Non-Singularity Conditions for Two Z-Matrix Types ()

Shinji Miura

Independent, Gifu, Japan.

**DOI: **10.4236/alamt.2014.42009
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Independent, Gifu, Japan.

A real square matrix whose non-diagonal elements are non-positive is called a Z-matrix. This paper shows a necessary and sufficient condition for non-singularity of two types of Z-matrices. The first is for the Z-matrix whose row sums are all non-negative. The non-singularity condition for this matrix is that at least one positive row sum exists in any principal submatrix of the matrix. The second is for the Z-matrix which satisfies where . Let be the ith row and the jth column element of , and be the jth element of . Let be a subset of which is not empty, and be the complement of if is a proper subset. The non-singularity condition for this matrix is such that or such that for . Robert Beauwens and Michael Neumann previously presented conditions similar to these conditions. In this paper, we present a different proof and show that these conditions can be also derived from theirs.

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Miura, S. (2014) Non-Singularity Conditions for Two Z-Matrix Types. *Advances in Linear Algebra & Matrix Theory*, **4**, 109-119. doi: 10.4236/alamt.2014.42009.

1. Introduction

A real square matrix whose non-diagonal elements are non-positive is called a Z-matrix. The purpose of this paper is to show a necessary and sufficient condition for non-singularity of two types of Z-matrices.

The first is the Z-matrix whose row sums are all non-negative. In this paper, we denote this as a Non-negative Sums Z-matrix (NSZ-matrix).

The second is the Z-matrix A which satisfies where. In this paper, we denote this as a Non-negative Product Z-matrix (NPZ-matrix).

The following relation exists between these matrices.

Theorem 1.1 An NSZ-matrix is equivalent to an NPZ-matrix where all elements of are the same number.

Proof. Let be a set of numbers, and be a positive vector with all elements equal to the same number.

If is an NSZ-matrix, the ith element of is. for because is an NSZ-matrix. Also, from the premise. Hence, for is satisfied.

Therefore, is an NPZ-matrix.

Conversely, consider that is an NPZ-matrix which satisfies for where. If we divide both sides of by, we obtain for. Thus,

is an NSZ-matrix. [Q. E. D.]

As Theorem 1.1 shows, the NSZ-matrix is a type of the NPZ-matrix. Therefore, if we can find a necessary and sufficient condition for non-singularity of the NPZ-matrix, we find the necessary and sufficient condition for non-singularity of the NSZ-matrix automatically. However, we will prove the latter condition first, and then address the former condition.

We first state the basic propositions of linear algebra used in this paper.

The determinant of a square matrix is denoted in this paper.

Theorem 1.2 Let be a Z-matrix. Take a real number which is equal to or more than all diagonal elements and construct the matrix where refers to the unit matrix. is a non-negative matrix.

Proof. The non-diagonal elements of are. As is a Z-matrix, for. On the other hand, the diagonal elements of B are. Since for by the premise, for. Therefore,. [Q. E. D.]

Theorem 1.3 A non-negative square matrix always has a non-negative eigenvalue. Let be the maximum non-negative eigenvalue of. Then there exists a non-negative eigenvector corresponding to.

If a Z-matrix satisfies, is called an M-matrix^{1}.

Theorem 1.4 An M-matrix is non-singular if and only if. In this case, and all elements of the inverse of an M-matrix are non-negative. In particular, all diagonal elements of the inverse are equal to or more than ^{2}.

Proof. As Theorems 1.3 and 1.4 are well known, we entrust the proof to another work^{3}. However, as it receives less attention that diagonal elements of are or more, we confirm this aspect.

Regarding the inverse of an M-matrix, is satisfied^{4}. Thus, if we set and, is satisfied. As is a non-negative matrix and, for. Therefore, we can obtain. [Q. E. D.]

Theorem 1.5 If the row sums of a square matrix are all zeroes,.

Proof. Let be the jth column vector of. We construct the linear combination. If the row sums of are all zeroes, when. Therefore, are linearly dependent. The determinant of a matrix whose column vectors are linearly dependent is zero^{5}. [Q. E. D.]

If holds for the square matrices and and a non-singular matrix, and are called similar to each other.

Theorem 1.6 When two matrices are similar, if one matrix is non-singular, the other is also non-singular.

Theorem 1.7 Similar matrices have the same eigenvalue^{6}.

Here, we define the notation for submatrices in this paper.

Let be a set of number of rows and columns of a square matrix and let be a subset of which is not empty. refers to a submatrix of whose row and column elements belong to. When is a proper subset, we define as the complement of. refers to a submatrix of whose row elements belong to and column elements belong to. Similarly, refers to a submatrix of whose row elements belong to and column elements belong to, and refers to a submatrix of whose row and column elements belong to. Clearly, and are principal submatrices. is itself.

Based on the above, we confirm the following basic proposition.

Theorem 1.8 If is a zero matrix, ^{7}.

2. A Necessary and Sufficient Condition for Non-Singularity of the Z-Matrix Whose Row Sums Are All Non-Negative

In this section, we discuss the non-singularity of the NSZ-matrix. We reconfirm that the NSZ-matrix is defined as the Z-matrix whose row sums are all non-negative. denotes an NSZ-matrix in this section.

Theorem 2.1 An NSZ-matrix is an M-matrix^{8}.

Proof. Take an NSZ-matrix and a real number which is equal to or more than all diagonal element, and construct the matrix. is a non-negative matrix from Theorem 1.2. Let be the i th row and the jth column element of and let be the jth element of a non-negative eigenvector of corresponding to. Moreover, let be a maximum of. Incidentally, multiple candidates of may exist. In that case, one can choose any of these. From the definition of eigenvalue and eigenvector,

is satisfied. On the other hand, for because is a maximum of. because is a non-negative matrix. From these conditions, is satisfied generally. Thus,.

Here, we confirm. Note that because it is an element of non-negative eigenvector. Therefore, if is zero, because is the maximum of_{.} However, the eigenvector is not a zero vector from its definition. Hence,.

Based on the above, we divide both sides of the formula by. Then, we can derive. Note that for because is an NSZ-matrix. Then,. It is obvious that satisfies the definition of an M-matrix. [Q. E. D.]

Theorem 2.2 An NSZ-matrix is non-singular if and only if it is a non-singular M-matrix.

Proof. It is obvious that an NSZ-matrix is non-singular if it is a non-singular M-matrix. Conversely, if an NSZ-matrix is non-singular, it is a non-singular M-matrix by Theorem 2.1. [Q. E. D.]

Considering Theorem 2.2, we see that finding a necessary and sufficient condition for non-singularity of the NSZ-matrix equates to finding a condition that it is a non-singular M-matrix. We will show this.

Theorem 2.3 All row sums of any principal submatrix of an NSZ-matrix are non-negative.

Proof. Regarding itself, this is obvious because of the definition of the NSZ-matrix. In the following, we show a proof for principal submatrices which are not itself.

Let be a proper subset of and be the complement of. As all row sums of are non-negative, holds for. Hence,. Because is the complement of, it is clear that for are non-diagonal elements of the Z-matrix A. Therefore, all of these are non-positive. Then, for is true. [Q. E. D.]

Theorem 2.4 If there exists at least one principal submatrix of an NSZ-matrix whose row sums are all zeroes, the matrix is singular.

Proof. If all row sums of itself, which is one of the principal submatrices of an NSZ-matrix, are zeroes, the proposition is derived from Theorem 1.5 immediately. In the following, we show a proof for principal submatrices which are not itself.

Choose which is a proper subset of, where the row sums of are all zeroes., which is the complement of, is also not empty.

Based on this, we will confirm that is a zero matrix. From the definition of an NSZ-matrix,

holds for. Then, for are obtained because for as defined. However, for because is a Z-matrix. For these two propositions to be compatible, must hold for. Therefore, is a zero matrix.

Then, ||holds by Theorem1.8. Because all row sums of are zeroes, | |by Theorem 1.5. Thus, |. [Q. E. D.]

Theorem 2.5 If an NSZ-matrix is non-singular, at least one positive row sum exists in any principal submatrix of the matrix.

Proof. Due to the contraposition of Theorem 2.4, if an NSZ-matrix is non-singular, there does not exist a principal submatrix whose row sums are all zeroes. Then, by Theorem 2.3, at least one positive row sum exists in any principal submatrix of the matrix. [Q. E. D.]

As a result of Theorem 2.5, a necessary condition for non-singularity of the NSZ-matrix is shown. We now prove this is also a sufficient condition.

Theorem 2.6 If at least one positive row sum exists in any principal submatrix of an NSZ-matrix, the matrix is a non-singular M-matrix.

For the proof of Theorem 2.6, we have to use inference. In the following section, we will set as an NSZ-matrix which has at least one positive row sum in any principal submatrix. Moreover, we take a real number which is equal to or more than all diagonal elements and construct the matrix. Note that, from Theorem 1.2, is a non-negative matrix. We now prove the following Lemmas.

Lemma 2.7 Any row sum of all principal submatrices of is equal to or less than.

Proof. It is obvious that for from the definition of. for by Theorem 2.3. Thus,. Therefore, for. [Q. E. D.]_{}

Lemma 2.8 Any principal submatrix of has at least one row sum which is less than.

Proof. If we take, such that from the premise. Hence, such that. Note that is true from the definition of. Thus, such that for. [Q. E. D.]

Then, we classify elements belonging to the number set.

According to Lemma 2.8, itself, which is one of the principal submatrices of, has at least one row sum which is less than. We choose which satisfy to belong to the set. from the premise.

If belong to, the classification is complete. In the following, we consider the case where which does not belong to. First, we prove the following Lemma.

Lemma 2.9 for.

Proof. By Lemma 2.7, for. Since, is not true from the definition of. Thus, for. [Q. E. D.]

We now consider the classification where that does not belong to. We define as the complement of. By Lemma 2.8, such that. We classify such i as belonging to the set.

If belong to or, the classification is complete. If which belongs to neither set, we execute the third classification.

Generally, classification steps are executed when which does not belong to any of.

In such a case, we define as the complement of. Then, such that by Lemma 2.8. We classify such as belonging to the set. The next Lemma is obvious from the past consideration.

Lemma 2.10 if that belongs to the complement of.

Then, the following Lemmas are derived.

Lemma 2.11 If for, then for.

Proof. It is obvious by the method to construct and mathematical induction. [Q. E. D.]

Lemma 2.12 If, then.

Proof. Without loss of generality, we suppose and prove the Lemma under this supposition. If

, belongs to the complement of by the definition of. Hence, does not belong to

where. That is,. [Q. E. D.]

From Lemmas 2.11 and 2.12, can be defined at most by. In short, the classification is finished in limited time. If it is finished within times, we derive the next Lemmas.

Lemma 2.13.

Proof. It is obvious that. Hence, we prove. We suppose that that belongs to the complement of. Then, from Lemma 2.10, but this contradicts the definition of. By reductio ad absurdum, belong to. [Q. E. D.]

Lemma 2.14 We define. If where is not empty, for.

Proof. Let be the complement of and be any element of where. By Lemma 2.9, holds. Then, is satisfied. holds by the definition of. Therefore, holds. Thus, we obtain. [Q. E. D.]

Note that because is a non-negative matrix, a non-negative eigenvector corresponding to exists by Theorem 1.3. However, refers to a maximum non-negative eigenvalue of. Let be the jth element of the non-negative eigenvector and be a maximum of.

Lemma 2.15 If,.

Proof. for from the definition of. Further, because is a non-negative matrix. From these two conditions, for.

On the other hand, is true from the definition of eigenvalue and eigenvector. From these, is derived.

Note that if, which is a maximum of the non-negative eigenvector, is zero, the eigenvector must be a zero vector. However, this contradicts the definition of eigenvector. Thus,. Then if we divide both sides of the former formula by, we derive. Note that from the premise.

From these two formulas, is derived. [Q. E. D.]

Lemma 2.16 Let be a natural number equal to or more than 2. If and, then

._{}

Proof. We define and as the complement of. since, and

by Lemma 2.11. Let such that for. As is an element of the eigenvector corresponding to eigenvalue, holds. for from the definition of and because is a non-negative matrix. Hence, for holds. Therefore, is satisfied. Further, by the definition of and. Accordingly, for holds. Hence, is satisfied. From the above results, we see that.

We divide the leftmost and rightmost sides of this formula by,

is derived. Note that is defined as the maximum of the elements in the non-negative eigenvector corresponding to. Moreover, does not belong to and belongs to. Hence,. Therefore,

. From the premise where and Lemma 2.14,. Accordingly,

holds. Hence, holds. As, we derive. Note that as by the premise, it does not belong to either. Therefore, by Lemma 2.9. is derived. [Q. E. D.]

Proof of Theorem 2.6. By Lemma 2.13, belong to any. By Lemmas 2.15 and 2.16, is true when belongs to any. From Theorem 1.4, is a non-singular M-matrix. [Q. E. D.]

Now, we can show a necessary and sufficient condition for non-singularity of the NSZ-matrix. We will also show a necessary and sufficient condition for singularity of the matrix.

Theorem 2.17 A necessary and sufficient condition for non-singularity of the Z-matrix whose row sums are all non-negative is that at least one positive row sum exists in any principal submatrix of the matrix.

Proof. Necessity is shown in Theorem 2.5. Sufficiency is derived from Theorem 2.6. [Q. E. D.]

Theorem 2.18 A necessary and sufficient condition for singularity of the Z-matrix whose row sums are all non-negative is that there exists at least one principal submatrix of the matrix whose row sums are all zeroes.

Proof. By the contraposition of Theorem 2.17, a necessary and sufficient condition for singularity of the NSZ-matrix is that at least one principal submatrix of the NSZ-matrix whose row sums are all non-positive exists. From Theorem 2.3, this means that there exists at least one principal submatrix of the NSZ-matrix whose row sums are all zeroes. [Q. E. D.]

3. A Necessary and Sufficient Condition for Non-Singularity of the Z-Matrix Which Has a Non-Negative Product with a Positive Vector

In this section, we discuss the non-singularity of the NPZ-matrix. We reconfirm that the NPZ-matrix is defined as the Z-matrix which satisfies where. denotes an NPZ-matrix in this section.

We construct the diagonal matrix whose i th diagonal element is the ith element of. As the diagonal elements of are all positive, its inverse exists. Note that is a diagonal matrix whose ith diagonal element is.

We subsequently construct a matrix which satisfies. A and V are similar to each other by the definition of matrix similarity.

Theorem 3.1 for.

Proof. As the ith row vector of is and the jth column vector of is

, the th element of is. Thus, the ith row vector of is. Further, the jth column vector of is. Thus, the th element of is. [Q. E. D.]

Theorem 3.2 is an NSZ-matrix.

Proof. for because is a Z-matrix, and because. Therefore, for. Then, for by Theorem 3.1. That is, is a Z-matrix.

Moreover, for because is an NPZ-matrix. If we divide both sides of this formula by, we obtain for. From Theorem 3.1, for is derived.

We have shown that is a Z-matrix and all row sums of are non-negative. Thus, satisfies the definition of an NSZ-matrix. [Q. E. D.]

Theorem 3.3 An NPZ-matrix is non-singular if and only if at least one positive row sum exists in any principal submatrix of.

Proof. is an NSZ-matrix from Theorem 3.2. Then, by Theorem 2.17, is non-singular if and only if at least one positive row sum exists in any principal submatrix of. Since and are similar to each other, this is also a necessary and sufficient condition for the non-singularity of by Theorem 1.6. [Q. E. D.]

Further, we will prove that this is also an equivalent condition that is a non-singular M-matrix.

Theorem 3.4 Take a real number which is equal to or more than all diagonal elements of and construct the matrix. Moreover, we construct. Then, and are similar to each other.

Proof.. [Q. E. D.]

Theorem 3.5 is a non-negative matrix.

Proof. by the definition of and from Theorem 3.1. Then,. Therefore, , which are diagonal elements of, are non-negative for. Then, , which are non-diagonal elements of, are non-negative for because is a Z-matrix by Theorem 3.2. Thus, is a non-negative matrix. [Q. E. D.]

Theorem 3.6 An NPZ-matrix is an M-matrix^{9}.

Proof. and are both non-negative matrices due to Theorems 1.2 and 3.5. Thus, they have maximums of non-negative eigenvalues both by Theorem 1.3. Let and each be a maximum of a non-negative eigenvalue. Because and are similar by Theorem 3.4, by Theorem 1.7. Note that is an NSZ-matrix from Theorem 3.2. Then, is an M-matrix from Theorem 2.1. Therefore, from the definition of an M-matrix. Then, because. We can confirm that A satisfies the definition of an M-matrix. [Q. E. D.]

Theorem 3.7 An NPZ-matrix is non-singular if and only if it is a non-singular M-matrix.

Proof. It is obvious that an NPZ-matrix is non-singular if it is a non-singular M-matrix. Conversely, if an NPZ-matrix is non-singular, it is a non-singular M-matrix by Theorem 3.6. [Q. E. D.]

By Theorem 3.3, we can find a necessary and sufficient condition for non-singularity of an NPZ-matrix. However, this condition is described with parts of which is similar to. It is not described with parts of and that are used in the definition of the NPZ-matrix. We will look for a condition described with such parts.

In the following, let be a subset of that is not empty, and be the complement of if is a proper subset. Then, the next Theorems hold.

Theorem 3.8 if and only if for where.

Proof. This is true because from Theorem 3.1 and because is a positive vector. [Q. E. D.]

Theorem 3.9 If is a proper subset, for.

Proof. Since is the complement of, for are non-diagonal elements of. Then, they are non-positive because is a Z-matrix. Moreover, because is a positive vector. Thus, for. [Q. E. D.]

Theorem 3.10 for where.

Proof. If, this is obvious because of the definition of the NPZ-matrix. In the following, we show a proof for.

By the definition of an NPZ-matrix, is satisfied. Hence, we obtain. Note that for holds from Theorem 3.9. Accordingly,. [Q. E. D.]

Theorem 3.11 Let be any element of. Then, or such that is a necessary and sufficient condition for.

Proof.

[Sufficiency] We prove this Theorem by dividing it into two cases.

(1) The case.

If, this is obvious. In the following, we show a proof for.

By the supposition, is satisfied. Therefore, we obtain. Note that if, from Theorem 3.9. Hence,.

(2) The case such that.

If, cannot be defined. Thus, this case is applied only when.

Let be one of which satisfies aij<0 , and H be a set which removes k from G. By the define tion of an NPZ-matrix, is satisfied. Then we obtain. Since we consider the case, and is true from the definition of an NPZ-matrix, we obtain. Then; in other words is true. Further, considering and, is satisfied by Theorem 3.9. Therefore, we obtain. [Q. E. D.]

[Necessity] By the definition of an NPZ-matrix, for and for holds. Thus, the negative proposition of “or such that” is “and for”. Further, as, the negative proposition of “” is “” from Theorem 3.10. Therefore, the contraposition of this Theorem is as follows. Let be any element of. If and for,.

When, this is obvious. When, if the supposition of the contraposition is satisfied,

is true for where. That is, the contraposition is true. [Q. E. D.]

Now, we can show a necessary and sufficient condition for non-singularity of the NPZ-matrix described with parts of A and. We reconfirm that the NPZ-matrix is defined as a Z-matrix which satisfies where. Let be a subset of the number set which is not empty, and be the complement of.

Theorem 3.12 Let the Z-matrix satisfy where. A necessary and sufficient condition for the non-singularity of is such that or such that

for ^{10}.

Proof. By Theorem 3.3, an NPZ-matrix is non-singular if and only if such that for. By referring to Theorems 3.8 and 3.11, this condition can be rewritten as such that or such that for. [Q. E. D.]

Theorem 3.13 Let the Z-matrix satisfy where. A necessary and sufficient condition for the singularity of is such that for and for

^{11}.

Proof. By the contraposition of Theorem 3.12, an NPZ-matrix is singular if and only if such that for and for. However, for and

for are satisfied by the definition of an NPZ-matrix. Therefore, this singularity condition means such that for and for. [Q. E. D.]

4. Derivation from the Conditions by Robert Beauwens and Michael Neumann

In fact, Robert Beauwens has already shown a condition which resembles what is shown in Theorem 2.17 as a necessary and sufficient condition for non-singularity of the NSZ-matrix. It is as follows.

First, we define the necessary concept.

If an -dimensional square matrix satisfies for, is called diagonally dominant.

Then, if a diagonally dominant matrix satisfies for, is called lower semi-strictly diagonally dominant.

In the following, a permutation of denotes by a permutation matrix. Then, if, a permutation of, is lower semi-strictly diagonally dominant, is called semi-strictly diagonally dominant.

Beauwens showed the following Theorem.

Theorem 4.1 Let the Z-matrix be diagonally dominant and have diagonal elements that are all non-negative. is a non-singular M-matrix if and only if it is semi-strictly diagonally dominant^{12}.

If is a Z-matrix, holds for all non-diagonal elements. If diagonal elements of are nonnegative, holds for all diagonal elements.

Thus, if is diagonally dominant and has diagonal elements that are all non-negative, all row sums of are non-negative. Conversely, if all row sums of the Z-matrix are non-negative, diagonal elements of it are all non-negative by Theorem 2.3, and it is obviously diagonally dominant.

Therefore, the matrix which Theorem 4.1 addresses is nothing but the NSZ-matrix defined in this paper. Further, if satisfies the premise of Theorem 4.1, can be rewritten as. Hence, Theorem 4.1 can be rewritten as follows.

Theorem 4.2 The NSZ-matrix is a non-singular M-matrix if and only if satisfies for or, a permutation of, satisfies for.

Considering Theorem 2.2, Theorem 4.2 can be also rewritten as follows.

Theorem 4.3 The NSZ-matrix is non-singular if and only if satisfies for or, a permutation of, satisfies for.

The Beauwens condition shown in Theorem 4.3 is equivalent to the condition shown in Theorem 2.17. However, before we prove this, we introduce another Theorem of Beauwens.

Theorem 4.4 The Z-matrix is a non-singular M-matrix if and only if there exists a vector

such that and for ^{13}.

Furthermore, Michael Neumann showed the next Theorem.

Theorem 4.5 Let be a Z-matrix and be a permutation of. is a non-singular Mmatrix if and only if there exists a vector such that and for ^{14}.

The matrix which Theorems 4.4 and 4.5 address is nothing but the NPZ-matrix defined in this paper. Considering also Theorem 3.7, the following Theorem can be derived.

Theorem 4.6 The NPZ-matrix is non-singular if and only if satisfies for or, a permutation of, satisfies for.

The condition for non-singularity of the NPZ-matrix shown in Theorem 3.12 is equivalent to the Beauwens-Neumann condition shown in Theorem 4.6. We now prove this.

Theorem 4.7 Let be an NPZ-matrix, and be a permutation of. such that

for is a necessary and sufficient condition that for or for.

Proof.

[Sufficiency] Based on the premise, such that. If we permute this with,

is satisfied. Next, let be a set which removes from. Based on the premise,

such that. If we permute this with, is satisfied. After this, in the range of, let be a set which removes from. Based on the premise, such that. If we permute this with, is satisfied. If these steps are executed to, holds for. [Q. E. D.]

[Necessity] is guaranteed by Theorem 3.10. Then, the contraposition of the proposition is as follows. If such that for, then such that and such that. We prove this contraposition.

Let be the maximum number of elements of F such that for. Naturally, holds. Then, we prove.

If the number of elements of is more than, cannot be the maximum number of elements of. Hence, there is no such possibility. Thus, if we define, holds.

Then, we prove by dividing it into two cases.

(1) The case.

In this case, is true. Moreover, is true from the premise. Thus,.

(2) The case.

Let be the relative complement of in. is true. Since from the premise,. because of Theorem 3.10. Therefore,. On the other hand, for because, and is a Z-matrix. Further, by the definition of the NPZ-matrix. Thus,. In order for these conditions to be compatible, we must have. Therefore,.

Then, is proved in any case. Thus, we obtain such that.

Next, we consider, a permutation of. Let be a permutated set of. Since such that for is premised on the contraposition, then such that for.

Let be the maximum number of elements of. Then, we can also prove similarly to the proof for. Thus, we also obtain such that. [Q. E. D.]

Theorem 4.8 Let be an NPZ-matrix, and be a permutation of. such that or such that for if and only if for or for.

Proof. This is derived from Theorems 3.11 and 4.7 immediately. [Q. E. D.]

Theorem 4.8 shows the equivalence between the two non-singularity conditions of the NPZ-matrix, the condition in Theorem 3.12 and the Beauwens-Neumann condition in Theorem 4.6. Theorem 3.12 is also derived from Theorems 4.6 and 4.8.

The equivalence between the two non-singularity conditions of the NSZ-matrix, the condition in Theorem 2.17 and the Beauwens condition in Theorem 4.3, can be also proved.

Theorem 4.9 Let be an NSZ-matrix, and be a permutation of. such that

for if and only if for or for.

Proof. By Theorem 1.1, is equivalent to an NPZ-matrix where all elements of are the same number. Therefore, considering Theorem 4.7 in the case all are equal to, such that

for if and only if for or for. If we divide and and by, we obtain this Theorem. [Q. E. D.]

If, means a row sum of a principal submatrix of. Thus, such that

for means that at least one positive row sum exists in any principal submatrix of. Hence, Theorem 2.17 can be also derived from Theorems 4.3 and 4.9.

NOTES

^{1}This definition is that given by Berman & Plemmons p. 133. Alexander Ostrowski, who used the concept M-matrix first, gave a different definition for M-matrix. Cf. Ostrowski p. 69, Berman & Plemmons p. 161. The definition of M-matrix given in Varga is also different. Cf. Varga p. 91.

^{2}Theorems 1.3 and 1.4 are often called Frobenius theorem after their discoverer, Georg Frobenius.

^{3}Cf. Berman & Plemmons pp. 6-7, p. 26, Nikaido pp. 101-102, Varga p. 51, p. 89.

^{4}Cf. Nikaido p. 96, Varga p. 89.

^{5}Cf. DeFranza & Gabliardi pp. 118-119.

^{6}Cf. Anton & Rorres pp. 305-306 for Theorems 1.6 and 1.7.

^{7}Cf. Bretscher p. 258.

^{8}This theorem is stated in Plemmons p. 248.

^{9}This theorem is stated in Berman & Plemmons p. 155.

^{10}In the case, this condition is merely such that.

^{11}In the case, this condition is merely for.

^{12}Cf. Beauwens pp. 110-111.

^{13}It is written in Plemmons p. 181, p. 183 and Berman & Plemmons p. 136, p. 162 that this theorem was first shown in Beauwens .

^{14}It is written in Berman & Plemmons p. 136, p. 162 that this theorem was first shown in Neumann .

Conflicts of Interest

The authors declare no conflicts of interest.

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