Some Properties of Reduced Biquaternion Tensors

Abstract

Compared to quaternions, reduced biquaternions satisfy the multiplication commutative rule and are widely employed in applications such as image processing, fuzzy recognition, image compression, and digital signal processing. However, there is little information available regarding reduced biquaternion tensors; thus, in this study, we investigate some properties of reduced biquaternion tensors. Firstly, we introduce the concept of reduced biquaternion tensors, propose the real and complex representations of reduced biquaternion tensors, and prove several fundamental theorems. Subsequently, we provide the definitions for the eigenvalues and eigentensors of reduced biquaternion tensors and present the Ger$\check{\mathrm{s}}$gorin theorem as it applies to their eigenvalues. Additionally, we establish the relationship between the reduced biquaternion tensor and its complex representation. Notably, the complex representation is a symmetry tensor, which significantly simplifies the process and complexity of solving for eigenvalues. Corresponding numerical examples are also provided in the paper. Furthermore, some special properties of eigenvalues of reduced biquaternion tensors are presented.

1. Introduction

The algebra of quaternions was introduced by mathematician William Hamilton in 1843 [1], and is noncommutative. The basic arithmetic properties of quaternions, including their real and complex representations, as well as the left and right eigenvalues and the corresponding eigenvectors of quaternion matrices, have been discussed for a long time. Notably, the Ger$\check{\mathrm{s}}$gorin theorem has been adapted for quaternion matrices ([2,3]). Additionally, Jia and Wang studied the quaternion matrix eigenvalue solving algorithms for applications in face recognition, image denoising, and image compression (e.g., [4,5,6,7]). Recent studies also presented findings on a system of Sylvester-type quaternion matrix equations (e.g., [8,9,10,11,12,13]). The theory of quaternions is constantly evolving, with algorithms being updated and refined, which play important roles in quantum mechanics, computer science, signal processing, color image processing, and more (e.g., [14,15,16,17]).

Reduced biquaternions were introduced by Segre in 1892 [18]. One of the differences from quaternions is their commutative property. This property generates great convenience and simplifies processes in both theoretical research and practical applications. Consequently, reduced biquaternions have significant applications in signal and image processing, control and system theory, and neural networks (e.g., [19,20,21,22,23]). It is important to study the theoretical structures and numerical calculations of reduced biquaternions. Scholars have conducted significant additional research on reduced biquaternions (e.g., [24,25,26,27,28,29,30]). For instance, three useful representations of reduced biquaternions (the $e_1-e_2$ form, matrix representation, and polar form) were discussed in [22]. Additionally, reduced quaternions can be expressed using two complex variables [31]. Research on the eigenvalues, corresponding eigenvectors, and singular value decomposition of the reduced biquaternion matries was detailed in [23].

A tensor represents a multidimensional array structure. Specifically, a vector is identified as a tensor of the first order, while a matrix is classified as a tensor of the second order. Tensors are higher-order generalizations of vectors and matrices, which effectively preserve the inherent structure and latent characteristics of multidimensional data. This characteristic is particularly valuable in fields like genetic chromosome series and medical data analysis (e.g., [32,33]), as well as video signal processing (e.g., [34,35]). Extensive research has been conducted on topics such as tensor completion, tensor recovery, and multilinear control systems (e.g., [36,37,38,39,40,41]).

To expand the concept of a quaternion matrix while completely preserving the structure of color channels, the concept of quaternion tensors was introduced in [42], and the Lanczos algorithm for eigenvalues of quaternion tensors and along with their relevant properties were proposed in [43]. However, there is no well-developed theory for the eigenvalues of reduced biquaternion tensors and their associated properties. Thus, extending reduced biquaternion matrices to tensors enriches the framework of quaternion algebra. The introduction of tensors brings new mathematical tools and methods, offering a more extensive approach to handling and analyzing quaternions.

Meanwhile, it is necessity to extend the theoretical knowledge of the reduced biquaternion tensors, which facilitates handling transformations and rotations in higher-dimensional spaces. This extension is crucial for accurately modeling and analyzing phenomena in four-dimensional and higher-dimensional spaces. In many practical scenarios, the complexity of data and systems often necessitates the use of higher-dimensional tensors rather than just matrices. This extension provides a more comprehensive approach to modeling and prediction in complex systems. Multilinear aspects treated through tensors are useful in the field of uncertainty and probability, proving beneficial in decision theory.

Motivated by the wide application of the reduced biquaternion tensors and in order to improve the theoretical development of the reduced biquaternion tensors, the structure of this paper is as follows: In Section 2, we present some fundamental theorems and related notations of reduced biquaternions. In Section 3, we investigate the real and complex representations of reduced biquaternion tensors and prove several fundamental theorems. In Section 4, we define the eigenvalues and corresponding eigentensors of the reduced biquaternion tensors, and present the Ger$\check{\mathrm{s}}$gorin theorem. Additionally, a numerical example is provided to demonstrate the relationship between the eigenvalues and eigentensors of reduced quaternion tensors and those of their complex representation tensors. This complex representation is a symmetry tensor, making the process of solving the eigenvalues and eigenvectors of the reduced biquaternion tensor more convenient. The detailed solution process is described in Algorithm 1.

Algorithm 1. Computing eigenvalues of reduced biquaternion tensors.

Input: Given a reduced biquaternion tensor ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$.

Output: Eigenvalues $\lambda$ and the corresponding eigentensors ${\mathcal{X}}_{RB}$. (1) Write the reduced biquaternion tensor ${\mathcal{A}}_{RB}$ as ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}$. (2) Compute ${\mathcal{B}}_1={\mathcal{A}}_1+{\mathcal{A}}_2,{\mathcal{B}}_2={\mathcal{A}}_1-{\mathcal{A}}_2$. (3) Compute the eignvalues and the corresponding eigentensors of complex tensors ${\mathcal{B}}_1,{\mathcal{B}}_2$. ${\lambda }_1,{\lambda }_2$ are the eigenvalues of ${\mathcal{B}}_1,{\mathcal{B}}_2$ and the corresponding eigentensors ${\mathcal{X}}_1,{\mathcal{X}}_2$ respectively. (4) Compute the eigenvalues of ${\mathcal{A}}_{RB}:\lambda ={\lambda }_1{e}_1+{\lambda }_2{e}_2$ and the eigentensors ${\mathcal{X}}_{RB}={\mathcal{X}}_1{e}_1+{\mathcal{X}}_2{e}_2$.

2. Notations and Preliminaries

In this section, we introduce some commonly used notations and present some foundational theorems about reduced biquaternion algebra to support this article.

In this paper, scalars are denoted using ordinary lowercase letters, where bold lowercase letters, $\mathbf{i},\mathbf{j},\mathbf{k}$ are used to represent imaginary units of reduced biquaternions. For matrices, which are tensors of the second order, we employ uppercase letters, A. Higher-order tensors (third-order or higher) are denoted using the Euler script, exemplified by $\mathcal{A}$. The symbols (${·)}^{-1}$, $\overline{(·)}$, ${(·)}^T$, and (${·)}^H$ correspond to inverse, conjugation, transpose, and conjugate transpose, respectively. The notations $\mathbb{R}$, $\mathbb{C}$, and ${\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n}$ are used to represent the real number field, the complex number field, and the sets of all nth order tensors with dimensions $I_1\times \cdots \times I_n$ defined over the reduced biquaternion algebra, respectively. $Z\left({\mathbb{H}}_{RB}\right)$ denotes the set of zero divisors of reduced biquaternion ${\mathbb{H}}_{RB}$.

In the context of tensors, a ‘square‘ tensor, denoted as ${\mathcal{D}}_{RB}=\left(d_{i_1\cdots i_n{i}_1\cdots i_n}\right)\in$${\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$, is termed a diagonal tensor if all its entries are zero except for $d_{i_1\cdots i_n{i}_1\cdots i_n}$. In the specific case, where $d_{i_1\cdots i_n{i}_1\cdots i_n}=1$, ${\mathcal{D}}_{RB}$ is defined as a unit tensor and represented by the symbol $\mathcal{I}$. Furthermore, a zero tensor, which conforms to the appropriate order, is simply represented by the numeral 0.

The set of reduced biquaternions is a 4-dimensional $\mathbb{R}$-algebra ([18,31]):

$${\mathbb{H}}_{RB}=\{q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k},q_0,q_1,q_2,q_3\in \mathbb{R}\},$$

(1)

where $\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy the following multiplication rules:

$${\mathbf{i}}^2={\mathbf{k}}^2=-1,{\mathbf{j}}^2=1,\mathbf{i}\mathbf{j}\mathbf{k}=-1,\mathbf{i}\mathbf{j}=\mathbf{j}\mathbf{i}=\mathbf{k},\mathbf{j}\mathbf{k}=\mathbf{k}\mathbf{j}=\mathbf{i},\mathbf{k}\mathbf{i}=\mathbf{i}\mathbf{k}=-\mathbf{j}.$$

(2)

In contrast to the conjugation of traditional quaternions, the conjugation of the reduced biquaternion $q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}$ is defined in three situations ([22,27]):

$$\begin{array}{c}\hfill {\overline{q}}^{\left(1\right)}=q_0-q_1\mathbf{i}+q_2\mathbf{j}-q_3\mathbf{k},\end{array}$$

(3)

$$\begin{array}{c}\hfill {\overline{q}}^{\left(2\right)}=q_0+q_1\mathbf{i}-q_2\mathbf{j}-q_3\mathbf{k},\end{array}$$

(4)

$$\begin{array}{c}\hfill {\overline{q}}^{\left(3\right)}=q_0-q_1\mathbf{i}-q_2\mathbf{j}+q_3\mathbf{k}.\end{array}$$

(5)

The norm of the reduced biquaternion $q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}$ is defined as

$$\begin{array}{cc}\hfill \left|\right|q\left|\right|& ={\left(q{\overline{q}}^{\left(1\right)}{\overline{q}}^{\left(2\right)}{\overline{q}}^{\left(3\right)}\right)}^{1/4}\hfill \\ & ={\left\{[{(q_0+q_2)}^2+{(q_1+q_3)}^2][{(q_0-q_2)}^2+{(q_1-q_3)}^2]\right\}}^{1/4}\ge 0.\hfill \end{array}$$

(6)

It is also worth noting that the algebra of reduced biquaternions is not a division algebra. There exist two special numbers: $e_1={\displaystyle \frac{1-\mathbf{j}}{2}},e_2={\displaystyle \frac{1+\mathbf{j}}{2}}$, which are both idempotent elements and divisors of zero. Any reduced biquaternion can be written as

$$q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}=(q_0+q_1\mathbf{i})+(q_2+q_3\mathbf{i})\mathbf{j}=a+b\mathbf{j}=(a+b)e_1+(a-b)e_2,$$

(7)

where $a=q_0+q_1\mathbf{i},b=q_2+q_3\mathbf{i},$ and $e_1^n=e_1^{n-1}=\cdots =e_1$, $e_2^n=e_2^{n-1}=\cdots =e_2$ [22].

Reduced biquaternion tensors are high-dimensional extensions of reduced biquaternion matrices. The following provides some characteristics of reduced biquaternion tensors.

The conjugate transpose of a reduced biquaternion tensor: ${\mathcal{A}}_{RB}=\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right)\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}$ is defined as

$${\mathcal{A}}_{RB}^{H_{\left(i\right)}}=\left({\overline{a}}_{j_1\cdots j_n{i}_1\cdots i_n}^{\left(i\right)}\right)\in {\mathbb{H}}_{RB}^{J_1\times \cdots \times J_n\times I_1\times \cdots \times I_n},i=1,2,3.$$

(8)

The Frobenius norm of ${\mathcal{A}}_{RB}$ is defined as

$${\left|\right|{\mathcal{A}}_{RB}\left|\right|}_F=\sqrt{{\sum }_{i_1=1}^{I_1}\cdots {\sum }_{j_n=1}^{J_n}\left|\right|a_{i_1\cdots j_n}{\left|\right|}^2}.$$

(9)

The operators on reduced buquaternion tensors can be defined in a usual way ([37,44]) as follows.

Definition 1.

For ${\mathcal{A}}_{RB}=\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right),{\mathcal{B}}_{RB}=\left(b_{i_1\cdots i_n{j}_1\cdots j_n}\right)\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},{\mathcal{C}}_{RB}=\left(c_{j_1\cdots j_n{k}_1\cdots k_n}\right)\in {\mathbb{H}}_{RB}^{J_1\times \cdots \times J_n\times K_1\times \cdots \times K_n}$, the addition of the reduced biquaternion tensors is defined by

$${({\mathcal{A}}_{RB}+{\mathcal{B}}_{RB})}_{i_1\cdots i_n{j}_1\cdots j_n}=(a_{i_1\cdots i_n{j}_1\cdots j_n}+b_{i_1\cdots i_n{j}_1\cdots j_n})\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},$$

(10)

and the scalar multiplication between a reduced biquaternion number $q\in {\mathbb{H}}_{RB}$ and a reduced biquaternion tensor is

$$q·{\mathcal{A}}_{RB}=(q·a_{i_1\cdots i_n{j}_1\cdots j_n})=(a_{i_1\cdots i_n{j}_1\cdots j_n}·q)={\mathcal{A}}_{RB}·q\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}.$$

(11)

The Einstein product of tensors ${\mathcal{A}}_{RB}$ and ${\mathcal{C}}_{RB}$ is defined by the operation * via

$${({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}_{i_1\cdots i_n{k}_1\cdots k_n}=\sum _{j_1\cdots j_n}{a}_{i_1\cdots i_n{j}_1\cdots j_n}{c}_{j_1\cdots j_n{k}_1\cdots k_n}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times K_1\times \cdots \times K_n}.$$

(12)

In a similar way ([37]), we define a transformation for tensor matricization as follows.

Definition 2.

The transformation f from reduced biquaternion tensors to reduced biquaternion matrices is given by

$$\begin{array}{cccc}f:\hfill & {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}\hfill & \to \hfill & {\mathcal{M}}_{I_1\cdots I_n,J_1\cdots J_n}\left({\mathbb{H}}_{RB}\right)\hfill \\ \hfill & {\left({\mathcal{A}}_{RB}\right)}_{i_1\cdots i_n{j}_1\cdots j_n}\hfill & \to \hfill & {\left(A_{RB}\right)}_{[i_1+{\sum }_{k=2}^n(i_k-1){\prod }_{s=1}^{k-1}{I}_s][j_1+{\sum }_{k=2}^n(j_k-1){\prod }_{s=1}^{k-1}{J}_s]}.\hfill \end{array}$$

(13)

Lemma 1.

For any ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},{\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{J_1\times \cdots J_n\times K_1\times \cdots \times K_n}$, the map f defined in Definition 2 has the following properties:

(i) 

The map f is a bijection. That is, there exists a bijective inverse map $f^{-1}$:

$$f^{-1}:{\mathcal{M}}_{I_1\cdots I_n,J_1\cdots J_n}\left({\mathbb{H}}_{RB}\right)\to {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}$$

(14)

(ii) 

The map f satisfies $f({\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB})=f\left({\mathcal{A}}_{RB}\right)·f\left({\mathcal{B}}_{RB}\right)$, where · refers to the usual matrix multiplication.

The proof of above lemma is similar to the transformation f for real tensors.

Example 1.

Given reduced biquaternion tensors ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{2\times 2\times 2\times 2}$ and ${\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{2\times 2\times 3\times 3}$.

$$\begin{array}{c}\mathcal{A}(:,:,1,1)=\left(\begin{array}{cc}6+2\mathbf{i}+\mathbf{j}-2\mathbf{k}& 3+2\mathbf{i}-3\mathbf{k}\\ 2-3\mathbf{i}+\mathbf{j}+5\mathbf{k}& -2+3\mathbf{i}-\mathbf{j}-5\mathbf{k}\end{array}\right),\\ \mathcal{A}(:,:,1,2)=\left(\begin{array}{cc}3+2\mathbf{i}+\mathbf{k}& 14+3\mathbf{j}+\mathbf{k}\\ 4+6\mathbf{i}+\mathbf{j}& 8+4\mathbf{i}-5\mathbf{j}-4\mathbf{k}\end{array}\right),\\ \mathcal{A}(:,:,2,1)=\left(\begin{array}{cc}2+3\mathbf{i}-4\mathbf{j}+5\mathbf{k}& 4+2\mathbf{i}-\mathbf{j}-4\mathbf{k}\\ 1+2\mathbf{j}-2\mathbf{k}& -3-2\mathbf{i}+3\mathbf{j}\end{array}\right),\\ \mathcal{A}(:,:,2,2)=\left(\begin{array}{cc}2+9\mathbf{i}+\mathbf{j}+5\mathbf{k}& 1+8\mathbf{i}+5\mathbf{j}+4\mathbf{k}\\ -3+5\mathbf{i}-3\mathbf{j}& 1+3\mathbf{j}+7\mathbf{k}\end{array}\right),\end{array}$$

(15)

applying the aforementioned transformation f, we transform reduced biquaternion tensor ${\mathcal{A}}_{RB}$ into the following matrix:

$${\mathcal{A}}_{RB}\stackrel{f}{\to }\left(\begin{array}{cccc}6+2\mathbf{i}+\mathbf{j}-2\mathbf{k}& 2+3\mathbf{i}-4\mathbf{j}+5\mathbf{k}& 3+2\mathbf{i}+\mathbf{k}& 2+9\mathbf{i}+\mathbf{j}+5\mathbf{k}\\ 2-3\mathbf{i}+\mathbf{j}+5\mathbf{k}& 1+2\mathbf{j}-2\mathbf{k}& 4+6\mathbf{i}+\mathbf{j}& -3+5\mathbf{i}-3\mathbf{j}\\ 3+2\mathbf{i}-3\mathbf{k}& 4+2\mathbf{i}-\mathbf{j}-4\mathbf{k}& 14+3\mathbf{j}+\mathbf{k}& 1+8\mathbf{i}+5\mathbf{j}+4\mathbf{k}\\ -2+3\mathbf{i}-\mathbf{j}-5\mathbf{k}& -3-2\mathbf{i}+3\mathbf{j}& 8+4\mathbf{i}-5\mathbf{j}-4\mathbf{k}& 1+3\mathbf{j}+7\mathbf{k}\end{array}\right.$$

(16)

${\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{2\times 2\times 3\times 3}:$

$$\begin{array}{c}\mathcal{B}(:,:,1,1)=\left(\begin{array}{cc}8-3\mathbf{j}& 3+9\mathbf{i}+3\mathbf{j}\\ 2+9\mathbf{j}& 2+3\mathbf{i}+\mathbf{j}+5\mathbf{k}\end{array}\right),\\ \mathcal{B}(:,:,1,2)=\left(\begin{array}{cc}3+9\mathbf{i}-3\mathbf{j}+\mathbf{k}& 3+9\mathbf{i}+3\mathbf{j}-3\mathbf{k}\\ 6+9\mathbf{j}+4\mathbf{k}& 2-3\mathbf{i}+\mathbf{j}+5\mathbf{k}\end{array}\right),\\ \mathcal{B}(:,:,1,3)=\left(\begin{array}{cc}8& 3-2\mathbf{i}+3\mathbf{j}\\ 2+9\mathbf{j}+\mathbf{k}& 2+\mathbf{j}-\mathbf{k}\end{array}\right),\\ \mathcal{B}(:,:,2,1)=\left(\begin{array}{cc}2+7\mathbf{i}-\mathbf{j}-7\mathbf{k}& 4-6\mathbf{i}+\mathbf{j}-4\mathbf{k}\\ 8+4\mathbf{i}+5\mathbf{j}+4\mathbf{k}& 6+2\mathbf{i}-3\mathbf{j}-2\mathbf{k}\end{array}\right),\\ \mathcal{B}(:,:,3,1)=\left(\begin{array}{cc}2+3\mathbf{i}-\mathbf{j}+5\mathbf{k}& 4-6\mathbf{i}+\mathbf{j}+4\mathbf{k}\\ 14+5\mathbf{j}-\mathbf{k}& -3-2\mathbf{i}+4\mathbf{j}\end{array}\right),\\ \mathcal{B}(:,:,2,2)=\left(\begin{array}{cc}4+9\mathbf{i}+\mathbf{j}-7\mathbf{k}& 4+6\mathbf{i}-\mathbf{j}-4\mathbf{k}\\ 2+\mathbf{i}-5\mathbf{j}+4\mathbf{k}& 6+8\mathbf{i}+2\mathbf{k}\end{array}\right),\\ \mathcal{B}(:,:,2,3)=\left(\begin{array}{cc}2+6\mathbf{i}-\mathbf{j}& 4-2\mathbf{i}+\mathbf{j}-4\mathbf{k}\\ 8+6\mathbf{j}+4\mathbf{k}& 6+4\mathbf{i}+3\mathbf{j}\end{array}\right),\\ \mathcal{B}(:,:,3,2)=\left(\begin{array}{cc}2+3\mathbf{i}+11\mathbf{j}+5\mathbf{k}& 5-6\mathbf{i}+2\mathbf{j}+4\mathbf{k}\\ 5-2\mathbf{j}-\mathbf{k}& 6+2\mathbf{i}+4\mathbf{j}\end{array}\right),\\ \mathcal{B}(:,:,3,3)=\left(\begin{array}{cc}2+3\mathbf{i}-\mathbf{j}+5\mathbf{k}& 4-\mathbf{i}+4\mathbf{k}\\ 2+5\mathbf{j}-\mathbf{k}& -3-2\mathbf{i}+4\mathbf{j}\end{array}\right),\end{array}$$

(17)

applying the aforementioned transformation f, we transform reduced biquaternion tensor ${\mathcal{B}}_{RB}$ into the following matrix,

$${\mathcal{B}}_{RB}\stackrel{f}{\to }\left(\begin{array}{cccc}8-3\mathbf{j}& 2+7\mathbf{i}-\mathbf{j}-7\mathbf{k}& 2+3\mathbf{i}-\mathbf{j}+5\mathbf{k}& 3+9\mathbf{i}-3\mathbf{j}+\mathbf{k}\\ 2+9\mathbf{j}& 8+4\mathbf{i}+5\mathbf{j}+4\mathbf{k}& 14+5\mathbf{j}-\mathbf{k}& 6+9\mathbf{j}\\ 3+9\mathbf{i}+3\mathbf{j}& 4-6\mathbf{i}+\mathbf{j}-4\mathbf{k}& 4-6\mathbf{i}+\mathbf{j}+4\mathbf{k}& 3+9\mathbf{i}+3\mathbf{j}-3\mathbf{k}\\ 2+3\mathbf{i}+\mathbf{j}-5\mathbf{k}& 6+2\mathbf{i}-3\mathbf{j}-2\mathbf{k}& -3-2\mathbf{i}+4\mathbf{j}& 2-3\mathbf{i}+1\mathbf{j}+5\mathbf{k}\end{array}\right.$$

(18)

$$\left.\begin{array}{ccccc}4+9\mathbf{i}+\mathbf{j}-7\mathbf{k}& 2+3\mathbf{i}+11\mathbf{j}+5\mathbf{k}& 8& 2+6\mathbf{i}-\mathbf{j}& 2+3\mathbf{i}-\mathbf{j}+5\mathbf{k}\\ 2+\mathbf{i}-5\mathbf{j}+4\mathbf{k}& 5-2\mathbf{j}-\mathbf{k}& 2+9\mathbf{j}+\mathbf{k}& 8+6\mathbf{j}+4\mathbf{k}& 2+5\mathbf{j}-\mathbf{k}\\ 4+6\mathbf{i}-\mathbf{j}-4\mathbf{k}& 5-6\mathbf{i}+2\mathbf{j}+4\mathbf{k}& 3-2\mathbf{i}+3\mathbf{j}& 4-2\mathbf{i}+\mathbf{j}-4\mathbf{k}& 4-\mathbf{i}+4\mathbf{k}\\ 6+8\mathbf{i}+2\mathbf{k}& 6+2\mathbf{i}+4\mathbf{j}& 2+\mathbf{j}-\mathbf{k}& 6+4\mathbf{i}+3\mathbf{j}& -3-2\mathbf{i}+4\mathbf{j}\end{array}\right).$$

We then compute ${\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB}$ and apply the transformation f to the new reduced biquaternion tensor to obtain the following equation.

$$\begin{array}{cc}f({\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB})\hfill & =\left(\begin{array}{ccc}-43+143\mathbf{i}-56\mathbf{j}+56\mathbf{k}& -24+144\mathbf{i}-5\mathbf{j}+\mathbf{k}& 64+80\mathbf{i}-35\mathbf{j}+147\mathbf{k}\\ -30-17\mathbf{i}-4\mathbf{j}+53\mathbf{k}& 119+24\mathbf{i}+6\mathbf{j}-29\mathbf{k}& 69-17\mathbf{i}+14\mathbf{j}+39\mathbf{k}\\ 37+170\mathbf{i}+26\mathbf{j}+46\mathbf{k}& 52-31\mathbf{i}+119\mathbf{j}-101\mathbf{k}& 150-56\mathbf{i}+33\mathbf{j}+9\mathbf{k}\\ -49+132\mathbf{i}+8\mathbf{j}-84\mathbf{k}& -14-53\mathbf{i}+54\mathbf{j}+16\mathbf{k}& 62-61\mathbf{i}-18\mathbf{j}-10\mathbf{k}\end{array}\right.\hfill \\ \hfill & \begin{array}{ccc}-24+188\mathbf{i}-25\mathbf{j}+86\mathbf{k}& -72+117\mathbf{i}-46\mathbf{j}-8\mathbf{k}& 71+86\mathbf{i}+51\mathbf{j}+157\mathbf{k}\\ 16+32\mathbf{i}-25\mathbf{j}+37\mathbf{k}& -8+67\mathbf{i}-63\mathbf{j}-32\mathbf{k}& 16+96\mathbf{i}-9\mathbf{j}+15\mathbf{k}\\ 74+180\mathbf{i}+71\mathbf{j}-24\mathbf{k}& -9+193\mathbf{i}-8\mathbf{j}-2\mathbf{k}& 115-5\mathbf{i}+96\mathbf{j}+112\mathbf{k}\\ -85+99\mathbf{i}+121\mathbf{j}-112\mathbf{k}& -101+103\mathbf{i}+63\mathbf{j}-25\mathbf{k}& 68-99\mathbf{i}-46\mathbf{j}+115\mathbf{k}\end{array}\hfill \\ \hfill & \left.\begin{array}{ccc}34+88\mathbf{i}+39\mathbf{j}+49\mathbf{k}& -30+160\mathbf{i}-19\mathbf{j}+121\mathbf{k}& 30+58\mathbf{i}+5\mathbf{j}+85\mathbf{k}\\ 56-17\mathbf{i}+32\mathbf{j}+61\mathbf{k}& 31+27\mathbf{i}-3\mathbf{j}-2\mathbf{k}& 26+3\mathbf{i}-14\mathbf{j}+62\mathbf{k}\\ 89-27\mathbf{i}+104\mathbf{j}+2\mathbf{k}& 88+38\mathbf{i}+68\mathbf{j}-2\mathbf{k}& 99-11\mathbf{i}+26\mathbf{j}+72\mathbf{k}\\ 34+11\mathbf{i}-19\mathbf{j}-38\mathbf{k}& 7+36\mathbf{i}+33\mathbf{j}-23\mathbf{k}& 83+7\mathbf{i}-42\mathbf{j}-39\mathbf{k}\end{array}\right)\hfill \\ \hfill & =f\left({\mathcal{A}}_{RB}\right)·f\left({\mathcal{B}}_{RB}\right).\hfill \end{array}$$

(19)

3. Real and Complex Representations for Reduced Biquaternion Tensors

In this section, we consider the real and complex representations of reduced biquaternion tensors and their associated properties. We first introduce the following useful $e_1-e_2$ form.

Definition 3.

($e_1-e_2$ form of reduced biquaternion tensor).

A reduced biquaternion tensor ${\mathcal{A}}_{RB}={\mathcal{A}}_0+{\mathcal{A}}_1\mathbf{i}+{\mathcal{A}}_2\mathbf{j}+{\mathcal{A}}_3\mathbf{k}$ can be written as

$$\begin{array}{cc}\hfill {\mathcal{A}}_{RB}& ={\mathcal{A}}_0+{\mathcal{A}}_1\mathbf{i}+({\mathcal{A}}_2+{\mathcal{A}}_3\mathbf{i})\mathbf{j}\hfill \\ & ={\mathcal{Q}}_1+{\mathcal{Q}}_2\mathbf{j}=({\mathcal{Q}}_1+{\mathcal{Q}}_2)e_1+({\mathcal{Q}}_1-{\mathcal{Q}}_2)e_2,\hfill \end{array}$$

(20)

where ${\mathcal{Q}}_1={\mathcal{A}}_0+{\mathcal{A}}_1\mathbf{i},{\mathcal{Q}}_2={\mathcal{A}}_2+{\mathcal{A}}_3\mathbf{i}.$

Definition 4.

Let ${\mathcal{A}}_{RB}={\mathcal{A}}_0+{\mathcal{A}}_1\mathbf{i}+{\mathcal{A}}_2\mathbf{j}+{\mathcal{A}}_3\mathbf{k}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}$. Then the real representation of ${\mathcal{A}}_{RB}$ is defined as follows:

$$\begin{array}{c}\hfill {\mathcal{A}}_{RB}^R=\left(\begin{array}{cccc}{\mathcal{A}}_0& -{\mathcal{A}}_1& {\mathcal{A}}_2& -{\mathcal{A}}_3\\ {\mathcal{A}}_1& {\mathcal{A}}_0& {\mathcal{A}}_3& {\mathcal{A}}_2\\ {\mathcal{A}}_2& -{\mathcal{A}}_3& {\mathcal{A}}_0& -{\mathcal{A}}_1\\ {\mathcal{A}}_3& {\mathcal{A}}_2& {\mathcal{A}}_1& {\mathcal{A}}_0\end{array}\right)\in {\mathbb{R}}^{4I_1\times \cdots \times 4I_n\times 4J_1\times \cdots \times 4J_n}.\end{array}$$

(21)

The properties of the real representations of reduced biquaternion tensors are given as follows:

Proposition 1.

Let ${\mathcal{A}}_{RB},{\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},{\mathcal{C}}_{RB}\in {\mathbb{H}}_{RB}^{J_1\times \cdots \times J_n\times K_1\times \cdots \times K_n}$, and $a\in \mathbb{R}$. Then,

$$\begin{array}{cc}\hfill & \left(1\right){({\mathcal{A}}_{RB}+{\mathcal{B}}_{RB})}^R={\mathcal{A}}_{RB}^R+{\mathcal{B}}_{RB}^R.\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill & \left(2\right){\left(a{\mathcal{A}}_{RB}\right)}^R=a{\mathcal{A}}_{RB}^R.\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill & \left(3\right){({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^R={\mathcal{A}}_{RB}^R{\mathcal{C}}_{RB}^R.\hfill \end{array}$$

(24)

The proof for (3) above is given as follows and other properties can be proven similarly.

Proof. 

$$\begin{array}{cc}\hfill {\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB}=& ({\mathcal{A}}_0+{\mathcal{A}}_1\mathbf{i}+{\mathcal{A}}_2\mathbf{j}+{\mathcal{A}}_3\mathbf{k})\ast ({\mathcal{C}}_0+{\mathcal{C}}_1\mathbf{i}+{\mathcal{C}}_2\mathbf{j}+{\mathcal{C}}_3\mathbf{k})\hfill \\ \hfill =& ({\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3)\hfill \\ \hfill & +({\mathcal{A}}_0\ast {\mathcal{C}}_1+{\mathcal{A}}_1\ast {\mathcal{C}}_0+{\mathcal{A}}_2\ast {\mathcal{C}}_3+{\mathcal{A}}_3\ast {\mathcal{C}}_2)\mathbf{i}\hfill \\ \hfill & +({\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1)\mathbf{j}\hfill \\ \hfill & +({\mathcal{A}}_0\ast {\mathcal{C}}_3+{\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1+{\mathcal{A}}_3\ast {\mathcal{C}}_0)\mathbf{k}.\hfill \end{array}$$

(25)

Then

$$\begin{array}{c}{({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^R=\hfill \\ \left(\begin{array}{cc}{\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3& -{\mathcal{A}}_0\ast {\mathcal{C}}_1-{\mathcal{A}}_1\ast {\mathcal{C}}_0-{\mathcal{A}}_2\ast {\mathcal{C}}_3-{\mathcal{A}}_3\ast {\mathcal{C}}_2\\ {\mathcal{A}}_0\ast {\mathcal{C}}_1+{\mathcal{A}}_1\ast {\mathcal{C}}_0+{\mathcal{A}}_2\ast {\mathcal{C}}_3+{\mathcal{A}}_3\ast {\mathcal{C}}_2& {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3\\ {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1& -{\mathcal{A}}_0\ast {\mathcal{C}}_3-{\mathcal{A}}_1\ast {\mathcal{C}}_2-{\mathcal{A}}_2\ast {\mathcal{C}}_1-{\mathcal{A}}_3\ast {\mathcal{C}}_0\\ {\mathcal{A}}_0\ast {\mathcal{C}}_3+{\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1+{\mathcal{A}}_3\ast {\mathcal{C}}_0& {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1\end{array}\right.\hfill \\ \left.\begin{array}{cc}{\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1& -{\mathcal{A}}_0\ast {\mathcal{C}}_3-{\mathcal{A}}_1\ast {\mathcal{C}}_2-{\mathcal{A}}_2\ast {\mathcal{C}}_1-{\mathcal{A}}_3\ast {\mathcal{C}}_0\\ {\mathcal{A}}_0\ast {\mathcal{C}}_3+{\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1+{\mathcal{A}}_3\ast {\mathcal{C}}_0& {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1\\ {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3& -{\mathcal{A}}_0\ast {\mathcal{C}}_1-{\mathcal{A}}_1\ast {\mathcal{C}}_0-{\mathcal{A}}_2\ast {\mathcal{C}}_3-{\mathcal{A}}_3\ast {\mathcal{C}}_2\\ {\mathcal{A}}_0\ast {\mathcal{C}}_1+{\mathcal{A}}_1\ast {\mathcal{C}}_0+{\mathcal{A}}_2\ast {\mathcal{C}}_3+{\mathcal{A}}_3\ast {\mathcal{C}}_2& {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3\end{array}\right).\hfill \end{array}$$

(26)

and

$$\begin{array}{c}{\mathcal{A}}_{RB}^R{\mathcal{C}}_{RB}^R=\left(\begin{array}{cccc}{\mathcal{A}}_0& -{\mathcal{A}}_1& {\mathcal{A}}_2& -{\mathcal{A}}_3\\ {\mathcal{A}}_1& {\mathcal{A}}_0& {\mathcal{A}}_3& {\mathcal{A}}_2\\ {\mathcal{A}}_2& -{\mathcal{A}}_3& {\mathcal{A}}_0& -{\mathcal{A}}_1\\ {\mathcal{A}}_3& {\mathcal{A}}_2& {\mathcal{A}}_1& {\mathcal{A}}_0\end{array}\right)\left(\begin{array}{cccc}{\mathcal{C}}_0& -{\mathcal{C}}_1& {\mathcal{C}}_2& -{\mathcal{C}}_3\\ {\mathcal{C}}_1& {\mathcal{C}}_0& {\mathcal{C}}_3& {\mathcal{C}}_2\\ {\mathcal{C}}_2& -{\mathcal{C}}_3& {\mathcal{C}}_0& -{\mathcal{C}}_1\\ {\mathcal{C}}_3& {\mathcal{C}}_2& {\mathcal{C}}_1& {\mathcal{C}}_0\end{array}\right)=\hfill \\ \left(\begin{array}{cc}{\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3& -{\mathcal{A}}_0\ast {\mathcal{C}}_1-{\mathcal{A}}_1\ast {\mathcal{C}}_0-{\mathcal{A}}_2\ast {\mathcal{C}}_3-{\mathcal{A}}_3\ast {\mathcal{C}}_2\\ {\mathcal{A}}_0\ast {\mathcal{C}}_1+{\mathcal{A}}_1\ast {\mathcal{C}}_0+{\mathcal{A}}_2\ast {\mathcal{C}}_3+{\mathcal{A}}_3\ast {\mathcal{C}}_2& {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3\\ {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1& -{\mathcal{A}}_0\ast {\mathcal{C}}_3-{\mathcal{A}}_1\ast {\mathcal{C}}_2-{\mathcal{A}}_2\ast {\mathcal{C}}_1-{\mathcal{A}}_3\ast {\mathcal{C}}_0\\ {\mathcal{A}}_0\ast {\mathcal{C}}_3+{\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1+{\mathcal{A}}_3\ast {\mathcal{C}}_0& {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1\end{array}\right.\hfill \\ \left.\begin{array}{cc}{\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1& -{\mathcal{A}}_0\ast {\mathcal{C}}_3-{\mathcal{A}}_1\ast {\mathcal{C}}_2-{\mathcal{A}}_2\ast {\mathcal{C}}_1-{\mathcal{A}}_3\ast {\mathcal{C}}_0\\ {\mathcal{A}}_0\ast {\mathcal{C}}_3+{\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1+{\mathcal{A}}_3\ast {\mathcal{C}}_0& {\mathcal{A}}_0\ast {\mathcal{C}}_2-{\mathcal{A}}_1\ast {\mathcal{C}}_3+{\mathcal{A}}_2\ast {\mathcal{C}}_0-{\mathcal{A}}_3\ast {\mathcal{C}}_1\\ {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3& -{\mathcal{A}}_0\ast {\mathcal{C}}_1-{\mathcal{A}}_1\ast {\mathcal{C}}_0-{\mathcal{A}}_2\ast {\mathcal{C}}_3-{\mathcal{A}}_3\ast {\mathcal{C}}_2\\ {\mathcal{A}}_0\ast {\mathcal{C}}_1+{\mathcal{A}}_1\ast {\mathcal{C}}_0+{\mathcal{A}}_2\ast {\mathcal{C}}_3+{\mathcal{A}}_3\ast {\mathcal{C}}_2& {\mathcal{A}}_0\ast {\mathcal{C}}_0-{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2-{\mathcal{A}}_3\ast {\mathcal{C}}_3\end{array}\right).\hfill \end{array}$$

(27)

□

Thus, ${({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^R={\mathcal{A}}_{RB}^R{\mathcal{C}}_{RB}^R$.

Definition 5.

Given ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},{\mathcal{A}}_1,{\mathcal{A}}_2\in {\mathbb{C}}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}$, the complex representation of ${\mathcal{A}}_{RB}$ is defined as follows:

$$\begin{array}{c}\hfill {\mathcal{A}}_{RB}^C=\left(\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{A}}_2\\ {\mathcal{A}}_2& {\mathcal{A}}_1\end{array}\right)\in {\mathbb{C}}^{2I_1\times \cdots \times 2I_n\times 2J_1\times \cdots \times 2J_n}.\end{array}$$

(28)

Note that the complex tensor is uniquely determined by ${\mathcal{A}}_{RB}$.

The properties of the complex representations for reduced biquaternion tensors are given as follows:

Proposition 2.

Let ${\mathcal{A}}_{RB},{\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n},{\mathcal{C}}_{RB}\in {\mathbb{H}}_{RB}^{J_1\times \cdots \times J_n\times K_1\times \cdots \times K_n}$. Then

$$\begin{array}{cc}\hfill & \left(1\right){\left({\mathcal{A}}_{RB}^C\right)}^T={\mathcal{A}}_{RB}^C.\hfill \end{array}$$

(29)

$$\begin{array}{cc}\hfill & \left(2\right){\left(a{\mathcal{A}}_{RB}\right)}^C=a{\mathcal{A}}_{RB}^C,a\in \mathbb{C}.\hfill \end{array}$$

(30)

$$\begin{array}{cc}\hfill & \left(3\right){\left({\mathcal{A}}_{RB}^{-1}\right)}^C={\left({\mathcal{A}}_{RB}^C\right)}^{-1},if{\mathcal{A}}_{RB}^{-1}exists.\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & \left(4\right){({\mathcal{A}}_{RB}+{\mathcal{B}}_{RB})}^C={\mathcal{A}}_{RB}^C+{\mathcal{B}}_{RB}^C.\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill & \left(5\right){({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^C={\mathcal{A}}_{RB}^C\ast {\mathcal{C}}_{RB}^C.\hfill \end{array}$$

(33)

The proof for (5) above is given as follows and other properties can be proven similarly.

Proof. 

Note that

$$\begin{array}{cc}\hfill {\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB}& =({\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j})\ast ({\mathcal{C}}_1+{\mathcal{C}}_2\mathbf{j})\hfill \\ \hfill & ={\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2+({\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1)\mathbf{j},\hfill \end{array}$$

(34)

then

$${({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^C=\left(\begin{array}{cc}{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2& {\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1\\ {\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1& {\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2\end{array}\right),$$

(35)

and

$$\begin{array}{cc}\hfill {\mathcal{A}}_{RB}^C\ast {\mathcal{C}}_{RB}^C& =\left(\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{A}}_2\\ {\mathcal{A}}_2& {\mathcal{A}}_1\end{array}\right)\left(\begin{array}{cc}{\mathcal{C}}_1& {\mathcal{C}}_2\\ {\mathcal{C}}_2& {\mathcal{C}}_1\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2& {\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1\\ {\mathcal{A}}_1\ast {\mathcal{C}}_2+{\mathcal{A}}_2\ast {\mathcal{C}}_1& {\mathcal{A}}_1\ast {\mathcal{C}}_1+{\mathcal{A}}_2\ast {\mathcal{C}}_2\end{array}\right).\hfill \end{array}$$

(36)

Hence, ${({\mathcal{A}}_{RB}\ast {\mathcal{C}}_{RB})}^C={\mathcal{A}}_{RB}^C\ast {\mathcal{C}}_{RB}^C$. □

Theorem 1.

Let ${\mathcal{A}}_{RB}$ be a reduced biquaternion tensor. Then ${\left({\mathcal{A}}_{RB}^{H_{\left(1\right)}}\right)}^C={\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(1\right)}}$. But, in general, ${\left({\mathcal{A}}_{RB}^{H_{\left(2\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(2\right)}},{\left({\mathcal{A}}_{RB}^{H_{\left(3\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(3\right)}}$, where $H_{\left(i\right)},i=1,2,3$ represent the conjugate transpose of ${\mathcal{A}}_{RB}$.

Proof. 

For any reduced biquaternion q, we can write

$$q=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}=(q_0+q_1\mathbf{i})+(q_2+q_3\mathbf{i})\mathbf{j}=a+b\mathbf{j}.$$

(37)

And the three types of conjugation are as follows:

$$\begin{array}{c}\hfill q^{\left(1\right)}=q_0-q_1\mathbf{i}+q_2\mathbf{j}-q_3\mathbf{k}=(q_0-q_1\mathbf{i})+(q_2-q_3\mathbf{i})\mathbf{j}=\overline{a}+\overline{b}\mathbf{j},\end{array}$$

(38)

$$\begin{array}{c}\hfill q^{\left(2\right)}=q_0+q_1\mathbf{i}-q_2\mathbf{j}-q_3\mathbf{k}=(q_0+q_1\mathbf{i})-(q_2+q_3\mathbf{i})\mathbf{j}=a-b\mathbf{j},\end{array}$$

(39)

$$\begin{array}{c}\hfill q^{\left(3\right)}=q_0-q_1\mathbf{i}-q_2\mathbf{j}+q_3\mathbf{k}=(q_0-q_1\mathbf{i})-(q_2-q_3\mathbf{i})\mathbf{j}=\overline{a}-\overline{b}\mathbf{j}.\end{array}$$

(40)

Then, for the reduced biquaternion tensor ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}$, for which complex representation is defined in (5), and the conjugation transpose is

$$\begin{array}{c}\hfill {\left({\mathcal{A}}_{RB}^C\right)}^H=\left(\begin{array}{cc}{\left(\overline{{\mathcal{A}}_1}\right)}^T& {\left(\overline{{\mathcal{A}}_2}\right)}^T\\ {\left(\overline{{\mathcal{A}}_2}\right)}^T& {\left(\overline{{\mathcal{A}}_1}\right)}^T\end{array}\right).\end{array}$$

(41)

For ${\mathcal{A}}_{RB}^{H_{\left(1\right)}}={\left(\overline{{\mathcal{A}}_1}\right)}^T+\mathbf{j}{\left(\overline{{\mathcal{A}}_2}\right)}^T$,

$$\begin{array}{c}\hfill {\left({\mathcal{A}}_{RB}^{H_{\left(1\right)}}\right)}^C=\left(\begin{array}{cc}{\left(\overline{{\mathcal{A}}_1}\right)}^T& {\left(\overline{{\mathcal{A}}_2}\right)}^T\\ {\left(\overline{{\mathcal{A}}_2}\right)}^T& {\left(\overline{{\mathcal{A}}_1}\right)}^T\end{array}\right),\end{array}$$

(42)

thus ${\left({\mathcal{A}}_{RB}^{H_{\left(1\right)}}\right)}^C={\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(1\right)}}$. □

Next, we give a counterexample to prove that ${\left({\mathcal{A}}_{RB}^{H_{\left(2\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(2\right)}},{\left({\mathcal{A}}_{RB}^{H_{\left(3\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(3\right)}}.$

Example 2.

Given ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{2\times 2\times 2\times 2}:$

$$\begin{array}{cc}\mathcal{A}(:,:,1,1)=\left(\begin{array}{cc}1+\mathbf{i}+\mathbf{j}+2\mathbf{k}& \mathbf{j}+5\mathbf{k}\\ 2+2\mathbf{i}+4\mathbf{j}+\mathbf{k}& 1+2\mathbf{i}+\mathbf{j}+\mathbf{k}\end{array}\right),& \mathcal{A}(:,:,2,1)=\left(\begin{array}{cc}3+2\mathbf{k}& 1+\mathbf{i}\\ 2+\mathbf{i}+4\mathbf{j}& 1+\mathbf{i}+\mathbf{j}\end{array}\right),\\ \mathcal{A}(:,:,1,2)=\left(\begin{array}{cc}2& 2+\mathbf{i}+3\mathbf{k}\\ 4+3\mathbf{i}+3\mathbf{j}+2\mathbf{k}& 1+3\mathbf{i}+\mathbf{j}+6\mathbf{k}\end{array}\right),& \mathcal{A}(:,:,2,2)=\left(\begin{array}{cc}1+3\mathbf{i}& 1+\mathbf{j}\\ 2+2\mathbf{j}+2\mathbf{k}& 2\mathbf{i}\end{array}\right).\end{array}$$

(43)

${\mathcal{A}}_{RB}$ can be written as ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}$, and ${\mathcal{A}}_1,{\mathcal{A}}_2$ are given as follows:

$$\begin{array}{cc}{\mathcal{A}}_1(:,:,1,1)=\left(\begin{array}{cc}1+\mathbf{i}& 0\\ 2+2\mathbf{i}& 1+2\mathbf{i}\end{array}\right),& {\mathcal{A}}_1(:,:,2,1)=\left(\begin{array}{cc}3& 1+\mathbf{i}\\ 2+\mathbf{i}& 1+\mathbf{i}\end{array}\right),\\ {\mathcal{A}}_1(:,:,1,2)=\left(\begin{array}{cc}2& 2+\mathbf{i}\\ 4+3\mathbf{i}& 1+3\mathbf{i}\end{array}\right),& {\mathcal{A}}_1(:,:,2,2)=\left(\begin{array}{cc}1+3\mathbf{i}& 1\\ 2& 2\mathbf{i}\end{array}\right).\end{array}$$

(44)

$$\begin{array}{cc}{\mathcal{A}}_2(:,:,1,1)=\left(\begin{array}{cc}1+2\mathbf{i}& 1+5\mathbf{i}\\ 4+\mathbf{i}& 1+\mathbf{i}\end{array}\right),& {\mathcal{A}}_2(:,:,2,1)=\left(\begin{array}{cc}2\mathbf{i}& 0\\ 4& 1\end{array}\right),\\ {\mathcal{A}}_2(:,:,1,2)=\left(\begin{array}{cc}0& 3\mathbf{i}\\ 3+2\mathbf{i}& 1+6\mathbf{i}\end{array}\right),& {\mathcal{A}}_2(:,:,2,2)=\left(\begin{array}{cc}0& 1\\ 2+2\mathbf{i}& 0\end{array}\right).\end{array}$$

(45)

Clearly,

$$\left(\begin{array}{cc}{\left({\mathcal{A}}_1\right)}^T& -{\left({\mathcal{A}}_2\right)}^T\\ -{\left({\mathcal{A}}_2\right)}^T& {\left({\mathcal{A}}_1\right)}^T\end{array}\right)\ne \left(\begin{array}{cc}{\left(\overline{{\mathcal{A}}_1}\right)}^T& {\left(\overline{{\mathcal{A}}_2}\right)}^T\\ {\left(\overline{{\mathcal{A}}_2}\right)}^T& {\left(\overline{{\mathcal{A}}_1}\right)}^T\end{array}\right),$$

(46)

that is, ${\left({\mathcal{A}}_{RB}^{H_{\left(2\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(2\right)}}$.

$$\left(\begin{array}{cc}{\left(\overline{{\mathcal{A}}_1}\right)}^T& -{\left(\overline{{\mathcal{A}}_2}\right)}^T\\ -{\left(\overline{{\mathcal{A}}_2}\right)}^T& {\left(\overline{{\mathcal{A}}_1}\right)}^T\end{array}\right)\ne \left(\begin{array}{cc}{\left(\overline{{\mathcal{A}}_1}\right)}^T& {\left(\overline{{\mathcal{A}}_2}\right)}^T\\ {\left(\overline{{\mathcal{A}}_2}\right)}^T& {\left(\overline{{\mathcal{A}}_1}\right)}^T\end{array}\right),$$

(47)

Thus, ${\left({\mathcal{A}}_{RB}^{H_{\left(3\right)}}\right)}^C\ne {\left({\mathcal{A}}_{RB}^C\right)}^{H_{\left(3\right)}}$.

Theorem 2.

Let ${\mathcal{A}}_{RB},{\mathcal{B}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$. If ${\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB}=\mathcal{I}$, then ${\mathcal{B}}_{RB}\ast {\mathcal{A}}_{RB}=\mathcal{I}$.

Proof. 

According to Definition 2 and Lemma 1, we know that the above function f accomplishes a one-to-one correspondence between reduced biquaternion tensor elements and matrix elements; meanwhile, $f({\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB})=f\left({\mathcal{A}}_{RB}\right)·f\left({\mathcal{B}}_{RB}\right)$. Therefore, in this theorem we can conclude that

$$f({\mathcal{A}}_{RB}\ast {\mathcal{B}}_{RB})=f\left({\mathcal{A}}_{RB}\right)·f\left({\mathcal{B}}_{RB}\right)=f\left({\mathcal{B}}_{RB}\right)·f\left({\mathcal{A}}_{RB}\right)=f({\mathcal{B}}_{RB}\ast {\mathcal{A}}_{RB})=I.$$

(48)

Thus, ${\mathcal{B}}_{RB}\ast {\mathcal{A}}_{RB}=\mathcal{I}.$ □

4. The Eigenvalues and Corresponding Eigentensors of Reduced Biquaternion Tensors

The eigenvalues and corresponding eigenvtensors of reduced biquaternion matrices were proposed in [23].

Definition 6.

(Eigenvalue and eigentensor of a reduced biquaternion matrix) λ is an eigenvalue of a reduced biquaternion matrix $A_{RB}$ and X is the corresponding eigentensors if they satisfy

$$A_{RB}X=\lambda X=X\lambda .$$

(49)

In this section, we present the definition of eigenvalues and eigentensors for reduced biquaternion tensors. Due to the commutativity of their multiplication rule, unlike traditional quaternion tensors, where left and right eigenvalues differ, the left and right eigenvalues of reduced biquaternion tensors are the same. The definition is given as follows:

Definition 7.

A reduced biquaternion λ is said to be an eigenvalue of ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times J_1\times \cdots \times J_n}$ provided that

$${\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}={\mathcal{X}}_{RB}\lambda =\lambda {\mathcal{X}}_{RB},$$

(50)

for some nonzero reduced biquaternion tensor ${\mathcal{X}}_{RB}\in {\mathbb{H}}_{RB}^{J_1\times \cdots \times J_n}$, which is called the corresponding eigentensor of λ.

We define the set of $\sigma \left({\mathcal{A}}_{RB}\right)=\{\lambda \in {\mathbb{H}}_{RB}|{\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}=\lambda {\mathcal{X}}_{RB}\}$ as its spectrum.

The Ger$\check{\mathrm{s}}$gorin theorem for quaternion matrices was proposed in [45]. Inspired by this, we attempt to present the Ger$\check{\mathrm{s}}$gorin theorem for reduced biquaternion tensors as follows:

Theorem 3.

(Ger$\check{\mathrm{s}}$gorin theorem for a reduced biquaternion tensor).

For a given reduced biquaternion tensor ${\mathcal{A}}_{RB}=\left(a_{i_1\cdots i_n{i}_1\cdots i_n}\right)\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$, the eigenvalues of ${\mathcal{A}}_{RB}$ are located in

$$\sigma \left({\mathcal{A}}_{RB}\right)\subseteq \bigcup _{i_1\cdots i_n}\left\{\lambda \in {\mathbb{H}}_{RB}:\left|\right|\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n}\left|\right|\le \underset{I=i_1\cdots i_n}{R_I}\left({\mathcal{A}}_{RB}\right)\right\},$$

(51)

where $\underset{I=i_1\cdots i_n}{R_I}\left({\mathcal{A}}_{RB}\right)={\sum }_{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}\left|\right|\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right)\left|\right|.$

Proof. 

Let $\lambda \in {\mathbb{H}}_{RB}$ be the eigenvalue of ${\mathcal{A}}_{RB}$, and $\mathcal{X}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n}$ be the corresponding eigentensor. We write

$${\mathcal{A}}_{RB}\ast \mathcal{X}=\lambda \mathcal{X},$$

(52)

in the componentwise form:

$$\sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ 1\le j_i\le I_i,i=1,\cdots n\end{array}}{a}_{k_1\cdots k_n{j}_1\cdots j_n}{x}_{j_1\cdots j_n}=\lambda \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ 1\le j_i\le I_i,i=1,\cdots n\end{array}}{\delta }_{k_1{j}_1}\cdots {\delta }_{k_n{j}_n}{x}_{j_1\cdots j_n},$$

(53)

where $1\le k_i\le I_i,i=1,\cdots ,n$, and ${\delta }_{k_m{j}_m}$, $m=1,\cdots n$ is expressed as follows:

$${\delta }_{k_m{j}_m}=\left\{\begin{array}{cc}1,\hfill & \mathrm{if}{k}_m=j_m.\hfill \\ 0,\hfill & \mathrm{if}{k}_m\ne j_m.\hfill \end{array}\right.$$

(54)

Since $\mathcal{X}\ne 0$, we choose $x_{i_1\cdots i_n}$ such that $\left|\right|x_{i_1\cdots i_n}\left|\right|\ge \left|\right|x_{j_1\cdots j_n}\left|\right|$ for $1\le j_i\le I_i$, $i=1,\cdots n$. Thus, for the index $i_1,\cdots i_n$, we have

$$\begin{array}{cc}\hfill \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ 1\le j_i\le I_i,i=1,\cdots n\end{array}}{a}_{i_1\cdots i_n{j}_1\cdots j_n}{x}_{j_1\cdots j_n}& =\lambda \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ 1\le j_i\le I_i,i=1,\cdots n\end{array}}{\delta }_{i_1{j}_1}\cdots {\delta }_{i_n{j}_n}{x}_{j_1\cdots j_n}\hfill \\ \hfill & =\lambda x_{i_1\cdots i_n}.\hfill \end{array}$$

(55)

Equivalently,

$$(\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n})x_{i_1\cdots i_n}=\sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}{a}_{i_1\cdots i_n{j}_1\cdots j_n}{x}_{j_1\cdots j_n}.$$

(56)

Then,

$$\left|\right|(\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n})x_{i_1\cdots i_n}\left|\right|=\left|\right|\sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}{a}_{i_1\cdots i_n{j}_1\cdots j_n}{x}_{j_1\cdots j_n}\left|\right|.$$

(57)

Next,

$$\left|\right|\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n}\left|\right|\left|\right|x_{i_1\cdots i_n}\left|\right|\le \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}\left|\right|\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right)x_{j_1\cdots j_n}\left|\right|,$$

(58)

$$\left|\right|\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n}\left|\right|\left|\right|x_{i_1\cdots i_n}\left|\right|\le \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}\left|\right|\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right)\left|\right|\left|\right|x_{i_1\cdots i_n}\left|\right|,$$

(59)

divided by $\left|\right|x_{i_1\cdots i_n}\left|\right|$, it is clear that

$$\left|\right|\lambda -a_{i_1\cdots i_n{i}_1\cdots i_n}\left|\right|\le \sum _{\begin{array}{c}{j}_1,\cdots ,j_n\\ j_1\ne i_1,\cdots ,j_n\ne i_n\end{array}}\left|\right|\left(a_{i_1\cdots i_n{j}_1\cdots j_n}\right)\left|\right|.$$

(60)

□

Here are some properties of eigenvalues and eigentensors of reduced biquaternion tensors.

Theorem 4.

The reduced biquaternion tensor ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}$ has eigenvalue ${\lambda }_1+{\lambda }_2\mathbf{j}$ with the corresponding eigentensor ${\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j}$ if and only if ${\mathcal{B}}_1={\mathcal{A}}_1+{\mathcal{A}}_2,{\mathcal{B}}_2={\mathcal{A}}_1-{\mathcal{A}}_2$ have eigenvalues ${\lambda }_1+{\lambda }_2$ and ${\lambda }_1-{\lambda }_2$ with the corresponding eigentensors ${\mathcal{X}}_1+{\mathcal{X}}_2,{\mathcal{X}}_1-{\mathcal{X}}_2$, respectively. We note that ${\lambda }_1+{\lambda }_2\mathbf{j}=({\lambda }_1+{\lambda }_2)e_1+({\lambda }_1-{\lambda }_2)e_2,{\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j}=({\mathcal{X}}_1+{\mathcal{X}}_2)e_1+({\mathcal{X}}_1-{\mathcal{X}}_2)e_2$.

Proof. 

Note that ${\mathcal{A}}_{RB}={\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j}={\mathcal{B}}_1{e}_1+{\mathcal{B}}_2{e}_2$.

From ${\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}=\lambda {\mathcal{X}}_{RB}$, we have

$$\begin{array}{cc}\hfill ({\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j})\ast ({\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j})& ={\mathcal{A}}_1\ast {\mathcal{X}}_1+{\mathcal{A}}_2\ast {\mathcal{X}}_2+({\mathcal{A}}_1\ast {\mathcal{X}}_2+{\mathcal{A}}_2\ast {\mathcal{X}}_1)\mathbf{j}\hfill \\ \hfill & =({\lambda }_1+{\lambda }_2\mathbf{j})({\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j})\hfill \\ \hfill & ={\lambda }_1{\mathcal{X}}_1+{\lambda }_2{\mathcal{X}}_2+({\lambda }_1{\mathcal{X}}_2+{\lambda }_2{\mathcal{X}}_1)\mathbf{j}.\hfill \end{array}$$

(61)

Then, we can obtain

$$\left\{\begin{array}{c}{\mathcal{A}}_1\ast {\mathcal{X}}_1+{\mathcal{A}}_2\ast {\mathcal{X}}_2={\lambda }_1{\mathcal{X}}_1+{\lambda }_2{\mathcal{X}}_2\\ {\mathcal{A}}_1\ast {\mathcal{X}}_2+{\mathcal{A}}_2\ast {\mathcal{X}}_1={\lambda }_1{\mathcal{X}}_2+{\lambda }_2{\mathcal{X}}_1.\end{array}\right.$$

(62)

Adding and subtracting the above expressions yields the following equations:

$$({\mathcal{A}}_1+{\mathcal{A}}_2)\ast ({\mathcal{X}}_1+{\mathcal{X}}_2)=({\lambda }_1+{\lambda }_2)({\mathcal{X}}_1+{\mathcal{X}}_2),$$

(63)

$$({\mathcal{A}}_1-{\mathcal{A}}_2)\ast ({\mathcal{X}}_1-{\mathcal{X}}_2)=({\lambda }_1-{\lambda }_2)({\mathcal{X}}_1-{\mathcal{X}}_2).$$

(64)

Thus,

$$\begin{array}{c}\hfill {\mathcal{B}}_1\ast ({\mathcal{X}}_1+{\mathcal{X}}_2)=({\lambda }_1+{\lambda }_2)({\mathcal{X}}_1+{\mathcal{X}}_2),\end{array}$$

(65)

$$\begin{array}{c}\hfill {\mathcal{B}}_2\ast ({\mathcal{X}}_1-{\mathcal{X}}_2)=({\lambda }_1-{\lambda }_2)({\mathcal{X}}_1-{\mathcal{X}}_2).\end{array}$$

(66)

We can conclude that ${\lambda }_1+{\lambda }_2$ is the eigenvalue of ${\mathcal{B}}_1$ with the corresponding eigentensor ${\mathcal{X}}_1+{\mathcal{X}}_2$, and ${\lambda }_1-{\lambda }_2$ is the eigenvalue of ${\mathcal{B}}_2$ with the corresponding eigentensor ${\mathcal{X}}_1-{\mathcal{X}}_2.$

On the other hand, suppose that the eigenvalues and eigentensors of ${\mathcal{B}}_1,{\mathcal{B}}_2$ are as follows:

$$\begin{array}{c}\hfill {\mathcal{B}}_1\ast ({\mathcal{X}}_1+{\mathcal{X}}_2)=({\lambda }_1+{\lambda }_2)({\mathcal{X}}_1+{\mathcal{X}}_2),\end{array}$$

(67)

$$\begin{array}{c}\hfill {\mathcal{B}}_2\ast ({\mathcal{X}}_1-{\mathcal{X}}_2)=({\lambda }_1-{\lambda }_2)({\mathcal{X}}_1-{\mathcal{X}}_2).\end{array}$$

(68)

Then, for ${\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}$, we have

$$\begin{array}{cc}\hfill ({\mathcal{A}}_1+{\mathcal{A}}_2\mathbf{j})\ast ({\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j})& =({\mathcal{B}}_1{e}_1+{\mathcal{B}}_2{e}_2)\ast (({\mathcal{X}}_1+{\mathcal{X}}_2)e_1+({\mathcal{X}}_1-{\mathcal{X}}_2)e_2)\hfill \\ \hfill & ={\mathcal{B}}_1\ast ({\mathcal{X}}_1+X_2)e_1+{\mathcal{B}}_2\ast ({\mathcal{X}}_1-{\mathcal{X}}_2)e_2\hfill \\ \hfill & =({\lambda }_1+{\lambda }_2)({\mathcal{X}}_1+{\mathcal{X}}_2)e_1+({\lambda }_1-{\lambda }_2)({\mathcal{X}}_1-{\mathcal{X}}_2)e_2\hfill \\ \hfill & =({\lambda }_1+{\lambda }_2\mathbf{j})({\mathcal{X}}_1+{\mathcal{X}}_2\mathbf{j}).\hfill \end{array}$$

(69)

Hence, the converse holds. □

Remark 1.

Unlike the complex representation of a quaternion tensor, the complex representation of the reduced biquaternion tensor is a symmetry tensor. This symmetry is crucial for establishing the relationship between the eigenvalues of the reduced biquaternion tensor and those of its complex representation, as discussed in the aforementioned theorem. Consequently, by computing for the eigenvalues of the symmetry tensor elements ${\mathcal{B}}_1$ and ${\mathcal{B}}_2$, we can further compute the eigenvalues of the reduced biquaternion tensor.

From the above theorem, we have the following algorithm to compute the eignevalues of a redueced biquaternion tensor.

Now we give an example to show how Algorithm 1 works.

Example 3.

Given ${\mathcal{A}}_{RB}$ as in Example 1, then ${\mathcal{B}}_1={\mathcal{A}}_1+{\mathcal{A}}_2,{\mathcal{B}}_2={\mathcal{A}}_1-{\mathcal{A}}_2$.

$$\begin{array}{cc}{\mathcal{B}}_1(:,:,1,1)=\left(\begin{array}{cc}2+3\mathbf{i}& 1+5\mathbf{i}\\ 6+3\mathbf{i}& 2+3\mathbf{i}\end{array}\right),& {\mathcal{B}}_1(:,:,2,1)=\left(\begin{array}{cc}3+2\mathbf{i}& 1+\mathbf{i}\\ 6+\mathbf{i}& 2+\mathbf{i}\end{array}\right),\\ {\mathcal{B}}_1(:,:,1,2)=\left(\begin{array}{cc}2& 2+4\mathbf{i}\\ 7+5\mathbf{i}& 2+9\mathbf{i}\end{array}\right),& {\mathcal{B}}_1(:,:,2,2)=\left(\begin{array}{cc}1+3\mathbf{i}& 2\\ 4+2\mathbf{i}& 2\mathbf{i}\end{array}\right).\end{array}$$

(70)

$$\begin{array}{cc}{\mathcal{B}}_2(:,:,1,1)=\left(\begin{array}{cc}-\mathbf{i}& -1-5\mathbf{i}\\ -2+\mathbf{i}& \mathbf{i}\end{array}\right),& {\mathcal{B}}_2(:,:,2,1)=\left(\begin{array}{cc}3-2\mathbf{i}& 1+\mathbf{i}\\ -2+\mathbf{i}& \mathbf{i}\end{array}\right),\\ {\mathcal{B}}_2(:,:,1,2)=\left(\begin{array}{cc}2& 2-2\mathbf{i}\\ 1+\mathbf{i}& -3\mathbf{i}\end{array}\right),& {\mathcal{B}}_2(:,:,2,2)=\left(\begin{array}{cc}1+3\mathbf{i}& 0\\ -2\mathbf{i}& 2\mathbf{i}\end{array}\right).\end{array}$$

(71)

We compute the eigenvalues and corresponding eigentensors of ${\mathcal{B}}_1,{\mathcal{B}}_2$ by using matlab.

Below are the eigenvalues of ${\mathcal{B}}_1:$

$$\begin{array}{c}\hfill {\lambda }_{11}=10.1341+9.2778\mathbf{i},{\lambda }_{12}=-4.3854-0.7536\mathbf{i},\\ \hfill {\lambda }_{13}=-0.0436+1.3059\mathbf{i},{\lambda }_{14}=4.2950+0.1699\mathbf{i},\end{array}$$

(72)

and the corresponding eigentensors are

$$\begin{array}{c}{\mathcal{VB}}_{11}=\left(\begin{array}{cc}0.3524+0.1036\mathbf{i}& 0.2788+0.2154\mathbf{i}\\ 0.7620+0.0000\mathbf{i}& 0.2671+0.2982\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{12}=\left(\begin{array}{cc}-0.3113-0.2831\mathbf{i}& -0.1471+0.4042\mathbf{i}\\ 0.0696-0.1116\mathbf{i}& 0.7878+0.0000\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{13}=\left(\begin{array}{cc}-0.3715-0.1079\mathbf{i}& 0.0773+0.0917\mathbf{i}\\ 0.7039+0.0000\mathbf{i}& -0.3914+0.4328\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{14}=\left(\begin{array}{cc}0.6064+0.0000\mathbf{i}& -0.3004+0.3407\mathbf{i}\\ 0.0846-0.4653\mathbf{i}& -0.2704-0.3595\mathbf{i}\end{array}\right).\end{array}$$

(73)

The following are the eigenvalues of ${\mathcal{B}}_2.$

$$\begin{array}{c}\hfill {\lambda }_{21}=3.8361-4.2600\mathbf{i},{\lambda }_{22}=-4.2949-2.1725\mathbf{i},\\ \hfill {\lambda }_{23}=0.0925+4.9117\mathbf{i},{\lambda }_{24}=0.3664+1.5208\mathbf{i},\end{array}$$

(74)

and their eigentensors are

$$\begin{array}{c}{\mathcal{VB}}_{21}=\left(\begin{array}{cc}0.2855+0.3600\mathbf{i}& 0.8304+0.0000\mathbf{i}\\ -0.0396-0.0497\mathbf{i}& 0.2387-0.1958\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{22}=\left(\begin{array}{cc}0.5499+0.0000\mathbf{i}& 0.0527+0.5421\mathbf{i}\\ -0.1804-0.4920\mathbf{i}& -0.2783+0.2212\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{23}=\left(\begin{array}{cc}0.5917+0.0000\mathbf{i}& -0.3407+0.2769\mathbf{i}\\ -0.2983+0.3737\mathbf{i}& 0.4564-0.1424\mathbf{i}\end{array}\right),\\ {\mathcal{VB}}_{24}=\left(\begin{array}{cc}-0.1940-0.1225\mathbf{i}& 0.1103-0.1658\mathbf{i}\\ 0.1943-0.6278\mathbf{i}& 0.6898+0.0000\mathbf{i}\end{array}\right).\end{array}$$

(75)

According to Algorithm 1, the eigenvalues and corresponding eigentensors of ${\mathcal{A}}_{RB}$ are as follows:

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\lambda }_1=6.9851+2.5089\mathbf{i}+3.149\mathbf{j}+6.7689\mathbf{k},\\ {\mathcal{X}}_1=\left(\begin{array}{cc}0.31895+0.2318\mathbf{i}+0.03345\mathbf{j}-0.1282\mathbf{k}& 0.5546+0.1077\mathbf{i}-0.2758\mathbf{j}+0.1077\mathbf{k}\\ 0.3612-0.02485\mathbf{i}+0.4008\mathbf{j}+0.02485\mathbf{k}& 0.2529+0.0512\mathbf{i}+0.0142\mathbf{j}+0.247\mathbf{k}\end{array}\right),\end{array}\right.\end{array}$$

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\lambda }_2=-4.3401-1.463\mathbf{i}-0.04525\mathbf{j}+0.70945\mathbf{k},\\ {\mathcal{X}}_2=\left(\begin{array}{cc}0.1193-0.14155\mathbf{i}-0.4306\mathbf{j}-0.14155\mathbf{k}& -0.0472+0.47315\mathbf{i}-0.0999\mathbf{j}-0.06895\mathbf{k}\\ -0.0554-0.3018\mathbf{i}+0.125\mathbf{j}+0.1902\mathbf{k}& 0.25475+0.1106\mathbf{i}+0.53305\mathbf{j}-0.1106\mathbf{k}\end{array}\right),\end{array}\right.\end{array}$$

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\lambda }_3=0.02445+3.1088\mathbf{i}-0.06805\mathbf{j}-1.8029\mathbf{k},\\ {\mathcal{X}}_3=\left(\begin{array}{cc}0.1101-0.05395\mathbf{i}-0.4816\mathbf{j}-0.05395\mathbf{k}& -0.1317+0.1843\mathbf{i}+0.209\mathbf{j}-0.0926\mathbf{k}\\ 0.2028+0.18685\mathbf{i}+0.5011\mathbf{j}-0.18685\mathbf{k}& 0.0325+0.1452\mathbf{i}-0.4239\mathbf{j}+0.2876\mathbf{k}\end{array}\right),\end{array}\right.\end{array}$$

(76)

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{\lambda }_4=2.3307+0.84535\mathbf{i}+1.9643\mathbf{j}-0.67545\mathbf{k},\\ {\mathcal{X}}_4=\left(\begin{array}{cc}0.2062-0.06125\mathbf{i}+0.4002\mathbf{j}+0.06125\mathbf{k}& -0.09505+0.08745\mathbf{i}-0.20535\mathbf{j}+0.25325\mathbf{k}\\ 0.13945-0.54655\mathbf{i}-0.05485\mathbf{j}+0.08125\mathbf{k}& 0.2097-0.17975\mathbf{i}-0.4801\mathbf{j}-0.17975\mathbf{k}\end{array}\right).\end{array}\right.\end{array}$$

Theorem 5.

Assume that the reduced biquaternion tensor ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$ satisfies ${\mathcal{A}}_{RB}^n=a\mathcal{I}$, where $a=z_1+z_2\mathbf{j}$, with $z_1,z_2\in \mathbb{C}$, n is a positive integer. If $a=z_1+z_2\mathbf{j}\in {\mathbb{H}}_{RB}\setminus Z\left({\mathbb{H}}_{RB}\right)$, then

$$\sigma \left({\mathcal{A}}_{RB}\right)=\left\{{\displaystyle \frac{{\left(\sqrt[n]{z_1+z_2}\right)}_s+{\left(\sqrt[n]{z_1-z_2}\right)}_t}{2}}+{\displaystyle \frac{{\left(\sqrt[n]{z_1+z_2}\right)}_s-{\left(\sqrt[n]{z_1-z_2}\right)}_t}{2}}\mathbf{j}\right\},s,t=0,1,\cdots ,n-1.$$

(77)

We recall that when $z\in \mathbb{C}\setminus \left\{0\right\}$, the nth roots of $z=r{e}^{\mathbf{i}\theta }$ with $0\le \theta \le 2\pi$, $r>0$ in complex number field are

$${\left(\sqrt[n]{z}\right)}_t=r^{{\displaystyle \frac{1}{n}}}\left(cos{\displaystyle \frac{\theta +2t\pi }{n}}+sin{\displaystyle \frac{\theta +2t\pi }{n}}\right),t=0,1,2\cdots ,n-1.$$

(78)

Proof. 

Suppose that $(\lambda ,\mathcal{X})$ is an eigenpair of ${\mathcal{A}}_{RB}\in {\mathbb{H}}_{RB}^{I_1\times \cdots \times I_n\times I_1\times \cdots \times I_n}$. For ${\mathcal{A}}_{RB}^n\ast {\mathcal{X}}_{RB}$, we have

$$\begin{array}{cc}\hfill {\mathcal{A}}_{RB}^n\ast {\mathcal{X}}_{RB}& ={\mathcal{A}}_{RB}^{n-1}\ast {\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}={\mathcal{A}}_{RB}^{n-1}\ast {\mathcal{X}}_{RB}\lambda \hfill \\ & =\cdots ={\mathcal{A}}_{RB}\ast {\mathcal{X}}_{RB}{\lambda }^{n-1}={\mathcal{X}}_{RB}{\lambda }^n=a{\mathcal{X}}_{RB}.\hfill \end{array}$$

(79)

Thus,

$${\lambda }^n=a,sincea=z_1+z_2\mathbf{j}\in {\mathbb{H}}_{RB}\setminus Z\left({\mathbb{H}}_{RB}\right).$$

(80)

Hence, based on the solution of ${\lambda }^n=0$ given in [25], the set of eigenvalues are

$$\sigma \left({\mathcal{A}}_{RB}\right)=\left\{{\displaystyle \frac{{\left(\sqrt[n]{z_1+z_2}\right)}_s+{\left(\sqrt[n]{z_1-z_2}\right)}_t}{2}}+{\displaystyle \frac{{\left(\sqrt[n]{z_1+z_2}\right)}_s-{\left(\sqrt[n]{z_1-z_2}\right)}_t}{2}}\mathbf{j}\right\},s,t=0,1,\cdots ,n-1.$$

(81)

□

5. Conclusions

This article is devoted to extending the theoretical knowledge of the reduced biquaternion tensors. Firstly, we presented the real and complex representations of reduced biquaternion tensors and their corresponding properties under the Einstein product. Secondly, we introduced the concepts of eigenvalues for reduced biquaternion tensors and provided specific eigenvalues for some special cases. Finally, we established the relationship between the eigenvalues of the complex representation and the eigenvalues of the reduced biquaternion tensors itself, and presented the Ger$\check{\mathrm{s}}$gorin theorem for the eigenvalues of the reduced biquaternion tensors. Additionally, a numerical example was provided to demonstrate the eigenvalues and eigentensors of reduced biquaternion tensors, which will provide a theoretical foundation for utilizing reduced biquaternion tensors in image processing and facial recognition, such as extracting principal features through principal component analysis. Of course, as the processing and application of high-dimensional data become increasingly common in the future, offering theoretical contributions within the tensor framework is both necessary and valuable.