $\left[n\right]$ denote the set $\{1,2,\dots ,n\}$. For any $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_n\left(S\right)$, if $A\le B$ in the sense of (1), then

$$A\le B⟺(\forall i,j\in \left[n\right])a_{ij}\le b_{ij}.$$

By (2), for any $A,B,X\in M_n\left(S\right)$,

$$A\le B⟹A+X\le B+X,XA\le XB\mathrm{and}AX\le BX.$$

Denote by $E_n$ the $n\times n$ matrix with all entries 1. Obviously, $Z_n\le A\le E_n$ for all $A\in M_n\left(S\right)$. An $n\times n$ matrix is called a permutation matrix if it is formed from the identity matrix $I_n$ by reordering its columns and/or rows. Let $\mathrm{Sym}\left(\right[n\left]\right)$ denote the symmetric group on $\left[n\right]$. Then for any $n\times n$ permutation matrix $P=\left(p_{ij}\right)$, there exists a unique $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ such that $p_{i{i}^{\sigma }}=1$ for all $i\in \left[n\right]$ and other entries of P are all 0. We usually denote P by $P_{\sigma }$. Clearly, $P_{\sigma }{P}_{{\sigma }^{-1}}=P_{{\sigma }^{-1}}{P}_{\sigma }=I_n$. Thus the inverse matrix of $P_{\sigma }$ is $P_{{\sigma }^{-1}}$, i.e., $P_{\sigma }^{-1}=P_{{\sigma }^{-1}}$. For $A\in M_n\left(S\right)$, denote by $A^T$ the transpose of A and by $A^k$ the iterated product $AA\cdots A$, in which A occurs k times. Since S is commutative, it follows that ${\left(A^T\right)}^k={\left(A^k\right)}^T$ for any $k\in \left[n\right]$.

Let T be a linear transformation on $M_n\left(S\right)$, that is, for any $A,B\in M_n\left(S\right)$ and $x\in S$,

$$T(A+B)=T\left(A\right)+T\left(B\right)\mathrm{and}T\left(xA\right)=xT\left(A\right).$$

Suppose that $\mathcal{P}\subseteq M_n\left(S\right)$ and that ∼ is an equivalence relation on $M_n\left(S\right)$. We say that T preserves $\mathcal{P}$ if $A\in \mathcal{P}$ implies that $T\left(A\right)\in \mathcal{P}$ and say that T strongly preserves $\mathcal{P}$ if $A\in \mathcal{P}$ if and only if $T\left(A\right)\in \mathcal{P}$. We say that T preserves ∼ if $A\sim B$ implies that $T\left(A\right)\sim T\left(B\right)$ and say that T strongly preserves ∼ if $A\sim B$ if and only if $T\left(A\right)\sim T\left(B\right)$.

3. Transitive Closures

Recall that a digraph (see [20]) is a pair $D=(V,E)$ where V is a nonempty finite set whose elements are called the vertices of D, and E is a subset of $V\times V$ whose elements are called the arcs of D. Any arc of the form $(i,i)$ is a loop. A sequence $\pi =(i_1,i_2,\dots ,i_p)$ of vertices in D is a path (in D) if $p=1$ or $p>1$ and $(i_s,i_{s+1})\in E$ for all $s\in [p-1]$. We also say that $\pi$ is a path from $i_1$ to $i_p$. The transitive closure of the digraph D is a digraph which contains an arc $(i,j)$ whenever there is a path from i to j in D.

Let $A=\left(a_{ij}\right)\in M_n\left(S\right)$. The digraph associated with A is denoted by $D_A=(V\left(A\right),E\left(A\right))$, where $V\left(A\right)=\left[n\right]$ is the vertex set and $E\left(A\right)=\{(i,j)\in \left[n\right]\times \left[n\right]\mid a_{ij}\ne 0\}$ is the arc set. Define a mapping $\mathrm{\Gamma }:M_n\left(S\right)\to M_n\left(S\right)$ by

$$\mathrm{\Gamma }\left(A\right)=\sum _{k=1}^n{A}^k.$$

We shall verify later that the transitive closure of digraph $D_A$ is $D_{\mathrm{\Gamma }\left(A\right)}$. Motivated by this result, we say that $\mathrm{\Gamma }$ is the transitive closure operator on $M_n\left(S\right)$, and say that $\mathrm{\Gamma }\left(A\right)$ is the transitive closure of A. Firstly, we have

Proposition 1.

Suppose that $A=\left(a_{ij}\right)\in M_n\left(S\right)$ and $\mathrm{\Gamma }$ be defined as above. Let $\mathrm{\Gamma }\left(A\right)=\left(a_{ij}^{★}\right)$. Then for any $i,j\in \left[n\right]$,

$$a_{ij}^{★}=a_{ij}+\sum _{k_1\in \left[n\right]\setminus \{i,j\}}{a}_{i{k}_1}{a}_{k_{1}j}+\cdots +\sum _{k_1,\dots ,k_{n-1}\in \left[n\right]\setminus \{i,j\}\mathit{are}\mathit{distinct}}{a}_{i{k}_1}{a}_{k_1{k}_2}\cdots a_{k_{n-1}j}.$$

Proof. 

Notice that $\mathrm{\Gamma }\left(A\right)=A+A^2+\cdots +A^n$. Then for any $i,j\in \left[n\right]$,

$$a_{ij}^{★}=a_{ij}+\sum _{i_1\in \left[n\right]}{a}_{i{i}_1}{a}_{i_{1}j}+\cdots +\sum _{i_1,\dots ,i_{n-1}\in \left[n\right]}{a}_{i{i}_1}{a}_{i_1{i}_2}\cdots a_{i_{n-1}j}.$$

We deduce that the conclusion holds by (1) and (3). □

Proposition 2.

Let $A\in M_n\left(S\right)$ and $\mathrm{\Gamma }$ be defined as above. Then $A^k\le \mathrm{\Gamma }\left(A\right)$ for all $k\ge 1$. Moreover, $\mathrm{\Gamma }\left(A\right)={\sum }_{k=1}^m{A}^k$ for all $m\ge n$.

Proof. 

Let $A=\left(a_{ij}\right)$. If $1\le k\le n$, then $A^k\le \mathrm{\Gamma }\left(A\right)$ by the definition of $\mathrm{\Gamma }$. Now suppose that $k\ge n+1$. Let $A^k=\left(a_{ij}^{\left(k\right)}\right)$ and $\mathrm{\Gamma }\left(A\right)=\left(a_{ij}^{★}\right)$. Then for any $i,j\in \left[n\right]$,

$$a_{ij}^{\left(k\right)}=\sum _{i_1,i_2,\cdots ,i_{k-1}\in \left[n\right]}{a}_{i{i}_1}{a}_{i_1{i}_2}\cdots a_{i_{k-1}j}.$$

Since $k\ge n+1$, we deduce that $i,i_1,i_2,\dots ,i_{k-1}$ are not pairwise different in $\left[n\right]$. By (3), $a_{ij}^{\left(k\right)}\le a_{ij}^{★}$. Thus $A^k\le \mathrm{\Gamma }\left(A\right)$. Hence $\mathrm{\Gamma }\left(A\right)={\sum }_{k=1}^m{A}^k$ for all $m\ge n$. □

Proposition 3.

Let $A\in M_n\left(S\right)$. Then the transitive closure of $D_A$ is $D_{\mathrm{\Gamma }\left(A\right)}$.

Proof. 

Let $A=\left(a_{ij}\right)$ and $\mathrm{\Gamma }\left(A\right)=\left(a_{ij}^{★}\right)$. Then

$$D_A=(\left[n\right],E\left(A\right))\mathrm{and}{D}_{\mathrm{\Gamma }\left(A\right)}=(\left[n\right],E\left(\mathrm{\Gamma }\left(A\right)\right)),$$

where

$$E\left(A\right)=\{(i,j)\in \left[n\right]\times \left[n\right]\mid a_{ij}\ne 0\},E\left(\mathrm{\Gamma }\left(A\right)\right)=\{(i,j)\in \left[n\right]\times \left[n\right]\mid a_{ij}^{★}\ne 0\}.$$

Suppose that $u,v\in \left[n\right]$. If there exist $u_1,\dots ,u_{k-1}\in \left[n\right]$ such that $(u,u_1,\dots ,u_{k-1},v)$ is a path in $D_A$, then $(u,u_1),\dots ,(u_{k-1},v)\in E\left(A\right)$ and so $a_{u{u}_1},\dots ,a_{u_{k-1}v}\ne 0$. Thus $a_{u{u}_1}\dots a_{u_{k-1}v}\ne 0$ since S has no zero divisors. Let $A^k=\left(a_{ij}^k\right)$. Then $a_{uv}^k\ne 0$ since S is anti-negative. By Proposition 2, $A^k\le \mathrm{\Gamma }\left(A\right)$ and so $a_{uv}^{★}\ne 0$. Hence $(u,v)\in E\left(\mathrm{\Gamma }\right(A\left)\right)$. This shows that the transitive closure of the digraph $D_A$ is $D_{\mathrm{\Gamma }\left(A\right)}$. □

Denote by $M_n^c$ the set of images of matrices under $\mathrm{\Gamma }$, i.e., $M_n^c=\{\mathrm{\Gamma }\left(A\right)\mid A\in M_n\left(S\right)\}$. As a consequence of Proposition 2, we immediately have

Corollary 1.

Let $A\in M_n\left(S\right)$. Then $A\in M_n^c$ if and only if $A^2\le A$.

By Corollary 1, it is easily seen that $A\in M_n^c$ if and only if $A^k\le A$ for all $k\ge 1$. Thus $\mathrm{\Gamma }\left(A\right)=A$ for any $A\in M_n^c$, that is, the restriction of $\mathrm{\Gamma }$ to $M_n^c$ is the identity mapping. In particular, $\mathrm{\Gamma }\left(I_n\right)=I_n$. Notice that $I_n$ is a permutation matrix; it is natural to consider the images of all permutation matrices under $\mathrm{\Gamma }$. In general, for any permutation matrix $P_{\sigma }\in M_n\left(S\right)$, write $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ as a product of m disjoint permutation cycles with lengthes $r_1,r_2,\dots ,r_m$, that is

$$\sigma =(i_1\cdots i_{r_1})(i_{r_1+1}\cdots i_{r_1+r_2})\cdots (i_{r_1+\cdots +r_{m-1}+1}\cdots i_n),$$

where $\{i_1,\dots ,i_n\}=\left[n\right]$ and $r_1+r_2+\cdots +r_m=n$. Let us denote

$$N_1=\{i_1,\dots ,i_{r_1}\},N_2=\{i_{r_1+1},\dots ,i_{r_1+r_2}\},N_m=\{i_{r_1+\cdots +r_{m-1}+1},\dots ,i_n\}.$$

Let $\mathrm{\Gamma }\left(P_{\sigma }\right)=\left(x_{ij}\right)$. Since $\mathrm{\Gamma }\left(P_{\sigma }\right)=P_{\sigma }+P_{{\sigma }^2}+\cdots +P_{{\sigma }^n}$, we deduce that

$$x_{ij}=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}(i,j)\in \bigcup _{k=1}^m{N}_k\times N_k,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

That is, for any $k\in \left[m\right]$, the $(N_k,N_k)$-submatrix of $\mathrm{\Gamma }\left(P_{\sigma }\right)$ is an $r_k\times r_k$ matrix with all entries 1, and for any $k,\mathsf{\ell }\in \left[m\right]$ with $k\ne \mathsf{\ell }$, the $(N_k,N_{\mathsf{\ell }})$-submatrix of $\mathrm{\Gamma }\left(P_{\sigma }\right)$ is an $r_k\times r_{\mathsf{\ell }}$ matrix with all entries 0. Now we immediately have

Proposition 4.

Let $P_{\sigma }$ be a permutation matrix in $M_n\left(S\right)$. Then $\mathrm{\Gamma }\left(P_{\sigma }\right)=E_n$ if and only if σ is a permutation cycle of order n.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a linear transformation and $\mathrm{\Gamma }$ be the transitive closure operator. We say that T commutes with $\mathrm{\Gamma }$ if $\mathrm{\Gamma }\left(T\right(A\left)\right)=T\left(\mathrm{\Gamma }\right(A\left)\right)$ for any $A\in M_n\left(S\right)$. Denote by ${\rho }^c$ the kernel of $\mathrm{\Gamma }$, that is,

$$A{\rho }^{c}B⟺\mathrm{\Gamma }\left(A\right)=\mathrm{\Gamma }\left(B\right).$$

Clearly, ${\rho }^c$ is an equivalence relation on $M_n\left(S\right)$.

Proposition 5.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a linear transformation. Then T commutes with $\mathrm{\Gamma }$ if and only if T preserves $M_n^c$ and preserves ${\rho }^c$.

Proof. 

Suppose that T commutes with $\mathrm{\Gamma }$. Let $A\in M_n^c$. Then $\mathrm{\Gamma }\left(A\right)=A$ and so $T\left(A\right)=T\left(\mathrm{\Gamma }\left(A\right)\right)=\mathrm{\Gamma }\left(T\left(A\right)\right)\in M_n^c$. Thus T preserves $M_n^c$. Now suppose that $A,B\in M_n\left(S\right)$ and $A{\rho }^{c}B$. Then $\mathrm{\Gamma }\left(A\right)=\mathrm{\Gamma }\left(B\right)$ and so $T\left(\mathrm{\Gamma }\right(A\left)\right)=T\left(\mathrm{\Gamma }\right(B\left)\right)$. Hence $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(B\left)\right)$ and so $T\left(A\right){\rho }^{c}T\left(B\right)$. This shows that T preserves ${\rho }^c$.

Conversely, suppose that T preserves $M_n^c$ and preserves ${\rho }^c$. Let $A\in M_n\left(S\right)$. Notice that $\mathrm{\Gamma }\left(A\right)\in M_n^c$. Then $T\left(\mathrm{\Gamma }\left(A\right)\right)\in M_n^c$ and so $\mathrm{\Gamma }\left(T\right(\mathrm{\Gamma }\left(A\right)\left)\right)=T\left(\mathrm{\Gamma }\right(A\left)\right)$. Since $A{\rho }^c\mathrm{\Gamma }\left(A\right)$, we have $T\left(A\right){\rho }^{c}T\left(\mathrm{\Gamma }\left(A\right)\right)$ and so $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(\mathrm{\Gamma }\left(A\right)\left)\right)$. Thus $\mathrm{\Gamma }\left(T\right(A\left)\right)=T\left(\mathrm{\Gamma }\right(A\left)\right)$. This shows that T commutes with $\mathrm{\Gamma }$. □

4. Surjective Linear Preservers of ${\mathit{M}}_{\mathit{n}}^{\mathit{c}}$

In this section, we consider surjective linear transformations on $M_n\left(S\right)$ that preserve $M_n^c$. Let $E_{i,j}$ denote the $n\times n$ matrix with 1 in the $(i,j)$-th position and 0 elsewhere. It is easily seen that

$$E_{i,j}{E}_{k,\mathsf{\ell }}=\left\{\begin{array}{cc}{E}_{i,\mathsf{\ell }},\hfill & \mathrm{if}j=k,\hfill \\ 0,\hfill & \mathrm{if}j\ne k.\hfill \end{array}\right.$$

For $A=\left(a_{ij}\right)\in M_n\left(S\right)$, we have $A={\sum }_{i,j\in \left[n\right]}{a}_{ij}{E}_{i,j}$. Let T be a linear transformation on $M_n\left(S\right)$. Then $T\left(A\right)={\sum }_{i,j\in \left[n\right]}{a}_{ij}T\left(E_{i,j}\right)$. This implies that T is completely determined by the images $T\left(E_{i,j}\right)$. Beasley and Guterman have given a characterization of surjective linear transformations on matrices over anti-negative semirings without zero divisors (see [3] (Theorem 2.14)).

Lemma 1

([3], Theorem 2.14). Let R be an anti-negative semiring without zero divisors and $T:M_{m,n}\left(R\right)\to M_{m,n}\left(R\right)$ be a linear transformation. Then the following statements are equivalent:

(i) 

T is bijective;

(ii) 

T is surjective;

(iii) 

There exists a permutation $\sigma \in Sym\left(\right[m]\times [n\left]\right)$ and units $x_{i,j}\in Z\left(R\right)$ such that $T\left(E_{i,j}\right)=x_{i,j}{E}_{{(i,j)}^{\sigma }}$ for all $i\in \left[m\right],j\in \left[n\right]$, where $Z\left(R\right)$ is the center of R.

Notice that the incline S which we consider in this paper is commutative, anti-negative and has no zero divisors. Also, we can verify that S contains the unique unit 1 (in fact, if $a\in S$ and $ab=1$, then $a=a1=a(1+b)=a+ab=a+1=1$ and so $b=1$). Now we have the following result which is a special case of Lemma 1.

Lemma 2.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a linear transformation. Then the following statements are equivalent:

(i) 

T is bijective;

(ii) 

T is surjective;

(iii) 

There exists $\sigma \in \mathit{Sym}\left(\right[n]\times [n\left]\right)$ such that $T\left(E_{i,j}\right)=E_{{(i,j)}^{\sigma }}$ for all $i,j\in \left[n\right]$.

Let T be a surjective linear transformation on $M_n\left(S\right)$. By Lemma 2, $T\left(E_n\right)=E_n$. In particular, if $n=1$, then $M_1\left(S\right)=S$ and $T\left(1\right)=1$. Thus T is the identity transformation on $M_1\left(S\right)$. In the following, we always assume that $n\ge 2$. Let

$${\mathcal{D}}_n=\{E_{i,i}\mid i\in \left[n\right]\}\mathrm{and}{\mathcal{E}}_n=\{E_{i,j}\mid i,j\in \left[n\right]\mathrm{and}i\ne j\}.$$

Then ${\mathcal{D}}_n$ and ${\mathcal{E}}_n$ are both finite sets. For $A=\left(a_{ij}\right)\in M_n\left(S\right)$, let us denote

$$A_{i,\ast }=\sum _{\mathsf{\ell }\in \left[n\right]}{a}_{i\mathsf{\ell }}{E}_{i,\mathsf{\ell }}\mathrm{and}{A}_{\ast ,i}=\sum _{\mathsf{\ell }\in \left[n\right]}{a}_{\mathsf{\ell }i}{E}_{\mathsf{\ell },i}.$$

Proposition 6.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$. Then $T\left(I_n\right)=I_n$. Moreover, $T\left({\mathcal{D}}_n\right)={\mathcal{D}}_n$ and $T\left({\mathcal{E}}_n\right)={\mathcal{E}}_n$.

Proof. 

By Lemma 2, there exists $\sigma \in \mathrm{Sym}\left(\right[n]\times [n\left]\right)$ such that

$$T\left(I_n\right)=T(E_{1,1}+\cdots +E_{n,n})=E_{{(1,1)}^{\sigma }}+\cdots +E_{{(n,n)}^{\sigma }}.$$

This implies that $T\left(I_n\right)$ has exactly n entries equal to 1, and all other entries are 0. We claim that $T\left(I_n\right)=\left(y_{ij}\right)$ is a permutation matrix. Otherwise, there are two possibilities of $T\left(I_n\right)$.

Case 1. ${\left(T\left(I_n\right)\right)}_{p,\ast }=Z_n$ for some $p\in \left[n\right]$. Then there exists $q\in \left[n\right]\setminus \left\{p\right\}$ such that $y_{qk},y_{q\mathsf{\ell }}\ne 0$ for some $k,\mathsf{\ell }\in \left[n\right]$ with $k\ne \mathsf{\ell }$. Since T is surjective, by Lemma 2, $E_{p,q}=T\left(E_{s,t}\right)$ for some $s,t\in \left[n\right]$ with $s\ne t$. Let $A=I_n+E_{s,t}$. Then $A^2=A$ and so $A\in M_n^c$. Let $T\left(A\right)=\left(z_{ij}\right)$ and ${\left(T\left(A\right)\right)}^2=\left(w_{ij}\right)$. Then $z_{pq}=1$ and $z_{pu}=0$ for all $u\in \left[n\right]\setminus \left\{q\right\}$. Also, $w_{pk}=w_{p\mathsf{\ell }}=1$. Notice that $k\ne \mathsf{\ell }$. Then $T\left(A\right)\notin M_n^c$ by Corollary 1. This contradicts that T preserves $M_n^c$.

Case 2. ${\left(T\left(I_n\right)\right)}_{\ast ,p}=Z_n$ for some $p\in \left[n\right]$. Then there exists $q\in \left[n\right]\setminus \left\{p\right\}$ such that $y_{kq},y_{\mathsf{\ell }q}\ne 0$ for some $k,\mathsf{\ell }\in \left[n\right]$ with $k\ne \mathsf{\ell }$. Since T is surjective, by Lemma 2, $E_{q,p}=T\left(E_{s,t}\right)$ for some $s,t\in \left[n\right]$ with $s\ne t$. Let $A=I_n+E_{s,t}$. Then $A^2=A$ and so $A\in M_n^c$. A similar argument as in Case 1 establishes that $T\left(A\right)\notin M_n^c$, a contradiction.

From the above discussion, $T\left(I_n\right)$ is a permutation matrix. Since $I_n\in M_n^c$, we have $T\left(I_n\right)\in M_n^c$ and so ${\left(T\left(I_n\right)\right)}^2\le T\left(I_n\right)$. Thus $T\left(I_n\right)=I_n$. Moreover, notice that ${\mathcal{D}}_n$, ${\mathcal{E}}_n$ are both finite sets. Then $T\left({\mathcal{D}}_n\right)={\mathcal{D}}_n$ and $T\left({\mathcal{E}}_n\right)={\mathcal{E}}_n$. □

Lemma 3.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$, $i,j\in \left[n\right]$ and $i\ne j$. Then the following statements hold:

(i) 

If $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{i,j}\right)=E_{s,t}$, then either $s=p,t\ne p$ or $s\ne p,t=p$;

(ii) 

If $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{j,i}\right)=E_{s,t}$, then either $s=p,t\ne p$ or $s\ne p,t=p$.

Proof. 

Suppose that $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{i,j}\right)=E_{s,t}$. If $s=p$, then $t\ne p$ by Proposition 6. If $s\ne p$, we assert $t=p$. Otherwise, assume that $t\ne p$ and $T\left(E_{u,v}\right)=E_{t,s}$. Then $u\ne v$ and $(u,v)\ne (i,j)$. We consider the following two possibilities of u:

Case 1. $u=j$. Firstly, we show that $v=i$. Otherwise, if $v\ne i$, let $A=E_{i,j}+E_{j,v}+E_{i,v}$. Then $A\in M_n^c$. Since T preserves $M_n^c$, we have $T\left(A\right)\in M_n^c$ and so ${\left(T\left(A\right)\right)}^2\le T\left(A\right)$. Suppose that $T\left(E_{i,v}\right)=E_{h,k}$, $h\ne k$, then $T\left(A\right)=E_{s,t}+E_{t,s}+E_{h,k}$. Notice that the $(t,t)$-entry of $T\left(A\right)$ is 0 and the $(t,t)$-entry of $T{\left(A\right)}^2$ is 1; this contradicts with ${\left(T\left(A\right)\right)}^2\le T\left(A\right)$. Thus $v=i$. Now let

$$B=E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j}\mathrm{and}T\left(E_{j,j}\right)=E_{q,q},q\ne p.$$

Then $B\in M_n^c$ and $T\left(B\right)=E_{p,p}+E_{s,t}+E_{t,s}+E_{q,q}$. Since T preserves $M_n^c$, we have $T\left(B\right)\in M_n^c$ and so ${\left(T\left(B\right)\right)}^2\le T\left(B\right)$. Thus $p=s$ or $p=t$. It is a contradiction.

Case 2. $u\ne j$. Firstly, we shall show that $v\ne i$. Otherwise, if $v=i$, let $A=E_{i,j}+E_{u,i}+E_{u,j}$. Then $A\in M_n^c$. Since T preserves $M_n^c$, we have $T\left(A\right)\in M_n^c$ and so ${\left(T\left(A\right)\right)}^2\le T\left(A\right)$. Suppose that $T\left(E_{u,j}\right)=E_{h,k}$, $h\ne k$, then $T\left(A\right)=E_{s,t}+E_{t,s}+E_{h,k}$. Notice that the $(t,t)$-entry of $T\left(A\right)$ is 0 and the $(t,t)$-entry of $T{\left(A\right)}^2$ is 1; this contradicts with ${\left(T\left(A\right)\right)}^2\le T\left(A\right)$. Thus $v\ne i$. Now let

$$B=E_{i,i}+E_{i,j}+E_{u,v}+E_{i,v}\mathrm{and}T\left(E_{i,v}\right)=E_{q,r},q\ne r.$$

Then $B\in M_n^c$ and $T\left(B\right)=E_{p,p}+E_{s,t}+E_{t,s}+E_{q,r}$. Since T preserves $M_n^c$, we have $T\left(B\right)\in M_n^c$ and so ${\left(T\left(B\right)\right)}^2\le T\left(B\right)$. Thus $p=s$ or $p=t$. It is a contradiction.

We have shown (i) holds. A similar argument as in (i) establishes (ii). □

Lemma 4.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$; $i,j,k\in \left[n\right]$ are pairwise different. Then the following statements hold:

(i) 

If $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{i,k}\right)=E_{s,t}$, then either $s=p,t\ne q$ or $s\ne p,t=q$;

(ii) 

If $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{k,j}\right)=E_{s,t}$, then either $s=p,t\ne q$ or $s\ne p,t=q$.

Proof. 

Suppose that $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{i,k}\right)=E_{s,t}$, then $p\ne q$ and $s\ne t$ by Proposition 6. If $s=p$, then $t\ne q$ since T is bijective. If $s\ne p$, we assert $t=q$. Otherwise, if $t\ne q$, let $A=E_{i,i}+E_{i,j}+E_{i,k}$, then $A\in M_n^c$. Since T preserves $M_n^c$, we have $T\left(A\right)\in M_n^c$. Notice that $T\left(E_{i,j}\right)=E_{p,q}$, then $T\left(E_{i,i}\right)=E_{p,p}$ or $T\left(E_{i,i}\right)=E_{q,q}$ by Lemma 3.

If $T\left(E_{i,i}\right)=E_{p,p}$, then $T\left(A\right)=E_{p,p}+E_{p,q}+E_{s,p}\notin M_n^c$, a contradiction. If $T\left(E_{i,i}\right)=E_{q,q}$, then $T\left(A\right)=E_{q,q}+E_{p,q}+E_{q,t}\notin M_n^c$, a contradiction. Thus, $s\ne p,t=q$ and (i) holds. The proof of (ii) follows by a similar argument as in (i). □

Lemma 5.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$ and $T\left(E_{i,i}\right)=E_{p,p}$. Then

(i) 

$T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$ or $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,p}$;

(ii) 

$T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$ or $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{p,\ast }$.

Proof. 

(i) Suppose $T\left(E_{i,j}\right)=E_{s,t}$, $j\in \left[n\right]\setminus \left\{i\right\}$. Then $s=p,t\ne p$ or $s\ne p,t=p$. If $n=2$, then (i) holds. Now suppose that $n\ge 3$. For any $k\in \left[n\right]\setminus \{i,j\}$, let $T\left(E_{i,k}\right)=E_{u,v}$, then $u\ne v$. By Lemmas 3 and 4, if $s=p,t\ne p$, then $u=p$; if $s\ne p,t=p$, then $v=p$. Thus (i) holds.

(ii) Suppose $T\left(E_{j,i}\right)=E_{s,t}$, $j\in \left[n\right]\setminus \left\{i\right\}$. Then $s=p,t\ne p$ or $s\ne p,t=p$. If $n=2$, then (i) holds. Now suppose that $n\ge 3$. For any $k\in \left[n\right]\setminus \{i,j\}$, let $T\left(E_{k,i}\right)=E_{u,v}$, then $u\ne v$. By Lemmas 3 and 4, if $s=p,t\ne p$, then $u=p$; if $s\ne p,t=p$, then $v=p$. Thus (ii) holds. □

Lemma 6.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$ and $i\in \left[n\right]$. Then the following statements hold:

(i) 

If $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$, then for any $j\in \left[n\right]\setminus \left\{i\right\}$, there exists $h\in \left[n\right]\setminus \left\{p\right\}$ such that $T{\left(E_n\right)}_{j,\ast }={\left(E_n\right)}_{h,\ast }$;

(ii) 

If $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$, then for any $j\in \left[n\right]\setminus \left\{i\right\}$, there exists $h\in \left[n\right]\setminus \left\{p\right\}$ such that $T{\left(E_n\right)}_{\ast ,j}={\left(E_n\right)}_{\ast ,h}$;

(iii) 

If $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,p}$, then for any $j\in \left[n\right]\setminus \left\{i\right\}$, there exists $h\in \left[n\right]\setminus \left\{p\right\}$ such that $T{\left(E_n\right)}_{j,\ast }={\left(E_n\right)}_{\ast ,h}$;

(iv) 

If $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{p,\ast }$, then for any $j\in \left[n\right]\setminus \left\{i\right\}$, there exists $h\in \left[n\right]\setminus \left\{p\right\}$ such that $T{\left(E_n\right)}_{\ast ,j}={\left(E_n\right)}_{h,\ast }$.

Proof. 

We shall show that (i) holds. The proofs of (ii), (iii) and (iv) are similar to those of (i). If $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$, then $T\left(E_{i,i}\right)=E_{p,p}$. For any $j\in \left[n\right]\setminus \left\{i\right\}$, let $T\left(E_{j,j}\right)=E_{h,h}$, $h\ne p$. We assert $T{\left(E_n\right)}_{j,\ast }={\left(E_n\right)}_{h,\ast }$. Otherwise, $T{\left(E_n\right)}_{j,\ast }={\left(E_n\right)}_{\ast ,h}$ by Lemma 5. Thus there exist $s,t\in \left[n\right]$ such that $T\left(E_{i,s}\right)=E_{p,h}$ and $T\left(E_{j,t}\right)=E_{p,h}$. This contradicts that T is bijective. Thus (i) holds. □

Lemma 7.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving $M_n^c$. Then for any $i\in \left[n\right]$,

(i) 

$T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$ if and only if $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$;

(ii) 

$T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,p}$ if and only if $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{p,\ast }$.

Proof. 

We shall show that (i) holds. The proof of (ii) is similar to that of (i). Suppose that $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$, then $T\left(E_{i,i}\right)=E_{p,p}$. By Lemma 5 (ii), $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$ or $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{p,\ast }$. If $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{p,\ast }$, then for $q\in \left[n\right]\setminus \left\{p\right\}$, there exist $s,u\in \left[n\right]\setminus \left\{i\right\}$ such that $T\left(E_{i,s}\right)=E_{p,q}$ and $T\left(E_{u,i}\right)=E_{p,q}$. This contradicts that T is bijective. Thus $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$.

Conversely, if $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,p}$, then $T\left(E_{i,i}\right)=E_{p,p}$. By Lemma 5 (i), $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$ or $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,p}$. If $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,p}$, then for $q\in \left[n\right]\setminus \left\{p\right\}$, there exist $s,u\in \left[n\right]\setminus \left\{i\right\}$ such that $T\left(E_{s,i}\right)=E_{q,p}$ and $T\left(E_{i,u}\right)=E_{q,p}$. This contradicts that T is bijective. Thus $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{p,\ast }$. □

Lemma 8.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a transformation. If there exists $\sigma \in \mathit{Sym}\left(\right[n\left]\right)$ such that $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(S\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(S\right)$, then T is a surjective linear transformation strongly preserving $M_n^c$.

Proof. 

Suppose that $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(S\right)$. It is easily seen that T is linear and surjective. Moreover, for any $A\in M_n\left(S\right)$,

$$A\in M_n^c\iff A^2\le A\iff P_{\sigma }^{-1}{A}^2{P}_{\sigma }\le P_{\sigma }^{-1}A{P}_{\sigma }\iff {\left(T\left(A\right)\right)}^2\le T\left(A\right)\iff T\left(A\right)\in M_n^c.$$

Thus T is a surjective linear transformation strongly preserving $M_n^c$.

Now suppose that $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(S\right)$. Notice that $P_{\sigma }^T=P_{\sigma }^{-1}$. Then $T\left(A\right)={\left(P_{\sigma }^{-1}A{P}_{\sigma }\right)}^T$ and so T is a linear surjective transformation. Since $A\in M_n^c$ if and only if $A^T\in M_n^c$, we deduce that T strongly preserves $M_n^c$. □

Theorem 1.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a surjective linear transformation. Then the following statements are equivalent:

(i) 

T preserves $M_n^c$;

(ii) 

T strongly preserves $M_n^c$;

(iii) 

There exists $\sigma \in \mathit{Sym}\left(\right[n\left]\right)$ such that either $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(S\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(S\right)$.

Proof. 

By Lemma 8, (iii) implies (ii). (ii) implies (i) is clear. We shall show that (i) implies (iii). Suppose that T is a surjective linear transformation preserving $M_n^c$. If $n=1$, then T is the identity transformation on $M_1\left(S\right)=S$. Thus (i) ⇒ (iii). Now we consider $n\ge 2$. Let $T\left(E_{k,k}\right)=E_{p,p}$. By Lemma 5, we consider the following two cases of $T{\left(E_n\right)}_{k,\ast }$.

Case 1. $T{\left(E_n\right)}_{k,\ast }={\left(E_n\right)}_{p,\ast }$. By Lemma 6 (i), (ii) and Lemma 7 (i), for any $i\in \left[n\right]$, there exists $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ such that $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{i^{\sigma },\ast }$ and $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{\ast ,i^{\sigma }}$. Thus for any $i,j\in \left[n\right]$, $T\left(E_{i,j}\right)=E_{i^{\sigma },j^{\sigma }}$. Hence for any $A=\left(a_{ij}\right)\in M_n\left(S\right)$,

$$T\left(A\right)=\sum _{i,j\in \left[n\right]}{a}_{ij}T\left(E_{i,j}\right)=\sum _{i,j\in \left[n\right]}{a}_{ij}{E}_{i^{\sigma },j^{\sigma }}=P_{\sigma }^{-1}A{P}_{\sigma }.$$

Case 2. $T{\left(E_n\right)}_{k,\ast }={\left(E_n\right)}_{\ast ,p}$. By Lemma 6 (iii), (iv) and Lemma 7 (ii), for any $i\in \left[n\right]$, there exists $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ such that $T{\left(E_n\right)}_{i,\ast }={\left(E_n\right)}_{\ast ,i^{\sigma }}$ and $T{\left(E_n\right)}_{\ast ,i}={\left(E_n\right)}_{i^{\sigma },\ast }$. Thus for any $i,j\in \left[n\right]$, $T\left(E_{i,j}\right)=E_{j^{\sigma },i^{\sigma }}$. Hence for any $A=\left(a_{ij}\right)\in M_n\left(S\right)$,

$$T\left(A\right)=\sum _{i,j\in \left[n\right]}{a}_{ij}T\left(E_{i,j}\right)=\sum _{i,j\in \left[n\right]}{a}_{ij}{E}_{j^{\sigma },i^{\sigma }}=P_{\sigma }^{-1}{A}^T{P}_{\sigma }.$$

□

Theorem 2.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a surjective linear transformation. Then T preserves $M_n^c$ if and only if T commutes with $\mathrm{\Gamma }$.

Proof. 

By Proposition 5, the converse holds. Now suppose that T is a surjective linear transformation preserving $M_n^c$. By Theorem 1, there exists $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ such that $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(S\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(S\right)$. In both cases, we can verify that $\mathrm{\Gamma }\left(T\right(A\left)\right)=T\left(\mathrm{\Gamma }\right(A\left)\right)$. Thus T commutes with $\mathrm{\Gamma }$. □

5. Equivalent Statements

In this section, we consider surjective linear transformations that preserve ${\rho }^c$, and then provide some equivalent statements to determine whether a surjective linear transformation preserves $M_n^c$. Notice that every surjective linear transformation on $M_1\left(S\right)$ is the identity transformation. In the following, we always assume that $n\ge 2$.

Proposition 7.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving ${\rho }^c$. Then the following statements hold:

(i) 

For any cycle $\sigma \in \mathit{Sym}\left(\right[n\left]\right)$ of order n, there exists a cycle $\delta \in Sym\left(\right[n\left]\right)$ of order n such that $T\left(P_{\sigma }\right)=P_{\delta }$;

(ii) 

$T\left({\mathcal{D}}_n\right)={\mathcal{D}}_n$ and $T\left({\mathcal{E}}_n\right)={\mathcal{E}}_n$. Moreover, $T\left(I_n\right)=I_n$.

Proof. 

(i) Let $\sigma \in \mathrm{Sym}\left(\right[n\left]\right)$ be a cycle of order n. Then $\mathrm{\Gamma }\left(P_{\sigma }\right)=E_n=\mathrm{\Gamma }\left(E_n\right)$ and so $P_{\sigma }{\rho }^c{E}_n$. Since T preserves ${\rho }^c$, we have $T\left(P_{\sigma }\right){\rho }^{c}T\left(E_n\right)$ and so $\mathrm{\Gamma }\left(T\left(P_{\sigma }\right)\right)=\mathrm{\Gamma }\left(T\left(E_n\right)\right)$. By Lemma 2, $T\left(E_n\right)=E_n$. Thus $\mathrm{\Gamma }\left(T\left(P_{\sigma }\right)\right)=E_n$. Clearly, $T\left(P_{\sigma }\right)$ is an $n\times n$ matrix with exactly n entries 1, and all other entries 0. By Proposition 4, we deduce that $T\left(P_{\sigma }\right)=P_{\delta }$ for some cycle $\delta \in \mathrm{Sym}\left(\right[n\left]\right)$ of order n.

(ii) Let $\mathcal{C}=\{P_{\sigma }\in M_n\left(S\right)|\sigma \mathrm{be}\mathrm{the}\mathrm{cycle}\mathrm{of}\mathrm{order}n\mathrm{in}\mathrm{Sym}\left(\left[n\right]\right)\}$. Then T is bijective on $\mathcal{C}$ by (i). Notice that ${\sum }_{i\ne j}{E}_{i,j}={\sum }_{P_{\sigma }\in \mathcal{C}}{P}_{\sigma }$. Then

$$T(\sum _{i\ne j}{E}_{i,j})=T(\sum _{P_{\sigma }\in \mathcal{C}}{P}_{\sigma })=\sum _{P_{\sigma }\in \mathcal{C}}{P}_{\sigma }=\sum _{i\ne j}{E}_{i,j}.$$

By Lemma 2, $T\left({\mathcal{D}}_n\right)={\mathcal{D}}_n$ and $T\left({\mathcal{E}}_n\right)={\mathcal{E}}_n$. Moreover, $T\left(I_n\right)=I_n$. □

Lemma 9.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving ${\rho }^c$, $i,j\in \left[n\right]$ and $i\ne j$. Then the following statements hold:

(i) 

If $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{i,j}\right)=E_{s,t}$, then either $s=p,t\ne p$ or $s\ne p,t=p$;

(ii) 

If $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{j,i}\right)=E_{s,t}$, then either $s=p,t\ne p$ or $s\ne p,t=p$.

Proof. 

Suppose that $T\left(E_{i,i}\right)=E_{p,p}$ and $T\left(E_{i,j}\right)=E_{s,t}$. If $s=p$, then $t\ne p$ by Proposition 7. If $s\ne p$, we assert $t=p$. In fact, let $T\left(E_{j,i}\right)=E_{u,v}$ and $A=E_{i,j}+E_{j,i}$. Then $u\ne v$ and $\mathrm{\Gamma }\left(A\right)=E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j}$. Since $A{\rho }^c\mathrm{\Gamma }\left(A\right)$ and T preserves ${\rho }^c$, we have $T\left(A\right){\rho }^{c}T\left(\mathrm{\Gamma }\left(A\right)\right)$ and so $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(\mathrm{\Gamma }\left(A\right)\left)\right)$. Clearly, $T\left(\mathrm{\Gamma }\right(A\left)\right)\le \mathrm{\Gamma }\left(T\right(A\left)\right)$. Notice that

$$\mathrm{\Gamma }\left(T\left(A\right)\right)=\mathrm{\Gamma }(E_{s,t}+E_{u,v})=E_{s,t}+E_{u,v}+E_{s,t}{E}_{u,v}+E_{u,v}{E}_{s,t},$$

$$T\left(\mathrm{\Gamma }\left(A\right)\right)=E_{p,p}+E_{s,t}+E_{u,v}+E_{h,h},\mathrm{where}T\left(E_{j,j}\right)=E_{h,h}.$$

We have $s=v,t=u=p$ and so (i) holds. The proof of (ii) follows by a similar argument as in (i). □

Proposition 7 tells us that if T is a surjective linear transformation preserving ${\rho }^c$, then T is bijective on ${\mathcal{D}}_n$ and ${\mathcal{E}}_n$, respectively. As a consequence of Lemma 9, we have

Corollary 2.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving ${\rho }^c$, $i,j\in \left[n\right]$ and $i\ne j$. Then the following statements hold:

(i) 

If $T\left(E_{i,j}\right)=E_{s,t}$ and $T\left(E_{j,i}\right)=E_{u,v}$, then $s=v,t=u$;

(ii) 

If $T\left(E_{i,j}\right)=E_{s,t}$ and $T\left(E_{u,v}\right)=E_{t,s}$, then $i=v,j=u$.

Lemma 10.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving ${\rho }^c$, $i,j,k\in \left[n\right]$ are pairwise different. Then the following statements hold:

(i) 

If $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{i,k}\right)=E_{s,t}$, then either $s=p,t\ne q$ or $s\ne p,t=q$;

(ii) 

If $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{k,j}\right)=E_{s,t}$, then either $s=p,t\ne q$ or $s\ne p,t=q$.

Proof. 

Suppose that $T\left(E_{i,j}\right)=E_{p,q}$ and $T\left(E_{i,k}\right)=E_{s,t}$, then $p\ne q$ and $s\ne t$ by Proposition 7. If $s=p$, then $t\ne q$ since T is bijective. If $s\ne p$, we assert $t=q$. Otherwise, if $t\ne q$, let $A=E_{k,i}+E_{i,j}$. Then $\mathrm{\Gamma }\left(A\right)=E_{k,i}+E_{i,j}+E_{k,j}$. By Corollary 2, $T\left(E_{k,i}\right)=E_{t,s}$. Then $T\left(A\right)=E_{t,s}+E_{p,q}$. Since $A{\rho }^c\mathrm{\Gamma }\left(A\right)$ and T preserves ${\rho }^c$, we have $T\left(A\right){\rho }^{c}T\left(\mathrm{\Gamma }\left(A\right)\right)$ and so $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(\mathrm{\Gamma }\left(A\right)\left)\right)$. Clearly, $T\left(\mathrm{\Gamma }\right(A\left)\right)\le \mathrm{\Gamma }\left(T\right(A\left)\right)$. Notice that

$$\mathrm{\Gamma }\left(T\left(A\right)\right)=T\left(A\right)=E_{t,s}+E_{p,q},$$

$$T\left(\mathrm{\Gamma }\left(A\right)\right)=E_{t,s}+E_{p,q}+T\left(E_{k,j}\right),T\left(E_{k,j}\right)\notin \{E_{t,s},E_{p,q}\}.$$

It is a contradiction. Thus $t=q$ and so (i) holds. The proof of (ii) follows by a similar argument as in (i). □

As a consequence of Lemmas 9 and 10, we have

Corollary 3.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a surjective linear transformation preserving ${\rho }^c$, $k\in \left[n\right]\setminus \{i,j\}$. Then the following statements hold:

(i) 

If $T\left(E_{p,q}\right)=E_{i,j}$ and $T\left(E_{s,t}\right)=E_{i,k}$, then either $s=p,t\ne q$ or $s\ne p,t=q$;

(ii) 

If $T\left(E_{p,q}\right)=E_{j,i}$ and $T\left(E_{s,t}\right)=E_{k,i}$, then either $s=p,t\ne q$ or $s\ne p,t=q$.

Theorem 3.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a surjective linear transformation. Then T preserves ${\rho }^c$ if and only if T commutes with $\mathrm{\Gamma }$.

Proof. 

By Proposition 5, the converse holds. Now suppose that T is a surjective linear transformation preserving ${\rho }^c$. We shall show that T preserves $M_n^c$. Let $A=\left(a_{ij}\right)\in M_n^c$ and $T\left(A\right)=\left(b_{ij}\right)$. Then for any $i,j\in \left[n\right]$, we have

$$E_{i,j}=T\left(E_{p,q}\right)\iff b_{ij}=a_{pq}.$$

(4)

Clearly, the $(i,j)$-entry of $T{\left(A\right)}^2$ is ${\sum }_{k=1}^n{b}_{ik}{b}_{kj}$. Let $T\left(E_{p,q}\right)=E_{i,j}$. If $k\notin \{i,j\}$, then $E_{i,k}=T\left(E_{p,t}\right)$ or $E_{i,k}=T\left(E_{t,q}\right)$ by Corollary 3. Moreover, if $E_{i,k}=T\left(E_{p,t}\right)$, then $E_{k,j}=T\left(E_{t,q}\right)$; if $E_{i,k}=T\left(E_{t,q}\right)$, then $E_{k,j}=T\left(E_{p,t}\right)$. By (4), we have $b_{ik}{b}_{kj}=a_{pt}{a}_{tq}$. Since $A^2\le A$ and $b_{ij}=a_{pq}$, we have

$$\sum _{k=1}^n{b}_{ik}{b}_{kj}=b_{ii}{b}_{ij}+b_{ij}{b}_{jj}+\sum _{k\ne i,j}{b}_{ik}{b}_{kj}\le b_{ij}+\sum _{t\ne p,q}{a}_{pt}{a}_{tq}\le b_{ij}+a_{pq}=b_{ij}.$$

Hence $T{\left(A\right)}^2\le T\left(A\right)$ and so $T\left(A\right)\in M_n^c$. Thus T preserves $M_n^c$. By Proposition 5, T commutes with $\mathrm{\Gamma }$. □

We summarize the results of Theorems 1–3 in a single theorem as follows.

Theorem 4.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be a surjective linear transformation. Then the following statements are equivalent:

(i) 

T preserves $M_n^c$;

(ii) 

T preserves ${\rho }^c$;

(iii) 

T commutes with $\mathrm{\Gamma }$;

(iv) 

T strongly preserves $M_n^c$;

(v) 

T strongly preserves ${\rho }^c$;

(vi) 

There exists $\sigma \in Sym\left(\right[n\left]\right)$ such that either $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(S\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(S\right)$.

In the remainder of this section, we provide examples to show that there exist non-surjective linear transformations that preserve $M_n^c$, commute with $\mathrm{\Gamma }$ or preserve ${\rho }^c$.

Example 1.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ be defined by $T\left(A\right)=Z_n$ for all $A\in M_n\left(S\right)$. Clearly, T is linear and non-surjective. We can verify that T commutes with $\mathrm{\Gamma }$, preserves $M_n^c$ and preserves ${\rho }^c$.

Example 2.

Let $n=2$ and $T:M_2\left(S\right)\to M_2\left(S\right)$ be a transformation such that $T\left(A\right)=a_{12}{E}_{1,2}$ for all $A=\left(a_{ij}\right)\in M_2\left(S\right)$. Then T is a non-surjective linear transformation commuting with $\mathrm{\Gamma }$. Consequently, T preserves $M_2^c$ and ${\rho }^c$, respectively.

It is clear that T is linear and non-surjective. For any $A=\left(a_{ij}\right)\in M_2\left(S\right)$, by a simple calculation, we obtain that

$$\mathrm{\Gamma }\left(A\right)=\left(\begin{array}{cc}{a}_{11}+a_{12}{a}_{21}& a_{12}\\ a_{21}& a_{22}+a_{21}{a}_{12}\end{array}\right).$$

Thus $T\left(\mathrm{\Gamma }\left(A\right)\right)=a_{12}{E}_{1,2}=\mathrm{\Gamma }\left(T\left(A\right)\right)$. This shows that T commutes with $\mathrm{\Gamma }$. By Proposition 5, T preserves $M_2^c$ and ${\rho }^c$, respectively.

Example 3.

Let $n=2$ and $T:M_2\left(S\right)\to M_2\left(S\right)$ be a transformation such that $T\left(A\right)=a_{12}{E}_{1,2}+a_{21}{E}_{2,1}$ for all $A=\left(a_{ij}\right)\in M_2\left(S\right)$. Then T is a non-surjective linear transformation preserving ${\rho }^c$. However, T does not preserve $M_2^c$ or commute with $\mathrm{\Gamma }$.

Obviously, T is linear and non-surjective. Suppose that $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_2\left(S\right)$ and $A{\rho }^{c}B$. Then $\mathrm{\Gamma }\left(A\right)=\mathrm{\Gamma }\left(B\right)$. We deduce that $a_{12}=b_{12}$ and $a_{21}=b_{21}$. Thus $T\left(A\right)=a_{12}{E}_{1,2}+a_{21}{E}_{2,1}=b_{12}{E}_{1,2}+b_{21}{E}_{2,1}=T\left(B\right)$ and so $T\left(A\right){\rho }^{c}T\left(B\right)$. This shows that T preserves ${\rho }^c$. Notice that $E_2\in M_2^c$ but $T\left(E_2\right)=E_{1,2}+E_{2,1}\notin M_2^c$. Then T does not preserve $M_2^c$. By Proposition 5, T does not commute with $\mathrm{\Gamma }$.

Example 4.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a transformation such that $T\left(A\right)=a_{11}{E}_{1,1}$ for all $A=\left(a_{ij}\right)\in M_n\left(S\right)$. Then T is a non-surjective linear transformation preserving $M_n^c$. However, T does not preserve ${\rho }^c$ or commute with $\mathrm{\Gamma }$.

It is clear that T is linear and non-surjective. Notice that $a_{11}{E}_{1,1}\in M_n^c$ for any $A=\left(a_{ij}\right)\in M_n\left(S\right)$. Then T preserves $M_n^c$. Now let $A=E_{1,2}+E_{2,1}$. Then $T\left(\mathrm{\Gamma }\left(A\right)\right)=E_{1,1}$ and $\mathrm{\Gamma }\left(T\left(A\right)\right)=Z_n$. Thus $T\left(\mathrm{\Gamma }\right(A\left)\right)\ne \mathrm{\Gamma }\left(T\right(A\left)\right)$, and so T does not commute with $\mathrm{\Gamma }$. By Proposition 5, T does not preserve ${\rho }^c$.

Example 5.

Let $S=\mathbb{B}$ and $T:M_n\left(\mathbb{B}\right)\to M_n\left(\mathbb{B}\right)$ ($n\ge 2$) be a transformation such that $T\left(A\right)=A^{\circ }$ for all $A=\left(a_{ij}\right)\in M_n\left(\mathbb{B}\right)$, where $A^{\circ }={\sum }_{i\ne j}{a}_{ij}{E}_{i,j}$. Then T is a non-surjective linear transformation preserving ${\rho }^c$. However, T does not preserve $M_n^c$ or commute with $\mathrm{\Gamma }$.

Clearly, T is linear and non-surjective. Suppose that $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_n\left(\mathbb{B}\right)$. Let $\mathrm{\Gamma }\left(A\right)=\left(a_{ij}^{★}\right)$ and $\mathrm{\Gamma }\left(B\right)=\left(b_{ij}^{★}\right)$. If $A{\rho }^{c}B$, then $\mathrm{\Gamma }\left(A\right)=\mathrm{\Gamma }\left(B\right)$ and so $a_{ij}^{★}=b_{ij}^{★}$ for all $i,j\in \left[n\right]$. We shall show that $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(B\left)\right)$, that is, $\mathrm{\Gamma }\left(A^{\circ }\right)=\mathrm{\Gamma }\left(B^{\circ }\right)$. Now let $A^{\circ }=\left(a_{ij}^{\circ }\right)$, $B^{\circ }=\left(b_{ij}^{\circ }\right)$, $\mathrm{\Gamma }\left(A^{\circ }\right)=\left(x_{ij}\right)$, and $\mathrm{\Gamma }\left(B^{\circ }\right)=\left(y_{ij}\right)$. Notice that $\mathbb{B}=\{0,1\}$. We consider the following two cases of $a_{ij}^{★}$.

Case 1. $a_{ij}^{★}=0$. Then $b_{ij}^{★}=0$. Since $A^{\circ }\le A$ and $B^{\circ }\le B$, we have $\mathrm{\Gamma }\left(A^{\circ }\right)\le \mathrm{\Gamma }\left(A\right)$ and $\mathrm{\Gamma }\left(B^{\circ }\right)\le \mathrm{\Gamma }\left(B\right)$. Then $x_{ij}=y_{ij}=0$.

Case 2. $a_{ij}^{★}=1$. Then $b_{ij}^{★}=1$. Let

$$H=\sum _{k_1\in \left[n\right]\setminus \{i,j\}}{a}_{i{k}_1}{a}_{k_{1}j}+\cdots +\sum _{k_1,\dots ,k_{n-1}\in \left[n\right]\setminus \{i,j\}\mathrm{are}\mathrm{distinct}}{a}_{i{k}_1}{a}_{k_1{k}_2}\cdots a_{k_{n-1}j}$$

and

$$L=\sum _{k_1\in \left[n\right]\setminus \{i,j\}}{b}_{i{k}_1}{b}_{k_{1}j}+\cdots +\sum _{k_1,\dots ,k_{n-1}\in \left[n\right]\setminus \{i,j\}\mathrm{are}\mathrm{distinct}}{b}_{i{k}_1}{b}_{k_1{k}_2}\cdots b_{k_{n-1}j}.$$

By Proposition 1, $a_{ij}^{★}=a_{ij}+H$ and $b_{ij}^{★}=b_{ij}+L$. Since $a_{ij}^{★}=b_{ij}^{★}$, we have $a_{ij}+H=b_{ij}+L$. We claim that $H=1$ if and only if $L=1$. In fact, if $H=1$, then there exist distinct $k_1,\dots ,k_s\in \left[n\right]\setminus \{i,j\}$ such that $a_{i{k}_1}{a}_{k_1{k}_2}\cdots a_{k_{s}j}=1$. Hence $a_{i{k}_1}=a_{k_1{k}_2}=\cdots =a_{k_{s}j}=1$. Since $A\le \mathrm{\Gamma }\left(A\right)$, we have $a_{i{k}_1}^{★}=a_{k_1{k}_2}^{★}=\cdots =a_{k_{s}j}^{★}=1$ and so $b_{i{k}_1}^{★}=b_{k_1{k}_2}^{★}=\cdots =b_{k_{s}j}^{★}=1$. Thus $b_{i{k}_1}^{★}{b}_{k_1{k}_2}^{★}\cdots b_{k_{s}j}^{★}=1$. By (3), we deduce that $L=1$. Similarly, if $L=1$, then $H=1$. Moreover, $H=L$ since $\mathbb{B}=\{0,1\}$. For any $i,j\in \left[n\right]$, $x_{ij}=a_{ij}^{\circ }+H$ and $y_{ij}=b_{ij}^{\circ }+L$. If $i=j$, then $a_{ij}^{\circ }=b_{ij}^{\circ }=0$ and so $x_{ij}=H=L=y_{ij}$. If $i\ne j$, then $a_{ij}^{\circ }=a_{ij}$ and $b_{ij}^{\circ }=b_{ij}$. Thus $x_{ij}=a_{ij}^{★}=b_{ij}^{★}=y_{ij}$.

We have shown that $\mathrm{\Gamma }\left(T\right(A\left)\right)=\mathrm{\Gamma }\left(T\right(B\left)\right)$. Thus $T\left(A\right){\rho }^{c}T\left(B\right)$ and so T preserves ${\rho }^c$. Notice that $E_n\in M_n^c$ but $T\left(E_n\right)=E_n^{\circ }\notin M_n^c$. Then T does not preserve $M_n^c$. By Proposition 5, T does not commute with $\mathrm{\Gamma }$.

6. Strong Linear Preservers of ${\mathit{M}}_{\mathit{n}}^{\mathit{c}}$

In this section, we stipulate that for any $a,b\in S$, $a\le b$ or $b\le a$. That is, the partial order ≤ defined by (1) is a total order on incline S. We shall consider linear transformations strongly preserving $M_n^c$ on $M_n\left(S\right)$. In particular, we characterize the strong linear preservers on $M_2\left(\mathcal{R}\right)$, $M_2\left(\mathcal{F}\right)$ and $M_n\left(\mathbb{B}\right)$.

Now we give some notations that will be used later. If $a\le b$ and $a\ne b$, we denote $a<b$. Let $Y\subseteq S$. If $a\in Y$ and $b\le a$ for any $b\in Y$, we denote $a=\max Y$; if $a\in Y$ and $a\le b$ for any $b\in Y$, we denote $a=\min Y$. Let $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_n\left(S\right)$. Denote $A-B=\left(c_{ij}\right)$, where

$$c_{ij}=\left\{\begin{array}{cc}0,\hfill & \mathrm{if}{b}_{ij}\ne 0,\hfill \\ a_{ij},\hfill & \mathrm{if}{b}_{ij}=0.\hfill \end{array}\right.$$

Notice that $M_1\left(S\right)=M_1^c=S$. Take $a\in S$ and define $T:M_1\left(S\right)\to M_1\left(S\right)$ by $T\left(x\right)=ax$ for any $x\in M_1\left(S\right)$. It is clear that T strongly preserves $M_1^c$. In the following, we consider $n\ge 2$.

Proposition 8.

Let $T:M_n\left(S\right)\to M_n\left(S\right)$ ($n\ge 2$) be a linear transformation strongly preserving $M_n^c$. Then for any $i,j\in \left[n\right]$, there is no $(s,t)\ne (i,j)$ such that $T\left(E_{i,j}\right)\le T\left(E_{s,t}\right)$. Consequently, $T\left(E_{i,j}\right)\ne Z_n$ and $T\left(E_{i,j}\right)\ne E_n$.

Proof. 

Assume that $T\left(E_{i,j}\right)\le T\left(E_{s,t}\right)$ for some $(s,t)\ne (i,j)$. If $n\ge 3$ or $i=j$, then $E_n-E_{i,j}\notin M_n^c$. Notice that $T(E_n-E_{i,j})=T\left(E_n\right)\in M_n^c$. This contradicts that T strongly preserves $M_n^c$. If $n=2$ and $i\ne j$, then $E_{s,t}=E_{j,i}$ or $E_{s,t}=E_{s,s}$ ($s\in \{i,j\}$). Hence $E_{i,j}+E_{j,i}+E_{s,s}\notin M_2^c$ and $E_{j,i}+E_{s,s}\in M_2^c$. Notice that $T(E_{i,j}+E_{j,i}+E_{s,s})=T(E_{j,i}+E_{s,s})$, a contradiction.

Thus for any $i,j\in \left[n\right]$, there is no $(s,t)\ne (i,j)$ such that $T\left(E_{i,j}\right)\le T\left(E_{s,t}\right)$. Clearly, $T\left(E_{i,j}\right)\ne Z_n$ and $T\left(E_{i,j}\right)\ne E_n$. □

We now consider strong linear preservers of $M_2^c$ on $M_2\left(S\right)$. Firstly, it is easily verified that

Proposition 9.

Let $A=\left(a_{ij}\right)\in M_2\left(S\right)$. Then $A\in M_2^c\iff a_{12}{a}_{21}\le \min \{a_{11},a_{22}\}$.

Let T be a linear transformation strongly preserving $M_2^c$. In the following, we denote

$$T\left(E_{1,1}\right)=\left(\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right),T\left(E_{1,2}\right)=\left(\begin{array}{cc}{b}_{11}& b_{12}\\ b_{21}& b_{22}\end{array}\right),$$

$$T\left(E_{2,1}\right)=\left(\begin{array}{cc}{c}_{11}& c_{12}\\ c_{21}& c_{22}\end{array}\right),T\left(E_{2,2}\right)=\left(\begin{array}{cc}{d}_{11}& d_{12}\\ d_{21}& d_{22}\end{array}\right).$$

Notice that $E_{1,1},E_{1,2},E_{2,1},E_{2,2}\in M_2^c$. Then $T\left(E_{1,1}\right),T\left(E_{1,2}\right),T\left(E_{2,1}\right),T\left(E_{2,2}\right)\in M_2^c$. By Proposition 9,

$$a_{12}{a}_{21}\le \min \{a_{11},a_{22}\},b_{12}{b}_{21}\le \min \{b_{11},b_{22}\},$$

$$c_{12}{c}_{21}\le \min \{c_{11},c_{22}\},d_{12}{d}_{21}\le \min \{d_{11},d_{22}\}.$$

Moreover, we have the following conclusions.

Lemma 11.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then the following statements hold:

(i) 

$\min \{a_{11},a_{22}\}=0$, $\min \{a_{12},a_{21}\}=0$;

(ii) 

$\min \{b_{11},b_{22}\}=0$, $\min \{b_{12},b_{21}\}=0$;

(iii) 

$\min \{c_{11},c_{22}\}=0$, $\min \{c_{12},c_{21}\}=0$;

(iv) 

$\min \{d_{11},d_{22}\}=0$, $\min \{d_{12},d_{21}\}=0$.

Proof. 

(i) Assume that $r=\min \{a_{11},a_{22}\}\ne 0$. Then $E_{1,1}+r{E}_{1,2}+r{E}_{2,1}\notin M_2^c$ and

$$\begin{array}{ccc}\hfill (a_{12}+r{b}_{12}+r{c}_{12})(a_{21}+r{b}_{21}+r{c}_{21})& \le & a_{12}{a}_{21}+r\hfill \\ & =& r\hfill \\ & \le & \min \{a_{11}+r{b}_{11}+r{c}_{11},a_{22}+r{b}_{22}+r{c}_{22}\}.\hfill \end{array}$$

Thus $T(E_{1,1}+r{E}_{1,2}+r{E}_{2,1})\in M_2^c$. This contradicts that T strongly preserves $M_2^c$. Hence $\min \{a_{11},a_{22}\}=0$. Moreover, $0\le a_{12}{a}_{21}\le \min \{a_{11},a_{22}\}=0$ and so $a_{12}{a}_{21}=0$. Since S contains no zero divisors, we deduce that $\min \{a_{12},a_{21}\}=0$.

(ii) Assume that $r=\min \{b_{11},b_{22}\}\ne 0$. Then $E_{1,2}+r{E}_{2,1}\notin M_2^c$ and

$$\begin{array}{ccc}\hfill (b_{12}+r{c}_{12})(b_{21}+r{c}_{21})& \le & b_{12}{b}_{21}+r\hfill \\ & =& r\hfill \\ & \le & \min \{b_{11}+r{c}_{11},b_{22}+r{c}_{22}\}.\hfill \end{array}$$

Thus $T(E_{1,2}+r{E}_{2,1})\in M_2^c$. This contradicts that T strongly preserves $M_2^c$. Hence $\min \{b_{11},b_{22}\}=0$. Moreover, $0\le b_{12}{b}_{21}\le \min \{b_{11},b_{22}\}=0$ and so $b_{12}{b}_{21}=0$. Since S contains no zero divisors, we deduce that $\min \{b_{12},b_{21}\}=0$.

(iii) Assume that $r=\min \{c_{11},c_{22}\}\ne 0$. Then $r{E}_{1,2}+E_{2,1}\notin M_2^c$ and

$$\begin{array}{ccc}\hfill (r{b}_{12}+c_{12})(r{b}_{21}+c_{21})& \le & c_{12}{c}_{21}+r\hfill \\ & =& r\hfill \\ & \le & \min \{r{b}_{11}+c_{11},r{b}_{22}+c_{22}\}.\hfill \end{array}$$

Thus $T(r{E}_{1,2}+E_{2,1})\in M_2^c$. This contradicts that T strongly preserves $M_2^c$. Hence $\min \{c_{11},c_{22}\}=0$. Moreover, $0\le c_{12}{c}_{21}\le \min \{c_{11},c_{22}\}=0$ and so $c_{12}{c}_{21}=0$. Since S contains no zero divisors, we deduce that $\min \{c_{12},c_{21}\}=0$.

(iv) Assume that $r=\min \{d_{11},d_{22}\}\ne 0$. Then $E_{2,2}+r{E}_{1,2}+r{E}_{2,1}\notin M_2^c$ and

$$\begin{array}{ccc}\hfill (d_{12}+r{b}_{12}+r{c}_{12})(d_{21}+r{b}_{21}+r{c}_{21})& \le & d_{12}{d}_{21}+r\hfill \\ & =& r\hfill \\ & \le & \min \{d_{11}+r{b}_{11}+r{c}_{11},d_{22}+r{b}_{22}+r{c}_{22}\}.\hfill \end{array}$$

Thus $T(E_{2,2}+r{E}_{1,2}+r{E}_{2,1})\in M_2^c$. This contradicts that T strongly preserves $M_2^c$. Hence $\min \{d_{11},d_{22}\}=0$. Moreover, $0\le d_{12}{d}_{21}\le \min \{d_{11},d_{22}\}=0$ and so $d_{12}{d}_{21}=0$. Since S contains no zero divisors, we deduce that $\min \{d_{12},d_{21}\}=0$. □

Lemma 12.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then either (I) $b_{21}=c_{12}=0,b_{12}\ne 0,c_{21}\ne 0$ or (II) $b_{12}=c_{21}=0,b_{21}\ne 0,c_{12}\ne 0$.

Proof. 

By Lemma 11 (iii) and (iv),

$$b_{12}\ne 0,b_{21}=0\mathrm{or}{b}_{12}=0,b_{21}\ne 0\mathrm{or}{b}_{12}=b_{21}=0$$

$$c_{12}\ne 0,c_{21}=0\mathrm{or}{c}_{12}=0,c_{21}\ne 0\mathrm{or}{c}_{12}=c_{21}=0.$$

Assume that the conclusion does not hold. Then $b_{12}+c_{12}=0$ or $b_{21}+c_{21}=0$. Thus

$$0=(b_{12}+c_{12})(b_{21}+c_{21})\le \min \{b_{11}+c_{11},b_{22}+c_{22}\}$$

and so $T(E_{1,2}+E_{2,1})\in M_2^c$. Notice that $E_{1,2}+E_{2,1}\notin M_2^c$. This contradicts that T strongly preserves $M_2^c$. □

Lemma 13.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then $a_{12}=a_{21}=0,d_{12}=d_{21}=0$. Moreover, either (I) $a_{11}\ne 0$, $a_{12}=a_{21}=a_{22}=0$, $d_{11}=d_{12}=d_{21}=0$, $d_{22}\ne 0$ or (II) $a_{22}\ne 0$, $a_{11}=a_{12}=a_{21}=0$, $d_{12}=d_{21}=d_{22}=0$, $d_{11}\ne 0$.

Proof. 

Firstly, let $t=\max \{a_{12},a_{21}\}$, we assert $t=0$. Otherwise, if $t\ne 0$, then

$$E_{1,1}+t{E}_{1,2},E_{1,1}+t{E}_{2,1}\in M_2^c\mathrm{and}{E}_{1,1}+t{E}_{1,2}+t{E}_{2,1}\notin M_2^c.$$

Since T preserves $M_2^c$, we have $T(E_{1,1}+t{E}_{1,2}),T(E_{1,1}+t{E}_{2,1})\in M_2^c$. Thus

$$(a_{12}+t{b}_{12})(a_{21}+t{b}_{21})\le \min \{a_{11}+t{b}_{11},a_{22}+t{b}_{22}\}$$

(5)

and

$$(a_{12}+t{c}_{12})(a_{21}+t{c}_{21})\le \min \{a_{11}+t{c}_{11},a_{22}+t{c}_{22}\}.$$

(6)

By Lemma 12, there are two cases to consider:

Case 1. $b_{12}\ne 0,c_{21}\ne 0,b_{21}=c_{12}=0$. By (5) and (6), we have

$$a_{12}{a}_{21}+t{b}_{12}{a}_{21}\le \min \{a_{11}+t{b}_{11},a_{22}+t{b}_{22}\},$$

$$a_{12}{a}_{21}+t{a}_{12}{c}_{21}\le \min \{a_{11}+t{c}_{11},a_{22}+t{c}_{22}\}.$$

Since $t=\max \{a_{12},a_{21}\}$, we deduce that

$$\begin{array}{ccc}\hfill (a_{12}+t{b}_{12}+t{c}_{12})(a_{21}+t{b}_{21}+t{c}_{21})& =& (a_{12}+t{b}_{12})(a_{21}+t{c}_{21})\hfill \\ & =& a_{12}{a}_{21}+t{b}_{12}{a}_{21}+t{a}_{12}{c}_{21}+t^2{b}_{12}{c}_{21}\hfill \\ & =& a_{12}{a}_{21}+t{b}_{12}{a}_{21}+t{a}_{12}{c}_{21}\hfill \\ & \le & \min \{a_{11}+t{b}_{11}+t{c}_{11},a_{22}+t{b}_{22}+t{c}_{22}\}\hfill \end{array}$$

and so $T(E_{1,1}+t{E}_{1,2}+t{E}_{2,1})\in M_2^c$. Notice that $E_{1,1}+t{E}_{1,2}+t{E}_{2,1}\notin M_2^c$. This contradicts that T strongly preserves $M_2^c$.

Case 2. $b_{21}\ne 0,c_{12}\ne 0,b_{12}=c_{21}=0$. By (5) and (6), we have

$$a_{12}{a}_{21}+t{a}_{12}{b}_{21}\le \min \{a_{11}+t{b}_{11},a_{22}+t{b}_{22}\},$$

$$a_{12}{a}_{21}+t{c}_{12}{a}_{21}\le \min \{a_{11}+t{c}_{11},a_{22}+t{c}_{22}\}.$$

Since $t=\max \{a_{12},a_{21}\}$, we deduce that

$$\begin{array}{ccc}\hfill (a_{12}+t{b}_{12}+t{c}_{12})(a_{21}+t{b}_{21}+t{c}_{21})& =& (a_{12}+t{c}_{12})(a_{21}+t{b}_{21})\hfill \\ & =& a_{12}{a}_{21}+t{c}_{12}{a}_{21}+t{a}_{12}{b}_{21}+t^2{c}_{12}{b}_{21}\hfill \\ & =& a_{12}{a}_{21}+t{c}_{12}{a}_{21}+t{a}_{12}{b}_{21}\hfill \\ & \le & \min \{a_{11}+t{b}_{11}+t{c}_{11},a_{22}+t{b}_{22}+t{c}_{22}\}\hfill \end{array}$$

and so $T(E_{1,1}+t{E}_{1,2}+t{E}_{2,1})\in M_2^c$. Notice that $E_{1,1}+t{E}_{1,2}+t{E}_{2,1}\notin M_2^c$. This contradicts that T strongly preserves $M_2^c$.

From the above discussion, $t=0$ and so $a_{12}=a_{21}=0$. Secondly, let $u=\max \{d_{12},d_{21}\}$, we assert $u=0$. Otherwise, if $u\ne 0$, then

$$E_{2,2}+u{E}_{1,2},E_{2,2}+u{E}_{2,1}\in M_2^c\mathrm{and}{E}_{2,2}+u{E}_{1,2}+u{E}_{2,1}\notin M_2^c.$$

Since T preserves $M_2^c$, we have $T(E_{2,2}+u{E}_{1,2}),T(E_{2,2}+u{E}_{2,1})\in M_2^c$. By Lemma 12, $b_{12}\ne 0,c_{21}\ne 0,b_{21}=c_{12}=0$ or $b_{21}\ne 0,c_{12}\ne 0,b_{12}=c_{21}=0$. A similar discussion as above shows that $T(E_{2,2}+u{E}_{1,2}+u{E}_{2,1})\in M_2^c$. This contradicts that T strongly preserves $M_2^c$. Thus $u=0$ and so $d_{12}=d_{21}=0$.

Finally, by Lemma 11, we have

$$a_{11}\ne 0,a_{12}=a_{21}=a_{22}=0\mathrm{or}{a}_{11}=a_{12}=a_{21}=0,a_{22}\ne 0,$$

$$d_{11}\ne 0,d_{12}=d_{21}=d_{22}=0\mathrm{or}{d}_{11}=d_{12}=d_{21}=0,d_{22}\ne 0.$$

Assume that $a_{11}\ne 0$, $a_{12}=a_{21}=a_{22}=0$, $d_{11}\ne 0$, $d_{12}=d_{21}=d_{22}=0$ or $a_{11}=a_{12}=a_{21}=0$, $a_{22}\ne 0$, $d_{11}=d_{12}=d_{21}=0$, $d_{22}\ne 0$. Then $T\left(E_{1,1}\right)\le T\left(E_{2,2}\right)$ or $T\left(E_{2,2}\right)\le T\left(E_{1,1}\right)$. By Proposition 8, it’s a contradiction. Thus the conclusion holds. □

Lemma 14.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then $b_{11}=b_{22}=c_{11}=c_{22}=0$. Moreover, either (I) $b_{12}\ne 0$, $b_{11}=b_{21}=b_{22}=0$, $c_{21}\ne 0$, $c_{11}=c_{12}=c_{22}=0$ or (II) $b_{21}\ne 0$, $b_{11}=b_{12}=b_{22}=0$, $c_{12}\ne 0$, $c_{11}=c_{21}=c_{22}=0$.

Proof. 

Firstly, let $t=\max \{b_{11},c_{11}\}$, we assert $t=0$. Otherwise, if $t\ne 0$, by Lemma 13, there are two cases to consider:

Case 1. $a_{11}\ne 0,a_{12}=a_{21}=a_{22}=0,d_{11}=d_{12}=d_{21}=0,d_{22}\ne 0$. Then we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{11}+b_{11}+c_{11}& b_{12}+c_{12}\\ b_{21}+c_{21}& b_{22}+c_{22}+d_{22}\end{array}\right)$$

$$T(t{E}_{1,2}+t{E}_{2,1}+E_{2,2})=\left(\begin{array}{cc}t{b}_{11}+t{c}_{11}& t{b}_{12}+t{c}_{12}\\ t{b}_{21}+t{c}_{21}& t{b}_{22}+t{c}_{22}+d_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$ and so

$$(b_{12}+c_{12})(b_{21}+c_{21})\le \min \{a_{11}+b_{11}+c_{11},b_{22}+c_{22}+d_{22}\}.$$

Notice that $t=\max \{b_{11},c_{11}\}=b_{11}+c_{11}$; we deduce that

$$(t{b}_{12}+t{c}_{12})(t{b}_{21}+t{c}_{21})\le \min \{t{b}_{11}+t{c}_{11},t{b}_{22}+t{c}_{22}+d_{22}\}.$$

Thus $T(t{E}_{1,2}+t{E}_{2,1}+E_{2,2})\in M_2^c$. However, $t{E}_{1,2}+t{E}_{2,1}+E_{2,2}\notin M_2^c$. It is a contradiction.

Case 2. $a_{22}\ne 0,a_{11}=a_{12}=a_{21}=0,d_{12}=d_{21}=d_{22}=0,d_{11}\ne 0$. Then we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{b}_{11}+c_{11}+d_{11}& b_{12}+c_{12}\\ b_{21}+c_{21}& a_{22}+b_{22}+c_{22}\end{array}\right)$$

$$T(E_{1,1}+t{E}_{1,2}+t{E}_{2,1})=\left(\begin{array}{cc}t{b}_{11}+t{c}_{11}& t{b}_{12}+t{c}_{12}\\ t{b}_{21}+t{c}_{21}& a_{22}+t{b}_{22}+t{c}_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$ and so

$$(b_{12}+c_{12})(b_{21}+c_{21})\le \min \{b_{11}+c_{11}+d_{11},a_{22}+b_{22}+c_{22}\}.$$

Notice that $t=\max \{b_{11},c_{11}\}=b_{11}+c_{11}$; we deduce that

$$(t{b}_{12}+t{c}_{12})(t{b}_{21}+t{c}_{21})\le \min \{t{b}_{11}+t{c}_{11},a_{22}+t{b}_{22}+t{c}_{22}\}.$$

Thus $T(E_{1,1}+t{E}_{1,2}+t{E}_{2,1})\in M_2^c$. However, $E_{1,1}+t{E}_{1,2}+t{E}_{2,1}\notin M_2^c$. It is a contradiction. From the above discussion, we have $t=0$ and so $b_{11}=c_{11}=0$.

Secondly, let $u=\max \{b_{22},c_{22}\}$, we assert $u=0$. Otherwise, if $u\ne 0$, by Lemma 13, there are also two cases to consider:

Case 1. $a_{11}\ne 0,a_{12}=a_{21}=a_{22}=0,d_{11}=d_{12}=d_{21}=0,d_{22}\ne 0$. Then we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{11}+b_{11}+c_{11}& b_{12}+c_{12}\\ b_{21}+c_{21}& b_{22}+c_{22}+d_{22}\end{array}\right)$$

$$T(E_{1,1}+u{E}_{1,2}+u{E}_{2,1})=\left(\begin{array}{cc}{a}_{11}+u{b}_{11}+u{c}_{11}& u{b}_{12}+u{c}_{12}\\ u{b}_{21}+u{c}_{21}& u{b}_{22}+u{c}_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$ and so

$$(b_{12}+c_{12})(b_{21}+c_{21})\le \min \{a_{11}+b_{11}+c_{11},b_{22}+c_{22}+d_{22}\}.$$

Notice that $u=\max \{b_{22},c_{22}\}=b_{22}+c_{22}$; we deduce that

$$(u{b}_{12}+u{c}_{12})(u{b}_{21}+u{c}_{21})\le \min \{a_{11}+u{b}_{11}+u{c}_{11},u{b}_{22}+u{c}_{22}\}.$$

Thus $T(E_{1,1}+u{E}_{1,2}+u{E}_{2,1})\in M_2^c$. However, $E_{1,1}+u{E}_{1,2}+u{E}_{2,1}\notin M_2^c$. It is a contradiction.

Case 2. $a_{22}\ne 0,a_{11}=a_{12}=a_{21}=0,d_{12}=d_{21}=d_{22}=0,d_{11}\ne 0$. Then we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{b}_{11}+c_{11}+d_{11}& b_{12}+c_{12}\\ b_{21}+c_{21}& a_{22}+b_{22}+c_{22}\end{array}\right)$$

$$T(u{E}_{1,2}+u{E}_{2,1}+E_{2,2})=\left(\begin{array}{cc}u{b}_{11}+u{c}_{11}+d_{11}& u{b}_{12}+u{c}_{12}\\ u{b}_{21}+u{c}_{21}& u{b}_{22}+u{c}_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$ and so

$$(b_{12}+c_{12})(b_{21}+c_{21})\le \min \{b_{11}+c_{11}+d_{11},a_{22}+b_{22}+c_{22}\}.$$

Notice that $u=\max \{b_{22},c_{22}\}=b_{22}+c_{22}$; we deduce that

$$(u{b}_{12}+u{c}_{12})(u{b}_{21}+u{c}_{21})\le \min \{u{b}_{11}+u{c}_{11}+d_{11},u{b}_{22}+u{c}_{22}\}.$$

Thus $T(u{E}_{1,2}+u{E}_{2,1}+E_{2,2})\in M_2^c$. However, $u{E}_{1,2}+u{E}_{2,1}+E_{2,2}\notin M_2^c$. It is a contradiction. From the above discussion, we have $u=0$ and so $b_{22}=c_{22}=0$.

Finally, by Lemma 12, the conclusion holds. □

Lemma 15.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then either (I) $a_{11}=d_{22}\ne 0$, $a_{12}=a_{21}=a_{22}=0$, $d_{11}=d_{12}=d_{21}=0$ or (II) $a_{22}=d_{11}\ne 0$, $a_{11}=a_{12}=a_{21}=0$, $d_{12}=d_{21}=d_{22}=0$.

Proof. 

By Lemma 13, there are two cases to consider:

Case 1. $a_{11}\ne 0,d_{22}\ne 0,a_{12}=a_{21}=a_{22}=0,d_{11}=d_{12}=d_{21}=0$.

We assert $a_{11}=d_{22}$. Otherwise, without loss of generality, assume that $d_{22}<a_{11}$. By Lemma 14, if $b_{12}\ne 0,c_{21}\ne 0,b_{11}=b_{21}=b_{22}=0,c_{11}=c_{12}=c_{22}=0$, then

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{11}& b_{12}\\ c_{21}& d_{22}\end{array}\right),$$

$$T(d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{d}_{22}{a}_{11}& a_{11}{b}_{12}\\ c_{21}& d_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $b_{12}{c}_{21}\le \min \{a_{11},d_{22}\}$ and so $\left(a_{11}{b}_{12}\right)c_{21}\le \min \{d_{22}{a}_{11},d_{22}\}$. Thus $T(d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$. Notice that $d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2}\notin M_2^c$, a contradiction.

If $b_{21}\ne 0,c_{12}\ne 0,b_{11}=b_{12}=b_{22}=0,c_{11}=c_{21}=c_{22}=0$, then

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{11}& c_{12}\\ b_{21}& d_{22}\end{array}\right),$$

$$T(d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{d}_{22}{a}_{11}& c_{12}\\ a_{11}{b}_{21}& d_{22}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $b_{21}{c}_{12}\le \min \{a_{11},d_{22}\}$ and so $c_{12}\left(a_{11}{b}_{21}\right)\le \min \{d_{22}{a}_{11},d_{22}\}$. Thus $T(d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$. Notice that $d_{22}{E}_{1,1}+a_{11}{E}_{1,2}+E_{2,1}+E_{2,2}\notin M_2^c$, a contradiction.

Case 2. $a_{22}\ne 0,d_{11}\ne 0,a_{11}=a_{12}=a_{21}=0,d_{12}=d_{21}=d_{22}=0$.

We assert $a_{22}=d_{11}$. Otherwise, without loss of generality, assume that $d_{11}<a_{22}$. By Lemma 14, if $b_{12}\ne 0,c_{21}\ne 0,b_{11}=b_{21}=b_{22}=0,c_{11}=c_{12}=c_{22}=0$, then

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{22}& b_{12}\\ c_{21}& d_{11}\end{array}\right),$$

$$T(d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{d}_{11}{a}_{22}& a_{22}{b}_{12}\\ c_{21}& d_{11}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $b_{12}{c}_{21}\le \min \{a_{22},d_{11}\}$ and so $\left(a_{22}{b}_{12}\right)c_{21}\le \min \{d_{11}{a}_{22},d_{11}\}$. Thus $T(d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$. Notice that $d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2}\notin M_2^c$, a contradiction.

If $b_{21}\ne 0,c_{12}\ne 0,b_{11}=b_{12}=b_{22}=0,c_{11}=c_{21}=c_{22}=0$, then

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{a}_{22}& c_{12}\\ b_{21}& d_{11}\end{array}\right),$$

$$T(d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}{d}_{11}{a}_{22}& c_{12}\\ a_{22}{b}_{21}& d_{11}\end{array}\right).$$

Since $E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\in M_2^c$, we have $c_{12}{b}_{21}\le \min \{a_{22},d_{11}\}$ and so $c_{12}\left(a_{22}{b}_{21}\right)\le \min \{d_{11}{a}_{22},d_{11}\}$. Thus $T(d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2})\in M_2^c$. Notice that $d_{11}{E}_{1,1}+a_{22}{E}_{1,2}+E_{2,1}+E_{2,2}\notin M_2^c$, a contradiction.

From the above discussion, the conclusion holds. □

Theorem 5.

Let $T:M_2\left(S\right)\to M_2\left(S\right)$ be a linear transformation strongly preserving $M_2^c$. Then T satisfies one of the following conditions:

(i) 

$T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne 0$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$;

(ii) 

$T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne 0$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$;

(iii) 

$T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne 0$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$;

(iv) 

$T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne 0$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

Proof. 

By Lemmas 14 and 15, there are four cases to consider:

Case 1. $a_{11}=d_{22}\ne 0$, $a_{12}=a_{21}=a_{22}=0$, $d_{11}=d_{12}=d_{21}=0$, $b_{12}\ne 0$, $c_{21}\ne 0$, $b_{11}=b_{21}=b_{22}=0$, $c_{11}=c_{12}=c_{22}=0$.

Let $a_{11}=d_{22}=a$. Then $T\left(E_{1,1}\right)=a{E}_{1,1}$, $T\left(E_{2,2}\right)=a{E}_{2,2}$, $T\left(E_{1,2}\right)=b_{12}{E}_{1,2}$, $T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$. Moreover, we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}a& b_{12}\\ c_{21}& a\end{array}\right),$$

$$T(b_{12}{E}_{1,1}+a{E}_{1,2}+E_{2,1}+c_{21}{E}_{2,2})=\left(\begin{array}{cc}a{b}_{12}& a{b}_{12}\\ c_{21}& a{c}_{21}\end{array}\right).$$

Since T strongly preserves $M_2^c$, we deduce that $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

Case 2. $a_{11}=d_{22}\ne 0$, $a_{12}=a_{21}=a_{22}=0$, $d_{11}=d_{12}=d_{21}=0$, $b_{21}\ne 0$, $c_{12}\ne 0$, $b_{11}=b_{12}=b_{22}=0$, $c_{11}=c_{21}=c_{22}=0$.

Let $a_{11}=d_{22}=a$. Then $T\left(E_{1,1}\right)=a{E}_{1,1}$, $T\left(E_{2,2}\right)=a{E}_{2,2}$, $T\left(E_{1,2}\right)=b_{21}{E}_{2,1}$, $T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$. Moreover, we have

$$T(E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}a& c_{12}\\ b_{21}& a\end{array}\right),$$

$$T(c_{12}{E}_{1,1}+a{E}_{1,2}+E_{2,1}+b_{21}{E}_{2,2})=\left(\begin{array}{cc}a{c}_{12}& a{c}_{12}\\ b_{21}& a{b}_{21}\end{array}\right).$$

Since T strongly preserves $M_2^c$, we deduce that $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

Case 3. $a_{22}=d_{11}\ne 0$, $a_{12}=a_{21}=a_{11}=0$, $d_{22}=d_{12}=d_{21}=0$, $b_{12}\ne 0$, $c_{21}\ne 0$, $b_{11}=b_{21}=b_{22}=0$, $c_{11}=c_{12}=c_{22}=0$.

Let $a_{22}=d_{11}=a$. Then $T\left(E_{1,1}\right)=a{E}_{2,2}$, $T\left(E_{2,2}\right)=a{E}_{1,1}$, $T\left(E_{1,2}\right)=b_{12}{E}_{1,2}$, $T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$. A similar discussion to Case 1 shows that $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

Case 4. $a_{22}=d_{11}\ne 0$, $a_{12}=a_{21}=a_{11}=0$, $d_{22}=d_{12}=d_{21}=0$, $b_{21}\ne 0$, $c_{12}\ne 0$, $b_{11}=b_{12}=b_{22}=0$, $c_{11}=c_{21}=c_{22}=0$.

Let $a_{22}=d_{11}=a$. Then $T\left(E_{1,1}\right)=a{E}_{2,2}$, $T\left(E_{2,2}\right)=a{E}_{1,1}$, $T\left(E_{1,2}\right)=b_{21}{E}_{2,1}$, $T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$. A similar discussion to Case 2 shows that $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$. □

As an application of Theorem 5, we characterize strong linear preservers of transitive closure on $M_2\left(\mathcal{R}\right)$, $M_2\left(\mathcal{F}\right)$ and $M_n\left(\mathbb{B}\right)$. Recall that for $A=\left(a_{ij}\right),B=\left(b_{ij}\right)\in M_n\left(S\right)$, $A\circ B$ denotes the Hadamard (or Schur) product, i.e., the $(i,j)$-entry of $A\circ B$ is $a_{ij}{b}_{ij}$.

Theorem 6.

Let $T:M_2\left(\mathcal{R}\right)\to M_2\left(\mathcal{R}\right)$ be a linear transformation. Then T strongly preserves $M_2^c$ if and only if there exist $\sigma \in Sym\left(\right[2\left]\right)$ and $X=\left(x_{ij}\right)\in M_2^c$ with $x_{12}{x}_{21}=x_{11}=x_{22}$ such that $T\left(A\right)=X\circ \left(P_{\sigma }^{-1}A{P}_{\sigma }\right)$ for all $A\in M_2\left(\mathcal{R}\right)$ or $T\left(A\right)=X\circ \left(P_{\sigma }^{-1}{A}^T{P}_{\sigma }\right)$ for all $A\in M_2\left(\mathcal{R}\right)$.

Proof. 

First we show the sufficiency. Suppose that $W=\left(w_{ij}\right)\in M_2\left(\mathcal{R}\right)$ and $T\left(W\right)=\left(v_{ij}\right)$. Let $x=x_{12}{x}_{21}=x_{11}=x_{22}$. Then it is easily verified that

$$v_{12}{v}_{21}=x{w}_{12}{w}_{21},v_{11}=x{w}_{11},v_{22}=x{w}_{22}\mathrm{or}{v}_{12}{v}_{21}=x{w}_{12}{w}_{21},v_{11}=x{w}_{22},v_{22}=x{w}_{11}.$$

Moreover, since the multiplication in $\mathcal{R}$ is the usual arithmetic sum, we deduce that

$$W\in M_2^c\iff w_{11}{w}_{22}\le \min \{w_{12},w_{21}\}\iff x{w}_{12}{w}_{21}\le \min \{x{w}_{11},x{w}_{22}\}\iff T\left(W\right)\in M_2^c.$$

Thus T strongly preserves $M_2^c$.

Now we show the necessity. Suppose that T strongly preserves $M_2^c$. Notice that the additive identity element is $-\infty$ and the multiplicative identity element is 0. By Theorem 5, there are four cases to consider:

Case 1. $T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne -\infty$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

We assert $b_{12}{c}_{21}=a$. Otherwise, $b_{12}{c}_{21}<a$. Since $b_{12}{c}_{21}$ is the arithmetic sum of $b_{12}$ and $c_{21}$, we have $t=b_{12}{c}_{21}-a<0$ and so $t+a=b_{12}{c}_{21}$, that is, $ta=b_{12}{c}_{21}$. Notice that

$$T(t{E}_{1,1}+E_{1,2}+E_{2,1}+E_{2,2})=\left(\begin{array}{cc}ta& b_{12}\\ c_{21}& a\end{array}\right)\in M_2^c.$$

However, $t{E}_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}\notin M_2^c$. This contradicts that T strongly preserves $M_2^c$. Thus $b_{12}{c}_{21}=a$. Let $x_{11}=x_{22}=a$ and $x_{12}=b_{12},x_{21}=c_{21}$. Then $X=\left(x_{ij}\right)\in M_2^c$. Moreover, we can verify that $T\left(A\right)=X\circ A$ for all $A\in M_2\left(\mathcal{R}\right)$.

Case 2. $T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne -\infty$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

A similar discussion to Case 1 shows that $b_{21}{c}_{12}=a$. Let $x_{11}=x_{22}=a$ and $x_{12}=c_{12}$, $x_{21}=b_{21}$. Then $X=\left(x_{ij}\right)\in M_2^c$. Moreover, we can verify that $T\left(A\right)=X\circ A^T$ for all $A\in M_2\left(\mathcal{R}\right)$.

Case 3. $T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne -\infty$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

A similar discussion to Case 1 shows that $b_{12}{c}_{21}=a$. Let $x_{11}=x_{22}=a$ and $x_{12}=b_{12}$, $x_{21}=c_{21}$. Then $X=\left(x_{ij}\right)\in M_2^c$. Moreover, we can verify that $T\left(A\right)=X\circ \left(P_{\sigma }^{-1}{A}^T{P}_{\sigma }\right)$ for all $A\in M_2\left(\mathcal{R}\right)$, where $\sigma =\left(12\right)$.

Case 4. $T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne -\infty$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

A similar discussion to Case 1 shows that $b_{21}{c}_{12}=a$. Let $x_{11}=x_{22}=a$ and $x_{12}=c_{12}$, $x_{21}=b_{21}$. Then $X=\left(x_{ij}\right)\in M_2^c$. Moreover, we can verify that $T\left(A\right)=X\circ \left(P_{\sigma }^{-1}A{P}_{\sigma }\right)$ for all $A\in M_2\left(\mathcal{R}\right)$, where $\sigma =\left(12\right)$. □

Theorem 7.

Let $T:M_2\left(\mathcal{F}\right)\to M_2\left(\mathcal{F}\right)$ be a linear transformation. Then T strongly preserves $M_2^c$ if and only if there exists $\sigma \in Sym\left(\right[2\left]\right)$ such that $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_2\left(\mathcal{F}\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_2\left(\mathcal{F}\right)$.

Proof. 

By Lemma 8, the sufficiency holds. Now we show the necessity. By Theorem 5, there are four cases to consider:

Case 1. $T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne 0$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

We assert $a=1$. In fact, notice that $a^2=a$ in $\mathcal{F}$; we have

$$T\left(\begin{array}{cc}a& 1\\ 1& 1\end{array}\right)=\left(\begin{array}{cc}{a}^2& b_{12}\\ c_{21}& a\end{array}\right)=\left(\begin{array}{cc}a& b_{12}\\ c_{21}& a\end{array}\right)=T\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)\in M_2^c.$$

Since T strongly preserves $M_2^c$, we deduce that $\left(\begin{array}{cc}a& 1\\ 1& 1\end{array}\right)\in M_2^c$. Thus $a=1$. Notice that $a\le \min \{b_{12},c_{21}\}$; we have $b_{12}=c_{21}=a=1$. Thus $T\left(A\right)=A$ for all $A\in M_2\left(\mathcal{F}\right)$.

Case 2. $T\left(E_{1,1}\right)=a{E}_{1,1},T\left(E_{2,2}\right)=a{E}_{2,2},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne 0$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

A similar discussion to Case 1 shows that $b_{21}=c_{12}=a=1$. Thus $T\left(A\right)=A^T$ for all $A\in M_2\left(\mathcal{F}\right)$.

Case 3. $T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{12}{E}_{1,2},T\left(E_{2,1}\right)=c_{21}{E}_{2,1}$, $a\ne 0$ and $b_{12}{c}_{21}\le a\le \min \{b_{12},c_{21}\}$.

A similar discussion to Case 1 shows that $b_{12}=c_{21}=a=1$. Thus $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_2\left(\mathcal{F}\right)$, where $\sigma =\left(12\right)$.

Case 4. $T\left(E_{1,1}\right)=a{E}_{2,2},T\left(E_{2,2}\right)=a{E}_{1,1},T\left(E_{1,2}\right)=b_{21}{E}_{2,1},T\left(E_{2,1}\right)=c_{12}{E}_{1,2}$, $a\ne 0$ and $b_{21}{c}_{12}\le a\le \min \{b_{21},c_{12}\}$.

A similar discussion to Case 1 shows that $b_{21}=c_{12}=a=1$. Thus $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_2\left(\mathcal{F}\right)$, where $\sigma =\left(12\right)$. □

Theorem 8.

Let $T:M_n\left(\mathbb{B}\right)\to M_n\left(\mathbb{B}\right)$ be a linear transformation. Then T strongly preserves $M_n^c$ if and only if there exists $\sigma \in Sym\left(\right[n\left]\right)$ such that $T\left(A\right)=P_{\sigma }^{-1}A{P}_{\sigma }$ for all $A\in M_n\left(\mathbb{B}\right)$ or $T\left(A\right)=P_{\sigma }^{-1}{A}^T{P}_{\sigma }$ for all $A\in M_n\left(\mathbb{B}\right)$.

Proof. 

By Lemma 8, the sufficiency holds. We need to show the necessity. Notice that $\mathbb{B}=\{0,1\}$. If $n=2$, then T is bijective on $\left\{E_{i,j}\right|i,j\in \left[2\right]\}$ by Theorem 5. Thus T is surjective by Lemma 2. Consequently, the necessity holds by Theorem 1.

Now we consider $n\ge 3$. We first show that $T\left(E_{i,j}\right)=E_{k,l}$ for any $i,j\in \left[n\right]$. Otherwise, it follows from Proposition 8 that there exist $i,j\in \left[n\right]$ such that $E_{k,l}+E_{p,q}\le T\left(E_{i,j}\right)$ for some $(k,l)\ne (p,q)$. Let

$$T\left(E_n\right)-T\left(E_{i,j}\right)=\sum _{(s,t)\in \mathrm{\Delta }}{E}_{s,t},\mathrm{where}\mathrm{\Delta }\subseteq \left[n\right]\times \left[n\right].$$

Then $\left|\mathrm{\Delta }\right|\le n^2-2$. For any $(s,t)\in \mathrm{\Delta }$, select a unique $(s^{\prime },t^{\prime })\in \left[n\right]\times \left[n\right]$ such that $E_{s,t}\le T\left(E_{s^{\prime },t^{\prime }}\right)$. Let ${\mathrm{\Delta }}^{\prime }$ be the set of such elements $(s^{\prime },t^{\prime })$. Then $|{\mathrm{\Delta }}^{\prime }|\le |\mathrm{\Delta }|\le n^2-2$. Take $(u,v)\in \left[n\right]\times \left[n\right]\setminus ({\mathrm{\Delta }}^{\prime }\cup \left\{(i,j)\right\})$, then

$$T(E_n-E_{u,v})=T\left(E_n\right)\in M_n^c.$$

Since $n\ge 3$, we have $E_n-E_{u,v}\notin M_n^c$. This contradicts that T strongly preserves $M_n^c$. Thus $T\left(E_{i,j}\right)=E_{k,l}$ for any $i,j\in \left[n\right]$. By Proposition 8, $T\left(E_{i,j}\right)\ne T\left(E_{p,q}\right)$ for any $(i,j)\ne (p,q)$. Thus T is bijective on $\left\{E_{i,j}\right|i,j\in \left[n\right]\}$. By Lemma 2, T is surjective on $M_n\left(\mathbb{B}\right)$. Hence the necessity holds by Theorem 1. □

7. Conclusions and Future Work

In this paper, we have systematically studied transitive closure linear preservers for square matrices over a class of commutative inclines without zero divisors. We introduced the transitive closure operator $\mathrm{\Gamma }$ and studied its relation to the digraph transitive closure (Proposition 3). Our main results provide a complete classification of surjective linear preservers of transitive closures (Theorem 1), establish the equivalence between preserving transitive closures, commuting with $\mathrm{\Gamma }$, and preserving the kernel of $\mathrm{\Gamma }$ (Theorem 4), and characterize strong linear preservers of transitive closure on $2\times 2$ matrices over certain inclines (Theorems 6 and 7). As an important special case, we showed that strong linear preservers of transitive closure on $n\times n$ Boolean matrices are precisely the transformations given by permutation similarity or permutation similarity combined with transposition (Theorem 8). Several examples (Examples 1–5) demonstrate that the surjectivity condition in our main characterizations is essential.

The results presented in this paper are restricted to the following setting: the underlying semiring is a commutative incline without zero divisors; the linear preservers considered in Section 4 and Section 5 are surjective. Whether these conditions can be relaxed remains an open question.

In the future work, it would be interesting to explore whether similar characterizations hold for non-commutative inclines or more general semirings. Additionally, Examples 1–5 show that non-surjective linear transformations may also preserve transitive closures, or commute with $\mathrm{\Gamma }$ or preserve the kernel of $\mathrm{\Gamma }$, but a complete classification of such non-surjective linear preservers is still lacking, and we hope that the results and examples presented in this paper will provide useful insights for further studies on linear preservers of matrices over semirings and incline algebras.