The Minimum-Norm Least Squares Solutions to Quaternion Tensor Systems

Abstract

In this paper, we investigate the minimum-norm least squares solution to a quaternion tensor system ${\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1={\mathcal{C}}_1,$ ${\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3={\mathcal{C}}_2,$ ${\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_M{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_M{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_M{\mathcal{F}}_2=\mathcal{D}$ by using the Moore–Penrose inverses of block tensors. As an application, we discuss the quaternion tensor system $\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{C},$ $\mathcal{E}{\ast }_N\mathcal{X}{\ast }_M\mathcal{F}=\mathcal{D}$ for minimum-norm least squares reducible solutions. To illustrate the results, we present an algorithm and a numerical example.

1. Introduction

Previous research on the classic systems of matrix equations may go back to Cecioni [1] on

$$AX=C,XB=D.$$

(1)

During the past few decades, a great deal of research has been done on various generalized systems of (1) over complex number field $\mathbb{C}$ and quaternions $\mathbb{H}$ (see, for example, [2,3]). As a natural extension of a matrix, an N-mode tensor $\mathcal{A}=\left(a_{i_1\cdots i_N}\right)\left(1\le i_j\le I_j,j=1,\dots ,N\right)$ is a multidimensional array. In this paper, we consider solutions to some quaternion tensor systems.

Recall that a quaternion, introduced by Hamilton [4], is an associative and non-commutative division algebra over a real number field $\mathbb{R}$ and is generally represented in the form:

$$\mathbb{H}=\left\{q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}\mid {\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=\mathbf{ijk}=-1,q_0,q_1,q_2,q_3\in \mathbb{R}\right\}.$$

We refer the reader to [5,6,7,8] for quaternions and some applications. Nowadays, quaternion tensors are often used to solve problems in quantum mechanics, control theory, linear modeling, etc. [9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25].

It is well known that there are several definitions of tensor products. In this paper, we focus on the so-called “Einstein product”. Given two tensors $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_P\times K_1\times \cdots \times K_N},$ $\mathcal{B}\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_M}$, their Einstein product $\mathcal{A}{\ast }_N\mathcal{B}\in {\mathbb{H}}^{I_1\times \cdots \times I_P\times J_1\times \cdots \times J_M}$ [26] is defined as

$${\left(\mathcal{A}{\ast }_N\mathcal{B}\right)}_{i_1\dots i_P{j}_1\dots j_M}=\sum _{k_1,\dots ,k_N=1}^{K_1,\dots ,K_N}{a}_{i_1\dots i_P{k}_1\dots k_N}{b}_{k_1\dots k_N{j}_1\dots j_M}.$$

For simplicity, we will denote the above summation by ${\sum }_{k_1\dots k_N}$. When $N=P=M=1$, A and B are quaternion matrices, and their Einstein product is the usual matrix product.

Solving quaternion tensor systems has recently been explored. For example, He [27] gave the necessary and sufficient condition for their existence and the expression for the general solution to a quaternion tensor system

$$\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{C},\mathcal{X}{\ast }_N\mathcal{B}=\mathcal{D},$$

(2)

where $\mathcal{A},\mathcal{C}\in {\mathbb{H}}^{J_1\times \cdots \times J_N\times I_1\times \cdots \times I_N},$ $\mathcal{B},\mathcal{D}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times L_1\times \cdots \times L_N}$ are given. Since reducible matrices can be applied in many areas such as stochastic processes, random walks in graphs, and the connection of directed graphs [28,29,30], researchers have already started to investigate the reducible solutions of systems of matrix equations and tensor equations. For example, Nie et al. [2] presented the necessary and sufficient condition for a system of matrix equations to have a reducible solution and gave the representation of such a solution if it is possible. Xie and Wang [31] investigated the solvable condition and the expression of the reducible solution to the classical quaternion tensor equation

$$\mathcal{A}{\ast }_N\mathcal{X}{\ast }_N\mathcal{B}=\mathcal{C},$$

(3)

where $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times L_1\times \cdots \times L_N},$ $\mathcal{B}\in {\mathbb{H}}^{L_1\times \cdots \times L_N\times J_1\times \cdots \times J_N}$, $\mathcal{C}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}$ are given. In addition, they also derived the solvable condition and the expression of the general solution to the quaternion tensor equation

$${\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1{\ast }_M{\mathcal{B}}_1+{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2{\ast }_M{\mathcal{B}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3{\ast }_M{\mathcal{B}}_2=\mathcal{C},$$

(4)

where ${\mathcal{A}}_1,{\mathcal{B}}_1,{\mathcal{A}}_2,{\mathcal{B}}_2,$ and $\mathcal{C}$ are given tensors with appropriate dimensions.

Motivated by above results and the applications of tensor equations, in this paper, we consider the minimum-norm least squares solution to the system of quaternion tensor equations:

$$\left\{\begin{array}{c}{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1={\mathcal{C}}_1\hfill \\ {\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3={\mathcal{C}}_2\hfill \\ {\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_M{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_M{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_M{\mathcal{F}}_2=\mathcal{D}\hfill \end{array},\right.$$

(5)

where $\mathcal{A},\mathcal{C},\mathcal{E}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times L_1\times \cdots \times L_N},\mathcal{F}\in {\mathbb{H}}^{L_1\times \cdots \times L_N\times J_1\times \cdots \times J_N},\mathcal{D}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}$, and ${\mathcal{A}}_1, {\mathcal{C}}_1, {\mathcal{E}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N}, {\mathcal{A}}_2, {\mathcal{C}}_2, {\mathcal{E}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},$, ${\mathcal{F}}_1\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_N},{\mathcal{F}}_2\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times J_1\times \cdots \times J_N}$ are given, and the minimum-norm least squares reducible solution to the system of quaternion tensor equations:

$$\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{C},\mathcal{E}{\ast }_N\mathcal{X}{\ast }_M\mathcal{F}=\mathcal{D}.$$

(6)

As special cases, we also give the the minimum-norm least squares reducible solution to Equation (3) and the minimum-norm least squares solution to Equation (4).

This paper is organized as follows. In Section 2, we introduce some notation and review some results that will be used throughout the paper. In Section 3, we discuss the forms of the M–P inverses of block tensors. In Section 4, we present the minimum-norm least square solutions to systems (4) and (5). The least squares reducible solutions to systems (3) and (6) are also obtained. Finally, we develop an algorithm and calculate a numerical example to show the accuracy of our results in Section 5.

2. Preliminaries

We first recall some definitions and fix some notation. It is well known that the transpose of a tensor can be defined in many ways, according to the different partitions of its indices. To avoid potential confusion, we use different index letters to indicate the transpose of a tensor. That is, given a tensor $\mathcal{A}=\left(a_{i_1\dots i_N{j}_1\dots j_M}\right)\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_M},$ the conjugate transpose of $\mathcal{A}$ is defined as ${\mathcal{A}}^{\ast }=\left(b_{j_1\dots j_M{i}_1\dots i_N}\right)\in {\mathbb{H}}^{J_1\times \cdots \times J_M\times I_1\times \cdots \times I_N}$ hereafter, where $b_{j_1\dots j_M{i}_1\dots i_N}={\overline{a}}_{i_1\dots i_N{j}_1\dots j_M}$. In particular, when $\mathcal{A}$ is a real tensor, ${\mathcal{A}}^{\ast }$ is called the transpose of $\mathcal{A}$ and is simply denoted by ${\mathcal{A}}^{\mathrm{T}}$. Throughout this paper, we use the Einstein product for the tensor product and $\mathcal{I}$ for the identity tensor with appropriate dimension. Thus, the inverse of a tensor $\mathcal{A}$ is a tensor $\mathcal{B}$ satisfying $\mathcal{A}{\ast }_N\mathcal{B}=\mathcal{B}{\ast }_N\mathcal{A}=\mathcal{I}$. Finally, the Frobenius norm of a tensor $\mathcal{A}$ is defined as ${\parallel \mathcal{A}\parallel }_F={\left({\sum }_{a_{i_1\cdots i_N{j}_1\cdots j_M}}{\left|a_{i_1\cdots i_N{j}_1\cdots j_M}\right|}^2\right)}^{1/2}$.

The Moore–Penrose inverses of tensors via the Einstein product have been discussed over the complex number field and quaternions (see, for example, [22,31,32]).

Definition 1.

Let $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}.$ A tensor $\mathcal{X}\in {\mathbb{H}}^{J_1\times \cdots \times J_N\times I_1\times \cdots \times I_N}$ satisfying

(i) 

$\mathcal{A}{\ast }_N\mathcal{X}{\ast }_N\mathcal{A}=\mathcal{A}$;

(ii) 

$\mathcal{X}{\ast }_N\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{X}$;

(iii) 

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

(iv) 

${\left(\mathcal{X}{\ast }_N\mathcal{A}\right)}^{\ast }=\mathcal{X}{\ast }_N\mathcal{A}$

is called a Moore–Penrose inverse of $\mathcal{A}$, abbreviated as the M–P inverse, denoted by ${\mathcal{A}}^{†}$.

It has been shown that the M–P inverse of a tensor $\mathcal{A}$ exists and is unique. Two projectors about the M–P inverse are defined as: ${\mathcal{L}}_{\mathcal{A}}=\mathcal{I}-{\mathcal{A}}^{†}{\ast }_N\mathcal{A}$ and ${\mathcal{R}}_{\mathcal{A}}=\mathcal{I}-\mathcal{A}{\ast }_N{\mathcal{A}}^{†}$. The following properties of quaternion tensors will be used in the following sections.

Proposition 1

(Xie and Wang [31]). Let $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N}$. Then,

(1) 

${\left({\mathcal{A}}^{†}\right)}^{†}=\mathcal{A}$;

(2) 

${\left({\mathcal{A}}^{†}\right)}^{\ast }={\left({\mathcal{A}}^{\ast }\right)}^{†}$;

(3) 

${\left({\mathcal{A}}^{\ast }{\ast }_N\mathcal{A}\right)}^{†}={\mathcal{A}}^{†}{\ast }_N{\left({\mathcal{A}}^{\ast }\right)}^{†},{\left(\mathcal{A}{\ast }_N{\mathcal{A}}^{\ast }\right)}^{†}={\left({\mathcal{A}}^{\ast }\right)}^{†}{\ast }_N{\mathcal{A}}^{†};$

(4) 

${\mathcal{A}}^{†}{\ast }_N{\mathcal{R}}_{\mathcal{A}}=0$ and ${\mathcal{R}}_{\mathcal{A}}{\ast }_N\mathcal{A}=0$;

(5) 

$\mathcal{A}{\ast }_N{\mathcal{L}}_{\mathcal{A}}=0$ and ${\mathcal{L}}_{\mathcal{A}}{\ast }_N{\mathcal{A}}^{†}=0$.

For the same reason as block matrices in matrix theory, the block techniques for tensors play important roles in tensor theory. Next, we extend the concept of row block tensors given in [22] to the quaternion case.

Definition 2.

Let $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_M},\mathcal{B}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_M}$. The row block tensor consisted of $\mathcal{A}$ and $\mathcal{B}$ is defined as

$$\left[\begin{array}{cc}\mathcal{A}& \mathcal{B}\end{array}\right]\in {\mathbb{H}}^{{\alpha }^N\times {\beta }_1\times \cdots {\beta }_M},$$

where ${\alpha }^N=I_1\times \cdots \times I_N$ and ${\beta }_i=J_i+K_i(i=1,\dots ,M)$,

$${\left[\begin{array}{cc}\mathcal{A}& \mathcal{B}\end{array}\right]}_{i_1\cdots i_N{l}_1\cdots l_M}=\left\{\begin{array}{cc}{a}_{i_1\cdots i_N{l}_1\cdots l_M},\hfill & i_1\cdots i_N\in \left[I_1\right]\times \cdots \times \left[I_N\right],l_1\cdots l_M\in \left[J_1\right]\times \cdots \times \left[J_M\right]\hfill \\ b_{i_1\cdots i_N{l}_1\cdots l_M},\hfill & i_1\cdots i_N\in \left[I_1\right]\times \cdots \times \left[I_N\right],l_1\cdots l_M\in {\gamma }_1\times \cdots \times {\gamma }_M\hfill \\ 0,\hfill & \mathit{else}\hfill \end{array}\right.,$$

where ${\gamma }_i=\left\{J_i+1,\dots ,J_i+K_i\right\}(i=1,\dots ,M)$.

The following properties of block tensors are presented in [22] for complex tensors. It is easy to check that they also hold for quaternion tensors.

Proposition 2.

The following equalities hold:

(1) 

$\left[\begin{array}{cc}{\mathcal{A}}_1\hfill & {\mathcal{B}}_1\hfill \\ {\mathcal{A}}_2\hfill & {\mathcal{B}}_2\hfill \end{array}\right]{\ast }_M\left[\begin{array}{c}\mathcal{C}\hfill \\ \mathcal{D}\hfill \end{array}\right]=\left[\begin{array}{c}{\mathcal{A}}_1{\ast }_M\mathcal{C}+{\mathcal{B}}_1{\ast }_M\mathcal{D}\hfill \\ {\mathcal{A}}_2{\ast }_M\mathcal{C}+{\mathcal{B}}_2{\ast }_M\mathcal{D}\hfill \end{array}\right]\in {\mathbb{H}}^{{\rho }_1\times \cdots \times {\rho }_N\times {\alpha }^N}$;

(2) 

$\left[\begin{array}{cc}\mathcal{G}\hfill & \mathcal{H}\hfill \end{array}\right]{\ast }_N\left[\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{B}}_1\\ {\mathcal{A}}_2& {\mathcal{B}}_2\end{array}\right]=\left[\begin{array}{cc}\mathcal{G}{\ast }_N{\mathcal{A}}_1+\mathcal{H}{\ast }_N{\mathcal{A}}_2\hfill & \mathcal{G}{\ast }_N{\mathcal{B}}_1+\mathcal{H}{\ast }_N{\mathcal{B}}_2\hfill \end{array}\right]$ is belong to ${\mathbb{H}}^{S_1\times \cdots \times S_N\times {\beta }_1\times \cdots \times {\beta }_M}$, where $\mathcal{C}\in {\mathbb{H}}^{J_1\times \cdots \times J_M\times I_1\times \cdots \times I_N},$ $\mathcal{D}\in {\mathbb{H}}^{K_1\times \cdots \times K_M\times I_1\times \cdots \times I_N},$ $\mathcal{G}\in {\mathbb{H}}^{S_1\times \cdots \times S_N\times I_1\times \cdots \times I_N},$ $\mathcal{H}\in {\mathbb{H}}^{S_1\times \cdots \times S_N\times L_1\times \cdots \times L_N},$ ${\alpha }^N=I_1\times \cdots \times I_N$, and ${\rho }_i=I_i+L_i,i=1,\dots ,N;{\beta }_j=J_j+K_j,j=$ $1,\dots ,M$.

Using above notations for block tensors and properties, we can give the complex and real representations of a quaternion tensor as follows.

Let $\mathcal{A}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_M}$ with ${\mathcal{A}}_1,{\mathcal{A}}_2\in {\mathbb{C}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_M}$. The complex vector representation of $\mathcal{A}$ is ${\mathsf{\Phi }}_{\mathcal{A}}=\left[\begin{array}{cc}{\mathcal{A}}_1\hfill & {\mathcal{A}}_2\hfill \end{array}\right]$, and its complex representation is $f\left(\mathcal{A}\right)=\left[\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{A}}_2\\ -\overline{{\mathcal{A}}_2}& \overline{{\mathcal{A}}_1}\end{array}\right]\in {\mathbb{C}}^{2I_1\times \cdots \times 2I_N\times 2J_1\times \cdots \times 2J_M}$. Furthermore, let the real part and the imaginary part of a complex tensor $\mathcal{P}$ be denoted by $\Re \mathcal{P}$ and $\Im \mathcal{P}$, respectively. Then, $\mathcal{A}$ can be expressed as $\mathcal{A}=\Re {\mathcal{A}}_1+\Im {\mathcal{A}}_1\mathbf{i}+\Re {\mathcal{A}}_2\mathbf{j}+\Im {\mathcal{A}}_2\mathbf{k}$, and the real vector representation of $\mathcal{A}$ is $\overrightarrow{\mathcal{A}}=\left[\begin{array}{cccc}Re{\mathcal{A}}_1\hfill & Im{\mathcal{A}}_1\hfill & Re{\mathcal{A}}_2\hfill & Im{\mathcal{A}}_2\hfill \end{array}\right]$.

Different papers might use different orderings of the columns for a tensor unfolding. In this paper, we will use the transformation presented in Brazell et al. [9] and He et al. [32], which transforms a quaternion tensor into a quaternion matrix.

Definition 3.

The unfolding transformation

$$\begin{array}{ccc}\theta :{\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}& \to & {\mathbb{M}}_{(I_1·I_2\cdots I_{N-1}·I_N)\times (J_1·J_2\cdots J_{N-1}·J_N)}\left(\mathbb{H}\right)\\ \mathcal{A}& \to & A\end{array}$$

is defined by

$${(\mathcal{A})}_{i_1{i}_2\cdots i_n{j}_1{j}_2\cdots j_n}\stackrel{\theta }{\to }{(A)}_{\left[i_1+{\displaystyle \sum }_{k=2}^N\left(i_k-1\right){\displaystyle \prod }_{s=1}^{k-1}{I}_s\right]\left[j_1+{\displaystyle \sum }_{k=2}^N\left(j_k-1\right){\displaystyle \prod }_{s=1}^{k-1}{J}_s\right]}.$$

Using above transformation θ, we give the definition of a permutation tensor which is similar to [31] (Definition 4).

Definition 4.

A tensor $\mathcal{K}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times I_1\times \cdots \times I_N}$ is called a θ-permutation tensor (simply called a “permutation tensor" hereafter if no confusion arises) if the unfolding $\theta \left(\mathcal{K}\right)$ of $\mathcal{K}$ is a permutation matrix.

As [32] shows, θ is a one-to-one correspondence and preserves the Einstein tensor products, that is, $\theta \left(\mathcal{A}{\ast }_N\mathcal{B}\right)=\theta \left(\mathcal{A}\right)\theta \left(\mathcal{B}\right)$. Therefore, a permutation tensor $\mathcal{A}$ is invertible, since $\theta \left(\mathcal{A}\right)$ is invertible.

Definition 5.

A tensor $\mathcal{A}\in {\mathbb{H}}^{{\alpha }_1\times \cdots \times {\alpha }_N\times {\alpha }_1\times \cdots \times {\alpha }_N}$ is said to be $\mathcal{K}$-reducible if there exists a permutation tensor $\mathcal{K}$ such that $\mathcal{A}$ is permutationally similar to a triangular (upper or lower) block tensor, that is,

$$\mathcal{A}=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{B}}_1& {\mathcal{B}}_2\\ 0& {\mathcal{B}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1},$$

where ${\mathcal{B}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times I_1\times \cdots \times I_N},{\mathcal{B}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N},{\mathcal{B}}_3\in {\mathbb{H}}^{J_1\times \cdots \times J_N\times J_1\times \cdots \times J_N}$, and $\mathcal{K}\in$ ${\mathbb{H}}^{{\alpha }_1\times \cdots \times {\alpha }_N\times {\alpha }_1\times \cdots \times {\alpha }_N}\left({\alpha }_i=I_i+J_i,i=1,\dots ,N\right)$.

3. The M–P Inverses of Block Tensors

The solutions of tensor equations in the next section will be given in terms of block tensors. In this section, we will discuss the M–P inverses of block tensors. First, we recall the following lemma regrading the vec operator acting on a complex representation Φ in [33].

Lemma 1.

Let $\mathcal{A}={\mathcal{A}}_1+{\mathcal{A}}_{2}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}, \mathcal{B}={\mathcal{B}}_1+{\mathcal{B}}_{2}j\in {\mathbb{H}}^{J_1\times \cdots \times J_N\times L_1\times \cdots \times L_M}$, $\mathcal{C}={\mathcal{C}}_1+{\mathcal{C}}_{2}j\in {\mathbb{H}}^{L_1\times \cdots \times L_M\times K_1\times \cdots \times K_M}$. Set

$$K_{4{\gamma }^N}=\left[\begin{array}{cccc}{I}_{{\gamma }^N\times {\gamma }^N}& i{I}_{{\gamma }^N\times {\gamma }^N}& 0& 0\\ 0& 0& I_{{\gamma }^N\times {\gamma }^N}& i{I}_{{\gamma }^N\times {\gamma }^N}\\ I_{{\gamma }^N\times {\gamma }^N}& -i{I}_{{\gamma }^N\times {\gamma }^N}& 0& 0\\ 0& 0& I_{{\gamma }^N\times {\gamma }^N}& -i{I}_{{\gamma }^N\times {\gamma }^N}\end{array}\right],$$

where

$${\gamma }^N=J_1\times \cdots \times J_N,I_{{\gamma }^N\times {\gamma }^N}=\left[\begin{array}{cccc}\mathcal{I}& 0& \cdots & 0\\ 0& \mathcal{I}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \mathcal{I}\end{array}\right],\mathcal{I}\in {\mathbb{R}}^{L_1\times \cdots \times L_M\times L_1\times \cdots \times L_M}.$$

Then, we have

$$vec\left({\mathsf{\Phi }}_{\mathcal{A}{\ast }_N\mathcal{B}{\ast }_M\mathcal{C}}\right)=\left[{\mathcal{A}}_1•f{\left(\mathcal{C}\right)}^{\mathrm{T}}{\mathcal{A}}_2•f{\left(\mathcal{C}\mathbf{j}\right)}^{\ast }\right]{\ast }_M{\mathcal{K}}_{K_{4{\gamma }^N}}{\ast }_{M}vec\left(\overrightarrow{\mathcal{B}}\right),$$

where $\mathcal{A}•f\left(\mathcal{C}\right)=\left[\begin{array}{cc}\mathcal{A}\otimes {\mathcal{C}}_1& \mathcal{A}\otimes {\mathcal{C}}_2\hfill \\ \mathcal{A}\otimes \left(-{\overline{\mathcal{C}}}_2\right)& \mathcal{A}\otimes {\overline{\mathcal{C}}}_1\hfill \end{array}\right]$, and $f\left(\mathcal{C}\mathbf{j}\right)=\left[\begin{array}{cc}-{\mathcal{C}}_2& {\mathcal{C}}_1\\ -\overline{{\mathcal{C}}_1}& -\overline{{\mathcal{C}}_2}\end{array}\right]$.

We have obtained the following lemma about the M–P inverse of a column block tensor in [33] and will use it to prove Proposition 3.

Lemma 2.

Let ${\mathcal{T}}_1,{\mathcal{T}}_2\in {\mathbb{R}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N},\mathcal{A}=\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right].$ Then,

$${\mathcal{A}}^{†}={\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]}^{†}=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\mathcal{H}}^{\mathrm{T}}\right],$$

where

$$\begin{array}{l}\mathcal{H}={\mathcal{R}}^{†}+\left(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\right){\ast }_N\mathcal{Z}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\ast }_N{\left({\mathcal{T}}_1^{†}\right)}^{\mathrm{T}}{\ast }_N\left(\mathcal{I}-{\mathcal{T}}_2^{\mathrm{T}}{\ast }_N{\mathcal{R}}^{†}\right),\hfill \\ \mathcal{R}=\left(\mathcal{I}-{\mathcal{T}}_1^{†}{\ast }_N{\mathcal{T}}_1\right){\ast }_N{\mathcal{T}}_2^{\mathrm{T}},\\ \mathcal{Z}={(\mathcal{I}+(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\left){\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\ast }_N{\left({\mathcal{T}}_1^{†}\right)}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_2^{\mathrm{T}}{\ast }_N\right(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R})}^{-1}.\end{array}$$

In the following proposition, we give the general form of the M–P inverse of a tensor with three sub-blocks. We will apply it to describe the solutions in next section. One technique in this proof is using the quaternion SVD ([32]). For a tensor $\mathcal{A}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}$, the singular value decomposition (SVD) of $\mathcal{A}$ has the following form:

$$\mathcal{A}=\mathcal{U}{\ast }_N\mathcal{B}{\ast }_N{\mathcal{V}}^{\ast },$$

where $\mathcal{B}\in {\mathbb{R}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}$ is diagonal and $\mathcal{U}\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times I_1\cdots \times I_N}$ and $\mathcal{V}\in {\mathbb{H}}^{J_1\times \cdots \times J_N\times J_1\times \cdots \times J_N}$ are unitary tensors such that ${\mathcal{U}}^{\ast }{\ast }_N\mathcal{U}=\mathcal{I}$ and ${\mathcal{V}}^{\ast }{\ast }_N\mathcal{V}=\mathcal{I}$.

Proposition 3.

Given ${\mathcal{P}}_1\in {\mathbb{R}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N},{\mathcal{Q}}_1\in {\mathbb{R}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N}$, and ${\mathcal{Y}}_1\in {\mathbb{R}}^{I_1\times \cdots \times I_N\times L_1\times \cdots \times L_N}$, let $\mathcal{A}=\left[\begin{array}{ccc}{\mathcal{P}}_1\hfill & {\mathcal{Q}}_1\hfill & {\mathcal{Y}}_1\hfill \end{array}\right]$. Then, the M–P inverse of row block tensor $\mathcal{A}$ can be expressed as

$$\begin{array}{ccc}\hfill {\mathcal{A}}^{†}& ={\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{Q}}_1& {\mathcal{Y}}_1\end{array}\right]}^{†}\hfill & \hfill =\left[\begin{array}{c}{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right){\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_2\end{array}\right],\end{array}$$

(7)

where

$$\begin{array}{l}{\mathcal{H}}_1={\mathcal{R}}_1^{†}+\left(\mathcal{I}-{\mathcal{R}}_1^{†}{\ast }_N{\mathcal{R}}_1\right){\ast }_N{\mathcal{Z}}_1{\ast }_N{\mathcal{Q}}_1^T{\ast }_N{\mathcal{P}}_1^{T†}{\ast }_N{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{R}}_1^{†}\right),\\ {\mathcal{H}}_2={\mathcal{R}}_2^{†}+\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right){\ast }_N{\mathcal{Z}}_2{\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{R}}_2^{†}\right),\\ {\mathcal{R}}_1=(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}){\ast }_N{\mathcal{Q}}_1,\\ {\mathcal{R}}_2=(\mathcal{I}-{\mathcal{S}}_1{\ast }_N{\mathcal{S}}_1^{†}){\ast }_N{\mathcal{Y}}_1,\\ {\mathcal{S}}_1^{†}=\left[\begin{array}{c}{\mathcal{P}}_1^{†}-{\mathcal{P}}_1^{†}{\ast }_N{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\\ {\mathcal{H}}_1,\end{array}\right], \mathit{for} {\mathcal{S}}_1=\left[\begin{array}{cc}{\mathcal{P}}_1\hfill & {\mathcal{Q}}_1\hfill \end{array}\right],\\ {\mathcal{Z}}_1={\left[\mathcal{I}+(\mathcal{I}-{\mathcal{R}}_1^{†}{\ast }_N{\mathcal{R}}_1){\ast }_N{\mathcal{Q}}_1^T{\ast }_N{\mathcal{P}}_1^{T†}{\ast }_N{\mathcal{P}}_1^{†}{\ast }_N{\mathcal{Q}}_1{\ast }_N(\mathcal{I}-{\mathcal{R}}_1^{†}{\ast }_N{\mathcal{R}}_1)\right]}^{-1},\\ {\mathcal{Z}}_2={\left[\mathcal{I}+(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2){\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2)\right]}^{-1}.\end{array}$$

(8)

Proof of Proposition 3.

Let $\mathcal{X}$ be the right-hand side of (7). We only need to show that $\mathcal{X}$ satisfies the conditions in Definition 1.

We start by proving the existence of ${\mathcal{Z}}_1$ and ${\mathcal{Z}}_2$ in (8). Let ${\mathcal{K}}_1={\mathcal{P}}_1^{†}{\ast }_N{\mathcal{Q}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_1^{†}{\ast }_N{\mathcal{R}}_1\right)$ and ${\mathcal{K}}_2={\mathcal{P}}_2^{†}{\ast }_N{\mathcal{Q}}_2{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)$. Then, we have the SVD of ${\mathcal{K}}_1={\mathcal{U}}_1{\ast }_N{\mathcal{M}}_1{\ast }_N{\mathcal{V}}_1^{\ast }$, where ${\mathcal{U}}_1\in {\mathbb{R}}^{J_1\times \cdots \times J_N\times J_1\times \cdots \times J_N},{\mathcal{V}}_1\in {\mathbb{R}}^{K_1\times \cdots \times K_N\times K_1\times \cdots \times K_N}$ are unitary tensors and ${\mathcal{M}}_1\in {\mathbb{R}}^{J_1\times \cdots \times J_N\times K_1\times \cdots \times K_N}$ is diagonal. We write $\left(\mathcal{I}+{\mathcal{K}}_1^T{\ast }_N{\mathcal{K}}_1\right)={\mathcal{V}}_1{\ast }_N\left(\mathcal{I}+{\mathcal{M}}_1^T{\ast }_N{\mathcal{M}}_1\right){\ast }_N{\mathcal{V}}_1^{\ast }$. Since the diagonal tensor $\left(\mathcal{I}+{\mathcal{K}}_1^T{\ast }_N{\mathcal{K}}_1\right)$ has nonzero diagonal entries, we know that ${\mathcal{Z}}_1$ exists. Similarly, ${\mathcal{Z}}_2$ exists. Next, we prove that $\mathcal{X}$ satisfies the four conditions in Definition 1.

We first prove that ${\left(\mathcal{A}{\ast }_N\mathcal{X}\right)}^T=\mathcal{A}{\ast }_N\mathcal{X}$. It is easy to verify that ${\mathcal{S}}_1{\ast }_N{\mathcal{S}}_1^{†}={\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}$, and so ${\mathcal{R}}_2=\left(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+\right.$ $\left.{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}\right){\ast }_N{\mathcal{Y}}_1$. For any tensor $\mathcal{Q}$, by Proposition 1, we have

$$\left(\mathcal{I}-\mathcal{Q}{\ast }_N{\mathcal{Q}}^{†}\right){\ast }_N\mathcal{Q}=0,{\mathcal{Q}}^{†}{\ast }_N\left(\mathcal{I}-\mathcal{Q}{\ast }_N{\mathcal{Q}}^{†}\right)=0,$$

and it follows that

$${\mathcal{R}}_1{\ast }_N{\mathcal{H}}_1={\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†},{\mathcal{R}}_2{\ast }_N{\mathcal{H}}_2={\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2^{†}.$$

Therefore, by Proposition 2, we have

$$\begin{array}{cc}\hfill \mathcal{A}{\ast }_N\mathcal{X}=& {\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right){\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\hfill \\ & +{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)+{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\hfill \\ \hfill =& {\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+\left(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}\right){\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\hfill \\ & +\left[\left(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}\right)-\left(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}\right){\ast }_N{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right]{\ast }_N{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\hfill \\ \hfill =& {\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}+{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2^{†}.\hfill \end{array}$$

Since ${\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†},{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}$ and ${\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2^{†}$ are symmetric, we have ${\left(\mathcal{A}{\ast }_N\mathcal{X}\right)}^T=\mathcal{A}{\ast }_N\mathcal{X}$.

Next we prove that $\mathcal{A}{\ast }_N\mathcal{X}{\ast }_N\mathcal{A}=\mathcal{A}$. For any tensor $\mathcal{Q}$, it is obvious that $\left(\mathcal{I}-\mathcal{Q}{\ast }_N{\mathcal{Q}}^{†}\right)$ is idempotent. Then, we have

$${\mathcal{R}}_1^{†}{\ast }_N{\mathcal{P}}_1=0,{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{S}}_1=0,{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Q}}_1=0,{\mathcal{R}}_1^{†}{\ast }_N{\mathcal{Q}}_1={\mathcal{R}}_1^{†}{\ast }_N{\mathcal{R}}_1.$$

(9)

Based on this, we can derive that $\left(\mathcal{I}-{\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}-{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}\right)$ is idempotent and ${\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Y}}_1={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2$. Thus,

$$\begin{array}{cc}\hfill \mathcal{A}{\ast }_N\mathcal{X}{\ast }_N\mathcal{A}& =\left({\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}+{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2^{†}\right){\ast }_N\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{Q}}_1& {\mathcal{Y}}_1\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}{\ast }_N{\mathcal{Q}}_1+{\mathcal{R}}_1& \left({\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}\right){\ast }_N{\mathcal{Y}}_1+{\mathcal{R}}_2\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{Q}}_1& {\mathcal{Y}}_1\end{array}\right]=\mathcal{A}.\hfill \end{array}$$

To prove that $\mathcal{X}{\ast }_N\mathcal{A}{\ast }_N\mathcal{X}=\mathcal{X}$, we set $\mathcal{B}=\left({\mathcal{P}}_1{\ast }_N{\mathcal{P}}_1^{†}+{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_1^{†}+{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2^{†}\right)$. It is easy to verify that

$$\begin{array}{c}{\mathcal{R}}_1{\ast }_N{\mathcal{R}}_2^{†}=0,{\mathcal{P}}_1^{†}{\ast }_N\mathcal{B}={\mathcal{P}}_1^{†},{\mathcal{R}}_1^{†}{\ast }_N\mathcal{B}={\mathcal{R}}_1^{†},{\mathcal{R}}_2^{†}{\ast }_N\mathcal{B}={\mathcal{R}}_2^{†},\\ {\mathcal{H}}_1{\ast }_N\mathcal{B}={\mathcal{H}}_1,{\mathcal{S}}_1^{†}{\ast }_N\mathcal{B}={\mathcal{S}}_1^{†},{\mathcal{H}}_2{\ast }_N\mathcal{B}={\mathcal{H}}_2.\end{array}$$

It follows from Proposition 2 that

$$\begin{array}{cc}\hfill \mathcal{X}{\ast }_N\mathcal{A}{\ast }_N\mathcal{X}& =\left[\begin{array}{c}{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right){\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_2\end{array}\right]{\ast }_N\mathcal{B}\hfill \\ & =\left[\begin{array}{c}{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right){\ast }_N\left(\mathcal{B}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_1{\ast }_N\left(\mathcal{B}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_2\end{array}\right]\hfill \\ & =\left[\begin{array}{c}{\mathcal{P}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Q}}_1{\ast }_N{\mathcal{H}}_1\right){\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_2\end{array}\right]=\mathcal{X}.\hfill \end{array}$$

Finally, we prove that ${\left(\mathcal{X}{\ast }_N\mathcal{A}\right)}^T=\mathcal{X}{\ast }_N\mathcal{A}$. By Proposition 2, we know that $\mathcal{X}{\ast }_N\mathcal{A}$ can be written as

$$\begin{array}{cc}\hfill \mathcal{X}{\ast }_N\mathcal{A}& =\left[\begin{array}{c}{\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right)\\ {\mathcal{H}}_2\end{array}\right]{\ast }_N\left[\begin{array}{cc}{\mathcal{S}}_1& {\mathcal{Y}}_1\end{array}\right]\hfill \\ & =\left[\begin{array}{cc}{\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right){\ast }_N{\mathcal{S}}_1& {\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right){\ast }_N{\mathcal{Y}}_1\\ {\mathcal{H}}_2{\ast }_N{\mathcal{S}}_1& {\mathcal{H}}_2{\ast }_N{\mathcal{Y}}_1\end{array}\right]\hfill \\ & \triangleq \left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right].\hfill \end{array}$$

Next, we need to prove that $a_1,b_1,a_2,b_2$ are symmetric. Set

$$\begin{array}{c}{\mathcal{U}}_2={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right),\hfill \\ {\mathcal{V}}_2=\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right){\ast }_N{\mathcal{Z}}_2{\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}={\mathcal{Z}}_2{\ast }_N{\mathcal{U}}_2^T.\hfill \end{array}$$

Here, ${\mathcal{Z}}_2={\left[\mathcal{I}-{\mathcal{U}}_2^T{\ast }_N{\mathcal{U}}_2\right]}^{-1}={\mathcal{Z}}_2^T$. Then,

$${\mathcal{P}}_1^{T†}{\ast }_N{\mathcal{P}}_1^{†}{\ast }_N{\mathcal{P}}_1={\mathcal{P}}_1^{T†},{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1={\mathcal{S}}_1^{T†},$$

(10)

$${\mathcal{Z}}_2^{-1}{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)=\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right){\ast }_N{\mathcal{Z}}_2^{-1},$$

and

$$\begin{array}{cc}\hfill {\mathcal{V}}_2{\ast }_N{\mathcal{U}}_2& =\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right){\ast }_N{\mathcal{Z}}_2{\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)\hfill \\ \hfill & ={\mathcal{Z}}_2{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right){\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)\hfill \\ \hfill & ={\mathcal{Z}}_2{\ast }_N{\mathcal{U}}_2^T{\ast }_N{\mathcal{U}}_2\hfill \\ \hfill & ={\left[\mathcal{I}-{\mathcal{U}}_2^T{\ast }_N{\mathcal{U}}_2\right]}^{-1}{\ast }_N{\mathcal{U}}_2^T{\ast }_N{\mathcal{U}}_2\hfill \\ \hfill & =\mathcal{I}-\mathcal{Z}.\hfill \end{array}$$

(11)

Because $\left(\mathcal{I}-{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2\right)$ is idempotent, using (9) and (10), we obtain

$$\begin{array}{cc}\hfill a_1& ={\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right){\ast }_N{\mathcal{S}}_1\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2{\ast }_N{\mathcal{S}}_1\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left[{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{S}}_1+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N\left({\mathcal{S}}_1-{\mathcal{Y}}_1{\ast }_N{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{S}}_1\right)\right]\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2\right){\mathcal{Z}}_2{\ast }_N{\mathcal{Y}}_1^T{\ast }_N{\mathcal{S}}_1^{T†}{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{S}}_1-{\mathcal{U}}_2{\ast }_N{\mathcal{Z}}_2{\ast }_N{\mathcal{U}}_2^T.\hfill \end{array}$$

Similarly, we have

$$a_2={\mathcal{H}}_2{\ast }_N{\mathcal{S}}_1={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{S}}_1+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N\left({\mathcal{S}}_1-{\mathcal{Y}}_1{\ast }_N{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{S}}_1\right)={\mathcal{V}}_2.$$

Using ${\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Y}}_1={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2$ and (11),

$$\begin{array}{cc}\hfill b_1& ={\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2\right){\ast }_N{\mathcal{Y}}_1\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N{\mathcal{H}}_2{\ast }_N{\mathcal{Y}}_1\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left[{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Y}}_1+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N\left({\mathcal{Y}}_1-{\mathcal{Y}}_1{\ast }_N{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Y}}_1\right)\right]\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1-{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left[{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)\right]\hfill \\ & ={\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2\right)-{\mathcal{U}}_2{\ast }_N{\mathcal{V}}_2{\ast }_N{\mathcal{U}}_2\hfill \\ & ={\mathcal{U}}_2{\ast }_N\left(\mathcal{I}-{\mathcal{V}}_2{\ast }_N{\mathcal{U}}_2\right)={\mathcal{U}}_2{\ast }_N{\mathcal{Z}}_2={\mathcal{V}}_2^T.\hfill \end{array}$$

Similarly to the above,

$$\begin{array}{cc}\hfill b_2& ={\mathcal{H}}_2{\ast }_N{\mathcal{Y}}_1\hfill \\ & ={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{Y}}_1+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N\left(\mathcal{I}-{\mathcal{Y}}_1{\ast }_N{\mathcal{R}}_2^{†}\right){\ast }_N{\mathcal{Y}}_1\hfill \\ & ={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2+{\mathcal{V}}_2{\ast }_N{\mathcal{S}}_1^{†}{\ast }_N{\mathcal{Y}}_1{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_2{\ast }_N{\mathcal{R}}_2\right)\hfill \\ & ={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2+{\mathcal{U}}_2{\ast }_N{\mathcal{V}}_2={\mathcal{R}}_2^{†}{\ast }_N{\mathcal{R}}_2+{\mathcal{U}}_2{\ast }_N{\mathcal{Z}}_2{\ast }_N{\mathcal{U}}_2^T.\hfill \end{array}$$

Thus, ${\left(\mathcal{X}{\ast }_N\mathcal{A}\right)}^T=\mathcal{X}{\ast }_N\mathcal{A}$ is proved.

Now we have proved that $\mathcal{X}$ satisfies all four conditions in Definition 1. Therefore, $\mathcal{X}$ is the M–P inverse of $\mathcal{A}$. □

4. The Minimum-Norm Least Squares Solutions

In the first part of this section, we consider the minimum-norm least squares solution of tensor system (5). The minimum-norm least squares solution of tensor Equation (4) will be given as a special case. For the coefficient tensors in system (5), we fix some notation as follows:

$$\begin{array}{ccc}\hfill & {\mathcal{A}}_1={\mathcal{A}}_{11}+{\mathcal{A}}_{12}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{A}}_2={\mathcal{A}}_{21}+{\mathcal{A}}_{22}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{C}}_1={\mathcal{C}}_{11}+{\mathcal{C}}_{12}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{C}}_2={\mathcal{C}}_{21}+{\mathcal{C}}_{22}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{E}}_1={\mathcal{E}}_{11}+{\mathcal{E}}_{12}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{E}}_2={\mathcal{E}}_{21}+{\mathcal{E}}_{22}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{F}}_1={\mathcal{F}}_{11}+{\mathcal{F}}_{12}j\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_N},\hfill & {\mathcal{F}}_2={\mathcal{F}}_{21}+{\mathcal{F}}_{22}j\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times J_1\times \cdots \times J_N},\hfill \\ \hfill & \mathcal{D}={\mathcal{D}}_1+{\mathcal{D}}_{2}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N}.\hfill \end{array}$$

(12)

Moreover, we set

$$\begin{array}{c}\mathcal{P}=\left[\begin{array}{cc}{\mathcal{A}}_{11}•f^T\left({\mathcal{I}}_1\right)& {\mathcal{A}}_{12}•f^{\ast }\left({\mathcal{I}}_{1}j\right)\\ 0& 0\\ {\mathcal{E}}_{11}•f^T\left({\mathcal{F}}_1\right)& {\mathcal{E}}_{12}•f^{\ast }\left({\mathcal{F}}_{1}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\\ \mathcal{Q}=\left[\begin{array}{cc}0& 0\\ {\mathcal{A}}_{11}•f^T\left({\mathcal{I}}_1\right)& {\mathcal{A}}_{12}•f^{\ast }\left({\mathcal{I}}_{1}j\right)\\ {\mathcal{E}}_{11}•f^T\left({\mathcal{F}}_2\right)& {\mathcal{E}}_{12}•f^{\ast }\left({\mathcal{F}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\\ \mathcal{Y}=\left[\begin{array}{cc}0& 0\\ {\mathcal{A}}_{21}•f^T\left({\mathcal{I}}_1\right)& {\mathcal{A}}_{22}•f^{\ast }\left({\mathcal{I}}_{1}j\right)\\ {\mathcal{E}}_{21}•f^T\left({\mathcal{F}}_2\right)& {\mathcal{E}}_{22}•f^{\ast }\left({\mathcal{F}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\end{array}$$

$$\begin{array}{c}{\mathcal{P}}_1=Re\mathcal{P},{\mathcal{P}}_2=Im\mathcal{P},{\mathcal{Q}}_1=Re\mathcal{Q},{\mathcal{Q}}_2=Im\mathcal{Q},{\mathcal{Y}}_1=Re\mathcal{Y},{\mathcal{Y}}_2=Im\mathcal{Y},\end{array}$$

$$\begin{array}{c}{\mathcal{T}}_1=\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{Q}}_1& {\mathcal{Y}}_1\end{array}\right],{\mathcal{T}}_2=\left[\begin{array}{ccc}{\mathcal{P}}_2& {\mathcal{Q}}_2& {\mathcal{Y}}_2\end{array}\right],\\ \mathcal{G}=\left[\begin{array}{c}vec\left(Re{\mathsf{\Phi }}_{{\mathcal{C}}_1}\right)\\ vec\left(Re{\mathsf{\Phi }}_{{\mathcal{C}}_2}\right)\\ vec\left(Re{\mathsf{\Phi }}_{\mathcal{D}}\right)\\ vec\left(Im{\mathsf{\Phi }}_{{\mathcal{C}}_1}\right)\\ vec\left(Im{\mathsf{\Phi }}_{{\mathcal{C}}_2}\right)\\ vec\left(Im{\mathsf{\Phi }}_{\mathcal{D}}\right)\end{array}\right].\end{array}$$

(13)

From Lemma 2, let

$$\mathcal{H}={\mathcal{R}}^{†}+\left(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\right){\ast }_N\mathcal{V}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\ast }_N{\left({\mathcal{T}}_1^{†}\right)}^{\mathrm{T}}{\ast }_N\left(\mathcal{I}-{\mathcal{T}}_2^{\mathrm{T}}{\ast }_N{\mathcal{R}}^{†}\right).$$

(14)

In general, tensor system (5) may have many solutions or no solution at all. Thus, it is useful to find the minimum-norm least squares solution $[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L]$ of (5) such that

$${∥\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]∥}_F^2=\underset{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\in H_L}{min}\left\{{∥{\mathcal{X}}_1∥}_F^2+{∥{\mathcal{X}}_2∥}_F^2+{∥{\mathcal{X}}_3∥}_F^2\right\},$$

where

$$\begin{array}{cc}\hfill H_L=& \left\{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\mid {∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1^L-{\mathcal{C}}_1∥}_F^2+{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2^L+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3^L-{\mathcal{C}}_2∥}_F^2\right.\hfill \\ \hfill & +{∥{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1^L{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2^L{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3^L{\ast }_N{\mathcal{F}}_2-\mathcal{D}∥}_F^2\hfill \\ \hfill & =\underset{{\mathcal{X}}_i}{min}\left\{{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1-{\mathcal{C}}_1∥}_F^2+{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3-{\mathcal{C}}_2∥}_F^2\right.\hfill \\ \hfill & \left.\left.+{∥{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2-\mathcal{D}∥}_F^2\right\}\right\}.\hfill \end{array}$$

(15)

Theorem 1.

With the above notation, the minimum-norm least squares solution $[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L]$ of (5) is determined by

$$\begin{array}{cc}\hfill H_L=\{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\mid \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=& \left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†},{\mathcal{H}}^T\right]{\ast }_N\mathcal{G}\hfill \\ \hfill & \left.+\left(\mathcal{I}-{\mathcal{T}}_1^{†}{\ast }_N{\mathcal{T}}_1-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N\mathcal{W}\right\},\hfill \end{array}$$

where $\mathcal{W}$ is an arbitrary real tensor with appropriate dimension.

Proof of Theorem 1.

The following calculations are based on Lemmas 1 and 2, and Proposition 3.

$$\begin{array}{cc}\hfill \underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \{\parallel {\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1-{\mathcal{C}}_1{\parallel }_F^2+\parallel {\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3-{\mathcal{C}}_2{\parallel }_F^2\hfill \\ \hfill & +\parallel {\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2-\mathcal{D}{\parallel }_F^2\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \{{∥{\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1-{\mathcal{C}}_1}∥}_F^2+{∥{\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3-{\mathcal{C}}_2}∥}_F^2\hfill \\ \hfill & +{∥{\mathsf{\Phi }}_{{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2-\mathcal{D}}∥}_F^2\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \{{∥vec\left({\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1-{\mathcal{C}}_1}\right)∥}_F^2+{∥vec\left({\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3-{\mathcal{C}}_2}\right)∥}_F^2\hfill \\ \hfill & +{∥vec\left({\mathsf{\Phi }}_{{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2-\mathcal{D}}\right)∥}_F^2\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \{{∥vec\left({\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1}\right)-vec\left({\mathsf{\Phi }}_{{\mathcal{C}}_1}\right)∥}_F^2+{∥vec\left({\mathsf{\Phi }}_{{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2}\right)+vec\left({\mathsf{\Phi }}_{{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3}\right)-vec\left({\mathsf{\Phi }}_{{\mathcal{C}}_2}\right)∥}_F^2\hfill \\ \hfill & +{∥vec\left({\mathsf{\Phi }}_{{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1}\right)+vec\left({\mathsf{\Phi }}_{{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2}\right)+vec\left({\mathsf{\Phi }}_{{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2}\right)-vec\left({\mathsf{\Phi }}_{\mathcal{D}}\right)∥}_F^2\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \left\{{∥\left[\begin{array}{cc}{\mathcal{A}}_{11}•f^T\left(\mathcal{I}\right)& {\mathcal{A}}_{12}•f^{\ast }\left(\mathcal{I}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)vec\left({\mathsf{\Phi }}_{{\mathcal{C}}_2}\right)∥}_F^2\right.\hfill \\ \hfill & +∥\left[\begin{array}{cc}{\mathcal{A}}_{11}•f^T\left(\mathcal{I}\right)& {\mathcal{A}}_{21}•f^{\ast }\left(\mathcal{I}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\hfill \\ \hfill & +{\left[\begin{array}{cc}{\mathcal{A}}_{21}•f^T\left(\mathcal{I}\right)& {\mathcal{A}}_{22}•f^{\ast }\left(\mathcal{I}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_3}\right)-vec\left({\mathsf{\Phi }}_{{\mathcal{C}}_2}\right)∥}_F^2\hfill \\ \hfill & +∥\left[\begin{array}{cc}{\mathcal{E}}_{11}•f^T\left({\mathcal{F}}_1\right)& {\mathcal{E}}_{12}•f^{\ast }\left({\mathcal{F}}_{1}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\hfill \\ \hfill & +\left[\begin{array}{cc}{\mathcal{E}}_{11}•f^T\left({\mathcal{F}}_2\right)& {\mathcal{E}}_{21}•f^{\ast }\left({\mathcal{F}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\hfill \\ \hfill & \left.+{\left[\begin{array}{cc}{\mathcal{E}}_{21}•f^T\left({\mathcal{F}}_2\right)& {\mathcal{E}}_{22}•f^{\ast }\left({\mathcal{F}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_3}\right)-vec\left({\mathsf{\Phi }}_{\mathcal{D}}\right)∥}_F^2\right\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \left\{{∥\mathcal{P}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)+\mathcal{Q}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_2}\right)+\mathcal{Y}{\ast }_{N}vec\left(\overrightarrow{{\mathcal{X}}_3}\right)-\mathcal{G}∥}_F^2\right\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \left\{{∥\left[\begin{array}{ccc}{\mathcal{P}}_1& {\mathcal{Q}}_1& {\mathcal{Y}}_1\\ {\mathcal{P}}_2& {\mathcal{Q}}_2& {\mathcal{Y}}_2\end{array}\right]{\ast }_N\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]-\mathcal{G}∥}_F^2\right\}\hfill \\ \hfill =\underset{{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3}{min}& \left\{{∥\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]{\ast }_N\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]-\mathcal{G}∥}_F^2\right\}.\hfill \end{array}$$

Using the results of the real tensor equation in [22], we have

$$\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]={\left[\begin{array}{c}{\mathcal{T}}_1\hfill \\ {\mathcal{T}}_2\hfill \end{array}\right]}^{†}{\ast }_N\mathcal{G}+\left(\mathcal{I}-{\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]}^{†}{\ast }_N\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]\right){\ast }_N\mathcal{W},$$

(16)

where $\mathcal{W}$ is an arbitrary real tensor over $\mathbb{R}$. A simple calculation gives

$$\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†},{\mathcal{H}}^T\right]{\ast }_N\mathcal{G}+\left(\mathcal{I}-{\mathcal{T}}_1^{†}{\ast }_N{\mathcal{T}}_1-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N\mathcal{W}.$$

(17)

□

Using Lemma 2, we have the following solvability condition.

Corollary 1.

The quaternion tensor system (5) is solvable if and only if

$$\left(\mathcal{I}-\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]\ast {\left[\begin{array}{c}{\mathcal{T}}_1\\ {\mathcal{T}}_2\end{array}\right]}^{†}\right){\ast }_N\mathcal{G}=\left[\begin{array}{cc}{\mathcal{S}}_{11}& {\mathcal{S}}_{12}\\ {\mathcal{S}}_{12}^T& {\mathcal{S}}_{22}\end{array}\right]{\ast }_N\mathcal{G}=0,$$

(18)

where

$$\begin{array}{c}{\mathcal{S}}_{11}=\mathcal{I}-{\mathcal{T}}_1{\ast }_N{\mathcal{T}}_1^{†}+{\mathcal{T}}_1^{T†}{\ast }_N{\mathcal{T}}_2^T{\ast }_N\mathcal{Z}{\ast }_N\left(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\right),\hfill \\ {\mathcal{S}}_{12}=-{\mathcal{T}}_1^{T†}{\ast }_N{\mathcal{T}}_2^T{\ast }_N\mathcal{Z}{\ast }_N\left(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\right),\hfill \\ {\mathcal{S}}_{22}=\mathcal{Z}{\ast }_N\left(\mathcal{I}-{\mathcal{R}}^{†}{\ast }_N\mathcal{R}\right).\hfill \end{array}$$

In this case, the solution satisfies

$$\begin{array}{cc}\hfill H_G=\{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\mid \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=& \left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†},{\mathcal{H}}^T\right]{\ast }_N\mathcal{G}\hfill \\ \hfill & \left.+\left(\mathcal{I}-{\mathcal{T}}_1^{†}{\ast }_N{\mathcal{T}}_1-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N\mathcal{W}\right\},\hfill \end{array}$$

where $\mathcal{W}$ is an arbitrary real tensor with appropriate dimension.

Furthermore, the uniqueness of the solution of (5) can be obtained as follows.

Corollary 2.

The tensor system (5) has a unique solution $\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]\in H_L$, which satisfies

$$\begin{array}{c}\hfill \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3^L}\right)\end{array}\right]=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\mathcal{H}}^{\mathrm{T}}\right]{\ast }_N\mathcal{G}.\end{array}$$

(19)

Proof of Corollary 2.

The solution set ${\mathbb{H}}_L$ is a nonempty and closed convex set. Thus, it must have a unique solution $\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]\in H_L$. It follows that

$$\begin{array}{cc}& {∥\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]∥}_F^2={∥{\mathcal{X}}_1^L∥}_F^2+{∥{\mathcal{X}}_2^L∥}_F^2+{∥{\mathcal{X}}_3^L∥}_F^2\hfill \\ \hfill =& \underset{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\in H_L}{min}\left\{{∥{\mathcal{X}}_1∥}_F^2+{∥{\mathcal{X}}_2∥}_F^2+{∥{\mathcal{X}}_3∥}_F^2\right\}\hfill \\ \hfill =& \underset{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\in H_L}{min}\left\{{∥vec\left({\overrightarrow{\mathcal{X}}}_1\right)∥}_F^2+{∥vec\left({\overrightarrow{\mathcal{X}}}_2\right)∥}_F^2+{∥vec\left({\overrightarrow{\mathcal{X}}}_3\right)∥}_F^2\right\}\hfill \\ \hfill =& \underset{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\in H_L}{min}\left\{{∥\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]∥}_F^2\right\}.\hfill \end{array}$$

Hence, when $\mathcal{W}=0$, $\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]$ is the minimum-norm least squares solution of system (5), which satisfies

$$\begin{array}{c}\hfill \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3^L}\right)\end{array}\right]=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\mathcal{H}}^{\mathrm{T}}\right]{\ast }_N\mathcal{G}.\end{array}$$

□

As a special case, we give the solution of tensor Equation (4). For the coefficient tensors in (4), let

$$\begin{array}{cc}\hfill & {\mathcal{A}}_1={\mathcal{A}}_{11}+{\mathcal{A}}_{12}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},{\mathcal{A}}_2={\mathcal{A}}_{21}+{\mathcal{A}}_{22}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{B}}_1={\mathcal{B}}_{11}+{\mathcal{B}}_{12}j\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_N},{\mathcal{B}}_2={\mathcal{B}}_{21}+{\mathcal{B}}_{22}j\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times J_1\times \cdots \times J_N},\hfill \\ \hfill & \mathcal{C}={\mathcal{C}}_1+{\mathcal{C}}_{2}j\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times J_1\times \cdots \times J_N},\hfill \end{array}$$

(20)

and set

$$\begin{array}{c}{\mathcal{P}}_{01}=\left[\begin{array}{cc}{\mathcal{A}}_{11}•f^T\left({\mathcal{B}}_1\right)& {\mathcal{A}}_{12}•f^{\ast }\left({\mathcal{B}}_{1}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\\ {\mathcal{Q}}_{01}=\left[\begin{array}{cc}{\mathcal{A}}_{11}•f^T\left({\mathcal{B}}_2\right)& {\mathcal{A}}_{12}•f^{\ast }\left({\mathcal{B}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\\ {\mathcal{Y}}_{01}=\left[\begin{array}{cc}{\mathcal{A}}_{21}•f^T\left({\mathcal{B}}_2\right)& {\mathcal{A}}_{22}•f^{\ast }\left({\mathcal{B}}_{2}j\right)\end{array}\right]{\ast }_N{K}_{4{\gamma }^N},\end{array}$$

$$\begin{array}{c}{\mathcal{T}}_{01}=\left[\begin{array}{ccc}Re{\mathcal{P}}_{01}& Re{\mathcal{Q}}_{01}& Re{\mathcal{Y}}_{01}\end{array}\right],{\mathcal{T}}_{02}=\left[\begin{array}{ccc}Im{\mathcal{P}}_{01}& Im{\mathcal{Q}}_{01}& Im{\mathcal{Y}}_{01}\end{array}\right],\end{array}$$

(21)

$$\begin{array}{c}{\mathcal{R}}_{01}=\left(\mathcal{I}-{\mathcal{T}}_{01}^{†}{\ast }_N{\mathcal{T}}_{01}\right){\ast }_N{\mathcal{T}}_{02}^{\mathrm{T}},\hfill \\ {\mathcal{Z}}_{01}={\left(\mathcal{I}+\left(\mathcal{I}-{\mathcal{R}}_{01}^{†}{\ast }_N{\mathcal{R}}_{01}\right){\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†}{\ast }_N{\left({\mathcal{T}}_{01}^{†}\right)}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_{02}^{\mathrm{T}}{\ast }_N\left(\mathcal{I}-{\mathcal{R}}_{01}^{†}{\ast }_N{\mathcal{R}}_{01}\right)\right)}^{-1},\hfill \\ {\mathcal{H}}_{01}={\mathcal{R}}_{01}^{†}+\left(\mathcal{I}-{\mathcal{R}}_{01}^{†}{\ast }_N{\mathcal{R}}_{01}\right){\ast }_N{\mathcal{Z}}_{01}{\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†}{\ast }_N{\left({\mathcal{T}}_{01}^{†}\right)}^{\mathrm{T}}{\ast }_N\left(\mathcal{I}-{\mathcal{T}}_{02}^{\mathrm{T}}{\ast }_N{\mathcal{R}}_{01}^{†}\right).\hfill \end{array}$$

Theorem 2.

With the above notation, the minimum-norm least squares solution $[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L]$ of tensor Equation (4) is determined by

$$\begin{array}{cc}\hfill H_{L1}=& \{\left[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3\right]\mid {∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{B}}_1+{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{B}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{B}}_2-\mathcal{C}∥}_F^2\hfill \\ \hfill & \left.=\underset{\widetilde{{\mathcal{X}}_i}}{min}{∥{\mathcal{A}}_1{\ast }_N\widetilde{{\mathcal{X}}_1}{\ast }_N{\mathcal{B}}_1+{\mathcal{A}}_1{\ast }_N\widetilde{{\mathcal{X}}_2}{\ast }_N{\mathcal{B}}_2+{\mathcal{A}}_2{\ast }_N\widetilde{{\mathcal{X}}_3}{\ast }_N{\mathcal{B}}_2-\mathcal{C}∥}_F^2\right\}\hfill \\ \hfill =& \{[{\mathcal{X}}_1,{\mathcal{X}}_2,{\mathcal{X}}_3]\mid \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=\left[{\mathcal{T}}_{01}^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†},{\mathcal{H}}_{01}^T\right]{\ast }_{N}vec\left({\mathsf{\Phi }}_{\mathcal{C}}\right)\hfill \\ \hfill & +\left(\mathcal{I}-{\mathcal{T}}_{01}^{†}{\ast }_N{\mathcal{T}}_{01}-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N{\mathcal{W}}_1\},\hfill \end{array}$$

(22)

where ${\mathcal{W}}_1$ is an arbitrary real tensor with appropriate dimension.

The minimum-norm least squares solution $\left[{\mathcal{X}}_1^L,{\mathcal{X}}_2^L,{\mathcal{X}}_3^L\right]\in H_{L1}$ satisfies

$$\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2^L}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3^L}\right)\end{array}\right]=\left[{\mathcal{T}}_{01}^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†},{\mathcal{H}}_{01}^T\right]{\ast }_{N}vec\left({\mathsf{\Phi }}_{\mathcal{C}}\right).$$

As applications of the minimum-norm least squares solutions of (5) and (4), in the second part of this section we consider the minimum-norm least squares reducible solutions of system (6) and Equation (3). We write the reducible solution $\mathcal{X}\in {\mathbb{H}}^{L_1\times \cdots \times L_N\times L_1\times \cdots \times L_N}$ in the following form:

$$\mathcal{X}=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ 0& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1},$$

(23)

where $\mathcal{K}$ is a permutation tensor and ${\mathcal{X}}_1\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times K_1\times \cdots \times K_N}$, ${\mathcal{X}}_2\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times Q_1\times \cdots \times Q_N}$, ${\mathcal{X}}_3\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times Q_1\times \cdots \times Q_N}$, $K_i+Q_i=L_i,i=1,\cdots ,N$. Next, for the coefficient tensors in (6), we denote their products in the following block tensors:

$$\begin{array}{cccc}\hfill & \mathcal{A}{\ast }_N\mathcal{K}=\left[\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{A}}_2\end{array}\right],\hfill & {\mathcal{A}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{A}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & \mathcal{C}{\ast }_N\mathcal{K}=\left[\begin{array}{cc}{\mathcal{C}}_1& {\mathcal{C}}_2\end{array}\right],\hfill & {\mathcal{C}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{C}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & \mathcal{E}{\ast }_N\mathcal{K}=\left[\begin{array}{cc}{\mathcal{E}}_1& {\mathcal{E}}_2\end{array}\right],\hfill & {\mathcal{E}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},\hfill & {\mathcal{E}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{K}}^{-1}{\ast }_N\mathcal{F}=\left[\begin{array}{c}{\mathcal{F}}_1\\ {\mathcal{F}}_2\end{array}\right],\hfill & {\mathcal{F}}_1\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_N},\hfill & {\mathcal{F}}_2\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times J_1\times \cdots \times J_N}.\hfill \end{array}$$

(24)

Substituting (23) and (24) into tensor system (6), we compare the corresponding tensors on both sides, and thus obtain a system in the form of (5). Therefore, we can use our previous results on (5) to find the minimum-norm least squares reducible solution ${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R& {\mathcal{X}}_2^R\\ O& {\mathcal{X}}_3^R\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}$ such that

$${∥{\mathcal{X}}^R∥}_F^2=\underset{\mathcal{X}\in H_R}{min}\left\{{\parallel \mathcal{X}\parallel }_F^2\right\}=\underset{X\in H_R}{min}\left\{{∥{\mathcal{X}}_1∥}_F^2+{∥{\mathcal{X}}_2∥}_F^2+{∥{\mathcal{X}}_3∥}_F^2\right\},$$

where

$$\begin{array}{cc}\hfill H_R=& \left\{\left.\mathcal{X}=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ O& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\right|{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1^R-{\mathcal{C}}_1∥}_F^2+{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2^R+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3^R-{\mathcal{C}}_2∥}_F^2\right.\hfill \\ & +{∥{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1^R{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2^R{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3^R{\ast }_N{\mathcal{F}}_2-\mathcal{D}∥}_F^2\hfill \\ & =\underset{{\mathcal{X}}_i}{min}\left\{{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1-{\mathcal{C}}_1∥}_F^2+{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3-{\mathcal{C}}_2∥}_F^2\right.\hfill \\ & +\left.\left.{∥{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{F}}_1+{\mathcal{E}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{F}}_2+{\mathcal{E}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{F}}_2-\mathcal{D}∥}_F^2\right\}\right\}.\hfill \end{array}$$

Theorem 3.

With above notation for coefficient tensors $\mathcal{A}$, $\mathcal{C}$, $\mathcal{E}$, $\mathcal{F}$, $\mathcal{D}$ in (6), ${\mathcal{T}}_1$, ${\mathcal{T}}_2$, $\mathcal{G}$ in (13) and permutation tensor $\mathcal{K}$, we have

$$\begin{array}{cc}\hfill H_R=& \left\{\left.\mathcal{X}=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ O& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\right|\right.\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†},{\mathcal{H}}^T\right]{\ast }_N\mathcal{G}\hfill \\ & \left.+\left(\mathcal{I}-{\mathcal{T}}_1^{†}{\ast }_N{\mathcal{T}}_1-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N\mathcal{W}\right\},\hfill \end{array}$$

where $\mathcal{W}$ is an arbitrary real tensor with appropriate dimension. The tensor system (6) has the minimum-norm least squares reducible solution ${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R& {\mathcal{X}}_2^R\\ O& {\mathcal{X}}_3^R\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\in H_R$, which satisfies

$$\begin{array}{c}\hfill \left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1^R}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2^R}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3^R}\right)\end{array}\right]=\left[{\mathcal{T}}_1^{†}-{\mathcal{H}}^{\mathrm{T}}{\ast }_N{\mathcal{T}}_2{\ast }_N{\mathcal{T}}_1^{†}{\mathcal{H}}^{\mathrm{T}}\right]{\ast }_N\mathcal{G}.\end{array}$$

In particular, we can obtain the minimum-norm least squares reducible solution of tensor Equation (3). Let

$$\begin{array}{ccc}\hfill & \mathcal{A}{\ast }_N\mathcal{K}=\left[\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{A}}_2\end{array}\right],\hfill & {\mathcal{A}}_1\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times K_1\times \cdots \times K_N},{\mathcal{A}}_2\in {\mathbb{H}}^{I_1\times \cdots \times I_N\times Q_1\times \cdots \times Q_N},\hfill \\ \hfill & {\mathcal{K}}^{-1}{\ast }_N\mathcal{B}=\left[\begin{array}{c}{\mathcal{B}}_1\\ {\mathcal{B}}_2\end{array}\right],\hfill & {\mathcal{B}}_1\in {\mathbb{H}}^{K_1\times \cdots \times K_N\times J_1\times \cdots \times J_N},{\mathcal{B}}_2\in {\mathbb{H}}^{Q_1\times \cdots \times Q_N\times J_1\times \cdots \times J_N}.\hfill \end{array}$$

(25)

Theorem 4.

With the above notation for coefficient tensors in Equation (3), and ${\mathcal{T}}_{01}$, ${\mathcal{T}}_{02}$, and $vec\left({\mathsf{\Phi }}_{\mathcal{C}}\right)$ in (21), we have

$$\begin{array}{cc}\hfill H_{R1}=& \left.\left\{\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ O& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\right.\right|{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1^R{\ast }_N{\mathcal{B}}_1+{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2^R{\ast }_N{\mathcal{B}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3^R{\ast }_N{\mathcal{B}}_2-\mathcal{C}∥}_F^2\hfill \\ & =\underset{{\mathcal{X}}_i}{min}{∥{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_1{\ast }_N{\mathcal{B}}_1+{\mathcal{A}}_1{\ast }_N{\mathcal{X}}_2{\ast }_N{\mathcal{B}}_2+{\mathcal{A}}_2{\ast }_N{\mathcal{X}}_3{\ast }_N{\mathcal{B}}_2-\mathcal{C}∥}_F^2\}\hfill \\ \hfill =& \left\{\left.\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ O& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\right|\right.\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3}\right)\end{array}\right]=\left[{\mathcal{T}}_{01}^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†},{\mathcal{H}}_{01}^T\right]{\ast }_{N}vec\left({\mathsf{\Phi }}_{\mathcal{C}}\right)\hfill \\ & \left.+\left(\mathcal{I}-{\mathcal{T}}_{01}^{†}{\ast }_N{\mathcal{T}}_{01}-\mathcal{R}{\ast }_N{\mathcal{R}}^{†}\right){\ast }_N{\mathcal{W}}_1\right\},\hfill \end{array}$$

where ${\mathcal{W}}_1$ is an arbitrary real tensor with appropriate dimension. The minimum-norm least squares reducible solution ${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R\hfill & {\mathcal{X}}_2^R\hfill \\ O\hfill & X_3^R\hfill \end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\in H_{R1}$ satisfies

$$\left[\begin{array}{c}vec\left(\overrightarrow{{\mathcal{X}}_1^R}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_2^R}\right)\\ vec\left(\overrightarrow{{\mathcal{X}}_3^R}\right)\end{array}\right]=\left[{\mathcal{T}}_{01}^{†}-{\mathcal{H}}^T{\ast }_N{\mathcal{T}}_{02}{\ast }_N{\mathcal{T}}_{01}^{†},{\mathcal{H}}_{01}^T\right]{\ast }_{N}vec\left({\mathsf{\Phi }}_{\mathcal{C}}\right).$$

5. Algorithm and Numerical Example

Based on the methods discussed in Section 4, we develop the following Algorithm 1 to solve tensor system (6). We also present a numerical example to show the accuracy of our method.

Algorithm 1: Finding the minimum-norm least squares reducible solution of system (6)

Input: the coefficient tensors $\mathcal{A},$ $\mathcal{C},$ $\mathcal{E},$ $\mathcal{F},$ and $\mathcal{D}$ in (6), and permutation tensor $\mathcal{K}\in {\mathbb{H}}^{L_1\times \cdots \times L_N\times L_1\times \cdots \times L_N}$. Compute the ${\mathcal{A}}_1$, ${\mathcal{A}}_2$, ${\mathcal{C}}_1$, ${\mathcal{C}}_2$, ${\mathcal{E}}_1$, ${\mathcal{E}}_2$, ${\mathcal{F}}_1$, ${\mathcal{F}}_2$, $\mathcal{D}$ in (24). Compute ${\mathcal{T}}_1,$ ${\mathcal{T}}_2,$ $\mathcal{R},$ $\mathcal{Z},$ $\mathcal{H},$ ${\mathcal{S}}_{11},$ ${\mathcal{S}}_{12},$ ${\mathcal{S}}_{22}$, and $\mathcal{G}$, which are defined in Section 3 and Section 4. If (18) holds, we compute $\left[{\mathcal{X}}_1^R,{\mathcal{X}}_2^R,{\mathcal{X}}_3^R\right]\in H_G$ and obtain the least squares reducible solution ${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R& {\mathcal{X}}_2^R\\ O& {\mathcal{X}}_3^R\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}.$ Otherwise, go to next step. Compute $\left[{\mathcal{X}}_1^R,{\mathcal{X}}_2^R,{\mathcal{X}}_3^R\right]\in H_L$. Then we obtain the least squares reducible solution ${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R& {\mathcal{X}}_2^R\\ O& {\mathcal{X}}_3^R\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}.$

We implemented the above algorithm in MATLAB. The codes are available upon request. The following example shows that our algorithm works well.

Example 1.

Let $I_1=I_2=J_1=J_2=Q_1=Q_2=K_1=K_2=2$. Given tensor $\mathcal{A}$, $\mathcal{E}$, $\mathcal{F}$ in system (6) and permutation tensor $\mathcal{K}\in {\mathbb{H}}^{4\times 4\times 4\times 4}$:

$$\begin{array}{cc}\mathcal{A}(:,:,1,1)=\left[\begin{array}{cc}\mathbf{j}+\mathbf{k}& \mathbf{i}+\mathbf{j}\\ 0& \mathbf{i}+\mathbf{j}\end{array}\right],& \mathcal{A}(:,:,2,1)=\left[\begin{array}{cc}1+\mathbf{i}& \mathbf{i}\\ 1& 1+\mathbf{j}\end{array}\right],\\ \mathcal{A}(:,:,1,2)=\left[\begin{array}{cc}\mathbf{j}& \mathbf{i}+\mathbf{j}\\ \mathbf{k}& \mathbf{i}+\mathbf{j}+\mathbf{k}\end{array}\right],& \mathcal{A}(:,:,2,2)=\left[\begin{array}{cc}\mathbf{j}+\mathbf{k}& \mathbf{i}+\mathbf{j}\\ 1+\mathbf{j}+\mathbf{k}& 1+\mathbf{j}\end{array}\right],\\ \mathcal{A}(:,:,1,3)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{j}& 1+\mathbf{i}+\mathbf{j}+\mathbf{k}\\ 0& \mathbf{i}+\mathbf{k}\end{array}\right],& \mathcal{A}(:,:,2,3)=\left[\begin{array}{cc}\mathbf{k}& 1+\mathbf{i}\\ 1+\mathbf{j}& 1+\mathbf{j}\end{array}\right],\\ \mathcal{A}(:,:,1,4)=\left[\begin{array}{cc}1+\mathbf{i}+\mathbf{j}+\mathbf{k}& 1+\mathbf{i}+\mathbf{j}\\ 1+\mathbf{i}+\mathbf{j}& \mathbf{k}\end{array}\right],& \mathcal{A}(:,:,2,4)=\left[\begin{array}{cc}1+\mathbf{j}& \mathbf{i}+\mathbf{j}\\ 1+\mathbf{j}+\mathbf{k}& \mathbf{i}\end{array}\right],\\ \mathcal{A}(:,:,3,1)=\mathcal{A}(:,:,4,1)=0,& \mathcal{A}(:,:,3,2)=\mathcal{A}(:,:,4,2)=0,\\ \mathcal{A}(:,:,3,3)=\mathcal{A}(:,:,4,3)=0,& \mathcal{A}(:,:,3,4)=\mathcal{A}(:,:,4,4)=0,\\ \mathcal{E}(:,:,1,1)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{j}& \mathbf{k}\\ 1+\mathbf{k}& \mathbf{i}+\mathbf{j}+\mathbf{k}\end{array}\right],& \mathcal{E}(:,:,2,1)=\left[\begin{array}{cc}1+\mathbf{i}+\mathbf{j}& \mathbf{j}\\ 1+\mathbf{i}+\mathbf{j}& 1+\mathbf{j}+\mathbf{k}\end{array}\right],\\ \mathcal{E}(:,:,1,2)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{j}& \mathbf{i}+\mathbf{j}\\ \mathbf{i}+\mathbf{k}& 1+\mathbf{k}\end{array}\right],& \mathcal{E}(:,:,2,2)=\left[\begin{array}{cc}1+\mathbf{j}& 0\\ 1+\mathbf{i}+\mathbf{j}& 1+\mathbf{i}+\mathbf{k}\end{array}\right],\\ \mathcal{E}(:,:,1,3)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{k}& 1+\mathbf{j}\\ \mathbf{i}+\mathbf{k}& \mathbf{j}+\mathbf{k}\end{array}\right],& \mathcal{E}(:,:,2,3)=\left[\begin{array}{cc}1+\mathbf{i}& 0\\ \mathbf{i}& 1+\mathbf{i}+\mathbf{k}\end{array}\right],\\ \mathcal{E}(:,:,1,4)=\left[\begin{array}{cc}1& 1+\mathbf{i}+\mathbf{j}+\mathbf{k}\\ 1+\mathbf{i}+\mathbf{j}+\mathbf{k}& 1+\mathbf{k}\end{array}\right],& \mathcal{E}(:,:,2,4)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{k}& 1+\mathbf{i}+\mathbf{j}+\mathbf{k}\\ 1+\mathbf{j}+\mathbf{k}& 1+\mathbf{i}+\mathbf{j}\end{array}\right],\\ \mathcal{E}(:,:,3,1)=\mathcal{E}(:,:,4,1)=0,& \mathcal{E}(:,:,3,2)=\mathcal{E}(:,:,4,2)=0,\\ \mathcal{E}(:,:,3,3)=\mathcal{E}(:,:,4,3)=0,& \mathcal{E}(:,:,3,4)=\mathcal{E}(:,:,4,4)=0,\\ \mathcal{F}(:,:,1,1)=\left[\begin{array}{cccc}1& 0& 0& 0\\ 1& 1& 0& 0\\ 0& 0& \mathbf{j}& 0\\ 0& 0& 1+\mathbf{j}& 1\end{array}\right],& \mathcal{F}(:,:,2,1)\left[\begin{array}{cccc}1+\mathbf{j}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& \mathbf{j}\\ 0& 0& 1& 1+\mathbf{j}\end{array}\right],\\ \mathcal{F}(:,:,1,2)=\left[\begin{array}{cccc}\mathbf{j}& 0& 0& 0\\ \mathbf{j}& 1& 0& 0\\ 0& 0& 1& 1\\ 0& 0& \mathbf{j}& 1+\mathbf{j}\end{array}\right],& \mathcal{F}(:,:,2,2)\left[\begin{array}{cccc}1& \mathbf{j}& 0& 0\\ 1& 1+\mathbf{j}& 0& 0\\ 0& 0& 1& 1\\ 0& 0& \mathbf{j}& \mathbf{j}\end{array}\right],\end{array}$$

and

$$\begin{array}{cc}\mathcal{K}(:,:,1,1)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,2,1)=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,3,1)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,4,1)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,1,2)=\left[\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,2,2)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,3,2)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,4,2)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,1,3)=\left[\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,2,3)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\end{array}$$

$$\begin{array}{cc}\mathcal{K}(:,:,3,3)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,4,3)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,1,4)=\left[\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,2,4)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],\\ \mathcal{K}(:,:,3,4)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right],& \mathcal{K}(:,:,4,4)=\left[\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right].\end{array}$$

We set the solution as follows:

$$\mathcal{X}=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1& {\mathcal{X}}_2\\ 0& {\mathcal{X}}_3\end{array}\right]{\ast }_N{\mathcal{K}}^{-1},$$

where

$$\begin{array}{cc}{\mathcal{X}}_1(:,:,1,1)=\left[\begin{array}{cc}\mathbf{j}+\mathbf{k}& 1\\ 0& \mathbf{i}\end{array}\right],& {\mathcal{X}}_1(:,:,2,1)=\left[\begin{array}{cc}1+\mathbf{j}& 1+\mathbf{k}\\ 0& 1+\mathbf{i}+\mathbf{k}\end{array}\right],\\ {\mathcal{X}}_1(:,:,1,2)=\left[\begin{array}{cc}1+\mathbf{i}+\mathbf{k}& \mathbf{i}+\mathbf{j}\\ 1& \mathbf{k}\end{array}\right],& {\mathcal{X}}_1(:,:,2,2)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{k}& 1+\mathbf{i}\\ 1+\mathbf{i}+\mathbf{k}& \mathbf{k}\end{array}\right],\\ {\mathcal{X}}_2(:,:,1,1)=\left[\begin{array}{cc}1& \mathbf{j}\\ \mathbf{i}& \mathbf{i}+\mathbf{j}\end{array}\right],& {\mathcal{X}}_2(:,:,2,1)=\left[\begin{array}{cc}1& \mathbf{i}\\ 0& 0\end{array}\right],\\ {\mathcal{X}}_2(:,:,1,2)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{k}& \mathbf{j}+\mathbf{k}\\ 1+\mathbf{j}+\mathbf{k}& 1+\mathbf{j}\end{array}\right],& {\mathcal{X}}_2(:,:,2,2)=\left[\begin{array}{cc}1+\mathbf{i}& 1+\mathbf{k}\\ \mathbf{j}& 1+\mathbf{i}+\mathbf{k}\end{array}\right],\\ {\mathcal{X}}_3(:,:,1,1)=\left[\begin{array}{cc}1+\mathbf{k}& \mathbf{i}\\ \mathbf{i}& 1\end{array}\right],& {\mathcal{X}}_3(:,:,2,1)=\left[\begin{array}{cc}1+\mathbf{i}+\mathbf{k}& \mathbf{i}+\mathbf{j}\\ 0& \mathbf{i}+\mathbf{j}+\mathbf{k}\end{array}\right],\\ {\mathcal{X}}_3(:,:,1,2)=\left[\begin{array}{cc}\mathbf{i}+\mathbf{j}+\mathbf{k}& \mathbf{k}\\ 1+\mathbf{k}& 1+\mathbf{j}\end{array}\right],& {\mathcal{X}}_3(:,:,2,2)=\left[\begin{array}{cc}\mathbf{j}+\mathbf{k}& \mathbf{i}+\mathbf{j}+\mathbf{k}\\ 1+\mathbf{j}+\mathbf{k}& \mathbf{i}\end{array}\right].\end{array}$$

$\mathcal{C}$and$\mathcal{D}$in (6) can be obtained from the above assumptions. Using the Algorithm 1, we have$N_1=2.1936\times {10}^{-11}$and the minimum-norm least squares reducible solution${\mathcal{X}}^R=\mathcal{K}{\ast }_N\left[\begin{array}{cc}{\mathcal{X}}_1^R& {\mathcal{X}}_2^R\\ 0& {\mathcal{X}}_3^R\end{array}\right]{\ast }_N{\mathcal{K}}^{-1}\in {\mathcal{H}}_R$such that

$$\begin{array}{c}\hfill \parallel {\mathcal{X}}^R{-\mathcal{X}\parallel }_F^2=7.6082\times {10}^{-12}.\end{array}$$

6. Conclusions

We obtained the minimum-norm least squares solution for system (5) by using the Moore–Penrose inverses of block tensors. In terms of applications, we derived the the minimum-norm least squares reducible solution for system (6). In addition, we used an algorithm and a numerical example to verify the main results of this paper. It is worth noting that the main results of (5) and (6) can be used for solving other quaternion tensor systems.