An Improved Diagonal Transformation Algorithm for the Maximum Eigenvalue of Zero Symmetric Nonnegative Matrices

Abstract

The irreducibility of nonnegative matrices is an important condition for the diagonal transformation algorithm to succeed. In this paper, we introduce zero symmetry to replace the irreducibility of nonnegative matrices and propose an improved diagonal transformation algorithm for finding the maximum eigenvalue without any partitioning. The improved algorithm retains all of the benefits of the diagonal transformation algorithm while having fewer computations. Numerical examples are reported to show the efficiency of the proposed algorithm. As an application, the improved algorithm is used to check whether a zero symmetric matrix is an H-matrix.

1. Introduction

The issues of the maximum eigenvalues of nonnegative matrices are among the most interesting and important problems in matrix analysis and engineering [1]. Some researchers directly estimated the bounds of the maximum eigenvalues according to the nature of the nonnegative irreducible matrices [2,3,4,5,6,7,8]. Based on a geometric symmetrization of the powers of a matrix, Szyld et al. [8] presented a sequence of lower bounds for the spectral radius. By computing the sums for each row, Li et al. [5] presented a new method to finding the two-sided bounds for the spectral radius. Adam et al. [2] proposed some new bounds for the spectral radius of a nonnegative matrix with an average of the two-row sums. However, it is difficult to find the maximum eigenvalue by means of estimation methods. Therefore, some efficient algorithms for computing the eigenpairs of irreducible nonnegative matrices have been proposed [9,10,11,12]. Based on the diagonal transformation, Bunse et al. [9] obtained the maximum eigenvalue by constructing a similar matrix sequence and obtained the minimum row sum $\left\{r^k\right\}$ and the maximum row sum $\left\{R^k\right\}$. Duan et al. [10] pointed out that the diagonal transformation algorithm (Algorithm 1) was convergent for irreducible nonnegative matrices. By eliminating redundant operations, Wen et al. [12] improved the diagonal transformation algorithm so that it quickly caught the maximum eigenvalue for irreducible nonnegative matrices. Unfortunately, a major problem is that the diagonal transformation techniques above may not be convergent for reducible nonnegative matrices. To stimulate this question, we want to replace irreducibility with some symmetry conditions for nonnegative matrices, since symmetry plays an essential part in physics and mathematics [1,13]. To this end, we introduce zero symmetric reducible nonnegative matrices, which can be thought of as combinations of irreducible principal submatrices. Based on these properties, we propose an improved diagonal transformation algorithm (Algorithm 2) for computing the largest eigenvalue without any partitioning and show that the algorithm is convergent for zero symmetric nonnegative matrices. These constitute the main motivations of this article.

The remainder of this paper is organized as follows. In Section 2, some definitions and preliminary results are recalled. In Section 3, we establish the improved diagonal transformation algorithm and prove that the algorithm is convergent. In Section 4, we give Algorithm 3 to verify whether a zero symmetric matrix is an H-matrix. Numerical examples are provided to illustrate the obtained results.

2. Preliminaries

We start this section with the preliminary results [1] and introduce zero symmetric matrices.

Definition 1. 

(1) A matrix $A\in {\mathbb{R}}^{n\times n}$ is said to be reducible if there exists a permutation matrix $P\in {\mathbb{R}}^{n\times n}$ such that

$$PA{P}^{\top }=\left(\begin{array}{cc}{A}_{11}& 0\\ A_{21}& A_{22}\end{array}\right),$$

where $A_{11}\in {\mathbb{R}}^{r\times r}, A_{22}\in {\mathbb{R}}^{n-r\times n-r}, A_{21}\in {\mathbb{R}}^{n-r\times r}, 1\le r\le n-1$, and $0\in {\mathbb{R}}^{r\times n-r}$ is the zero matrix. If any permutation matrix does not exist, then A is called irreducible.

(2) The spectral radius of A is denoted by

$$\rho \left(A\right)=max\left\{\right|\lambda |:\lambda \in \sigma (A\left)\right\},$$

where $\sigma \left(A\right)$ represents the set of all eigenvalues on A.

Obviously, $\rho \left(A\right)$ is the maximum eigenvalue when A is nonnegative.

Definition 2. 

(i) $A=\left(a_{ij}\right)\in {\mathbb{R}}^{n\times n}$ is called the following:

(1) symmetric if $a_{ij}=a_{ji}$;

(2) zero symmetric if

$$a_{ij}=0\Rightarrow a_{ji}=0 \mathit{and} a_{ji}=0\Rightarrow a_{ij}=0.$$

Clearly, if A is symmetric, then A is zero symmetric. Conversely, the result may not hold.

Definition 3. 

Let $A\in {\mathbb{R}}^{n\times n}$ and $I\subset \{1,\dots ,n\}$ with $\left|I\right|=r$. $A_I$ is an r-dimensional submatrix of A consisting of $r^2$ elements, defined by

$$A_I=\left(a_{ij}\right),i,j\in I,$$

where $\left|I\right|$ is the number of elements of I.

For a nonnegative irreducible matrix $A=A_0=\left(a_{ij}^0\right),$ we can compute its row sum $a_i^0={\displaystyle \sum _{j=1}^n}{a}_{ij}^0,(i=1,2,\dots ,n)$ to find the minimum row sum $r^0$ and the maximum sum row $R^0$. Clearly, $r^0\le \rho \left(A\right)\le R^0$ [1]. According to [10], we set $D_0=diag(a_1^0,a_2^0,\dots ,a_n^0) \mathrm{and} A_{k+1}=\left(a_{ij}^{k+1}\right)=D_k^{-1}{A}_k{D}_k.$ Therefore, we find the serial of similar matrices $\left\{A_k\right\},$ the serial of the minimum row sum $\left\{r^k\right\}$, and the serial of the maximum row sum $\left\{R^k\right\}$. It can be verified that

$$r^0\le \dots r^k\le r^{k+1}\dots \le \rho \left(A\right)\le \dots R^{k+1}\le R^k\dots \le R^0.$$

Based on these properties, Duan et al. [10] established a diagonal transformation algorithm and the following convergent theorem for irreducible nonnegative matrices as follows (Algorithm 1):

Algorithm 1

Step 0.  Given a nonnegative irreducible matrix $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ and $ϵ>0.$ Step 1.  Let $A=A_0=I+B=\left(a_{ij}^0\right).$ Step 2.  Compute $$a_i^k=\sum _{j=1}^n{a}_{ij}^k,r^k=\underset{1\le i\le n}{min}{a}_i^k,R^k=\underset{1\le i\le n}{max}{a}_i^k.$$ If $(R^k-r^k)<ϵ$, go to Step 4. Step 3.  Compute $D_k=diag(a_1^k,a_2^k,\dots ,a_n^k)$ and update $A_{k+1}=D_k^{-1}{A}_k{D}_k,$ go to Step 2. Step 4.  $\rho \left(B\right)=\frac{1}{2}(r^k+R^k)-1.$ Stop.

Theorem 1 

(Theorem 2.5 of [10]). Let B be a nonnegative irreducible matrix. For any $\lambda >0$, it is obvious that $A=\lambda I+B={\left(a_{ij}\right)}_{n\times n}$ is a nonnegative irreducible matrix with positive diagonal elements. If $\rho \left(B\right)$ is the maximum eigenvalue of B for the sequences $\left\{r^k\right\}, \left\{R^k\right\}$ generated by Algorithm 1, then it holds that

$$\underset{k\to \infty }{lim}{r}_k-\lambda =\underset{k\to \infty }{lim}{R}_k-\lambda =\rho \left(B\right).$$

3. Improved Diagonal Transformation Algorithm for the Maximum Eigenvalue

In this section, we give the improved diagonal transformation algorithm for computing the maximum eigenvalue of zero symmetric reducible nonnegative matrices. First, we investigate the properties of zero symmetric reducible nonnegative matrices.

Lemma 1. 

Let $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ be a zero symmetric reducible nonnegative matrix. Then, the following results hold:

(1) There exists a partition $\{I_1,I_2,\dots ,I_s\}$ of $\{1,2,\dots ,n\}$ such that each induced matrix $B_{I_i}$ is either an irreducible matrix or a zero matrix for $i=1,2,\dots ,s;$

(2) $\rho \left(B\right)=\underset{1\le i\le s}{max}\rho \left(B_{I_i}\right).$

Proof. 

(1) Since B is reducible, by Definition 2, we can find a partition $\{I_1, I_2\}$ of $\{1,\dots ,n\}$ such that $b_{st}=0$ for any $s\in I_1 and t\in I_2$. Since $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ is zero symmetric, we obtain $b_{ts}=0$ for any $t\in I_2 and s\in I_1.$ If both $B_{I_1}$ and $B_{I_2}$ are irreducible, then we are done. Otherwise, we can repeat the above analysis for any reducible block(s) obtained above. In this way, since $\{1,\dots ,n\}$ is a finite set, we can arrive at the desired result.

(2) Since B is a zero symmetric reducible nonnegative matrix, we deduce that $B_{I_i}$ represents the principal submatrices for $i=1,2,\dots ,s$. Following Theorem 5.3 of [1], we obtain

$$\rho \left(B\right)=max\{\rho \left(B_{I_1}\right),\rho \left(B_{I_2}\right),\dots ,\rho \left(B_{I_s}\right)\},$$

which implies the desired results.    □

In general, partitioning a large-size reducible nonnegative matrix to be irreducible nonnegative is expensive. Hence, it is important to find an algorithm to compute the maximum eigenvalues without any partitioning. Fortunately, we observe that $\left\{R^k\right\}$ is monotonously decreasing and convergent for the diagonal transformation algorithm. Following the algorithm technology in [12], we state the improved diagonal transformation algorithm as follows (Algorithm 2):

Algorithm 2

Step 0  Given a zero symmetric nonnegative matrix $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ and $ϵ>0.$ Step 1  Let $A=A_0=I+B={\left(a_{ij}^0\right)}_{n\times n}.$ Step 2  Compute $p^k=(p_1^k,\cdots ,p_n^k)$ and $R^k$, where $p_i^k=\sum _{j=1}^n{a}_{ij}^k,R^k=\underset{1\le i\le n}{max}{p}_i^k.$ If $(R^{k-1}-R^k)<ϵ.$ Go to Step 4. Step 3  Update $a_{ij}^{k+1}=\frac{p_j^k}{p_i^k}{a}_{ij}^k.$ Go to Step 2. Step 4  $\rho \left(B\right)=R^k-1.$ Stop.

In what follows, we show the properties of the sequence $\left\{R^k\right\}$.

Lemma 2. 

Suppose that $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ is a zero symmetric reducible nonnegative matrix, and the sequence $\left\{R^k\right\}$ is generated by Algorithm 2. Then, $\left\{R^k\right\}$ is monotonously decreasing and bounded to the findings below.

Proof. 

According to Algorithm 2, it is easy to check $R^k\ge 1$. Now, we show that $\left\{R^k\right\}$ is a decreasing sequence. From the definition of $p_i^{k+1},$ it holds that

$$p_i^{k+1}=\sum _{j=1}^n{a}_{ij}^{k+1}=\sum _{j=1}^n\frac{p_j^k}{p_i^k}{a}_{ij}^k=\frac{1}{p_i^k}\sum _{j=1}^n{a}_{ij}^k{p}_j^k,1\le i\le n.$$

It follows from $p_j^k\le \underset{1\le i\le n}{max}{p}_i^k=R^k$ that

$$p_i^{k+1}\le \frac{R^k}{p_i^k}\sum _{j=1}^n{a}_{ij}^k=R^k,1\le i\le n,$$

which implies

$$\underset{1\le i\le n}{max}{p}_i^{k+1}=R^{k+1}\le R^k.$$

Consequently, $\left\{R^k\right\}$ is monotonously decreasing and bounded to the findings below.    □

The following theorem demonstrates that the sequences $\left\{R^k\right\}$ generated by Algorithm 2 are convergent for zero symmetric reducible nonnegative matrices.

Theorem 2. 

Suppose that $B=\left(b_{ij}\right)\in {\mathbb{R}}^{n\times n}$ is a zero symmetric reducible nonnegative matrix and the sequence $\left\{R_k\right\}$ is generated by Algorithm 2. Then, we have

$${\displaystyle \underset{k\to \infty }{lim}{R}^k=\rho \left(A\right)=\rho \left(B\right)+1}.$$

Proof. 

From Lemma 2, we obtain that $\left\{R^k\right\}$ is monotonously decreasing and bounded to the findings below. Thus, there exists $\overline{R}$ such that $R^k\to \overline{R}$ when $k\to \infty .$ Next, we show that $\overline{R}=\rho \left(A\right)$. Since B is zero symmetric, without loss of generality, we assume that A is stated as follows:

$$A=\left(\begin{array}{cccc}{A}_{I_1}& 0& \cdots & 0\\ 0& A_{I_2}& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & A_{I_s}\end{array}\right),$$

(1)

where each block $A_{I_i}=B_{I_i}+I_{I_i}(i=1,2,\dots ,s)$ is either an irreducible matrix or a unit matrix. It follows from Lemma 1 that

$$\rho \left(A\right)=max\{\rho \left(A_{I_1}\right),\rho \left(A_{I_2}\right),\cdots ,\rho \left(A_{I_s}\right)\}.$$

We now break up the argument into three cases.

Case 1: A has a unique maximum eigenvalue. Without loss of generality, we can assume

$$\rho \left(A\right)=\rho \left(A_{I_1}\right)>\underset{i=2,\cdots ,s}{max}\rho \left(A_{I_i}\right).$$

(2)

Let $\left\{R^k\right\}$ be the maximum row sum of matrix A. Let $\rho \left(A_{I_i}\right)$ and $R_i^k$ be the maximum eigenvalue and the maximum row sum of $A_{I_i}$ generated by Algorithm 2, respectively. It follows from Equation (2) that

$${\alpha }_i=\rho \left(A_{I_1}\right)-\rho \left(A_{I_i}\right),\forall i=2,\cdots ,s.$$

Obviously, ${\alpha }_i>0$ from Equation (2). Taking into account that each $A_{I_i}(i=1,2,\dots ,s)$ is irreducible, from Theorem 1, we have

$$\underset{k\to \infty }{lim}{R}_i^k=\rho \left(A_{I_i}\right)(i=1,2,\cdots ,s).$$

For $0<ϵ<\underset{2\le i\le s}{min}{\alpha }_i,\exists K,$ when $k>K$, one has

$$|R_i^k-\rho \left(A_{I_i}\right)|<ϵ,$$

which implies

$$\rho \left(A_{I_i}\right)+ϵ>R_i^k, \forall i=2,\dots ,s.$$

(3)

Since $\rho \left(A_{I_1}\right)-\rho \left(A_{I_i}\right)={\alpha }_i$ and $0<ϵ<{\alpha }_i,$ it holds that

$$\rho \left(A_{I_1}\right)>\rho \left(A_{I_i}\right)+ϵ, \forall i=2,\dots ,s.$$

(4)

Recalling Equations (3) and (4), for $k>K$, we obtain

$$R_1^k\ge \rho \left(A_{I_1}\right)>R_i^k, \forall i=2,\dots ,s,$$

(5)

It follows from $R^k=max\{R_1^k,R_2^k,\cdots ,R_s^k\}$ and Equation (5) that

$$R_1^k=R^k,\forall k>K.$$

By Theorem 1, we have

$$\rho \left(A_{I_1}\right)=\underset{k\to \infty }{lim}{R}_1^k=\underset{k\to \infty }{lim}{R}^k=\rho \left(A\right)=\rho \left(B\right)+1.$$

Case 2: A has two maximum eigenvalues. Without loss of generality, we can assume

$$\rho \left(A\right)=\rho \left(A_{I_1}\right)=\rho \left(A_{I_2}\right)>\underset{i=3,\cdots ,s}{max}\rho \left(A_{I_i}\right).$$

By a similar argument from Equation (5), there exists K with $k>K$ such that

$$R_1^k>R_i^k \mathrm{or} R_2^k>R_i^k,\forall i=3,\cdots ,s.$$

For $k>K,$ we deduce $R_1^k=R^k$ or $R_2^k=R^k.$ It follows from Theorem 1 that

$$\underset{k\to \infty }{lim}{R}_1^k=\rho \left(A_{I_1}\right) \mathrm{or} \underset{k\to \infty }{lim}{R}_2^k=\rho \left(A_{I_2}\right).$$

Hence, we have

$$\rho \left(A_{I_1}\right)=\underset{k\to \infty }{lim}{R}_1^k=\underset{k\to \infty }{lim}{R}^k=\rho \left(A\right) \mathrm{or} \rho \left(A_{I_2}\right)=\underset{k\to \infty }{lim}{R}_2^k=\underset{k\to \infty }{lim}{R}^k=\rho \left(A\right).$$

Case 3: A has more than two maximum eigenvalues. We repeat the process above and can obtain the same convergent conclusions.    □

Remark 1. 

Algorithm 2 has three nice properties: (1) the convergence property is guaranteed; (2) it has fewer calculations since there is no need to calculate $\left\{r^k\right\}$ and $A_{k+1}=D_k^{-1}{A}_k{D}_k;$ (3) it obtains the maximum eigenvalue without any partitioning.

In the following, we report some numerical results to show that the above new algorithm is efficient. For Algorithm 2, we stop the iteration as long as $|R^k-R^{k-1}|\le ϵ,$ where $ϵ={10}^{-10}$.

All tested zero symmetric reducible nonnegative matrices were generated as follows. Give an integer p, and randomly generate three zero symmetric matrices $B_{I_1},B_{I_2},$ and $B_{I_3}$. Let $I_1=\{1,2,\dots ,p\}, I_2=\{p+1,\dots ,2p\}$, and $I_3=\{2p+1,\dots ,3p\}$. Then, define $B=(B_{I_1},B_{I_2},B_{I_3})$ with other elements being zero, where $n=3p.$ Clearly, B is reducible. Algorithm 2 was implemented in MATLAB (R2015b), and all the numerical computations were conducted using an Intel 3.60-GHz computer with 8 GB of RAM. The numerical results are reported in Table 1, where $\rho$ is the maximum eigenvalue obtained by the algorithm and cpu(s) denotes the total computer time in seconds. Meanwhile, the cpu time is the average of 10 instances for each n. From Table 1, we see that Algorithm 2 is convergent and efficient.

Table 1. Numerical comparisons of Algorithm 2, Wen’s algorithm, and Duan’s algorithm.

	Algorithm 2		Wen’s Algorithm of [12]		Duan’s Algorithm of [10]
n	$\mathbf{\lambda }$	cpu(s)	$\mathbf{\lambda }$	cpu(s)	$\mathbf{\lambda }$	cpu(s)
300	56.1780	0.2496	56.1780	0.4220	56.1780	0.7021
600	100.7233	0.8714	100.7233	1.2132	100.7233	2.1438
900	152.9686	1.9056	152.9686	3.4469	152.9686	7.3729
1200	201.5574	3.8674	201.5574	6.6561	201.5574	10.7564
1500	251.2686	7.9249	251.2686	12.6122	251.2686	20.5346
1800	300.3427	12.5756	300.3427	19.4923	300.3427	39.1875
2100	351.0464	26.2648	351.0464	38.9256	351.0464	51.8204
2400	401.8630	38.5693	401.8630	52.4726	401.8630	86.4629
2700	451.1935	48.2369	452.1935	68.7222	452.1935	120.3801
3000	501.3178	61.8363	501.3178	99.5643	501.3178	146.5896

4. Identifying an $\mathbf{H}$-Matrix

Identifying an H-matrix, especially for large-scale matrices, has important theory and application value in numerical algebra and matrix analysis. A number of effective algorithms for identifying an H-matrix have been presented [14,15,16,17,18]. In this section, we shall identify whether a zero symmetric reducible matrix is an H-matrix by Algorithm 3 without any partitioning.

Definition 4. 

A matrix$A=\left(a_{ij}\right)\in {\mathbb{R}}^{n\times n}$is said to be the following:

(1) 

Strictly diagonally dominant if

$$|a_{ii}|>\sum _{j=1,j\ne i}^n\left|a_{ij}\right|,1\le i\le n.$$

(2) 

An H-matrix if there exists a diagonal matrix D with positive diagonal elements such that$AD$is strictly diagonally dominant;

(3) 

An M-matrix if there exists a nonnegative matrix B with the spectral radius$\rho \left(B\right)$such that$A=sI-B,$where$s>\rho \left(B\right)$.

Definition 5. 

The comparison matrix of $A=\left(a_{ij}\right)\in {\mathbb{R}}^{n\times n}$ is the matrix $M\left(A\right)$ with the elements

$$m_{ij}=\left\{\begin{array}{cc}|a_{ii}|\hfill & if i=j=1,2,\dots ,n\hfill \\ -|a_{ij}|\hfill & if i,j=1,2,\dots ,n,i\ne j.\hfill \end{array}\right.$$

Consequently, a matrix $A\in {\mathbb{R}}^{n\times n}$ is an H-matrix if and only if its comparison matrix is an M-matrix [1,15].

Theorem 3. 

Let $A=\left(a_{ij}\right)\in {\mathbb{R}}^{n\times n}$ with nonzero diagonal entries. Then, A is an H-matrix if and only if $\rho \left(\right|I-D_A^{-1}A\left|\right)<1,$ where $D_A$ denotes the diagonal matrix with the same dimensions and diagonal entries as A, and $|I-D_A^{-1}A|$ represents the matrix composed of absolute values of each element.

Proof. 

For necessity, we decompose the comparison matrix $M\left(A\right)$ of A as follows:

$$M\left(A\right)=D_{M\left(A\right)}-E,$$

where $D_{M\left(A\right)}$ denotes the diagonal matrix of the same dimensions and diagonal entries as $M\left(A\right)$ and E is a nonnegative matrix. Taking into account that the diagonal entry is a nonzero of A, we obtain

$$|I-D_A^{-1}A|=|I-D_{M\left(A\right)}^{-1}M\left(A\right)|.$$

(6)

Since A is an H-matrix, then $D_A^{-1}A$ is an H-matrix. Meanwhile, $D_{M\left(A\right)}^{-1}M\left(A\right)$ is the comparison matrix of $D_A^{-1}A.$ Thus, $D_{M\left(A\right)}^{-1}M\left(A\right)$ is an M-matrix. It follows from Equation (6) that

$$\rho \left(\right|I-D_A^{-1}A\left|\right)=\rho \left(\right|I-D_{M\left(A\right)}^{-1}M\left(A\right)\left|\right)<1.$$

For sufficiency, since $a_{ii}\ne 0$ for all $1\le i\le n$ and $\rho \left(\right|I-D_A^{-1}A\left|\right)<1,$ from Equation (6), one has

$$\rho \left(\right|I-D_{\mathcal{M}\left(A\right)}^{-1}\mathcal{M}\left(A\right)\left|\right)<1,$$

which implies $D_{\mathcal{M}\left(A\right)}^{-1}\mathcal{M}\left(A\right)$ is an M-matrix. Since $D_{M\left(A\right)}^{-1}M\left(A\right)$ is the comparison matrix of $D_A^{-1}A,$ then $D_A^{-1}A$ is an H-matrix. Thus, A is an H-matrix.    □

From Theorem 3, we can identify whether A is an H-matrix by computing $\rho \left(\right|I-D_A^{-1}A\left|\right).$ Indeed, $|I-D_A^{-1}A|$ has the following form:

$$|I-D_A^{-1}A|=\left(\begin{array}{cccc}0& \frac{|a_{12}|}{|a_{11}|}& \cdots & \frac{|a_{1n}|}{|a_{11}|}\\ \frac{|a_{21}|}{|a_{22}|}& 0& \cdots & \frac{|a_{2n}|}{|a_{22}|}\\ ⋮& ⋮& \ddots & ⋮\\ \frac{|a_{n1}|}{|a_{nn}|}& \frac{|a_{n2}|}{|a_{nn}|}& \cdots & 0\end{array}\right)=B.$$

It is clear that B is nonnegative and zero symmetric when A is zero symmetric. We define

$$C^{\left(k\right)}=I+B^{\left(k\right)},k=0,1,2,\cdots ,$$

where $B^{\left(k\right)}={\left(D^{(k-1)}\right)}^{-1}{B}^{(k-1)}{D}^{(k-1)}$ and $B^{\left(0\right)}=B.$ Thus, $B^{\left(k\right)}$ and $B^{(k-1)}$ are similar, and so are $B^{\left(k\right)}$ and $B^{\left(0\right)}$; that is, $B^{\left(k\right)}$ has the same eigenvalues. Furthermore, the  maximum eigenvalue $\rho \left(B\right)=\rho \left(C\right)-1$ holds. Consequently, if $\rho \left(C\right)<2$, then A is an H-matrix.

In the following, we propose Algorithm 3 to judge whether A is an H-matrix:

Algorithm 3

Step 0.  Given a zero symmetric matrix $A=\left(a_{ij}\right)\in {\mathbb{R}}^{n\times n}$ with $a_{ii}\ne 0.$ Step 1.  Set $C=I+|I-D_A^{-1}A|$ and compute $\rho \left(C\right)$ by Algorithm 2. Step 2.  If $\rho \left(C\right)<2,$ Output “A is an H-matrix”. Otherwise, Output “A is not an H-matrix”.

Remark 2. 

In Algorithm 3, set

$$c_i^{\left(k\right)}=\sum _{j=1}^n\left|c_{ij}^{\left(k\right)}\right|, R^{\left(k\right)}=\underset{1\le i\le n}{max}{c}_i^{\left(k\right)}.$$

From $C^{\left(k\right)}=I+B^{\left(k\right)},k=0,1,2,\cdots ,$ we have

$$\rho \left(B\right)+1\le \cdots \le R^{\left(2\right)}\le R^{\left(1\right)}.$$

If $R^{\left(k\right)}<2$, then A is an H-matrix. Therefore, it is not necessary to calculate the exact maximum eigenvalue in some cases.

The following examples illustrate the efficiency of Algorithm 3.

Example 1. 

Consider a zero symmetric reducible matrix A defined by

$$A=\left(\begin{array}{cccc}3& -2& 1& 0\\ -1& 3& 1& 0\\ 3& 2& -4& 0\\ 0& 0& 0& 1\end{array}\right).$$

It is easy to see that $C=I+|I-D_A^{-1}A|$ as follows:

$$C=\left(\begin{array}{cccc}1& \frac{2}{3}& \frac{1}{3}& 0\\ \frac{1}{3}& 1& \frac{1}{3}& 0\\ \frac{3}{4}& \frac{1}{2}& 1& 0\\ 0& 0& 0& 1\end{array}\right).$$

We may compute $R^{\left(3\right)}=1.9628$ by Algorithm 2. Therefore, A is an H-matrix.

Example 2. 

Consider the irreducible matrix A of the example in [15,19], defined by

$$A=\left(\begin{array}{ccccc}-1& 1.146392& 0& 0& 0\\ 0.5& -1& 0& -0.6& 0\\ 0& -0.1& 1& 0& 0.5\\ 0& 0.5& 0& 1& -0.5\\ -0.2& 0.1& 0.3& 0& -1\end{array}\right).$$

We compute $C=I+|I-D_A^{-1}A|$ as follows:

$$C=\left(\begin{array}{ccccc}1& 1.146392& 0& 0& 0\\ 0.5& 1& 0& 0.6& 0\\ 0& 0.1& 1& 0& 0.5\\ 0& 0.5& 0& 1& 0.5\\ 0.2& 0.1& 0.3& 0& 1\end{array}\right).$$

Under Algorithm 3, A is not an H-matrix since $\rho \left(C\right)=2.0000.$ Compared with Algorithm $\mathbb{AH}2$ in [15], which requires 37 iterations to verify that A is not an H-matrix, Algorithm 3 requires only 20 iterations and takes $0.0002$ s.

5. Conclusions

In this paper, we proposed the improved diagonal transformation algorithm to compute the maximum eigenvalue of zero symmetric nonnegative matrices, which inherited all of the advantages of the diagonal transformation algorithm while having fewer computations. In addition, the improved algorithm can also deal with irreducible matrices more efficiently than the diagonal transformation algorithms of [10,12]. Based on the improved diagonal transformation algorithm, we established Algorithm 3 to judge whether a zero symmetric matrix is an H-matrix quickly. Further studies can be considered to develop a diagonal transformation algorithm to compute eigenvalue problems of multi-linear algebra.