T-Eigenvalues of Third-Order Quaternion Tensors

Abstract

In this paper, theories, algorithms and properties of eigenvalues of quaternion tensors via the t-product termed T-eigenvalues are explored. Firstly, we define the T-eigenvalue of quaternion tensors and provide an algorithm to compute the right T-eigenvalues and the corresponding T-eigentensors, along with an example to illustrate the efficiency of our algorithm by comparing it with other methods. We then study some inequalities related to the right T-eigenvalues of Hermitian quaternion tensors, providing upper and lower bounds for the right T-eigenvalues of the sum of a pair of Hermitian tensors. We further generalize the Weyl theorem from matrices to quaternion third-order tensors. Additionally, we explore estimations related to right T-eigenvalues, extending the Geršgorin theorem for matrices to quaternion third-order tensors.

1. Introduction

Quaternion algebra was first introduced by Hamilton [1], and it is a noncommutative division ring. One of its most important properties is that multiplication does not satisfy the commutative law, which leads to many situations that hold for complex numbers not being valid in quaternion algebra. However, quaternions play a crucial role in various fields, including computer graphics, signal processing, image processing, face recognition, etc. (e.g., [2,3,4,5,6,7]).The concept of tensors is extended from that of matrices, and tensors can serve as valuable resources for handling high-dimensional data, such as image processing [8], computer vision [9] and analysis of medical data [10]. In recent years, numerous scholars have studied tensors and their various applications [11,12,13,14,15,16]. To the best of our knowledge, the eigenvalues of matrices are widely used in various fields, such as facial recognition, image processing, image denoising, image clustering, etc. (e.g., [17,18,19]). As an extension from matrices, eigenvalues of tensors have begun to receive more and more attention among scholars, and theories and applications about different kinds of eigenvalues of tensors have been explored, such as E-eigenvalue, H-eigenvalue, Z-eigenvalue, etc. (e.g., [20,21]). Recently, with the advent of the tensor Einstein product [22] and the t-product [23], there has been a growing interest in the eigenvalues of tensors associated with these operations. For instance, He et al. have already studied theories and algorithms of eigenvalues of tensors via the Einstein product on quaternion realm and have given applications of color video processing [24,25]. Very recently, Liu et al. [26] have proposed eigenvalues of tensors under the t-product named T-eigenvalues, and have extended Weyl’s theorem and Cauchy’s interlacing theorem from matrix case to tensor case. Chen et al. [27] concentrated on perturbation theory regarding the tensor T-eigenvalues, and give three extensions from matrix domain to tensor domain: the Geršgorin theorem [28], the Bauer–Fike theorem [29] and Kahan theorem [30]. Recent research on t-products has primarily concentrated on tensors within real and complex number fields, whereas the study of quaternion algebra has been comparatively limited. To address this gap in research, this paper seeks to explore the theoretical foundations, computational methods and distinctive characteristics of T-eigenvalues in the realm of third-order quaternion tensors.

The arrangement of the remaining sections of the paper is as follows: In Section 2, we provide the basic symbols and definitions that will be used later. In Section 3, we present the computation of right T-eigenvalues and their associated T-eigentensors for third-order quaternion tensors under the T-product, and we compare our algorithm with other methods in terms of computational error and CPU time to demonstrate the efficiency of our method. In Section 4, we investigate some important inequalities related to the right T-eigenvalues for Hermitian quaternion tensors. In Section 5, we explore estimations concerning the right T-eigenvalues.

2. Preliminaries

In this section, some basic notations and definitions are given. First, we begin with some notations that will be used in the following sections. In this paper, the real number field, the complex number field and the quaternion algebra are symbolized by $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{Q}$, respectively. The scalars under real number field or complex number field are denoted by lowercase characters such as q, and we use bold minuscule letters to denote vectors, like $\mathbf{q}$. For matrices, we choose bold uppercase characters to symbolize them, for example, $\mathbf{Q}$. For tensors, we utilize Euler script letters to denote them, taking the expression $\mathcal{Q}$ for example. In the context of quaternion algebra, these symbols are accompanied by a breve, i.e., $\breve{q},\breve{\mathbf{q}},\breve{\mathbf{Q}}$ and $\breve{\mathcal{Q}}$, respectively.

Quaternion algebra [1] is a noncommutative division ring. For any given quaternion, here we use the bold lowercase letters $\mathbf{i},\mathbf{j}$ and $\mathbf{k}$ to denote the three imaginary units of any given quaternion. Each element $\breve{q}$ of the quaternion algebra can be denoted as $\breve{q}=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}$, where $q_0,q_1,q_2,q_3\in \mathbb{R}$ and $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ satisfy that ${\mathbf{i}}^2={\mathbf{j}}^2={\mathbf{k}}^2=-1$, $\mathbf{i}\mathbf{j}=-\mathbf{j}\mathbf{i}=\mathbf{k}$, $\mathbf{j}\mathbf{k}=-\mathbf{k}\mathbf{j}=\mathbf{i}$ and $\mathbf{k}\mathbf{i}=-\mathbf{i}\mathbf{k}=\mathbf{j}$. The modulus of the above quaternion number can be expressed as $|\breve{q}|=\sqrt{q_0^2+q_1^2+q_2^2+q_3^2}$, and a quaternion $\breve{q}=q_0+q_1\mathbf{i}+q_2\mathbf{j}+q_3\mathbf{k}$ is called unit pure quaternion provided that $q_0=0$ and $\sqrt{q_1^2+q_2^2+q_3^2}=1$. The conjugate of the above quaternion $\breve{q}$ is $\overline{\breve{q}}=q_0-q_1\mathbf{i}-q_2\mathbf{j}-q_3\mathbf{k}$. For any nonzero quaternion $\breve{q}$, there is an inverse ${\breve{q}}^{-1}={\displaystyle \frac{\overline{\breve{q}}}{|\breve{q}{|}^2}}$. For two quaternions $\breve{x}=x_0+x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}$ and $\breve{y}=y_0+y_1\mathbf{i}+y_2\mathbf{j}+y_3\mathbf{k}$, the dot product of $\breve{x}$ and $\breve{y}$ is defined as $\breve{x}·\breve{y}={\sum }_{i=0}^3{x}_i{y}_i$, and $\breve{x}\perp \breve{y}$ if $\breve{x}·\breve{y}=0$, i.e., $x_0{y}_0+x_1{y}_1+x_2{y}_2+x_3{y}_3=0$. We use $di{m}_{\mathbb{Q}}\left(\breve{\mathbf{S}}\right)$ to denote of the dimension of a subspace $\breve{\mathbf{S}}$ of ${\mathbb{Q}}^{n\times 1}$, which is considered as a right quaternion vector space, and we use $Spa{n}_{\mathbb{Q}}\{{\breve{\mathbf{u}}}_1,\cdots ,{\breve{\mathbf{u}}}_m\}:=\{{\breve{\mathbf{u}}}_1{\breve{x}}_1+\cdots +{\breve{\mathbf{u}}}_m{\breve{x}}_m:{\breve{x}}_1,\cdots ,{\breve{x}}_m\in \mathbb{Q}\}$ to symbolize the subspace spanned by ${\breve{\mathbf{u}}}_1,\cdots ,{\breve{\mathbf{u}}}_m\in {\mathbb{Q}}^{n\times 1}$. Given a quaternion vector $\breve{\mathbf{x}}$, the norm on ${\mathbb{Q}}^{n\times 1}$ is denoted by $\Vert \breve{\mathbf{x}}{\Vert }_{\mathbb{Q}}=\sqrt{{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}}$ [31]. For a set of $n\times 1$ quaternion vectors {${\breve{x}}_1,{\breve{x}}_2,\cdots ,{\breve{x}}_n$}, we say they are orthonormal if ${\breve{x}}_i^H{\breve{x}}_j={\delta }_{ij}$, where ${\delta }_{ij}$ is the Kronecker delta, i.e., ${\delta }_{ij}=1$ if $i=j$, and ${\delta }_{ij}=0$ if $i\ne j$. For more details about quaternion algebra, we propose that the readers read the latest book about quaternions [31].

Similarly, we can use this expression $\breve{\mathbf{Q}}={\mathbf{Q}}_0+{\mathbf{Q}}_1\mathbf{i}+{\mathbf{Q}}_2\mathbf{j}+{\mathbf{Q}}_3\mathbf{k}$ to denote a quaternion matrix $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{m\times n}$, where each ${\mathbf{Q}}_i\in {\mathbb{R}}^{m\times n}$. The conjugate transpose of quaternion matrix $\breve{\mathbf{Q}}=\left({\breve{q}}_{ij}\right)={\mathbf{Q}}_0+{\mathbf{Q}}_1\mathbf{i}+{\mathbf{Q}}_2\mathbf{j}+{\mathbf{Q}}_3\mathbf{k}$ is ${\breve{\mathbf{Q}}}^H=\left({\overline{\breve{q}}}_{ji}\right)={\mathbf{Q}}_0^T-{\mathbf{Q}}_1^T\mathbf{i}-{\mathbf{Q}}_2^T\mathbf{j}-{\mathbf{Q}}_3^T\mathbf{k}$. The Kronecker product for two matrices $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{m\times n}$ and $\breve{\mathbf{S}}\in {\mathbb{Q}}^{s\times t}$ is denoted by $\breve{\mathbf{Q}}\otimes \breve{\mathbf{S}}$, which is the $\left(ms\right)\times \left(nt\right)$ quaternion matrix with the following structure:

$$\breve{\mathbf{Q}}\otimes \breve{\mathbf{S}}=\left(\begin{array}{cccc}{\breve{q}}_{11}\breve{\mathbf{S}}& {\breve{q}}_{12}\breve{\mathbf{S}}& \cdots & {\breve{q}}_{1n}\breve{\mathbf{S}}\\ {\breve{q}}_{21}\breve{\mathbf{S}}& {\breve{q}}_{22}\breve{\mathbf{S}}& \cdots & {\breve{q}}_{2n}\breve{\mathbf{S}}\\ ⋮& ⋮& \ddots & ⋮\\ {\breve{q}}_{m1}\breve{\mathbf{S}}& {\breve{q}}_{m2}\breve{\mathbf{S}}& \cdots & {\breve{q}}_{mn}\breve{\mathbf{S}}\end{array}\right).$$

In this paper, we focus on third-order quaternion tensors, which only have three dimensions. A third-order quaternion tensor with $n_1{n}_2{n}_3$ elements can be denoted by $\breve{\mathcal{Q}}=\left({\breve{q}}_{ijk}\right)\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$, where $1\le i\le n_1$, $1\le j\le n_2$, $1\le k\le n_3$. For a third-order quaternion tensor $\breve{\mathcal{Q}}=\left({\breve{q}}_{ijk}\right)\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$, its Frobenius norm is denoted as ${\Vert \breve{\mathcal{Q}}\Vert }_F=\sqrt{{\displaystyle \sum _{i=1}^{n_1}}{\displaystyle \sum _{j=1}^{n_2}}{\displaystyle \sum _{k=1}^{n_3}}{\left|{\breve{q}}_{ijk}\right|}^2}$. For a third-order quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$, the i-th frontal slice is referred to as ${\breve{\mathbf{Q}}}_{\left(i\right)}$ in this paper, which can also be represented by $\breve{\mathcal{Q}}(:,:,i)$ using MATLAB R2022a notation to describe algorithms. Additionally, we will introduce the following operators related to the quaternion third-order tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times n\times p}$ that are frequently used in the following sections, i.e., $unfold(·)$, $bcirc(·)$ [23,32]:

$$unfold\left(\breve{\mathcal{Q}}\right)=\left(\begin{array}{c}{\breve{\mathbf{Q}}}_{\left(1\right)}\\ {\breve{\mathbf{Q}}}_{\left(2\right)}\\ ⋮\\ {\breve{\mathbf{Q}}}_{\left(p\right)}\end{array}\right),bcirc\left(\breve{\mathcal{Q}}\right)=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_{\left(1\right)}& {\breve{\mathbf{Q}}}_{\left(p\right)}& \cdots & {\breve{\mathbf{Q}}}_{\left(2\right)}\\ {\breve{\mathbf{Q}}}_{\left(2\right)}& {\breve{\mathbf{Q}}}_{\left(1\right)}& \cdots & {\breve{\mathbf{Q}}}_{\left(3\right)}\\ ⋮& ⋮& \ddots & ⋮\\ {\breve{\mathbf{Q}}}_{\left(p\right)}& {\breve{\mathbf{Q}}}_{(p-1)}& \cdots & {\breve{\mathbf{Q}}}_{\left(1\right)}\end{array}\right),$$

(1)

where ${\breve{\mathbf{Q}}}_{\left(i\right)}$ is the i-th frontal slice of tensor $\breve{\mathcal{Q}}$ with a size of $m\times n$, $i=1,2,\cdots ,p$. It is noteworthy that $fold(unfold\left(\breve{\mathcal{Q}}\right))=\breve{\mathcal{Q}}$ and ${bcirc}^{-1}(bcirc\left(\breve{\mathcal{Q}}\right))=\breve{\mathcal{Q}}$.

A quaternion discrete Fourier transform matrix ${\breve{\mathbf{F}}}_{\breve{\mu }}\in {\mathbb{Q}}^{m\times m}$ has the following structure [33]:

$${\breve{\mathbf{F}}}_{\breve{\mu }}={\displaystyle \frac{1}{\sqrt{m}}}\left(\begin{array}{cccc}1& 1& \cdots & 1\\ 1& \breve{\omega }& \cdots & {\breve{\omega }}^{m-1}\\ ⋮& ⋮& \ddots & ⋮\\ 1& {\breve{\omega }}^{m-1}& \cdots & {\breve{\omega }}^{(m-1)(m-1)}\end{array}\right),$$

(2)

where $\breve{\omega }=e^{{\displaystyle \frac{-2\pi \breve{\mu }}{m}}}$, and $\breve{\mu }$ is a unit pure quaternion. According to [33], the quaternion discrete Fourier transform matrix ${\breve{\mathbf{F}}}_{\breve{\mu }}$ is a unitary matrix satisfying

$${\breve{\mathbf{F}}}_{\breve{\mu }}^H{\breve{\mathbf{F}}}_{\breve{\mu }}={\breve{\mathbf{F}}}_{\breve{\mu }}{\breve{\mathbf{F}}}_{\breve{\mu }}^H={\mathbf{I}}_m,$$

(3)

$${\left({\breve{\mathbf{F}}}_{\breve{\mu }}^H\right)}^2={\left({\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^2=\mathbf{M},$$

(4)

where

$$\mathbf{M}=\left(\begin{array}{ccccc}1& 0& \cdots & 0& 0\\ 0& 0& \cdots & 0& 1\\ 0& 0& \cdots & 1& 0\\ ⋮& ⋮& \ddots & ⋮& ⋮\\ 0& 1& \cdots & 0& 0\end{array}\right).$$

(5)

Next, we give some basic definitions about third-order quaternion tensors that will be utilized in the subsequent sections:

Definition 1 

(T-product [23,32]). Let $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$ and $\breve{\mathcal{P}}\in {\mathbb{Q}}^{n_2\times n_4\times n_3}$, then we call the following expression the t-product of $\breve{\mathcal{Q}}$ and $\breve{\mathcal{P}}$:

$$\begin{array}{c}\hfill \breve{\mathcal{Q}}★\breve{\mathcal{P}}:=fold(bcirc\left(\breve{\mathcal{Q}}\right)unfold\left(\breve{\mathcal{P}}\right))\in {\mathbb{Q}}^{n_1\times n_4\times n_3}.\end{array}$$

(6)

Definition 2 

(F-square tensor [34]). A tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n\times n\times m}$ is referred to as an F-square tensor provided that each of its frontal slices are square matrices.

Definition 3 

(Identity tensor [23]). We call a third-order tensor $\mathcal{I}\in {\mathbb{Q}}^{n\times n\times m}$ an identity tensor if its first frontal slice is an identity matrix with the rest being zero matrices.

Definition 4 

(Conjugate transpose of a third-order quaternion tensor [23]). Given $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$, the conjugate transpose of $\breve{\mathcal{Q}}$ is represented by ${\breve{\mathcal{Q}}}^H$, which is obtained by conjugating and transposing every frontal slice of $\breve{\mathcal{Q}}$, then reversing the order from 2 to $n_3$.

Definition 5 

(Right T-eigenvalue and T-eigentensor). Given $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, the right T-eigenvalue is a quaternion number ${\breve{\lambda }}_r$ satisfying that

$$\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\breve{\lambda }}_r,$$

where $\breve{\mathcal{X}}\in {\mathbb{Q}}^{m\times 1\times n}$ is a nonzero quaternion tensor named T-eigentensor of $\breve{\mathcal{Q}}$ corresponding to ${\breve{\lambda }}_r$. We use ${\sigma }_r\left(\breve{\mathcal{Q}}\right)$ to represent the set containing all the right T-eigenvalues of $\breve{\mathcal{Q}}$.

It is clear that for each right T-eigenvalue ${\breve{\lambda }}_r$ of $\breve{\mathcal{Q}}$, we have the following observations:

$$bcirc\left(\breve{\mathcal{Q}}\right)unfold\left(\breve{\mathcal{X}}\right)=unfold\left(\breve{\mathcal{X}}\right){\breve{\lambda }}_r,$$

that is to say, the right T-eigenvalues of $\breve{\mathcal{Q}}$ clearly correspond to the right eigenvalues belonging to the quaternion matrix $bcirc\left(\breve{\mathcal{Q}}\right)$, and the relationship holds in the reverse direction as well.

Definition 6 

(Hermitian quaternion tensors). A quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$ is called a Hermitian tensor provided that ${\breve{\mathcal{Q}}}^H=\breve{\mathcal{Q}}$.

3. The Right T-Eigenvalues of Third-Order Quaternion Tensors

In this section, we consider the right T-eigenvalues of third-order quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{H}}^{m\times m\times n}$. The following is a list of several useful lemmas:

Lemma 1 

([35]). Given four quaternion matrices $\breve{\mathbf{P}}$, $\breve{\mathbf{Q}}$, $\breve{\mathbf{U}}$ and $\breve{\mathbf{V}}$ of appropriate dimensions, then the following properties of the Kronecker product hold:

1. 

${(\breve{\mathbf{P}}\otimes \breve{\mathbf{Q}})}^H={\breve{\mathbf{P}}}^H\otimes {\breve{\mathbf{Q}}}^H$;

2. 

$(\breve{\mathbf{P}}\otimes \breve{\mathbf{Q}})(\breve{\mathbf{U}}\otimes \breve{\mathbf{V}})=\left(\breve{\mathbf{P}}\breve{\mathbf{U}}\right)\otimes \left(\breve{\mathbf{Q}}\breve{\mathbf{V}}\right)$.

where “⊗” denotes the Kronecker product.

Lemma 2 

([32,34]). Given a third-order quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n_1\times n_2\times n_3}$, the subsequent statements are true:

1. 

$bcirc(\breve{\mathcal{Q}}★\breve{\mathcal{P}})=bcirc\left(\breve{\mathcal{Q}}\right)bcirc\left(\breve{\mathcal{P}}\right)$, where $\breve{\mathcal{P}}\in {\mathbb{Q}}^{n_2\times n_4\times n_3}$.

2. 

$bcirc\left({\breve{\mathcal{Q}}}^H\right)={(bcirc\left(\breve{\mathcal{Q}}\right))}^H$.

3. 

${(\breve{\mathcal{Q}}★\breve{\mathcal{P}})}^H={\breve{\mathcal{P}}}^H★{\breve{\mathcal{Q}}}^H$.

We have the following theorems.

Lemma 3 

([33]). Suppose a quaternion circulant matrix $\breve{\mathbf{Q}}={\mathbf{Q}}_0+{\mathbf{Q}}_1\breve{\mu }+{\mathbf{Q}}_2\breve{\alpha }+{\mathbf{Q}}_3\breve{\beta }\in {\mathbb{Q}}^{m\times m}$ is in a three-axis system $(\breve{\mu },\breve{\alpha },\breve{\beta })$, where $\breve{\mu }$ is any given unit pure quaternion, $\breve{\alpha }$ is a unit pure quaternion satisfying $\breve{\alpha }\perp \breve{\mu }$ and $\breve{\beta }=\breve{\alpha }\breve{\mu }$. Then, there exists a permuted quaternion Fourier transform matrix $\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}$ that can block-diagonalize $\breve{\mathbf{Q}}$, that is,

$$\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}{\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_s\end{array}\right),$$

(7)

where $\mathbf{P}$ is a permutation matrix obtained by exchanging of the i-th and $(m-i+3)$-th columns of a $m\times m$ identity matrix for $i=3,4,\cdots ,{\displaystyle \frac{m+1}{2}}$ when m is odd or for $i=3,4,\cdots ,{\displaystyle \frac{m}{2}}$ when m is even. ${\breve{\mathbf{Q}}}_1$ is of size $1\times 1$, while the other diagonal block is of size $2\times 2$ when m is odd. In the final diagonal block for an even m, there is an additional 1-by-1 matrix, in addition to the previously mentioned diagonal blocks.

Theorem 1. 

Given a quaternion circulant matrix $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{m\times m}$, the matrix $\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}{\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H$ has the same right eigenvalues as matrix $\breve{\mathbf{Q}}$, where $\mathbf{P}$ is a permutation matrix defined as in Lemma 3.

Proof. 

Suppose ${\breve{\lambda }}_r$ is one right eigenvalue of $\breve{\mathbf{Q}}$, and $\breve{\eta }$ is the corresponding eigenvector, i.e.,

$$\breve{\mathbf{Q}}\breve{\eta }=\breve{\eta }{\breve{\lambda }}_r,\breve{\eta }\ne \mathbf{0}.$$

(8)

By left-multiplying $\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}$ from both sides of Equation (8), we have that

$$\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}\breve{\eta }=\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\eta }{\breve{\lambda }}_r,\breve{\eta }\ne \mathbf{0},$$

(9)

i.e.,

$$\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}{\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\eta }=\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\eta }{\breve{\lambda }}_r,\breve{\eta }\ne \mathbf{0}.$$

(10)

The above equation holds true because ${\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)={\mathbf{I}}_m$.

Let $\breve{\xi }=\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\eta }$; it is easy to find that $\breve{\xi }\ne \mathbf{0}$, and from Equation (10) we have that

$$\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}{\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H\breve{\xi }=\breve{\xi }{\breve{\lambda }}_r,\breve{\xi }\ne \mathbf{0}.$$

(11)

That is to say, ${\breve{\lambda }}_r$ is also one right eigenvalue of $\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\mathbf{Q}}{\left[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\right]}^H$, and $\breve{\xi }=\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\breve{\eta }$ is an eigenvector corresponding to ${\breve{\lambda }}_r$.    □

Corollary 1. 

Given a quaternion F-square tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, the matrix $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m]$ has the same right eigenvalues as matrix $bcirc\left(\breve{\mathcal{Q}}\right)$, where $\mathbf{P}$ is a permutation matrix defined by swapping the t-th and $(n-t+3)$-th columns of an identity matrix for $t=3,\cdots ,{\displaystyle \frac{n+1}{2}}$ if n is odd or for $t=3,\cdots ,{\displaystyle \frac{n}{2}}$ if n is even.

Theorem 2. 

Assume that $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{n\times n}=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_t\end{array}\right)$, where each ${\breve{\mathbf{Q}}}_s$ is a quaternion matrix. Then, we have that the right eigenvalues of $\breve{\mathbf{Q}}$ are the union of right eigenvalues of quaternion matrices ${\breve{\mathbf{Q}}}_s$, that is, ${\sigma }_r\left(\breve{\mathbf{Q}}\right)={\cup }_{s=1}^t{\sigma }_r\left({\breve{\mathbf{Q}}}_s\right)$.

Proof. 

Suppose that the size of each matrix ${\breve{\mathbf{Q}}}_s$ is $n_s\times n_s$, and ${\breve{\lambda }}_s$ is one right eigenvalue belonging to ${\breve{\mathbf{Q}}}_s$ with the corresponding eigenvector ${\breve{\mathbf{x}}}_s\ne \mathbf{0}$. Then, we have that

$$\breve{\mathbf{Q}}\left(\begin{array}{c}{\breve{\mathbf{x}}}_1\\ \mathbf{0}\\ ⋮\\ \mathbf{0}\end{array}\right)=\left(\begin{array}{c}{\breve{\mathbf{x}}}_1\\ \mathbf{0}\\ ⋮\\ \mathbf{0}\end{array}\right){\breve{\lambda }}_1,i.e.,\breve{\mathbf{Q}}{\breve{\mathbf{X}}}_1={\breve{\mathbf{X}}}_1{\breve{\lambda }}_1,{\breve{\mathbf{X}}}_1=\left(\begin{array}{c}{\breve{\mathbf{x}}}_1\\ \mathbf{0}\\ ⋮\\ \mathbf{0}\end{array}\right)\ne \mathbf{0},$$

$$\breve{\mathbf{Q}}\left(\begin{array}{c}\mathbf{0}\\ {\breve{\mathbf{x}}}_2\\ ⋮\\ \mathbf{0}\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ {\breve{\mathbf{x}}}_2\\ ⋮\\ \mathbf{0}\end{array}\right){\breve{\lambda }}_2,i.e.,\breve{\mathbf{Q}}{\breve{\mathbf{X}}}_2={\breve{\mathbf{X}}}_2{\breve{\lambda }}_2,{\breve{\mathbf{X}}}_2=\left(\begin{array}{c}\mathbf{0}\\ {\breve{\mathbf{x}}}_2\\ ⋮\\ \mathbf{0}\end{array}\right)\ne \mathbf{0},$$

$$⋮$$

$$\breve{\mathbf{Q}}\left(\begin{array}{c}\mathbf{0}\\ \mathbf{0}\\ ⋮\\ {\breve{\mathbf{x}}}_t\end{array}\right)=\left(\begin{array}{c}\mathbf{0}\\ \mathbf{0}\\ ⋮\\ {\breve{\mathbf{x}}}_t\end{array}\right){\breve{\lambda }}_t,i.e.,\breve{\mathbf{Q}}{\breve{\mathbf{X}}}_t={\breve{\mathbf{X}}}_t{\breve{\lambda }}_t,{\breve{\mathbf{X}}}_t=\left(\begin{array}{c}\mathbf{0}\\ \mathbf{0}\\ ⋮\\ {\breve{\mathbf{x}}}_t\end{array}\right)\ne \mathbf{0}.$$

Thus, it is easy to see that ${\breve{\lambda }}_s$ is one right eigenvalue of $\breve{\mathbf{Q}}$ with the correlated eigenvector ${\breve{\mathbf{X}}}_s$, and the right eigenvalues of the block-diagonal quaternion matrix $\breve{\mathbf{Q}}$ consist of the right eigenvalues of each diagonal block ${\breve{\mathbf{Q}}}_s$, $s=1,2,\cdots ,t$.    □

Theorem 3. 

Given an F-square tensor $\breve{\mathcal{Q}}={\mathcal{Q}}_0+{\mathcal{Q}}_1\breve{\mu }+{\mathcal{Q}}_2\breve{\alpha }+{\mathcal{Q}}_3\breve{\beta }\in {\mathbb{Q}}^{m\times m\times n}$ is in a three-axis system $(\breve{\mu },\breve{\alpha },\breve{\beta })$, where $\breve{\mu }$ is any given unit pure quaternion, $\breve{\alpha }$ is a unit pure quaternion satisfying $\breve{\alpha }\perp \breve{\mu }$ and $\breve{\beta }=\breve{\alpha }\breve{\mu }$. Thus, the right T-eigenvalues of $\breve{\mathcal{Q}}$ are composed of right eigenvalues of the diagonal block matrices of $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]$, where $\mathbf{P}$ is a permutation matrix by swapping i-th and $(n-i+3)$-th columns of a $n\times n$ identity matrix for $i=3,4,\cdots ,{\displaystyle \frac{n+1}{2}}$ if n is odd and for $i=3,4,\cdots ,{\displaystyle \frac{n}{2}}$ if n is even; ${\breve{\mathbf{F}}}_{\breve{\mu }}$ is the quaternion discrete Fourier transform matrix.

Proof. 

It is shown by Ng et al. [33] that a permuted quaternion discrete Fourier transform matrix can transform a quaternion circulant matrix into a block-diagonal form, and further it is extended to the following case:

Given a third-order F-square tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, then according to [33], we have that

$$[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_t\end{array}\right)\equiv D_Q,$$

(12)

where $t={\displaystyle \frac{n}{2}}+1$ if n is even, $t={\displaystyle \frac{n+1}{2}}$ if n is odd. $\mathbf{P}$ is a permutation matrix by swapping the i-th and the $(n-i+3)$-th columns of an identity matrix for $i=3,4,\cdots ,{\displaystyle \frac{n+1}{2}}$ if n is odd or for $i=3,4,\cdots ,{\displaystyle \frac{n}{2}}$ if n is even. It is worth noting that $D_Q$ is a block-diagonal quaternion matrix with ${\breve{\mathbf{Q}}}_1$ being $m\times m$ while other diagonal blocks are $2m\times 2m$ if n is odd, and in the final diagonal block for an even m, there is an additional $m\times m$ matrix, in addition to the previously mentioned diagonal blocks.

It is noted that the proofs for both the even and odd cases bear similarities. For the sake of simplicity, only the odd case will be considered here.

Suppose that ${\breve{\lambda }}_i$ is one right eigenvalue of diagonal block matrix ${\breve{\mathbf{Q}}}_i$ with the corresponding nonzero eigenvector ${\breve{\mathbf{x}}}_i$. Then by Theorem 2, we have that ${\breve{\lambda }}_i$ is also one right eigenvalue of $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]$ with the corresponding eigenvector ${\breve{\mathbf{X}}}_i={(\mathbf{0},\cdots ,{\breve{\mathbf{x}}}_i,\cdots ,\mathbf{0})}^T$, that is,

$$[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]{\breve{\mathbf{X}}}_i={\breve{\mathbf{X}}}_i{\breve{\lambda }}_i.$$

(13)

After left-multiplying both sides of Equation (13) by matrix $[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]$, we can determine that

$$bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]{\breve{\mathbf{X}}}_i=[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]{\breve{\mathbf{X}}}_i{\breve{\lambda }}_i.$$

(14)

Let ${\breve{\mathbf{Y}}}_i=[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]{\breve{\mathbf{X}}}_i$, then it is easy to see that ${\breve{\mathbf{Y}}}_i\ne 0$. Thus we have that ${\breve{\lambda }}_r$ is one right T-eigenvalue belonging to $\breve{\mathcal{Q}}$, and the corresponding T-eigentensor is $fold\left({\breve{\mathbf{Y}}}_i\right)$. This can be proven by the combination of the proof of Theorems 1 and 2, that is, ${\sigma }_r\left(\breve{\mathcal{Q}}\right)={\sigma }_r\left(D_Q\right)={\cup }_{s=1}^t{\sigma }_r\left({\breve{\mathbf{Q}}}_s\right).$    □

Theorem 4. 

Given a right T-eigenvalue ${\breve{\lambda }}_r$ of the quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$ with a T-eigentensor $\breve{\mathcal{X}}$ related to ${\breve{\lambda }}_r$, then for every nonzero quaternion number $\breve{p}$, ${\breve{p}}^{-1}{\breve{\lambda }}_r\breve{p}$ is again a right T-eigenvalue of $\breve{\mathcal{Q}}$ with the corresponding T-eigentensor $\breve{\mathcal{X}}\breve{p}$.

Proof. 

Note that $\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\breve{\lambda }}_r$, thus we have that

$$\breve{\mathcal{Q}}★\left(\breve{\mathcal{X}}\breve{p}\right)=(\breve{\mathcal{Q}}★\breve{\mathcal{X}})\breve{p}=\left(\breve{\mathcal{X}}{\breve{\lambda }}_r\right)\breve{p}=\breve{\mathcal{X}}\breve{p}\left({\breve{p}}^{-1}{\breve{\lambda }}_r\breve{p}\right).$$

   □

Regarding a Hermitian quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, the following theorems hold.

Theorem 5. 

Given $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, then $\breve{\mathcal{Q}}$ is a Hermitian tensor $\iff {(bcirc\left(\breve{\mathcal{Q}}\right))}^H=bcirc\left(\breve{\mathcal{Q}}\right)$.

Theorem 6. 

The right T-eigenvalues of a quaternion Hermitian tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$ all belong to the real number field.

Proof. 

It is evident that if $bcirc\left(\breve{\mathcal{Q}}\right)=bcirc\left({\breve{\mathcal{Q}}}^H\right)=bcirc{\left(\breve{\mathcal{Q}}\right)}^H$ from Lemma 2, then $bcirc\left(\breve{\mathcal{Q}}\right)$ is a quaternion Hermitian matrix. Suppose that $\breve{\lambda }$ is one right T-eigenvalue belonging to $\breve{\mathcal{Q}}$ with the corresponding eigentensor $\breve{\mathcal{X}}$, i.e., $\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}\breve{\lambda }$. That is,

$$bcirc\left(\breve{\mathcal{Q}}\right)unfold\left(\breve{\mathcal{X}}\right)=unfold\left(\breve{\mathcal{X}}\right)\breve{\lambda }.$$

Let $\breve{\mathbf{X}}=unfold\left(\breve{\mathcal{X}}\right)$, and from the above equation it is easy to see that

$${\breve{\mathbf{X}}}^H{(bcirc\left(\breve{\mathcal{Q}}\right))}^H=\overline{\breve{\lambda }}{\breve{\mathbf{X}}}^H.$$

Due to the fact that $bcirc\left(\breve{\mathcal{Q}}\right)=bcirc{\left(\breve{\mathcal{Q}}\right)}^H$, we have

$${\breve{\mathbf{X}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)=\overline{\breve{\lambda }}{\breve{\mathbf{X}}}^H.$$

We multiply $\breve{\mathbf{X}}$ from the right side of the above equation, then we obtain this equation:

$${\breve{\mathbf{X}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{X}}=\overline{\breve{\lambda }}{\breve{\mathbf{X}}}^H\breve{\mathbf{X}},$$

and one step closer, we have

$${\breve{\mathbf{X}}}^H\breve{\mathbf{X}}\breve{\lambda }=\overline{\breve{\lambda }}{\breve{\mathbf{X}}}^H\breve{\mathbf{X}},$$

i.e., 

$$\breve{\lambda }\parallel \breve{\mathbf{X}}{\parallel }_{\mathbb{Q}}^2=\overline{\breve{\lambda }}{\parallel \breve{\mathbf{X}}\parallel }_{\mathbb{Q}}^2.$$

Then, we have $\breve{\lambda }=\overline{\breve{\lambda }}$.    □

Proposition 1. 

Given a Hermitian quaternion third-order tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$, then each diagonal block in Equation (12) is a Hermitian matrix.

Proof. 

According to Theorem 3, it is easy to see that

$$\{[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]{\}}^H=[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc{\left(\breve{\mathcal{Q}}\right)}^H[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)],$$

(15)

$$\begin{array}{cc}\hfill & =[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)],\hfill \end{array}$$

(16)

which means that $\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_t\end{array}\right)\equiv D_Q$ is a Hermitian matrix, i.e., $D_Q^H=D_Q$.

Then, we have ${\breve{\mathbf{Q}}}_i^H={\breve{\mathbf{Q}}}_i,i=1,2,\cdots t$.    □

Very recently, Wu et al. [36] proposed a complex representation matrix (CRM) of third-order quaternion tensors under Qt-product, which provides a way to map quaternion operators to their complex counterparts, providing a promising avenue for generalizing classical tensor algorithms to the quaternion case. By this, the computational complexity generated by extra imaginary parts of quaternions can be reduced. Inspired by [36], here we consider complex representation matrix to compute right T-eigenvalues of third-order quaternion tensors. First, we revisit the complex representation operator for a quaternion matrix.

Definition 7 

(Complex representation operator for a quaternion matrix [31]). Given a quaternion matrix $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{m\times n}$ in its Cayley–Dickson (CD) form $\breve{\mathbf{Q}}={\mathbf{Q}}_1+{\mathbf{Q}}_2\mathbf{j}$, where ${\mathbf{Q}}_i\in {\mathbb{C}}^{m\times n},i=1,2.$ $\Phi :{\mathbb{Q}}^{m\times n}⟶{\mathbb{C}}^{2m\times 2n}$ is an operator defined as follows:

$$\Phi \left(\breve{\mathbf{Q}}\right)=\left(\begin{array}{cc}{\mathbf{Q}}_1& {\mathbf{Q}}_2\\ -\overline{{\mathbf{Q}}_2}& \overline{{\mathbf{Q}}_1}\end{array}\right).$$

Definition 8 

(Complex representation matrix of third-order quaternion tensors under t-product). Consider $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times n\times p}$ and the corresponding block-diagonal matrix $D_Q$ in (12). Then, $\Psi :{\mathbb{Q}}^{m\times n\times p}⟶{\mathbb{C}}^{2mp\times 2np}$ is an operator defined as $\Psi \left(\breve{\mathcal{Q}}\right)=\left(\begin{array}{cccc}\Phi \left({\breve{\mathbf{Q}}}_1\right)& & & \\ & \Phi \left({\breve{\mathbf{Q}}}_2\right)& & \\ & & \ddots & \\ & & & \Phi \left({\breve{\mathbf{Q}}}_t\right)\end{array}\right),$ where Φ is the complex representation operator for quaternion matrices.

Based on the proof of the theorems above, next we provide an algorithm for calculating the right T-eigenvalues belonging to a quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$.

Upon the constructed algorithm, next we give a numerical experiment and compare our method with the command eig in Quaternion tool box [37] for computing the right eigenvalues of $bcirc\left(\breve{\mathcal{Q}}\right)$, i.e., the right T-eigenvalues of $\breve{\mathcal{Q}}$. For each right T-eigenvalue ${\breve{\lambda }}_i$ and the corresponding T-eigentensor ${\breve{\mathcal{X}}}_i$ of $\breve{\mathcal{Q}}$, we define the residual by $\Vert \breve{\mathcal{Q}}★{\breve{\mathcal{X}}}_i-{\breve{\mathcal{X}}}_i{\breve{\lambda }}_i{\Vert }_F$. Then, we define the error for calculating right T-eigenvalues of $\breve{\mathcal{Q}}$ as follows: Error $=\sqrt{{\displaystyle \sum _i}{\Vert \breve{\mathcal{Q}}★{\breve{\mathcal{X}}}_i-{\breve{\mathcal{X}}}_i{\breve{\lambda }}_i\Vert }_F^2}$.

Example 1. 

Suppose $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{3\times 3\times 3}$ with the following structure:

$$\breve{\mathcal{Q}}(:,:,1)=\left(\begin{array}{ccc}-0.5244& -0.4052-0.7897\mathbf{i}-0.1849\mathbf{j}-0.2486\mathbf{k}& 0.2136-0.1193\mathbf{i}-0.0896\mathbf{j}+0.0811\mathbf{k}\\ -0.4052+0.7897\mathbf{i}+0.1849\mathbf{j}+0.2486\mathbf{k}& 0.5731& 0.1266-0.1104\mathbf{i}-0.1398\mathbf{j}-0.6492\mathbf{k}\\ 0.2136+0.1193\mathbf{i}+0.0896\mathbf{j}-0.0811\mathbf{k}& 0.1266+0.1104\mathbf{i}+0.1398\mathbf{j}+0.6492\mathbf{k}& 0.9014\end{array}\right),$$

$$\breve{\mathcal{Q}}(:,:,2)=\left(\begin{array}{ccc}1.4770-0.1757\mathbf{i}+0.9177\mathbf{j}-0.2357\mathbf{k}& 0.0746+0.2922\mathbf{i}+1.4866\mathbf{j}-0.3078\mathbf{k}& -0.4387-0.2311\mathbf{i}-0.7924\mathbf{j}+0.2718\mathbf{k}\\ 0.3402-0.6744\mathbf{i}+0.2752\mathbf{j}-0.1544\mathbf{k}& 1.0954+0.3069\mathbf{i}-0.9133\mathbf{j}-0.9556\mathbf{k}& -0.7820-0.6189\mathbf{i}+1.0518\mathbf{j}-0.4706\mathbf{k}\\ -0.7543+0.5253\mathbf{i}-1.4818\mathbf{j}-0.2781\mathbf{k}& 1.4399+0.6250\mathbf{i}+0.9021\mathbf{j}+0.0340\mathbf{k}& 0.4441+0.0945\mathbf{i}-0.1625\mathbf{j}+0.4622\mathbf{k}\end{array}\right),$$

$$\breve{\mathcal{Q}}(:,:,3)=\left(\begin{array}{ccc}-1.4770+0.1757\mathbf{i}-0.9177\mathbf{j}+0.2357\mathbf{k}& 0.3402+0.6744\mathbf{i}-0.2752\mathbf{j}+0.1544\mathbf{k}& -0.7543-0.5253\mathbf{i}+1.4818\mathbf{j}+0.2781\mathbf{k}\\ 0.0746-0.2922\mathbf{i}-1.4866\mathbf{j}+0.3078\mathbf{k}& 1.0954-0.3069\mathbf{i}+0.9133\mathbf{j}+0.9556\mathbf{k}& 1.4399-0.6250\mathbf{i}-0.9021\mathbf{j}-0.0340\mathbf{k}\\ -0.4387+0.2311\mathbf{i}+0.7924\mathbf{j}-0.2718\mathbf{k}& -0.7820+0.6189\mathbf{i}-1.0518\mathbf{j}+0.4706\mathbf{k}& 0.4441-0.0945\mathbf{i}+0.1625\mathbf{j}-0.4622\mathbf{k}\end{array}\right).$$

By Algorithm 1, we can compute its right T-eigenvalues and the corresponding T-eigentensors as follows:

$$\left\{\begin{array}{c}{\lambda }_1=-4.0017\hfill \\ {\breve{\mathcal{X}}}_1=\left(\begin{array}{ccc}-0.5555& -0.5555& -0.5555\\ 0.0057-0.0367\mathbf{i}-0.0565\mathbf{j}+0.0140\mathbf{k}& 0.0057-0.0367\mathbf{i}-0.0565\mathbf{j}+0.0140\mathbf{k}& 0.0057-0.0367\mathbf{i}-0.0565\mathbf{j}+0.0140\mathbf{k}\\ -0.1004+0.0764\mathbf{i}-0.0393\mathbf{j}-0.0503\mathbf{k}& -0.1004+0.0764\mathbf{i}-0.0393\mathbf{j}-0.0503\mathbf{k}& -0.1004+0.0764\mathbf{i}-0.0393\mathbf{j}-0.0503\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_2=4.6672\hfill \\ {\breve{\mathcal{X}}}_2=\left(\begin{array}{ccc}-0.1238& -0.1238& -0.1238\\ 0.0202-0.0588\mathbf{i}+0.3867\mathbf{j}-0.1894\mathbf{k}& 0.0202-0.0588\mathbf{i}+0.3867\mathbf{j}-0.1894\mathbf{k}& 0.0202-0.0588\mathbf{i}+0.3867\mathbf{j}-0.1894\mathbf{k}\\ 0.1526-0.1986\mathbf{i}+0.1967\mathbf{j}+0.1654\mathbf{k}& 0.1526-0.1986\mathbf{i}+0.1967\mathbf{j}+0.1654\mathbf{k}& 0.1526-0.1986\mathbf{i}+0.1967\mathbf{j}+0.1654\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_3=0.4097\hfill \\ {\breve{\mathcal{X}}}_3=\left(\begin{array}{ccc}-0.0972& -0.0972& -0.0972\\ -0.0580+0.2848\mathbf{i}-0.1696\mathbf{j}+0.1614\mathbf{k}& -0.0580+0.2848\mathbf{i}-0.1696\mathbf{j}+0.1614\mathbf{k}& -0.0580+0.2848\mathbf{i}-0.1696\mathbf{j}+0.1614\mathbf{k}\\ 0.3797-0.1839\mathbf{i}-0.0259\mathbf{j}+0.0767\mathbf{k}& 0.3797-0.1839\mathbf{i}-0.0259\mathbf{j}+0.0767\mathbf{k}& 0.3797-0.1839\mathbf{i}-0.0259\mathbf{j}+0.0767\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_4=-5.4785\hfill \\ {\breve{\mathcal{X}}}_4=\left(\begin{array}{ccc}-0.0885-0.0476\mathbf{i}-0.1642\mathbf{j}+0.2118\mathbf{k}& 0.0443-0.2085\mathbf{i}+0.1675\mathbf{j}-0.0919\mathbf{k}& 0.0443+0.2561\mathbf{i}-0.0034\mathbf{j}-0.1199\mathbf{k}\\ -0.2962+0.0151\mathbf{i}+0.2791\mathbf{j}+0.2094\mathbf{k}& -0.1037-0.1208\mathbf{i}-0.1905\mathbf{j}-0.3848\mathbf{k}& 0.3999+0.1057\mathbf{i}-0.0886\mathbf{j}+0.1754\mathbf{k}\\ -0.0230-0.1590\mathbf{i}-0.0848\mathbf{j}-0.0932\mathbf{k}& 0.1800+0.0722\mathbf{i}+0.0638\mathbf{j}-0.0020\mathbf{k}& -0.1570+0.0868\mathbf{i}+0.0210\mathbf{j}+0.0952\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_5=-3.6183\hfill \\ {\breve{\mathcal{X}}}_5=\left(\begin{array}{ccc}-0.2079-0.1533\mathbf{i}+0.0128\mathbf{j}+0.1405\mathbf{k}& 0.1040-0.0912\mathbf{i}+0.0365\mathbf{j}-0.2572\mathbf{k}& 0.1040+0.2444\mathbf{i}-0.0493\mathbf{j}+0.1167\mathbf{k}\\ 0.2022-0.1117\mathbf{i}+0.1590\mathbf{j}+0.0069\mathbf{k}& -0.1281+0.2330\mathbf{i}+0.0809\mathbf{j}-0.0377\mathbf{k}& -0.0740-0.1213\mathbf{i}-0.2398\mathbf{j}+0.0308\mathbf{k}\\ -0.0155+0.3049\mathbf{i}+0.1364\mathbf{j}-0.2371\mathbf{k}& -0.0944+0.0266\mathbf{i}-0.3470\mathbf{j}+0.1951\mathbf{k}& 0.1098-0.3315\mathbf{i}+0.2106\mathbf{j}+0.0421\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_6=5.7744\hfill \\ {\breve{\mathcal{X}}}_6=\left(\begin{array}{ccc}-0.1234+0.1455\mathbf{i}-0.2358\mathbf{j}+0.0903\mathbf{k}& 0.0617-0.2975\mathbf{i}+0.0286\mathbf{j}+0.0838\mathbf{k}& 0.0617+0.1520\mathbf{i}+0.2072\mathbf{j}-0.1741\mathbf{k}\\ 0.0260-0.2498\mathbf{i}+0.0963\mathbf{j}-0.2124\mathbf{k}& 0.1699+0.2923\mathbf{i}-0.0164\mathbf{j}-0.0539\mathbf{k}& -0.1960-0.0425\mathbf{i}-0.0799\mathbf{j}+0.2663\mathbf{k}\\ -0.0339+0.0895\mathbf{i}-0.1100\mathbf{j}+0.3073\mathbf{k}& -0.1265-0.2703\mathbf{i}+0.1469\mathbf{j}-0.0708\mathbf{k}& 0.1604+0.1808\mathbf{i}-0.0369\mathbf{j}-0.2364\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_7=-0.0916\hfill \\ {\breve{\mathcal{X}}}_7=\left(\begin{array}{ccc}0.1887+0.1837\mathbf{i}-0.1470\mathbf{j}-0.0367\mathbf{k}& -0.0944-0.0526\mathbf{i}+0.0577\mathbf{j}+0.2781\mathbf{k}& -0.0944-0.1311\mathbf{i}+0.0893\mathbf{j}-0.2414\mathbf{k}\\ 0.0430-0.0062\mathbf{i}+0.0704\mathbf{j}-0.2568\mathbf{k}& 0.0748+0.1882\mathbf{i}-0.1390\mathbf{j}+0.1116\mathbf{k}& -0.1178-0.1820\mathbf{i}+0.0686\mathbf{j}+0.1452\mathbf{k}\\ -0.1055-0.0479\mathbf{i}-0.1379\mathbf{j}-0.3685\mathbf{k}& 0.3299+0.0865\mathbf{i}-0.1441\mathbf{j}+0.1765\mathbf{k}& -0.2244-0.0386\mathbf{i}+0.2820\mathbf{j}+0.1920\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_8=3.2685\hfill \\ {\breve{\mathcal{X}}}_8=\left(\begin{array}{ccc}0.3407-0.0823\mathbf{i}-0.1201\mathbf{j}+0.2024\mathbf{k}& -0.1704+0.0502\mathbf{i}+0.3728\mathbf{j}+0.0881\mathbf{k}& -0.1704+0.0321\mathbf{i}-0.2527\mathbf{j}-0.2905\mathbf{k}\\ -0.1337+0.0152\mathbf{i}-0.1322\mathbf{j}+0.0583\mathbf{k}& 0.0962-0.1697\mathbf{i}+0.0208\mathbf{j}-0.0223\mathbf{k}& 0.0374+0.1545\mathbf{i}+0.1114\mathbf{j}-0.0360\mathbf{k}\\ -0.1836+0.2650\mathbf{i}+0.0897\mathbf{j}+0.0641\mathbf{k}& -0.1176-0.2115\mathbf{i}-0.2371\mathbf{j}-0.0362\mathbf{k}& 0.3012-0.0535\mathbf{i}+0.1474\mathbf{j}-0.0279\mathbf{k}\end{array}\right),\hfill \end{array}\right.$$

$$\left\{\begin{array}{c}{\lambda }_9=1.9206\hfill \\ {\breve{\mathcal{X}}}_9=\left(\begin{array}{ccc}0.3396-0.0729\mathbf{i}+0.0816\mathbf{j}-0.0087\mathbf{k}& -0.1698+0.2513\mathbf{i}+0.1611\mathbf{j}+0.0969\mathbf{k}& -0.1698-0.1785\mathbf{i}-0.2427\mathbf{j}-0.0882\mathbf{k}\\ 0.1663-0.1670\mathbf{i}+0.2986\mathbf{j}-0.0659\mathbf{k}& -0.1819+0.2830\mathbf{i}+0.0503\mathbf{j}-0.1826\mathbf{k}& 0.0156-0.1160\mathbf{i}-0.3489\mathbf{j}+0.1167\mathbf{k}\\ 0.2151-0.0615\mathbf{i}+0.0081\mathbf{j}+0.0827\mathbf{k}& -0.1222+0.1010\mathbf{i}+0.1756\mathbf{j}+0.0314\mathbf{k}& -0.0929-0.0395\mathbf{i}-0.1837\mathbf{j}-0.1141\mathbf{k}\end{array}\right).\hfill \end{array}\right.$$

Algorithm 1: Calculation for the right T-eigenvalues of a third-order quaternion tensor $\breve{\mathcal{Q}}$

Input: One quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{m\times m\times n}$. Output: Right T-eigenvalues $\breve{\lambda }$ and T-eigentensors $\breve{\mathcal{X}}$ such that

$\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}\breve{\lambda }$.

step1: Construct the following block-diagonalization:

$[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]=diag({\breve{\mathbf{Q}}}_1,\cdots ,{\breve{\mathbf{Q}}}_t)\equiv D_Q$.

step2: Construct the complex representation matrix:

$\Psi \left(\breve{\mathcal{Q}}\right)=diag(\Phi \left({\breve{\mathbf{Q}}}_1\right),\cdots ,\Phi \left({\breve{\mathbf{Q}}}_t\right))$.

step3: Calculate the right eigenvalues ${\breve{\lambda }}_i$ and corresponding eigenvectors ${\breve{\mathbf{x}}}_i$ of each diagonal block ${\breve{\mathbf{Q}}}_i$ of $D_Q$ through $\Phi \left({\breve{\mathbf{Q}}}_i\right)$ by the method in [38]. step4: Compute the eigenvector ${\breve{\mathbf{X}}}_i={(0,\cdots ,{\breve{\mathbf{x}}}_i,\cdots ,0)}^T$ of $D_Q$ corresponding to eigenvalue ${\breve{\lambda }}_i$. step5: Compute the eigenvector ${\breve{\mathbf{Y}}}_i=[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m]{\breve{\mathbf{X}}}_i$ of $bcirc\left(\breve{\mathcal{Q}}\right)$ corresponding to eigenvalue ${\breve{\lambda }}_i$. step6: Compute the T-eigentensor $\breve{\mathcal{X}}=fold\left({\breve{\mathbf{Y}}}_i\right)$.

In this example, the Error of Algorithm 1 is $1.0931\times {10}^{-14}$ and the CPU time is 0.0122253 s, compared with the Error $1.7225\times {10}^{-14}$ and the CPU time 0.0409140 s of the method based on the command eig in Quaternion tool box [37], which means that our method for computing the right T-eigenvalues is faster while maintaining better computational accuracy.

4. Inequalities for Right T-Eigenvalues of Hermitian Quaternion Tensors

Next, we examine the inequalities for right T-eigenvalues belonging to Hermitian quaternion third-order tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n\times n\times m}$, and we assume that the right T-eigenvalues ${\lambda }_i\left(\breve{\mathcal{Q}}\right)$ are arranged in nondecreasing order, i.e., ${\lambda }_1\left(\breve{\mathcal{Q}}\right)\le {\lambda }_2\left(\breve{\mathcal{Q}}\right)\le \cdots \le {\lambda }_{nm}\left(\breve{\mathcal{Q}}\right)$.

Fact 1 

([31]). Vectors ${\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_t\in {\mathbb{R}}^{n\times 1}$ are linearly independent over $\mathbb{R}$ if and only if they are linearly independent over $\mathbb{Q}$, i.e., $di{m}_{R}Span\{{\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_t\}$ is equal to $di{m}_{Q}Span\{{\mathbf{v}}_1,{\mathbf{v}}_2,\cdots ,{\mathbf{v}}_t\}$.

Lemma 4. 

Given a Hermitian matrix $\breve{\mathbf{S}}\in {\mathbb{Q}}^{n\times n}$, let ${\breve{\mathbf{u}}}_1,{\breve{\mathbf{u}}}_2,\cdots ,{\breve{\mathbf{u}}}_n$ be orthonormal eigenvectors of $\breve{\mathbf{S}}$ associated with real right eigenvalues ${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$, respectively.

Assume that $\breve{\mathbf{W}}$ is spanned by the above eigenvectors, i.e., $\breve{\mathbf{W}}=Spa{n}_{\mathbb{Q}}\{{\breve{\mathbf{u}}}_p,\cdots ,{\breve{\mathbf{u}}}_q\},1\le p\le q\le n$. Thus for any unit $\breve{\mathbf{x}}\in \breve{\mathbf{W}}$, we have

$${\lambda }_p\left(\breve{\mathbf{S}}\right)\le {\breve{\mathbf{x}}}^H\breve{\mathbf{S}}\breve{\mathbf{x}}\le {\lambda }_q\left(\breve{\mathbf{S}}\right).$$

(17)

Proof. 

Let $\breve{\mathbf{x}}={\breve{\mathbf{u}}}_p{\breve{x}}_p+\cdots +{\breve{\mathbf{u}}}_q{\breve{x}}_q$, then ${\breve{\mathbf{x}}}^H={\overline{\breve{x}}}_p{\breve{\mathbf{u}}}_p^H+\cdots +{\overline{\breve{x}}}_q{\breve{\mathbf{u}}}_q^H$. One step closer, we have

$$\begin{array}{cc}\hfill {\breve{\mathbf{x}}}^H\breve{\mathbf{S}}\breve{\mathbf{x}}& ={\breve{x}}^H\left[\breve{\mathbf{S}}({\breve{\mathbf{u}}}_p{\breve{x}}_p+\cdots +{\breve{\mathbf{u}}}_q{\breve{x}}_q)\right] \hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill & ={\breve{\mathbf{x}}}^H({\breve{\mathbf{u}}}_p{\lambda }_p{\breve{x}}_p+\cdots +{\breve{\mathbf{u}}}_q{\lambda }_q{\breve{x}}_q) \hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & ={\lambda }_p{\breve{\mathbf{x}}}^H{\breve{\mathbf{u}}}_p{\breve{x}}_p+\cdots +{\lambda }_q{\breve{\mathbf{x}}}^H{\breve{\mathbf{u}}}_q{\breve{x}}_q \hfill \end{array}$$

(20)

$$\begin{array}{cc}\hfill & ={\lambda }_p({\overline{\breve{x}}}_p{\breve{\mathbf{u}}}_p^H+\cdots +{\overline{\breve{x}}}_q{\breve{\mathbf{u}}}_q^H){\breve{\mathbf{u}}}_p{\breve{x}}_p+\cdots +{\lambda }_q({\overline{\breve{x}}}_p{\breve{\mathbf{u}}}_p^H+\cdots +{\overline{\breve{x}}}_q{\breve{\mathbf{u}}}_q^H){\breve{\mathbf{u}}}_q{\breve{x}}_q\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill & ={\lambda }_p|{\breve{x}}_p{|}^2+\cdots +{\lambda }_q{\left|{\breve{x}}_q\right|}^2. \hfill \end{array}$$

(22)

Since ${\sum }_{i=p}^q{\left|{\breve{\mathbf{x}}}_i\right|}^2=1$, then it is easy to determine that ${\lambda }_p\left(\breve{\mathbf{S}}\right)\le {\breve{\mathbf{x}}}^H\breve{\mathbf{S}}\breve{\mathbf{x}}\le {\lambda }_q\left(\breve{\mathbf{S}}\right)$. □

Lemma 5 

(Rayleigh–Ritz). Consider a third-order Hermitian tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n\times n\times m}$. Then,

$${\lambda }_{min}\left(\breve{\mathcal{Q}}\right)=\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{min}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}},$$

and

$${\lambda }_{max}\left(\breve{\mathcal{Q}}\right)=\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{max}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}}.$$

Proof. 

The eigenvectors of $bcirc\left(\breve{\mathcal{Q}}\right)$ form an orthonormal basis for ${\mathbb{Q}}^{nm}$. By Lemma 4, it is easy to observe that

$${\lambda }_{min}\left(\breve{\mathcal{Q}}\right)={\breve{\mathbf{u}}}_1^{H}bcirc\left(\breve{\mathcal{Q}}\right){\breve{\mathbf{u}}}_{1}and{\lambda }_{max}\left(\breve{\mathcal{Q}}\right)={\breve{\mathbf{u}}}_n^{H}bcirc\left(\breve{\mathcal{Q}}\right){\breve{\mathbf{u}}}_n.$$

□

The following theorem yields the lower and upper bounds for right T-eigenvalues of $\breve{\mathcal{Q}}+\breve{\mathcal{S}}$.

Theorem 7. 

Given two Hermitian tensors $\breve{\mathcal{Q}}$ and $\breve{\mathcal{S}}$ of the same size $n\times n\times m$, then the following hold:

1. 

${\lambda }_i\left(\breve{\mathcal{Q}}\right)+{\lambda }_1\left(\breve{\mathcal{S}}\right)\le {\lambda }_i(\breve{\mathcal{Q}}+\breve{\mathcal{S}})\le {\lambda }_i\left(\breve{\mathcal{Q}}\right)+{\lambda }_{nm}\left(\breve{\mathcal{S}}\right)$;

2. 

${\lambda }_{nm}(\breve{\mathcal{Q}}-\breve{\mathcal{S}})+{\lambda }_1\left(\breve{\mathcal{S}}\right)\le {\lambda }_{nm}\left(\breve{\mathcal{Q}}\right)$.

Proof. 

• Let $\breve{\mathbf{x}}\in {\mathbb{Q}}^{nm}$ be a unit vector. Then, we have

  $${\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}}+\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{min}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{S}}\right)\breve{\mathbf{x}}\le {\breve{\mathbf{x}}}^{H}bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{S}})\breve{\mathbf{x}}\le {\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}}+\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{max}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{S}}\right)\breve{\mathbf{x}}.$$
  Then,

  $${\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}}+{\lambda }_1(bcirc\left(\breve{\mathcal{S}}\right))\le {\breve{\mathbf{x}}}^{H}bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{S}})\breve{\mathbf{x}}\le {\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}}+{\lambda }_{nm}(bcirc\left(\breve{\mathcal{S}}\right)),$$

  which gives the desired inequality.
• $\breve{\mathcal{Q}}=(\breve{\mathcal{Q}}-\breve{\mathcal{S}})+\breve{\mathcal{S}}$; according to Lemma 5, we have

  $$\begin{array}{cc}\hfill {\lambda }_{max}\left(\breve{\mathcal{Q}}\right)& =\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{max}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\mathbf{x}} \hfill \end{array}$$

  (23)

  $$\begin{array}{cc}\hfill & =\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{max}{\breve{\mathbf{x}}}^H[(bcirc\left(\breve{\mathcal{Q}}\right)-bcirc\left(\breve{\mathcal{S}}\right))+bcirc\left(\breve{\mathcal{S}}\right)]\breve{\mathbf{x}} \hfill \end{array}$$

  (24)

  $$\begin{array}{cc}\hfill & =\underset{{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1}{max}{\breve{\mathbf{x}}}^H\left[(bcirc\left(\breve{\mathcal{Q}}\right)-bcirc\left(\breve{\mathcal{S}}\right))\right]\breve{\mathbf{x}}+\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{max}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{S}}\right)\breve{\mathbf{x}}\hfill \end{array}$$

  (25)

  $$\begin{array}{cc}\hfill & \ge \underset{{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1}{max}{\breve{\mathbf{x}}}^H\left[(bcirc\left(\breve{\mathcal{Q}}\right)-bcirc\left(\breve{\mathcal{S}}\right))\right]\breve{\mathbf{x}}+\underset{\begin{array}{c}{\breve{\mathbf{x}}}^H\breve{\mathbf{x}}=1\end{array}}{min}{\breve{\mathbf{x}}}^{H}bcirc\left(\breve{\mathcal{S}}\right)\breve{\mathbf{x}}\hfill \end{array}$$

  (26)

  $$\begin{array}{cc}\hfill & ={\lambda }_{max}(\breve{\mathcal{Q}}-\breve{\mathcal{S}})+{\lambda }_{min}\left(\breve{\mathcal{S}}\right), \hfill \end{array}$$

  (27)

  which gives to ${\lambda }_{nm}(\breve{\mathcal{Q}}-\breve{\mathcal{S}})+{\lambda }_1\left(\breve{\mathcal{S}}\right)\le {\lambda }_{nm}\left(\breve{\mathcal{Q}}\right)$.

□

Next, we extend the Weyl theorem from matrices to third-order Hermitian tensors, and only the right T-eigenvalues are considered.

Lemma 6 

(Weyl [39]). Assume that $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{m\times m}$ and $\breve{\mathbf{R}}\in {\mathbb{Q}}^{m\times m}$ are both Hermitian matrices, and let ${\left\{{\lambda }_r\right\}}_{r=1}^m$, ${\left\{{\mu }_r\right\}}_{r=1}^m$, ${\left\{{\nu }_r\right\}}_{r=1}^m$ be the right eigenvalues of $\breve{\mathbf{Q}}$, $\breve{\mathbf{R}}$ and $\breve{\mathbf{Q}}+\breve{\mathbf{R}}$, respectively. All of them are arranged in nondecreasing order, then

$${\nu }_r\le {\lambda }_{r+s}+{\mu }_{m-s},s=0,1,\cdots ,m-r.$$

(28)

For each $r=1,2,\cdots ,m$, the equality holds if and only if in the case that there exists $\breve{\mathbf{x}}\ne \mathbf{0}$, satisfying

$$\breve{\mathbf{Q}}\breve{\mathbf{x}}=\breve{\mathbf{x}}{\lambda }_{r+s},\breve{\mathbf{R}}\breve{\mathbf{x}}=\breve{\mathbf{x}}{\mu }_{m-s},(\breve{\mathbf{Q}}+\breve{\mathbf{R}})\breve{\mathbf{x}}=\breve{\mathbf{x}}{\nu }_r.$$

Similarly, we have that

$${\lambda }_{r-s+1}+{\mu }_s\le {\nu }_r,s=1,\cdots ,r.$$

(29)

For each $r=1,2,\cdots ,m$, the equality holds if and only if in the case that there exists $\breve{\mathbf{x}}\ne \mathbf{0}$ such that

$$\breve{\mathbf{Q}}\breve{\mathbf{x}}=\breve{\mathbf{x}}{\lambda }_{r-s+1},\breve{\mathbf{R}}\breve{\mathbf{x}}=\breve{\mathbf{x}}{\mu }_s,(\breve{\mathbf{Q}}+\breve{\mathbf{R}})\breve{\mathbf{x}}=\breve{\mathbf{x}}{\nu }_r.$$

If there is no common right eigenvector between $\breve{\mathbf{Q}}$, $\breve{\mathbf{R}}$ and $\breve{\mathbf{Q}}+\breve{\mathbf{R}}$, then each inequality in (28) and (29) is a strict inequality.

Theorem 8. 

Given two quaternion Hermitian tensors $\breve{\mathcal{Q}}$ and $\breve{\mathcal{R}}$ with the identical size $n\times n\times m$, let ${\left\{{\lambda }_r\right\}}_{i=1}^{nm}$, ${\left\{{\mu }_r\right\}}_{r=1}^{nm}$, ${\left\{{\nu }_r\right\}}_{r=1}^{nm}$ be the right T-eigenvalues of $\breve{\mathcal{Q}}$, $\breve{\mathcal{R}}$ and $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$, respectively. All of them are arranged in nondecreasing order, then

$${\nu }_r\le {\lambda }_{r+s}+{\mu }_{nm-s},s=0,1,\cdots ,nm-r.$$

(30)

For each $r=1,2,\cdots ,nm$, the equality holds if and only if inthe case that there exists a nonzero tensor $\breve{\mathcal{X}}$, where $\breve{\mathcal{X}}=fold\left([{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)\right]\breve{\mathbf{x}})$ $(\breve{\mathbf{x}}\ne 0)$ such that

$$\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\lambda }_{r+s},\breve{\mathcal{R}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\mu }_{nm-s},(\breve{\mathcal{Q}}+\breve{\mathcal{R}})★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\nu }_r,$$

where $\mathbf{P}$ is a permutation matrix obtained by a series of multiplications of elementary permutation matrices like in Theorem 3, and ${\breve{\mathbf{F}}}_{\breve{\mu }}$ is the discrete quaternion Fourier matrix. Similarly, we have that

$${\lambda }_{r-s+1}+{\mu }_s\le {\nu }_r,s=1,\cdots ,r.$$

(31)

For each $r=1,2,\cdots ,nm$, the equality holds if and only if in the case that there exists nonzero $\breve{\mathcal{X}}\in {\mathbb{Q}}^{n\times 1\times m}$ such that

$$\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\lambda }_{r-s+1},\breve{\mathcal{R}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\mu }_s,(\breve{\mathcal{Q}}+\breve{\mathcal{R}})★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\nu }_r.$$

If there is no common T-eigentensor between $\breve{\mathcal{Q}}$, $\breve{\mathcal{R}}$ and $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$, then each inequality in (30) and (31) is a strict inequality.

Proof. 

For simplicity, only the case where m is odd is taken into consideration and the even case can be similarly discussed.

It is not difficult to see that $bcirc\left(\breve{\mathcal{Q}}\right)$ and $bcirc\left(\breve{\mathcal{R}}\right)$ are both Hermitian matrices, as $\breve{\mathcal{Q}}$ and $\breve{\mathcal{R}}$ are Hermitian tensors. Then, so is $bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}})$, i.e., $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$ is a Hermitian tensor. According to the definition of T-eigenvalue, it is easy to see that the right T-eigenvalues of $\breve{\mathcal{Q}}$, $\breve{\mathcal{R}}$ and $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$ are right eigenvalues of Hermitian matrices $bcirc\left(\breve{\mathcal{Q}}\right)$, $bcirc\left(\breve{\mathcal{R}}\right)$ and $bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}})$, respectively. Then, inequalities (30) and (31) hold immediately according to Lemma 6. And the equalities in (30) and (31) hold if and only if there exists one vector $\breve{\alpha }\ne \mathbf{0}$ such that

$$\left\{\begin{array}{c}bcirc\left(\breve{\mathcal{Q}}\right)\breve{\alpha }=\breve{\alpha }{\lambda }_{r+s}\hfill \\ bcirc\left(\breve{\mathcal{R}}\right)\breve{\alpha }=\breve{\alpha }{\mu }_{nm-s}\hfill \\ (bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}}))\breve{\alpha }=\breve{\alpha }{\nu }_r\hfill \end{array}\right.$$

(32)

That is to say, $\breve{\alpha }$ is the common eigenvector of $bcirc\left(\breve{\mathcal{Q}}\right)$, $bcirc\left(\breve{\mathcal{R}}\right)$ and $bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}})$ associated with right eigenvalues ${\lambda }_{r+s}$, ${\mu }_{nm-s}$ and ${\nu }_r$. From Theorem 3, we have that $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]\breve{\alpha }$ is the common eigenvector of $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]$, $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc\left(\breve{\mathcal{R}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]$ and $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}})[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]$, i.e., $D_Q$, $D_R$ and $D_{Q+R}$, with the corresponding right eigenvalues ${\lambda }_{r+s}$, ${\mu }_{nm-s}$ and ${\nu }_r$. For convenience in notation, we denote ${\breve{\mathbf{C}}}_t$ as the t-th diagonal block of $D_{Q+R}$.

Suppose that $\breve{\beta }=[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]\breve{\alpha }$ has the following partition:

$${\breve{\beta }}^H=[\underset{1\times n}{{\breve{\beta }}_1^H},\underset{1\times 2n}{{\breve{\beta }}_2^H},\cdots ,\underset{1\times 2n}{{\breve{\beta }}_{(m+1)/2}^H}],$$

where ${\breve{\beta }}_1\in {\mathbb{Q}}^{n\times 1}$ and ${\breve{\beta }}_t\in {\mathbb{Q}}^{2n\times 1}$, $2\le t\le (m+1)/2$. Based on the fact that $D_Q$, $D_R$ and $D_{Q+R}$ are all block-diagonal matrices with the same block structure, and according to Theorem 3, it is easy to see that ${\lambda }_{r+s}$ must be a right eigenvalue belonging to some diagonal block in $D_Q$ and ${\mu }_{nm-s}$ a right eigenvalue belonging to some diagonal block in $D_R$, similarly, ${\nu }_r$ be a right eigenvalue associated with some diagonal block in $D_{Q+R}$. Consequently, there exist three index sets denoted as ${\mathbb{I}}_Q$, ${\mathbb{I}}_R$ and ${\mathbb{I}}_{Q+R}$, which are subsets of $\{1,2,\cdots ,{\displaystyle \frac{m+1}{2}}\}$ that satisfy the following equations:

$$\left\{\begin{array}{cc}{\breve{\mathbf{Q}}}_t{\breve{\beta }}_t={\breve{\beta }}_t{\lambda }_{r+s}\hfill & \mathrm{when}t\in {\mathbb{I}}_Q\hfill \\ {\breve{\mathbf{Q}}}_t{\breve{\beta }}_t\ne {\breve{\beta }}_t{\lambda }_{r+s}\hfill & \mathrm{when}t\notin {\mathbb{I}}_Q\hfill \end{array}\right.,$$

$$\left\{\begin{array}{cc}{\breve{\mathbf{R}}}_t{\breve{\beta }}_t={\breve{\beta }}_t{\mu }_{nm-s}\hfill & \mathrm{when}t\in {\mathbb{I}}_R\hfill \\ {\breve{\mathbf{R}}}_t{\breve{\beta }}_t\ne {\breve{\beta }}_t{\mu }_{nm-s}\hfill & \mathrm{when}t\notin {\mathbb{I}}_R\hfill \end{array}\right.,$$

$$\left\{\begin{array}{cc}{\breve{\mathbf{C}}}_t{\breve{\beta }}_t={\breve{\beta }}_t{\nu }_r\hfill & \mathrm{when}t\in {\mathbb{I}}_{Q+R}\hfill \\ {\breve{\mathbf{C}}}_t{\breve{\beta }}_t\ne {\breve{\beta }}_t{\nu }_r\hfill & \mathrm{when}t\notin {\mathbb{I}}_{Q+R}\hfill \end{array}\right..$$

Since $\breve{\beta }$ is a common eigenvector of $D_Q$, $D_R$ and $D_{Q+R}$, we have $\mathbb{I}\equiv {\mathbb{I}}_Q\cap {\mathbb{I}}_R\cap {\mathbb{I}}_{Q+R}\ne \varnothing$ and ${\breve{\beta }}_t\ne \mathbf{0}$ when $t\in \mathbb{I}$. It is easy to find that $\breve{\mathbf{x}}=({\mathbf{0}}_1,{\breve{\beta }}_t,{\mathbf{0}}_2)\in {\mathbb{Q}}^{nm\times 1}$ is also a common eigenvector of $D_Q$, $D_R$ and $D_{Q+R}$ when $t\in \mathbb{I}$, where ${\mathbf{0}}_1$ and ${\mathbf{0}}_2$ are zero vectors of an appropriate size. It is worth pointing out that the column vector $\breve{\mathbf{x}}$ has the same row partition with $D_Q$, $D_R$ and $D_{Q+R}$. This yields that $[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]\breve{\mathbf{x}}$ is a common eigenvector of block circulant matrices $bcirc\left(\breve{\mathcal{Q}}\right)$, $bcirc\left(\breve{\mathcal{R}}\right)$ and $bcirc(\breve{\mathcal{Q}}+\breve{\mathcal{R}})$. Consequently, $\breve{\mathcal{Q}}$, $\breve{\mathcal{R}}$ and $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$ have a common T-eigentensor $fold\left([{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)\right]\breve{\mathbf{x}})$.

Conversely, suppose that $\breve{\mathcal{Q}}$, $\breve{\mathcal{R}}$ and $\breve{\mathcal{Q}}+\breve{\mathcal{R}}$ share a common T-eigentensor $\breve{\mathcal{X}}\in {\mathbb{Q}}^{n\times 1\times m}$ of the form $\breve{\mathcal{X}}=fold\left([{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)\right]\breve{\mathbf{x}})$ $(\breve{\mathbf{x}}\ne 0)$, which satisfies the following equations:

$$\breve{\mathcal{Q}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\lambda }_{r+s},\breve{\mathcal{R}}★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\mu }_{nm-s},(\breve{\mathcal{Q}}+\breve{\mathcal{R}})★\breve{\mathcal{X}}=\breve{\mathcal{X}}{\nu }_r.$$

(33)

Then, it is simple to confirm that the nonzero vector $[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]\breve{\mathbf{x}}$ satisfies all the equalities in (32). The equality of (31) can be similarly proved by applying the proof of (30) to $-\breve{\mathcal{Q}}$, $-\breve{\mathcal{R}}$ and $-(\breve{\mathcal{Q}}+\breve{\mathcal{R}})$; we will not elaborate it here. □

5. The Estimation of Right T-Eigenvalues of Quaternion Tensors

To the best of our knowledge, the $Ger\check{s}gorin$ disc theorem is a key result used to estimate the spectra of square matrices. This section is devoted to the localization of the right T-eigenvalues of a third-order quaternion tensor under the tensor–tensor product. Next, we extend the $Ger\check{s}gorin$ theorem from matrices to third-order quaternion tensors via the t-product.

Lemma 7 

([28]). Given a quaternion matrix $\breve{\mathbf{Q}}=\left({\breve{q}}_{ij}\right)$ $\in {\mathbb{Q}}^{n\times n}$, let $R_i={\sum }_{j=1,j\ne i}^n\left|{\breve{q}}_{ij}\right|,$ $i=1,2,\cdots ,n$. For each right eigenvalue ${\lambda }_r$ belonging to $\breve{\mathbf{Q}}$, there exists a nonzero quaternion $\breve{\mu }$ that satisfies that ${\breve{\mu }}^{-1}{\breve{\lambda }}_r\breve{\mu }$ (which is also a right eigenvalue) lies in the union of the $Ger\check{s}gorin$ balls $\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}|\le R_i\}$. Specifically, if ${\lambda }_r$ is a real number, it is located in a $Ger\check{s}gorin$ ball.

Theorem 9 

($Ger\check{s}gorin$ theorem for right T-eigenvalues of third-order quaternion tensors). Given a third-order quaternion tensor $\breve{\mathcal{Q}}$ $\in {\mathbb{Q}}^{n\times n\times m}$, the following hold:

• Case 1: If m is odd, let $\mathbf{P}$ be a permutation matrix by swapping i-th and $(m-i+3)$-th columns of an identity matrix for $i=3,4,\cdots ,{\displaystyle \frac{m+1}{2}}$. Then, $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]=diag({\breve{\mathbf{Q}}}_1,{\breve{\mathbf{Q}}}_2,\cdots ,{\breve{\mathbf{Q}}}_{(m+1)/2})\equiv D_Q$, where ${\breve{\mathbf{Q}}}_k=\left({\breve{q}}_{ij}^{\left(k\right)}\right)$, $i,j=1,2,\cdots ,n$ when $k=1$; $i,j=1,2,\cdots ,2n$ when $k\ge 2$. Let $R_i^{\left(1\right)}={\sum }_{j=1,j\ne i}^n\left|{\breve{q}}_{ij}^{\left(1\right)}\right|,$ $i=1,2,\cdots ,n$ and $R_i^{\left(k\right)}={\sum }_{j=1,j\ne i}^{2n}\left|{\breve{q}}_{ij}^{\left(k\right)}\right|,$ $i=1,2,\cdots ,2n$, $k\ge 2$. Then, for each right T-eigenvalue ${\breve{\lambda }}_r$ of $\breve{\mathcal{Q}}$, there exists a quaternion $\breve{\mu }$ such that ${\breve{\mu }}^{-1}{\breve{\lambda }}_r\breve{\mu }$ (which is also a right T-eigenvalue) lies in the union of $mn$ closed $Ger\check{s}gorin$ balls ${\cup }_{i=1}^n\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(1\right)}|\le R_i^{\left(1\right)}\}\bigcup {\cup }_{k=2}^{(m+1)/2}{\cup }_{i=1}^{2n}\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(k\right)}|\le R_i^{\left(k\right)}\}$. Specifically, if ${\lambda }_r$ is a real T-eigenvalue of $\breve{\mathcal{A}}$, it is contained in a $Ger\check{s}gorin$ ball.
• Case 2: If m is even, let $\mathbf{P}$ be a permutation matrix by swapping i-th and $(m-i+3)$-th columns of an identity matrix for $i=3,4,\cdots ,{\displaystyle \frac{m}{2}}$. Then, $[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]=diag({\breve{\mathbf{Q}}}_1,{\breve{\mathbf{Q}}}_2,\cdots ,{\breve{\mathbf{Q}}}_{m/2+1})\equiv D_Q$, where ${\breve{\mathbf{Q}}}_k=\left({\breve{q}}_{ij}^{\left(k\right)}\right)$, $i,j=1,2,\cdots ,n$ when $k=1$ and $k=m/2+1$; $i,j=1,2,\cdots ,2n$ when $k\ge 2$ and $k\ne (m/2+1)$. Let $R_i^{\left(1\right)}={\sum }_{j=1,j\ne i}^n\left|{\breve{q}}_{ij}^{\left(1\right)}\right|,$ $i=1,2,\cdots ,n$, $R_i^{(m/2+1)}={\sum }_{j=1,j\ne i}^n\left|{\breve{q}}_{ij}^{(m/2+1)}\right|,$ $i=1,2,\cdots ,n$ and $R_i^{\left(k\right)}={\sum }_{j=1,j\ne i}^{2n}\left|{\breve{q}}_{ij}^{\left(k\right)}\right|,$ $i=1,2,\cdots ,2n$, $k\ge 2$ and $k\ne (m/2+1)$. Then, for each right T-eigenvalue ${\lambda }_r$ belonging to $\breve{\mathcal{Q}}$, there exists a quaternion $\breve{\mu }$ such that ${\breve{\mu }}^{-1}{\breve{\lambda }}_r\breve{\mu }$ (which is also a right T-eigenvalue) lies in the union of $mn$ closed $Ger\check{s}gorin$ balls ${\cup }_{i=1}^n\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(1\right)}|\le R_i^{\left(1\right)}\}\bigcup {\cup }_{i=1}^n\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{(m/2+1)}|\le R_i^{(m/2+1)}\}\bigcup {\cup }_{k=2}^{m/2}{\cup }_{i=1}^{2n}\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(k\right)}|\le R_i^{\left(k\right)}\}$. Specifically, if ${\lambda }_r$ is a real T-eigenvalue of $\breve{\mathcal{Q}}$, it is contained in a $Ger\check{s}gorin$ ball.

Proof. 

For simplicity, we only consider the odd case here as the even case can be similarly proved.

As shown in [33], a quaternion block circulant matrix $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{nm\times nm}$ can be block-diagonalized into a block-diagonal matrix with n-by-n blocks and 2n-by-2n blocks. That is, given a tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n\times n\times m}$, then we have

$$[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_n]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_n)]=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_{(m+1)/2}\end{array}\right)\equiv D_Q,$$

(34)

where ${\breve{\mathbf{Q}}}_1=\left({\breve{q}}_{ij}^{\left(1\right)}\right)$, $i,j=1,2,\cdots ,n$ is of size $n\times n$, while ${\breve{\mathbf{Q}}}_k=\left({\breve{q}}_{ij}^{\left(k\right)}\right)$, $i,j=1,2,\cdots ,2n$ is of size $2n\times 2n$, $k=2,3,\cdots ,(m+1)/2$.

Applying Theorem 3 and Lemma 7, it is easy to obtain the desired result. □

Remark 1. 

The even case can be similarly proved as the odd case, except that there exist one more n-by-n block at the right bottom of the block-diagonal matrix $D_Q$, which we do not elaborate further here.

Example 2. 

Consider a third-order Hermitian tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{3\times 3\times 3}$ of the following structure:

$$\breve{\mathcal{Q}}(:,:,1)=\left(\begin{array}{ccc}1& \mathbf{i}+\mathbf{j}& \mathbf{i}\\ -\mathbf{i}-\mathbf{j}& 2& 4\mathbf{k}\\ -\mathbf{i}& -4\mathbf{k}& 3\end{array}\right),$$

$$\breve{\mathcal{Q}}(:,:,2)=\left(\begin{array}{ccc}4& \mathbf{i}& 2\mathbf{k}\\ 3\mathbf{j}& 5& \mathbf{i}+\mathbf{j}\\ \mathbf{i}+3\mathbf{k}& \mathbf{j}+\mathbf{k}& 6\end{array}\right),$$

$$\breve{\mathcal{Q}}(:,:,3)=\left(\begin{array}{ccc}4& -3\mathbf{j}& -\mathbf{i}-3\mathbf{k}\\ -\mathbf{i}& 5& -\mathbf{j}-\mathbf{k}\\ -2\mathbf{k}& -\mathbf{i}-\mathbf{j}& 6\end{array}\right).$$

By Algorithm 1, the right T-eigenvalues of $\breve{\mathcal{Q}}$ are 6.7446, 11.9286, 17.3269, 6.4356, 3.5744, −2.2721, −11.2869, −9.0741 and −5.3769. According to Theorem 9, we have that the right T-eigenvalues of $\breve{\mathcal{Q}}$ are contained within the union of the nine closed discs listed below:

$$C_1=\{z\in \mathbb{Q}:|z-9|\le 3.8284\},C_2=\{z\in \mathbb{Q}:|z-12|\le 5.9907\},$$

$$C_3=\{z\in \mathbb{Q}:|z-15|\le 4.1623\},C_4=\{z\in \mathbb{Q}:|z+3|\le 10.4700\},$$

$$C_5=\{z\in \mathbb{Q}:|z+3|\le 11.7952\},C_6=\{z\in \mathbb{Q}:|z+3|\le 15.1362\},$$

$$C_7=\{z\in \mathbb{Q}:|z+3|\le 13.0092\},C_8=\{z\in \mathbb{Q}:|z+3|\le 12.6964\},$$

$$C_9=\{z\in \mathbb{Q}:|z+3|\le 11.6958\}.$$

Lemma 8 

([28]). Consider a quaternion matrix $\breve{\mathbf{Q}}\in {\mathbb{Q}}^{n\times n}$, if there exists t connected $Ger\check{s}gorin$ balls, each of which is disjointed with others. Then, there exists at least t right eigenvalues of $\breve{\mathbf{Q}}$.

Theorem 10. 

Given a third-order quaternion tensor $\breve{\mathcal{Q}}\in {\mathbb{Q}}^{n\times n\times p}$, if there are m connected $Ger\check{s}gorin$ balls that are disjoint with other $Ger\check{s}gorin$ balls, then the connected region must contain at least m right T-eigenvalues of the third-order quaternion tensor (some of which may be counted multiple times).

Proof. 

Suppose an F-square tensor $\breve{\mathcal{Q}}={\mathcal{Q}}_0+{\mathcal{Q}}_1\breve{\mu }+{\mathcal{Q}}_2\breve{\alpha }+{\mathcal{Q}}_3\breve{\beta }\in {\mathbb{Q}}^{m\times m\times n}$ is in a three-axis system $(\breve{\mu },\breve{\alpha },\breve{\beta })$, where $\breve{\mu }$ is any given unit pure quaternion, $\breve{\alpha }$ is a unit pure quaternion satisfying $\breve{\alpha }\perp \breve{\mu }$ and $\breve{\beta }=\breve{\alpha }\breve{\mu }$. As the cases for even n and odd n are highly similar, we consider only the odd case. By the above proofs, we know that there exists an $n\times n$ quaternion Fourier transform matrix ${\breve{\mathbf{F}}}_{\breve{\mu }}$ and a $n\times n$ permutation matrix $\mathbf{P}$ such that the following equation holds:

$$[\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)\otimes {\mathbf{I}}_m]bcirc\left(\breve{\mathcal{Q}}\right)[{\left(\mathbf{P}{\breve{\mathbf{F}}}_{\breve{\mu }}\right)}^H\otimes {\mathbf{I}}_m)]=\left(\begin{array}{cccc}{\breve{\mathbf{Q}}}_1& & & \\ & {\breve{\mathbf{Q}}}_2& & \\ & & \ddots & \\ & & & {\breve{\mathbf{Q}}}_{{\displaystyle \frac{n+1}{2}}}\end{array}\right)\equiv D_Q,$$

(35)

where $\mathbf{P}$ is obtained by swapping the t-th and $(n-t+3)$-th columns of a $n\times n$ identity matrix for $t=3,\cdots ,{\displaystyle \frac{n+1}{2}}$, and ${\breve{\mathbf{Q}}}_1=\left({\breve{q}}_{ij}^{\left(1\right)}\right)\in {\mathbb{Q}}^{m\times m}$ while ${\breve{\mathbf{Q}}}_k=\left({\breve{q}}_{ij}^{\left(k\right)}\right)\in {\mathbb{Q}}^{2m\times 2m}$, $k=2,3,\cdots ,{\displaystyle \frac{n+1}{2}}$. From Theorem 3, we know that ${\sigma }_r\left(\breve{\mathcal{Q}}\right)={\sigma }_r\left(D_Q\right)={\cup }_{s=1}^{{\displaystyle \frac{n+1}{2}}}{\sigma }_r\left({\breve{\mathbf{Q}}}_s\right).$ Employing Lemma 7 yields the conclusion that for each right T-eigenvalue $\breve{\lambda }$ of $\breve{\mathcal{Q}}$ there exists a quaternion $\breve{q}\ne 0$ such that ${\breve{q}}^{-1}\breve{\lambda }\breve{q}$ (which is again a right T-eigenvalue) is located in the union of those $Ger\check{s}gorin$ balls $\{{\breve{p}}^{-1}\breve{\lambda }\breve{p}:0\ne \breve{p}\in \mathbb{Q}\}\cap \{{\cup }_{i=1}^m\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(1\right)}|\le R_i^{\left(1\right)}\}\cup {\cup }_{k=2}^{{\displaystyle \frac{(n+1)}{2}}}{\cup }_{i=1}^{2m}\{z\in \mathbb{Q}:|z-{\breve{q}}_{ii}^{\left(k\right)}|\le R_i^{\left(k\right)}\left\}\right\}$, where $R_i^{\left(1\right)}={\sum }_{j=1,j\ne i}^n\left|{\breve{q}}_{ij}^{\left(1\right)}\right|,$ $i=1,2,\cdots ,m$ and $R_i^{\left(k\right)}={\sum }_{j=1,j\ne i}^{2m}\left|{\breve{q}}_{ij}^{\left(k\right)}\right|,$ $i=1,2,\cdots ,2m$, $k=2,3,\cdots ,{\displaystyle \frac{n+1}{2}}$. The desired conclusion then can be obtained by applying the result of Lemma 8 to $D_Q$ in Equation (35). □

6. Conclusions and Prospects

In this paper, we explore the right T-eigenvalues of third-order quaternion tensors, and we give an algorithm to compute it, along with a specific example, which involves comparison with other methods in terms of computational error and CPU time to demonstrate the better accuracy and speed of our algorithm. We prove that the right T-eigenvalues are real numbers, and give the upper and lower bounds for the right T-eigenvalues of the sum of a pair of Hermitian tensors. Moreover, we extend the Weyl theorem and $Ger\check{s}gorin$ theorem from matrices to third-order Hermitian quaternion tensors. However, there is still potential for progress in our research article. Recently, Miao et al. [40] have generalized the tensor–tensor product to higher-order cases, which is a flexible transform-based tensor product, i.e., Qt-product. Currently, research on the eigenvalues of higher-order quaternion tensors is relatively scarce. This finding opens new avenues for our future research. In the future, we hope to conduct more research on the eigenvalues of higher-order quaternion tensors.