Weighted Moore-Penrose inverses of arbitrary-order tensors

TL;DR

This paper investigates the properties and representations of weighted Moore-Penrose inverses of arbitrary-order tensors using Einstein product, including their decompositions, cancellation properties, and conditions for reverse-order law.

Contribution

It introduces a new algebraic approach to characterize and compute weighted Moore-Penrose inverses of tensors, extending existing tensor inverse theory.

Findings

Derived singular value and full-rank decompositions for tensors.

Established properties and representations of weighted Moore-Penrose inverses.

Provided conditions for the reverse-order law to hold for these inverses.

Abstract

Within the field of multilinear algebra, inverses and generalized inverses of tensors based on the Einstein product have been investigated over the past few years. In this paper, we explore the singular value decomposition and full-rank decomposition of arbitrary-order tensors using {\it reshape} operation. Applying range and null space of tensors along with the reshape operation; we further study the Moore-Penrose inverse of tensors and their cancellation properties via the Einstein product. Then we discuss weighted Moore-Penrose inverses of arbitrary-order tensors using such product. Following a specific algebraic approach, a few characterizations and representations of these inverses are explored. In addition to this, we obtain a few necessary and sufficient conditions for the reverse-order law to hold for weighted Moore-Penrose inverses of arbitrary-order tensors.

Equations

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

$$(\mathcal{A}{)}_{ijij}$$

$$(\mathcal{A}{)}_{i_1...i_N{j}_1...j_N}$$

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

$$rsh(\mathcal{A})=A=reshape(\mathcal{A},I_1\cdots I_M,J_1\cdots J_N)$$

$$rs{h}^{-1}(A)=\mathcal{A}=reshape(A,I_1,\cdots ,I_M,J_1,\cdots ,J_N).$$

$$rshrank(\mathcal{A})=rank(rsh(\mathcal{A})).$$

$$rsh(\mathcal{A}{\ast }_N\mathcal{B})=rsh(\mathcal{A})rsh(\mathcal{B})=AB\in {\mathbb{C}}^{I_1\cdots I_M\times K_1\cdots K_L}$$

$$\mathcal{A}{\ast }_N\mathcal{B}=rs{h}^{-1}(AB),i.e.,rs{h}^{-1}(AB)=rs{h}^{-1}(A){\ast }_{N}rs{h}^{-1}(B).$$

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

$${\mathcal{P}}^{1/2}=\mathcal{U}{\ast }_N{\mathcal{D}}^{1/2}{\ast }_N{\mathcal{U}}^{\ast }$$

$$\mathcal{N}(\mathcal{A})=\{\mathcal{X}:\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{O}\in {\mathbb{C}}^{I_1\times \cdots \times I_M}\},and\mathfrak{R}(\mathcal{A})=\{\mathcal{A}{\ast }_N\mathcal{X}:\mathcal{X}\in {\mathbb{C}}^{J_1\times \cdots \times J_N}\}$$

$$\mathcal{A}=\mathcal{U}{\ast }_M\mathcal{D}{\ast }_N{\mathcal{V}}^{\ast }$$

$$(\mathcal{D})_{i_{1}\cdots i_{M}j_{1}\cdots j_{N}}=\left\{\,\IEEEeqnarraybox[][c]{l?s}\IEEEstrut\sigma_{I}>0,&if{I}={J}\in{\{1,2,...,r\}},\\ 0,otherwise\IEEEstrut\right.$$

$$(D)_{I,J}=\left\{\,\IEEEeqnarraybox[][c]{l?s}\IEEEstrut\sigma_{I}>0,&if{I}={J}\in{\{1,2,...,r\}},\\ 0,otherwise\IEEEstrut\right.$$

$$\mathcal{A}$$

$$(1)\mathcal{A}{\ast }_N\mathcal{X}{\ast }_M\mathcal{A}=\mathcal{A};$$

$$(2)\mathcal{X}{\ast }_M\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{X};$$

$$(3)(\mathcal{A}{\ast }_N\mathcal{X}{)}^{\ast }=\mathcal{A}{\ast }_N\mathcal{X};$$

$$(4)(\mathcal{X}{\ast }_M\mathcal{A}{)}^{\ast }=\mathcal{X}{\ast }_M\mathcal{A},$$

$$\mathfrak{R}(\mathcal{A}{\ast }_N\mathcal{B}{\ast }_L{\mathcal{B}}^{†})=\mathfrak{R}(\mathcal{A}{\ast }_N\mathcal{B}),$$

$$\mathcal{B}{\ast }_N(\mathcal{A}{\ast }_N\mathcal{X}{\ast }_N\mathcal{B}{)}^{-1}{\ast }_N\mathcal{A}={\mathcal{X}}^{-1}$$

$${\mathcal{X}}^{†}=\mathcal{B}{\ast }_R(\mathcal{A}{\ast }_M\mathcal{X}{\ast }_N\mathcal{B}{)}^{†}{\ast }_L\mathcal{A}$$

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

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

$$({\mathcal{A}}^{†}{)}_{ij11}=\left(\begin{array}{cc}0& -1\\ 0& 0\end{array}\right),({\mathcal{A}}^{†}{)}_{ij21}=\left(\begin{array}{cc}-1& -1\\ 1& 0\end{array}\right),({\mathcal{A}}^{†}{)}_{ij12}=\left(\begin{array}{cc}-1& -1\\ 0& 0\end{array}\right),({\mathcal{A}}^{†}{)}_{ij22}=\left(\begin{array}{cc}0& -1\\ 0& -1\end{array}\right).$$

$${\mathcal{A}}^{†}{\ast }_2\mathcal{A}{\ast }_2\mathcal{X}=\mathcal{X}\text{ and }\mathcal{X}{\ast }_2\mathcal{B}{\ast }_2{\mathcal{B}}^{†}=\mathcal{X},$$

$$\mathcal{B}{\ast }_2(\mathcal{A}{\ast }_2\mathcal{X}{\ast }_2\mathcal{B}{)}^{†}{\ast }_2\mathcal{A}\ne {\mathcal{X}}^{†},$$

$$(\mathcal{B}{\ast }_2(\mathcal{A}{\ast }_2\mathcal{X}{\ast }_2\mathcal{B}{)}^{†}{\ast }_2\mathcal{A}{)}_{ij11}=\left(\begin{array}{cc}1& 1\\ \frac{1}{2}& \frac{1}{2}\end{array}\right),(\mathcal{B}{\ast }_2(\mathcal{A}{\ast }_2\mathcal{X}{\ast }_2\mathcal{B}{)}^{†}{\ast }_2\mathcal{A}{)}_{ij21}=\left(\begin{array}{cc}0& 0\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right),$$

$$(\mathcal{B}{\ast }_2(\mathcal{A}{\ast }_2\mathcal{X}{\ast }_2\mathcal{B}{)}^{†}{\ast }_2\mathcal{A}{)}_{ij12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),(\mathcal{B}{\ast }_2(\mathcal{A}{\ast }_2\mathcal{X}{\ast }_2\mathcal{B}{)}^{†}{\ast }_2\mathcal{A}{)}_{ij22}=\left(\begin{array}{cc}0& -1\\ 0& 0\end{array}\right),$$

$$({\mathcal{X}}^{†}{)}_{ij11}=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),({\mathcal{X}}^{†}{)}_{ij21}=\left(\begin{array}{cc}0& 0\\ -\frac{1}{2}& \frac{1}{2}\end{array}\right),({\mathcal{X}}^{†}{)}_{ij12}=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),({\mathcal{X}}^{†}{)}_{ij22}=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$$

$${\mathcal{X}}^{†}=(\mathcal{A}{\ast }_M\mathcal{X}{)}^{†}{\ast }_L\mathcal{A}and{\mathcal{X}}^{†}=\mathcal{B}{\ast }_R(\mathcal{X}{\ast }_N\mathcal{B}{)}^{†}aresatisfied.$$

Full text

1 Introductions

1.1 Background and motivation

Tensors or hypermatrix are multidimensional generalizations of vectors and matrices, and have attracted tremendous interest in recent years (see kolda ; Marcar ; loan ; Qil05 ; Sho13 ). Indeed, multilinear systems are closely related to tensors and such systems are encountered in a number of fields of practical interest, i.e., signal processing (see LalmV00 ; SidLaP17 ; CopBol89 ), scientific computing (see BeyMo05 ; ShiWeL13 ; BraliNT13 ), data mining Chew07 , data compression and retrieval of large structured data (see SilL08 ; MaoChLY18 ). Further, the Moore-Penrose inverse of tensors plays an important role in solving such multilinear systems (see bm ; JinBa17 ; ma2019 ) and the reverse-order law for the Moore-Penrose inverses of tensors yields a class of interesting problems that are fundamental in the theory of generalized inverses of tensors (see PanRad18 ; JR_rev ). In view of these, multilinear algebra is drawing more and more attention from researchers (see JinBa17 ; Bader:2006 ; Marcar ; Kru77 ; LalmV00 ), specifically, the recent findings in (see bm ; BraliNT13 ; weit2 ; PanRad18 ; psmv18 ; sun ), motivate us to study this subject in the framework of arbitrary-order tensors.

Let $\mathbb{C}^{I_{1}\times\cdots\times I_{N}}(\mathbb{R}^{I_{1}\times\cdots\times I_{N}})$ be the set of order $N$ and dimension $I_{1}\times\cdots\times I_{N}$ tensors over the complex (real) field $\mathbb{C}(\mathbb{R})$. Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}}$ be a multiway array with $N$-th order tensor, and $I_{1},I_{2},\cdots,I_{N}$ be dimensions of the first, second,$\cdots$, $N$th way, respectively. Indeed, a matrix is a second order tensor, and a vector is a first order tensor. We denote ${{\mathbb{R}}}^{m\times n}$ to be the set of all ${m\times n}$ matrices with real entries. 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. Each entry of $\mathcal{A}$ is denoted by $a_{i_{1}...i_{N}}$. The Einstein product (see ein ) $\mathcal{A}{*_{N}}\mathcal{B}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$ of tensors $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ is defined by the operation ${*_{N}}$ via

[TABLE]

The Einstein product is not commutative but associative, and distributes with respect to tensor addition. Further, cancellation does not work but there is a multiplicative identity tensor $\mathcal{I}$. This type of product of tensors is used in the study of the theory of relativity ein and also used in the area of continuum mechanics lai .

On the other hand, one of the most successful developments in the world of linear algebra is the concept of Singular Value Decomposition (SVD) of matrices BenGr74 . This concept gives us important information about a matrix such as its rank, an orthonormal basis for the column or row space, and reduction to a diagonal form Tian04 . Recently this concept is also used in low rank matrix approximations Gra04 ; ishteva2011best ; ye2005generalized . Since tensors are natural multidimensional generalizations of matrices, there are many applications involving arbitrary-order tensors. Further, the problem of decomposing tensors is approached in a variety of ways by extending the SVD, and extensive studies have exposed many aspects of such decomposition and its applications ( see for example CheQZ17 ; kolda ; Kru77 ; LalmV00 ; SidLaP17 ; LIANG2018 ). However, the existing framework of SVD of tensors appears to be insufficient and/or inadequate in several situations.

The aim of this paper is to present a proper generalization of the SVD of arbitrary-order tensors under Einstein tensor product. In fact, the existing form BraliNT13 of the SVD is well suited for square tensors, which is defined as follows:

Definition 1

(Definition 2.8, BraliNT13 ): The transformation defined as;

$f:\mathbb{T}_{I,J,I,J}(\mathbb{R})\longrightarrow\mathbb{M}_{IJ,IJ}(\mathbb{R})$ with $f(\mathcal{A})=A$ and defined component wise as;

[TABLE]

where $\mathbb{T}_{I,J,I,J}(R)=\{\mathcal{A}\in\mathcal{R}^{I\times J\times I\times J}~{}:~{}det(f(\mathcal{A}))\neq 0\}$. In general for any even order tensor, the transformation is defined as; $f:\mathbb{T}_{I_{1},...,I_{N},I_{1},...,J_{N}}(\mathbb{R})\longrightarrow\mathbb{M}_{I_{1}...I_{N},J_{1}...J_{N}}(\mathbb{R})$

[TABLE]

Using the above Definition and Theorem 3.17 in BraliNT13 , we obtain the SVD of a tensor $\mathcal{A}\in\mathbb{R}^{I\times J\times I\times J}$, which can be extended only to any square tensor, i.e., for $\mathcal{A}\in\mathbb{R}^{I_{1}\times I_{2}\times\cdots\times I_{N}\times I_{1}\times I_{2}\times\cdots\times I_{N}}$. Extension of the SVD for an arbitrary-order tensor using this method BraliNT13 is impossible, since $f$ is not a homomorphism for even-order and/or arbitrary-order tensors. In fact, the Einstein product is not defined for the following two even-order tensors, $\mathcal{A}\in\mathbb{R}^{I_{1}\times I_{2}\times J_{1}\times J_{2}}$ and $\mathcal{B}\in\mathbb{R}^{I_{1}\times I_{2}\times J_{1}\times J_{2}}$, i.e., $\mathcal{A}*_{2}\mathcal{B}$ is not defined. Therefore, our aim in this paper is to find the SVD for any arbitrary order tensors using reshape operation, which is discussed in the next section.

In addition, recently there has been increasing interest in analyzing inverses and generalized inverses of tensors based on different tensor products (seeSahBe20 ; Wei18 ; JinBa17 ; BraliNT13 ; sun ). The representations and properties of the ordinary tensor inverse were introduced in BraliNT13 . This interpretation is extended to the Moore-Penrose inverse of tensors in sun and investigated for a few characterizations of different generalized inverses of tensors via Einstein product in bm . Appropriately, Behera and Mishra bm posed the open question: “Does there exist a full rank decomposition of tensors ? If so, can this be used to compute the Moore-Penrose inverse of a tensor”? It is worth mentioning that Liang and Zheng LIANG2018 investigated this question and discuss the computation of Moore-Pensore inverse of tensors using full rank decomposition. Here we extend this work for weighted Moore-Pensore inverse using reshape operation. When investigating on the weighted Moore-Pensore inverse, we find some interesting characterization, which is discussed in the next Section.

Recently, Panigrahy and Mishra pami1 investigated the Moore-Penrose inverse of a product of two tensors via Einstein product. Using such theory of Einstein product, Stanimirovic, et al. psmv18 also introduced some basic properties of the range and null space of multidimensional arrays, and the effective definition of the tensor rank, termed as reshaping rank. In this respect, Panigrahy et al. PanRad18 obtained a few necessary and sufficient conditions for the reverse order law for the Moore-Penrose inverses of tensors, which can be used to simplify various tensor expressions that involve inverses of tensor products dingwei . Since then, many authors investigate the reverse order law for various classes of generalized inverses of tensors chetheory ; JR_rev ; Mispa18 . At the same time, the representations of the weighted Moore-Penrose inverse weit2 of an even-order tensor was introduced via the Einstein product. In this context, we focus our attention on exploring some characterizations and representation of weighted Moore-Penrose inverses of arbitrary-order tensors.

In this paper, we study the weighted Moore-Penrose inverse of an arbitrary-order tensor. This study can lead to the enhancement of the computation of SVD and full rank decomposition of arbitrary-order tensor using reshape operation. With that in mind, we discuss some identities involving the weighted Moore-Penrose inverses of tensors and then obtain a few necessary and sufficient conditions of the reverse order law for the weighted Moore-Penrose inverses of arbitrary-order tensors via the Einstein product.

1.2 Outline

We organize the paper as follows: In the next subsection, we introduce some notations and definitions which are helpful in proving the main results of this paper. In Section 2, we provide the main results of the paper. In order to do so, we introduce SVD and full rank decomposition of an arbitrary-order tensor using reshape operation. Within this framework, the Moore-Penrose and the generalized weighted Moore-Penrose inverse for arbitrary-order tensor is defined. Furthermore, we obtain several identities involving the weighted Moore-Penrose inverses of tensors via Einstein product. The Section 3 contains a few necessary and sufficient conditions of the reverse-order law for the weighted Moore-Penrose inverses of tensors.

1.3 Notations and definitions

For convenience, we first briefly explain a few essential facts about the Einstein product of tensors, which are found in bm ; BraliNT13 ; sun . For a tensor $\mathcal{A}=(a_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}})\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, the tensor $\mathcal{B}=(b_{{j_{1}}...{j_{N}}{i_{1}}...{i_{M}}})\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$ is said to be conjugate transpose of $\mathcal{A}$, if $b_{{j_{1}}...{j_{N}}{i_{1}}...{i_{M}}}=\overline{a}_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}$ and $\mathcal{B}$ is denoted by $\mathcal{A}^{*}$. When $b_{{j_{1}}...{j_{N}}{i_{1}}...{i_{M}}}={a}_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}$, $\mathcal{B}$ is the transpose of $\mathcal{A}$, denoted by $\mathcal{A}^{T}$. The definition of a diagonal tensor was introduced as a square tensor in BraliNT13 (Definition 3.12) and as an even-order tensor in sun . Here we introduce arbitrary-order diagonal tensor, as follows:

Definition 2

A tensor $\mathcal{D}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ with entries $d_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}$ is called a diagonal tensor if $d_{{i_{1}}...{i_{M}}{j_{1}}...{j_{N}}}=0$, when

${\left[i_{1}+\sum_{k=2}^{M}({i_{k}-1})\prod_{l=1}^{k-1}I_{l}]\neq[j_{1}+\sum_{k=2}^{N}(j_{k}-1)\prod_{l=1}^{k-1}J_{l}\right]}$.

Now we recall the definition of an identity tensor below.

Definition 3

(Definition 3.13, BraliNT13 ) A tensor $\mathcal{I}_{N}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ with entries $(\mathcal{I}_{N})_{i_{1}i_{2}\cdots i_{N}j_{1}j_{2}\cdots j_{N}}=\prod_{k=1}^{N}\delta_{i_{k}j_{k}}$, where

[TABLE]

is called a unit tensor or identity tensor.

Note that throughout the paper, we denote $\mathcal{I}_{M}$,$\mathcal{I}_{L}$ and $\mathcal{I}_{R}$ as identity tensors in the space $\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ and $\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$ respectively. Further, a tensor $\mathcal{O}$ denotes the zero tensor if all the entries are zero. 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 $\mathcal{A}=\mathcal{A}^{*}$ and skew-Hermitian if $\mathcal{A}=-\mathcal{A}^{*}$. Subsequently, 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 $\mathcal{A}{*_{N}}\mathcal{A}^{*}=\mathcal{A}^{*}{*_{N}}\mathcal{A}=\mathcal{I}_{N}$, and idempotent if $\mathcal{A}{*_{N}}\mathcal{A}=\mathcal{A}.$ In the case of tensors of real entries, Hermitian, skew-Hermitian and unitary tensors are called symmetric (see Definition 3.16, BraliNT13 ), skew-symmetric and orthogonal (see Definition 3.15, BraliNT13 ) tensors, respectively. Next we present the definition of the reshape operation, which was introduced earlier in psmv18 . This is a more general way of rearranging the entries in a tensor (it is also a standard Matlab function), as follows:

Definition 4

(Definition 3.1, psmv18 ): The 1-1 and onto reshape map, rsh, is defined as: $rsh:\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}\longrightarrow\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$ with

[TABLE]

where $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and the matrix $A\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$. Further, the inverse reshaping is the mapping defined as: $rsh^{-1}:\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}\longrightarrow\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ with

[TABLE]

where the matrix $A\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$ and the tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$.

Further, Lemma 3.2 in psmv18 defined the rank of a tensor, $\mathcal{A}$, denoted by $rshrank(\mathcal{A})$ as

[TABLE]

Continuing this research, Stanimirovic et. al. in psmv18 discussed the homomorphism properties of the rsh function, as follows,

Lemma 1

(Lemma 3.1 psmv18 ): Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ be given tensors. Then

[TABLE]

where $A=rsh(\mathcal{A})\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}},B=rsh(\mathcal{B})\in\mathbb{C}^{J_{1}\cdots J_{N}\times K_{1}\cdots K_{L}}$.

An immediate consequence of the above Lemma is the following:

[TABLE]

Existence of SVD of any square tensor is discussed in BraliNT13 . Using this framework, Jun and Wei weit2 defined Hermitian positive definite tensors, as follows:

Definition 5

(Definition 1, weit2 ) For $\mathcal{P}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$, if there exists a unitary tensor $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ such that

[TABLE]

where $\mathcal{D}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ is a diagonal tensor with positive diagonal entries, then $\mathcal{P}$ is said to be a Hermitian positive definite tensor.

Further, Jun and Wei weit2 defined the square root of a Hermitian positive definite tensor, $\mathcal{P}$, as follows:

[TABLE]

where $\mathcal{D}^{1/2}$ is the diagonal tensor, which obtained from $\mathcal{D}$ by taking the square root of all its diagonal entries. Notice that $\mathcal{P}^{1/2}$ is always non-singular and its inverse is denoted by $\mathcal{P}^{-1/2}$.

we now recall the definition of the range and the null space of arbitrary order tensors.

Definition 6

( Definition 2.1,psmv18 ): The null space and the range space of a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ are defined as follows:

[TABLE]

It is easily seen that $\mathcal{N}(A)$ is a subspace of $\mathbb{C}^{J_{1}\times\cdots\times J_{N}}$ and $\mathfrak{R}(\mathcal{A})$ is a subspace of $\mathbb{C}^{I_{1}\times\cdots\times I_{M}}$. In particular, $\mathcal{N}(\mathcal{A})=\{\mathcal{O}\}$ if and only if $\mathcal{A}$ is left invertiable via $*_{M}$ operation and $\mathfrak{R}(\mathcal{A})=\mathbb{C}^{I_{1}\times\cdots\times I_{M}}$ if and only if $\mathcal{A}$ is right invertiable via $*_{N}$ operation.

2 Main results

Mathematical modelling of problems in science and engineering typically involves solving multilinear systems; this becomes particularly challenging for problems having an arbitrary-order tensor. However, the existing framework on Moore-Penrose inverses of arbitrary-order tensor appears to be insufficient and/or inappropriate. It is thus of interest to study the theory of Moore-Penrose inverse of an arbitrary-order tensor via the Einstein product.

2.1 Moore-Penrose inverses

One of the most widely used methods is the SVD to compute Moore-Penrose inverse. Here we present a generalization of the SVD via the Einstein product.

Lemma 2

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ with $rshrank(\mathcal{A})=r$. Then the SVD for tensor $\mathcal{A}$ has the form

[TABLE]

where $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ are unitary tensors, and $\mathcal{D}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ is a diagonal tensor, defined by

[TABLE]

where $I=[i_{1}+\sum_{k=2}^{M}({i_{k}-1})\prod_{l=1}^{k-1}I_{l}]$ and ${J}=[j_{1}+\sum_{k=2}^{N}(j_{k}-1)\prod_{l=1}^{k-1}J_{l}]$.

Proof

Let $A=rsh(\mathcal{A})\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$. In the context of the SVD of the matrix $A$, one can write $A=UDV^{*}$, where $U\in\mathbb{C}^{I_{1}\cdots I_{M}\times I_{1}\cdots I_{M}}$ and $V\in\mathbb{C}^{J_{1}\cdots J_{N}\times J_{1}\cdots J_{N}}$ are unitary matrices and $D\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$ is a diagonal matrix with

[TABLE]

From relations (7) and (8), we can write

[TABLE]

where $\mathcal{U}=rsh^{-1}(U),\mathcal{V}=rsh^{-1}(V)$ and $\mathcal{D}=rsh^{-1}(D)$. Further, $\mathcal{U}{*_{M}}\mathcal{U}^{*}=rsh^{-1}(UU^{*})=rsh^{-1}(I)=\mathcal{I}_{M}$ and $\mathcal{V}{*_{N}}\mathcal{V}^{*}=rsh^{-1}(VV^{*})=rsh^{-1}(I)=\mathcal{I}_{N}$ gives $\mathcal{A}=\mathcal{U}{*_{M}}\mathcal{D}{*_{N}}\mathcal{V}^{*}$, where $\mathcal{U}$ and $\mathcal{V}$ are unitary tensors and $\mathcal{D}$ diagonal tensor.

Remark 1

The authors of in LIANG2018 proved the Theorem 3.2 for a square tensor. Here we proved for an arbitrary-order tensor. Specifically, the decomposition is followed from the homomorphism property of reshape map.

Continuing this study, we recall the definition of the Moore-Penrose inverse of tensors in $\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times...\times J_{N}}$ via the Einstein product, which was introduced in LIANG2018 for arbitrary-order.

Definition 7

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

[TABLE]

is called the Moore-Penrose inverse of $\mathcal{A}$, and is denoted by $\mathcal{A}^{{\dagger}}$.

Similarly, to the proof of Theorem 3.2 in sun , we have the existence and uniqueness of the Moore-Penrose inverse of an arbitrary-order tensor in $\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times...\times J_{N}}$, as follows.

Theorem 2.1

The Moore-Penrose inverse of an arbitrary-order tensor, $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ exists and is unique.

By straightforward derivation, the following results can be obtained, which also hold (Lemma 2.3 and Lemma 2.6) in bm for even-order tensor.

Lemma 3

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

1. (a)

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

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

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

From psmv18 , we present the relation of range space of multidimensional arrays which will be used to prove next Lemma.

Lemma 4

(Lemma2.2, psmv18 ) Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{B}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times K_{1}\times\cdots\times K_{L}}$. Then $\mathfrak{R}(\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})$ if and only if there exists $\mathcal{U}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ such that $\mathcal{B}=\mathcal{A}{*_{N}}\mathcal{U}$.

We now discuss the important relation between range and Moore-Penrose inverse of an arbitrary order tensor, which are mostly used in various section of this paper.

Lemma 5

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

(a) $\mathfrak{R}(\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})\Leftrightarrow\mathcal{A}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{B}=\mathcal{B}$,

(b) $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{B})\Leftrightarrow\mathcal{A}*_{N}\mathcal{A}^{\dagger}=\mathcal{B}*_{L}\mathcal{B}^{\dagger}$,

(c) $\mathfrak{R}(\mathcal{A})=\mathfrak{R}[(\mathcal{A}^{\dagger})^{*}]$ and $\mathfrak{R}(\mathcal{A}^{*})=\mathfrak{R}(\mathcal{A}^{\dagger})$.*

Proof

(a) Using the fact that $\mathfrak{R}(\mathcal{A}*_{N}\mathcal{U})\subseteq\mathfrak{R}(\mathcal{A})$ for two tensors $\mathcal{A}$ and $\mathcal{U}$ in appropriate order, one can conclude $\mathfrak{R}(\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})$ from $\mathcal{A}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{B}=\mathcal{B}$. Applying Lemma 4, we conclude $\mathcal{B}=\mathcal{A}*_{N}\mathcal{P}$ from $\mathfrak{R}(\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})$, where $\mathcal{P}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Hence, $\mathcal{A}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{B}=\mathcal{A}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{A}*_{N}\mathcal{P}=\mathcal{B}.$

(b) From (a), we have $\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{B})$ if and only if $\mathcal{A}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{B}=\mathcal{B}$ and $\mathcal{B}*_{L}\mathcal{B}^{\dagger}*_{M}\mathcal{A}=\mathcal{A}$ which implies $\mathcal{B}^{\dagger}=\mathcal{B}^{\dagger}{*_{M}}\mathcal{A}{*_{N}}\mathcal{A}^{\dagger}$. Then $\mathcal{A}*_{N}\mathcal{A}^{\dagger}=\mathcal{B}*_{L}\mathcal{B}^{\dagger}*_{M}\mathcal{A}*_{N}\mathcal{A}^{\dagger}=\mathcal{B}*_{L}\mathcal{B}^{\dagger}$.

(c) Using the Lemma 3 [(b),(c)], one can conclude that $\mathfrak{R}(\mathcal{A})\subseteq\mathfrak{R}[(\mathcal{A}^{\dagger})^{*}]$ and $\mathfrak{R}[(\mathcal{A}^{\dagger})^{*}]\subseteq\mathfrak{R}(\mathcal{A})$ respectively. This follows $\mathfrak{R}(\mathcal{A})=\mathfrak{R}[(\mathcal{A}^{\dagger})^{*}]$. Further, replacing $\mathcal{A}$ by $\mathcal{A}^{*}$ and using the fact $(\mathcal{A}^{*})^{\dagger}=(\mathcal{A}^{\dagger})^{*}$ we obtain $\mathfrak{R}(\mathcal{A}^{*})=\mathfrak{R}(\mathcal{A}^{\dagger})$.

Using the fact that $\mathfrak{R}(\mathcal{A}*_{N}\mathcal{B})\subseteq\mathfrak{R}(\mathcal{A})$ for two tensor $\mathcal{A}$ and $\mathcal{B}$ and the Definition-7, we get,

[TABLE]

where $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Now using the method as in the proof of Lemma 5, one can prove the next Lemma.

Lemma 6

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

(a) $\mathfrak{R}(\mathcal{B}^{*})\subseteq\mathfrak{R}(\mathcal{A}^{*})\Leftrightarrow\mathcal{B}*_{N}\mathcal{A}^{\dagger}*_{M}\mathcal{A}=\mathcal{B}$,

(b) $\mathfrak{R}(\mathcal{A}^{*})=\mathfrak{R}(\mathcal{B}^{*})\Leftrightarrow\mathcal{A}^{\dagger}*_{M}\mathcal{A}=\mathcal{B}^{\dagger}*_{L}\mathcal{B},$

(c) $\mathfrak{R}(\mathcal{A}*_{N}\mathcal{B}^{\dagger})=\mathfrak{R}(\mathcal{A}*_{N}\mathcal{B}^{*})$.*

Consider $\mathcal{A},~{}\mathcal{B},~{}\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{N}\times I_{1}\times\cdots\times I_{N}}$ and all are invertible, the following equation

[TABLE]

is called the cancellation property of product of tensors $(\mathcal{A},\mathcal{B},\mathcal{X})$. When the ordinary inverse is replaced by generalized inverse with suitable order, this cancellation property is not true in general. In this context, we concentrate to characterize all triples $(\mathcal{A},\mathcal{B},\mathcal{X})$ which satisfy

[TABLE]

where $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$. The first result obtained below deals with the necessary condition of this properties.

Lemma 7

*Let $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$.

If $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$, then ${\mathcal{X}=\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}}~{}~{}and~{}~{}{\mathcal{X}=\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger}}$.*

Proof

Let, $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$. It is quite obvious that $\mathcal{X}^{\dagger}=\mathcal{X}^{\dagger}{*_{M}}\mathcal{A}^{\dagger}{*_{L}}\mathcal{A}=\mathcal{B}{*_{R}}\mathcal{B}^{\dagger}{*_{N}}\mathcal{X}^{\dagger}$. Hence, from Lemma 5[(a),(c)] and Lemma 6(a), we have $\mathfrak{R}[(\mathcal{X}^{\dagger})^{*}]\subseteq\mathfrak{R}(\mathcal{A}^{*}),$

$i.e,\mathfrak{R}(\mathcal{X})\subseteq\mathfrak{R}(\mathcal{A}^{\dagger})$ and $\mathfrak{R}(\mathcal{X}^{\dagger})\subseteq\mathfrak{R}(\mathcal{B}),i.e.,\mathfrak{R}(\mathcal{X}^{*})\subseteq\mathfrak{R}[(\mathcal{B}^{\dagger})^{*}]$. Which implies $\mathcal{X}=\mathcal{A}^{\dagger}{*_{L}}\mathcal{A}{*_{M}}\mathcal{X}$ and $\mathcal{X}=\mathcal{X}{*_{N}}\mathcal{B}{*_{R}}\mathcal{B}^{\dagger}$.

The following example shows that converse of the above theorem is not true in general.

Example 1

Consider tensors $\mathcal{A}=(a_{ijkl})_{1\leq i,j,k,l\leq 2}\in\mathbb{R}^{2\times 2\times 2\times 2}$, $\mathcal{B}=\mathcal{A}^{*}$ and $\mathcal{X}=(x_{ijkl})_{1\leq i,j,k,l\leq 2}\in\mathbb{R}^{2\times 2\times 2\times 2}$ such that

[TABLE]

and

[TABLE]

Then,

[TABLE]

Thus we have

[TABLE]

But

[TABLE]

where

[TABLE]

[TABLE]

However, the converse of Lemma 7 holds under the assumption of additional condition which is stated below.

Lemma 8

Let $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$. If ${\mathcal{X}=\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}}{=\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger}}$ along with the condition $\mathcal{K}=\mathcal{A}^{\dagger}{*_{L}}(\mathcal{A}{*_{M}}\mathcal{X}){*_{N}}(\mathcal{A}{*_{M}}\mathcal{X})^{\dagger}{*_{L}}\mathcal{A}$ and $\mathcal{L}=\mathcal{B}{*_{R}}(\mathcal{X}{*_{N}}\mathcal{B})^{\dagger}{*_{M}}(\mathcal{X}{*_{N}}\mathcal{B}){*_{R}}\mathcal{B}^{\dagger}$ are Hermitian, Then $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$.

Proof

Let $\mathcal{W}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$.

Now, $\mathcal{X}*_{N}\mathcal{W}*_{M}\mathcal{X}\\ =(\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger})*_{N}\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}*_{M}(\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger})$

= $\mathcal{A}^{\dagger}*_{L}[(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})]*_{R}\mathcal{B}^{\dagger}\\ =\mathcal{A}^{\dagger}*_{L}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})*_{R}\mathcal{B}^{\dagger}=\mathcal{X}$.

Further, $\mathcal{W}*_{M}\mathcal{X}*_{N}\mathcal{W}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}=\mathcal{W}$.

Again $\mathcal{K}=\mathcal{X}*_{N}\mathcal{W}$ and $\mathcal{L}=\mathcal{W}*_{M}\mathcal{X}$ are Hermitian. Hence $\mathcal{W}=\mathcal{X}^{\dagger}$.

From Lemma 7 It is clear that, If $\mathcal{X}^{\dagger}=\mathcal{B}{*_{R}}(\mathcal{A}{*_{M}}\mathcal{X}{*_{N}}\mathcal{B})^{\dagger}{*_{L}}\mathcal{A}$, Then $\mathfrak{R}(\mathcal{A}{*_{M}}\mathcal{X})=\mathfrak{R}(\mathcal{A}{*_{M}}\mathcal{X}{*_{N}}\mathcal{B})$, which implies that $(\mathcal{A}{*_{M}}\mathcal{X}{*_{N}}\mathcal{B}){*_{R}}(\mathcal{A}{*_{M}}\mathcal{X}{*_{N}}\mathcal{B})^{\dagger}=\mathcal{A}{*_{M}}\mathcal{X}{*_{N}}(\mathcal{A}{*_{M}}\mathcal{X})^{\dagger}$. It is easy to verify that $\mathcal{X}{*_{N}}\mathcal{X}^{\dagger}=\mathcal{K}$ and $\mathcal{X}^{\dagger}{*_{M}}\mathcal{X}=\mathcal{L}$, and both $\mathcal{K}$ and $\mathcal{L}$ both are Hermitian. Therefore, a necessary and sufficient condition for the cancellation law can be stated as:

Theorem 2.2

*Let $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$.

$\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$ if and only if ${\mathcal{X}=\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}}{=\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger}}$ and both $\mathcal{K}=\mathcal{A}^{\dagger}{*_{L}}(\mathcal{A}{*_{M}}\mathcal{X}){*_{N}}(\mathcal{A}{*_{M}}\mathcal{X})^{\dagger}{*_{L}}\mathcal{A}$ and $\mathcal{L}=\mathcal{B}{*_{R}}(\mathcal{X}{*_{N}}\mathcal{B})^{\dagger}{*_{M}}(\mathcal{X}{*_{N}}\mathcal{B}){*_{R}}\mathcal{B}^{\dagger}$ are Hermitian.*

We now proceed to discuss a few necessary and sufficient conditions for the cancellation law.

Corollary 1

Let $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$, and $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$ if and only if both the equations

[TABLE]

Proof

By taking $\mathcal{B}=\mathcal{I}$ in Theorem 2.2, we have $\mathcal{X}^{\dagger}=(\mathcal{A}*_{M}\mathcal{X})^{\dagger}*_{L}\mathcal{A}$ if and only if $\mathcal{A}^{\dagger}*_{L}(\mathcal{A}{*_{M}}\mathcal{X}){*_{N}}(\mathcal{A}{*_{M}}\mathcal{X})^{\dagger}*_{L}\mathcal{A}$ is Hermitian and $\mathcal{X}=\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}$. Similarly with the special case $\mathcal{A}=\mathcal{I}$ in Theorem 2.2, we get $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{X}*_{N}\mathcal{B})^{\dagger}$ if and only if $\mathcal{B}{*_{R}}(\mathcal{X}{*_{N}}\mathcal{B})^{\dagger}{*_{M}}(\mathcal{X}{*_{N}}\mathcal{B}){*_{R}}\mathcal{B}^{\dagger}$ is Hermitian and $\mathcal{X}=\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger}$. Using the fact of Theorem2.2 one can prove the required result.

Using Lemma 5 and Lemma 6 in the corollary 1 one obtain the following result.

Theorem 2.3

Let $\mathcal{X}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{A}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times H_{1}\times\cdots\times H_{R}}$, then

[TABLE]

if and only if

[TABLE]

Proof

Suppose that $\mathcal{X}^{\dagger}=\mathcal{B}*_{R}(\mathcal{A}*_{M}\mathcal{X}*_{N}\mathcal{B})^{\dagger}*_{L}\mathcal{A}$. Then from Corollary 1,

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

Now, $\mathfrak{R}(\mathcal{X})=\mathfrak{R}[(\mathcal{X}^{\dagger})^{*}]=\mathfrak{R}[\mathcal{A}^{*}*_{L}\{(\mathcal{A}*_{M}\mathcal{X})^{\dagger}\}^{*}]=\mathfrak{R}(\mathcal{A}^{*}*_{L}\mathcal{A}*_{M}\mathcal{X})$ and $\mathfrak{R}(\mathcal{X}^{*})=\mathfrak{R}(\mathcal{X}^{\dagger})=\mathfrak{R}[\mathcal{B}*_{R}(\mathcal{X}*_{N}\mathcal{B})^{\dagger}]=\mathfrak{R}[\mathcal{B}*_{R}(\mathcal{X}*_{N}\mathcal{B})^{*}]=\mathfrak{R}(\mathcal{B}*_{R}\mathcal{B}^{*}*_{N}\mathcal{X}^{*})$.

Therefore, $\mathfrak{R}(\mathcal{X}*_{N}\mathcal{X}^{*}*_{M}\mathcal{A}^{*})\subseteq\mathfrak{R}(\mathcal{X})=\mathfrak{R}(\mathcal{A}^{*}*_{L}\mathcal{A}*_{M}\mathcal{X})\subseteq\mathfrak{R}(\mathcal{A}^{*})$, i.e., $\mathfrak{R}(\mathcal{X}*_{N}\mathcal{X}^{*}*_{M}\mathcal{A}^{*})\subseteq\mathfrak{R}(\mathcal{A}^{*})$ and $\mathfrak{R}(\mathcal{A}^{*}{*_{L}}\mathcal{A}{*_{M}}\mathcal{X})\subseteq\mathfrak{R}(\mathcal{X})$ implies that $(\mathcal{A}*_{M}\mathcal{X})^{\dagger}=\mathcal{X}^{\dagger}*_{M}\mathcal{A}^{\dagger}$.

$\mathfrak{R}(\mathcal{X})\subseteq\mathfrak{R}(\mathcal{A}^{\dagger})$ implies $\mathcal{A}^{\dagger}*_{L}\mathcal{A}*_{M}\mathcal{X}=\mathcal{X}$. Similarly from $\mathfrak{R}(\mathcal{X}^{*})=\mathfrak{R}(\mathcal{B}{*_{R}}\mathcal{B}^{*}{*_{N}}\mathcal{X}^{*})$ it follows that $(\mathcal{X}*_{N}\mathcal{B})^{\dagger}=\mathcal{B}^{\dagger}*_{N}\mathcal{X}^{\dagger}$, $\mathcal{X}=\mathcal{X}*_{N}\mathcal{B}*_{R}\mathcal{B}^{\dagger}$.

Conversely, Using Lemma 5(c), Lemma6(a) in the fact

$\mathfrak{R}[(\mathcal{X}^{\dagger})^{*}]=\mathfrak{R}(\mathcal{X})\subseteq\mathfrak{R}(\mathcal{A}^{\dagger})=\mathfrak{R}(\mathcal{A}^{*})$ and $\mathfrak{R}(\mathcal{X}^{\dagger})=\mathfrak{R}(\mathcal{X}^{*})\subseteq\mathfrak{R}[(\mathcal{B}^{\dagger})^{*}]=\mathfrak{R}(\mathcal{B}).$

One have $\mathcal{X}^{\dagger}=\mathcal{X}^{\dagger}{*_{M}}\mathcal{A}^{\dagger}{*_{L}}\mathcal{A}$ and $\mathcal{X}^{\dagger}=\mathcal{B}{*_{R}}\mathcal{B}^{\dagger}{*_{N}}\mathcal{X}^{\dagger}$.

Now, $(\mathcal{A}*_{M}\mathcal{X})^{\dagger}*_{L}\mathcal{A}=\mathcal{X}^{\dagger}*_{M}\mathcal{A}^{\dagger}*_{L}\mathcal{A}=\mathcal{X}^{\dagger}$ and $\mathcal{B}*_{R}(\mathcal{X}*_{N}\mathcal{B})^{\dagger}=\mathcal{B}*_{R}\mathcal{B}^{\dagger}*_{N}\mathcal{X}^{\dagger}=\mathcal{X}^{\dagger}$. then by Corollary1 proof is done.

From the above theorem one can conclude the necessary and sufficient condition for cancellation law in terms of range.

Lemma 9

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

[TABLE]

if and only if

[TABLE]

2.2 Weighted Moore-Penrose inverse

Weighted Moore-Penrose inverse of even-order tensor, $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{K}\times J_{1}\times\cdots\times J_{K}}$ was introduced in weit2 , very recently. Here we have discussed weighted Moore-Penrose inverse for an arbitrary-order tensor via Einstein product, which is a special case of generalized weighted Moore-Penrose inverse.

Definition 8

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$,and a pair of invertible and Hermitian tensors $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$. A tensor $\mathcal{Y}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$ is said to be the generalized weighted Moore-Penrose inverse of $\mathcal{A}$ with respect to $\mathcal{M}$ and $\mathcal{N}$, if $\mathcal{Y}$ satisfies the following four tensor equations

[TABLE]

In particular, when both $\mathcal{M},~{}\mathcal{N}$ are Hermitian positive definite tensors, the tensor $\mathcal{Y}$ is called the weighted Moore-Penrose inverse of $\mathcal{A}$ and denote by $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger}$.

However, the generalized weighted Moore-Penrose inverse $\mathcal{Y}$ does not always exist for any tensor $\mathcal{A}$, as shown below with an example.

Example 2

Consider tensors $~{}\mathcal{A}=(a_{ijk})\in\mathbb{R}^{\overline{2\times 3}\times\overline{2}}$ and $\mathcal{M}=(a_{ijkl})\in\mathbb{R}^{\overline{2\times 3}\times\overline{2\times 3}}$ with $\mathcal{N}=(n_{ij})\in\mathbb{R}^{\overline{2}\times\overline{2}}$ such that

[TABLE]

with

[TABLE]

[TABLE]

Then we have

[TABLE]

This shows $\mathcal{A}^{T}*_{2}\mathcal{M}*_{2}\mathcal{A}$ is not invertible. Consider the generalized weighted Moore-Penrose inverse $\mathcal{Y}\in\mathbb{R}^{\overline{2}\times\overline{2\times 3}}$ of the given tensor $\mathcal{A}$ is exist, then using relation (1) and relation (3) of Definition 8, we have

[TABLE]

Since $(\mathcal{A}^{T}*_{2}\mathcal{A})^{-1}*_{1}\mathcal{A}^{T}*_{2}\mathcal{A}=\mathcal{I}$, then $\mathcal{A}$ is left cancellable, now (16) becomes

[TABLE]

this follows that $\mathcal{A}^{T}*_{2}\mathcal{M}*_{2}\mathcal{A}$ is invertiable, which is a contradiction.

At this point one may be interested to know when does the generalized weighted Moore-Penrose inverse exist ? The answer to this question is explained in the following theorem.

Theorem 2.4

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. If both $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ 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. Then generalized weighted Moore-Penrose inverse of an arbitrary-order tensor $\mathcal{A}$ exists and is unique, i.e., there exist a unique tensor $\mathcal{X}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times I_{1}\times\cdots\times I_{M}}$, such that,

[TABLE]

where $\mathcal{M}^{1/2}$ and $\mathcal{N}^{1/2}$ are square roots of $\mathcal{M}$ and $\mathcal{N}$ respectively, satisfy all four relations of Definition 8.

One can prove the above theorem, using Theorem 1 in weit2 and Theorem 2.1. Further, it is know that identity tensors are always Hermitian and positive definite, therefore, for any $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{A}^{\dagger}_{\mathcal{I}_{M},\mathcal{I}_{N}}$ exists and $\mathcal{A}^{\dagger}_{\mathcal{I}_{M},\mathcal{I}_{N}}=\mathcal{A}^{\dagger}$, which is called the Moore-Penrose inverse of $\mathcal{A}$. Specifically, if we take $\mathcal{M}=\mathcal{I}_{M}$ or $\mathcal{N}=\mathcal{I}_{N}$ in Eq.(18), then the following identities are hold.

Corollary 2

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

[TABLE]

Using the Definition 8 and following the Lemma 2 in weit2 one can write $(\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger})_{\mathcal{N},\mathcal{M}}^{\dagger}=\mathcal{A}$ and $(\mathcal{A}_{\mathcal{M},\mathcal{N}}^{\dagger})^{*}=(\mathcal{A}^{*})^{\dagger}_{\mathcal{N}^{-1},\mathcal{M}^{-1}}$, where $\mathcal{A}$ is any arbitrary-order tensor.

Now we define weighted conjugate transpose of a arbitrary-order tensor, as follows.

Definition 9

Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ are Hermitian positive definite tensors, the weighted conjugate transpose of $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ is denoted by $\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}}$ and defined as $\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}}=\mathcal{N}^{-1}*_{N}\mathcal{A}^{*}*_{M}\mathcal{M}$.

Next we present the properties of the weighted conjugate transpose of any arbitrary-order tensor, $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, as follows.

Lemma 10

*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 Hermitian positive definite tensors $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{P}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{N}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$.Then

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

(b) $(\mathcal{A}{*_{N}}\mathcal{B})^{\#}_{\mathcal{P},\mathcal{M}}=\mathcal{B}^{\#}_{\mathcal{P},\mathcal{N}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}}$.*

Adopting the result of Lemma 10(b) and the definition of the weighted Moore-Penrose inverse, we can write the following identities.

Lemma 11

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

1. (a)

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

$\mathcal{A}=\mathcal{A}{*_{N}}\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}}{*_{M}}(\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}})^{{\dagger}}_{\mathcal{N},\mathcal{M}}=(\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}})^{{\dagger}}_{\mathcal{N},\mathcal{M}}{*_{N}}\mathcal{A}^{\#}_{\mathcal{N},\mathcal{M}}{*_{M}}\mathcal{A};$** 3. (c)

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

Using the Lemma 3.17 in PanRad18 on two invertible tensors $\mathcal{B}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{C}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$, one can write the following identities

[TABLE]

where $\mathcal{A}$ is the arbitrary-order tensor, i.e., $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. By Eq.(19) and Corollary 2, we get following results.

Lemma 12

*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 a pair of Hermitian positive definite tensors. Then

(a) $\mathcal{A}_{\mathcal{M},\mathcal{I}_{N}}^{\dagger}*_{M}\mathcal{A}=(\mathcal{M}^{1/2}*_{M}\mathcal{A})^{\dagger}*_{M}\mathcal{M}^{1/2}*_{M}\mathcal{A}=\mathcal{A}^{\dagger}*_{M}\mathcal{A}$,

(b) $\mathcal{A}*_{N}\mathcal{A}_{\mathcal{I}_{M},\mathcal{N}}^{\dagger}=\mathcal{A}*_{N}\mathcal{N}^{-1/2}*_{N}(\mathcal{A}*_{N}\mathcal{N}^{-1/2})^{\dagger}=\mathcal{A}*_{N}\mathcal{A}^{\dagger}$,

(c) $(\mathcal{A}_{\mathcal{M},\mathcal{I}_{N}}^{\dagger})^{*}=\mathcal{M}^{1/2}*_{M}[\mathcal{A}^{*}*_{M}\mathcal{M}^{1/2}]^{\dagger}$,

(d) $(\mathcal{A}_{\mathcal{I}_{M},\mathcal{N}}^{\dagger})^{*}=(\mathcal{N}^{-1/2}*_{N}\mathcal{A}^{*})^{\dagger}*_{N}\mathcal{N}^{-1/2}$.*

The considerable amount of conventional and important facts with the properties concerning the range space of arbitrary-order tensor, the following theorem obtains the well-formed result.

Theorem 2.5

Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{K_{1}\times\cdots\times k_{L}\times K_{1}\times\cdots\times K_{L}}$ be a pair of Hermitian positive definite tensors. Then

[TABLE]

Proof

Let $\mathcal{X}=(\mathcal{U}^{\dagger})^{*}*_{N}\mathcal{V}$ and $\mathcal{Y}=\mathcal{U}*_{N}(\mathcal{V}^{\dagger})^{*}$. From Lemma 6(c) we get

[TABLE]

Now, using Lemma 6(b) and Lemma 5(b) along with the fact $(\mathcal{V}^{\dagger})^{*}=\mathcal{V}*_{L}\mathcal{V}^{\dagger}*_{N}(\mathcal{V}^{\dagger})^{*}$, we obtain,

[TABLE]

Replacing $\mathcal{U}$ and $\mathcal{V}$ by $\mathcal{M}^{1/2}{*_{M}}\mathcal{U}$ and $\mathcal{V}*_{L}\mathcal{N}^{-1/2}$ respectively on the above result, we get

[TABLE]

Substituting the above result in Eq.(18) we get the desired result.

Further, in connection with range space of arbitrary-order tensor, the following Theorem collects some useful identities of weighted Moore-Penrose inverses.

Theorem 2.6

Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}},~{}~{}\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and  $\mathcal{W}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times H_{1}\times\cdots\times H_{R}}$. If $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$, where $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$ are Hermitian positive definite tensors. Then

1. (a)

$\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{X}^{\dagger}_{\mathcal{I}_{N},\mathcal{N}}{*_{N}}\mathcal{V}{*_{L}}\mathcal{Y}^{\dagger}_{\mathcal{M},\mathcal{I}_{L}}$, where $\mathcal{X}=(\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{V}^{\dagger})^{\dagger}{*_{M}}\mathcal{A}$ and $\mathcal{Y}=\mathcal{A}{*_{R}}(\mathcal{V}^{\dagger}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}$; 2. (b)

$\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{X}^{\dagger}_{\mathcal{I}_{L},\mathcal{N}}{*_{L}}\mathcal{V}^{*}{*_{N}}\mathcal{V}{*_{L}}\mathcal{V}^{*}{*_{N}}\mathcal{Y}^{\dagger}_{\mathcal{M},\mathcal{I}_{N}}$, where $\mathcal{X}=[\mathcal{U}{*_{N}}(\mathcal{V}^{\dagger})^{*}]^{\dagger}{*_{M}}\mathcal{A}$ and $\mathcal{Y}=\mathcal{A}{*_{R}}[(\mathcal{V}^{\dagger})^{*}{*_{L}}\mathcal{W}]^{\dagger}$.

Proof

(a) From Eq.(18) we have

[TABLE]

where $\mathcal{U}_{1}=\mathcal{M}^{1/2}{*_{M}}\mathcal{U}$ and $\mathcal{W}_{1}=\mathcal{W}{*_{R}}\mathcal{N}^{-1/2}$. On the other hand, by Eq. (12), we have

[TABLE]

Also by Lemma 5(c), we get

[TABLE]

Thus, by using Lemma 5[(a),(b)] and Lemma 6[(a),(b)], we have

[TABLE]

Replacing $\mathcal{U}$ by $\mathcal{U}_{1}$ and $\mathcal{W}$ by $\mathcal{W}_{1}$ in Eq.(20) and then using Lemma 12[(a),(b)], we get

[TABLE]

(b) Following the Lemma 3(b) and Eq. (12), we get

$\mathfrak{R}(\mathcal{A})=\mathfrak{R}(\mathcal{Y})$ and $\mathfrak{R}(\mathcal{X}^{*})=\mathfrak{R}[(\mathcal{V}^{*}{*_{N}}\mathcal{V}*_{L}\mathcal{W})^{*}*_{L}(\mathcal{U}*_{N}(\mathcal{V}^{\dagger})^{*})^{*}*_{M}\{(\mathcal{U}*_{N}(\mathcal{V}^{\dagger})^{*})^{*}\}^{\dagger}]=\mathfrak{R}(\mathcal{A}^{*})$. Also by using Lemma 5(c)

[TABLE]

Using Lemma 5[(a),(b)] and Lemma 6[(a),(b)], we obtain

[TABLE]

Let $\mathcal{U}_{1}=\mathcal{M}^{1/2}{*_{M}}\mathcal{U},~{}~{}\mathcal{W}_{1}=\mathcal{W}{*_{R}}\mathcal{N}^{-1/2}$ and $\mathcal{A}_{1}=\mathcal{U}_{1}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}_{1}$. Then using Eq. (21) and the Lemma 12 [(a),(b)], we can write

[TABLE]

Therefore, $\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{N}^{-1/2}{*_{R}}\mathcal{A}_{1}^{\dagger}{*_{M}}\mathcal{M}^{1/2}=\mathcal{X}^{\dagger}_{\mathcal{I}_{L},\mathcal{N}}{*_{L}}\mathcal{V}^{*}{*_{N}}\mathcal{V}{*_{L}}\mathcal{V}^{*}{*_{N}}\mathcal{Y}^{\dagger}_{\mathcal{M},\mathcal{I}_{N}}$.

Theorem 2.7

Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{W}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times H_{1}\times\cdots\times H_{R}}$. Also let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$ be a pair of Hermitian positive definite tensors. Then

1. (a)

$(\mathcal{U}*_{N}\mathcal{V}*_{L}\mathcal{W})_{\mathcal{M},\mathcal{N}}^{{\dagger}}=[(\mathcal{U}_{\mathcal{M},\mathcal{I}_{N}}^{{\dagger}})^{*}*_{N}\mathcal{V}*_{L}\mathcal{W}]_{\mathcal{M}^{-1},\mathcal{N}}^{{\dagger}}*_{M}(\mathcal{U}_{\mathcal{M},\mathcal{I}_{N}}^{{\dagger}})^{*}*_{N}\mathcal{V}*_{L}(\mathcal{W}_{\mathcal{I}_{L},\mathcal{N}}^{{\dagger}})^{*}{*_{R}}[\mathcal{U}*_{N}\mathcal{V}*_{L}(\mathcal{W}_{\mathcal{I}_{L},\mathcal{N}}^{{\dagger}})^{*}]_{\mathcal{M},\mathcal{N}^{-1}}^{{\dagger}}$; 2. (b)

$(\mathcal{U}*_{N}\mathcal{V}*_{L}\mathcal{W})_{\mathcal{M},\mathcal{N}}^{{\dagger}}=[\{(\mathcal{U}*_{N}\mathcal{V})_{\mathcal{M},\mathcal{I}_{L}}^{{\dagger}}\}^{*}*_{L}\mathcal{W}]_{\mathcal{M}^{-1},\mathcal{N}}^{{\dagger}}*_{M}[(\mathcal{U}*_{N}\mathcal{V})_{\mathcal{M},\mathcal{I}_{L}}^{{\dagger}}]^{*}*_{L}\mathcal{V}^{{\dagger}}*_{N}[(\mathcal{V}*_{L}\mathcal{W})_{\mathcal{I}_{N},\mathcal{N}}^{{\dagger}}]^{*}*_{R}[\mathcal{U}*_{N}\{(\mathcal{V}*_{L}\mathcal{W})_{\mathcal{I}_{N},\mathcal{N}}^{{\dagger}}\}^{*}]_{\mathcal{M},\mathcal{N}^{-1}}^{{\dagger}}$.

Proof

(a) Let $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$, $\mathcal{X}=(\mathcal{U}^{\dagger})^{*}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$ and $\mathcal{Y}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}(\mathcal{W}^{\dagger})^{*}$.

Using Lemma 6(c), we get

[TABLE]

Further, using the fact that $\mathcal{W}^{\dagger}=\mathcal{W}^{\dagger}*_{N}(\mathcal{W}^{\dagger})^{*}*_{N}\mathcal{W}^{*}$,and Lemma 5(b) and Lemma 6(b), we can write,

[TABLE]

Using the above result to $[({\mathcal{M}^{1/2}*_{M}\mathcal{U}){*_{N}}\mathcal{V}{*_{L}}(\mathcal{W}{*_{R}}\mathcal{N}^{-1/2})}]^{\dagger}$ and following the Lemma 12[(c),(d)] we get,

[TABLE]

(b) Let $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$, $\mathcal{X}=[(\mathcal{U}*_{N}\mathcal{V})^{\dagger}]^{*}*_{L}\mathcal{W}$ and $\mathcal{Y}=\mathcal{U}*_{N}[(\mathcal{V}*_{L}\mathcal{W})^{\dagger}]^{*}$. Using Lemma6(c), we get

[TABLE]

From Lemma 3(a), one can write $[(\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}]^{*}=(\mathcal{V}{*_{L}}\mathcal{W}){*_{R}}(\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}{*_{N}}[(\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}]^{*}$.

Now using Lemma 5(b) and Lemma 6(b), we obtain

[TABLE]

Replacing $\mathcal{U}$ and $\mathcal{W}$ by $\mathcal{M}^{1/2}{*_{M}}\mathcal{U}$ and $\mathcal{W}{*_{R}}\mathcal{N}^{-1/2}$ respectively on the above result, we have

[TABLE]

Substituting the above result in Eq.(18) one can get the desired result.

Theorem 2.8

Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{W}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times H_{1}\times\cdots\times H_{R}}$. If $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$, and $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$ be a pair of Hermitian positive definite tensors, Then

[TABLE]

where $\mathcal{X}=\mathcal{U}^{\dagger}{*_{M}}\mathcal{A}$ and $\mathcal{Y}=\mathcal{A}{*_{R}}\mathcal{W}^{\dagger}$.

Proof

Let $\mathcal{U}_{1}=\mathcal{M}^{1/2}{*_{M}}\mathcal{U}$ and $\mathcal{W}_{1}=\mathcal{W}{*_{R}}\mathcal{N}^{-1/2}$. It is known, from Eq.(18),

[TABLE]

Now using Eq.(12) we have $\mathfrak{R}(\mathcal{X}^{*})=\mathfrak{R}[(\mathcal{U}*_{N}\mathcal{V}*_{L}\mathcal{W})^{*}*_{N}(\mathcal{U}^{*})^{\dagger}]=\mathfrak{R}(\mathcal{A}^{*})$ and $\mathfrak{R}(\mathcal{Y})=\mathfrak{R}(\mathcal{A})$. Also, from Lemma 5 (c), we have

[TABLE]

Using Lemma 5[(a),(b)] and Lemma 6[(a),(b)], we get

[TABLE]

Using Lemma12[(a),(b)]) one can conclude

[TABLE]

[TABLE]

Hence the proof is complete.

By Lemma 12[(a),(b)] and Eq. (18), and Eq. (22) we have

Corollary 3

*Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{W}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times H_{1}\times\cdots\times H_{R}}$. Let $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W}$. Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$, $\mathcal{P}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{Q}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ are Hermitian positive definite tensors. Then the weighted Moore-Penrose inverse of $\mathcal{A}$ with respect to $\mathcal{M}$ and $\mathcal{N}$ satisfies the following identities:

(a) $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{{\dagger}}=(\mathcal{U}_{\mathcal{I}_{M},\mathcal{P}}^{{\dagger}}{*_{M}}\mathcal{A})_{\mathcal{P},\mathcal{N}}^{{\dagger}}{*_{N}}\mathcal{V}{*_{L}}(\mathcal{A}{*_{R}}\mathcal{W}_{\mathcal{Q},\mathcal{I}_{R}}^{{\dagger}})_{\mathcal{M},\mathcal{Q}}^{{\dagger}},$

(b) $\mathcal{A}_{\mathcal{M},\mathcal{N}}^{{\dagger}}=[(\mathcal{U}*_{N}\mathcal{V}*_{L}\mathcal{V}_{\mathcal{P},\mathcal{I}_{L}}^{{\dagger}})_{\mathcal{M},\mathcal{P}}^{{\dagger}}*_{M}\mathcal{A}]_{\mathcal{P},\mathcal{N}}^{{\dagger}}*_{N}\mathcal{V}*_{L}[\mathcal{A}{*_{R}}(\mathcal{V}_{\mathcal{I}_{N},\mathcal{Q}}^{{\dagger}}*_{N}\mathcal{V}*_{L}\mathcal{W})_{\mathcal{Q},\mathcal{N}}^{{\dagger}}]_{\mathcal{M},\mathcal{Q}}^{{\dagger}}$.*

2.3 The full rank decomposition

The tensors and their decompositions originally appeared in 1927, Hit27 . The idea of decomposing a tensor as a product of tensors with a more desirable structure may well be one of the most important developments in numerical analysis such as the implementation of numerically efficient algorithms and the solution of multilinear systems kolda ; Che2018 ; Marcar ; Kolda01 . As part of this section, we focus on the full rank decomposition of a tensor. Unfortunately, It is very difficult to compute tensor rank. But the authors of in psmv18 introduced an useful and effective definition of the tensor rank, termed as reshaping rank. With the help of reshaping rank, We present one of our important results, full rank decomposition of an arbitrary-order tensor.

Theorem 2.9

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$. Then there exist a left invertible tensor $\mathcal{F}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}$ and a right invertible tensor $\mathcal{G}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}$ such that

[TABLE]

where $rshrank(\mathcal{F})=rshrank(\mathcal{G})=rshrank(\mathcal{A})=r=H_{1}\cdots H_{R}$. This is called the full rank decomposition of the tensor $\mathcal{A}$.

Proof

Let the matrix $A=rsh(\mathcal{A})\in\mathbb{C}^{I_{1}\cdots I_{M}\times J_{1}\cdots J_{N}}$. Then we have, rank(A) = r. Suppose that the matrix $A$ has a full rank decompositions, as follows,

[TABLE]

where $F\in\mathbb{C}^{I_{1}\cdots I_{M}\times H_{1}\cdots H_{R}}$ is a full column rank matrix and $G\in\mathbb{C}^{H_{1}\cdots H_{R}\times J_{1}\cdots J_{N}}$ is a full row rank matrix. From Eq. (8) and Eq. (24) we obtain

[TABLE]

where $\mathcal{F}=rsh^{-1}(F)\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}\textnormal{~{} and~{}}\mathcal{G}=rsh^{-1}(G)\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}.$ it follows that

[TABLE]

where $\mathcal{F}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}$ is the left invertible tensor and $\mathcal{G}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}$ is a right invertible tensor.

The prove of the above theorem was proved earlier (see Lemma 2.3(a), LIANG2018 ) indifferent way. Here, we have provided another proof without using reshape operation. Further, the authors of in LIANG2018 computed the Moore-Penrose inverse of a tensor using full rank decomposition of tensors. as follows

Lemma 13

(Theorem 3.7, LIANG2018 ) If the full rank decomposition of a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ is given as Theorem 2.9, then

[TABLE]

Now, the following theorem expressed the weighted Moore-Penrose inverse of a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ in form of the ordinary tensor inverse.

Theorem 2.10

If the full rank decomposition of a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ is given by Eq. (23), then the weighted Moore-Penrose inverse of $\mathcal{A}$ can be written as

[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.

Proof

From Eq. (18), we have

[TABLE]

where $\mathcal{B}=(\mathcal{M}^{1/2}{*_{M}}\mathcal{F}){*_{R}}(\mathcal{G}{*_{N}}\mathcal{N}^{-1/2})$,  and  $\mathcal{M}~{}\&~{}\mathcal{N}$ are Hermitian positive definite tensors. Here $\mathcal{B}$ is in the form of full rank decomposition, as both $\mathcal{M}^{1/2}$ and $\mathcal{N}^{1/2}$ are invertible. Now, from Lemma 13, we get

[TABLE]

Therefore, $\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{N}^{-1}{*_{N}}\mathcal{G}^{*}{*_{R}}(\mathcal{F}^{*}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}^{-1}{*_{N}}\mathcal{G}^{*})^{-1}{*_{R}}\mathcal{F}^{*}{*_{M}}\mathcal{M}$.

In particular when the arbitrary-order tensor, $\mathcal{A}$ is either left invertible or right invertible, we have the following results.

Corollary 4

Let a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ has the full rank decomposition.

1. (a)

If the tensor $\mathcal{A}$ is left invertible, then $\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{N}^{-1}{*_{N}}(\mathcal{A}^{*}{*_{M}}\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}^{-1})^{-1}{*_{N}}\mathcal{A}^{*}{*_{M}}\mathcal{M}$; 2. (b)

If the tensor $\mathcal{A}$ is right invertible, then $\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{N}^{-1}{*_{N}}\mathcal{A}^{*}{*_{M}}(\mathcal{M}{*_{M}}\mathcal{A}{*_{N}}\mathcal{N}^{-1}{*_{N}}\mathcal{A}^{*})^{-1}{*_{M}}\mathcal{M}$.

It is easy to see that the full rank factorizations of a tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ are not unique: if $\mathcal{A}=\mathcal{F}*_{R}\mathcal{G}$ is one full rank factorization, where $\mathcal{F}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}$ is the left invertible tensor and $\mathcal{G}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}$ is the right invertible tensor, then there exist a invertible tensor $\mathcal{P}$ of appropriate size, such that, $\mathcal{A}=(\mathcal{F}*_{R}\mathcal{P})*_{R}(\mathcal{P}^{-1}*_{R}\mathcal{G})$ is another full rank factorization. The following Theorem represents the result.

Theorem 2.11

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ with $rshrank(\mathcal{A})=r=H_{1}H_{2}\cdots H_{R}$. Then $\mathcal{A}$ has infinitely many full rank decompositions. However if $\mathcal{A}$ has two full rank decompositions, as follows,

[TABLE]

where $\mathcal{F},~{}\mathcal{F}_{1}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}$ and $\mathcal{G},~{}\mathcal{G}_{1}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}$, then there exists an invertible tensor $\mathcal{B}$ such that

[TABLE]

Moreover,

[TABLE]

Proof

Suppose the tensor, $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ has two full rank decompositions, as follows,

[TABLE]

where $\mathcal{F},~{}\mathcal{F}_{1}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times H_{1}\times\cdots\times H_{R}}$ and $\mathcal{G},~{}\mathcal{G}_{1}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times J_{1}\times\cdots\times J_{N}}$. Then

[TABLE]

Substituting $\mathcal{M}=\mathcal{I}_{M}$ and $\mathcal{N}=\mathcal{I}_{N}$ in Corollary 4(b) we have, $\mathcal{G}_{1}{*_{N}}\mathcal{G}_{1}^{\dagger}=\mathcal{I}_{R}$.

Therefore, $\mathcal{F}_{1}=\mathcal{F}{*_{R}}(\mathcal{G}{*_{N}}\mathcal{G}_{1}^{\dagger})$, similarly we can find $\mathcal{G}_{1}=(\mathcal{F}_{1}^{\dagger}{*_{M}}\mathcal{F}){*_{R}}\mathcal{G}.$

Let $rsh(\mathcal{G})=G=reshape(\mathcal{G},r,J_{1}\cdots J_{N})$ and $rsh(\mathcal{G}_{1})=G_{1}=reshape(\mathcal{G}_{1},r,J_{1}\cdots J_{N})$. Then $rsh(\mathcal{G}{*_{N}}\mathcal{G}_{1}^{\dagger})=GG_{1}^{\dagger}\in\mathbb{C}^{r\times r}$ and

[TABLE]

Hence $GG_{1}^{\dagger}$ is invertible as it has full rank. This concluded $\mathcal{G}{*_{N}}\mathcal{G}_{1}^{\dagger}=rsh^{-1}(GG_{1}^{\dagger})$ is invertible. Similarly $\mathcal{F}_{1}^{\dagger}{*_{M}}\mathcal{F}$ is also invertible. Let $\mathcal{B}=\mathcal{G}{*_{N}}\mathcal{G}_{1}^{\dagger}$ and $\mathcal{C}=\mathcal{F}_{1}^{\dagger}{*_{M}}\mathcal{F}$. Then

[TABLE]

is equivalent to $\mathcal{C}=\mathcal{B}^{-1}$. Therefore

[TABLE]

Further

[TABLE]

Similarly $\mathcal{G}_{1}^{\dagger}=\mathcal{G}^{\dagger}{*_{R}}\mathcal{B}$.

3 Reverse order law

In this section, we present various necessary and sufficient conditions of the reverse-order law for the weighted Moore-Penrose inverses of tensors. The first result obtained below addresses the sufficient condition for reverse-order law of tensor.

Theorem 3.1

Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}},$ and $\mathcal{N}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ be a pair of Hermitian positive definite tensors. If $\mathfrak{R}(\mathcal{B})=\mathfrak{R}(\mathcal{A}^{*})$, Then

[TABLE]

Proof

Let $\mathcal{X}=\mathcal{A}^{\dagger}*_{M}\mathcal{A}*_{N}\mathcal{B}$ and $\mathcal{Y}=\mathcal{A}*_{N}\mathcal{B}*_{L}\mathcal{B}^{\dagger}$. By using Lemma 5(c), we get

[TABLE]

Similarly, from Eq. (12) we have

[TABLE]

Further, from Lemma 5[(a), (b)] and Lemma 6[(a), (b)], we obtain

[TABLE]

i.e.,

[TABLE]

Let $\mathcal{A}_{1}=\mathcal{M}^{1/2}{*_{M}}\mathcal{A}$ and $\mathcal{B}_{1}=\mathcal{B}{*_{L}}\mathcal{N}^{-1/2}$. Using the Lemma 12[(a),(b)], we get

[TABLE]

Now, replacing $\mathcal{A}$ and $\mathcal{B}$ by $\mathcal{A}_{1}$ and $\mathcal{B}_{1}$ respectively on Eq.(28), we get

[TABLE]

Thus from corollary 2, we can conclude

[TABLE]

From the given condition and Lemma 5[(b),(c)], we have $\mathcal{B}{*_{L}}\mathcal{B}^{\dagger}=\mathcal{A}^{\dagger}{*_{M}}\mathcal{A}$, i.e.,

[TABLE]

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

Further, using Theorem 3.30 in PanRad18 one can write a necessary and sufficient condition for reverse order law for arbitrary-order tensors, i.e., for $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. Then $(\mathcal{A}{*_{N}}\mathcal{B})^{{\dagger}}=\mathcal{B}^{{\dagger}}{*_{N}}\mathcal{A}^{{\dagger}}$ if and only if

[TABLE]

Now, utilizing the above result and the fact of Lemma 5[(a),(c)], we conclude a beautiful result for necessary and sufficient condition for Moore-Penrose inverse of arbitrary-order tensor, as follows.

Lemma 14

*Let $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ and $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$. The Reverse order law hold for Moore-Penrose inverse, i.e.,$(\mathcal{A}*_{N}\mathcal{B})^{\dagger}=\mathcal{B}^{\dagger}*_{N}\mathcal{A}^{\dagger}$ if and only if 

$\mathfrak{R}(\mathcal{A}^{*}*_{M}\mathcal{A}*_{N}\mathcal{B})\subseteq\mathfrak{R}(\mathcal{B})$ and $\mathfrak{R}(\mathcal{B}*_{L}\mathcal{B}^{*}*_{N}\mathcal{A}^{*})\subseteq\mathfrak{R}(\mathcal{A}^{*})$.*

The primary result of this paper is presented next under the impression of the properties of range space of arbitrary-order tensor.

Theorem 3.2

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

[TABLE]

if and only if

[TABLE]

Proof

From equation (18), we have $(\mathcal{A}*_{N}\mathcal{B})_{\mathcal{M},\mathcal{N}}^{\dagger}=\mathcal{B}_{\mathcal{P},\mathcal{N}}^{\dagger}*_{N}\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}$ if and only if

[TABLE]

is equivalent to, if and only if

[TABLE]

where $\tilde{\mathcal{A}}=\mathcal{M}^{1/2}*_{M}\mathcal{A}*_{N}\mathcal{P}^{-1/2}$ and $\tilde{\mathcal{B}}=\mathcal{P}^{1/2}*_{N}\mathcal{B}*_{L}\mathcal{N}^{-1/2}$. From Lemma 14, we have

[TABLE]

if and only if

[TABLE]

which equivalently if and only if

[TABLE]

Hence, $(\mathcal{A}*_{N}\mathcal{B})_{\mathcal{M},\mathcal{N}}^{\dagger}=\mathcal{B}_{\mathcal{P},\mathcal{N}}^{\dagger}*_{N}\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}$ if and only if

[TABLE]

This completes the proof.

As a corollary to Theorem 3.2, we present another reverse order law for the weighted Moore-Penrose inverse of arbitrary-order tensor.

Corollary 5

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

[TABLE]

if and only if

[TABLE]

Proof

From Theorem 3.2, Eq.(29) and Lemma 5(a), we have $(\mathcal{A}*_{N}\mathcal{B})_{\mathcal{M},\mathcal{N}}^{\dagger}=\mathcal{B}_{\mathcal{P},\mathcal{N}}^{\dagger}*_{N}\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}$ if and only if

[TABLE]

and

[TABLE]

i.e., if and only if

[TABLE]

i.e., if and only if

[TABLE]

This completes the proof.

Theorem 3.3

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

[TABLE]

if and only if

[TABLE]

Proof

Suppose, $(\mathcal{A}{*_{N}}\mathcal{B})^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{B}^{\dagger}_{\mathcal{P},\mathcal{N}}{*_{N}}\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{P}}.$ Now one can write

[TABLE]

Further we can write

[TABLE]

[TABLE]

[TABLE]

Hence,

[TABLE]

By similar arguments one can also show that,$(\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{\dagger}_{\mathcal{P},\mathcal{N}})^{\dagger}_{\mathcal{M},\mathcal{P}}=\mathcal{B}{*_{L}}\mathcal{B}^{\dagger}_{\mathcal{P},\mathcal{N}}{*_{N}}\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{P}}$.

Conversely, For proving converse, first we prove a identity.

From Eq.(18) and Eq.(28) we have,

[TABLE]

Further, using the given hypothesis and above identity, we can write

[TABLE]

This completes the proof.

In the next theorem, we develop the characterization for the weighted Moore-Penrose inverse of the product of arbitrary-order tensors $\mathcal{A}$ and $\mathcal{B}$, as follows.

Theorem 3.4

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

[TABLE]

where $\mathcal{A}_{1}=\mathcal{A}*_{N}\mathcal{B}_{1}*_{L}(\mathcal{B}_{1})_{\mathcal{P},\mathcal{N}}^{\dagger}$ and $\mathcal{B}_{1}=\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}*_{M}\mathcal{A}*_{N}\mathcal{B}$.

Proof

[TABLE]

[TABLE]

[TABLE]

From (31) and (3), we have

[TABLE]

Therefore,

[TABLE]

Hence,

[TABLE]

Let $\mathcal{X}=\mathcal{A}*_{N}\mathcal{B}$ and $\mathcal{Y}=(\mathcal{B}_{1})^{\dagger}_{\mathcal{P},\mathcal{N}}*_{N}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}$. Using (3) and (32) we obtain

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Hence, $\mathcal{X}^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{Y}$, i.e., $(\mathcal{A}*_{N}\mathcal{B})^{\dagger}_{\mathcal{M},\mathcal{N}}=(\mathcal{B}_{1})^{\dagger}_{\mathcal{P},\mathcal{N}}*_{N}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}$.

We shall present the following example as a confirmation of the above Theorem.

Example 3

Let $\mathcal{A}_{1}=\mathcal{A}*_{2}\mathcal{B}_{1}*_{1}(\mathcal{B}_{1})_{\mathcal{P},\mathcal{N}}^{\dagger}$ and $\mathcal{B}_{1}=\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}*_{1}\mathcal{A}*_{2}\mathcal{B}$, where $~{}\mathcal{A}=(a_{ijk})\in\mathbb{R}^{3\times 2\times 4},~{}~{}\mathcal{B}=(b_{ijk})\in\mathbb{R}^{2\times 4\times 3},~{}~{}\mathcal{M}=(m_{ij})\in\mathbb{R}^{3\times 3},~{}~{}\mathcal{N}=(n_{ij})\in\mathbb{R}^{3\times 3}$ and $\mathcal{P}=(p_{ijkl})\in\mathbb{R}^{2\times 4\times 2\times 4}$ such that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Then $\mathcal{A}_{1}=(\tilde{a}_{ijk})\in\mathbb{R}^{3\times 2\times 4}$, $\mathcal{B}_{1}=(\tilde{b}_{ijk})\in\mathbb{R}^{2\times 4\times 3},~{}~{}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}=(x_{ijk})\in\mathbb{R}^{2\times 4\times 3}$

and $(\mathcal{B}_{1})^{\dagger}_{\mathcal{P},\mathcal{N}}=(y_{ijk})\in\mathbb{R}^{3\times 2\times 4}$ such that

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Thus

[TABLE]

Hence $(\mathcal{A}*_{N}\mathcal{B})^{\dagger}_{\mathcal{M},\mathcal{N}}=(\mathcal{B}_{1})^{\dagger}_{\mathcal{P},\mathcal{N}}*_{N}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}$

Further, using the Lemma 4 in weit2 on an arbitrary-order tensor $\mathcal{A}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$ with Hermitian positive definite tensors $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ one can write the following identity

[TABLE]

Using the above identity, a sufficient condition for the reverse order law for weighted Moore-Penrose inverse of tensor is presented next.

Corollary 6

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

[TABLE]

Then

[TABLE]

Proof

From Theorem3.4 we have, $(\mathcal{A}*_{N}\mathcal{B})^{\dagger}_{\mathcal{M},\mathcal{N}}=(\mathcal{B}_{1})_{\mathcal{P},\mathcal{N}}^{\dagger}*_{N}(\mathcal{A}_{1})_{\mathcal{M},\mathcal{P}}^{\dagger}$, where

$\mathcal{A}_{1}=\mathcal{A}*_{N}\mathcal{B}_{1}*_{L}(\mathcal{B}_{1})_{\mathcal{P},\mathcal{N}}^{\dagger}$ and $\mathcal{B}_{1}=\mathcal{A}_{\mathcal{M},\mathcal{P}}^{\dagger}*_{M}\mathcal{A}*_{N}\mathcal{B}$.

From Eq.(33) and given hypothesis, we have

[TABLE]

So there exists $\mathcal{P}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ such that $\mathcal{B}=\mathcal{A}^{\dagger}_{\mathcal{M},\mathcal{N}}*_{M}\mathcal{A}*_{N}\mathcal{P}$. Now,

[TABLE]

Hence $\mathcal{A}_{1}=\mathcal{A}*_{N}\mathcal{B}*_{L}\mathcal{B}^{\dagger}_{\mathcal{N},\mathcal{P}}$.

Further, we have, $\mathcal{N}(\mathcal{B}^{\#}_{\mathcal{N}^{1/2},\mathcal{P}^{1/2}})\subseteq\mathcal{N}(\mathcal{A})$, which is equivalent to

[TABLE]

Then from Lemma5(a), we have

[TABLE]

which equivalently

[TABLE]

that is

[TABLE]

i.e.

[TABLE]

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

We next present another characterization of the product of arbitrary-order tensors, as follows,

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 $\mathcal{B}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$.

Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$, $\mathcal{N}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times K_{1}\times\cdots\times K_{L}}$ and $\mathcal{P}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times J_{1}\times\cdots\times J_{N}}$ are three Hermitian positive definite tensors. Then*

[TABLE]

where $\mathcal{A}_{1}=\mathcal{A}{*_{N}}\mathcal{B}{*_{L}}\mathcal{B}^{\dagger}_{\mathcal{P},\mathcal{I}_{L}}$ and $\mathcal{B}_{1}=(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}{*_{M}}\mathcal{A}_{1}{*_{N}}\mathcal{B}.$

Proof

Let $\mathcal{X}=\mathcal{A}{*_{N}}\mathcal{B}$ and $\mathcal{Y}=(\mathcal{B}_{1})^{\dagger}_{\mathcal{P},\mathcal{N}}{*_{N}}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}$. Now we have

[TABLE]

Now, using Eq. (34), we obtain

[TABLE]

Further, using the following relations

[TABLE]

we have

[TABLE]

It concludes that,

[TABLE]

From the relations (35)-(38) validates $\mathcal{Y}=\mathcal{X}^{\dagger}_{\mathcal{M},\mathcal{N}}$. Hence $(\mathcal{A}*_{N}\mathcal{B})_{\mathcal{M},\mathcal{N}}^{{\dagger}}=(\mathcal{B}_{1})^{\dagger}_{\mathcal{\mathcal{P},\mathcal{N}}}{*_{N}}(\mathcal{A}_{1})^{\dagger}_{\mathcal{M},\mathcal{P}}.$ This completes the proof.

The significance of the properties of range and null space of arbitrary-order tensors, the last result achieved the sufficient condition for the triple reverse order law of tensor.

Theorem 3.6

*Let $\mathcal{U}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times J_{1}\times\cdots\times J_{N}}$, $\mathcal{V}\in\mathbb{C}^{J_{1}\times\cdots\times J_{N}\times K_{1}\times\cdots\times K_{L}}$ and

$\mathcal{W}\in\mathbb{C}^{K_{1}\times\cdots\times K_{L}\times H_{1}\times\cdots\times H_{R}}$. Let $\mathcal{M}\in\mathbb{C}^{I_{1}\times\cdots\times I_{M}\times I_{1}\times\cdots\times I_{M}}$ and $\mathcal{N}\in\mathbb{C}^{H_{1}\times\cdots\times H_{R}\times H_{1}\times\cdots\times H_{R}}$ be a pair of Hermitian positive definite tensors. If*

[TABLE]

Then

[TABLE]

Proof

Let $\mathcal{A}=\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W},~{}~{}\mathcal{W}_{1}=(\mathcal{U}{*_{N}}\mathcal{V})^{\dagger}{*_{M}}\mathcal{A}$ and $\mathcal{U}_{1}=\mathcal{A}{*_{R}}(\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}$. From Eq.(12), we get

[TABLE]

Also from Lemma 5(c), we get

[TABLE]

Applying Lemma 5[(a),(b)] and Lemma 6[(a),(b)], we have

[TABLE]

which is equivalent to

[TABLE]

Replacing $\mathcal{U}$ and $\mathcal{W}$ by $\mathcal{M}^{1/2}{*_{M}}\mathcal{U}$ and $\mathcal{W}{*_{R}}\mathcal{N}^{-1/2}$ in Eq.(39) along with using Eq.(18) and Lemma 12[(a),(b)] we have

[TABLE]

Applying Lemma 5[(a),(c)] and Lemma 6(a) in the given condition, we get

[TABLE]

Hence, $(\mathcal{U}{*_{N}}\mathcal{V}{*_{L}}\mathcal{W})^{\dagger}_{\mathcal{M},\mathcal{N}}=\mathcal{W}^{\dagger}_{\mathcal{I}_{L},\mathcal{N}}{*_{L}}\mathcal{V}^{\dagger}{*_{N}}\mathcal{U}^{\dagger}_{\mathcal{M},\mathcal{I}_{N}}.$

4 Conclusion

In this paper, a novel SVD and full rank-decomposition of arbitrary-order tensors using reshape operation is developed. Using this decomposition, we have studied the Moore-Penrose and general weighted Moore-Penrose inverse for arbitrary-order tensors via the Einstein product. We have also added some results on the range and null spaces to the existing theory. Then we discuss a few characterizations of cancellation properties for Moore-Penrose inverse of tensors. In addition to these, we have discussed the reverse-order laws for weighted Moore-Penrose inverses. In the future, it will be more interesting to express additional identities of weighted Moore-Penrose inverse in terms of the ordinary Moore-Penrose inverse for arbitrary-order tensor.

**Acknowledgments

**The first and the third authors are grateful to the Mohapatra Family Foundation and the College of Graduate Studies, University of Central Florida, Orlando, for their financial support for this research. In addition to this the first author is grateful for the partially supported by Science and Engineering Research Board (SERB), Department of Science and Technology, India, under the Grant No. EEQ/2017/000747.