Linear Operators That Preserve Arctic Ranks of Boolean Matrices

Abstract

We study some properties of arctic rank of Boolean matrices. We compare the arctic rank with Boolean rank and term rank of a given Boolean matrix. Furthermore, we obtain some characterizations of linear operators that preserve arctic rank on Boolean matrix space.

1. Introduction and Preliminaries

There are many research articles on the linear operators which preserve certain matrix properties over some matrix spaces (see [1,2,3,4,5]). However, there are few articles for the characterizations of the linear operators that preserve arctic rank of matrices [3,4]. In this paper, we study some properties of arctic rank of Boolean matrices. We compare the arctic rank with Boolean rank and term rank of a given Boolean matrix. Furthermore, we obtain some characterizations of linear operators that preserve arctic rank on Boolean matrix space.

The Boolean algebra consists of the set $\mathbb{B}=\{0,1\}$ equipped with two binary operations, addition and multiplication. Two operations are defined as usual except that $1+1=1$.

Every matrix whose entries are in $\mathbb{B}$ is called a Boolean matrix, and we write ${\mathbb{M}}_{m,n}$ for the set of all $m\times n$ Boolean matrices. The usual definitions for addition and multiplication of matrices over fields are applied to ${\mathbb{M}}_{m,n}$. If $m=n$, we use ${\mathbb{M}}_n$ instead of ${\mathbb{M}}_{n,n}$. We denote the $n\times n$ identity matrix by $I_n$, and the $m\times n$ matrix all of whose entries are 1 by $J_{m,n}$. The zero matrices of any size are denoted by O. $A^t$ is the transpose of $A\in {\mathbb{M}}_{m,n}$.

A Boolean matrix in ${\mathbb{M}}_{m,n}$ is called a cell [4] if it has exactly one nonzero entry. We denote the cell whose nonzero entry is in $(i,j)$-th position by $E_{i,j}$. The cell $E_{i,j}$ is called diagonal or off-diagonal according as $i=j$ or $i\ne j$.

Every element in ${\mathbb{B}}^n={\mathbb{M}}_{n,1}$ is called a vector, and it is represented as lowercase and boldface letter. That is, $\mathbf{x}\in {\mathbb{B}}^n$ is a column vector, and ${\mathbf{x}}^t$ is the row vector for $\mathbf{x}$. Let $\left|\mathbf{x}\right|$ denote the number of nonzero entries in $\mathbf{x}$. In particular, ${\mathbf{e}}_{n|i}\in {\mathbb{B}}^n$ with $|{\mathbf{e}}_{n|i}|=1$ is the vector whose only nonzero entry is in i-th position for $i=1,\dots ,n$.

The Boolean rank [1] of nonzero $A\in {\mathbb{M}}_{m,n}$, $\mathrm{br}\left(A\right)$, is the least integer k such that A is the sum of k Boolean matrices of rank one. That is, $\mathrm{br}\left(A\right)$ is the least integer k for which there exist nonzero vectors ${\mathbf{x}}_1,\cdots ,{\mathbf{x}}_k\in {\mathbb{B}}^m$ and ${\mathbf{y}}_1,\cdots ,{\mathbf{y}}_k\in {\mathbb{B}}^n$ such that

$$A={\mathbf{x}}_1{\mathbf{y}}_1^t+\cdots +{\mathbf{x}}_k{\mathbf{y}}_k^t.$$

Thus, if $A\in {\mathbb{M}}_{m,n}$ is a Boolean rank-1 matrix, then $A=\mathbf{x}{\mathbf{y}}^t$ for some nonzero vectors $\mathbf{x}\in {\mathbb{B}}^m$ and $\mathbf{y}\in {\mathbb{B}}^n$. It is easy to verity that these vectors $\mathbf{x}$ and $\mathbf{y}$ are uniquely determined by A. Every nonzero Boolean matrix $A\in {\mathbb{M}}_{m,n}$ can be written as a sum of Boolean rank-1 matrices. Here, the sum is called a decomposition of A. As an easy example, ${\displaystyle \sum _{i=1}^n}{\mathbf{a}}_i{\mathbf{e}}_{n|i}^t$ is a decomposition of A, where ${\mathbf{a}}_i\in {\mathbb{B}}^m$ is the i-th column of A.

The arctic rank ([6] of nonzero $A\in {\mathbb{M}}_{m,n}$, $\mathrm{ar}\left(A\right)$, is the least value of

$$\frac{1}{2}\left(\right|{\mathbf{x}}_1|+|{\mathbf{y}}_1|+\cdots +|{\mathbf{x}}_l|+|{\mathbf{y}}_l\left|\right)$$

over all decompositions, ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$, of A. Evidently, $\mathrm{ar}\left(E_{i,j}\right)=1$ for every cell $E_{i,j}$ since ${\mathbf{e}}_{m|i}{\mathbf{e}}_{n|j}^t$ is the unique decomposition of $E_{i,j}$, and $\mathrm{ar}\left(J_{m,n}\right)=\frac{1}{2}(m+n)$ since ${\mathbf{j}}_m{\mathbf{j}}_n^t$ is the unique decomposition of $J_{m,n}$, where ${\mathbf{j}}_m\in {\mathbb{B}}^m$ and ${\mathbf{j}}_n\in {\mathbb{B}}^n$ are vectors all of whose entries are 1. Clearly, ${\displaystyle \sum _{i=1}^n}{\mathbf{e}}_{n|i}{\mathbf{e}}_{n|i}^t$ is the unique decomposition of $I_n$, and hence $\mathrm{ar}\left(I_n\right)=\frac{1}{2}{\displaystyle \sum _{i=1}^n}\left(\right|{\mathbf{e}}_{n|i}|+|{\mathbf{e}}_{n|i}\left|\right)=n$. In above definition, the value of l providing optimal decomposition of A can be greater than the Boolean rank of A, see Example 2.6 in [6]. Note that the arctic rank of $A\in {\mathbb{M}}_{m,n}$ may not be an integer, and may be much larger than $max\{m,n\}$. For example, consider the $3\times 4$ Boolean matrix $A=\left[\begin{array}{cccc}0& 1& 1& 1\\ 1& 1& 1& 1\\ 1& 1& 1& 1\end{array}\right]$. As shown in Proposition 1 below, we have $\mathrm{ar}\left(A\right)=\frac{9}{2}$. Thus, the arctic rank of A is not integer and is larger than 4. In general, for any nonzero $A\in {\mathbb{M}}_{m,n}$, the value of $\mathrm{ar}\left(A\right)$ is of the form $\mathrm{ar}\left(A\right)=\frac{k}{2}$, where $k\ge 2$ is a positive integer. It is well-known [6] that the arctic rank of a Boolean matrix in ${\mathbb{M}}_{m,n}$ cannot be grater than $\frac{1}{2}(n+m{2}^{m-1})$.

In this research, we show that the arctic rank of a matrix is equal or greater than both the Boolean rank (Theorem 1) and the term rank (Theorem 2). This implies that the arctic rank of a matrix is an upper bound of these matrix functions. So, arctic rank can be used to estimate the Boolean rank or the term rank of a Boolean matrix. In addition, in the linear preserver problems, this research is the beginning of the arctic rank preserver problems over integers or real field that is more complex than the Boolean case. So, this paper can also help in revitalizing the research on linear preserver problems.

2. Basic Results for Arctic Ranks of Boolean Matrices

In this section, we study the basic behavior of the arctic rank of Boolean matrices. In Boolean matrix space, we compare the arctic rank with the Boolean rank and the term rank.

Theorem 1.

If $A\in {\mathbb{M}}_{m,n}$ is nonzero, then $\mathrm{br}\left(A\right)\le \mathrm{ar}\left(A\right)$.

Proof. 

Let ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ be an optimal decomposition of A such that $\mathrm{ar}\left(A\right)=\frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)$. Since ${\mathbf{x}}_i$ and ${\mathbf{y}}_i$ are nonzero, we have $|{\mathbf{x}}_i|\ge 1$ and $|{\mathbf{y}}_i|\ge 1$ for all $i=1,\dots ,l$. If $\mathrm{br}\left(A\right)=k$, then $l\ge k$ by the definition of Boolean rank. It follows that $\mathrm{br}\left(A\right)=k=\frac{1}{2}{\displaystyle \sum _{i=1}^k}(1+1)\le \frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)=\mathrm{ar}\left(A\right)$. □

Two cells $E_{i,j}$ and $E_{i^{\prime },j^{\prime }}$ in ${\mathbb{M}}_{m,n}$ are called collinear if either $i=i^{\prime }$ or $j=j^{\prime }$. This terminology can be extended to cells more than two cells.

Lemma 1.

Let $A\in {\mathbb{M}}_{m,n}$ be nonzero. Then the followings hold:

(1)

if $A^{\prime }$ is a Boolean matrix which is obtained from A by deleting some zero rows or columns, then $\mathrm{ar}\left(A^{\prime }\right)=\mathrm{ar}\left(A\right);$

(2)

$\mathrm{ar}\left(A^t\right)=\mathrm{ar}\left(A\right)$;

(3)

if $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$ are permutation matrices, then $\mathrm{ar}\left(PAQ\right)=\mathrm{ar}\left(A\right)$;

(4)

if A is a sum of $h(\ge 2)$ cells which are collinear, then $\mathrm{ar}\left(A\right)=\frac{1+h}{2}$.

Proof. 

(1)

It is an easy exercise to check.

(2)

Note that ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ is a decomposition of A if and only if ${\displaystyle \sum _{i=1}^l}{\mathbf{y}}_i{\mathbf{x}}_i^t$ is a decomposition of $A^t$. This implies $\mathrm{ar}\left(A^t\right)=\mathrm{ar}\left(A\right)$.

(3)

First, suppose that ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ is a decomposition of A. Then

$${\displaystyle \sum _{i=1}^l}\left(P{\mathbf{x}}_i\right){\left(Q^t{\mathbf{y}}_i\right)}^t={\displaystyle \sum _{i=1}^l}\left(P{\mathbf{x}}_i{\mathbf{y}}_i^{t}Q\right)=P\left({\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t\right)Q=PAQ$$

is a decomposition of $PAQ$. Next, suppose that ${\displaystyle \sum _{i=1}^{l^{\prime }}}{\mathbf{u}}_i{\mathbf{v}}_i^t$ is a decomposition of $PAQ$. Since P and Q are permutation matrices, we have $P^{t}P=I_m$ and $Q{Q}^t=I_n$. It follows that

$${\displaystyle \sum _{i=1}^{l^{\prime }}}\left(P^t{\mathbf{u}}_i\right){\left(Q{\mathbf{v}}_i\right)}^t={\displaystyle \sum _{i=1}^{l^{\prime }}}\left(P^t{\mathbf{u}}_i{\mathbf{v}}_i^t{Q}^t\right)=P^t\left({\displaystyle \sum _{i=1}^{l^{\prime }}}{\mathbf{u}}_i{\mathbf{v}}_i^t\right)Q^t=P^t\left(PAQ\right)Q^t=A$$

is a decomposition of A. From these two inclusions, we see that the set of decompositions of $PAQ$ is equal to that of A. This shows $\mathrm{ar}\left(PAQ\right)=\mathrm{ar}\left(A\right)$.

(4)

Clearly $\mathrm{br}\left(A\right)=1$. By (2) and (3), without loss of generality we assume that $A=E_{1,1}+E_{1,2}+\cdots +E_{1,h}$. Then for the vector $\mathbf{y}={[1\cdots 10\cdots 0]}^t\in {\mathbb{B}}^n$ with $\left|\mathbf{y}\right|=h$, ${\mathbf{e}}_{m|1}{\mathbf{y}}^t$ is the unique decomposition of A. Thus $\mathrm{ar}\left(A\right)=\frac{1}{2}\left(\right|{\mathbf{e}}_{m|1}|+|\mathbf{y}\left|\right)=\frac{1+h}{2}$.

□

Lemma 2.

Let $A,B\in {\mathbb{M}}_{m,n}$ be nonzero. Then $\mathrm{ar}(A+B)\le \mathrm{ar}\left(A\right)+\mathrm{ar}\left(B\right)$.

Proof. 

If ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ and ${\displaystyle \sum _{i=1}^{l^{\prime }}}{\mathbf{u}}_i{\mathbf{v}}_i^t$ are decompositions of A and B, respectively, then their sum is a decomposition of $A+B$. This shows $\mathrm{ar}(A+B)\le \mathrm{ar}\left(A\right)+\mathrm{ar}\left(B\right)$. □

For $A\in {\mathbb{M}}_{m,n}$, let $\left|A\right|$ denote the number of nonzero entries in A.

Corollary 1.

If $A\in {\mathbb{M}}_{m,n}$ is nonzero, then $\mathrm{ar}\left(A\right)\le \left|A\right|$.

Proof. 

Let A be a sum of h cells $E_1,\dots ,E_h$ so that $A={\displaystyle \sum _{i=1}^h}{E}_i$. Since $\mathrm{ar}\left(E_i\right)=1$ for all $i=1,\dots ,h$, by Lemma 2, we have $\mathrm{ar}\left(A\right)\le {\displaystyle \sum _{i=1}^h}\mathrm{ar}\left(E_i\right)={\displaystyle \sum _{i=1}^h}1=h=\left|A\right|$. □

For Boolean matrices $A\in {\mathbb{M}}_{m,n}$ and $B\in {\mathbb{M}}_{m,k}$, their concatenation is denoted by $\left[A\right|B]$ so that it is an $m\times (n+k)$ Boolean matrix. Then by Lemma 1, we have $\mathrm{ar}\left(A\right|O)=\mathrm{ar}(A)$ and $\mathrm{ar}\left(O\right|B)=\mathrm{ar}(B)$.

From now on, we will assume that $2\le m\le n$ unless specified otherwise. Consider $A=\left[a_{i,j}\right]\in {\mathbb{M}}_{m,n}$ for which $a_{i,j}=0$ if $i=j=1$ and $a_{i,j}=1$ otherwise. Then for $B=E_{2,1}+\cdots +E_{m,1}$, we can write $A=B+\left[\mathbf{0}\right|J_{m,n-1}]$, where $\mathbf{0}$ is the zero vector in ${\mathbb{B}}^m$. Hence by Lemmas 1 and 2, we have

$$\begin{array}{ccc}\hfill \mathrm{ar}\left(A\right)& \le & \mathrm{ar}\left(B\right)+\mathrm{ar}\left(\mathbf{0}\right|J_{m,n-1})=\mathrm{ar}\left(B\right)+\mathrm{ar}\left(J_{m,n-1}\right)\hfill \\ & =& \frac{1+(m-1)}{2}+\frac{m+(n-1)}{2}=m+\frac{n-1}{2}.\hfill \end{array}$$

In fact, $\mathrm{ar}\left(A\right)=m+\frac{n-1}{2}$ by the below Proposition 1.

If $A=\left[a_{i,j}\right]$ and $B=\left[b_{i,j}\right]$ are Boolean matrices in ${\mathbb{M}}_{m,n}$, we say that A dominates [5] B, written $B⊑A$, if $a_{i,j}=0$ implies $b_{i,j}=0$ for all i and j. Equivalently, $B⊑A$ if and only if $A+B=A$.

Proposition 1.

For $A=\left[a_{i,j}\right]\in {\mathbb{M}}_{m,n}$, if $\left|A\right|=mn-1$, then $\mathrm{ar}\left(A\right)=m+\frac{n-1}{2}$.

Proof. 

By Lemma 1, we may assume that $a_{1,1}$ is the unique zero entry of A. For any decomposition, ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$, of A, consider the value $\alpha =\frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)$. Let I be a subset of $\{1,\dots ,l\}$ such that $j\in I$ if the first entry of ${\mathbf{x}}_j$ is 1. Without loss of generality, we assume that $I=\{1,\dots ,k\}$ with $1\le k<l$. In this situation, we let ${\mathbf{p}}_1={\mathbf{x}}_1+\cdots +{\mathbf{x}}_k$, ${\mathbf{q}}_1={\mathbf{y}}_1+\cdots +{\mathbf{y}}_k$, ${\mathbf{p}}_2={\mathbf{x}}_{k+1}+\cdots +{\mathbf{x}}_l$ and ${\mathbf{q}}_2={\mathbf{y}}_{k+1}+\cdots +{\mathbf{y}}_l$. Then we have

$$\frac{1}{2}\left(\right|{\mathbf{p}}_1|+|{\mathbf{q}}_1|+|{\mathbf{p}}_2|+|{\mathbf{q}}_2\left|\right)\le \alpha \mathrm{and}A⊑{\mathbf{p}}_1{\mathbf{q}}_1^t+{\mathbf{p}}_2{\mathbf{q}}_2^t.$$

Note that the first entry of ${\mathbf{p}}_2$ is 0 by definitions of I and ${\mathbf{p}}_2$. This implies that the $(1,1)$th entry of ${\mathbf{p}}_2{\mathbf{q}}_2^t$ is 0. Suppose that the $(1,1)$th entry of ${\mathbf{p}}_1{\mathbf{q}}_1^t$ is 1. Then the first entry of ${\mathbf{q}}_1$ is 1, and hence there exists an index $i\in I$ such that the first entry of ${\mathbf{y}}_i$ is 1. The first entry of ${\mathbf{x}}_i$ is 1 by definition of I, and thus the $(1,1)$-th entry of ${\mathbf{x}}_i{\mathbf{y}}_i^t$ is 1. This contradicts $a_{1,1}=0$. That is, the $(1,1)$-th entry of ${\mathbf{p}}_1{\mathbf{q}}_1^t$ is 0, and hence the $(1,1)$-th entry of ${\mathbf{p}}_1{\mathbf{q}}_1^t+{\mathbf{p}}_2{\mathbf{q}}_2^t$ is 0. This implies ${\mathbf{p}}_1{\mathbf{q}}_1^t+{\mathbf{p}}_2{\mathbf{q}}_2^t⊑A$ since $a_{1,1}$ is the unique zero entry of A. Consequently, $A={\mathbf{p}}_1{\mathbf{q}}_1^t+{\mathbf{p}}_2{\mathbf{q}}_2^t$, and we conclude that the value $\alpha$ can be optimal when $l=2$. Let $\mathbf{p}{\mathbf{q}}^t+\mathbf{u}{\mathbf{v}}^t$ be an optimal decomposition of A so that $\mathrm{ar}\left(A\right)=\frac{1}{2}\left(\right|\mathbf{p}|+\left|\mathbf{q}\right|+\left|\mathbf{u}\right|+\left|\mathbf{v}\right|)$. Put $\mathbf{p}={[p_1{p}_2\cdots p_m]}^t$, $\mathbf{q}={[q_1{q}_2\cdots q_n]}^t$, $\mathbf{u}={[u_1{u}_2\cdots u_m]}^t$ and $\mathbf{v}={[v_1{v}_2\cdots v_n]}^t$. If $p_1=u_1=0$ or $p_1=u_1=1$, then we can easily check that $\mathbf{p}{\mathbf{q}}^t+\mathbf{u}{\mathbf{v}}^t$ is not equal to A. Thus we can say that $p_1=1$ and $u_1=0$ so that

$$\mathbf{p}{\mathbf{q}}^t+\mathbf{u}{\mathbf{v}}^t=\left[\begin{array}{c}1\\ p_2\\ ⋮\\ p_m\end{array}\right][q_1{q}_2\cdots q_n]+\left[\begin{array}{c}0\\ u_2\\ ⋮\\ u_m\end{array}\right][v_1{v}_2\cdots v_n].$$

This implies $q_1=0$, $q_2=\cdots =q_n=1$, $u_2=\cdots =u_m=1$ and $v_1=1$. Then $p_i+v_j=1$ for every $i=2,\dots ,m$ and $j=2,\dots ,n$. In this situation, the optimal is when $p_i=1$ and $v_j=0$ for every $i=2,\dots ,m$ and $j=2,\dots ,n$. That is, we have $\mathbf{p}={[11\cdots 1]}^t$, $\mathbf{q}={[01\cdots 1]}^t$, $\mathbf{u}={[01\cdots 1]}^t$ and $\mathbf{v}={[10\cdots 0]}^t$. It follows that $\mathrm{ar}\left(A\right)=\frac{1}{2}\left(\right|\mathbf{p}|+\left|\mathbf{q}\right|+\left|\mathbf{u}\right|+\left|\mathbf{v}\right|)=\frac{1}{2}[m+(n-1)+(m-1)+1]=m+\frac{n-1}{2}$. □

If A and B are Boolean matrices of suitable size, then their direct sum is denoted by $A\oplus B$ so that $A\oplus B=\left[\begin{array}{cc}A& O\\ O& B\end{array}\right]$. In this case, we can easily check that $\mathrm{ar}(A\oplus B)=\mathrm{ar}\left(A\right)+\mathrm{ar}\left(B\right)$.

The term rank [2] of nonzero $A\in {\mathbb{M}}_{m,n}$, $\mathrm{tr}\left(A\right)$, is the minimum number of lines that needed to include all nonzero entries of A. Here, a line of A is a row or column of A. It follows from $2\le m\le n$ that

$$1\le \mathrm{tr}\left(A\right)\le m$$

for all nonzero $A\in {\mathbb{M}}_{m,n}$. It is clear that if $A\in {\mathbb{M}}_{m,n}$ has term rank h, then there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$, such that $E_{1,1}+E_{2,2}+\cdots +E_{h,h}⊑PAQ$.

Theorem 2.

If $A\in {\mathbb{M}}_{m,n}$ is nonzero, then $\mathrm{tr}\left(A\right)\le \mathrm{ar}\left(A\right)$.

Proof. 

If $\mathrm{tr}\left(A\right)=h$, without loss of generality, we assume that $E_{1,1}+E_{2,2}+\cdots +E_{h,h}⊑A$. Let ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ be a decomposition of A. Then for any $r\in \{1,\dots ,h\}$, there exist $j\in \{1,\dots ,l\}$ such that $E_{r,r}⊑{\mathbf{x}}_j{\mathbf{y}}_j^t$. It follows from $\mathrm{ar}\left(E_{r,r}\right)=1$ that $\mathrm{ar}\left(A\right)\ge h$, and hence $\mathrm{tr}\left(A\right)\le \mathrm{ar}\left(A\right)$. □

For $k=1,\frac{3}{2},2,\frac{5}{2},3,\cdots$, let $\mathcal{A}R\left(k\right)$ denote the set of Boolean matrices in ${\mathbb{M}}_{m,n}$ whose arctic rank is k. Let ${\mathcal{A}R}^{*}\left(k\right)$ be the set of members in $\mathcal{A}R\left(k\right)$ with fewest number of nonzero entries. That is, if $A\in {\mathcal{A}R}^{*}\left(k\right)$, then $\left|A\right|\le \left|X\right|$ for all $X\in \mathcal{A}R\left(k\right)$. Notice that any member in $\mathcal{A}R\left(1\right)$ must be a cell, and hence ${\mathcal{A}R}^{*}\left(1\right)=\mathcal{A}R\left(1\right)$. Any member in $\mathcal{A}R\left(\frac{3}{2}\right)$ must be a sum of two cells that are collinear, and thus ${\mathcal{A}R}^{*}\left(\frac{3}{2}\right)=\mathcal{A}R\left(\frac{3}{2}\right)$. If $k\ge 2$, then ${\mathcal{A}R}^{*}\left(k\right)$ is not equal to $\mathcal{A}R\left(k\right)$. For example, consider two members $A=I_2\oplus O$ and $B=J_{2,2}\oplus O$ in $\mathcal{A}R\left(2\right)$. Then $A\in {\mathcal{A}R}^{*}\left(2\right)$, while $B\notin {\mathcal{A}R}^{*}\left(2\right)$.

If $1\le h\le m$, using Theorem 2, we can obtain the specific structure of elements in ${\mathcal{A}R}^{*}\left(h\right)$ as following:

Proposition 2.

Let $1\le h\le m$. Then for $A\in {\mathbb{M}}_{m,n}$, we have $\left|A\right|=\mathrm{tr}\left(A\right)=h$ if and only if $A\in {\mathcal{A}R}^{*}\left(h\right)$.

Proof. 

Assume $\left|A\right|=\mathrm{tr}\left(A\right)=h$. Then there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$ such that $PAQ=I_h\oplus O$. Thus by Lemma 1, $\mathrm{ar}\left(A\right)=\mathrm{ar}\left(PAQ\right)=\mathrm{ar}\left(I_h\right)=h$. If $A\notin {\mathcal{A}R}^{*}\left(h\right)$, then $\left|X\right|\le h-1$ for all $X\in {\mathcal{A}R}^{*}\left(h\right)$. Then, by Corollary 1, $\mathrm{ar}\left(X\right)\le h-1$ and thus $X\notin \mathcal{A}R\left(h\right)$, a contradiction. Hence we must have $A\in {\mathcal{A}R}^{*}\left(h\right)$.

Conversely, assume $A\in {\mathcal{A}R}^{*}\left(h\right)$. Then $\mathrm{ar}\left(A\right)=h$ and so $\mathrm{tr}\left(A\right)\le h$ by Theorem 2. Consider $B=I_h\oplus O\in {\mathbb{M}}_{m,n}$ with $\left|B\right|=\mathrm{tr}\left(B\right)=h$. Then $B\in {\mathcal{A}R}^{*}\left(h\right)$ and thus $\left|A\right|=\left|B\right|=h$. Hence we can write $A=E_1+E_2+\cdots +E_h$ for distinct cells $E_1,\dots ,E_h$. If $\mathrm{tr}\left(A\right)<h$, then there exist two cells in $\{E_1,\dots ,E_h\}$ that are collinear. Without loss of generality, we assume that $E_1$ and $E_2$ are collinear. Then, by Lemma 2,

$$h=\mathrm{ar}\left(A\right)\le \mathrm{ar}(E_1+E_2)+{\displaystyle \sum _{i=3}^h}\mathrm{ar}\left(E_i\right)=\frac{3}{2}+(h-2)=h-\frac{1}{2}$$

which is impossible. Hence we must have $\mathrm{tr}\left(A\right)=h$. □

Lemma 3.

For $A\in {\mathbb{M}}_{m,n}$ and $1\le h\le m$, let $\mathrm{tr}\left(A\right)=h$. Then for $1\le r\le m-1$, if $\left|A\right|=h+r$ and A dominates $r+1$ cells that are collinear, then $A\in \mathcal{A}R(h+\frac{1}{2}r)$.

Proof. 

Since $\mathrm{tr}\left(A\right)=h$, we can assume $B=E_{1,1}+\cdots +E_{h,h}⊑A$. Suppose $\left|A\right|=h+r$ with $1\le r\le m-1$. From $B⊑A$ and $\mathrm{tr}\left(A\right)=h$, we can write $A=B+E_1+\cdots +E_r$ for distinct off-diagonal cells $E_1,\dots ,E_r$. Since A dominates $r+1$ cells that are collinear, it follows that $E_1,\dots ,E_r$ are collinear and they must be collinear to a cell in $\{E_{1,1},\dots ,E_{h,h}\}$. Without loss of generality, we can write $E_i$ as $E_i=E_{1,j_i}$ for all $i=1,\dots ,r$ and $2\le j_1<j_2<\cdots <j_r\le n$ so that $A=E_{1,1}+\cdots +E_{h,h}+E_{1,j_1}+\cdots +E_{1,j_r}$. Let ${\mathbf{f}}_1\in {\mathbb{B}}^n$ be the vector whose all nonzero entries are located in the first and $j_i$th positions for all $i=1,\dots ,r$. Then $|{\mathbf{f}}_1|=r+1$ and we see that ${\mathbf{e}}_{m|1}{\mathbf{f}}_1^t+{\displaystyle \sum _{i=2}^h}{\mathbf{e}}_{m|i}{\mathbf{e}}_{n|i}^t$ is an optimal decomposition of A. Therefore, we have

$$\mathrm{ar}\left(A\right)=\frac{1}{2}((1+r+1)+2(h-1))=h+\frac{1}{2}r,$$

and hence $A\in \mathcal{A}R(h+\frac{1}{2}r)$. □

Using Lemma 3 and Theorem 2, we can obtain the specific structure of elements in ${\mathcal{A}R}^{*}(h+\frac{1}{2})$ with $1\le h\le m$ as following:

Proposition 3.

Let $1\le h\le m$. Then for $A\in {\mathbb{M}}_{m,n}$, we have $\left|A\right|=h+1$ and $\mathrm{tr}\left(A\right)=h$ if and only if $A\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$.

Proof. 

Assume that $\left|A\right|=h+1$ and $\mathrm{tr}\left(A\right)=h$. Then A dominates two cells which are collinear. Thus by Lemma 3, $A\in \mathcal{A}R(h+\frac{1}{2})$. If $A\notin {\mathcal{A}R}^{*}(h+\frac{1}{2})$, then $\left|X\right|\le h$ for all $X\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$. Then, by Corollary 1, $\mathrm{ar}\left(X\right)\le h$ and thus $X\notin \mathcal{A}R(h+\frac{1}{2})$, a contradiction. Hence we must have $A\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$.

Conversely, assume $A\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$. Then $\mathrm{tr}\left(A\right)\le h+\frac{1}{2}$ by Theorem 2. Since $\mathrm{tr}\left(A\right)$ is an integer, we have $\mathrm{tr}\left(A\right)\le h$. Consider $B=(I_h\oplus O)+E_{1,2}\in {\mathbb{M}}_{m,n}$ with $\mathrm{tr}\left(B\right)=h$ and $\left|B\right|=h+1$. Then $B\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$ and thus $\left|A\right|=\left|B\right|=h+1$. Hence we can write $A=E_1+\cdots +E_h+E_{h+1}$ for distinct cells $E_1,\dots ,E_h,E_{h+1}$. If $\mathrm{tr}\left(A\right)<h$, then there exist three cells in $\{E_1,\dots ,E_{h+1}\}$ that are collinear, or there exist four cells in $\{E_1,\dots ,E_{h+1}\}$ such that two cells are collinear and the others are collinear. For the first case, we can assume that $E_1,E_2,E_3$ are collinear. Then by Lemmas 1 and 2,

$$h+\frac{1}{2}=\mathrm{ar}\left(A\right)\le \mathrm{ar}(E_1+E_2+E_3)+{\displaystyle \sum _{i=4}^{h+1}}\mathrm{ar}\left(E_i\right)=\frac{1+3}{2}+(h-2)=h$$

which is impossible. For the second case, we can assume that $E_1,E_2$ are collinear and $E_3,E_4$ are collinear. Again by Lemma 2, $h+\frac{1}{2}=\mathrm{ar}\left(A\right)\le \mathrm{ar}(E_1+E_2)+\mathrm{ar}(E_3+E_4)+{\displaystyle \sum _{i=5}^{h+1}}\mathrm{ar}\left(E_i\right)=\frac{3}{2}+\frac{3}{2}+(h-3)=h$ which is impossible. Hence we must have $\mathrm{tr}\left(A\right)=h$. □

The sum of four cells in ${\mathbb{M}}_{m,n}$ is called a rectangle if the sum is of the form $E_{i,j}+E_{i,j^{\prime }}+E_{i^{\prime },j}+E_{i^{\prime },j^{\prime }}$. Equivalently, $A\in {\mathbb{M}}_{m,n}$ is a rectangle if and only if there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$, such that $PAQ=E_{1,1}+E_{1,2}+E_{2,1}+E_{2,2}=J_{2,2}\oplus O$. This shows that $\mathrm{ar}\left(A\right)=2$ for every rectangle $A\in {\mathbb{M}}_{m,n}$.

Proposition 4.

For $2\le h\le m$, let $A\in {\mathbb{M}}_{m,n}$ with $\mathrm{tr}\left(A\right)=h$ and $\left|A\right|=h+2$. Then $A\in \mathcal{A}R\left(h\right)$ if and only if A dominates a rectangle.

Proof. 

Assume $A\in \mathcal{A}R\left(h\right)$ and let ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ be an optimal decomposition of A. Then we have $\mathrm{ar}\left(A\right)=\frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)$, equivalently, ${\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)=2h$. Since ${\mathbf{x}}_i$ and ${\mathbf{y}}_i$ are nonzero, we have $|{\mathbf{x}}_i|\ge 1$ and $|{\mathbf{y}}_i|\ge 1$ for all $i=1,\dots ,l$. Hence the above equality is impossible if $l>h$. If $l=h$, then $|{\mathbf{x}}_i|=|{\mathbf{y}}_i|=1$ for all $i=1,\dots ,h$, equivalently, ${\mathbf{x}}_i{\mathbf{y}}_i^t$ is a cell for all $i=1,\dots ,h$. This implies $\left|A\right|\le h$, a contradiction to the fact that $\left|A\right|=h+2$. Hence $l\le h-1$. Since $\mathrm{tr}\left(A\right)=h$ and $\left|A\right|=h+2$, it follows that $\mathrm{br}\left(A\right)\ge h-1$ and so $l\ge h-1$. Thus we must have $l=h-1$ so that

$$\left(\right|{\mathbf{x}}_1|+|{\mathbf{y}}_1\left|\right)+\left(\right|{\mathbf{x}}_2|+|{\mathbf{y}}_2\left|\right)+\cdots +\left(\right|{\mathbf{x}}_{h-1}|+|{\mathbf{y}}_{h-1}\left|\right)=2h.$$

For this equality, three cases arise:

(1)

$|{\mathbf{x}}_i|=3$, $|{\mathbf{y}}_i|=1$ (or $|{\mathbf{x}}_i|=1$, $|{\mathbf{y}}_i|=3$) for some i, and $|{\mathbf{x}}_j|=|{\mathbf{y}}_j|=1$ for all $j\in \{1,\dots ,h-1\}\setminus \left\{i\right\}$;

(2)

$|{\mathbf{x}}_i|+|{\mathbf{y}}_i|=|{\mathbf{x}}_j|+|{\mathbf{y}}_j|=3$ for some distinct $i,j$, and $|{\mathbf{x}}_k|=|{\mathbf{y}}_k|=1$ for all $k\in \{1,\dots ,h-1\}\setminus \{i,j\}$;

(3)

$|{\mathbf{x}}_i|=|{\mathbf{y}}_i|=2$ for some i, and $|{\mathbf{x}}_j|=|{\mathbf{y}}_j|=1$ for all $j\in \{1,\dots ,h-1\}\setminus \left\{i\right\}$.

If (1) holds, then ${\mathbf{x}}_i{\mathbf{y}}_i^t$ is a sum of three cells, and ${\mathbf{x}}_j{\mathbf{y}}_j^t$ is a cell for all $j\in \{1,\dots ,h-1\}\setminus \left\{i\right\}$. These imply that ${\displaystyle \sum _{i=1}^{h-1}}{\mathbf{x}}_i{\mathbf{y}}_i^t$ dominates at most $3+(h-2)=h+1$ cells, a contradiction to the fact that $\left|A\right|=h+2$. If (2) holds, then both ${\mathbf{x}}_i{\mathbf{y}}_i^t$ and ${\mathbf{x}}_j{\mathbf{y}}_j^t$ are a sum of two cells, and ${\mathbf{x}}_k{\mathbf{y}}_k^t$ is a cell for all $k\in \{1,\dots ,h-1\}\setminus \{i,j\}$. These imply that ${\displaystyle \sum _{i=1}^{h-1}}{\mathbf{x}}_i{\mathbf{y}}_i^t$ dominates at most $4+(h-3)=h+1$ cells, a contradiction to the fact that $\left|A\right|=h+2$. Consequently, (3) must hold. In this case, there exist nonzero entries $x_{i_1},x_{i_2}$ of ${\mathbf{x}}_i$, and nonzero entries $y_{i_1},y_{i_2}$ of ${\mathbf{y}}_i$. These imply that A dominates the rectangle ${\mathbf{x}}_i{\mathbf{y}}_i^t=E_{i_1,i_1}+E_{i_1,i_2}+E_{i_2,i_1}+E_{i_2,i_2}$.

Conversely, assume that A dominates a rectangle. Since $\mathrm{tr}\left(A\right)=h$, we can assume $E_{1,1}+E_{2,2}+\cdots +E_{h,h}⊑A$. Thus, the rectangle must be of the form $E_{i,i}+E_{i,j}+E_{j,i}+E_{j,j}$ for distinct $i,j\in \{1,\dots ,h\}$. Without loss of generality, we assume that $i=1$ and $j=2$ so that $A=J_{2,2}\oplus I_{h-2}\oplus O$. Thus we have $\mathrm{ar}\left(A\right)=\mathrm{ar}\left(J_{2,2}\right)+\mathrm{ar}\left(I_{h-2}\right)=2+(h-2)=h$ and so $A\in \mathcal{A}R\left(h\right)$. □

3. Linear Preservers of Arctic Ranks 1 and $\frac{k}{2}$ with $k\ge 4$

An operator T on ${\mathbb{M}}_{m,n}$ is a map from ${\mathbb{M}}_{m,n}$ into ${\mathbb{M}}_{m,n}$. For an operator T on ${\mathbb{M}}_{m,n}$, we say that it is linear if $T\left(O\right)=O$ and $T(A+B)=T\left(A\right)+T\left(B\right)$ for all A and B. In particular, every linear operator on ${\mathbb{M}}_{m,n}$ is completely determined by its behavior on the set of cells in ${\mathbb{M}}_{m,n}$.

Let T be an operator on ${\mathbb{M}}_{m,n}$. Then we say that:

(1)

T preserves arctic rank k if $T\left(X\right)\in \mathcal{A}R\left(k\right)$ for all $X\in \mathcal{A}R\left(k\right)$;

(2)

T preserves arctic rank if $\mathrm{ar}\left(T\right(X\left)\right)=\mathrm{ar}\left(X\right)$ for all $X\in {\mathbb{M}}_{m,n}$.

If T is an operator on ${\mathbb{M}}_{m,n}$, then it is called a $(P,Q)$-operator if there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$, such that $T\left(X\right)=PXQ$ for all X, or $m=n$ and $T\left(X\right)=P{X}^{t}Q$ for all X. It is obvious that every $(P,Q)$-operator on ${\mathbb{M}}_{m,n}$ is linear.

Theorem 3.

If T is a $(P,Q)$-operator on ${\mathbb{M}}_{m,n}$, then T preserves arctic rank.

Proof. 

Since T is a $(P,Q)$-operator, if $X\in {\mathbb{M}}_{m,n}$, then $T\left(X\right)$ is of the form either $T\left(X\right)=PXQ$ or $T\left(X\right)=P{X}^{t}Q$ for some permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$. By Lemma 1, $\mathrm{ar}\left(PXQ\right)=\mathrm{ar}\left(X\right)$ and $\mathrm{ar}\left(P{X}^{t}Q\right)=\mathrm{ar}\left(X\right)$ since $\mathrm{ar}\left(X^t\right)=\mathrm{ar}\left(X\right)$. These two results imply $\mathrm{ar}\left(T\right(X\left)\right)=\mathrm{ar}\left(X\right)$. Thus, T preserves arctic rank. □

Proposition 5.

For $2\le h\le m$, let $A\in {\mathbb{M}}_{m,n}$ with $\mathrm{tr}\left(A\right)=h$ and $\left|A\right|=h+2$. If A does not dominate a rectangle, then $\mathrm{ar}\left(A\right)\ge h+1$.

Proof. 

Since $\mathrm{tr}\left(A\right)=h$, we have $\mathrm{ar}\left(A\right)\ge h$ by Theorem 2. Suppose that A does not dominate a rectangle. Then by Proposition 4, $A\notin \mathcal{A}R\left(h\right)$ and thus $\mathrm{ar}\left(A\right)\ge h+\frac{1}{2}$.

If $\mathrm{ar}\left(A\right)=h+\frac{1}{2}$, then there is an optimal decomposition ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ of A such that $\mathrm{ar}\left(A\right)=\frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)$, equivalently, ${\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)=2h+1$. By a similar argument in the proof of Proposition 4, we see that $l=h$. That is, we have

$$\left(\right|{\mathbf{x}}_1|+|{\mathbf{y}}_1\left|\right)+\left(\right|{\mathbf{x}}_2|+|{\mathbf{y}}_2\left|\right)+\cdots +\left(\right|{\mathbf{x}}_h|+|{\mathbf{y}}_h\left|\right)=2h+1.$$

Then, there must be an index $i\in \{1,\dots ,h\}$ such that $|{\mathbf{x}}_i|+|{\mathbf{y}}_i|=3$ and $|{\mathbf{x}}_j|+|{\mathbf{y}}_j|=2$ for all $j\in \{1,\dots ,h\}\setminus \left\{i\right\}$. These mean that ${\mathbf{x}}_i{\mathbf{y}}_i^t$ is a sum of two cells, and ${\mathbf{x}}_j{\mathbf{y}}_j^t$ is a cell for all $j\in \{1,\dots ,h\}\setminus \left\{i\right\}$. That is, ${\displaystyle \sum _{i=1}^h}{\mathbf{x}}_i{\mathbf{y}}_i^t$ dominates at most $2+(h-1)=h+1$ cells, a contradiction to the fact that $\left|A\right|=h+2$. Therefore, $\mathrm{ar}\left(A\right)\ge h+1$. □

Example 1.

Consider a linear operator T on ${\mathbb{M}}_2$ defined by

$$T\left(\left[\begin{array}{cc}{x}_{1,1}& x_{1,2}\\ x_{2,1}& x_{2,2}\end{array}\right]\right)=\left[\begin{array}{cc}{x}_{1,1}+x_{1,2}& 0\\ 0& x_{2,1}+x_{2,2}\end{array}\right]$$

for every $\left[x_{i,j}\right]\in {\mathbb{M}}_2$. Since T maps cells to cells, T preserves arctic rank 1. Note that all members of $\mathcal{A}R\left(2\right)$ are $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$, $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]$ and $\left[\begin{array}{cc}1& 1\\ 1& 1\end{array}\right]$. All of them are mapped to $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. Thus T preserves arctic rank 2. For the member $E_{1,1}+E_{1,2}\in \mathcal{A}R\left(\frac{3}{2}\right)$, we have $\mathrm{ar}\left(T(E_{1,1}+E_{1,2})\right)=\mathrm{ar}\left(E_{1,1}\right)=1$. Therefore T does not preserve arctic rank $\frac{3}{2}$ and hence T is not a $(P,Q)$-operator by Theorem 3.

Example 2.

Consider a liner operator T on ${\mathbb{M}}_{2,3}$ defined by

$$T\left(\left[\begin{array}{ccc}{x}_{1,1}& x_{1,2}& x_{1,3}\\ x_{2,1}& x_{2,2}& x_{2,3}\end{array}\right]\right)=\left[\begin{array}{ccc}{x}_{1,1}& x_{1,2}& x_{1,3}\\ x_{2,2}& x_{2,1}& x_{2,3}\end{array}\right]$$

for every $\left[x_{i,j}\right]\in {\mathbb{M}}_{2,3}$. Obviously, T preserves arctic rank 1. Clearly, $\mathrm{ar}\left(J_{2,3}\right)=\frac{5}{2}$ and $T\left(J_{2,3}\right)=J_{2,3}$. Now, we can easily check that if A is a member of $\mathcal{A}R\left(\frac{5}{2}\right)$ different from $J_{2,3}$, then $\left|A\right|=3$ and $\mathrm{tr}\left(A\right)=2$. There are 18 such members, and each of them is mapped to a member in $\mathcal{A}R\left(\frac{5}{2}\right)$. Thus, T preserves arctic rank $\frac{5}{2}$. For the member $E_{1,1}+E_{2,2}\in \mathcal{A}R\left(2\right)$, we have $\mathrm{ar}\left(T(E_{1,1}+E_{2,2})\right)=\mathrm{ar}(E_{1,1}+E_{2,1})=\frac{3}{2}$. Therefore, T does not preserve arctic rank 2, and hence T is not a $(P,Q)$-operator by Theorem 3.

Henceforth, in order to study linear preservers of arctic ranks of Boolean matrices, we need the condition, $3\le m\le n$. For a linear operator T on ${\mathbb{M}}_{m,n}$, it is clear that if there exists a member $A\in {\mathcal{A}R}^{*}\left(k\right)$ such that $\left|T\right(A\left)\right|<\left|A\right|$, then T does not preserve arctic rank k.

Let ${\mathcal{E}}_{m,n}=\left\{E_{i,j}\mid 1\le i\le m,1\le j\le n\right\}$. That is, ${\mathcal{E}}_{m,n}$ is the set of all cells in ${\mathbb{M}}_{m,n}$.

Lemma 4.

For $3\le m\le n$, let T be a linear operator on ${\mathbb{M}}_{m,n}$ which preserves arctic ranks 1 and $\frac{k}{2}$ with $4\le k\le 2m+1$. Then T is bijective on ${\mathcal{E}}_{m,n}$.

Proof. 

Since T preserves arctic rank 1, it shows that T maps cells to cells. Now we will show that T is bijective on ${\mathcal{E}}_{m,n}$, equivalently T is injective on ${\mathcal{E}}_{m,n}$ since ${\mathcal{E}}_{m,n}$ is finite.

Case 1: $k=2h$ for some $2\le h\le m$. That is, T preserves arctic rank h. Suppose that T is not injective on ${\mathcal{E}}_{m,n}$. Then $T\left(E_{i,j}\right)=T\left(E_{i^{\prime },j^{\prime }}\right)$ for distinct cells $E_{i,j}$ and $E_{i^{\prime },j^{\prime }}$. If these cells are not collinear, then there is a member $A\in {\mathcal{A}R}^{*}\left(h\right)$ such that $E_{i,j}+E_{i^{\prime },j^{\prime }}⊑A$. Then, since $\left|T\right(A\left)\right|<\left|A\right|$, we have $T\left(A\right)\notin \mathcal{A}R\left(h\right)$, a contradiction. Therefore $E_{i,j}$ and $E_{i^{\prime },j^{\prime }}$ must be collinear. Without loss of generality we may assume that $T\left(E_{1,1}\right)=T\left(E_{1,2}\right)$ (or $T\left(E_{1,1}\right)=T\left(E_{2,1}\right)$). Since $E_{1,1}+E_{2,2}+\cdots +E_{h,h}\in {\mathcal{A}R}^{*}\left(h\right)$ and T preserves arctic rank h, by Proposition 2, we can write $T\left(E_{i,i}\right)=E_{i,i}$ for all $i=1,\dots ,h$. $T\left(E_{1,2}\right)=E_{1,1}$ since $T\left(E_{1,1}\right)=T\left(E_{1,2}\right)$.

Let $h=2$ so that T preserves arctic rank 2. Since $T(E_{1,2}+E_{2,2})=E_{1,1}+E_{2,2}$, it follows from $E_{1,2}+E_{2,2}+E_{1,3}+E_{2,3}\in \mathcal{A}R\left(2\right)$ that $T(E_{1,3}+E_{2,3})$ is either $E_{1,1}+E_{2,2}$ or $E_{1,2}+E_{2,1}$. From $E_{1,1}+E_{1,2}+E_{1,3}\in \mathcal{A}R\left(2\right)$ and its image, we have $T\left(E_{1,3}\right)=E_{a,b}$ with $a,b\ge 2$. If we combine these two results, we must have $T\left(E_{1,3}\right)=E_{2,2}$. Then, for $E_{1,3}+E_{2,2}\in \mathcal{A}R\left(2\right)$, we have $T(E_{1,3}+E_{2,2})=E_{2,2}\in \mathcal{A}R\left(1\right)$, a contradiction to the fact that T preserves arctic rank 2. Hence, T is injective on ${\mathcal{E}}_{m,n}$ for the case of $h=2$.

Now, let $h\ge 3$. Consider $\mathsf{\Lambda }=\left[\begin{array}{ccc}0& 1& 1\\ 0& 1& 1\\ 0& 0& 1\end{array}\right]\in {\mathbb{M}}_3$. Then by Lemma 1 and Proposition 1, $\mathrm{ar}\left(\mathsf{\Lambda }\right)=\mathrm{ar}\left({\mathsf{\Lambda }}^t\right)=\mathrm{ar}\left(\left[\begin{array}{ccc}1& 1& 0\\ 1& 1& 1\end{array}\right]\right)=2+\frac{3-1}{2}=3$. Hence $\mathsf{\Lambda }\oplus I_{h-3}\oplus O\in \mathcal{A}R\left(h\right)$ and thus $T(\mathsf{\Lambda }\oplus I_{h-3}\oplus O)\in \mathcal{A}R\left(h\right)$. Since $T\left(E_{i,i}\right)=E_{i,i}$ for all $i=2,\dots ,h$ and $T\left(E_{1,2}\right)=E_{1,1}$, by Proposition 4, we have either $T(E_{1,3}+E_{2,3})⊑E_{1,1}+E_{2,2}+\cdots +E_{h,h}$ or there exist distinct $i,j\in \{1,\dots ,h\}$ such that $T(E_{1,3}+E_{2,3})=E_{i,j}+E_{j,i}$. However, if we consider the member

$$E_{1,1}+E_{1,2}+E_{1,3}+E_{2,2}+\cdots +E_{h-1,h-1}\in \mathcal{A}R\left((h-1)+\frac{1}{2}\times 2\right)=\mathcal{A}R\left(h\right)$$

by Lemma 3, it follows from $T(E_{1,1}+E_{1,2}+E_{2,2}+\cdots +E_{h-1,h-1})=E_{1,1}+E_{2,2}+\cdots +E_{h-1,h-1}$ that $T\left(E_{1,3}\right)=E_{c,d}$ with $c,d\ge h$. Combining these results, we see that $c=d=h$ and hence $T\left(E_{1,3}\right)=E_{h,h}$. By a parallel argument, we have $T\left(E_{2,3}\right)=E_{h,h}$. If $h=3$, then $E_{1,1}+E_{1,3}+E_{2,3}+E_{3,3}\in \mathcal{A}R\left(3\right)$ by Lemma 3, while $T(E_{1,1}+E_{1,3}+E_{2,3}+E_{3,3})=E_{1,1}+E_{3,3}\in \mathcal{A}R\left(2\right)$, a contradiction to the fact that T preserves arctic rank 3. If $h>3$, then $E_{1,3}$ and $E_{h,h}$ are not collinear, and hence we can choose a member $B\in {\mathcal{A}R}^{*}\left(h\right)$ with $E_{1,3}+E_{h,h}⊑B$. Then $T\left(B\right)\notin \mathcal{A}R\left(h\right)$ since $\left|T\right(B\left)\right|<\left|B\right|$. This contradicts that T preserves arctic rank h. Thus T is injective on ${\mathcal{E}}_{m,n}$.

Case 2: $k=2h+1$ for some $2\le h\le m$. That is, T preserves arctic rank $h+\frac{1}{2}$. If T is not injective on ${\mathcal{E}}_{m,n}$, then $T\left(E\right)=T\left(F\right)$ for distinct cells E and F. Now, we can take $C\in {\mathbb{M}}_{m,n}$ dominating $E+F$, such that $\left|C\right|=h+1$ and $\mathrm{tr}\left(C\right)=h$. Then $C\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$ by Proposition 3, while $T\left(C\right)\notin \mathcal{A}R(h+\frac{1}{2})$ since $\left|T\right(C\left)\right|<\left|C\right|$. This contradicts that T preserves arctic rank $h+\frac{1}{2}$. Thus T is injective on ${\mathcal{E}}_{m,n}$. □

Proposition 6.

Let A be a Boolean matrix in ${\mathbb{M}}_{m,n}$, where $3\le m\le n$.

(1)

If $\left|A\right|=6$, then $\mathrm{ar}\left(A\right)=\frac{5}{2}$ if and only if there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$, such that $PAQ=J_{2,3}\oplus O$ or $PAQ=J_{3,2}\oplus O;$

(2)

If $\left|A\right|=7$, then $\mathrm{ar}\left(A\right)=\frac{7}{2}$ if and only if there exist permutation matrices $P\in {\mathbb{M}}_m$ and $Q\in {\mathbb{M}}_n$, such that $PAQ=(J_{2,3}\oplus O)+E$ or $PAQ=(J_{3,2}\oplus O)+E$, where E is a cell which is not dominated by $J_{2,3}\oplus O$ or $J_{3,2}\oplus O$.

Proof. 

(1) Suppose $\mathrm{ar}\left(A\right)=\frac{5}{2}$. Let ${\displaystyle \sum _{i=1}^l}{\mathbf{x}}_i{\mathbf{y}}_i^t$ be an optimal decomposition of A so that $\mathrm{ar}\left(A\right)=\frac{1}{2}{\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)$, equivalently, ${\displaystyle \sum _{i=1}^l}\left(\right|{\mathbf{x}}_i|+|{\mathbf{y}}_i\left|\right)=5$. Since ${\mathbf{x}}_i$ and ${\mathbf{y}}_i$ are nonzero vectors, $l\ge 3$ is impossible. If $l=2$, then $\left(\right|{\mathbf{x}}_1|+|{\mathbf{y}}_1\left|\right)+\left(\right|{\mathbf{x}}_2|+|{\mathbf{y}}_2\left|\right)=5$. This implies that the only one in $|{\mathbf{x}}_1|,|{\mathbf{y}}_1|,|{\mathbf{x}}_2|,|{\mathbf{y}}_2|$ has the value 2, and the others have the value 1. Then $\left|A\right|\le 3$ which is impossible. Hence we must have $l=1$ so that $|{\mathbf{x}}_1|+|{\mathbf{y}}_1|=5$. By an easy examination, we have ($|{\mathbf{x}}_1|=2$ and $|{\mathbf{y}}_1|=3$) or ($|{\mathbf{x}}_1|=3$ and $|{\mathbf{y}}_1|=2$). This is a result which we desire. The converse is obvious since $\mathrm{ar}\left(J_{2,3}\right)=\mathrm{ar}\left(J_{3,2}\right)=\frac{5}{2}$.

(2) The proof is similar to (1). □

A Boolean matrix L in ${\mathbb{M}}_{m,n}$ is called a line matrix if $L={\displaystyle \sum _{s=1}^n}{E}_{i,s}$ for some $i\in \{1,\dots ,m\}$, or $L={\displaystyle \sum _{t=1}^m}{E}_{t,j}$ for some $j\in \{1,\dots ,n\}$; $R_i={\displaystyle \sum _{s=1}^n}{E}_{i,s}$ is an i-th row matrix and $C_j={\displaystyle \sum _{t=1}^m}{E}_{t,j}$ is a j-th column matrix.

Lemma 5.

For $3\le m\le n$ and $2\le h\le m$, let T be a linear operator on ${\mathbb{M}}_{m,n}$ which preserves arctic ranks 1 and h. Then the image of each line matrix is dominated by a line matrix.

Proof. 

Since T preserves arctic ranks 1 and h, by Lemma 4, T is bijective on ${\mathcal{E}}_{m,n}$. Suppose that the image of some line matrix is not dominated by a line matrix. Then by the bijection of T on ${\mathcal{E}}_{m,n}$, there exist two cells E and F which are not collinear such that $T\left(E\right)$ and $T\left(F\right)$ are collinear. Now, we can choose a member $A\in {\mathcal{A}R}^{*}\left(h\right)$ with $\left|A\right|=h$ such that A dominates E and F. Then we have $\left|T\right(A\left)\right|=h$ and $\mathrm{tr}\left(T\right(A\left)\right)\le h-1$ since $T\left(E\right)$ and $T\left(F\right)$ are collinear. Hence by Proposition 2, $T\left(A\right)\notin \mathcal{A}R\left(h\right)$, a contradiction to the fact that T preserves arctic rank h. Therefore, the result follows. □

Lemma 6.

For $3\le m\le n$ and $2\le h\le m$, let T be a linear operator on ${\mathbb{M}}_{m,n}$ which preserves arctic ranks 1 and $h+\frac{1}{2}$. Then the image of each line matrix is dominated by a line matrix.

Proof. 

Since T preserves arctic ranks 1 and $h+\frac{1}{2}$, by Lemma 4, T is bijective on ${\mathcal{E}}_{m,n}$. Suppose that the image of some line matrix is not dominated by a line matrix. Then there exist two cells E and F such that $\mathrm{tr}(E+F)=1$ and $\mathrm{tr}\left(T\right(E+F\left)\right)=2$. Without loss of generality, we assume that $E=E_{1,1}$ and $F=E_{1,2}$ so that $T\left(E_{1,1}\right)$ and $T\left(E_{1,2}\right)$ are not collinear.

First, let $h=2$ so that T preserves arctic rank $\frac{5}{2}$. If we consider the member $J_{2,3}\oplus O\in \mathcal{A}R\left(\frac{5}{2}\right)$ and its image, by Proposition 6(1), we can write $T(J_{2,3}\oplus O)=J_{2,3}\oplus O$ (or $J_{3,2}\oplus O$). Since $T\left(E_{1,1}\right)$ and $T\left(E_{1,2}\right)$ are not collinear, without loss of generality, we assume that $T\left(E_{1,1}\right)=E_{1,1}$, and $T\left(E_{1,2}\right)=E_{2,2}$. This implies that there are two cells $E^{\prime }$ and $F^{\prime }$ in $\Omega =\{E_{1,3},E_{2,1},E_{2,2},E_{2,3}\}$ such that $T(E^{\prime }+F^{\prime })=E_{1,2}+E_{1,3}$. By an easy examination, we see that if X is a sum of $E_{1,1}$ and two cells in $\Omega$, then $\mathrm{ar}\left(X\right)=\frac{5}{2}$. It follows that $E_{1,1}+E^{\prime }+F^{\prime }\in \mathcal{A}R\left(\frac{5}{2}\right)$, while

$$T(E_{1,1}+E^{\prime }+F^{\prime })=E_{1,1}+E_{1,2}+E_{1,3}\in \mathcal{A}R\left(2\right)$$

by Lemma 1, a contradiction to the fact that T preserves arctic rank $\frac{5}{2}$. Hence, the result holds for the case of $h=2$.

Next, if $h=3$, then T preserves arctic rank $\frac{7}{2}$. If we use Proposition 6(2), by a similar argument as the case of $h=2$, we have the same result for the case of $h=3$.

Now, let $h\ge 4$. In this case, $m\ge 4$ and T preserves arctic rank $h+\frac{1}{2}$. Since $A=E_{1,1}+E_{1,2}+E_{2,2}+\cdots +E_{h,h}\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$ and T is bijective on ${\mathcal{E}}_{m,n}$, by Proposition 3, $T\left(A\right)\in {\mathcal{A}R}^{*}(h+\frac{1}{2})$ with $\left|T\right(A\left)\right|=h+1$ and $\mathrm{tr}\left(T\right(A\left)\right)=h$. Since $\mathrm{tr}\left(T(E_{1,1}+E_{1,2})\right)=2$, we can find $B\in {\mathbb{M}}_{m,n}$ which is the sum of h cells in $\{E_{1,1},E_{1,2},E_{2,2},\dots ,E_{h,h}\}$ such that B dominates $E_{1,1}+E_{1,2}$, and $\mathrm{tr}\left(T\right(B\left)\right)=h$. Let $C=B+E_{1,3}+E_{1,4}$. We note that $\mathrm{tr}\left(C\right)=h-1$, $\left|C\right|=h+2$ and C dominates 4 cells that are collinear. Thus, by Lemma 3, $\mathrm{ar}\left(C\right)=(h-1)+\frac{1}{2}·3=h+\frac{1}{2}$ and thus $T\left(C\right)\in \mathcal{A}R(h+\frac{1}{2})$. We have $\left|T\right(C\left)\right|=h+2$ and $\mathrm{tr}\left(T\right(C\left)\right)\ge h$ since $T\left(B\right)⊑T\left(C\right)$ and $\mathrm{tr}\left(T\right(B\left)\right)=h$. Now, if $\mathrm{tr}\left(T\right(C\left)\right)\ge h+1$, then $\mathrm{ar}\left(T\right(C\left)\right)\ge h+1$ by Theorem 2, and if $\mathrm{tr}\left(T\right(C\left)\right)=h$, then $\mathrm{ar}\left(T\right(C\left)\right)=h$ or $\mathrm{ar}\left(T\right(C\left)\right)\ge h+1$ by Propositions 4 and 5. These contradict that T preserves arctic rank $h+\frac{1}{2}$. Hence the result follows. □

Now, we are ready to prove the main result:

Theorem 4.

Let T be a linear operator on ${\mathbb{M}}_{m,n}$ with $3\le m\le n$. Then the followings are equivalent:

(1)

T preserves arctic rank;

(2)

T preserves arctic ranks 1 and $\frac{k}{2}$ with $4\le k\le 2m+1$;

(3)

T is a $(P,Q)$-operator.

Proof. 

It is obvious that (1) implies (2). (3) implies (1) by Theorem 3. To show that (2) implies (3), suppose that T preserves arctic ranks 1 and $\frac{k}{2}$ with $4\le k\le 2m+1$. By Lemma 4, T is bijective on ${\mathcal{E}}_{m,n}$, and by Lemmas 5 and 6, the image of each line matrix is dominated by a line matrix. Thus we have either $T\left(R_1\right)⊑R_i$ for some i, or $T\left(R_1\right)⊑C_j$ for some j.

Case 1: $T\left(R_1\right)⊑R_i$. In this case, $T\left(R_1\right)=R_i$ since T is bijective on ${\mathcal{E}}_{m,n}$. Suppose that there exist a column matrix $C_j$ such that $T\left(C_j\right)$ is dominated by a row matrix, say that $T\left(C_j\right)⊑R_k$. Since $E_{1,j}⊑R_1$ and $E_{1,j}⊑C_j$, we must have $k=i$, and hence $T(R_1+C_j)=R_i$. This contradicts that T is bijective on ${\mathcal{E}}_{m,n}$. Thus, the image of each column matrix is dominated by a column matrix. Furthermore, by the bijection of T on ${\mathcal{E}}_{m,n}$, we conclude that the image of each column matrix is a column matrix. By a similar argument, the image of each row matrix is a row matrix. Thus, there exist permutations $\sigma$ and $\tau$ of $\{1,\dots ,m\}$ and $\{1,\dots ,n\}$, respectively such that $T\left(R_i\right)=R_{\sigma \left(i\right)}$ and $T\left(C_j\right)=C_{\tau \left(j\right)}$ for all i and j. Let P and Q be permutation matrices corresponding to $\sigma$ and $\tau$, respectively. Then we have

$$T\left(E_{i,j}\right)=E_{\sigma \left(i\right),\tau \left(j\right)}=P{E}_{i,j}Q$$

for all cells $E_{i,j}$. Let $X={\displaystyle \sum _{i=1}^m}{\displaystyle \sum _{j=1}^n}{x}_{i,j}{E}_{i,j}$ be any Boolean matrix in ${\mathbb{M}}_{m,n}$. Then by the action of T on the cells, we have $T\left(X\right)=PXQ$. Therefore T is a $(P,Q)$-operator.

Case 2: $T\left(R_1\right)⊑C_j$. In this case, we must have $m=n$ and $T\left(R_1\right)=C_j$ by the bijection of T on ${\mathcal{E}}_{m,n}$. By a parallel argument in Case 1, the image of each row matrix is a column matrix and the image of each column matrix is a row matrix. Furthermore, there exist permutation matrices P and Q in ${\mathbb{M}}_n$ such that $T\left(X\right)=P{X}^{t}Q$ for all X. Hence T is a $(P,Q)$-operator. □

4. Linear Preservers of Arctic Ranks 1 and $\frac{3}{2}$

Notice that the possible values of arctic ranks of nonzero Boolean matrices are $1,\frac{3}{2},2,\frac{5}{2},\cdots$. In previous section, we have characterized linear operator on ${\mathbb{M}}_{m,n}$ that preserve arctic ranks 1 and $\frac{k}{2}$ with $4\le k\le 2m+1$ except for $k=3$.

In this section, we investigate the characterization of linear operators on ${\mathbb{M}}_{m,n}$ that preserve arctic ranks 1 and $\frac{3}{2}$.

We remaind that $\mathcal{A}R\left(\frac{3}{2}\right)$ is the set of a sum of two cells in ${\mathbb{M}}_{m,n}$ that are collinear. For example, all members of $\mathcal{A}R\left(\frac{3}{2}\right)$ in ${\mathbb{M}}_2$ are $\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$, $\left[\begin{array}{cc}0& 0\\ 1& 1\end{array}\right]$, $\left[\begin{array}{cc}1& 0\\ 1& 0\end{array}\right]$ and $\left[\begin{array}{cc}0& 1\\ 0& 1\end{array}\right]$.

Example 3.

Define a linear operator T on ${\mathbb{M}}_2$ by

$$T\left(\left[\begin{array}{cc}{x}_{1,1}& x_{1,2}\\ x_{2,1}& x_{2,2}\end{array}\right]\right)=\left[\begin{array}{cc}{x}_{1,1}+x_{2,2}& x_{1,2}\\ x_{2,1}& 0\end{array}\right]$$

for every $\left[x_{i,j}\right]\in {\mathbb{M}}_2$. An easy observation implies that T preserves arctic ranks 1 and $\frac{3}{2}$, and the image of T is not a single line matrix.

Example 4.

Consider a linear operator T on ${\mathbb{M}}_{2,3}$ defined by

$$T\left(\left[\begin{array}{ccc}{x}_{1,1}& x_{1,2}& x_{1,3}\\ x_{2,1}& x_{2,2}& x_{2,3}\end{array}\right]\right)=\left[\begin{array}{ccc}{x}_{1,1}+x_{2,3}& x_{1,2}+x_{2,1}& x_{1,3}+x_{2,2}\\ 0& 0& 0\end{array}\right]$$

for every $\left[x_{i,j}\right]\in {\mathbb{M}}_{2,3}$. By a simple examination, we see that T preserves arctic ranks 1 and $\frac{3}{2}$. Furthermore, the image of T is the first row matrix.

Let T be an operator on ${\mathbb{M}}_{m,n}$. Then we say that T is line-injective in a line matrix L if $\mathrm{tr}\left(T\right(L\left)\right)=1$ and $T\left(E\right)\ne T\left(F\right)$ for all distinct cells $E,F⊑L$. It is obvious that if T is a $(P,Q)$-operator, then T is line-injective in any line matrix. We see that the operators, in the above two examples, are line-injective in any line matrix.

We can extend Example 4 as following: For a fixed value $k\in \{1,\dots ,m\}$, define a linear operator T on ${\mathbb{M}}_{m,n}$ by

$$T\left(E_{i,j}\right)=E_{k,r}$$

for all cell $E_{i,j}$, where $r\equiv i+j-1$ (mod n) and $1\le r\le n$. We can check that T preserves arctic ranks 1 and $\frac{3}{2}$. T is line-injective in any line matrix, and the image of T is the k-th row matrix.

Now, look again at the linear operator T on ${\mathbb{M}}_2$, in Example 3, which preserves arctic ranks 1 and $\frac{3}{2}$. We see that T is not a $(P,Q)$-operator, and the image of T is not a single line matrix. For a linear operator T on ${\mathbb{M}}_{m,n}$, if $2\le m\le n$ and $n\ge 3$, then the following holds:

Theorem 5.

For $2\le m\le n$ and $n\ge 3$, let T be a linear operator on ${\mathbb{M}}_{m,n}$ which preserves arctic ranks 1 and $\frac{3}{2}$. Then T is line-injective in any line matrix. Furthermore, either T is a $(P,Q)$-operator or the image of T is a single line matrix.

Proof. 

Since T preserves arctic rank 1, it shows that T maps cells to cells. Let $L\in {\mathbb{M}}_{m,n}$ be an arbitrary line matrix. If $\mathrm{tr}\left(T\right(L\left)\right)\ge 2$, then there exist two cells $E,F⊑L$ such that $\mathrm{tr}\left(T\right(E+F\left)\right)=2$. Then $\mathrm{ar}(E+F)=\frac{3}{2}$, while $\mathrm{ar}\left(T\right(E+F\left)\right)=2$, a contradiction to the fact that T preserves arctic rank $\frac{3}{2}$. Thus we have $\mathrm{tr}\left(T\right(L\left)\right)=1$. If $T\left(E^{\prime }\right)=T\left(F^{\prime }\right)$ for distinct cells $E^{\prime },F^{\prime }⊑L$, then $\mathrm{ar}(E^{\prime }+F^{\prime })=\frac{3}{2}$, while $\mathrm{ar}\left(T(E^{\prime }+F^{\prime })\right)=\mathrm{ar}\left(T\left(E^{\prime }\right)\right)=1$, a contradiction. Hence $T\left(E^{\prime }\right)\ne T\left(F^{\prime }\right)$ for all distinct cells $E^{\prime },F^{\prime }⊑L$. Consequently, T is line-injective in L. Since L is an arbitrary line matrix, T is line-injective in any line matrix.

That is, we have established that the image of each line matrix is dominated by a line matrix. Thus, if T is not a $(P,Q)$-operator, then there exist two-line matrices such that their images are dominated by the same line matrix. There are three possibilities:

Case 1: Suppose that the images of two row matrices are dominated by the same row matrix. Without loss of generality, we may assume that $T\left(R_1\right)⊑R_1$ and $T\left(R_2\right)⊑R_1$. In this case, we must have $T\left(R_1\right)=R_1$ since T is line-injective in $R_1$. Now we show that $T\left(J_{m,n}\right)=R_1$, equivalently, $T\left(E_{i,j}\right)⊑R_1$ for all $i=1,\dots ,m$ and $j=1,\dots ,n$. If $i=1$ or 2, the result is obvious. Thus we assume $i\ge 3$. Since $T(E_{1,j}+E_{2,j})⊑R_1$ and T is line-injective in $C_j$, we can write $T\left(E_{1,j}\right)=E_{1,1}$ and $T\left(E_{2,j}\right)=E_{1,2}$. If we consider the member $E_{1,j}+E_{i,j}\in \mathcal{A}R\left(\frac{3}{2}\right)$ and its image, we can write $T\left(E_{i,j}\right)=E_{1,a}$ for some $a\ge 3$, or $T\left(E_{i,j}\right)=E_{b,1}$ for some $b\ge 2$. If $T\left(E_{i,j}\right)=E_{b,1}$, then we see that $\mathrm{ar}(E_{2,j}+E_{i,j})=\frac{3}{2}$, while $\mathrm{ar}\left(T(E_{2,j}+E_{i,j})\right)=\mathrm{ar}(E_{1,2}+E_{b,1})=2$. This contradicts that T preserves arctic rank $\frac{3}{2}$, and hence $T\left(E_{i,j}\right)=E_{1,a}⊑R_1$. Since $E_{i,j}$ is an arbitrary cell, we conclude that the image of T is a single row matrix.

If the images of two-row matrices are dominated by the same column matrix, then we must have $m=n$ and, by a similar argument as above, the image of T is a single column matrix.

Case 2: Suppose that the images of two-column matrices are dominated by the same line matrix. Then by a parallel proof of Case 1, we see that the image of T is a single line matrix.

Case 3: Suppose that the images of a row matrix and a column matrix are dominated by the same row matrix. Without loss of generality, we may assume that $T\left(R_1\right)⊑R_1$ and $T\left(C_1\right)⊑R_1$. In this case, we have $T\left(R_1\right)=R_1$. Now, we show that $T\left(R_2\right)⊑R_1$, and then by Case 1, the image of T is a single row matrix. Since $T(E_{1,1}+E_{1,2}+E_{1,3})⊑R_1$ and T is line-injective in $R_1$, we can write $T\left(E_{1,1}\right)=E_{1,1}$, $T\left(E_{1,2}\right)=E_{1,2}$ and $T\left(E_{1,3}\right)=E_{1,3}$. If we consider the member $E_{1,2}+E_{2,2}\in \mathcal{A}R\left(\frac{3}{2}\right)$ and its image, we have $T\left(E_{2,2}\right)=E_{1,c}$ for some $c\ne 2$ or $T\left(E_{2,2}\right)=E_{d,2}$ for some $d\ge 2$. If $T\left(E_{2,2}\right)=E_{d,2}$, then by considering the member $E_{2,1}+E_{2,2}\in \mathcal{A}R\left(\frac{3}{2}\right)$ and its image, we see that $T\left(E_{2,1}\right)=E_{1,2}$ since $T\left(E_{2,1}\right)⊑R_1$. Then $T(E_{2,1}+E_{2,2})=E_{1,2}+E_{d,2}$ shows $T\left(R_2\right)⊑C_2$. Thus, we can write $T\left(E_{2,3}\right)=E_{e,2}$ for some $e\ne 1,d$. Now, for the member $E_{1,3}+E_{2,3}\in \mathcal{A}R\left(\frac{3}{2}\right)$, we have $\mathrm{ar}\left(T(E_{1,3}+E_{2,3})\right)=\mathrm{ar}(E_{1,3}+E_{e,2})=2$. This contradicts that T preserves arctic rank $\frac{3}{2}$, and thus we must have $T\left(E_{2,2}\right)=E_{1,c}$. Consequently, we have $T\left(R_2\right)⊑R_1$ since $T(E_{2,1}+E_{2,2})⊑R_1$.

If the images of a row matrix and a column matrix are dominated by the same column matrix, then we have $m=n$ and, by a similar argument as above, the image of T is a single column matrix. □

The converse of Theorem 5 is not true. For example, we define a linear operator T on ${\mathbb{M}}_{2,3}$ by

$$T\left(\left[\begin{array}{ccc}a& b& c\\ d& e& f\end{array}\right]\right)=\left[\begin{array}{ccc}a+e& a+b+e+f& a+b+c+d+e+f\\ 0& 0& 0\end{array}\right]$$

for every $\left[\begin{array}{ccc}a& b& c\\ d& e& f\end{array}\right]$. Then T is line-injective in any line matrix, and the image of T is the first row matrix. However, T preserves neither arctic rank 1 nor arctic rank $\frac{3}{2}$.

5. Conclusions

In this article, we compared the arctic rank with Boolean rank and term rank of a given Boolean matrix. We also obtained some characterizations of linear operators that preserve the arctic rank of Boolean matrices. That is, a linear operator T is a linear operator that preserves arctic rank on ${\mathbb{M}}_{m,n}$ with $3\le m\le n$ if and only if T preserves arctic ranks 1 and $\frac{k}{2}$ with $4\le k\le 2m+1$ if and only if T is a $(P,Q)$-operator. In the future, we shall apply these results to investigate the linear preserver problems over some semirings. We hope to apply these results to extend the previous results on characterizations of linear operators that preserve some matrix properties on matrix spaces.