On Extended Perron Complements of Nonnegative Irreducible γ-Diagonally and Product γ-Diagonally Dominant Matrices

Abstract

This study explores the γ-diagonal dominance and product γ-diagonal dominance properties of extended Perron complements. We prove that nonnegative irreducible matrices that are (strictly) γ-diagonally dominant maintain their nonnegativity, irreducibility, and (strictly) γ-diagonal dominance after applying the extended Perron complement operation. Similarly, (strictly) product γ-diagonally dominant nonnegative irreducible matrices also retain these properties under the same operation. Two numerical examples illustrate and validate these theoretical results.

1. Introduction

We begin by recalling essential notations and definitions. Let $R^{m\times n}$ ($C^{m\times n}$) denote the set of all $m\times n$ real (complex) matrices. For a matrix $A=\left(a_{ij}\right)\in C^{n\times n}(n\ge 2)$, we define the index set $N=\left\{1,2,\dots ,n\right\}$. For any $i,j\in N$, we denote the sum of the absolute values of all off-diagonal elements in the $i$-th row and $j$-th column of matrix $A$ by $R_i\left(A\right)$ and $C_j\left(A\right)$, respectively, that is

$$R_i\left(A\right)={\displaystyle \sum _{\begin{array}{l}k=1\\ k\ne i\end{array}}^n\left|a_{ik}\right|},C_j\left(A\right)={\displaystyle \sum _{\begin{array}{l}k=1\\ k\ne j\end{array}}^n\left|a_{kj}\right|}.$$

The diagonal dominance index sets are defined as

$$N_r\left(A\right)=\left\{i\left|\left|a_{ii}\right|>R_i\left(A\right),i\in N\right.\right\},N_c\left(A\right)=\left\{j\left|\left|a_{jj}\right|>C_j\left(A\right),j\in N\right.\right\}.$$

For any matrix $A=\left(a_{ij}\right)$, we denote $\left|A\right|=\left(\left|a_{ij}\right|\right)$. A matrix is called nonnegative if all its entries are nonnegative. The comparison matrix $\mu \left(A\right)=\left({\mu }_{ij}\right)$ of a square matrix $A$ is defined by

$${\mu }_{ij}=\left\{\begin{array}{c}\left|a_{ii}\right|,i=j,\\ -\left|a_{ij}\right|,i\ne j.\end{array}\right.$$

A matrix $A$ is an M-matrix if it can be expressed as $A=sI-P$, where $P$ is nonnegative, $s>\rho \left(P\right),$ and $\rho \left(P\right)$ denotes the spectral radius of $P$. It is well known that $A$ is an H-matrix if its comparison matrix $\mu \left(A\right)$ is an M-matrix [1].

A matrix $A=\left(a_{ij}\right)\in C^{n\times n}$ is γ-diagonally dominant (denoted $A\in D_n^{\gamma }$) if there exists $\gamma \in \left[0,1\right]$ such that

$$\left|a_{ii}\right|\ge \gamma R_i\left(A\right)+\left(1-\gamma \right)C_i\left(A\right),i\in N.$$

(1)

When strict inequalities hold in (1), $A$ is called a strictly γ-diagonally dominant matrix (denoted $A\in S{D}_n^{\gamma }$).

A matrix $A=\left(a_{ij}\right)\in C^{n\times n}$ is product γ-diagonally dominant (denoted $A\in P{D}_n^{\gamma }$) if there exists $\gamma \in \left[0,1\right]$ such that

$$\left|a_{ii}\right|\ge {\left[R_i\left(A\right)\right]}^{\gamma }{\left[C_i\left(A\right)\right]}^{1-\gamma },i\in N.$$

(2)

When strict inequalities hold in (2), $A$ is called a strictly product γ-diagonally dominant matrix (denoted $A\in SP{D}_n^{\gamma }$).

For nonempty index sets $\alpha ,\beta \subseteq N$, where $\beta =N\backslash \alpha$, we denote by $\left|\alpha \right|$ the cardinality of $\alpha$. The submatrix of $A\in C^{n\times n}$ with rows indexed by $\alpha$ and columns indexed by $\beta$ is written as $A\left(\alpha ,\beta \right)$, with $A\left(\alpha ,\alpha \right)$ abbreviated to $A\left(\alpha \right)$. When $A\left(\alpha \right)$ is nonsingular, the Schur complement of $A$ with respect to $A\left(\alpha \right)$ is defined as

$$A/\alpha =A\left(\beta \right)-A\left(\beta ,\alpha \right){\left[A\left(\alpha \right)\right]}^{-1}A\left(\alpha ,\beta \right).$$

(3)

Schur complements play a fundamental role in matrix theory [2,3], statistics [4], and applied mathematics [5]. Since the 1960s, extensive research has established their properties for various matrix classes, including positive definite matrices, M-matrices, inverse M-matrices [6,7], and totally nonnegative (TN) matrices [8]. Notably, the classes of positive definite matrices, M-matrices, and inverse M-matrices are all closed under Schur complementation.

Liu and Huang [9] and Liu et al. [10] demonstrated that Schur complements preserve strictly γ-diagonal dominance and strictly product γ-diagonal dominance under certain conditions. Pioneering work by Smith [11] and subsequent researchers established eigenvalue, singular value, and determinant bounds for Schur complements [12,13,14,15] with applications in numerical analysis convergence and matrix inequalities [16,17].

A valuable approach for addressing large-scale linear systems is through divide-and-conquer methods. Specifically, Meyer [18] demonstrated that the Perron eigenproblem can be decoupled into smaller subproblems through the technique known as Perron complementation, and he further introduced the concept of the Perron complement. For an $n\times n$ nonnegative and irreducible matrix $A$ and nonempty subset $\alpha \subset N$, the Perron complement of $A\left(\alpha \right)$ in $A$ is given by

$$P\left(A/\alpha \right)=A\left(\beta \right)+A\left(\beta ,\alpha \right){\left[\rho \left(A\right)I-A\left(\alpha \right)\right]}^{-1}A\left(\alpha ,\beta \right),\beta =N\backslash \alpha .$$

(4)

Meyer established key properties: if $A$ is nonnegative irreducible, then $P\left(A/\alpha \right)$ is nonnegative irreducible and satisfies $\rho \left(A\right)=\rho \left(P\left(A/\alpha \right)\right)$. Johnson and Xenophontos [19] studied the irreducibility and primitivity of Perron complements, with applications to graph compression, graph theory, and sparse matrix computations. Since then, the Perron complement has become a fundamental concept in matrix analysis, with a wide range of demonstrated applications. Researchers have leveraged the properties and interrelationships of the Perron complement to attain deeper insights into the structural features and intrinsic characteristics of the original matrix. Extensive studies on the Perron complement have been conducted across various matrix classes, including diagonally dominant matrices and H-matrices [20], Z-matrices [21], inverse M-matrices [22], inverse $N_0$-matrices [23], and totally nonnegative matrices [24]. Furthermore, the generalized Perron complements of diagonally dominant matrices and strictly generalized doubly diagonally dominant matrices are discussed in [25,26]. These investigations have not only enriched our understanding of matrix properties but also expanded the scope of applications of the Perron complements in numerous domains of mathematical research and applied sciences.

In 2001, Fallat and Neumann [27] introduced an important extension of the Perron complement. Consider a nonnegative irreducible matrix $A$. For any nonempty subset $\alpha \subset N$ and parameter $t\ge \rho \left(A\right)$, the extended Perron complement is defined as

$$P_t\left(A/\alpha \right)=A\left(\beta \right)+A\left(\beta ,\alpha \right){\left[tI-A\left(\alpha \right)\right]}^{-1}A\left(\alpha ,\beta \right),\beta =N\backslash \alpha .$$

(5)

Fallat and Neumann’s key theoretical advancement established that extended Perron complements preserve total nonnegativity for irreducible totally nonnegative matrices.

An essential observation reveals the fundamental connection between extended Perron complements and Schur complements. Let $A\in R^{n\times n}$ be nonnegative. For any nonempty subset $\alpha \subset N$ and $t\ge \rho \left(A\right)$, we obtain

$$\left(tI-A\right)/\alpha =tI-A\left(\beta \right)-A\left(\beta ,\alpha \right){\left[tI-A\left(\alpha \right)\right]}^{-1}A\left(\alpha ,\beta \right).$$

This immediately yields the identity

$$P_t\left(A/\alpha \right)=tI-\left(tI-A\right)/\alpha ,t\ge \rho \left(A\right),$$

demonstrating the profound relationship between extended Perron complements and classical Schur complementation.

Motivated by the theoretical results established in [9,10] and [27], this paper presents a comprehensive investigation into the extended Perron complements of nonnegative irreducible γ-diagonally dominant matrices and product γ-diagonally dominant matrices. The structure of the remaining sections is as follows:

In Section 2, we first establish two fundamental lemmas that serve as the theoretical foundation for our main results. Building upon these lemmas, we rigorously prove that nonnegativity, irreducibility, γ-diagonal dominance, and product γ-diagonal dominance are preserved under extended Perron complementation for nonnegative irreducible matrices. As a direct consequence, we further demonstrate that the strict versions of these dominance properties (strictly γ-diagonal dominance and strictly product γ-diagonal dominance) are likewise maintained under the same operation.

In Section 3, two well-designed numerical experiments are presented to verify our theoretical findings. The first experiment rigorously examines the closure property of γ-diagonal dominance under extended Perron complementation for nonnegative irreducible matrices. The second experiment systematically investigates the corresponding closure property for product γ-diagonally dominant matrices. Numerical results from both experiments show remarkable consistency with our theoretical predictions.

The concluding section summarizes the key theoretical results of this study.

2. Extended Perron Complements of Nonnegative Irreducible γ-Diagonally and Product γ-Diagonally Dominant Matrices

To lay the groundwork for our analysis, we start by reviewing several fundamental results from prior research. These foundational theorems not only provide the mathematical basis for our subsequent derivations but also serve as crucial building blocks for the main theoretical contributions we will present in later section.

Lemma 1

([7]). For any H-matrix $A$, the following inequality holds:

$${\left[\mu \left(A\right)\right]}^{-1}\ge \left|A^{-1}\right|.$$

Lemma 2

([28]). Let $A\in R^{n\times n}$ be a nonnegative matrix, and let $A_k$ be any principal submatrix of $A$. Then,

(i)

$\rho \left(A_k\right)\le \rho \left(A\right)$;

(ii)

If $A$ is irreducible and $A_k$ is a proper submatrix (i.e.,$A_k\ne A$), then the inequality is strict: $\rho \left(A_k\right)<\rho \left(A\right)$.

Lemma 3

([29]). Let $A\in R^{n\times n}$ be a nonnegative irreducible matrix, and let $\alpha \subset N$ be a nonempty proper subset. Then $P_{\lambda }\left(A/\alpha \right)$ is nonnegative and irreducible for any $\lambda >\rho \left(A\left(\alpha \right)\right).$

Lemma 4

([9]). Let $x>\tau >0$ and $y>\tau >0$ . For any $r\in \left[0,1\right]$ , the following inequality holds

$$x^r{y}^{1-r}-\tau \ge {\left(x-\tau \right)}^r{\left(y-\tau \right)}^{1-r}.$$

Lemma 5.

Let $A\in R^{n\times n}$ be a nonnegative irreducible matrix, and let $\alpha \subset N$ be a nonempty proper subset. Then for any $t\ge \rho \left(A\right)$, the following inequality holds

$${\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\ge \left|{\left[tI-A\left(\alpha \right)\right]}^{-1}\right|.$$

Proof. 

Since $A$ is nonnegative irreducible, and $\alpha$ is a nonempty proper subset of $N$, Lemma 2 implies that $\rho \left(A\right)>\rho \left(A\left(\alpha \right)\right)$. Consequently, for any $t\ge \rho \left(A\right)$, it follows that $t>\rho \left(A\left(\alpha \right)\right)$. By definition, $tI-A\left(\alpha \right)$ is an M-matrix. Moreover, since

$$\mu \left[tI-A\left(\alpha \right)\right]=tI-A\left(\alpha \right),$$

the matrix $\mu \left[tI-A\left(\alpha \right)\right]$ is also an M-matrix. Thus, $tI-A\left(\alpha \right)$ is an H-matrix, and the desired inequality follows directly from Lemma 1. □

Lemma 6.

Let $A=\left(a_{ij}\right)\in R^{n\times n}$ be a nonnegative irreducible matrix.

(i)

For any nonempty subset $\alpha =\left\{i_1,i_2,\dots ,i_p\right\}\subseteq N_r\left(A\right)$ with $p<n$, if $t\ge 2\rho \left(A\right),$ then for the complement set $\beta =N\backslash \alpha =\left\{j_1,j_2,\dots ,j_q\right\}$, the following holds:

$${\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left(\begin{array}{c}{\displaystyle \sum _{t=1}^q\left|a_{i_1{j}_t}\right|}\\ {\displaystyle \sum _{t=1}^q\left|a_{i_2{j}_t}\right|}\\ ⋮\\ {\displaystyle \sum _{t=1}^q\left|a_{i_p{j}_t}\right|}\end{array}\right)<\left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right).$$

(6)

(ii)

For any nonempty subset $\beta =\left\{j_1,j_2,\dots ,j_q\right\}\subseteq N_c\left(A\right)$ with $q<n$, if  $t\ge 2\rho \left(A\right),$  then for the complement set  $\alpha =N\backslash \beta =\left\{i_1,i_2,\dots ,i_p\right\}$, the following holds:

$${\left(\begin{array}{c}{\displaystyle \sum _{t=1}^p\left|a_{i_t{j}_1}\right|}\\ {\displaystyle \sum _{t=1}^p\left|a_{i_t{j}_2}\right|}\\ ⋮\\ {\displaystyle \sum _{t=1}^p\left|a_{i_t{j}_q}\right|}\end{array}\right)}^{\mathrm{T}}{\left\{\mu \left[tI-A\left(\beta \right)\right]\right\}}^{-1}<{\left(\begin{array}{c}1\\ 1\\ ⋮\\ 1\end{array}\right)}^{\mathrm{T}}.$$

(7)

Proof. 

We establish inequality (6), while inequality (7) can be proved analogously. Let

$$X=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_p\end{array}\right)={\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left(\begin{array}{c}{\displaystyle \sum _{t=1}^q\left|a_{i_1{j}_t}\right|}\\ {\displaystyle \sum _{t=1}^q\left|a_{i_2{j}_t}\right|}\\ ⋮\\ {\displaystyle \sum _{t=1}^q\left|a_{i_p{j}_t}\right|}\end{array}\right).$$

Equivalently, this can be expressed as

$$\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}X=\left(\begin{array}{c}{\displaystyle \sum _{t=1}^q\left|a_{i_1{j}_t}\right|}\\ {\displaystyle \sum _{t=1}^q\left|a_{i_2{j}_t}\right|}\\ ⋮\\ {\displaystyle \sum _{t=1}^q\left|a_{i_p{j}_t}\right|}\end{array}\right).$$

(8)

Let $x_s=\max \left\{x_1,x_2,\dots ,x_p\right\}$, where $x_t$ denotes the t-th coordinate of $X$. From (8), we derive

$$\begin{gathered} {\displaystyle \sum _{t=1}^q\left|a_{i_s{j}_t}\right|}=\left[t-a_{i_s{i}_s}\right]x_s-{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne s\end{array}}^p\left|a_{i_s{i}_t}\right|}{x}_t \\ \ge \left[t-a_{i_s{i}_s}\right]x_s-{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne s\end{array}}^p\left|a_{i_s{i}_t}\right|}{x}_s \\ =t{x}_s-\left(a_{i_s{i}_s}+{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne s\end{array}}^p\left|a_{i_s{i}_t}\right|}\right)x_s \\ =\left[t-{\displaystyle \sum _{t=1}^p\left|a_{i_s{i}_t}\right|}\right]x_s \end{gathered}$$

(9)

Given that $t\ge 2\rho \left(A\right)$ and $\alpha =\left\{i_1,i_2,\dots ,i_p\right\}\subseteq N_r\left(A\right)$, according to Lemma 2, we obtain

$$t\ge 2\rho \left(A\right)>2a_{i_s{i}_s}>a_{i_s{i}_s}+R_{i_s}\left(A\right)=a_{i_s{i}_s}+{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne s\end{array}}^p\left|a_{i_s{i}_t}\right|}+{\displaystyle \sum _{t=1}^q\left|a_{i_s{j}_t}\right|}={\displaystyle \sum _{t=1}^p\left|a_{i_s{i}_t}\right|}+{\displaystyle \sum _{t=1}^q\left|a_{i_s{j}_t}\right|}.$$

This implies

$$t-{\displaystyle \sum _{t=1}^p\left|a_{i_s{i}_t}\right|}>{\displaystyle \sum _{t=1}^q\left|a_{i_s{j}_t}\right|}\ge 0.$$

By applying inequality (9), we conclude

$$x_s\le {\displaystyle \frac{{\displaystyle \sum _{t=1}^q\left|a_{i_s{j}_t}\right|}}{t-{\displaystyle \sum _{t=1}^p\left|a_{i_s{i}_t}\right|}}}<1.$$

Thus, the result (6) is established. □

Theorem 1. 

Let $A\in D_n^{\gamma }$ be a nonnegative irreducible matrix. If

(i)

$N_r\left(A\right)\cap N_c\left(A\right)$  is nonempty,

(ii)

nonempty subset  $\alpha \subseteq N_r\left(A\right)\cap N_c\left(A\right)$,

(iii)

$t\ge 2\rho \left(A\right)$,

then the extended Perron complement  $P_t\left(A/\alpha \right)$ is nonnegative irreducible, and retains the γ-diagonally dominant property, that is,

$$P_t\left(A/\alpha \right)\in D_{n-\left|\alpha \right|}^{\gamma }$$

Proof. 

Since $A$ is nonnegative irreducible and $t\ge 2\rho \left(A\right)$, according to Lemma 2, we obtain

$$t\ge 2\rho \left(A\right)>\rho \left(A\right)>\rho \left(A\left(\alpha \right)\right).$$

By Lemma 3, $P_t\left(A/\alpha \right)$ is nonnegative irreducible. □

In what follows, we demonstrate the γ-diagonal dominance of $P_t\left(A/\alpha \right)$. Without loss of generality, we may assume that the index set $\alpha$ is given by $\alpha =\left\{i_1,i_2,\dots ,i_p\right\}$ with $0<\left|\alpha \right|=p<n$. We denote $\beta =N\backslash \alpha =\left\{j_1,j_2,\dots ,j_q\right\}$, where $q=n-p$. Let $P_t\left(A/\alpha \right)=\left(h_{ij}\right)$ and denote

$$R_{j_k}\left[P_t\left(A/\alpha \right)\right]={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|h_{j_k{j}_t}\right|},C_{j_k}\left[P_t\left(A/\alpha \right)\right]={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|h_{j_t{j}_k}\right|},$$

$${\eta }_{j_k}={\left(a_{i_1{j}_k},a_{i_2{j}_k},\dots ,a_{i_p{j}_k}\right)}^{\mathrm{T}},k=1,2,\dots ,q.$$

(10)

By the definition of extended Perron complement $P_t\left(A/\alpha \right)$, for any $j_k\in \beta$, the entries $h_{j_k{j}_k}$ satisfy the relation:

$$h_{j_k{j}_k}=a_{j_k{j}_k}+\left(a_{j_k{i}_1},a_{j_k{i}_2},\dots ,a_{j_k{i}_p}\right){\left[tI-A\left(\alpha \right)\right]}^{-1}{\eta }_{j_k}.$$

Consequently, we have

$$\left|h_{j_k{j}_k}\right|=\left|a_{j_k{j}_k}+\left(a_{j_k{i}_1},a_{j_k{i}_2},\dots ,a_{j_k{i}_p}\right){\left[tI-A\left(\alpha \right)\right]}^{-1}{\eta }_{j_k}\right|$$

$$\ge \left|a_{j_k{j}_k}\right|-\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right)\left|{\left[tI-A\left(\alpha \right)\right]}^{-1}\right|\left|{\eta }_{j_k}\right|.$$

By applying Lemma 5, we obtain

$$\left|h_{j_k{j}_k}\right|\ge \left|a_{j_k{j}_k}\right|-\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|.$$

(11)

Furthermore, we observe that

$$h_{j_k{j}_t}=a_{j_k{j}_t}+\left(a_{j_k{i}_1},a_{j_k{i}_2},\dots ,a_{j_k{i}_p}\right){\left[tI-A\left(\alpha \right)\right]}^{-1}{\eta }_{j_t}.$$

This leads to

$$\left|h_{j_k{j}_t}\right|=\left|a_{j_k{j}_t}+\left(a_{j_k{i}_1},a_{j_k{i}_2},\dots ,a_{j_k{i}_p}\right){\left[tI-A\left(\alpha \right)\right]}^{-1}{\eta }_{j_t}\right|$$

$$\le \left|a_{j_k{j}_t}\right|+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right)\left|{\left[tI-A\left(\alpha \right)\right]}^{-1}\right|\left|{\eta }_{j_t}\right|.$$

From Lemma 5, it follows that

$$\left|h_{j_k{j}_t}\right|\le \left|a_{j_k{j}_t}\right|+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_t}\right|.$$

(12)

Thus, we obtain

$$\begin{array}{c}{R}_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ ={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|h_{j_k{j}_t}\right|}+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ \le {\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|a_{j_k{j}_t}\right|}+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|{\eta }_{j_t}\right|}(\mathrm{by}(12))\hfill \\ +\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ ={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|a_{j_k{j}_t}\right|}+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}{\displaystyle \sum _{t=1}^q\left|{\eta }_{j_t}\right|}\hfill \\ ={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|a_{j_k{j}_t}\right|}+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left(\begin{array}{c}{\displaystyle \sum _{t=1}^q\left|a_{i_1{j}_t}\right|}\\ {\displaystyle \sum _{t=1}^q\left|a_{i_2{j}_t}\right|}\\ ⋮\\ {\displaystyle \sum _{t=1}^q\left|a_{i_p{j}_t}\right|}\end{array}\right)(\mathrm{by}(10))\hfill \\ <{\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|a_{j_k{j}_t}\right|}+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left(1,1,\dots ,1\right)}^{\mathrm{T}}(\mathrm{by}(6))\hfill \\ ={\displaystyle \sum _{\begin{array}{l}t=1\\ t\ne k\end{array}}^q\left|a_{j_k{j}_t}\right|}+{\displaystyle \sum _{t=1}^p\left|a_{j_k{i}_t}\right|}\hfill \\ =R_{j_k}\left(A\right).\hfill \end{array}$$

(13)

Following similar arguments, we derive

$$C_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|<C_{j_k}\left(A\right).$$

(14)

Since $A\in D_n^{\gamma }$, there exists $\gamma \in \left[0,1\right]$ such that

$$\left|a_{j_k{j}_k}\right|\ge \gamma R_{j_k}\left(A\right)+\left(1-\gamma \right)C_{j_k}\left(A\right).$$

This implies

$$\begin{array}{c}\left|a_{j_k{j}_k}\right|-\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ \ge \gamma R_{j_k}\left(A\right)+\left(1-\gamma \right)C_{j_k}\left(A\right)-\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ \ge \gamma R_{j_k}\left[P_t\left(A/\alpha \right)\right]+\gamma \left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ +\left(1-\gamma \right)C_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(1-\gamma \right)\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\hfill \\ -\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|(\mathrm{by}(13)\mathrm{and}(14))\hfill \\ =\gamma R_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(1-\gamma \right)C_{j_k}\left[P_t\left(A/\alpha \right)\right].\hfill \end{array}$$

(15)

By combining (11) and (15), we conclude there exists $\gamma \in \left[0,1\right]$ such that

$$\left|h_{j_k{j}_k}\right|\ge \gamma R_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(1-\gamma \right)C_{j_k}\left[P_t\left(A/\alpha \right)\right].$$

This demonstrates that $P_t\left(A/\alpha \right)\in D_{n-\left|\alpha \right|}^{\gamma }$, which completes the proof of the theorem.

Corollary 1.

Let  $A\in S{D}_n^{\gamma }$  be a nonnegative irreducible matrix. If

(i)

$N_r\left(A\right)\cap N_c\left(A\right)$  is nonempty,

(ii)

nonempty subset  $\alpha \subseteq N_r\left(A\right)\cap N_c\left(A\right)$,

(iii)

$t\ge 2\rho \left(A\right)$,

then the extended Perron complement  $P_t\left(A/\alpha \right)$  is nonnegative irreducible, and retains the strictly γ-diagonally dominant property, that is,

$$P_t\left(A/\alpha \right)\in S{D}_{n-\left|\alpha \right|}^{\gamma }$$

Theorem 2. 

Let $A\in P{D}_n^{\gamma }$ be a nonnegative irreducible matrix. If

(i)

$N_r\left(A\right)\cap N_c\left(A\right)$  is nonempty,

(ii)

nonempty subset  $\alpha \subseteq N_r\left(A\right)\cap N_c\left(A\right)$,

(iii)

$t\ge 2\rho \left(A\right)$,

then the extended Perron complement  $P_t\left(A/\alpha \right)$  is nonnegative irreducible, and retains the product γ-diagonally dominant property, that is,

$$P_t\left(A/\alpha \right)\in P{D}_{n-\left|\alpha \right|}^{\gamma }$$

Proof. 

Let $P_t\left(A/\alpha \right)=\left(h_{ij}\right)$ denote the extended Perron complement. Following the notions established in Theorem 1, we define $R_{j_k}\left[P_t\left(A/\alpha \right)\right],C_{j_k}\left[P_t\left(A/\alpha \right)\right]$, and ${\eta }_{j_k}$ for $k=1,2,\dots ,q$. Since $A\in P{D}_n^{\gamma }$, we may apply Lemma 4 along with inequalities (13) and (14) to obtain that there exists $\gamma \in \left[0,1\right]$ such that the following key inequality holds:

$$\begin{array}{c}\left|a_{j_k{j}_k}\right|\ge {\left[R_{j_k}\left(A\right)\right]}^{\gamma }{\left[C_{j_k}\left(A\right)\right]}^{1-\gamma }\hfill \\ \ge {\left\{R_{j_k}\left[P_t\left(A/\alpha \right)\right]+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\right\}}^{\gamma }\hfill \\ \times {\left\{C_{j_k}\left[P\left(A/\alpha \right)\right]+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|\right\}}^{1-\gamma }\hfill \\ \ge {\left\{R_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{\gamma }{\left\{C_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{1-\gamma }+\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right|.\hfill \end{array}$$

Consequently, we establish the inequality

$$\begin{gathered} \left|a_{j_k{j}_k}\right|-\left(\left|a_{j_k{i}_1}\right|,\left|a_{j_k{i}_2}\right|,\dots ,\left|a_{j_k{i}_p}\right|\right){\left\{\mu \left[tI-A\left(\alpha \right)\right]\right\}}^{-1}\left|{\eta }_{j_k}\right| \\ \ge {\left\{R_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{\gamma }{\left\{C_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{1-\gamma }. \end{gathered}$$

(16)

According to inequalities (11) and (16), we conclude there exists $\gamma \in \left[0,1\right]$ such that

$$\left|h_{j_k{j}_k}\right|\ge {\left\{R_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{\gamma }{\left\{C_{j_k}\left[P_t\left(A/\alpha \right)\right]\right\}}^{1-\gamma }.$$

The final inequality demonstrates that $P_t\left(A/\alpha \right)\in P{D}_{n-\left|\alpha \right|}^{\gamma }$, completing the proof. □

Corollary 2. 

Let $A\in SP{D}_n^{\gamma }$ be a nonnegative irreducible matrix. If

(i)

$N_r\left(A\right)\cap N_c\left(A\right)$  is nonempty,

(ii)

nonempty subset  $\alpha \subseteq N_r\left(A\right)\cap N_c\left(A\right)$,

(iii)

$t\ge 2\rho \left(A\right)$,

then the extended Perron complement  $P_t\left(A/\alpha \right)$  is nonnegative irreducible, and retains the strictly product γ-diagonally dominant property, that is,

$$P_t\left(A/\alpha \right)\in SP{D}_{n-\left|\alpha \right|}^{\gamma }$$

3. Numerical Examples

In this section, we provide two numerical examples to validate the theoretical results established in Theorems 1 and 2.

Example 3.1. 

We investigate a nonnegative irreducible matrix$A\in R^{5\times 5}$defined as

$$A=\left(\begin{array}{ccccc}5.2& 0.8& 1.3& 0.7& 1.5\\ 1.1& 6.4& 0.9& 1.6& 0.6\\ 0.5& 1.2& 4.1& 0.8& 1.7\\ 1.4& 0.7& 1.0& 7.3& 0.9\\ 0.8& 1.5& 0.6& 1.2& 5.9\end{array}\right).$$

The row and column sums of the off-diagonal elements are

$$\begin{gathered} R_1\left(A\right)=4.3,R_2\left(A\right)=4.2,R_3\left(A\right)=4.2,R_4\left(A\right)=4.0,R_5\left(A\right)=4.1, \\ {C}_1\left(A\right)=3.8,C_2\left(A\right)=4.2,C_3\left(A\right)=3.8,C_4\left(A\right)=4.3,C_5\left(A\right)=4.7. \end{gathered}$$

Through careful analysis, we obtain the following key observations:

$$\left|a_{ii}\right|>R_i\left(A\right),i=1,2,4,5,\left|a_{ii}\right|>C_i\left(A\right),i=1,2,3,4,5.$$

Therefore, we have

$$N_r\left(A\right)=\left\{1,2,4,5\right\},N_c\left(A\right)=\left\{1,2,3,4,5\right\}$$

with nonempty intersection:

$$N_r\left(A\right)\cap N_c\left(A\right)=\left\{1,2,4,5\right\}$$

When taking $\gamma =0.55$, our calculations demonstrate that

$$\left|a_{ii}\right|\ge 0.55R_i\left(A\right)+0.45C_i\left(A\right),i=1,2,3,4,5,$$

leading to the result that $A\in D_5^{0.55}$.

Selecting  $\alpha =\left\{2,4\right\}\subseteq N_r\left(A\right)\cap N_c\left(A\right)$, we obtain the complementary index set$\beta =N\backslash \alpha =\left\{1,3,5\right\}$. The corresponding matrix partitions are:

$$\begin{gathered} A\left(\alpha \right)=\left(\begin{array}{cc}6.4& 1.6\\ 0.7& 7.3\end{array}\right),A\left(\alpha ,\beta \right)=\left(\begin{array}{ccc}1.1& 0.9& 0.6\\ 1.4& 1.0& 0.9\end{array}\right), \\ A\left(\beta ,\alpha \right)=\left(\begin{array}{cc}0.8& 0.7\\ 1.2& 0.8\\ 1.5& 1.2\end{array}\right),A\left(\beta \right)=\left(\begin{array}{ccc}5.2& 1.3& 1.5\\ 0.5& 4.1& 1.7\\ 0.8& 0.6& 5.9\end{array}\right). \end{gathered}$$

Setting  $t=21\ge 2\rho \left(A\right)=20.3558$, we compute the extended Perron complement:

$$\begin{gathered} P_{21}\left(A/\alpha \right)=A\left(\beta \right)+A\left(\beta ,\alpha \right){\left[21I-A\left(\alpha \right)\right]}^{-1}A\left(\alpha ,\beta \right) \\ =\left(\begin{array}{ccc}5.3443& 1.4096& 1.5866\\ 0.6897& 4.2453& 1.8128\\ 1.0585& 0.7969& 6.0547\end{array}\right)=\left({a^{\prime }}_{vu}\right), \end{gathered}$$

where

$$\begin{gathered} R_1\left[P_{21}\left(A/\alpha \right)\right]=2.9962,R_2\left[P_{21}\left(A/\alpha \right)\right]=2.5025,R_3\left[P_{21}\left(A/\alpha \right)\right]=1.8554, \\ {C}_1\left[P_{21}\left(A/\alpha \right)\right]=1.7482,C_2\left[P_{21}\left(A/\alpha \right)\right]=2.2065,C_3\left[P_{21}\left(A/\alpha \right)\right]=3.3994. \end{gathered}$$

For  $\gamma =0.55$, following detailed computations, we establish that

$$\left|{a^{\prime }}_{ii}\right|\ge 0.55R_i\left[P_{21}\left(A/\alpha \right)\right]+0.45C_i\left[P_{21}\left(A/\alpha \right)\right],i=1,2,3,$$

confirming that $P_{21}\left(A/\alpha \right)\in D_3^{0.55}$ while preserving nonnegativity and irreducibility. These computational results provide concrete validation of Theorem 1.

Below, we provide the second example to validate Theorem 2.

Example 3.2. 

We investigate a nonnegative irreducible matrix $B\in R^{5\times 5}$ defined as

$$B=\left(\begin{array}{ccccc}4& 1.2& 0.9& 1.1& 0.8\\ 1& 5& 1.3& 0.7& 1\\ 0.8& 1.1& 4& 1.2& 0.9\\ 1.2& 0.8& 1& 6& 1\\ 0.9& 1.2& 0.7& 1.1& 4.5\end{array}\right).$$

The row and column sums of the off-diagonal elements are

$$\begin{gathered} R_1\left(B\right)=4,R_2\left(B\right)=4,R_3\left(B\right)=4,R_4\left(B\right)=4,R_5\left(B\right)=3.9, \\ {C}_1\left(B\right)=3.9,C_2\left(B\right)=4.3,C_3\left(B\right)=3.9,C_4\left(B\right)=4.1,C_5\left(B\right)=3.7. \end{gathered}$$

Through careful analysis, we obtain the following key observations:

$$\left|b_{ii}\right|>R_i\left(B\right),i=2,4,5,\left|b_{ii}\right|>C_i\left(B\right),i=1,2,3,4,5.$$

Therefore, we have

$$N_r\left(B\right)=\left\{2,4,5\right\},N_c\left(B\right)=\left\{1,2,3,4,5\right\}$$

with nonempty intersection:

$$N_r\left(B\right)\cap N_c\left(B\right)=\left\{2,4,5\right\}$$

When taking $\gamma =0.5$, our calculations demonstrate that

$$\left|b_{ii}\right|\ge \sqrt{R_i\left(B\right)C_i\left(B\right)},i=1,2,3,4,5,$$

leading to the result that $B\in P{D}_5^{0.5}$.

Selecting  $\alpha =\left\{5\right\}\subseteq N_r\left(B\right)\cap N_c\left(B\right)$, we obtain the complementary index set  $\beta =N\backslash \alpha =\left\{1,2,3,4\right\}$. The corresponding matrix partitions are

$$\begin{gathered} B\left(\alpha \right)=\left(4.5\right),B\left(\alpha ,\beta \right)=\left(\begin{array}{cccc}0.9& 1.2& 0.7& 1.1\end{array}\right), \\ B\left(\beta ,\alpha \right)=\left(\begin{array}{c}0.8\\ 1\\ 0.9\\ 1\end{array}\right),B\left(\beta \right)=\left(\begin{array}{cccc}4& 1.2& 0.9& 1.1\\ 1& 5& 1.3& 0.7\\ 0.8& 1.1& 4& 1.2\\ 1.2& 0.8& 1& 6\end{array}\right). \end{gathered}$$

Setting $t=18\ge 2\rho \left(A\right)=17.6266$, we compute the extended Perron complement:

$$\begin{gathered} P_{18}\left(B/\alpha \right)=B\left(\beta \right)+B\left(\beta ,\alpha \right){\left[18I-B\left(\alpha \right)\right]}^{-1}B\left(\alpha ,\beta \right) \\ =\left(\begin{array}{cccc}4.0533& 1.2711& 0.9415& 1.1652\\ 1.0667& 5.0889& 1.3519& 0.7815\\ 0.8600& 1.1800& 4.0467& 1.2733\\ 1.2667& 0.8889& 1.0519& 6.0815\end{array}\right)=\left({b^{\prime }}_{vu}\right), \end{gathered}$$

where

$$\begin{gathered} \begin{array}{l}{R}_1\left[P_{18}\left(B/\alpha \right)\right]=3.3778,R_2\left[P_{18}\left(B/\alpha \right)\right]=3.2001,\\ R_3\left[P_{18}\left(B/\alpha \right)\right]=3.3133,R_4\left[P_{18}\left(B/\alpha \right)\right]=3.2075,\end{array} \\ \begin{array}{l}{C}_1\left[P_{18}\left(B/\alpha \right)\right]=3.1934,C_2\left[P_{18}\left(B/\alpha \right)\right]=3.3400,\\ C_3\left[P_{18}\left(B/\alpha \right)\right]=3.3453,C_4\left[P_{18}\left(B/\alpha \right)\right]=3.2200.\end{array} \end{gathered}$$

For  $\gamma =0.5$, following detailed computations, we establish that

$$\left|{b^{\prime }}_{ii}\right|\ge \sqrt{R_i\left[P_{18}\left(B/\alpha \right)\right]C_i\left[P_{18}\left(B/\alpha \right)\right]},i=1,2,3,4.$$

confirming that $P_{18}\left(B/\alpha \right)\in P{D}_4^{0.5}$ while preserving nonnegativity and irreducibility. These computational results provide concrete validation of Theorem 2.

4. Conclusions

This study investigated the γ-diagonally dominant and product γ-diagonally dominant properties of extended Perron complements. We demonstrated that (strictly) γ-diagonally dominant nonnegative irreducible matrices retain their (strictly) γ-diagonal dominance, nonnegativity, and irreducibility under the extended Perron complement operation. Analogously, we established that (strictly) product γ-diagonally dominant nonnegative irreducible matrices preserve these properties under the same operation.