Generalized Inverses of Boolean Tensors via Einstein Product

TL;DR

This paper extends the theory of boolean matrices to boolean tensors using the Einstein product, introducing generalized inverses, characterizations, and tensor rank definitions to enhance multiway data analysis.

Contribution

It develops a new framework for generalized inverses of boolean tensors with the Einstein product, including characterizations, equivalence results, and a tensor rank concept.

Findings

Characterizations of generalized inverses for boolean tensors

Equivalence results on boolean tensors

Definition of tensor rank via space decomposition

Abstract

Applications of the theory and computations of boolean matrices are of fundamental importance to study a variety of discrete structural models. But the increasing ability of data collection systems to store huge volumes of multidimensional data, the boolean matrix representation of data analysis is not enough to represent all the information content of the multiway data in different fields. From this perspective, it is appropriate to develop an infrastructure that supports reasoning about the theory and computations. In this paper, we discuss the generalized inverses of the Boolean tensors with the Einstein product. Further, we elaborate on this theory by producing a few characterizations of different generalized inverses and several equivalence results on boolean tensors. In addition to these, we define the rank of a boolean tensor through space decomposition.

Equations

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{i_1...i_N{j}_1...j_M}=\sum _{k_1...k_N}{a}_{i_1...i_N{k}_1...k_N}{b}_{k_1...k_N{j}_1...j_M}.$$

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{i_1...i_N}=\sum _{k_1...k_N}{a}_{i_1...i_N{k}_1...k_N}{b}_{k_1...k_N}.$$

$$(\mathcal{A}+\mathcal{B}{)}_{i_1...i_N{k}_1...k_N}=a_{i_1...i_N{k}_1...k_N}+b_{i_1...i_N{k}_1...k_N}.$$

$${\delta }_{i_k{j}_k}=$$

$$\pi :=\left(\begin{array}{cccccccc}{i}_1& i_2& \cdots & i_N& j_1& j_2& \cdots & j_N\\ \pi (i_1)& \pi (i_2)& \cdots & \pi (i_N)& \pi (j_1)& \pi (j_2)& \cdots & \pi (j_N)\end{array}\right).$$

$${ϵ}_{i_k}{ϵ}_{j_k}=$$

$$t=i_N+\sum _{K=1}^{N-1}\left[(i_K-1)\prod _{L=K+1}^N{I}_L\right].$$

$$[\mathcal{A}\mathcal{B}{]}_{i_1\cdots i_N{l}_1\cdots l_M}=\{\begin{array}{ll}{a}_{i_1\cdots i_N{l}_1\cdots l_M},& i_1\cdots i_N\in [I_1]\times \cdots \times [I_N],l_1\cdots l_M\in [J_1]\times \cdots \times [J_M];\\ b_{i_1\cdots i_N{l}_1\cdots l_M},& i_1\cdots i_N\in [I_1]\times \cdots \times [I_N],l_1\cdots l_M\in {\Gamma }_1\times \cdots \times {\Gamma }_M;\\ 0,& \text{otherwise}.\end{array}$$

$$\left[\begin{array}[]{c}\mathcal{C}\\ \mathcal{D}\\ \end{array}\right]=[\mathcal{C}^{T}~{}\mathcal{D}^{T}]^{T}\in\mathbb{R}^{\beta_{1}\times\cdots\times\beta_{M}\times\alpha^{N}}.$$

$$\left[\begin{array}[]{c}\mathcal{A}_{1}~{}~{}\mathcal{B}_{1}\\ \mathcal{A}_{2}~{}~{}\mathcal{B}_{2}\\ \end{array}\right]\in\mathbb{R}^{\rho_{1}\times\cdots\times\rho_{N}\times\beta_{1}\times\cdots\times\beta_{M}},$$

$$\mathfrak{R}(\mathcal{A})=\left\{\mathcal{A}{\ast }_N\mathcal{X}:\mathcal{X}\in {\mathbb{R}}^{J_1\times \cdots \times J_N}\right\}\text{mbox}and\mathcal{N}(\mathcal{A})=\left\{\mathcal{X}:\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{O}\in {\mathbb{R}}^{I_1\times \cdots \times I_M}\right\}.$$

$$b_{{i_{1}}...{i_{N}}{j_{1}}...{j_{M}}}=\left\{\begin{array}[]{cc}1,&\mbox{ when }a_{i_{1}\cdots i_{N}j_{1}j_{2}\cdots j_{M}}=0.\\ 0,&\mbox{ when }a_{i_{1}\cdots i_{N}j_{1}j_{2}\cdots j_{M}}=1.\end{array}\right.$$

$$\overline{\mathcal{A}}=\sum _{k=1}^{\infty }{\mathcal{A}}^k.$$

$$\overline{(\mathcal{A}+\mathcal{B})}=\overline{(\overline{\mathcal{A}}{\ast }_N\overline{\mathcal{B}})}=\overline{(\overline{\mathcal{B}}{\ast }_N\overline{\mathcal{A}})}.$$

$$\overline{\left(\overline{\mathcal{A}}{\ast }_N\overline{\mathcal{B}}\right)}\ge \overline{\mathcal{A}+\mathcal{B}}.$$

$$\overline{\left(\overline{\mathcal{A}}{\ast }_N\overline{\mathcal{B}}\right)}\le \overline{\left(\overline{\mathcal{A}+\mathcal{B}}\right)}=\overline{\mathcal{A}+\mathcal{B}}.$$

$$a_{i_1\cdots i_M{j}_1\cdots j_N}\le b_{i_1\cdots i_M{j}_1\cdots j_N}.$$

$$b_{i_1\cdots i_M{j}_1\cdots j_N}=\sum _{k_1\cdots k_N}\sum _{l_1\cdots l_M}{a}_{i_1\cdots i_M{k}_1\cdots k_N}{a}_{l_1\cdots l_M{k}_1\cdots k_N}{a}_{l_1\cdots l_M{j}_1\cdots j_N}.$$

$$b_{i_1\cdots i_M{j}_1\cdots j_N}\ge (a_{i_1\cdots i_M{j}_1\cdots j_N}{)}^3=a_{i_1\cdots i_M{j}_1\cdots j_N}=1.$$

$$c_{i_{1}\cdots i_{N}j_{1}\cdots j_{n}}=\left\{\begin{array}[]{cc}1&\mbox{ if }a_{i_{1}\cdots i_{N}i_{1}\cdots i_{N}}=0\mbox{ or }b_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1\mbox{ for all }i_{k},~{}1\leq k\leq N,\\ 0&otherwise.\end{array}\right.$$

$$a_{ij11}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 1\end{array}\right),a_{ij12}=a_{ij13}=a_{ij21}=a_{ij22}=a_{ij23}=\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 1\end{array}\right),\text{mbox}and$$

$$b_{ij11}=b_{ij12}=b_{ij13}=b_{ij21}=b_{ij22}=b_{ij23}=\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\end{array}\right).$$

$$x_{ij11}=x_{ij12}=x_{ij13}=x_{ij21}=x_{ij22}=x_{ij23}=\left(\begin{array}{ccc}0& 1& 1\\ 1& 1& 0\end{array}\right),$$

$$y_{ij11}=y_{ij12}=y_{ij13}=y_{ij21}=y_{ij22}=y_{ij23}=\left(\begin{array}{ccc}1& 1& 1\\ 1& 1& 0\end{array}\right),\text{mbox}and$$

$$z_{ij11}=z_{ij12}=z_{ij13}=z_{ij21}=z_{ij22}=z_{ij23}=\left(\begin{array}{ccc}0& 1& 1\\ 1& 1& 1\end{array}\right).$$

$$tr(\mathcal{A}{\ast }_N{\mathcal{A}}^C)$$

$$tr(\mathcal{A}{\ast }_N{\mathcal{A}}^C)=tr({\mathcal{A}}^C{\ast }_N\mathcal{A}).$$

$$a_{ij11}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),a_{ij12}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),a_{ij21}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),a_{ij22}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right).$$

$$x_{ij11}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),x_{ij12}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),x_{ij21}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),x_{ij22}=\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right),$$

Full text

††journal: arxiv

1 Introduction

1.1 Background and motivation

The study of the Boolean matrices [1, 2, 27, 37] play an important role in linear algebra [6, 19, 36], combinatorics [10], graph theory [7] and network theory [24, 25]. However, this becomes particularly challenging to store huge volumes of multidimensional data. This potential difficulty can be easily overcome, thanks to tensors, which are natural multidimensional generalizations of matrices [21, 35]. Here the notion of tensors is different in physics and engineering (such as stress tensors) [31], which are generally referred to as tensor fields in mathematics [12]. However, it will be more appropriate if we study the Boolean tensors and the generalized inverses of Boolean tensors. Hence the generalized inverses of Boolean tensors will encounter in many branches of mathematics, including relations theory [34], logic, graph theory, lattice theory [8] and algebraic semigroup theory.

Recently, there has been increasing interest in studying inverses [9] and different generalized inverses of tensors based on the Einstein product [5, 17, 40, 41], and opened new perspectives for solving multilinear systems [21, 26]. In [17, 40], the authors have introduced some basic properties of the range and null space of multidimensional arrays. Further, in [40], it was discussed the adequate definition of the tensor rank, termed as reshaping rank. Corresponding representations of the weighted Moore-Penrose inverse introduced in [4, 16] and investigated a few characterizations in [33]. Though this work is focusing on the binary case; i.e., concentrating some interesting results based on the Boolean tensors and generalized inverses of Boolean tensors via the Einstein product. In many instances, the result in the general case does not immediately follow even though it is not difficult to conclude.

On the other hand, one of the most successful developments in the world of multilinear algebra is the concept of tensor decomposition [20, 21, 23]. This concept gives a clear and convenient way to implement all basic operations efficiently. Recently this concept is extended in Boolean tensors [14, 18, 38]. Further, the fast and scalable distributed algorithms for Boolean tensor decompositions were discussed in [29]. In addition to that, a few applications of these decompositions are discussed in [14, 28] for information extraction and clustering. At that same time, Brazell, et al. in [9] discussed decomposition of tensors from the isomorphic group structure on the influence of the Einstein Product and demonstrated that they are special cases of the canonical polyadic decomposition [11]. The vast work on decomposition on the tensors and its several applications in different areas of mathematics in the literature, and the recent works in [9, 41], motivate us to study the generalized inverses and space decomposition in the framework of Boolean tensors. This study leads to introduce the rank and the weight for the Boolean tensor with its application to generalized inverses.

1.2 Organization of the paper

The rest of the paper is organized as follows. In Section 2 we present some definitions, notations, and preliminary results, which are essential in proving the main results. The main results are discussed in Section 3. It has four subparts. In the first part, some identities are proved while the generalized inverses for Boolean tensor are discussed in the second part. The third part mainly focuses on weighted Moore-Penrose inverses. Space decomposition and its application to generalized inverses are discussed in the last part. Finally, the results along with a few questions are concluded in Section 4.

2 Preliminaries

We first introduce some basic definitions and notations which will be used throughout the article.

2.1 Definitions and terminology

For convenience, we first briefly explain some of the terminologies which will be used here onwards. The tensor notation and definitions are followed from the article [9, 41]. We refer $\mathbb{R}^{I_{1}\times\cdots\times I_{N}}$ as the set of order $N$ real tensors. Indeed, a matrix is a second order tensor, and a vector is a first order tensor. Let $\mathbb{R}^{I_{1}\times\cdots\times I_{N}}$ be the set of order $N$ and dimension $I_{1}\times\cdots\times I_{N}$ tensors over the real field $\mathbb{R}$. $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}}$ is a tensor with $N$-th order tensor, and each entry of $\mathcal{A}$ is denoted by $a_{i_{1}...i_{N}}$. Note that throughout the paper, tensors are represented in calligraphic letters like $\mathcal{A}$, and the notation $(\mathcal{A})_{i_{1}...i_{N}}=a_{i_{1}...i_{N}}$ represents the scalars. The Einstein product ([13]) $\mathcal{A}{*_{N}}\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}}$ of tensors $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times K_{1}\times\cdots\times K_{N}}$ and $\mathcal{B}\in\mathbb{R}^{K_{1}\times\cdots\times K_{N}\times J_{1}\times\cdots\times J_{M}}$ is defined by the operation ${*_{N}}$ via

[TABLE]

Specifically, if $\mathcal{B}\in\mathbb{R}^{K_{1}\times\cdots\times K_{N}}$, then $\mathcal{A}{*_{N}}\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}}$ and

[TABLE]

This product is discussed in the area of continuum mechanics [13] and the theory of relativity [22]. Further, the addition of two tensors $\mathcal{A},~{}\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times K_{1}\times\cdots\times K_{N}}$ is defined as

[TABLE]

For a tensor $~{}\mathcal{A}=(a_{{i_{1}}...{i_{N}}{j_{1}}...{j_{M}}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}},$ let $\mathcal{B}=(b_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}})\in\mathbb{R}^{J_{1}\times\cdots\times J_{M}\times I_{1}\times\cdots\times I_{N}}$, be the transpose of $\mathcal{A}$, where $b_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}=a_{j_{1}\cdots j_{M}i_{1}\cdots i_{N}}.$ The tensor $\mathcal{B}$ is denoted by $\mathcal{A}^{T}$. Also, we denote $\mathcal{A}^{T}=\left({a}_{{i_{1}}...{i_{N}}{j_{1}}...{j_{M}}}^{t}\right).$ The trace of a tensor $\mathcal{A}$ with entries $(\mathcal{A})_{{i_{1}}...{i_{N}}{j_{1}}...{j_{N}}}$, denoted by $tr(\mathcal{A})$, is defined as the sum of the diagonal entries, i.e., $tr(\mathcal{A})=\displaystyle\sum_{i_{1}\cdots i_{N}}a_{{i_{1}...i_{N}}{i_{1}...i_{N}}}.$ Further, a tensor $\mathcal{O}$ denotes the zero tensor if all the entries are zero. A tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is symmetric if $\mathcal{A}=\mathcal{A}^{T},$ and orthogonal if $\mathcal{A}{*_{M}}\mathcal{A}^{T}=\mathcal{A}^{T}{*_{N}}\mathcal{A}=\mathcal{I}$. Further, a tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is idempotent if $\mathcal{A}{*_{N}}\mathcal{A}=\mathcal{A}$. The definition of a diagonal tensor follows. Further, a tensor with entries $(\mathcal{D})_{{i_{1}}...{i_{N}}{j_{1}}...{j_{N}}}$ is called a diagonal tensor if $d_{{i_{1}}...{i_{N}}{j_{1}}...{j_{N}}}=0$ for $(i_{1},\cdots,i_{N})\neq(j_{1},\cdots,j_{N}).$ A few more notations and definitions are discussed below for defining generalized inverses of Boolean tensors. We first recall the definition of an identity tensor below.

Definition 2.1**.**

*(Definition 3.13, [9])

A tensor with entries $(\mathcal{I})_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=\prod_{k=1}^{N}\delta_{i_{k}j_{k}}$, where*

[TABLE]

is called a unit tensor or identity tensor.

The permutation tensor is defined as follows.

Definition 2.2**.**

Let $\pi$ be a permutation map on $(i_{1},i_{2},\cdots,i_{N},j_{1},j_{2},\cdots,j_{N})$ defined by

[TABLE]

A tensor $\mathcal{P}$ with entries $(\mathcal{P})_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=\prod_{k=1}^{N}\epsilon_{i_{k}}\epsilon_{j_{k}}$, where

[TABLE]

is called a permutation tensor.

Now we recall the block tensor as follows.

Definition 2.3**.**

[41]** For a tensor $\mathcal{A}=(a_{i_{1}...i_{N}j_{1}...j_{M}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}},\\ \mathcal{A}_{(i_{1}...i_{N}|:)}=(a_{i_{1}...i_{N}:...:})\in\mathbb{R}^{J_{1}\times\cdots\times J_{M}}$ is a subblock of $\mathcal{A}$. $Vec(\mathcal{A})$ is obtained by lining up all the subtensors in a column, and $t$-th subblock of $Vec(\mathcal{A})$ is $\mathcal{A}_{(i_{1}...i_{N}|:)}$, where

[TABLE]

Let $\mathcal{A}=(a_{i_{1}\cdots i_{N}j_{1}\cdots j_{M}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}}$ and $\mathcal{B}=(b_{i_{1}\cdots i_{N}k_{1}\cdots k_{M}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times K_{1}\times\cdots\times K_{M}}$. The row block tensor consisting of $\mathcal{A}$ and $\mathcal{B}$ is denoted by $[\mathcal{A}~{}\mathcal{B}]\in\mathbb{R}^{\alpha^{N}\times\beta_{1}\times\cdots\times\beta_{M}},$ where $\alpha^{N}=I_{1}\times\cdots\times I_{N},\beta_{i}=J_{i}+K_{i},i=1,\cdots,M$, and is defined by

[TABLE]

where $\Gamma_{i}=\{J_{i}+1,\cdots,J_{i}+K_{i}\},i=1,\cdots,M.$

Let $\mathcal{C}=(c_{j_{1}\cdots j_{M}i_{1}\cdots i_{N}})\in\mathbb{R}^{J_{1}\times\cdots\times J_{M}\times I_{1}\times\cdots\times I_{N}}$ and $\mathcal{D}=(d_{k_{1}\cdots k_{M}i_{1}\cdots i_{N}})\in\mathbb{R}^{K_{1}\times\cdots\times K_{M}\times I_{1}\times\cdots\times I_{N}}$. The column block tensor consisting of $\mathcal{C}$ and $\mathcal{D}$ is

[TABLE]

For $\mathcal{A}_{1}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}},\mathcal{B}_{1}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times K_{1}\times\cdots\times K_{M}},\mathcal{A}_{2}\in\mathbb{R}^{L_{1}\times\cdots\times L_{N}\times J_{1}\times\cdots\times J_{M}}$ and $\mathcal{B}_{2}\in\mathbb{R}^{L_{1}\times\cdots\times L_{N}\times K_{1}\times\cdots\times K_{M}}$, we denote $\tau_{1}=[\mathcal{A}_{1}~{}\mathcal{B}_{1}]$ and $\tau_{2}=[\mathcal{A}_{2}~{}\mathcal{B}_{2}]$ as the row block tensors. The column block tensor \left[\begin{array}[]{c}{\tau}_{1}\\ {\tau}_{2}\\ \end{array}\right] can be written as

[TABLE]

where $\rho_{i}=I_{i}+L_{i},i=1,\cdots,N;\beta_{j}=J_{j}+K_{j}$ and $j=1,\cdots,M.$

Definition 2.4**.**

*(Definition 2.1, [40])

The range space and null space of a tensor $\mathcal{A}\in\mathbb{R}^{{I_{1}}\times\cdots\times{I_{M}}\times{J_{1}}\times\cdots\times{J_{N}}}$ are defined as per the following:*

[TABLE]

The relation of range space for tensors is discussed in [40] as follows.

Lemma 2.5** (Lemma 2.2. [40]).**

Let $\mathcal{A}\in\mathbb{R}^{{I_{1}}\times\cdots\times{I_{M}}\times{J_{1}}\times\cdots\times{J_{N}}}$, $\mathcal{B}\in\mathbb{R}^{{I_{1}}\times\cdots\times{I_{M}}\times{K_{1}}\times\cdots\times{K_{L}}}.$ Then $\mathfrak{R}(\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})$ if and only if there exists $\mathcal{U}\in\mathbb{R}^{{J_{1}}\times\cdots\times{J_{N}}\times{K_{1}}\times\cdots\times{K_{L}}}$ such that $\mathcal{B}=\mathcal{A}{*_{N}}\mathcal{U}.$

The next subsection is discussed the Boolean tensor and some useful definitions

2.2 The Boolean tensor

The binary Boolean algebra $\mathfrak{B}$ consists of the set $\{0,1\}$ equipped with the operations of addition and multiplication defined as follows:

[TABLE]

[TABLE]

Definition 2.6**.**

Let $\mathcal{A}=(a_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ If $a_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}\in\{0,1\},$ then the tensor $\mathcal{A}$ is called Boolean tensor.

The addition and product of Boolean tensors are defined as in Eqs. (1) and (2) but addition and product of two entries will follow addition and product rule of Boolean algebra. The order relation for tensors is defined as follows.

Definition 2.7**.**

Let $\mathcal{A}=(a_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}\text{ and }~{}\mathcal{B}=(b_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}})~{}~{}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ Then $\mathcal{A}\leq\mathcal{B}$ if and only if $a_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}\leq b_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}$ for all $i_{s}$ and $j_{t}$ where $1\leq s\leq M$ and $1\leq t\leq N.$

We generalize the component-wise complement of the Boolean matrix [15] to Boolean tensors and defined below.

Definition 2.8**.**

Let $\mathcal{A}=(a_{{i_{1}}...{i_{N}}{j_{1}}...{j_{M}}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}}$ be a Boolean tensor. A tensor $\mathcal{B}=(b_{{i_{1}}...{i_{N}}{j_{1}}...{j_{M}}})\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}}$ is called component-wise complement of $\mathcal{A}$ if

[TABLE]

The tensor $\mathcal{B}$ and its entries respectively, denoted by $\mathcal{A}^{C}$ and $\left(a_{i_{1}\cdots i_{N}j_{1}\cdots j_{M}}^{c}\right).$

3 Main Results

In this section, we prove a few exciting results on tensors which are emphasized in the binary case. We divided this section into four folds. In the first part of this section, we discuss some identities on the Boolean tensors. Then, after having introduced some necessary ingredients, we study the generalized inverses of the Boolean tensor and some equivalence results to other generalized inverses in the second part. The existence and uniqueness of weighted Moore-Penrose inverses are discussed in the third part. The space decomposition and its connection to generalized inverses are presented in the final part.

3.1 Some identities on Boolean tensors

By the definition of Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}},$ we always get $\mathcal{A}+\mathcal{A}=\mathcal{A}.$ The infinite series of the Boolean tensor, $\displaystyle\sum_{k=1}^{\infty}\mathcal{A}^{k}$, is convergent and reduces to a finite series, since there are only finite number of Boolean tensors of the same order. Now we denote $\overline{\mathcal{A}}$ for the infinite series of the Boolean tensors, i.e.,

[TABLE]

Since $\mathcal{A}\leq\mathcal{A}+\mathcal{B}$ for any two Boolean tensor (suitable order for addition) $\mathcal{A}$ and $\mathcal{B}$, likewise $\mathcal{A}=\mathcal{A}+\mathcal{A}\geq\mathcal{A}+\mathcal{B}$ for any two Boolean tensor $\mathcal{A}\geq\mathcal{B}$. This is stated in the next result.

Theorem 3.1**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ Then $\mathcal{A}\geq\mathcal{B}$ if and only if $\mathcal{A}+\mathcal{B}=\mathcal{A}$.

If we consider $\mathcal{A}\geq\mathcal{I}$ in the above theorem, then it is easy to verify that $\mathcal{I}+\mathcal{A}+\cdots+\mathcal{A}^{n}=\mathcal{A}^{n}$ and hence we can have the following result as a corollary.

Corollary 3.1**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ and $\overline{\mathcal{A}}=\sum_{k=1}^{\infty}\mathcal{A}^{k}.$ If $\mathcal{A}\geq\mathcal{I},$ then there exist $n,$ such that

1. (a)

$\overline{\mathcal{A}}=\mathcal{A}^{n};$** 2. (b)

$\left(\overline{\mathcal{A}}\right)^{2}=\overline{\mathcal{A}};$** 3. (c)

$\overline{\left(\overline{\mathcal{A}}\right)}=\overline{\mathcal{A}}.$**

Using the above theorem, we now prove another result on the Boolean tensor. As follows,

Theorem 3.2**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ and $\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}},$ with $\mathcal{A}\geq\mathcal{I}$ and $\mathcal{B}\geq\mathcal{I}.$ Then

[TABLE]

Proof.

Since $\mathcal{A}\geq\mathcal{I}$ and $\mathcal{B}\geq\mathcal{I}.$ So $\overline{\mathcal{A}}\geq\mathcal{I}$ and $\overline{\mathcal{B}}\geq\mathcal{I}.$ Also we have $\overline{\mathcal{A}}\geq\mathcal{A}$ and $\overline{\mathcal{B}}\geq\mathcal{B}.$ Combining these results, we get $\overline{A}{*_{N}}\overline{B}\geq\mathcal{A}$ and $\overline{A}{*_{N}}\overline{B}\geq\mathcal{B}.$ Thus $\overline{A}{*_{N}}\overline{B}\geq\mathcal{A}+\mathcal{B}$ and hence

[TABLE]

Now $\overline{\mathcal{A}+\mathcal{B}}\geq\overline{\mathcal{A}}$ and $\overline{\mathcal{A}+\mathcal{B}}\geq\overline{\mathcal{B}}.$ By using Corollary 3.1 $(c)$, we get $\overline{\mathcal{A}}{*_{N}}\overline{\mathcal{B}}\leq\left(\overline{\mathcal{A}+\mathcal{B}}\right)^{2}=\overline{\mathcal{A}+\mathcal{B}}.$ From Corollary 3.1 $(b)$, we have

[TABLE]

From Eqs.(3) and (4), the proof is complete. ∎

If $\mathfrak{R}(\mathcal{B}^{T})=\mathfrak{R}(\mathcal{B}^{T}{*_{M}}\mathcal{A}^{T}),$ then there exist a tensor $\mathcal{U}$ such that $\mathcal{B}=\mathcal{U}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$ and hence, we obtain $\mathcal{B}{*_{M}}\mathcal{C}=\mathcal{U}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{M}}\mathcal{C}=\mathcal{U}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{M}}\mathcal{D}=\mathcal{B}{*_{M}}\mathcal{D}.$ This leads the following result.

Theorem 3.3**.**

*Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{M}}$,

$\mathcal{C}\in\mathbb{R}^{K_{1}\times\cdots\times K_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{D}\in\mathbb{R}^{K_{1}\times\cdots\times K_{M}\times J_{1}\times\cdots\times J_{N}}$ be Boolean tensors with

$\mathcal{A}{*_{N}}\mathcal{B}{*_{M}}\mathcal{C}=\mathcal{A}{*_{N}}\mathcal{B}{*_{M}}\mathcal{D}.$ If $\mathfrak{R}(\mathcal{B}^{T})=\mathfrak{R}(\mathcal{B}^{T}{*_{N}}\mathcal{A}^{T}),$ then $\mathcal{B}{*_{M}}\mathcal{C}=\mathcal{B}{*_{M}}\mathcal{D}.$*

Similar way, we can prove the following corollary.

Corollary 3.2**.**

*Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{M}}$,

$\mathcal{C}\in\mathbb{R}^{K_{1}\times\cdots\times K_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{D}\in\mathbb{R}^{K_{1}\times\cdots\times K_{M}\times I_{1}\times I_{2}\times\cdots\times I_{M}}$ be Boolean tensors with $\mathcal{C}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{D}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}.$ If $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{B}),$ then $\mathcal{C}{*_{M}}\mathcal{A}=\mathcal{D}{*_{M}}\mathcal{A}.$*

We now discuss the important result on a transpose of an arbitrary order Boolean tensor, as follows.

Lemma 3.4**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be any Boolean tensor. Then $\mathcal{A}\leq\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}.$

Proof.

Let $\mathcal{B}=\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}.$ We need to show that

[TABLE]

This inequality is trivial if ${a}_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}=0.$ Let us assume ${a}_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}=1.$ Now

[TABLE]

For $1\leq s\leq N,$ if $k_{s}=j_{s}$ and $l_{s}=i_{s},$ then

[TABLE]

Hence the proof is complete. ∎

Theorem 3.5**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}.$ Then the equation $\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{B}$ is solvable if and only if $\mathcal{X}=\mathcal{C},$ where

[TABLE]

Proof.

Let $\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{B}$ is solvable and $\mathcal{A}{*_{N}}\mathcal{X}=D.$ To claim $\mathcal{D}=\mathcal{B},$ it is enough to show $d_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1$ if and only if $b_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1.$ Let $d_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1.$ This implies $a_{i_{1}\cdots i_{N}p_{1}\cdots p_{N}}=1$ and $c_{p_{1}\cdots p_{N}j_{1}\cdots j_{N}}=1$ for some $p_{k}~{}~{}1\leq k\leq N.$ The condition $c_{p_{1}\cdots p_{N}j_{1}\cdots j_{N}}=1$ yields either $a_{i_{1}\cdots i_{N}p_{1}\cdots p_{N}}=0$ or $b_{p_{1}\cdots p_{N}j_{1}\cdots j_{N}}=1$ for all $p_{k}~{}~{}1\leq k\leq N.$ Since $a_{i_{1}\cdots i_{N}p_{1}\cdots p_{N}}=1$ which makes $b_{p_{1}\cdots p_{N}j_{1}\cdots j_{N}}=1$ for all $p_{k},~{}~{}1\leq k\leq N.$ Therefore $b_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1.$ Now if $b_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1,$ then $a_{i_{1}\cdots i_{N}r_{1}\cdots r_{N}}=1$ and $x_{r_{1}\cdots r_{N}j_{1}\cdots j_{N}}=1$ for some $r_{k},~{}~{}1\leq k\leq N.$ Suppose $c_{r_{1}\cdots r_{N}j_{1}\cdots j_{N}}=0.$ Then $a_{q_{1}\cdots q_{N}r_{1}\cdots r_{N}}=1$ and $b_{q_{1}\cdots q_{N}j_{1}\cdots j_{N}}=0$ for some $q_{k},~{}~{}1\leq k\leq N.$ Combining $a_{q_{1}\cdots q_{N}r_{1}\cdots r_{N}}=1$ and $x_{r_{1}\cdots r_{N}j_{1}\cdots j_{N}}=1$, we get $b_{q_{1}\cdots q_{N}j_{1}\cdots j_{N}}=1.$ Which is the contradiction. So $c_{r_{1}\cdots r_{N}j_{1}\cdots j_{N}}=1$ and hence $d_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=1.$ The converse part is trivial. ∎

In view of the Definition 2.8 the following theorem is true for Boolean tensors.

Preposition 3.3**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be a Boolean tensor, then

1. (a)

$(\mathcal{A}^{C})^{C}=\mathcal{A};$** 2. (b)

$(\mathcal{A}^{C})^{T}=(\mathcal{A}^{T})^{C}=\mathcal{A}^{CT}.$**

Remark 3.6**.**

In general $\mathcal{B}^{C}*_{N}\mathcal{A}^{C}\neq(\mathcal{A}*_{N}\mathcal{B})^{C}\neq\mathcal{A}^{C}*_{N}\mathcal{B}^{C}$ for any two tensor $\mathcal{A},~{}\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$

Example 3.7**.**

Consider two Boolean tensor $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ and $~{}\mathcal{B}=(b_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ such that

[TABLE]

[TABLE]

It is easy to verify $\mathcal{B}^{C}{*_{2}}\mathcal{A}^{C}\neq(\mathcal{A}{*_{2}}\mathcal{B})^{C}\neq\mathcal{A}^{C}{*_{2}}\mathcal{B}^{C}$, where $~{}(\mathcal{A}*_{2}\mathcal{B})^{C}=\mathcal{X}=(x_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ $~{}\mathcal{A}^{C}*_{2}\mathcal{B}^{C}=\mathcal{Y}=(y_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ and $~{}\mathcal{B}^{C}*_{2}\mathcal{A}^{C}=\mathcal{Z}=(z_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}},$ where

[TABLE]

[TABLE]

[TABLE]

The next result is the one of the important tool to prove trace of a Boolean tensor.

Theorem 3.8**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}.$ Then $\mathcal{A}{*_{N}}\mathcal{A}^{C}=\mathcal{O}$ if and only either $\mathcal{A}=\mathcal{O}$ or $\mathcal{A}=\mathcal{O}^{C}.$

Proof.

Since the converse part is trivial, it is enough to show the sufficient part only. Let $\mathcal{A}{*_{N}}\mathcal{A}^{C}=\mathcal{O}.$ Thus $\displaystyle\sum_{k_{1}\cdots k_{N}}a_{i_{1}\cdots i_{N}k_{1}\cdots k_{N}}a_{k_{1}\cdots k_{N}j_{1}\cdots j_{N}}^{c}=0$. This implies, $a_{i_{1}\cdots i_{N}k_{1}\cdots k_{N}}a_{k_{1}\cdots k_{N}j_{1}\cdots j_{N}}^{c}=0$ for all $k_{s},~{}1\leq s\leq N.$ Which again yields either $a_{i_{1}\cdots i_{N}k_{1}\cdots k_{N}}=0$ for all $i_{s},~{}k_{s}$ or $a_{k_{1}\cdots k_{N}j_{1}\cdots j_{N}}^{c}=0$ for all $j_{s},~{}k_{s},~{}1\leq s\leq N.$. Therefore either $\mathcal{A}=\mathcal{O}$ or $\mathcal{A}^{C}=\mathcal{O}.$ Hence completes the proof. ∎

Further, when $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}$ is symmetric Boolean tensor, one can write

[TABLE]

Hence, the tensors in the trace of a product of symmetric tensor and its complement can be switched without changing the result. This is stated in the next result.

Theorem 3.9**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}.$ If $\mathcal{A}$ is symmetric, then

[TABLE]

Remark 3.10**.**

In addition to the result of Theorem 3.9 one can write $tr(\mathcal{A}{*_{N}}\mathcal{A}^{C})=tr(\mathcal{A}^{C}{*_{N}}\mathcal{A})=0$ . Further, the symmetricity condition in Theorem 3.9 is only sufficient but not necessary.

One can verify the Remark 3.10 by the following example.

Example 3.11**.**

Let a Boolean tensor $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 2}\times{2\rtimes 2}}$ such that

[TABLE]

It is clear that $\mathcal{A}$ is not symmetric but $tr(\mathcal{A}*_{2}\mathcal{A}^{C})=tr(\mathcal{A}^{C}*_{2}\mathcal{A})=2$, where $~{}\mathcal{A}*_{2}\mathcal{A}^{C}=(x_{ijkl})\in\mathbb{R}^{{2\times 2}\times{2\rtimes 2}}$ and $~{}\mathcal{A}^{C}*_{2}\mathcal{A}=(y_{ijkl})\in\mathbb{R}^{{2\times 2}\times{2\rtimes 2}}$ with entries

[TABLE]

[TABLE]

Using the complement of a tensor, we now prove the following result.

Lemma 3.12**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},$ $\mathcal{B}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{C}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}$ be Boolean tensors. Then

[TABLE]

Proof.

$\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{C}\leq\mathcal{I}^{C}$ if and only $\sum_{j_{1}\cdots j_{N}}\sum_{k_{1}\cdots k_{L}}a_{{i_{1}\cdots i_{M}}{j_{1}\cdots j_{N}}}b_{{j_{1}\cdots j_{N}}{k_{1}\cdots k_{L}}}c_{{k_{1}\cdots k_{L}}{i_{1}\cdots i_{M}}}=0$ for all $i_{r},$ $1\leq r\leq M.$ This is equivalent to $a_{{i_{1}\cdots i_{M}}{j_{1}\cdots j_{N}}}b_{{j_{1}\cdots j_{N}}{k_{1}\cdots k_{L}}}c_{{k_{1}\cdots k_{L}}{i_{1}\cdots i_{M}}}=0$ for all $i_{r},$ $j_{s}$ and $k_{t},$ where $1\leq r\leq M,$ $1\leq s\leq N,$ $1\leq t\leq L.$ This in turn is true if and only

[TABLE]

Thus the proof is complete. ∎

Now we discuss the important result based on transpose and component-wise complement of an arbitrary order Boolean tensor, as follows.

Theorem 3.13**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ Then $\mathcal{X}{*_{M}}\mathcal{A}\leq\mathcal{B}$ if and only if $\mathcal{X}\leq\left(\mathcal{B}^{C}{*_{N}}\mathcal{A}^{T}\right)^{C}$, and $\mathcal{A}{*_{N}}\mathcal{X}\leq\mathcal{B}$ if and only if $\mathcal{X}\leq\left(\mathcal{A}^{T}{*_{M}}\mathcal{B}^{C}\right)^{C}$.

Proof.

Let $\mathcal{X}{*_{M}}\mathcal{A}\leq\mathcal{B}.$ This yields $\sum_{k_{1}\cdots k_{M}}x_{i_{1}\cdots i_{M}k_{1}\cdots k_{M}}a_{k_{1}\cdots k_{M}j_{1}\cdots j_{N}}\leq b_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}$ for all $i_{r},~{}(1\leq r\leq M)$ and $j_{s},~{}(1\leq s\leq N).$ This is equivalent to $x_{i_{1}\cdots i_{M}k_{1}\cdots k_{M}}a_{k_{1}\cdots k_{M}j_{1}\cdots j_{N}}\leq b_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}$ for all $i_{r}$ and $j_{s}$ and $k_{t}~{}(1\leq t\leq M).$ This in turns is true if and only if $x_{i_{1}\cdots i_{M}k_{1}\cdots k_{M}}a_{k_{1}\cdots k_{M}j_{1}\cdots j_{N}}b_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}^{c}=0,$ for all $j_{s}$ and $k_{t}.$ Which is equivalent to

$x_{i_{1}\cdots i_{M}k_{1}\cdots k_{M}}a_{k_{1}\cdots k_{M}j_{1}\cdots j_{N}}\{b_{j_{1}\cdots j_{N}i_{1}\cdots i_{M}}^{t}\}^{c}=0$ for all $j_{s}$ and $k_{t}.$ Summing over all $j_{s}$ and $k_{t},$ we get, $\sum_{k_{1}\cdots k_{M}}\sum_{j_{1}\cdots j_{N}}x_{i_{1}\cdots i_{M}k_{1}\cdots k_{M}}a_{k_{1}\cdots k_{M}j_{1}\cdots j_{N}}\{b_{j_{1}\cdots j_{N}i_{1}\cdots i_{M}}^{t}\}^{c}=0.$ This is true if and only $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\left(\mathcal{B}^{T}\right)^{C}\leq\mathcal{I}^{C}.$ By Preposition 3.3 $(a)$, this is equivalent to $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\left(\mathcal{B}^{C}\right)^{T}\leq\mathcal{I}^{C}.$ By Lemma 3.12, this in turns true if and only $\mathcal{X}^{C}\geq\left(\mathcal{A}{*_{N}}(\mathcal{B}^{C})^{T}\right)^{T},$ that is, if and only if $\mathcal{X}\leq\left(\mathcal{B}^{C}{*_{M}}\mathcal{A}^{T}\right)^{C}.$

This completes first part of the theorem. Similar way, we can show the second part of the theorem. ∎

Corollary 3.4**.**

Let $\mathcal{E}=\mathcal{O}^{C},$ where $\mathcal{O}$ is the zero tensor. Then the following statements are equivalent:

1. (a)

$\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{O};$** 2. (b)

$\mathcal{X}\leq\left(\left(\mathcal{A}{*_{N}}\mathcal{E}\right)^{T}\right)^{C};$** 3. (c)

$\mathcal{E}{*_{N}}\mathcal{X}\leq\left(\left(\mathcal{A}{*_{N}}\mathcal{E}\right)^{T}\right)^{C}.$**

The same result is also true for $\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{O}.$ Also the following corollary easily follow from Theorem 3.13.

Corollary 3.5**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{X}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}.$ Then $\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{B}$ has a solution if and only if $\mathcal{B}\leq\left(\mathcal{B}^{C}{*_{N}}\mathcal{A}^{T}\right)^{C}{*_{M}}\mathcal{A}.$

3.2 Generalized inverses of Boolean tensors

For the generalization of the generalized inverses of Boolean matrix [37], we introduce the definition of $\{i\}$-inverses $(i=1,2,3,4)$ and the Moore-Penrose inverse of Boolean tensors via the Einstein product, as follows.

Definition 3.14**.**

For any Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},$ consider the following equations in $\mathcal{X}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}:$

[TABLE]

Then $\mathcal{X}$ is called

1. (a)

a generalized inverse of $\mathcal{A}$ if it satisfies $(1)$ and denoted by $\mathcal{A}^{(1)}.$ 2. (b)

a reflexive generalized inverse of $\mathcal{A}$ if it satisfies $(1)$ and $(2)$, which is denoted by $\mathcal{A}^{(1,2)}.$ 3. (c)

a $\{1,3\}$ inverse of $\mathcal{A}$ if it satisfies $(1)$ and $(3)$, which is denoted by $\mathcal{A}^{(1,3)}.$ 4. (d)

a $\{1,4\}$ inverse of $\mathcal{A}$ if it satisfies $(1)$ and $(4)$, which is denoted by $\mathcal{A}^{(1,4)}.$ 5. (e)

the Moore-Penrose inverse of $\mathcal{A}$ if it satisfies all four conditions $[(1)-(4)]$, which is denoted by $\mathcal{A}^{\dagger}.$

The following remark and corollary are follows from the Definition 3.14.

Remark 3.15**.**

If $\mathcal{X}$ is the generalized inverse of a Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ then $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}$ is the reflexive generalized inverse of $\mathcal{A}.$

Corollary 3.6**.**

If $\mathcal{X}$ is the generalized inverse of a Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ Then

1. (a)

$\mathcal{X}^{T}$* is the generalized inverse of $\mathcal{A}^{T};$* 2. (b)

$(\mathcal{X}_{1}+\mathcal{X}_{2})$* is the generalized inverse of of a Boolean tensor $\mathcal{A}$ when $\mathcal{X}_{1}$ and $\mathcal{X}_{2}$ are two generalized inverse of $\mathcal{A}.$*

Thus the existence of generalized inverse of a Boolean tensor guarantees the existence of a reflexive generalized inverse. In addition to that, the Remark 3.15 and Corollary 3.6 (b) ensures that the existence of one-generalized inverse implies the existence of finite number generalized inverses. In view of the fact, we define the maximum generalized inverse of a Boolean tensor, as follow:

Definition 3.16**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ A tensor $\mathcal{X}$ is called maximum generalized inverse of $\mathcal{A}$ if $\mathcal{G}\leq\mathcal{X}$ for every generalized inverse $\mathcal{G}$ of $\mathcal{A}.$

Note that, the generalized inverse of a Boolean tensor need not be unique which explained in the next example.

Example 3.17**.**

Consider a Boolean tensor $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ with entries

[TABLE]

Then it can be easily verified that both tensors $~{}\mathcal{X}=(x_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ and $~{}\mathcal{Y}=(y_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ with entries

[TABLE]

[TABLE]

are satisfies the required condition of the Definition 3.14.

Foa a Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}},$ the number of generalized inverses are finite and the maximum number of generalized inverses is $2^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}.$ The next result assures the uniqueness and is true only for invertiable tensors.

Lemma 3.18**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be any Boolean tensor. If $\mathcal{A}$ is invertiable then $\mathcal{A}^{-1}$ is the only generalized inverse of $\mathcal{A}.$

Next, we discus the equivalence condition for consistent system and generalized inverse.

Theorem 3.19**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{X}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}.$ Then the followings are equivalent:

1. (a)

$\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}.$** 2. (b)

$\mathcal{X}{*_{M}}\mathcal{Y}$* is a solution of the tensor equation $\mathcal{A}{*_{N}}\mathcal{Z}=\mathcal{Y}$ whenever $\mathcal{Y}\in\mathfrak{R}(\mathcal{A}).$* 3. (c)

$\mathcal{A}{*_{N}}\mathcal{X}$* is idempotent and $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{X}).$* 4. (d)

$\mathcal{X}{*_{M}}\mathcal{A}$* is idempotent and $\mathfrak{R}(\mathcal{A^{T}})=\mathfrak{R}(\mathcal{A}^{T}{*_{N}}\mathcal{X}^{T}).$*

Proof.

First we will claim $(a)$ if and only if $(b).$ Let us assume $(a)$ holds and $\mathcal{Y}\in\mathfrak{R}(\mathcal{A}).$ Then there exists a Boolean tensor $\mathcal{Z}\in\mathbb{R}^{J_{1}\times J_{2}\times\cdots\times J_{N}}$ such that $\mathcal{A}{*_{N}}\mathcal{Z}=\mathcal{Y}.$ Now

[TABLE]

Therefore, $\mathcal{X}{*_{M}}\mathcal{Y}$ is a solution of $\mathcal{A}{*_{N}}\mathcal{Z}=\mathcal{Y}.$ Conversely assume $(b)$ is true. That is $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{Y}=\mathcal{Y}$ for all $\mathcal{Y}\in\mathfrak{R}(\mathcal{A}).$ Since $\mathcal{Y}\in\mathfrak{R}(\mathcal{A})$ which implies there exists $\mathcal{U}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}}$ such that $\mathcal{A}{*_{N}}\mathcal{U}=\mathcal{Y}.$ Thus $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{U}=\mathcal{A}{*_{N}}\mathcal{U}$ for all $\mathcal{U}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}}.$ Therefore $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}.$ Next we show the equivalence between $(a)$ and $(c).$ Clearly $(a)$ implies $\mathcal{A}{*_{N}}\mathcal{X}$ idempotent. Since $\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}$ and $\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{A}{*_{N}}{\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}},$ so by Lemma 2.5 $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{X}).$ Using the same idea, we can easily show the equivalence between $(a)$ and $(d).$ Hence completes the proof. ∎

Since $\mathcal{A}^{T}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}_{1}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}=\mathcal{A}^{T}{*_{M}}\mathcal{A}$, so by Theorem 3.3, $\mathcal{A}{*_{N}}\mathcal{X}_{1}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}=\mathcal{A}.$ Which leads the following corollary.

Corollary 3.7**.**

Let $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{A}).$ If $\mathcal{X}_{1}$ and $\mathcal{X}_{2}$ are generalized inverses of $\mathcal{A}^{T}{*_{M}}\mathcal{A}$ and $\mathcal{A}{*_{N}}\mathcal{A}^{T}$ respectively, then $\mathcal{X}_{1}{*_{N}}\mathcal{A}^{T}$ and $\mathcal{A}^{T}{*_{M}}\mathcal{X}_{2}$ are generalized inverse of $\mathcal{A}.$

Further, from the range conditions, if $\mathfrak{R}(\mathcal{A}^{T})\subseteq\mathfrak{R}(\mathcal{B}^{T})$ and $\mathfrak{R}(\mathcal{C})\subseteq\mathfrak{R}(\mathcal{B}).$ Then $\mathcal{A}=\mathcal{V}{*_{M}}\mathcal{B}$ and $\mathcal{C}=\mathcal{B}{*_{N}}\mathcal{U}$ for some tensors $\mathcal{U}$ and $\mathcal{V}.$ Now $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{C}=\mathcal{V}{*_{M}}\mathcal{B}{*_{N}}\mathcal{X}{*_{M}}\mathcal{B}{*_{N}}\mathcal{U}=\mathcal{V}{*_{M}}\mathcal{B}{*_{N}}\mathcal{U}$ which does not rely on $\mathcal{X}$. So it is invariant to the choice of $\mathcal{X}.$ So, we conclude this observation in the following corollary.

Corollary 3.8**.**

Let $\mathcal{A},$ $\mathcal{B}$ and $\mathcal{C}$ be suitable tensors such that $\mathfrak{R}(\mathcal{A}^{T})\subseteq\mathfrak{R}(\mathcal{B}^{T})$ and $\mathfrak{R}(\mathcal{C})\subseteq\mathfrak{R}(\mathcal{B}).$ If the generalized inverse of $\mathcal{B}$ exists, then $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{C}$ is invariant to $\mathcal{X},$ where $\mathcal{X}$ is the generalized inverse of $\mathcal{B}.$

To prove the next result, we define regular and singular of tensors, i.e., A tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots J_{N}},$ is called regular if the tensor equation $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}$ has a solution, otherwise called singular.

Theorem 3.20**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},$ $\mathcal{S}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}},$ and $\mathcal{T}\in\mathbb{R}^{J_{1}\times\cdots\times J_{M}\times J_{1}\times\cdots\times J_{N}}.$ If $\mathcal{S}$ and $\mathcal{T}$ are invertible, then the following are equivalent:

1. (a)

$\mathcal{A}$* is regular.* 2. (b)

$\mathcal{S}{*_{M}}\mathcal{A}{*_{N}}\mathcal{T}$* is regular.* 3. (c)

$\mathcal{A}^{T}$* is regular.* 4. (d)

$\mathcal{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{S}$* is regular.*

Based on the block tensor[41] and their properties, we have the following lemma.

Lemma 3.21**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ Then $\mathcal{A}$ is regular if and only if $\begin{bmatrix}\mathcal{A}&\mathcal{O}\\ \mathcal{O}&\mathcal{B}\end{bmatrix}$ is regular for all regular tensors $\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$

Proof.

Let $\mathcal{A}$ and $\mathcal{B}$ be regular tensors. Then there exist tensors $\mathcal{X}$ and $\mathcal{Y}$ such that $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}$ and $\mathcal{B}{*_{N}}\mathcal{Y}{*_{M}}\mathcal{B}=\mathcal{B}.$ Let $\mathcal{Z}=\begin{bmatrix}\mathcal{X}&\mathcal{O}\\ \mathcal{O}&\mathcal{Y}\\ \end{bmatrix}.$ Now

[TABLE]

Thus $\begin{bmatrix}\mathcal{A}&\mathcal{O}\\ \mathcal{O}&\mathcal{B}\\ \end{bmatrix}$ is regular. The converse part can be proved in the similar way. ∎

We now present another characterization of the generalized inverse of the Boolean tensor, as follows.

Theorem 3.22**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}.$ Then

[TABLE]

Proof.

Applying Theorem 3.13 repetitively, we get $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}\leq\mathcal{A}$ if and only if $\mathcal{X}{*_{M}}\mathcal{A}\leq\left(\mathcal{A}^{T}{*_{M}}\mathcal{A}^{C}\right)^{C}$, which equivalently if and only if

[TABLE]

∎

Using the Theorem 3.22, and the fact of transpose and component-wise complement of a Boolean tensor, we obtain an important result for finding the maximum generalized inverse of a Boolean tensor.

Corollary 3.9**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be regular. Then the following are holds

1. (a)

$\mathcal{A}=\mathcal{A}{*_{N}}\left(\mathcal{A}{*_{N}}\mathcal{A}^{CT}{*_{M}}\mathcal{A}\right)^{CT}{*_{M}}\mathcal{A};$** 2. (b)

$\left(\mathcal{A}{*_{N}}\mathcal{A}^{CT}{*_{M}}\mathcal{A}\right)^{CT}$* is the maximum generalized inverse of $\mathcal{A};$* 3. (c)

$\left(\mathcal{A}{*_{N}}\mathcal{A}^{CT}{*_{M}}\mathcal{A}\right)^{CT}{*_{M}}\mathcal{A}{*_{N}}\left(\mathcal{A}{*_{N}}\mathcal{A}^{CT}{*_{M}}\mathcal{A}\right)^{CT}$* is the maximum reflexive generalized inverse of $\mathcal{A}.$*

Next, we discuss some equivalence results between generalized and other inverses.

Theorem 3.23**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be any Boolean tensor, then the following statements are equivalent:

1. (a)

$\mathcal{A}^{(1,4)}$* exists.* 2. (b)

$\mathcal{A}^{(1)}$* exists and $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{A}^{T}).$* 3. (c)

$(\mathcal{A}{*_{N}}\mathcal{A}^{T})^{(1)}$* exists and $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T}$ for some tensor $\mathcal{X}.$*

Proof.

Consider $(a)$ is true and $\mathcal{A}^{(1,4)}=\mathcal{X}.$ Existence of $\mathcal{A}^{(1)}$ is trivial and hence $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{A}^{T}).$ Now we claim $(b)\Rightarrow(c).$ Let $\mathcal{A}^{(1)}$ exists and $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{A}^{T}).$ Then there exist a Boolean tensor $\mathcal{U}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ such that $\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{U}.$ Which implies $\mathcal{A}{*_{N}}\mathcal{A}^{T}=\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{U}{*_{N}}\mathcal{U}^{T}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}.$ So generalized inverse of $\mathcal{A}{*_{N}}\mathcal{A}^{T}$ exists. If we take $\mathcal{X}=\mathcal{A}^{T}{*_{N}}(\mathcal{A}{*_{N}}\mathcal{A}^{T})^{(1)},$ then

[TABLE]

Finally, we claim $(c)\Rightarrow(a).$ Let $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T}$. Taking transpose on both sides, we get $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}$, As

[TABLE]

Thus $\mathcal{X}=\mathcal{A}^{(1,4)}.$ Hence the proof is complete. ∎

Using the similar way, we can show the following theorem.

Theorem 3.24**.**

Let $\mathcal{A}$ be any Boolean tensor, then the following statements are equivalent:

1. (a)

$\mathcal{A}^{(1,3)}$* exists.* 2. (b)

$\mathcal{A}^{(1)}$* exists and $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{A}).$* 3. (c)

There exists a Boolean tensor $\mathcal{X}$ such that $\mathcal{A}^{T}=\mathcal{A}^{T}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}.$

We now discuss the characterization of Moore-Penrose inverse of Boolean tensors. The similar proof of Theorem 3.2 in [41], we have the uniqueness of the Moore-Penrose inverse of a Boolean tensor in $\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, as follows.

Lemma 3.25**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be any Boolean tensor. If the Moore-Penrose inverse of $\mathcal{A}$ exists then it is unique.

In the next lemma, we discuss an estimate of Moore-Penrose inverse a tensor, as follows.

Lemma 3.26**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be a Boolean tensor and suppose $\mathcal{A}$ admits a Moore-Penrose inverse. Then $\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}\leq\mathcal{A}$

Proof.

Let $\mathcal{B}=\mathcal{A}^{T}{*_{M}}\mathcal{A}$. Since $\mathcal{B}$ is a Boolean tensor of even order and there are finitely many Boolean tensors of same order, so there must exist positive integers $s,t\in\mathbb{N}$ such that $\mathcal{B}^{s}$ = $\mathcal{B}^{s+t}.$ Without loss of generality, we can assume that $s$ is the smallest positive integer for which $\mathcal{B}^{s}=\mathcal{B}^{s+t}$ for some $t\in\mathbb{N}.$ Now we will show $s=1.$ Suppose $s\geq 2$. Let $\mathcal{X}$ be the Moore-Penrose inverse of $\mathcal{A}$. Since $\mathcal{B}=\mathcal{A}^{T}{*_{M}}\mathcal{A}$ and $\mathcal{B}^{s}=\mathcal{B}^{s+t}$ which implies $\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s-1}=\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s+t-1}.$ Pre-multiplying both side $\mathcal{X}^{T}$ yields $\mathcal{A}{{*_{N}}}\mathcal{B}^{s-1}=\mathcal{A}{{*_{N}}}\mathcal{B}^{s+t-1},$ which implies $\mathcal{A}{*_{N}}\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s-2}=\mathcal{A}{*_{N}}\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s+t-2}.$ Further, pre-multiplying both side $\mathcal{X}$ yields $\mathcal{A}^{T}{*_{M}}\mathcal{X}^{T}{*_{N}}\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s-2}=\mathcal{A}^{T}{*_{M}}\mathcal{X}^{T}{*_{N}}\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{s+t-2},$ which implies $\mathcal{B}^{s-1}=\mathcal{B}^{s+t-1}.$

Thus, the minimality of $s$ is false and hence $s=1.$ Therefore $\mathcal{B}=\mathcal{B}^{t+1}$ for some $t\in\mathbb{N}.$ Again we have $\mathcal{B}=\mathcal{B}^{t+1}$, which implies $\mathcal{A}^{T}{{*_{M}}}\mathcal{A}=\mathcal{A}^{T}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{t}$. Pre-multiplying $\mathcal{X}^{T}$ both sides yields $\mathcal{A}{*_{N}}\mathcal{X}{{*_{M}}}\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{X}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{B}^{t}.$ Thus

[TABLE]

Applying Lemma 3.4 to $\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{N}}\mathcal{A}$ repetitively and combining Eq. (5), we obtain

[TABLE]

∎

Using the Lemma 3.4 and 3.26 one can obtain an interesting result on invertibility of Boolean tensor as follows,

Corollary 3.10**.**

A Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is invertible if and only if

[TABLE]

From the Definition 2.2, we obtain

[TABLE]

Similar way, we can also show $\mathcal{P}^{T}{*_{N}}\mathcal{P}=\mathcal{I}.$ Therefore, every permutation tensors are orthogonal and invertible. Adopting this result, we now present a characterization of the permutation tensor, as follows.

Preposition 3.11**.**

A Boolean tensor $\mathcal{A}$ has an inverse if and only if it is a permutation tensor.

Next result contains five equivalent conditions involving the existence of Moore-Penrose inverse of a Boolean tensor.

Theorem 3.27**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}.}$ be any tensor. Then the following statements are equivalent:

1. (i)

The Moore-Penrose inverse of $\mathcal{A}$ exists and unique. 2. (ii)

$\mathcal{A}{{*_{N}}}\mathcal{A}^{T}{{*_{N}}}\mathcal{A}\leq\mathcal{A}.$** 3. (iii)

$\mathcal{A}{{*_{N}}}\mathcal{A}^{T}{{*_{N}}}\mathcal{A}=\mathcal{A}.$** 4. (iv)

The Moore-Penrose inverse of $\mathcal{A}$ exists and equals $\mathcal{A}^{T}$. 5. (v)

There exist a tensor $\mathcal{G}$ such that $\mathcal{G}{{*_{N}}}\mathcal{A}{{*_{N}}}\mathcal{A}^{T}=\mathcal{A}^{T}$ and $\mathcal{A}^{T}{{*_{N}}}\mathcal{A}{{*_{N}}}\mathcal{G}=\mathcal{A}^{T}$.

Proof.

If $(i)$ holds then by Lemma 3.26 $(ii)$ holds. Also $(ii)\Rightarrow(iii)$ by Lemma 3.4. The statements $(iii)\Rightarrow(iv)$ and $(iv)\Rightarrow(i)$ are trivial by definition. Now we will show equivalence between $(i)$ and $(v)$. Suppose $(i)$ holds. If we take $\mathcal{G}=\mathcal{A}^{T}$ then $(v)$ hold. Conversely assume $(v)$ is true. To prove Moore-Penrose inverse of $A$ exists, first we show the following results:

$\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}=\mathcal{A}$

Since $\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T}$ which implies $\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{G}^{T}=\mathcal{A}.$ Pre multiplying $\mathcal{G}$ and post multiplying $\mathcal{A}^{T}$ both sides, we obtain $\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{G}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{T}.$ Thus $\mathcal{A}^{T}{*_{M}}\mathcal{G}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T}.$ Hence $\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}=\mathcal{A}.$

• 2.

$(\mathcal{G}{*_{M}}\mathcal{A})^{T}=\mathcal{A}^{T}{*_{M}}\mathcal{G}^{T}=\mathcal{G}{{*_{M}}}\mathcal{A}{{*_{N}}}\mathcal{A}^{T}{*_{M}}\mathcal{G}^{T}=\mathcal{G}{*_{M}}\mathcal{A}.$ Therefore $\mathcal{G}{*_{M}}\mathcal{A}$ is symmetric.

• 3.

$(\mathcal{A}{*_{N}}\mathcal{G})^{T}=\mathcal{G}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{G}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}=\mathcal{A}{*_{N}}\mathcal{G}.$ Thus $\mathcal{A}{*_{N}}\mathcal{G}$ is symmetric.

Now we will show the tensor $\mathcal{X}=\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}$ is the Moore-Penrose of $\mathcal{A}.$ Since

1. $\bullet$

$\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}=\mathcal{A}.$ 2. $\bullet$

$\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}=\mathcal{X}.$ 3. $\bullet$

$(\mathcal{A}{*_{N}}\mathcal{X})^{T}=(\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G})^{T}=(\mathcal{A}{*_{N}}\mathcal{G})^{T}{*_{M}}(\mathcal{A}{*_{N}}\mathcal{G})^{T}=\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}=\mathcal{A}{*_{N}}\mathcal{X}.$ 4. $\bullet$

$(\mathcal{X}{*_{M}}\mathcal{A})^{T}=(\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A})^{T}=(\mathcal{G}{*_{M}}\mathcal{A})^{T}{*_{N}}(\mathcal{G}{*_{M}}\mathcal{A})^{T}=\mathcal{G}{*_{M}}\mathcal{A}{*_{N}}\mathcal{G}{*_{M}}\mathcal{A}=\mathcal{X}{*_{M}}\mathcal{A}.$

Therefore, $\mathcal{X}$ is the Moore-Penrose inverse of $\mathcal{A}$ and By Lemma 3.25 it is unique. ∎

The reverse order law for the Moore-Penrose inverses of tensors yields a class of challenging problems that are fundamental research in the theory of generalized inverses. Research on reverse order law tensors has been very active recently [30, 32] but as per the above theorem it is trivially true in case of Boolean tensors.

Remark 3.28**.**

If Moore-Penrose inverses of $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}},$ and $\mathcal{A}{*_{N}}\mathcal{B}$ exists, then the reverse-order law for the Moore-Penrose inverse is always exists, i.e.,

[TABLE]

3.3 Weighted Moore-Penrose inverse

Utilizing the Einstein product, weighted Moore-Penrose inverse of even-order tensor and arbitrary-order tensor was introduced in [4, 16], very recently. This work motivate us to study weighted Moore-Penrose inverse for Boolean tensors.

Definition 3.29**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{M}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be three Boolean tensors. If a Boolean tensor $\mathcal{Z}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$ satisfying

[TABLE]

is called weighted Moore-Penrose inverse of $\mathcal{A}$ and it is denoted by $A^{\dagger}_{\mathcal{M},\mathcal{N}}.$

Note that, the weighted Moore-Penrose inverse need not be unique in general. This can be verified by the following example.

Example 3.30**.**

Let the Boolean tensor $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ be defined as in Example 3.17 with $\mathcal{N}=\mathcal{O}\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ and $\mathcal{M}=(m_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ such that

[TABLE]

Then it can be easily verified that both $~{}\mathcal{X}=(x_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$, $~{}\mathcal{Y}=(y_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ defined in Example 3.17 satisfies all conditions of Definition 3.29.

The uniqueness and existence of weighted Moore-Penrose inverse and some equivalent properties will be discussed in the next part of this subsection.

Theorem 3.31**.**

*Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},~{}~{}\mathcal{M}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}},$

$\mathcal{N}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be three Boolean tensors with $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{N})$ and $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}).$ If $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$ exists, then*

1. (a)

$\mathcal{A}{*_{N}}\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T};$** 2. (b)

$\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}{*_{M}}\mathcal{A}=\mathcal{A}^{T}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A};$** 3. (c)

$\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$* is unique.*

Proof.

Let $\mathcal{X}$ be a weighted Moore-Penrose inverse of $\mathcal{A}.$ Now

[TABLE]

This completes the proof of part $(a).$ Using the similar lines of part $(a)$ and relation $(3)$ of Definition 3.29, we can prove part $(b).$ Next we will claim the uniqueness of $A^{\dagger}_{\mathcal{M},\mathcal{N}}.$

Suppose there exists two weighted Moore-Penrose inverses (say $\mathcal{X}_{1}$ and $\mathcal{X}_{2}$) for $\mathcal{A}.$ Then

[TABLE]

Since $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{N}).$ Which implies there exists $\mathcal{U}$ such that $\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{U}=\mathcal{A}.$ Thus $\mathcal{X}_{1}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{U}=\mathcal{X}_{2}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{U}.$ Hence $\mathcal{X}_{1}{*_{M}}\mathcal{A}=\mathcal{X}_{2}{*_{M}}\mathcal{A}$. Therefore

[TABLE]

Now by using Eq. (7), we get

[TABLE]

Again as $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}).$ This implies there exists $\mathcal{V}^{T}$ such that $\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}{*_{M}}\mathcal{V}^{T}=\mathcal{A}^{T}.$ It leads $\mathcal{V}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}=\mathcal{A}.$ Thus $\mathcal{V}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}_{1}=\mathcal{V}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}_{2}.$ Hence $\mathcal{A}{*_{N}}\mathcal{X}_{1}=\mathcal{A}{*_{N}}\mathcal{X}_{2}$. Therefore

[TABLE]

Combining Eq. (7) and (8), we obtain $\mathcal{X}_{1}=\mathcal{X}_{2}$ and hence the proof is complete. ∎

The existence of weighted Moore-Penrose inverse is not trivial like other generalized inverses. The next theorem discusses the existence of weighted Moore-Penrose inverse.

Theorem 3.32**.**

*Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},~{}~{}\mathcal{M}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}},$

$\mathcal{N}\in\mathbb{R}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be three Boolean tensors with $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{N})$ and $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}).$ If $\mathcal{M}\geq\mathcal{I}$ and $\mathcal{N}\geq\mathcal{I},$ then $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$ exists if and only if any one of the following conditions holds:*

1. (a)

$\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}=\mathcal{A}.$** 2. (b)

$\mathcal{A}{*_{N}}\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}=\mathcal{A}.$** 3. (c)

$\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}{*_{M}}\mathcal{A}=\mathcal{A}.$** 4. (d)

$\mathcal{A}{*_{N}}\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}{*_{M}}\mathcal{A}=\mathcal{A}.$**

In particular, $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}=\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}.$

Proof.

Assume $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$ exists and let $\mathcal{X}=\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$. Let $\mathcal{B}=\mathcal{A}^{T}{*_{M}}\mathcal{A}$. Since for every Boolean tensor, there are finitely many Boolean tensors of same order, so there must exist positive integers $s,t\in\mathbb{N}$ such that

[TABLE]

Without loss of generality, we can assume that $s$ is the smallest positive integer for which Eq. (9) holds. Now we will claim $s=1.$ Suppose on contradiction, assume $s>1.$ Now using Eq. (9), and properties of weighted Mooore-Penrose inverse, we get

[TABLE]

Since $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{N}),$ which implies there exists a tensor $\mathcal{U}$ such that $\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{U}=\mathcal{A}.$ Now premultiplying $\mathcal{U}^{T}$ to Eq. (3.3) and using the properties $\mathcal{U}^{T}{*_{N}}\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T}$ and $\mathcal{A}^{T}{*_{M}}\mathcal{X}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T},$ we get

[TABLE]

Again, premultiplying $\mathcal{X}^{T}$ to Eq. (11) and using the symmetricity of $\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}$, we get $\mathcal{M}{*_{M}}{\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}(\mathcal{A}}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T})^{s-1}=\mathcal{M}{*_{M}}{\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}(\mathcal{A}}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T})^{s-1+t}.$ This gives

[TABLE]

Since $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{N}}\mathcal{M}^{T}),$ which implies there exists a tensor $\mathcal{Z}$ such that $\mathcal{Z}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}=\mathcal{A}.$ Premultiplying $\mathcal{Z}$ to Eq. (12) yields

[TABLE]

and contradicts the minimality of $s.$ Therefore

[TABLE]

Premultiplying Eq. (13) by $\mathcal{X},$ and using $(\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N})^{T}=\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N},$ we obtain

[TABLE]

Since $\mathcal{A}^{T}{*_{M}}\mathcal{X}^{T}{*_{N}}\mathcal{A}^{T}=\mathcal{A}^{T},$ we get

[TABLE]

Premultiplying Eq. (14) by a tensor $\mathcal{U}^{T}$ and using $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{A}{*_{N}}\mathcal{N}),$ we again obtain $\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}=\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}{*_{M}}(\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T})^{t}.$ Postmultiplying $\mathcal{Z}^{T}$ and applying $\mathfrak{R}(\mathcal{A}^{T})=\mathfrak{R}(\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}),$ we have

[TABLE]

Now

[TABLE]

Thus

[TABLE]

As $\mathcal{M}\geq\mathcal{I},\mathcal{N}\geq\mathcal{I}$, so by Lemma 3.4

[TABLE]

Postmultiplying $\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}$, we obtain

[TABLE]

Combining Eqs.(15), (16) and (17), we have

[TABLE]

Therefore

[TABLE]

and hence completes the proof of the condition $(b).$ By using Theorem 3.31, the other conditions are holds since

[TABLE]

Further, we will claim not only the four conditions holds but also $\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}.$ Let $\mathcal{X}=\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}.$ From Eq. (19), $\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}$ and

[TABLE]

Using Theorem 3.31, we show

[TABLE]

Therefore, $\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}$ is symmetric. Similarly, we can show $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}$ is symmetric. So $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}=\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}$. Next we will show the converse part. Let $\mathcal{A}{*_{N}}\mathcal{N}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}=\mathcal{A}.$ Since $\mathcal{M}\geq\mathcal{I}$ and $\mathcal{N}\geq\mathcal{I}$, so by Lemma 3.4,

[TABLE]

and hence

[TABLE]

Using the Eq. (20) and symmetricity of $\mathcal{A}{*_{N}}\mathcal{A}^{T}$, we obtain

[TABLE]

Similar argument yields,

[TABLE]

Using Eqs. (20)-(22), it can be easily verified that $\mathcal{X}=\mathcal{N}^{T}{*_{N}}\mathcal{A}^{T}{*_{M}}\mathcal{M}^{T}$ is satisfies all four conditions of the weighted Moore-Penrose inverse. Similarly, one can start from other conditions to verify the same. Thus the proof is complete. ∎

Remark 3.33**.**

The equality condition in Theorem 3.32 $(a)$ can be replaced by ${\bf{`\geq^{\prime}}}.$

3.4 Space Decomposition

Using the theory of Einstein product, we introduce the definition of the space decomposition for Boolean tensors, which generalizes the matrix space decomposition [37].

Definition 3.34**.**

Let $\mathcal{F}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{R}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}$ be two tensors with

[TABLE]

then the tensor $\mathcal{A}$ is called space decomposable and this decomposition is called a space decomposition of $\mathcal{A}$.

In connection with the fact of the above Definition 3.34 and Lemma 2.5, one can conclude the existence of a generalized inverse, as follows.

Theorem 3.35**.**

Let $\mathcal{A}=\mathcal{F}{*_{L}}\mathcal{R}$ be a space decomposition of $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},$ where $\mathcal{F}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{R}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}$ . Then $\mathcal{A}^{(1)}$ exists.

We now present one of our essential result which represents not only the existence of reflexive generalized inverse but also other inverses through this decomposition.

Theorem 3.36**.**

Let $\mathcal{X}$ be a generalized inverse of the Boolean tensor $\mathcal{A}.$ If $\mathcal{A}=\mathcal{F}{*_{L}}\mathcal{R}$ is a space decomposition of $\mathcal{A},$ where $\mathcal{F}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{R}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}.$ Then the following are holds:

1. (a)

$\mathcal{F}^{(1)}$* and $\mathcal{R}^{(1)}$ exists.* 2. (b)

$\mathcal{F}^{(1)}{*_{M}}\mathcal{F}=\mathcal{R}{*_{N}}\mathcal{R}^{(1)}.$** 3. (c)

$\mathcal{F}^{(1)}{*_{M}}\mathcal{A}=\mathcal{R}$* and $\mathcal{A}{*_{N}}\mathcal{R}^{(1)}=\mathcal{F}.$* 4. (d)

$\mathcal{R}^{(1)}{*_{M}}\mathcal{F}^{(1)}$* is a generalized inverse of $\mathcal{A}.$* 5. (e)

$\mathcal{R}{*_{N}}\mathcal{X}$* is a reflexive inverse of $\mathcal{F}$ and $\mathcal{X}{*_{M}}\mathcal{F}$ is a reflexive inverse of $\mathcal{R}.$*

Proof.

Since $\mathcal{X}$ is the generalized inverse of $\mathcal{A}.$ Then we have $\mathcal{A}{*_{N}}\mathcal{X}{*_{M}}\mathcal{A}=\mathcal{A},$ which implies

$\mathcal{F}{*_{L}}\mathcal{R}{*_{N}}\mathcal{X}{*_{M}}\mathcal{F}{*_{L}}\mathcal{R}=\mathcal{F}{*_{L}}\mathcal{R}=\mathcal{I}{*_{M}}\mathcal{F}{*_{L}}\mathcal{R}.$ Further, using Corollary 3.2, we get $\mathcal{F}{*_{L}}\mathcal{R}{*_{N}}\mathcal{X}{*_{M}}\mathcal{F}=\mathcal{F}.$ Thus $\mathcal{R}{*_{N}}\mathcal{X}$ is a generalized inverse of $\mathcal{F}.$ Similarly, one can determine $\mathcal{X}{*_{M}}\mathcal{F}$ is a generalized inverse of $\mathcal{R}$. Hence $(a)$ is proved. Now using the result $(a),$ one can prove $(b)$ and $(c).$ To prove $(d)$ we use the fact $(a)$ and obtain.

[TABLE]

Hence $\mathcal{R}^{(1)}{*_{M}}\mathcal{F}^{(1)}$ is a generalized inverse of $\mathcal{A}.$ In a similar manner, one can prove $(e)$ using the fact $\mathcal{R}{*_{N}}\mathcal{X}{*_{M}}\mathcal{F}{*_{L}}\mathcal{R}{*_{N}}\mathcal{X}=\mathcal{R}{*_{N}}\mathcal{X}$ and $\mathcal{X}{*_{M}}\mathcal{F}{*_{L}}\mathcal{R}{*_{N}}\mathcal{X}{*_{M}}\mathcal{F}=\mathcal{X}{*_{M}}\mathcal{F}.$ This completes the proof. ∎

In view of the above theorem one can draw a conclusion, as follows.

Remark 3.37**.**

Every generalized inverse of $\mathcal{A}$ need not of the form $\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}.$

We verify the Remark 3.37 with the following example.

Example 3.38**.**

Let $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ be a Boolean tensor with

[TABLE]

Consider $\mathcal{A}^{(1)}=(x_{ijkl})\in\mathbb{R}^{{2\times 3}\times{2\rtimes 3}}$ is a generalized inverse of $\mathcal{A}$ with

[TABLE]

In light of the Theorem 3.36 (e) one can conclude

[TABLE]

Therefore, every generalized inverse of $\mathcal{A}$ need not of the form $\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}.$

At this point one may be interested to know when does the generalized inverse of a Boolean tensor of the form $\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}$ ? The answer to this question is explained in the following Remark.

Remark 3.39**.**

If $\mathcal{X}$ is a reflexive inverse of a Boolean tensor $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{A}=\mathcal{F}{*_{L}}\mathcal{R}$ has a space decomposition, where $\mathcal{F}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{R}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}.$ Then every generalized inverse is of the form $\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}.$

Now considering the fact of Remark 3.39 and the observation of the Example 3.38, one can get the desired result.

Theorem 3.40**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be a Boolean tensor and $\mathcal{A}=\mathcal{F}{*_{L}}\mathcal{R}$ be a space decomposition of $\mathcal{A},$ where $\mathcal{F}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}},$ $\mathcal{R}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}.$ Assume that generalized inverse of either $\mathcal{F}$ reflexive or $\mathcal{R}$ reflexive. Then $\mathcal{X}$ is a reflexive generalized inverse of $\mathcal{A}$ if and only if $\mathcal{X}=\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}.$

Proof.

Consider the generalized inverse of $\mathcal{F}$ is reflexive. Taking into account of Theorem 3.36 $(d)$, we obtain $\mathcal{X}=\mathcal{R}^{(1)}{*_{L}}\mathcal{F}^{(1)}$, which is a generalized inverse of $\mathcal{A}$. Therefore, it is enough to show $\mathcal{X}{*_{M}}\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{X}.$ Now using Theorem 3.36 $(c)$, we get

[TABLE]

Conversely, let $\mathcal{X}$ be a reflexive inverse of $\mathcal{A}.$ Then by Theorem 3.36 $(e)$,

[TABLE]

∎

Remark 3.41**.**

If we drop the condition either $\mathcal{F}$ or $\mathcal{R}$ is reflexive generalized inverse of $\mathcal{A}$ in Theorem 3.40, then the theorem will not true in general.

In favour of the the Remark 3.41 we produce an example as follows.

Example 3.42**.**

Let $\mathcal{A}$ be the Boolean tensor defined in Example 3.38 and $\mathcal{A}=\mathcal{F}=\mathcal{R}.$ Since $\mathcal{A}{*_{2}}\mathcal{I}{*_{2}}\mathcal{A}=\mathcal{I}$ and $\mathcal{I}{*_{2}}\mathcal{A}{*_{2}}\mathcal{I}\neq\mathcal{I}$, it follows that $\mathcal{I}$ is the generalized inverse for both $\mathcal{F}$ and $\mathcal{G}$ but not reflexive. In view of the Theorem 3.40, one can conclude $\mathcal{R}^{(1)}{*_{2}}\mathcal{F}^{(1)}=\mathcal{I}$ is not a reflexive generalized inverse of $\mathcal{A}.$

In [3] and [39], the authors have defined the rank of a Boolean matrix through space decomposition. Next, we discuss the rank and weight of a Boolean tensors.

Definition 3.43**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be a Boolean tensor. If there exist a least positive integer, $r=K_{1}\times\cdots\times K_{L}$ such that the Boolean tensors $\mathcal{B}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{C}\in\mathbb{R}^{K_{1}\times\cdots\times K_{L}\times J_{1}\times\cdots\times J_{N}}$ satisfies $\mathcal{A}=\mathcal{B}{*_{L}}\mathcal{C}$. Then $r$ is called the Boolean rank of $\mathcal{A}$ and denoted by $r_{b}(\mathcal{A}).$

Example 3.44**.**

Consider a Boolean tensor $~{}\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{{2\times 2}\times{2\rtimes 2}}$ with entries

[TABLE]

There exist a least positive integer $r=2$ and two tensor $~{}\mathcal{B}=(b_{ijk})\in\mathbb{R}^{{2\times 2\times 2}}$ and $~{}\mathcal{C}=(c_{ijk})\in\mathbb{R}^{{2\times 2\times 2}}$ with entries

[TABLE]

such that $\mathcal{A}=\mathcal{B}*_{1}\mathcal{C}$. However, $r=1$ gives two matrices $B$ and $C$, which is impossible to get a tensor. Thus rank of the tensor is $2$.

On the other hand, the rank of the Boolean tensor is zero if it is zero tensor. Further, we have $\mathcal{A}=\mathcal{I}_{m}{*_{M}}\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{I}_{n}$, where $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. It is quite apparent that

[TABLE]

To prove the last result of this paper, we define weight of Boolean tensor as.

Definition 3.45**.**

The weight of Boolean tensor is denoted by $w(\mathcal{A})$ and defined as

[TABLE]

The existence of generalized inverse can be discussed through Boolean rank, as follows.

Theorem 3.46**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ be any tensor with $r_{b}(\mathcal{A})\leq 1.$ Then $\mathcal{A}$ is regular.

Proof.

It is trivial for $r_{b}(\mathcal{A})=0,$ as a consequence of the fact $\mathcal{O}$ tensor is always regular. Further, consider $r_{b}(\mathcal{A})=1$ and define a tensor $\mathcal{J},$ with no zero elements. Then there exist permutation tensors $P$ and $\mathcal{Q}$ such that $\mathcal{P}{*_{M}}\mathcal{A}{*_{N}}\mathcal{Q}=\begin{bmatrix}\mathcal{J}&\mathcal{O}\\ \mathcal{O}&\mathcal{O}\\ \end{bmatrix}.$ As $\mathcal{J}$ is regular, it implies $\mathcal{P}{*_{M}}\mathcal{A}{*_{N}}\mathcal{Q}$ is regular. In view of the Lemma 3.21 and Preposition 3.11 one can conclude $\mathcal{A}$ is regular ∎

It is clear, if the weight of a Boolean tensor is $1,$ then the rank is also $1$. In view of this we obtain the following result.

Corollary 3.12**.**

Let $\mathcal{A}\in\mathbb{R}^{I_{1}\times\cdots I_{M}\times J_{1}\times\cdots J_{N}}$ be any tensor with $w(\mathcal{A})\leq 1.$ Then $\mathcal{A}$ is regular.

4 Conclusion

In this paper, we have introduced generalized inverses $(\{i\}$-inverses $(i=1,2,3,4))$ with the Moore-Penrose inverse and weighted Moore-Penrose inverse for Boolean tensors via the Einstein product, which is a generalization of the generalized inverses of Boolean matrices. In addition to this, we have discussed their existence and uniqueness. This paper also provides some characterization through complement and its application to generalized inverses. Further, we explored the space decomposition for the Boolean tensors, at the same time, we have studied rank and the weight for the Boolean tensor. In particular, we limited our study for Boolean tensors with $r_{b}(\mathcal{A})\leq 1$ and $w(\mathcal{A})\leq 1$. Herewith left as open problems for future studies.

Problem: If the Boolean rank or weight of a tensor $\mathcal{A}$ is greater than 1, then under which conditions the Boolean tensor $\mathcal{A}$ is regular $?$

Additionally, it would be interesting to investigate more generalized inverses on the Boolean tensors; this work is currently underway.

**Acknowledgments

**This research work was supported by Science and Engineering Research Board (SERB), Department of Science and Technology, India, under the Grant No. EEQ/2017/000747.