# The Hadamard Product of a Nonsingular General H-Matrix and Its Inverse Transpose Is Diagonally Dominant

Authors: Rafael Bru; Maria T. Gassó; Isabel Giménez; José A. Scott
Journal: Journal of Applied Mathematics
 (2015)

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## Abstract

We study the combined matrix of a nonsingular H-matrix. These matrices can belong to two different H-matrices classes: the most common, invertible class, and one particular class named mixed class. Different results regarding diagonal dominance of the inverse matrix and the combined matrix of a nonsingular H-matrix belonging to the referred classes are obtained. We conclude that the combined matrix of a nonsingular H-matrix is always diagonally dominant and then it is an H-matrix. In particular, the combined matrix in the invertible class remains in the same class.


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## Body

## 1. Introduction

The Hadamard product of a nonsingular general H-matrix and its inverse transpose, that is, the combined matrix, has been studied in several works such as [1–3]. A complete study of the combined matrix showing its linear application can be seen in [4]. In the last decade, new properties of the combined matrix have been presented in [5, 6]. It is in this last reference where the name of combined matrix appears for the first time.It is well known that the sum by row or by column of the entries of the combined matrix of a nonsingular matrixA, C(A), is exactly equal to 1. Then, if C(A)≥O, the combined matrix is doubly stochastic. In [7, 8], the authors studied conditions under which the combined matrix of some classes of matrices is nonnegative. In particular, the authors have studied the nonnegativity of the combined matrix of totally positive (nonnegative) and totally negative (nonpositive) matrices and of sign regular matrices.The combined matrix has different applications. In a process control problem, ifA represents the relation among inputs and outputs, the combined matrix of A represents the relative gain array of the process. This interpretation was given in [9] and was applied in chemistry, for instance, in [10]. In mathematics, the combined matrix of A is used in [11] to compute the eigenvalues of A.Results involving the Hadamard product of H-matrices can be found in [12, 13]. The result where the combined matrix of a nonsingular M-matrix is also a nonsingular M-matrix was obtained by Fiedler in [1]. The same statement can be deduced from [14, 15] as Fiedler and Markham indicated in [3]. In this work we extend this result to nonsingular H-matrices. Firstly, we recall nonsingular H-matrices properties and their relations with diagonal dominance. In Section 3.1, it is proven that the combined matrix of an H-matrix of the invertible class is also an H-matrix of this class. Moreover, in Section 3.2, we obtain properties on diagonal dominance of the inverse matrix and the combined matrix of a nonsingular H-matrix belonging to the mixed class. So, we conclude that the combined matrix of a nonsingular H-matrix of the mixed class is also an H-matrix.


## 2. Notations and Definitions

In this paper, we work with square matrices of sizen×n. Matrices we are considering are real or complex. We will point out when we assume real matrices.Let the set of indicesN={1,2,…,n}. Given a matrix A=[aij] of order n, the symbol A(i∣j) denotes the principal submatrix of order n-1 that results after deleting row i and column j of A, whereas the symbol Aij denotes its determinant, Aij=det⁡A(i∣j).In order to introduce the concept of combined matrix, we recall thatA∘B represents theHadamard product or entrywise product of the matrices A and B: (1)A∘Bij=aijbij.Definition 1.
Thecombined matrix of the matrix A∈Cn×n is defined as (2)CA=A∘A-1T.As a consequence, the(i,j) entry of C(A) is given by (3)cij=aij-1i+jdet⁡Ai∣jdet⁡A=aij-1i+jAijdet⁡Aaccording to the previous notation.A class of matrices that allows us to obtain interesting properties of their combined matrices is the class of irreducible matrices. We recall that a matrixA isreducible if there exists a permutation matrix P such that PAPT is block triangular. That is, (4)PAPT=Q110Q21Q22,where the diagonal blocks Q11 and Q22 are square matrices. A matrix isirreducible when it is not reducible.Applying repeatedly the triangular decomposition to the resulting diagonal blocks, provided they are reducible, we obtain theFrobenius normal form of a reducible matrix [16]. In other words, we obtain the block triangular decomposition:(5)PAPT=R110⋯0R21R22⋯0⋮⋮⋱⋮Rn1Rn2⋯Rnn,where each diagonal block Rii is either an irreducible square block or a null block of size 1×1.The properties that we use in the study of the combined matrix of H-matrices are listed below (see [11]).Theorem 2.
LetA be a nonsingular n×n matrix and let C(A)=[cij] be its combined matrix. Then(1)
the sum by row and by column satisfies(6)∑jcij=1,∀i∈N,∑icij=1,∀j∈N;(2)
ifD is a nonsingular diagonal matrix,(7)CDA=CAD=CA;(3)
ifP and Q are permutation matrices,(8)CPAQ=PCAQ;(4)
ifA is a reducible matrix, then C(A) is a block diagonal matrix. More precisely, if PAPT=[Rij] is the Frobenius normal form of A, then(9)CA=PTCR110⋯00CR22⋯0⋮⋮⋱⋮00⋯CRkkP.Regarding the combined matrix of a reducible matrix, it is enough to study the combined matrix of each irreducible diagonal block.Next we are going to recall basic definitions related to H-matrices.Definition 3.
The matrixA=[aij]∈Cn×n is said to bediagonally dominant (DD) if(10)aii≥∑j≠iaij,i=1,2,…,n,and A is calledstrictly diagonal dominant (SDD) if inequalities (10) are strict.We are going to work withgeneral H-matrices. Traditionally, H-matrices were considered only in the case where their comparison matrices are nonsingular M-matrices. Nevertheless, as we can see in [17], there are nonsingular H-matrices with singular comparison matrix. For example, (11)A=1-111is a nonsingular H-matrix, but its comparison matrix is a singular M-matrix. For this reason, we have to consider singular M-matrices in order to obtain all possible cases of nonsingular H-matrices.Thecomparison matrix of a matrix A=[aij]∈Cn×n is defined as (12)MA=mij=2diag⁡A-A;therefore,(13)mij=-aij,ifi≠j,aij,ifi=j,i,j=1,2,…,n.Definition 4.
A real matrixA is anM-matrix if A=M(A) and(14)MA=sI-B,withB≥0,s≥ρB,where ρ(B) represents the spectral radius of B.Definition 5.
A matrixA∈Cn×n is called anH-matrix if it satisfies condition (14); that is, its comparison matrix is an M-matrix.When the last inequality in (14) is strict, it follows that M(A) is a nonsingular M-matrix. Otherwise, M(A) is singular.Analyzing the properties of nonsingular and singular H-matrices, a partition of the set of H-matrices in three classes is given in [17]:(1)
A belongs to theinvertible classHI if and only if M(A) is a nonsingular M-matrix.(2)
A belongs to themixed classHM if and only if M(A) is a singular M-matrix and its diagonal entries are not null.(3)
A belongs to thesingular classHS if and only if M(A) is an M-matrix with at least one null diagonal entry.The conclusions related to the singularity of each class are the following: all H-matrices inHI are nonsingular, in HM there are nonsingular and singular H-matrices, and all H-matrices in HS are singular and reducible. Moreover, if A is irreducible, then all singular H-matrices in HM are diagonally equivalent to the singular matrix M(A) [18].We study the combined matrix of nonsingular H-matrices. More precisely, we study the combined matrix of nonsingular H-matrices of both the invertible and the mixed classes.We now recall the generalized diagonal dominance definitions.Definition 6.
A matrixA=[aij]∈Cn×n is said to begeneralized diagonally dominant (GDD) if there exists a nonnegative diagonal and nonsingular matrix D of size n such that AD is diagonally dominant; that is,(15)aiidii≥∑j≠iaijdij,i=1,2,…,n,and A is calledgeneralized strictly diagonal dominant (GSDD) if the inequalities in (15) are strict.We have the following well known results (see [17, 19]).Theorem 7.
LetA∈Cn×n. Then,(1)
A∈HI if and only if it is GSDD;(2)
ifA is GDD, then it is an H-matrix;(3)
ifA∈HM and is irreducible, then it is GDD.


## 3. Combined Matrices of H-Matrices

In this section, we are going to extend to nonsingular and real H-matrices the following theorem that we have already commented on in the previous section (see [1]). All matrices considered in this section are real.Theorem 8.
The combined matrix of a nonsingular M-matrix is also a nonsingular M-matrix.In fact, we are going to prove the diagonal dominance of the combined matrix of a nonsingular H-matrix.


### 3.1. Combined Matrices of H-Matrices of the Invertible Class

Let us start with H-matrices of the invertible class.Theorem 9.
LetA∈HI. Then its combined matrix is strictly diagonal dominant.Proof.
SinceC(AD)=C(A), without loss of generality, we can suppose that A is SDD; therefore,(16)aii>∑j≠iaij,∀i∈N.
In addition, according to Theorem  2.5.12 in [11], we know that A-1=[αij] satisfies the strict inequalities(17)αii>αji,∀i,j∈N,j≠i.
We should notice that, with this same notation, the entries of the inverse matrix are equal to(18)αij=-1i+jAjidet⁡Aand inequality (17) can be expressed in the form(19)Aii>Aij,∀i,j∈N,j≠i.Then, taking into account inequalities (16) and (19), it results that (20)aiiAii>∑j≠iaijAii>∑j≠iaijAijand consequently (21)cii>∑j≠icij,where cij are the entries of C(A) (3). This implies that the combined matrix C(A) is SDD.Corollary 10.
IfA∈HI, then C(A)∈HI.Proof.
It is straightforward.It is well known that theHI class has different subclasses: α1- and α2-matrices, S-SDD, and so forth (see [20, 21]). Since all strictly diagonally dominant matrices belong to these subclasses, we can establish the following general corollary.Corollary 11.
LetS be a subclass of H-matrices of the invertible class that contains all SDD matrices; that is, (22)A∈Cn×n:AisSDD⊂S⊂HI.If A∈S, then C(A) belongs to the same subclass S.Proof.
It is obvious.It is worth noticing that this result can be obtained combining Theorems  3.1 and 3.5 of [12], where the authors work with the W-class of matrices satisfying condition (17).Note that we can prove Theorem8 as a consequence of Theorem 9.Corollary 12.
IfA is a nonsingular M-matrix, then C(A) is a nonsingular M-matrix.Proof.
Let us recall first thatA∈HI; then C(A) is SDD. Since A-1≥0, C(A) has the signs pattern of A; consequently C(A) is a nonsingular M-matrix.


### 3.2. Combined Matrices of Nonsingular H-Matrices of the Mixed Class

We extend the previous results to nonsingular H-matrices of the mixed class. For this purpose, we need to extend two results of [11] on SDD matrices to nonsingular DD matrices.We will denote bysgn⁡(x) the sign of the real number x.Lemma 13.
IfA is diagonally dominant and nonsingular, then (23)sgn⁡det⁡A=sgn⁡∏iaii.Proof.
SinceA is DD and nonsingular, then aii≠0 for all i∈N.
Let us suppose first thataii>0 for all i. Given ϵ∈[0,1[ we build the matrix(24)Aϵ=D+ϵA-D,where D=diag⁡(A).
SinceAϵ=ϵA+D(1-ϵ) is SDD because it is the sum of a DD matrix and a SDD matrix and diag⁡(Aϵ)=diag⁡(A)=D, then (see [11, page 125]) sgn⁡(det⁡Aϵ)=sgn⁡(a11a22⋯ann) and, therefore, det⁡Aϵ>0. Since the determinant is a multilinear function, it is continuous, and since limϵ→1-⁡Aϵ=A, then det⁡A≥0. Therefore,(25)sgn⁡det⁡A=sgn⁡a11a22⋯ann>0since A is nonsingular.
Now let us suppose that there are some negative diagonal entries. We build the diagonal matrix(26)D¯=sgn⁡diag⁡D=diag⁡sgn⁡aii.
Then, matrixB=AD¯ has its diagonal entries positive and is DD. Therefore, applying the first part of this proof to B we have det⁡B>0.
Sincedet⁡B=det⁡Adet⁡D¯=det⁡A∏i‍sgn⁡(aii), we conclude that (27)sgn⁡det⁡A=sgn⁡∏isgn⁡aii=sgn⁡∏iaii.We should notice that ifA is singular and DD, then the sign of its determinant may not match with the sign of the diagonal entries product. This means that it can occur that det⁡A=0, but ∏i‍aii≠0. Besides, if ∏i‍aii=0, then A would have at least one null row.Lemma13 can be extended to GDD matrices.Lemma 14.
If matrixA is generalized diagonally dominant and nonsingular, then (28)sgn⁡det⁡A=sgn⁡∏iaii.Proof.
It is enough to observe that ifD=diag⁡di is a diagonal, nonsingular, and nonnegative matrix that transforms AD in a diagonally dominant matrix, then det⁡(AD)=det⁡A∏i‍di and the diagonal entries of AD are equal to aiidi.The following result is essential for achieving our goals.Theorem 15.
Suppose thatA is diagonally dominant with nonzero diagonal entries. Then,(29)Aii≥Aij,∀i,j∈N.Proof.
Let us start supposing thataii>0 for all i∈N. In order to prove inequalities (29), let us consider, without loss of generality, that i=1 and j=2. We are going to prove that A11±A12≥0. Then (30)A11±A12=det⁡a22a23⋯a2na32a33⋯a3n⋮⋮⋱⋮an2an3⋯ann+det⁡±a21a23⋯a2n±a31a33⋯a3n⋮⋮⋱⋮±an1an3⋯ann=det⁡±a21+a22a23⋯a2n±a31+a32a33⋯a3n⋮⋮⋱⋮±an2+an2an3⋯ann=det⁡B.Let us prove that this auxiliary matrix B is DD. In its first row, since A is DD, we know that (31)a22=a22≥∑k≠2a2k;in particular, a22≥a21; therefore, (32)a22±a21=a22±a21≥a22-a21≥∑k≠1,2a2k.For the remaining rows, j≠1, it is true that (33)aj2±aj1+∑k≠1,2,jajk≤aj2+aj1+∑k≠1,2,jajk≤ajj.
SinceB is DD and, besides, all its diagonal entries are nonnegative, we can apply Lemma 13 and conclude that det⁡B≥0. Consequently A11±A12≥0; then A11=A11≥A12.
Finally, in the general case, when there are different signs in the diagonal entries, we build the sign matrixD¯=sgn⁡diag⁡D such that the matrix F=AD¯ is DD and its diagonal entries are all positive. Applying the result of the first part of the proof to the matrix F we conclude that Fii≥Fij for all j≠i. Finally, since(34)Fij=Aij∏j≠isgn⁡ajj,we conclude that Aii=Fii≥Fij=Aij, ∀i,j∈N, j≠i.An immediate consequence of this result is the extension of Theorem  2.5.12 of [11] to DD matrices.Theorem 16.
LetA be a nonsingular and diagonally dominant matrix. Then A-1=[αij] is diagonally dominant of its column entries. That is,(35)αii≥αji,∀i,j∈N.Proof.
We know that a nonsingular and DD matrix does not have null diagonal entries (see [17]). Then, we can apply Theorem 15 to matrix A and then inequalities (29) become inequalities (35).Theorem15 cannot be extended to GDD and to GSDD matrices because (AD)ii=AiiDii but (AD)ij=AijDjj. The following example illustrates this fact.Example 17.
The matrix(36)A=34-3410-314-24is GSDD, because when it is multiplied by the diagonal matrix D=diag⁡(1,1/2,1/4) we have the following SDD matrix: (37)AD=32-3445-3412-6·Though the inverse matrix of AD satisfies inequality (35), the matrix A-1 does not satisfy it since α12=-84/det⁡A and α22=-69/det⁡A; α13=18/det⁡A and α33=14/det⁡A. That is, α22<α12 and α33<α13.We are almost in position to extend the result on the combined matrix to nonsingular H-matrices of the mixed class.Theorem 18.
LetA∈HM be a nonsingular and irreducible matrix. Then C(A) is diagonally dominant.Proof.
Sincedet⁡A≠0, we have to prove (38)aiiAii≥∑j≠iaijAij,∀i∈N.Since A is an irreducible H-matrix of the mixed class, we know that A is GDD by Theorem 7. As C(AD)=C(A), we can suppose that A is DD; that is,(39)aii≥∑j≠iaij,∀i∈N.According to Theorem 15, (40)Aii≥Aij,∀i,j∈N.Taking into account this last inequality in (39) we obtain (38) and consequently the combined matrix is DD.Corollary 19.
IfA is a nonsingular H-matrix, then its combined matrix C(A) is an H-matrix.Proof.
Suppose thatA is written in Frobenius normal form as it is shown in (5). Let us work with its diagonal blocks. If a diagonal block of A belongs to HI, then its corresponding block in C(A) (see expression (9)) is SDD by Theorem 9 and belongs to HI by Corollary 10. Further, if an irreducible diagonal block of A is in HM, then its corresponding block in C(A) is DD by Theorem 18. Then C(A) is an H-matrix.Example 20.
Let us consider the reducible and nonsingular H-matrixA of the mixed class (41)A=21000-1100073312021546011-2written in normal form, where the first 2×2 diagonal block belongs to HI and the second 3×3 diagonal block belongs to HM.
The combined matrix is(42)CA=11328844000448800000126-182400-121202400183084.
Notice thatC(A) is block diagonal. Each diagonal block is the combined matrix of the corresponding diagonal block of A. Specifically, both blocks are SDD and then C(A)∈HI.In conclusion, in this work, we have proved that the combined matrix of a nonsingular H-matrix of the class eitherHI or HM is an H-matrix. Actually, this is an extension of a well known result for nonsingular M-matrices.


## 3.1. Combined Matrices of H-Matrices of the Invertible Class

Let us start with H-matrices of the invertible class.Theorem 9.
LetA∈HI. Then its combined matrix is strictly diagonal dominant.Proof.
SinceC(AD)=C(A), without loss of generality, we can suppose that A is SDD; therefore,(16)aii>∑j≠iaij,∀i∈N.
In addition, according to Theorem  2.5.12 in [11], we know that A-1=[αij] satisfies the strict inequalities(17)αii>αji,∀i,j∈N,j≠i.
We should notice that, with this same notation, the entries of the inverse matrix are equal to(18)αij=-1i+jAjidet⁡Aand inequality (17) can be expressed in the form(19)Aii>Aij,∀i,j∈N,j≠i.Then, taking into account inequalities (16) and (19), it results that (20)aiiAii>∑j≠iaijAii>∑j≠iaijAijand consequently (21)cii>∑j≠icij,where cij are the entries of C(A) (3). This implies that the combined matrix C(A) is SDD.Corollary 10.
IfA∈HI, then C(A)∈HI.Proof.
It is straightforward.It is well known that theHI class has different subclasses: α1- and α2-matrices, S-SDD, and so forth (see [20, 21]). Since all strictly diagonally dominant matrices belong to these subclasses, we can establish the following general corollary.Corollary 11.
LetS be a subclass of H-matrices of the invertible class that contains all SDD matrices; that is, (22)A∈Cn×n:AisSDD⊂S⊂HI.If A∈S, then C(A) belongs to the same subclass S.Proof.
It is obvious.It is worth noticing that this result can be obtained combining Theorems  3.1 and 3.5 of [12], where the authors work with the W-class of matrices satisfying condition (17).Note that we can prove Theorem8 as a consequence of Theorem 9.Corollary 12.
IfA is a nonsingular M-matrix, then C(A) is a nonsingular M-matrix.Proof.
Let us recall first thatA∈HI; then C(A) is SDD. Since A-1≥0, C(A) has the signs pattern of A; consequently C(A) is a nonsingular M-matrix.


## 3.2. Combined Matrices of Nonsingular H-Matrices of the Mixed Class

We extend the previous results to nonsingular H-matrices of the mixed class. For this purpose, we need to extend two results of [11] on SDD matrices to nonsingular DD matrices.We will denote bysgn⁡(x) the sign of the real number x.Lemma 13.
IfA is diagonally dominant and nonsingular, then (23)sgn⁡det⁡A=sgn⁡∏iaii.Proof.
SinceA is DD and nonsingular, then aii≠0 for all i∈N.
Let us suppose first thataii>0 for all i. Given ϵ∈[0,1[ we build the matrix(24)Aϵ=D+ϵA-D,where D=diag⁡(A).
SinceAϵ=ϵA+D(1-ϵ) is SDD because it is the sum of a DD matrix and a SDD matrix and diag⁡(Aϵ)=diag⁡(A)=D, then (see [11, page 125]) sgn⁡(det⁡Aϵ)=sgn⁡(a11a22⋯ann) and, therefore, det⁡Aϵ>0. Since the determinant is a multilinear function, it is continuous, and since limϵ→1-⁡Aϵ=A, then det⁡A≥0. Therefore,(25)sgn⁡det⁡A=sgn⁡a11a22⋯ann>0since A is nonsingular.
Now let us suppose that there are some negative diagonal entries. We build the diagonal matrix(26)D¯=sgn⁡diag⁡D=diag⁡sgn⁡aii.
Then, matrixB=AD¯ has its diagonal entries positive and is DD. Therefore, applying the first part of this proof to B we have det⁡B>0.
Sincedet⁡B=det⁡Adet⁡D¯=det⁡A∏i‍sgn⁡(aii), we conclude that (27)sgn⁡det⁡A=sgn⁡∏isgn⁡aii=sgn⁡∏iaii.We should notice that ifA is singular and DD, then the sign of its determinant may not match with the sign of the diagonal entries product. This means that it can occur that det⁡A=0, but ∏i‍aii≠0. Besides, if ∏i‍aii=0, then A would have at least one null row.Lemma13 can be extended to GDD matrices.Lemma 14.
If matrixA is generalized diagonally dominant and nonsingular, then (28)sgn⁡det⁡A=sgn⁡∏iaii.Proof.
It is enough to observe that ifD=diag⁡di is a diagonal, nonsingular, and nonnegative matrix that transforms AD in a diagonally dominant matrix, then det⁡(AD)=det⁡A∏i‍di and the diagonal entries of AD are equal to aiidi.The following result is essential for achieving our goals.Theorem 15.
Suppose thatA is diagonally dominant with nonzero diagonal entries. Then,(29)Aii≥Aij,∀i,j∈N.Proof.
Let us start supposing thataii>0 for all i∈N. In order to prove inequalities (29), let us consider, without loss of generality, that i=1 and j=2. We are going to prove that A11±A12≥0. Then (30)A11±A12=det⁡a22a23⋯a2na32a33⋯a3n⋮⋮⋱⋮an2an3⋯ann+det⁡±a21a23⋯a2n±a31a33⋯a3n⋮⋮⋱⋮±an1an3⋯ann=det⁡±a21+a22a23⋯a2n±a31+a32a33⋯a3n⋮⋮⋱⋮±an2+an2an3⋯ann=det⁡B.Let us prove that this auxiliary matrix B is DD. In its first row, since A is DD, we know that (31)a22=a22≥∑k≠2a2k;in particular, a22≥a21; therefore, (32)a22±a21=a22±a21≥a22-a21≥∑k≠1,2a2k.For the remaining rows, j≠1, it is true that (33)aj2±aj1+∑k≠1,2,jajk≤aj2+aj1+∑k≠1,2,jajk≤ajj.
SinceB is DD and, besides, all its diagonal entries are nonnegative, we can apply Lemma 13 and conclude that det⁡B≥0. Consequently A11±A12≥0; then A11=A11≥A12.
Finally, in the general case, when there are different signs in the diagonal entries, we build the sign matrixD¯=sgn⁡diag⁡D such that the matrix F=AD¯ is DD and its diagonal entries are all positive. Applying the result of the first part of the proof to the matrix F we conclude that Fii≥Fij for all j≠i. Finally, since(34)Fij=Aij∏j≠isgn⁡ajj,we conclude that Aii=Fii≥Fij=Aij, ∀i,j∈N, j≠i.An immediate consequence of this result is the extension of Theorem  2.5.12 of [11] to DD matrices.Theorem 16.
LetA be a nonsingular and diagonally dominant matrix. Then A-1=[αij] is diagonally dominant of its column entries. That is,(35)αii≥αji,∀i,j∈N.Proof.
We know that a nonsingular and DD matrix does not have null diagonal entries (see [17]). Then, we can apply Theorem 15 to matrix A and then inequalities (29) become inequalities (35).Theorem15 cannot be extended to GDD and to GSDD matrices because (AD)ii=AiiDii but (AD)ij=AijDjj. The following example illustrates this fact.Example 17.
The matrix(36)A=34-3410-314-24is GSDD, because when it is multiplied by the diagonal matrix D=diag⁡(1,1/2,1/4) we have the following SDD matrix: (37)AD=32-3445-3412-6·Though the inverse matrix of AD satisfies inequality (35), the matrix A-1 does not satisfy it since α12=-84/det⁡A and α22=-69/det⁡A; α13=18/det⁡A and α33=14/det⁡A. That is, α22<α12 and α33<α13.We are almost in position to extend the result on the combined matrix to nonsingular H-matrices of the mixed class.Theorem 18.
LetA∈HM be a nonsingular and irreducible matrix. Then C(A) is diagonally dominant.Proof.
Sincedet⁡A≠0, we have to prove (38)aiiAii≥∑j≠iaijAij,∀i∈N.Since A is an irreducible H-matrix of the mixed class, we know that A is GDD by Theorem 7. As C(AD)=C(A), we can suppose that A is DD; that is,(39)aii≥∑j≠iaij,∀i∈N.According to Theorem 15, (40)Aii≥Aij,∀i,j∈N.Taking into account this last inequality in (39) we obtain (38) and consequently the combined matrix is DD.Corollary 19.
IfA is a nonsingular H-matrix, then its combined matrix C(A) is an H-matrix.Proof.
Suppose thatA is written in Frobenius normal form as it is shown in (5). Let us work with its diagonal blocks. If a diagonal block of A belongs to HI, then its corresponding block in C(A) (see expression (9)) is SDD by Theorem 9 and belongs to HI by Corollary 10. Further, if an irreducible diagonal block of A is in HM, then its corresponding block in C(A) is DD by Theorem 18. Then C(A) is an H-matrix.Example 20.
Let us consider the reducible and nonsingular H-matrixA of the mixed class (41)A=21000-1100073312021546011-2written in normal form, where the first 2×2 diagonal block belongs to HI and the second 3×3 diagonal block belongs to HM.
The combined matrix is(42)CA=11328844000448800000126-182400-121202400183084.
Notice thatC(A) is block diagonal. Each diagonal block is the combined matrix of the corresponding diagonal block of A. Specifically, both blocks are SDD and then C(A)∈HI.In conclusion, in this work, we have proved that the combined matrix of a nonsingular H-matrix of the class eitherHI or HM is an H-matrix. Actually, this is an extension of a well known result for nonsingular M-matrices.



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