Reverse-order law for weighted Moore--Penrose inverse of tensors

TL;DR

This paper explores properties and conditions under which the reverse-order law holds for the weighted Moore--Penrose inverse of tensors, extending understanding of tensor inverses via Einstein products.

Contribution

It introduces new properties and sufficient conditions for the reverse-order law of the weighted Moore--Penrose inverse for tensors of arbitrary order.

Findings

New properties of the weighted Moore--Penrose inverse for tensors

Sufficient conditions for the reverse-order law in even-order square tensors

Characterizations of the reverse-order law for tensors of any order

Abstract

In this paper, we provide a few properties of the weighted Moore--Penrose inverse for an arbitrary order tensor via the Einstein product. We again obtain some new sufficient conditions for the reverse-order law of the weighted Moore--Penrose inverse for even-order square tensors. We then present several characterizations of the reverse-order law for tensors of arbitrary order.

Equations

$$\mathcal{X}{\ast }_N\mathcal{Y}{\ast }_M\mathcal{X}$$

$$\mathcal{Y}{\ast }_M\mathcal{X}{\ast }_N\mathcal{Y}$$

$$(\mathcal{X}{\ast }_N\mathcal{Y}{)}^H$$

$$(\mathcal{Y}{\ast }_M\mathcal{X}{)}^H$$

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{i_1\cdots i_M{k}_1\cdots k_L}=\sum _{j_1\cdots j_N}{a}_{i_1\cdots i_M{j}_1\cdots j_N}{b}_{j_1\cdots j_N{k}_1\cdots k_L}$$

$$\mathcal{P}=\mathcal{U}{\ast }_K\mathcal{D}{\ast }_K{\mathcal{U}}^H,$$

$${\mathcal{P}}^{1/2}=\mathcal{U}{\ast }_K{\mathcal{D}}^{1/2}{\ast }_K{\mathcal{U}}^H,$$

$$\mathcal{A}{\ast }_N\mathcal{X}{\ast }_M\mathcal{A}$$

$$\mathcal{X}{\ast }_M\mathcal{A}{\ast }_N\mathcal{X}$$

$$(\mathcal{M}{\ast }_M\mathcal{A}{\ast }_N\mathcal{X}{)}^H$$

$$(\mathcal{N}{\ast }_N\mathcal{X}{\ast }_M\mathcal{A}{)}^H$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}={\mathcal{N}}^{-1}{\ast }_N{\mathcal{A}}^H{\ast }_M\mathcal{M},$$

$$(\mathcal{A}+\mathcal{B}{)}_{\mathcal{M}\mathcal{N}}^{\#}$$

$$({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}{)}^H$$

$$({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}{)}_{\mathcal{N}\mathcal{M}}^{\#}$$

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{\mathcal{ML}}^{\#}$$

$$(\mathcal{A}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{)}_{\mathcal{MM}}^{\#}=\mathcal{A}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}.$$

$$({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}{)}_{\mathcal{NN}}^{\#}={\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}.$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}{\ast }_N\mathcal{B}{\ast }_L{\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{†}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}=\mathcal{B}{\ast }_L{\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{†}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}$$

$$\mathcal{B}{\ast }_L{\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{†}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}{\ast }_N\mathcal{B}={\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}{\ast }_N\mathcal{B}.$$

$${\mathcal{Q}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{P}={\mathcal{Q}}_{\mathcal{M}\mathcal{N}}^{†}.$$

$$\mathcal{P}{\ast }_N{\mathcal{Q}}_{\mathcal{M}\mathcal{N}}^{†}={\mathcal{Q}}_{\mathcal{M}\mathcal{N}}^{†}.$$

$$\mathcal{A}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$$\mathcal{B}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$${\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$$\mathcal{B}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$${\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$$\mathcal{A}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$$\mathcal{A}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N(\mathcal{B}{\ast }_N{\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†})$$

$$\mathcal{B}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$${\mathcal{B}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N({\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_N\mathcal{A})$$

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{\mathcal{ML}}^{†}={\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{†}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}$$

$${\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{†}{\ast }_M\mathcal{A}{\ast }_N\mathcal{B}{\ast }_L{\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{\#}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}$$

$$\mathcal{B}{\ast }_L{\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{†}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}{\ast }_M\mathcal{A}{\ast }_N\mathcal{B}$$

$$(\mathcal{A}{\ast }_N\mathcal{B}{)}_{\mathcal{ML}}^{\#}=(\mathcal{A}{\ast }_N\mathcal{B}{)}_{\mathcal{ML}}^{†}{\ast }_M(\mathcal{A}{\ast }_N\mathcal{B}){\ast }_L(\mathcal{A}{\ast }_N\mathcal{B}{)}_{\mathcal{ML}}^{\#},$$

$${\mathcal{B}}_{\mathcal{N}\mathcal{L}}^{\#}{\ast }_N{\mathcal{A}}_{\mathcal{M}\mathcal{N}}^{\#}$$

Full text

††journal: arXiv.org

1 Introduction

A tensor is a multidimensional array. An element of ${{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}}$ is an $N$th-order tensor. Here $I_{1},I_{2},\cdots,I_{N}$ are dimensions of the first, second, $\cdots$, $N$th way, respectively. The order of a tensor is the number of its dimensions. The scalars, vectors and matrices are respectively zeroth-order, first-order and second-order tensors. The tensors of order three or higher are known as the higher-order tensors. These are denoted by calligraphic letters like $\mathcal{A}$. An element of an $N$th order tensor $\mathcal{A}$ at $(i_{1},\cdots,i_{N})$th position is denoted by $a_{i_{1}\cdots i_{N}}$. For more details, we refer to the recent books [20, 21, 24] on tensors.

There has been active research on tensors for the past four decades. But, a little research contributions on the theory and applications of generalized inverses of tensors are in the literature. In fact, a generalized inverse called the Moore–Penrose inverse of an even-order tensor via the Einstein product was introduced first by Sun et al. [23] in 2016. Then the authors find the minimum-norm least-squares solution of some multilinear systems by using the notion of Moore–Penrose inverse. In the next year, Behera and Mishra [2] continued the same study and proposed different types of generalized inverses of tensors. In 2018, Panigrahy and Mishra [18] improved the definition of the Moore–Penrose inverse of an even-order tensor to a tensor of any order via the same product which also appeared in [16] and [22]. Panigrahy and Mishra [18] also introduced an extension of the Moore–Penrose inverse of a tensor called as the Product Moore–Penrose inverse. The definition of the Moore–Penrose inverse of an arbitrary order tensor is recalled below.

Definition 1.1**.**

*(Definition 3.3, [16] $\&$ Definition 1.1, [18])

Let $\mathcal{X}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. The tensor $\mathcal{Y}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$ satisfying the following four tensor equations:*

[TABLE]

is defined as the Moore–Penrose inverse of $\mathcal{X}$, and is denoted by $\mathcal{X}^{{\dagger}}$.

In the above definition, $(\cdot)^{H}$ denotes the conjugate transpose of $(\cdot)$ and ${*_{N}}$ denotes the Einstein product [6] of tensors, and is defined by

[TABLE]

for tensors $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. In the case of an even-order invertible tensor, Definition 1.1 coincides with the notion of the inverse which was first introduced by Brazell et al. [5]. They also showed that such an inverse can be computed using the singular value decomposition of the same tensor (see Lemma 3.1, [5]). The idea of introducing generalized inverses of tensors origins from the necessity of finding a solution of a given tensor multilinear system (see [2, 9, 10, 23]).

The equality $(\mathcal{A}{*_{N}}\mathcal{B})^{\dagger}=\mathcal{B}^{\dagger}{*_{N}}\mathcal{A}^{\dagger}$ for any two complex tensors $\mathcal{A}$ and $\mathcal{B}$ of arbitrary order, is called as the reverse-order law for the Moore–Penrose inverse of tensors. Also called as two term reverse-order law. If the above equality contains the Einstein product of three different tensors, then it is called as the triple reverse-order law (or three term reverse-order law). In the last section of [2], the authors proposed two open problems. The first one is about the reverse-order law for the Moore–Penrose inverse of tensors and the second one is about full rank factorization of tensors. In 2018, Panigrahy et al. [17] answered the first one for even-order tensors. Again, Panigrahy and Mishra [19] added more results to the same theory but for arbitrary order tensors. In the same year, Liang and Zheng [16] solved the other problem.

In 2017, Ji and Wei [10] introduced another extension of the Moore–Penrose inverse of an even-order tensor called weighted Moore–Penrose inverse and established the relation between the minimum-norm least-squares solution of a multilinear system and the weighted Moore–Penrose inverse. Very recently, Behera et al. [1] proposed a few identities involving the weighted Moore–Penrose inverses of tensors. They also provided a method of computation of the weighted Moore–Penrose inverse using full rank decomposition of a tensor which was introduced by Liang and Zheng [16]. Among other results, they attempted the problem of triple reverse-order law mentioned in the conclusion section of [18] for the weighted Moore–Penrose inverse. In this paper, our aim is to study the reverse-order law for the weighted Moore–Penrose inverse of tensors via the Einstein product.

In this context, the paper is organized as follows. Section 2 collects various useful definitions and results. A few properties of the weighted Moore–Penrose inverse of tensors are explained in Section 3. Section 4 contains all our main results and is devoted to the reverse-order law for the weighted Moore–Penrose inverse of tensors. It has two subsections, the first subsection contains a few necessary and sufficient conditions of reverse-order law for the square tensors while the second subsection is for arbitrary order tensors.

2 Prerequisites

Here, we collect some definitions and earlier results which will be used to prove the main results. We begin with the definition of a diagonal tensor. A tensor in ${{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ 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 tensor $\mathcal{I}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ with entries $(\mathcal{I})_{i_{1}\cdots i_{N}j_{1}\cdots j_{N}}=\prod_{k=1}^{N}\delta_{i_{k}j_{k}}$ is called an identity tensor if $\delta_{i_{k}j_{k}}=\begin{cases}1,\text{ if }i_{k}=j_{k}\\ 0,\text{ otherwise}\end{cases}$. The conjugate transpose of a tensor $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ is denoted by $\mathcal{A}^{H}$, and is defined as $(\mathcal{A}^{H})_{j_{1}\ldots j_{N}i_{1}\ldots i_{M}}=\overline{a}_{i_{1}\ldots i_{M}j_{1}\ldots j_{N}},$ where the over-line stands for the conjugate of $a_{i_{1}\ldots i_{M}j_{1}\ldots j_{N}}$. A tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is Hermitian if $\mathcal{A}=\mathcal{A}^{H}$ and skew-Hermitian if $\mathcal{A}=-\mathcal{A}^{H}$. Further, a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is unitary if $\mathcal{A}{*_{N}}\mathcal{A}^{H}=\mathcal{A}^{H}{*_{N}}\mathcal{A}=\mathcal{I}$, and idempotent if $\mathcal{A}{*_{N}}\mathcal{A}=\mathcal{A}.$ Ji and Wei [10] define the class of Hermitian positive definite tensors as below.

Definition 2.1**.**

(Definition 1, [10]) Let $\mathcal{P}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{K}\times I_{1}\times\cdots\times I_{K}}$, if there exists a unitary tensor $\mathcal{U}$ such that

[TABLE]

where $\mathcal{D}$ is a diagonal tensor with positive diagonal entries, then $\mathcal{P}$ is said to be Hermitian positive definite.

For a Hermitian positive definite tensor $\mathcal{P}$ in the Definition 2.1, let $\mathcal{D}^{1/2}$ be the diagonal tensor obtained from $\mathcal{D}$ by taking the square root of all its diagonal entries and define

[TABLE]

the square root of $\mathcal{P}$. Notice that $\mathcal{P}^{1/2}$ is always non-singular and its inverse is denoted by $\mathcal{P}^{-1/2}$. Moreover, $(\mathcal{P}^{1/2})^{H}=\mathcal{P}^{1/2}.$ In 2017, Ji and Wei [10] introduce the weighted Moore-Penrose inverse for even-order tensors, however, very recently Behera et al. [1] defines it for any tensor and the definition is recalled next.

Definition 2.2**.**

(Definition 2.2, [1]) Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, and $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. Then the unique tensor $\mathcal{X}$ in $\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$ satisfying

[TABLE]

is called the weighted Moore–Penrose inverse of tensor $\mathcal{A}$ and this unique $\mathcal{X}$ is denoted by $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

In particular, when $\mathcal{M}$ and $\mathcal{N}$ are both identity tensors, then $\mathcal{A}^{{\dagger}}_{\mathcal{I}_{\mathcal{M}}\mathcal{I}_{\mathcal{N}}}=\mathcal{A}^{{\dagger}}$, where $\mathcal{I}_{\mathcal{M}}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{I}_{\mathcal{N}}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ are identity tensors. The null space and the range space of $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ are defined to be $N(\mathcal{A})=\{\mathcal{X}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}}:\mathcal{A}{*_{N}}\mathcal{X}=\mathcal{O}\}$ and $R(\mathcal{A})=\{\mathcal{A}{*_{N}}\mathcal{X}:\mathcal{X}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}}\}$, respectively. Due to Ji and Wei, [10] we have the following property for weighted Moore–Penrose inverse of an even-order tensor.

Lemma 2.3**.**

(Lemma 2, [10]) Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{N}}$, and let $\mathcal{M}$ and $\mathcal{N}$ be Hermitian positive definite tensors in ${{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ and ${{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$, respectively. Then

• (i)

$(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{{\dagger}}_{\mathcal{N}\mathcal{M}}=\mathcal{A}$,

• (ii)

$(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{H}=(\mathcal{A}^{H})^{{\dagger}}_{\mathcal{N}^{-1}\mathcal{M}^{-1}}$.

3 Weighted conjugate transpose

Ji and Wei [10] defines the weighted conjugate transpose of a tensor whereas Behera et al. [1] defines it formally for any tensor as below.

Definition 3.1**.**

(Definition 2.8, [1]) The weighted conjugate transpose of the tensor $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, denoted by $\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$, is defined by

[TABLE]

where $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ are Hermitian positive definite tensors.

Using Definition 3.1, one can easily verify the following properties of the weighted conjugate transpose.

Lemma 3.2**.**

Let $\mathcal{A},\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. Then

[TABLE]

Like the conjugate transpose of a tensor, the weighted conjugate transpose also satisfies reverse-order law and is stated below.

Lemma 3.3**.**

(Lemma 2.9, [1]) Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then

[TABLE]

The only tensor whose Einstein product with its weighted conjugate transpose results zero tensor is zero tensor. This is stated in next result.

Theorem 3.4**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, and $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. Then either of $\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}=\mathcal{O}$ and $\mathcal{A}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{O}$ implies $\mathcal{A}=\mathcal{O}$.

During the investigation of the reverse-order law for the weighted Moore–Penrose inverse of tensors, we observe that the weighted conjugate transpose of the Einstein product of a tensor with its weighted Moore–Penrose inverse remains unaltered.

Theorem 3.5**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, and $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. Then

[TABLE]

Similarly, we have the following result.

Theorem 3.6**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, and $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. Then

[TABLE]

Next, we provide the right cancellation property of $\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$ for a tensor $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$.

Lemma 3.7**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{C}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. If $\mathcal{B}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{C}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$, then $\mathcal{B}{*_{N}}\mathcal{A}=\mathcal{C}{*_{N}}\mathcal{A}$.

Similarly, the left cancellation property is provided below.

Lemma 3.8**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{C}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ be Hermitian positive definite tensors. If $\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{C}$, then $\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{A}{*_{N}}\mathcal{C}$.

A sufficient condition for the commutativity of $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ and $\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{M}\mathcal{N}}$ is presented below.

Lemma 3.9**.**

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. If $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$, then $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ commutes with $\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}$.

Similarly, we can have the following result.

Lemma 3.10**.**

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. If $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$, then $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ commutes with $\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$.

Equivalent conditions for the commutativity of $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ and $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ are obtained next.

Lemma 3.11**.**

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in{{\mathbb{C}}}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then the commutativity of $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ and $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ is equivalent to either of the conditions

[TABLE]

and

[TABLE]

The next result provides an absorbing property of a tensor which coincides with its weighted conjugate transpose.

Lemma 3.12**.**

Let $\mathcal{P}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ such that $\mathcal{N}{*_{N}}\mathcal{P}=(\mathcal{N}{*_{N}}\mathcal{P})^{H}$. Also, let $\mathcal{M}$ and $\mathcal{N}$ be Hermitian positive definite tensors of appropriate size. If $\mathcal{P}{*_{N}}\mathcal{Q}=\mathcal{Q}$ for $\mathcal{Q}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times J_{1}\times\cdots\times J_{M}}$, then

[TABLE]

In a similar way, we have the following result.

Lemma 3.13**.**

Let $\mathcal{P}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ such that $\mathcal{N}{*_{N}}\mathcal{P}=(\mathcal{N}{*_{N}}\mathcal{P})^{H}$. Also, let $\mathcal{M}$ and $\mathcal{N}$ be Hermitian positive definite tensors of appropriate size. If $\mathcal{Q}{*_{N}}\mathcal{P}=\mathcal{Q}$ for $\mathcal{Q}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{M}\times I_{1}\times\cdots\times I_{N}}$, then

[TABLE]

4 Main results

4.1 Reverse-order law for square tensors

In this subsection, we provide some sufficient conditions of the reverse-order law for the weighted Moore–Penrose inverse of the Einstein product of two square tensors.

Theorem 4.1**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors. If

[TABLE]

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

Proof.

Suppose that Equations (13)-(16) hold. Let $\mathcal{X}=\mathcal{A}{*_{N}}\mathcal{B}$ and $\mathcal{Y}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$, then with the help of Equations (15) and (16), we have $\mathcal{X}{*_{N}}\mathcal{Y}{*_{N}}\mathcal{X}=\mathcal{X}$ and using Equations (13) and (14), we have $\mathcal{Y}{*_{N}}\mathcal{X}{*_{N}}\mathcal{Y}=\mathcal{Y}.$ Using Equation (13), we have $(\mathcal{M}{*_{N}}\mathcal{X}{*_{N}}\mathcal{Y})^{H}=[(\mathcal{M}{*_{N}}\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}){*_{N}}\mathcal{M}^{-1}{*_{N}}(\mathcal{M}{*_{N}}\mathcal{A}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})]^{H}$, which results $(\mathcal{M}{*_{N}}\mathcal{X}{*_{N}}\mathcal{Y})^{H}=\mathcal{M}{*_{N}}\mathcal{X}{*_{N}}\mathcal{Y}$ due to Equation (14). Using Equation (16) we have $(\mathcal{N}{*_{N}}\mathcal{Y}{*_{N}}\mathcal{X})^{H}=[\mathcal{N}{*_{N}}(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A})*_{N}\mathcal{N}^{-1}{*_{N}}(\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{B})]^{H}$, which results $(\mathcal{N}{*_{N}}\mathcal{Y}{*_{N}}\mathcal{X})^{H}=\mathcal{N}{*_{N}}\mathcal{Y}{*_{N}}\mathcal{X}$ due to Equation (15). Therefore, by Definition 2.2, we get $\mathcal{X}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{Y}$, i.e., $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$. ∎

We replace the conditions (13) and (14) of Theorem 4.1 by a single condition, and is presented next.

Theorem 4.2**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors. If

• (i)

$(\mathcal{M}{*_{N}}\mathcal{A}{*_{N}}\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{H}=\mathcal{M}{*_{N}}\mathcal{A}{*_{N}}\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$,

• (ii)

$\mathcal{B}{*_{N}}(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A})=(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}){*_{N}}\mathcal{B}$,

• (iii)

$\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A})=(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}){*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$,

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

Similarly, one can replace the conditions (15) and (16) of Theorem 4.1 by a single condition as follows.

Theorem 4.3**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors. If

• (i)

$\mathcal{A}{*_{N}}(\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}})=(\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}){*_{N}}\mathcal{A}$,

• (ii)

$\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}(\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}})=(\mathcal{B}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}){*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$,

• (iii)

$(\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}{*_{N}}\mathcal{B})^{H}=\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}{*_{N}}\mathcal{B}$,

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

The next result confirms that whenever the tensor $\mathcal{A}$ satisfies $\mathcal{M}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{M}{*_{N}}\mathcal{A})^{H}$, then (15) and (16) are sufficient to hold the reverse-order law.

Theorem 4.4**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors and $\mathcal{M}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{M}{*_{N}}\mathcal{A})^{H}$. If

[TABLE]

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

Similarly, conditions (13) and (14) are sufficient to hold the reverse-order law whenever $\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{N}{*_{N}}\mathcal{B})^{H}$.

Theorem 4.5**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors and $\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{N}{*_{N}}\mathcal{B})^{H}$. If

[TABLE]

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

Next, we provide an interesting result which confirms that only one condition among (13)-(16) is sufficient to hold the reverse-order law for the tensors $\mathcal{A}$ and $\mathcal{B}$ satisfying $\mathcal{M}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{M}{*_{N}}\mathcal{A})^{H}$ and $\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{N}{*_{N}}\mathcal{B})^{H}$.

Theorem 4.6**.**

Let $\mathcal{A}$, $\mathcal{B}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$. Also, let $\mathcal{M},~{}\mathcal{N}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ be Hermitian positive definite tensors and $\mathcal{M}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{M}{*_{N}}\mathcal{A})^{H}$ and $\mathcal{N}{*_{N}}\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}=(\mathcal{N}{*_{N}}\mathcal{B})^{H}$. If at least one of the following holds

[TABLE]

then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

4.2 Reverse-order law for arbitrary tensors

In this subsection, we provide some necessary and sufficient conditions of the reverse-order law for the weighted Moore–Penrose inverse of the Einstein product of two tensors. The very first result is proved in [1] using the range and null space of a tensor, however we present a new proof without using the notion of the range and null space of a tensor.

Theorem 4.7**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then

[TABLE]

if and only if

[TABLE]

Proof.

Suppose that $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$. We have

[TABLE]

which results

[TABLE]

Pre-multiply both sides with $\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{B}$, we get

[TABLE]

From which, we get

[TABLE]

Since $\mathcal{I}-\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ is idempotent and $(\mathcal{I}-\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A})^{\#}_{\mathcal{N}\mathcal{N}}=\mathcal{I}-\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$. So, the last equation can be written as

[TABLE]

Thus,

[TABLE]

Again,

[TABLE]

Now, pre-multiplying $\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$ to $\mathcal{A}{*_{N}}\mathcal{B}=(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{\#}_{\mathcal{N}\mathcal{M}}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$, we get

[TABLE]

which gives

[TABLE]

Since $\mathcal{I}-\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ is idempotent and $(\mathcal{I}-\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}})^{\#}_{\mathcal{N}\mathcal{N}}=\mathcal{I}-\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$. So

[TABLE]

Thus,

[TABLE]

Conversely, suppose that Equations (23) and (24) hold, i.e.,

[TABLE]

Pre-multiplying and post-multiplying Equation (23) by $\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ and $((\mathcal{A}{*_{N}}\mathcal{B})^{\#}_{\mathcal{M}\mathcal{L}})^{{\dagger}}_{\mathcal{L}\mathcal{M}}$, respectively, we get

[TABLE]

Taking the weighted conjugate transpose of Equation (24), we have

[TABLE]

Pre-multiplying and post-multiplying Equation (26) by $((\mathcal{A}{*_{N}}\mathcal{B})^{\#}_{\mathcal{M}\mathcal{L}})^{{\dagger}}_{\mathcal{L}\mathcal{M}}$ and $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$, respectively, we obtain

[TABLE]

In order to show $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$, we have to show that $\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ satisfies Definition 1.1, and is shown below. Using Equation (25), we have

[TABLE]

Using Lemma 3.9, we get $\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$, which on pre-multiplying by $\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ gives $\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$. Again, pre-multiplying $(\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}})^{\#}_{\mathcal{L}\mathcal{N}}$ and post-multiplying $(\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{\#}_{\mathcal{N}\mathcal{M}}$ yields

[TABLE]

Now, with the help of Equation (27), we get

[TABLE]

and using Equation (25), we get

[TABLE]

∎

Next, we provide a simpler characterization than that of Theorem 4.7 of the reverse-order law for the weighted Moore-Penrose inverse of the Einstein product of two tensors.

Theorem 4.8**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then

[TABLE]

if and only if

[TABLE]

In the last two Theorems 4.7 and 4.8, we need to check two conditions for the reverse-order law, however the next theorem needs only one condition to check.

Theorem 4.9**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ if and only if

[TABLE]

We next present another characterization of the reverse-order law.

Theorem 4.10**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ if and only if

[TABLE]

and

[TABLE]

Next theorem replaces the first condition of Theorem 4.7 by a new condition.

Theorem 4.11**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ if and only if

• (i)

$(\mathcal{L}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B})^{H}=\mathcal{L}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$,

• (ii)

$\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$.

Similarly, we replace the second condition of Theorem 4.7 by a new condition.

Theorem 4.12**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M}\in\mathcal{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathcal{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{L}\in\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be Hermitian positive definite tensors. Then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ if and only if

• (i)

$\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{B}{*_{L}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$,

• (ii)

$(\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}})^{H}=\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$.

It is interesting to note that the conditions $\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$ and $\mathcal{N}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}*_{N}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$ being Hermitian are weaker conditions than are $\mathcal{B}{*_{M}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}=\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}$ and $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}{*_{N}}\mathcal{B}{*_{M}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}=\mathcal{B}{*_{M}}\mathcal{B}^{\#}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{M}\mathcal{N}}$. The last result of this paper shows that the reverse-order law is a sufficient condition for the commutativity of $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ and $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$.

Theorem 4.13**.**

Let $\mathcal{A}\in{{\mathbb{C}}}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in{{\mathbb{C}}}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Also, let $\mathcal{M},~{}\mathcal{N},$ $\mathcal{L}$ be Hermitian positive definite tensors in $\mathcal{C}^{I_{1}\times I_{2}\times\cdots\times I_{M}\times I_{1}\times I_{2}\times\cdots\times I_{M}}$, $\mathcal{C}^{J_{1}\times J_{2}\times\cdots\times J_{N}\times J_{1}\times J_{2}\times\cdots\times J_{N}}$ and $\mathcal{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$, respectively. If $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}_{\mathcal{M}\mathcal{L}}=\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}{*_{N}}\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}$, then $\mathcal{A}^{{\dagger}}_{\mathcal{M}\mathcal{N}}{*_{M}}\mathcal{A}$ and $\mathcal{B}{*_{L}}\mathcal{B}^{{\dagger}}_{\mathcal{N}\mathcal{L}}$ commute.