Linear Jointly Disjointness-Preserving Maps Between Rectangular Matrix Spaces

Abstract

This paper studies pairs of linear maps that preserve the disjointness of matrices in rectangular matrix spaces. We present a complete characterization of all pairs of bijective linear maps that jointly preserve disjointness. Additionally, we apply these results to study maps that preserve specific matrix properties, such as the double-zero product property.

1. Introduction

The study of linear preserver problems began with the pioneering work of Frobenius [1], who examined linear maps that preserve the determinant on matrix spaces. He established that any linear map $\varphi :{\mathbf{M}}_n\left(\mathbb{C}\right)\to {\mathbf{M}}_n\left(\mathbb{C}\right)$ of $n\times n$ complex matrices that preserves the determinant, meaning that $det\left(A\right)=det\left(\varphi \right(A\left)\right)$, must take one of the forms of $A\mapsto PAQ$ or $A\mapsto P{A}^{T}Q$, where $P,Q\in {\mathbf{M}}_n$, and $det\left(PQ\right)=1$. Since then, this result has been extended in various directions and applied to topics such as rank preservers and zero product preservers (see, e.g., [2,3,4]).

In particular, some of the results in preserver problems on matrix spaces have been applied in quantum computing (see, e.g., [5,6,7,8,9,10,11,12,13]). In the mathematical framework of quantum computing, quantum states are density matrices, i.e., positive semidefinite matrices with trace one, and quantum channels (operators) are completely positive and trace-preserving linear maps. Unital qubit channels preserve the set of quantum states that form a basis of a qubit system, and Choi and Li characterized such maps in [11]. Additionally, Molnár and Szokol [12] gave a characterization of maps for quantum states which preserve the relative entropy. Furthermore, He, Yuan, and Hou [13] investigated surjective maps for quantum states that preserve the quantum entropy of convex combinations.

In recent years, the scope of preserver problems has naturally expanded to include the study of pairs of maps that are jointly preserving certain properties (see, e.g., [14,15,16,17,18,19,20,21,22]). For instance, Costara proved that if $\varphi$ and $\psi$ are maps from ${\mathbf{M}}_n\left(\mathbb{F}\right)$, which is the set of $n\times n$ matrices over a field $\mathbb{F}$ with at least $n^2+1$ distinct elements into itself, and one of them is surjective so that $det\left(\varphi \right(A)+\psi (B\left)\right)=det(A+B)$ for all $A,B\in {\mathbf{M}}_n\left(\mathbb{F}\right)$, then there exist $A_0\in {\mathbf{M}}_n\left(\mathbb{F}\right)$ and $P,Q\in {\mathbf{M}}_n\left(\mathbb{F}\right)$ with $det\left(PQ\right)=1$ such that either $\varphi \left(A\right)=P(A+A_0)Q$ and $\psi \left(A\right)=P(A-A_0)Q$ for all $A\in {\mathbf{M}}_n\left(\mathbb{F}\right)$, or $\varphi \left(A\right)=P{(A+A_0)}^{T}Q$ and $\psi \left(A\right)=P{(A-A_0)}^{T}Q$ for all $A\in {\mathbf{M}}_n\left(\mathbb{F}\right)$.

Our focus is on a pair of linear maps that jointly preserve disjointness, a versatile concept that is applicable across various mathematical frameworks. The notion of disjointness is often described by using terms such as d-homomorphisms [23], Lamperti operators [24], or separating maps [25], depending on the structure of the spaces that are involved. For example, in *-algebra, disjointness for a pair of elements a and b has been defined in at least three ways: (i) $ab=0$ (zero product); (ii) $a^{*}b=0$ (range orthogonality); and (iii) $a{b}^{*}=0$ (domain orthogonality). Leung, Tsai, and Wong [26] characterized linear disjointness preservers on $W^{*}$-algebras, and Liu, Chou, Liao, and Wong [27] extended this analysis to $A{W}^{*}$-algebras. Chebotar, Ke, Lee, and Wong [28] focused on mappings preserving zero products on $C^{*}$-algebras. In vector-value function spaces, a pair of functions is said to be disjoint if at least one of them takes the value 0 at every point in their domain. Lau and Wong [29] studied orthogonality and disjointness-preserving linear maps between Fourier and Fourier–Stieltjes algebras of locally compact groups. Alaminos, Extremera, and Villena [30] investigated zero-product-preserving maps of Lipschitz functions, while Leung and Wang [31] examined disjointness-preserving operators on differentiable function spaces.

Let ${\mathbf{M}}_{m\times n}$ denote the set of $m\times n$ matrices over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, and let ${\mathbf{M}}_n={\mathbf{M}}_{n\times n}$. Two matrices, $A,B\in {\mathbf{M}}_{m\times n}$, are said to be disjoint, denoted by

$$A\perp B,\mathrm{if}A{B}^{*}=0_m\mathrm{and}{A}^{*}B=0_n.$$

Here, the adjoint $A^{*}$ of a rectangular matrix A is its conjugate transpose $\overline{A^T}$. For real matrices, $A^{*}$ reduces to $A^T$, which is the transpose of A.

A pair of linear maps $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{p\times q}$ is said to jointly preserve disjointness if $\varphi \left(A\right)$ and $\psi \left(B\right)$ are disjoint whenever A and B are disjoint; that is,

$$A\perp B⟹\varphi \left(A\right)\perp \psi \left(B\right)$$

If $\varphi =\psi$, Li, Tsai, Wang, and Wong proved in [32] that a linear map $\varphi$ preserves disjointness; that is,

$$A\perp B⟹\varphi \left(A\right)\perp \varphi \left(B\right)$$

if and only if there are unitary matrices $U\in {\mathbf{M}}_p$ and $V\in {\mathbf{M}}_q$ and positive diagonal matrices $Q_1\in {\mathbf{M}}_r$ and $Q_2\in {\mathbf{M}}_s$ with $Q_1$ or $Q_2$ possibly being vacuous ($r=0$ or $s=0$), so that

$$\varphi \left(A\right)=U\left(\begin{array}{ccc}A\otimes Q_1& 0& 0\\ 0& A^T\otimes Q_2& 0\\ 0& 0& 0\end{array}\right)V\mathrm{for}\mathrm{all}A\in {\mathbf{M}}_{m\times n}.$$

Here, $A\otimes Q_1$ means the tensor product of A and $Q_1$.

In this paper, we focus on $\varphi$ and $\psi$, which are bijective linear maps on ${\mathbf{M}}_{m\times n}$ that jointly preserve disjointness. The situation is quite different if $\varphi$ and $\psi$ are not bijective. For example, let $\varphi ,\psi :{\mathbf{M}}_2\to {\mathbf{M}}_2$ be defined by $\varphi \left(A\right)=A{E}_{11}$ and $\psi \left(A\right)=A{E}_{22}$. Then, $\varphi$ and $\psi$ jointly preserve disjointness. In Section 2, we characterize the representation of $\varphi$ and $\psi$, which are bijective linear maps that jointly preserve disjointness. In Section 3, using the results of the main theorem, we further explore several related topics. First, we derive the specific structure of $\varphi$ in the special case where $\varphi =\psi$. Second, we establish that the inverse maps ${\varphi }^{-1}$ and ${\psi }^{-1}$ also jointly preserve disjointness. Lastly, we extend the discussion to maps that jointly preserve double-zero products.

2. Main Results

First, we assume that $\varphi$ and $\psi$ are bijective linear maps on ${\mathbf{M}}_{m\times n}$ and have the following main result:

Theorem 1.

Let $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{m\times n}$ be bijective linear maps. Then, ϕ and ψ jointly preserve disjointness if and only if there exist invertible matrices $P\in {\mathbf{M}}_m$, $Q\in {\mathbf{M}}_n$, and $\alpha \in \mathbb{F}$ so that either

$$\varphi \left(A\right)=PAQand\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}A{\left(Q^{*}\right)}^{-1}$$

(1)

or

$$\varphi \left(A\right)=P{A}^{T}Qand\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}{A}^T{\left(Q^{*}\right)}^{-1},ifm=n$$

(2)

for all $A\in {\mathbf{M}}_{m\times n}$.

The following example demonstrates that while the pair of linear bijective maps $\varphi$ and $\psi$ jointly preserve disjointness, neither $\varphi$ nor $\psi$ individually preserves disjointness.

Example 1.

Let $\varphi ,\psi :{\mathbf{M}}_2\to {\mathbf{M}}_2$ be defined by

$$\varphi \left(A\right)=\left(\begin{array}{cc}1& 0\\ 0& 2\end{array}\right)A\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right)and\psi \left(A\right)=\left(\begin{array}{cc}2& 0\\ 0& 1\end{array}\right)A\left(\begin{array}{cc}3& 0\\ 0& 2\end{array}\right).$$

Based on Theorem 1, ϕ and ψ are bijective linear maps that jointly preserve disjointness. Now consider the disjoint pair

$$A=\left(\begin{array}{cc}1& -1\\ 1& -1\end{array}\right)andB=\left(\begin{array}{cc}1& 1\\ -1& -1\end{array}\right).$$

Based on a direct calculation, we have the following:

$$\varphi {\left(A\right)}^{*}\varphi \left(B\right)=\left(\begin{array}{cc}-12& -18\\ 18& 27\end{array}\right)\ne 0_{2}and\psi {\left(A\right)}^{*}\psi \left(B\right)=\left(\begin{array}{cc}27& 18\\ -18& -12\end{array}\right)\ne 0_2.$$

Thus, $\varphi \left(A\right)\cancel{\perp }\varphi \left(B\right)$ and $\psi \left(A\right)\cancel{\perp }\psi \left(B\right)$.

Let $A\in {\mathbf{M}}_{m\times n}$. Denote the rank of A by $\mathrm{rank}A$, and define $S_A=\{B\in {\mathbf{M}}_{m\times n}:A\perp B\}$. It is straightforward to verify that $S_A$ is a subspace of ${\mathbf{M}}_{m\times n}$.

Lemma 1.

Let $A\in {\mathbf{M}}_{m\times n}$. If $\mathrm{rank}A=k$, then the dimension of $S_A$ is given by

$$\dim S_A=(m-k)(n-k).$$

Proof. 

Suppose that $\mathrm{rank}A=k$. By singular value decomposition, the unitary matrices $U\in {\mathbf{M}}_m$ and $V\in {\mathbf{M}}_n$ exist so that

$$A=U\left(\begin{array}{cc}{C}_1& 0\\ 0& 0_{(m-k)\times (n-k)}\end{array}\right)V^{*}$$

where $C_1\in {\mathbf{M}}_k$ is a positive diagonal matrix. For any $B\in S_A$,

$$\left(\begin{array}{cc}{C}_1& 0\\ 0& 0\end{array}\right){\left(U^{*}BV\right)}^{*}=\left(U^{*}AV\right){\left(U^{*}BV\right)}^{*}=U^{*}AV{V}^{*}{B}^{*}U=0_m$$

and

$${\left(\begin{array}{cc}{C}_1& 0\\ 0& 0\end{array}\right)}^{*}\left(U^{*}BV\right)={\left(U^{*}AV\right)}^{*}\left(U^{*}BV\right)=V^{*}{A}^{*}U{U}^{*}BV=0_n,$$

we can find that

$$B=U\left(\begin{array}{cc}{0}_{k\times k}& 0\\ 0& C_2\end{array}\right)V^{*}$$

for an arbitrary $C_2\in {\mathbf{M}}_{(m-k)\times (n-k)}$. Thus,

$$\dim S_A=(m-k)(n-k).$$

□

Lemma 2.

Let $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{m\times n}$ be bijective linear maps that jointly preserve disjointness. Then, for any $A\in {\mathbf{M}}_{m\times n}$ with $\mathrm{rank}A=1$, we have the following:

$$\mathrm{rank}\varphi \left(A\right)=1and\mathrm{rank}\psi \left(A\right)=1.$$

Proof. 

Since $\psi$ is linear, $\psi \left(S_A\right)$ is a subspace of ${\mathbf{M}}_{m\times n}$. Note that

$$S_{\varphi \left(A\right)}=\{B\in {\mathbf{M}}_{m\times n}:\varphi \left(A\right)\perp B\}.$$

For each $B\in S_A$, we have that $\psi \left(B\right)\in S_{\varphi \left(A\right)}$. This implies that $\psi \left(S_A\right)\subseteq S_{\varphi \left(A\right)}$ and $\dim \psi \left(S_A\right)\le \dim S_{\varphi \left(A\right)}$. Since $\psi$ is a bijective linear map, it follows that $\dim S_A=\dim \psi \left(S_A\right)$. Hence,

$$\dim S_A\le \dim S_{\varphi \left(A\right)}.$$

(3)

Now, suppose that $\mathrm{rank}A=1$ and $\mathrm{rank}\varphi \left(A\right)=k$. Based on Lemma 1, we have that

$$\dim S_A=(m-1)(n-1)\mathrm{and}\dim S_{\varphi \left(A\right)}=(m-k)(n-k)$$

Based on (3), we derive the following:

$$\dim S_A=(m-1)(n-1)\le \dim S_{\varphi \left(A\right)}=(m-k)(n-k).$$

This implies that $k=$ 0 or 1. Since $\varphi$ is bijective, we have that $k=1$, and $\varphi$ preserves rank-one matrices. The proof for $\psi$ is similar. □

In [4], it is shown that for any rank-one-preserving linear map $\Phi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{p\times q}$, there exist invertible matrices $P\in {\mathbf{M}}_p$ and $Q\in {\mathbf{M}}_q$ so that one of the following four alternatives holds:

(i)

$$\Phi \left(A\right)=PAQ,\mathrm{if}(p,q)=(m,n)$$

(4)

(ii)

$$\Phi \left(A\right)=P{A}^{T}Q,\mathrm{if}(p,q)=(n,m)$$

(5)

(iii)

$$\Phi \left(A\right)=P\left(\begin{array}{cc}\psi \left(A\right)& 0\end{array}\right)Q$$

(6)

where $\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{p\times 1}$ is a linear map so that $\psi \left(A\right)\ne 0$ for all $A\in {\mathbf{M}}_{m\times n}$ with rank 1.

(iv)

$$\Phi \left(A\right)=P\left(\begin{array}{c}\psi \left(A\right)\\ 0\end{array}\right)Q$$

(7)

where $\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{1\times q}$ is a linear map so that $\psi \left(A\right)\ne 0$ for all $A\in {\mathbf{M}}_{m\times n}$ with rank 1.

Proof of Theorem 1.

The ‘if’ part is obvious. We proceed to prove the ‘only if’ part. Based on Lemma 2, $\varphi$ and $\psi$ also preserve rank-one matrices. Since $\varphi$ and $\psi$ are bijective, neither $\varphi$ nor $\psi$ satisfies forms (6) or (7). First of all, we assume that $m\ne n$. Then, there exist invertible matrices $P,U\in {\mathbf{M}}_m$ and $Q,V\in {\mathbf{M}}_n$, so that

$$\varphi \left(A\right)=PAQ\mathrm{and}\psi \left(A\right)=UAV$$

for all $A\in {\mathbf{M}}_{m\times n}$. Let $\{E_{11},E_{12},\cdots ,E_{mn}\}$ be the standard basis for ${\mathbf{M}}_{m\times n}$. Write the following:

$$P=\left(\begin{array}{cccc}{p}_1& p_2& \cdots & p_m\end{array}\right),Q={\left(\begin{array}{cccc}{q}_1& q_2& \cdots & q_n\end{array}\right)}^{*},$$

$$U=\left(\begin{array}{cccc}{u}_1& u_2& \cdots & u_m\end{array}\right),V={\left(\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right)}^{*},$$

with $p_i,u_i\in {\mathbb{F}}^m$ and $q_i,v_i\in {\mathbb{F}}^n$ denoting the column vectors of P, U, Q, and V, respectively. For $i\ne k$, $j\ne l$, and

$$\varphi \left(E_{ij}\right)=P{E}_{ij}Q=p_i{q}_j^{*}\mathrm{and}\psi \left(E_{kl}\right)=U{E}_{kl}V=u_k{v}_l^{*}.$$

Since $E_{ij}\perp E_{kl}$, we have that $\varphi {\left(E_{ij}\right)}^{*}\psi \left(E_{kl}\right)=0_n$ and $\varphi \left(E_{ij}\right)\psi {\left(E_{kl}\right)}^{*}=0_m$. It follows that

$$\varphi {\left(E_{ij}\right)}^{*}\psi \left(E_{kl}\right)=\left(p_i^{*}{u}_k\right)q_j{v}_l^{*}=0_n$$

and

$$\varphi \left(E_{ij}\right)\psi {\left(E_{kl}\right)}^{*}=\left(q_j^{*}{v}_l\right)p_i{u}_k^{*}=0_m.$$

Thus,

$$p_i^{*}{u}_k=0\mathrm{and}{q}_j^{*}{v}_l=0.$$

Let $2\le i\le m$. For any nonzero $s_1,s_i,t_1,t_i\in \mathbb{F}$ with $\overline{s_1}{t}_1+\overline{s_i}{t}_i=0$, the matrices

$$Z_1=s_1{E}_{11}+s_i{E}_{i1}\mathrm{and}{Z}_2=t_1{E}_{12}+t_i{E}_{i2}$$

are disjoint, and so are the matrices

$$\varphi \left(Z_1\right)=P(s_1{E}_{11}+s_i{E}_{i1})Q=s_1{p}_1{q}_1^{*}+s_i{p}_i{q}_1^{*}$$

and

$$\psi \left(Z_2\right)=U(t_1{E}_{12}+t_i{E}_{i2})V=t_1{u}_1{v}_2^{*}+t_i{u}_i{v}_2^{*}.$$

Based on a direct calculation,

$$\varphi {\left(Z_1\right)}^{*}\psi \left(Z_2\right)={(s_1{p}_1{q}_1^{*}+s_i{p}_i{q}_1^{*})}^{*}(t_1{u}_1{v}_2^{*}+t_i{u}_i{v}_2^{*})=\overline{s_1}{t}_1(p_1^{*}{u}_1-p_i^{*}{u}_i)q_1{v}_2^{*}=0_n,$$

and then, $p_1^{*}{u}_1=p_i^{*}{u}_i$.

Let $2\le j\le n$. For any nonzero $x_1,x_j,y_1,y_j\in \mathbb{F}$ with $x_1\overline{y_1}+x_j\overline{y_j}=0$, the matrices

$$Z_3=x_1{E}_{11}+x_j{E}_{1j}\mathrm{and}{Z}_4=y_1{E}_{21}+y_j{E}_{2j}$$

are disjoint, and so are the matrices

$$\varphi \left(Z_3\right)=P(x_1{E}_{11}+x_j{E}_{1j})Q=x_1{p}_1{q}_1^{*}+x_j{p}_1{q}_j^{*}$$

and

$$\psi \left(Z_4\right)=U(y_1{E}_{21}+y_j{E}_{2j})V=y_1{u}_2{v}_1^{*}+y_j{u}_2{v}_j^{*}.$$

Based on a direct calculation,

$$\varphi \left(Z_3\right)\psi {\left(Z_4\right)}^{*}=(x_1{p}_1{q}_1^{*}+x_j{p}_1{q}_j^{*}){(y_1{u}_2{v}_1^{*}+y_j{u}_2{v}_j^{*})}^{*}=x_1\overline{y_1}(q_1^{*}{v}_1-q_j^{*}{v}_j)p_1{u}_2^{*}=0_m,$$

and then, $q_1^{*}{v}_1=q_j^{*}{v}_j$.

Thus, we obtain the following:

$$P^{*}U={\left(\begin{array}{cccc}{p}_1& p_2& \cdots & p_m\end{array}\right)}^{*}\left(\begin{array}{cccc}{u}_1& u_2& \cdots & u_m\end{array}\right)=k_1{I}_m,$$

and

$$Q{V}^{*}={\left(\begin{array}{cccc}{q}_1& q_2& \cdots & q_n\end{array}\right)}^{*}\left(\begin{array}{cccc}{v}_1& v_2& \cdots & v_n\end{array}\right)=k_2{I}_n,$$

where $k_1=p_1^{*}{u}_1$, and $k_2=q_1^{*}{v}_1$. Therefore, we conclude that $\varphi \left(A\right)$ and $\psi \left(A\right)$ can be expressed in the desired forms with $\alpha =k_1\overline{k_2}$.

Now, for $m=n$, we need to consider the following three cases:

Case 1—There exist invertible matrices $P,Q,U,$ and $V\in {\mathbf{M}}_m$ so that

$$\varphi \left(A\right)=PAQand\psi \left(A\right)=UAV.$$

Based on previous arguments, we obtain that

$$\varphi \left(A\right)=PAQ\mathrm{and}\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}A{\left(Q^{*}\right)}^{-1}$$

where $\alpha \in \mathbb{F}$ for all $A\in {\mathbf{M}}_m$.

Case 2—There exist invertible matrices $P,Q,U,$ and $V\in {\mathbf{M}}_m$ so that

$$\varphi \left(A\right)=P{A}^{T}Q\mathrm{and}\psi \left(A\right)=U{A}^{T}V.$$

Based on similar arguments to the previous ones, we obtain that

$$\varphi \left(A\right)=P{A}^{T}Q\mathrm{and}\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}{A}^T{\left(Q^{*}\right)}^{-1}$$

where $\alpha \in \mathbb{F}$ for all $A\in {\mathbf{M}}_m$.

Case 3—There exist invertible matrices $P,Q,U,$ and $V\in {\mathbf{M}}_m$ so that

$$\varphi \left(A\right)=PAQ\mathrm{and}\psi \left(A\right)=U{A}^{T}V$$

(or $\varphi \left(A\right)=P{A}^{T}Q$ and $\psi \left(A\right)=UAV$). Let $p_i,q_i,u_i,$ and $v_i\in {\mathbb{F}}^m$ be the column vectors of P, Q, U, and V, respectively. Since $E_{12}\perp E_{21}$, we have that $\varphi {\left(E_{12}\right)}^{*}\psi \left(E_{21}\right)=0_m$. It follows that

$$\varphi {\left(E_{12}\right)}^{*}\psi \left(E_{21}\right)=\left(p_1^{*}{u}_1\right)q_2{v}_2^{*}=0_m,\mathrm{and}{p}_1^{*}{u}_1=0.$$

Let $2\le l\le m$. Since $E_{11}\perp E_{2l}$, we have that $\varphi {\left(E_{11}\right)}^{*}\psi \left(E_{2l}\right)=0_m$. It follows that

$$\varphi {\left(E_{11}\right)}^{*}\psi \left(E_{2l}\right)=\left(p_1^{*}{u}_l\right)q_1{v}_2^{*}=0_m,\mathrm{and}{p}_1^{*}{u}_l=0.$$

Therefore,

$$p_1^{*}U=0_{m\times 1},\mathrm{and}{p}_1=0.$$

This contradicts the fact that $p_1$ is a column vector of the invertible matrix P. □

Remark 1.

In Theorem 1, if the condition that ϕ and ψ jointly preserve disjointness is replaced by the condition that

$$\varphi \left(A\right)\perp \psi \left(B\right)wheneverA\perp B$$

for all rank one $A,B\in {\mathbf{M}}_{m\times n}$, then the conclusion is also valid with the same proof.

In the case where $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{n\times m}$, as in the proof of Theorem 1 and using form (5), we have the result presented below.

Remark 2.

Let $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{n\times m}$ be bijective linear maps. Then, ϕ and ψ jointly preserve disjointness if and only if there exist invertible matrices $P\in {\mathbf{M}}_n$, $Q\in {\mathbf{M}}_m$, and $\alpha \in \mathbb{F}$ so that

$$\varphi \left(A\right)=P{A}^{T}Qand\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}{A}^T{\left(Q^{*}\right)}^{-1}$$

or

$$\varphi \left(A\right)=PAQand\psi \left(A\right)=\alpha {\left(P^{*}\right)}^{-1}A{\left(Q^{*}\right)}^{-1},ifm=n$$

for all $A\in {\mathbf{M}}_{m\times n}$.

In the rest of this section, we apply Theorem 1 to investigate several related topics. First, we derive the specific structure of $\varphi$ in the special case where $\varphi =\psi$.

Corollary 1.

Let $\varphi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{m\times n}$ be a bijective linear map. Then, ϕ preserves disjointness if and only if there exist unitary matrices $P\in {\mathbf{M}}_m$, $Q\in {\mathbf{M}}_n$, and $\alpha >0$ so that either

$$\varphi \left(A\right)=\alpha PAQ$$

or

$$\varphi \left(A\right)=\alpha P{A}^{T}Q,ifm=n$$

for all $A\in {\mathbf{M}}_{m\times n}$.

Proof. 

The ‘if’ part is obvious. We proceed to prove the ‘only if’ part. Based on Theorem 1 with $\varphi =\psi$, first, we consider the form of (1); there exist invertible matrices $P\in {\mathbf{M}}_m$, $Q\in {\mathbf{M}}_n$, and $\alpha \in \mathbb{F}$ so that

$$PAQ=\alpha {\left(P^{*}\right)}^{-1}A{\left(Q^{*}\right)}^{-1}.$$

Then,

$$P^{*}PAQ{Q}^{*}=\alpha A,$$

for all $A\in {\mathbf{M}}_{m\times n}$. For simplicity of notation, we write U and V instead of $P^{*}P$ and $Q{Q}^{*}$, respectively. Then,

$$UAV=\alpha A.$$

(8)

Note that U and V are Hermitian and invertible matrices. Notice that U is a positive definite, since ${\left(Px\right)}^{*}\left(Px\right)>0$ for any nonzero column vector $x\in {\mathbb{F}}^m$. Similarly, V is also a positive definite. Multiplying by $V^{-1}$ on both sides of (8), we obtain that $UA=\alpha A{V}^{-1}$. Taking the adjoint on both sides of (8), we obtain that $\overline{\alpha }{A}^{*}=V{A}^{*}U$. Thus, we obtain that

$$\overline{\alpha }UA{A}^{*}=UAV{A}^{*}U=\alpha A{A}^{*}U,$$

for all $A\in {\mathbf{M}}_{m\times n}$.

Now, we claim that $U=c{I}_m$ for some $c>0$. Let $U={\left(u_{ij}\right)}_{1\le i,j\le m}$. Since $\overline{\alpha }UA{A}^{*}=\alpha A{A}^{*}U$, by taking $A{A}^{*}=E_{kk}$ for $1\le k\le m$, we obtain that

$$u_{ij}=0,i\ne j\mathrm{and}\overline{\alpha }=\alpha .$$

Take ${\displaystyle A{A}^{*}=\sum _{i,j=1}^m{E}_{ij}}$. Since $UA{A}^{*}=A{A}^{*}U$ and U is a positive definite, we obtain that

$$u_{ii}=u_{jj}>0,i\ne j.$$

Similarly, $V=d{I}_n$ for some $d>0$.

Let $\widehat{P}={\displaystyle \frac{1}{\sqrt{c}}}P$ and $\widehat{Q}={\displaystyle \frac{1}{\sqrt{d}}}Q$. Then, $\widehat{P}$ and $\widehat{Q}$ are unitary. Based on form (1), we conclude that

$$\varphi \left(A\right)=PAQ=\sqrt{cd}\widehat{P}A\widehat{Q},$$

for all $A\in {\mathbf{M}}_{m\times n}$. If $m=n$, based on form (2) and using a similar method, we have that

$$\varphi \left(A\right)=P{A}^{T}Q=\sqrt{cd}\widehat{P}{A}^T\widehat{Q},$$

for all $A\in {\mathbf{M}}_m$. □

Second, under the bijectivity condition, we establish that the inverse maps ${\varphi }^{-1}$ and ${\psi }^{-1}$ also jointly preserve disjointness.

Corollary 2.

Let $\varphi ,\psi :{\mathbf{M}}_{m\times n}\to {\mathbf{M}}_{m\times n}$ be bijective linear maps that jointly preserve disjointness. Based on Theorem 1, there exist invertible matrices $P\in {\mathbf{M}}_m$, $Q\in {\mathbf{M}}_n$, and $\alpha \in \mathbb{F}$ so that

$${\varphi }^{-1}\left(A\right)=P^{-1}A{Q}^{-1}and{\psi }^{-1}\left(A\right)={\displaystyle \frac{1}{\alpha }}{P}^{*}A{Q}^{*}$$

or

$${\varphi }^{-1}\left(A\right)=P^{-1}{A}^T{Q}^{-1}and{\psi }^{-1}\left(A\right)={\displaystyle \frac{1}{\alpha }}{P}^{*}{A}^T{Q}^{*},ifm=n.$$

Again, based Theorem 1, we derive that ${\varphi }^{-1}$ and ${\psi }^{-1}$ also jointly preserve disjointness.

Lastly, we extend the discussion to maps that jointly preserve double-zero products. We assume bijective linear maps $\widehat{\varphi },\widehat{\psi }:{\mathbf{M}}_n\to {\mathbf{M}}_n$ that jointly preserve double-zero products; that is,

$$AB=BA=0_n⟹\widehat{\varphi }\left(A\right)\widehat{\psi }\left(B\right)=\widehat{\psi }\left(B\right)\widehat{\varphi }\left(A\right)=0_n$$

for all $A,B\in {\mathbf{M}}_n$. Set the following:

$$\varphi \left(A\right)=\widehat{\varphi }\left(A\right)\mathrm{and}\psi \left(B\right)={\left(\widehat{\psi }\left(B^{*}\right)\right)}^{*}.$$

Suppose that $A,B\in {\mathbf{M}}_n$, with $A{\left(B^{*}\right)}^{*}=0_n$, and $A^{*}{B}^{*}=0_n$. Then, $AB=BA=0_n$, and

$$\varphi {\left(A\right)}^{*}\psi \left(B^{*}\right)={\left(\widehat{\varphi }\left(A\right)\right)}^{*}{\left(\widehat{\psi }\left(B\right)\right)}^{*}={\left(\widehat{\psi }\left(B\right)\widehat{\varphi }\left(A\right)\right)}^{*}=0_n$$

and

$$\varphi \left(A\right){\left(\psi \left(B^{*}\right)\right)}^{*}=\widehat{\varphi }\left(A\right)\widehat{\psi }\left(B\right)=0_n.$$

Thus, $\varphi$ and $\psi$ are bijective linear maps that jointly preserve disjointness. So, we have the following remark:

Corollary 3.

Let $\widehat{\varphi },\widehat{\psi }:{\mathbf{M}}_n\to {\mathbf{M}}_n$ be bijective linear maps. Then, $\widehat{\varphi }$ and $\widehat{\psi }$ jointly preserve double-zero products if and only if there exist invertible matrices $P,Q\in {\mathbf{M}}_n$, and $\alpha \in \mathbb{F}$ so that

$$\widehat{\varphi }\left(A\right)=PAQand\widehat{\psi }\left(A\right)=\overline{\alpha }{Q}^{-1}A{P}^{-1}$$

or

$$\widehat{\varphi }\left(A\right)=P{A}^{T}Qand\widehat{\psi }\left(A\right)=\overline{\alpha }{Q}^{-1}{A}^T{P}^{-1}$$

for all $A\in {\mathbf{M}}_n$.

3. Conclusions

In this paper, we characterize the structure of the bijective linear maps $\varphi$ and $\psi$ that jointly preserve disjointness between rectangular matrix spaces. We demonstrate that $\varphi$ and $\psi$ are rank-one preservers. Based on the structure of linear rank-one-preserving maps, we derive Theorem 1. Notably, we extend the main result to maps that jointly preserve double-zero products, providing a foundation for the study of jointly preserving properties in linear maps.

Jointly disjointness-preserving problems that remove linearity or surjectivity are highly challenging. It would be interesting to relax these assumptions in our results. For instance, it would be valuable to investigate a version of Theorem 1 under the condition that $\varphi$ and $\psi$ jointly preserve disjointness, but that only one of the maps is surjective. Another direction involves studying the special case where $\varphi =\psi$ is a non-linear map that preserves disjointness. Additionally, one may consider linear or non-linear preservers for other types of matrices, such as Hermitian, symmetric, or skew-symmetric matrix spaces. Furthermore, extending the study to the linear maps $\varphi$ and $\psi$ that preserve other properties—such as the rank, norm, or spectrum—could offer new insights into their structural characteristics.