# A short survey of $Z$-matrices and new results on the subclass of
$F_{0}$-matrices

Samir Mondal and K.C. Sivakumar∗ Department of Mathematics, Indian Institute
of Technology Madras, Chennai 600036, India<EMAIL_ADDRESS>kcskumar@iitm.ac.in).

###### Abstract

A real square matrix $A$ of order $n\times n\leavevmode\nobreak\ (n\geq 3)$ is
called an $F_{0}$-matrix, if it is a $Z$-matrix (off-diagonal entries
nonpositive), all of whose principal submatrices of orders at most $n-2$ are
$M$-matrices while there is at least one principal submatrix of order $n-1$,
which is an $N_{0}$-matrix. An $M$-matrix is a $Z$-matrix with the property
that the real parts of all its eigenvalues are nonnegative. An $N_{0}$-matrix,
in turn, is characterized by the fact that it is an invertible $Z$-matrix
whose inverse is (entrywise) nonpositive. The first aim of this article is to
present a short survey of some subclasses of $Z$-matrices, pertinent to the
second objective, where new results concerning $F_{0}$-matrices are presented.

—————————————————————————————————————— This paper is dedicated, to Professor
M.T. Nair on the occasion of his 65th birthday. The authors wish him a long,
healthy, and productive life.

——————————————————————————————————————

Keywords: $Z$-matrix, $N_{0}$-matrix, $F_{0}$-matrix, Inverse $F_{0}$-matrix,
Circulant matrix, Linear complementarity problem.

AMS Subject Classifications: 15A48, 15A09.

##  1 Introduction

Let us recall the notion of an $F_{0}$-matrix. Let
$\mathbf{M}_{n}\,(\mathbb{R})$ denote the space of all real square matrices of
order $n\times n.$ A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is called a
$Z$-matrix, if all its off-diagonal entries are nonpositive.

######  Definition 1.1.

Let $A\in\mathbf{M}_{n}\,(\mathbb{R})$ be a $Z$-matrix. Then $A$ is called an
$F_{0}$-matrix if it satisfies the following conditions:
(1) all principal submatrices of orders at most $n-2$ are $M$-matrices.
(2) at least one principal submatrix of order $n-1$ is an $N_{0}$-matrix.

This concept was introduced by [11], and the nomenclature is due to [28], in
honour of Fan. In this article, we present some new results for this matrix
class. These will be presented after providing a short literature survey of
relevant subclasses of $Z$-matrices.

##  2 A survey of some subclasses of $Z$-matrices

In what follows, we undertake a short survey of some subclasses of
$Z$-matrices.

For a matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$, we use $R(A)$ to denote its
range space, $N(A)$ for the null space and $A^{T},$ for its transpose. The
value $\rho(.)$ denotes the spectral radius, whereas $\rho_{r}(B)$ stands for
the maximum of the spectral radii of all principal submatrices of $B$ of order
$r$. We refer to a matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ as a nonnegative
matrix if all its entries are nonnegative. This is denoted by $A\geq 0$. $A>0$
means that all the entries of $A$ are positive. This, in particular, applies
to vectors. In what follows, we collect a set of matrix classes, pertinent to
the discussion. These classes are denoted by a number of letters of the
alphabet, sometimes also going with subscripts.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is called a $Z$-matrix, if all the
off-diagonal entries of $A$ are nonpositive. Any $Z$-matrix $A$ may be
represented by $A=sI-B$, where $s$ is a real number and $B\geq 0.$ In any such
representation of $A$ as above, if $s\geq\rho(B),$ then $A$ is called an
$M$-matrix. A $Z$-matrix written as $A=sI-B$ is called an invertible
$M$-matrix, if $z>\rho(B).$ In such a case, it is well known that $A$ is
invertible and $A^{-1}\geq 0$. There is a variety of results that characterize
when a $Z$-matrix is an invertible $M$-matrix. The book [2] has fifty such
necessary and sufficient conditions and serves as an excellent resource for
this matrix class. Let us recall one such result. A $Z$-matrix $A$ is an
invertible $M$-matrix iff there exists $x>0$ such that $Ax>0.$ It follows from
this, for instance that, if $B$ is an irreducible stochastic matrix (either
all the column (row) sums equal one), and if $A$ is the $Z$-matrix given by
$A=sI-B$ with $s>1$, then $A$ is an invertible $M$-matrix. On the other hand,
if $A=I-B,$ then $A$ is a singular $M$-matrix. This is due to the fact that
$1$ is a simple eigenvalue of $B$ (by the Perron-Frobenius theorem) and that
$e,$ the ”all entries one” vector (whose dimension will be clear from the
context), is an eigenvector of $B$, so that $e$ is a nonzero vector in $N(A)$.

For $A\in\mathbb{C}^{n\times n}$, let ${\cal M}_{A}$ denote the comparison
matrix defined by

$({\cal M}_{A})_{ij}=\begin{cases}\leavevmode\nobreak\ \leavevmode\nobreak\
|a_{ii}|,&\leavevmode\nobreak\ {\textit{i}f}\leavevmode\nobreak\ i=j,\\\
-|a_{ij}|,&\leavevmode\nobreak\ {\textit{i}f}\leavevmode\nobreak\ i\neq
j.\end{cases}\\\ $ (2.1)

Note that ${\cal M}_{A}$ is a $Z$-matrix. $A$ is called an invertible
$H$-matrix, if ${\cal M}_{A}$ is an invertible $M$-matrix. Invertible
$H$-matrices have been extensively studied in the literature. A celebrated
result for invertible $H$-matrices is the following, proved in [24].

######  Theorem 2.1.

Let $A$ be an invertible $H$-matrix. Then the following coordinate wise
inequality holds:

$|A^{-1}|\leq{{\cal M}_{A}}^{-1}.$

The book [23] is a good source for results on invertible $H$-matrices. We
refer the reader to [25] for a number of recent results on invertible
$H$-matrices, which in turn, are motivated by those of $M$-matrices.

A nonnegative and nonsingular matrix is called an inverse $M$-matrix, if its
inverse is an $M$-matrix. Note that, since an invertible $M$-matrix has the
property that its inverse is nonnegative, we need this requirement of
nonnegativity, for a matrix to qualify to be an inverse $M$-matrix. We refer
the reader to [10], [13] and the monograph [14] for more details on inverse
$M$-matrices. We refer the reader also to the recent work [5].

If a matrix $A$ is such that $A-I$ is an invertible $M$-matrix, then $A$ is
invertible, and both $A$ and $I-A^{-1}$ are invertible $M$-matrices. This
result is due to Fan. This, as well as its converse were proved in [4, Theorem
5.1] and [15, Proposition 5.1]. Verbatim analogues of this result for singular
$M$-matrices were proved in [15], while the case of inverse $M$-matrices was
studied in [16, Theorem 2.1]. Results of this type for $H$-matrices are proved
in [4].

Next, let $A\in\mathbf{M}_{n}\,(\mathbb{R})$ be a $Z$-matrix given by
$A=sI-B$, with $B\geq 0$. Let $\rho_{n-1}(B)$ denote the maximum of the
spectral radii of the principal submatrices of $B$ of order
$(n-1)\times(n-1)$. If $s$ satisfies the inequalities
$\rho_{n-1}(B)<s<\rho(B),$ then $A$ is referred to as an $N$-matrix. This
matrix class was introduced and investigated in [7]. For instance, it is shown
that an $N$-matrix is precisely a $Z$-matrix with the property that its
determinant is negative, with each proper principal submatrix being an
invertible $M$-matrix. Let us recall that a proper principal submatrix of a
matrix is any principal submatrix other than the matrix itself. In [10], a
procedure of constructing an $N$-matrix, whose leading maximal principal
submatrix being an invertible $M$-matrix, was presented. Let us include it
here, as part of this exposition. Let $A$ be an invertible $M$-matrix so that
$A^{-1}\geq 0$. Set $q:=e^{T}A^{-1}e$, the sum of all the entries of $A^{-1}$,
so that $q>0.$ Set $t$ to be the minimum of all the column sums of $A^{-1}$ so
that $q>t$. Further, set $\gamma=-\frac{1}{q-t}$ and $\delta=-\frac{1}{q},$ so
that $\gamma<\delta<0.$ Finally, choose $\alpha\in(\gamma,\delta).$ If

$M=\begin{pmatrix}A&-e\\\ \alpha e^{t}&1\end{pmatrix},$

then $M$ is an invertible $Z$-matrix, ($M^{-1}<0$) and $M$ is an $N$-matrix.
For more details, we refer the reader to the discussion after Theorem 2.10,
[10].

Next, let us consider an extension of $N$-matrices. Let
$A\in\mathbf{M}_{n}\,(\mathbb{R})$ be a $Z$-matrix given by $A=sI-B$, with
$B\geq 0$. If $s$ satisfies the inequalities $\rho_{n-1}(B)\leq s<\rho(B),$
then $A$ is referred to as an $N_{0}$-matrix. This matrix class was proposed,
and several interesting properties were proved, in [11]. Note that every
$N$-matrix is an $N_{0}$-matrix and that the latter matrix class is the
topological closure of the former. Let us include an interesting extension of
the property of $N$-matrices, mentioned earlier, to the class of
$N_{0}$-matrices. A matrix $A$ is an $N_{0}$-matrix precisely if $A$ is a $Z$
matrix with the property that all proper principal submatrices are (not
necessarily invertible) $M$-matrices, with the determinant of $A$ being
negative [11, Lemma 2.1]. Here is another result: If $A$ is a $Z$-matrix, then
$A$ is an $N_{0}$-matrix iff $A^{-1}\leq 0$ and is irreducible [11, Theorem
2.7]. Note that a nonsingular matrix is irreducible iff its inverse is
irreducible. The following result concerning the spectrum of an $N_{0}$-matrix
is also interesting. Let $A$ be an $N_{0}$-matrix. Then $A$ has exactly one
negative eigenvalue [10, Theorem 2.1]. For further results, we refer the
reader to [27].

Let us turn our attention to the inverse class of $N_{0}$-matrices. A
nonpositive nonsingular matrix $A$ is called an inverse $N_{0}$-matrix, if
$A^{-1}$ is an $N_{0}$-matrix. This matrix class was introduced in [12] and
some characterizations were proved. We shall be interested in a sufficient
condition. In order to state this result, let us recall the following: Let
real numbers $a_{1},a_{2},\ldots,a_{n}$ be such that
$a_{n}>a_{n-1}>\ldots>a_{1}.$ A matrix $A:=(a_{ij})$ is called a matrix of
type $D$, if

$a_{ij}=\begin{cases}a_{i},&\leavevmode\nobreak\
{\textit{i}f}\leavevmode\nobreak\ i\leq j,\\\
a_{j},&{\textit{otherwise}}.\end{cases}\\\ $ (2.2)

A matrix of type $D$ of order $n\times n$ will be denoted by $D_{n}.$ For
instance,

$D_{4}=\begin{pmatrix}a_{1}&a_{1}&a_{1}&a_{1}\\\ a_{1}&a_{2}&a_{2}&a_{2}\\\
a_{1}&a_{2}&a_{3}&a_{3}\\\ a_{1}&a_{2}&a_{3}&a_{4}\end{pmatrix}.$

We also have the following recurrence relation:

$D_{n+1}=\begin{pmatrix}D_{n}&b\\\ b^{T}&a_{n+1}\end{pmatrix},$

where $b=(a_{1},a_{2},\ldots,a_{n})^{T}.$ Type $D$ matrices were introduced in
[17], where the following result was shown.

######  Theorem 2.2.

[17, Theorem 2.3] Let $A$ be a type $D$-matrix with $a_{1}>0.$ Then $A$ is an
inverse $M$-matrix with $A^{-1}$ being tridiagonal.

An analogue for inverse $N_{0}$-matrices was also proved.

######  Theorem 2.3.

[12, Theorem 2.4] Let $A$ be a type $D$-matrix with $a_{n}<0.$ Then $A$ is an
inverse $N_{0}$-matrix such that $A^{-1}$ is tridiagonal.

Let us turn our attention to the main object of investigation, viz.,
$F_{0}$-matrices (see Definition 1.1). Let’s recall a characterization of
$F_{0}$-matrices.

######  Theorem 2.4.

[6, Theorem 2.3] The matrix $A$ is an $F_{0}$-matrix if and only if $A=tI-B,$
with $B\geq 0$ and $\rho_{n-2}(B)\leq t<\rho_{n-1}(B)$ for $n\geq 3.$

Let $M\in\mathbf{M}_{n+1}\,(\mathbb{R})$ be an $F_{0}$-matrix. Then, one may
assume, without loss of generality (by permuting the rows and columns of $M$,
if necessary) that $M$ may be written as

$M=\begin{pmatrix}A&b\\\ c^{T}&d\end{pmatrix},$

where $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is an $N_{0}$-matrix,
$b,c\in\mathbb{R}^{n}$ are nonpositive and $d\in\mathbb{R}$ is nonnegative. We
will be making frequent use of this representation.

Here is an interesting result for this matrix class.

######  Theorem 2.5.

[10, Theorem 3.1] Let $A$ be a nonsingular $F_{0}$-matrix. Then $\rm det(A)<0$
and $A^{-1}$ is a $Z$-matrix with at least one positive diagonal entry.

The converse question was considered in [6].

######  Theorem 2.6.

[6, Theorem 2.4] Suppose that $M$ is a nonsingular $Z$-matrix. Then $M\in
F_{0}$ if and only if
$(1)$ ${\rm det}(M)<0,$
$(2)$ all principal minors of $M^{-1}$ of order at least two are nonpositive,
$(3)$ $M^{-1}$ has at least one positive diagonal entry.

There is much more information on nonsingular $F_{0}$-matrices. We only state
one relevant result. For more details, we refer to [28].

######  Theorem 2.7.

[28, Lemma 4.1] Let $A$ be a nonsingular $F_{0}$-matrix. Then all principal
minors of $A^{-1}$ of order at least two, are nonpositive.

Next, let us recall some results that are true for circulant matrices of order
$3\times 3$. Let

$P=\begin{pmatrix}0&1&0\\\ 0&0&1\\\ 1&0&0\end{pmatrix}$

and let $\cal C$ denote the class of all ($3\times 3$) circulant matrices so
that each $A\in\cal C$ is of the form

$\alpha_{0}I+\alpha_{1}P+\alpha_{2}P^{2},$ (2.3)

where $\alpha_{0},\alpha_{1}$ and $\alpha_{2}$ are real numbers.

The class $\cal C$ has the property that if $A\in\cal C$ and
$\alpha\in\mathbb{R},$ then $\alpha I+A,\alpha A\in\cal C.$ If $A\in\cal C$ is
nonsingular, then $A^{-1}\in\cal C.$ Let $A\in\cal C$ have the form as above.
Then $\alpha_{0}+\alpha_{1}+\alpha_{2}$ is an eigenvalue of $A$ with
$(1,1,1)^{T}$ as a corresponding eigenvector. Next, let
$\omega=-\frac{1}{2}+i\frac{\sqrt{3}}{2},$ with $i^{2}=-1.$ Then
$z_{A}:=\alpha_{0}+\alpha_{1}\omega+\alpha_{2}\omega^{2}$ is an eigenvalue of
$A,$ with $(1,\omega,\omega^{2})^{T},$ as an associated eigenvector. Of
course, the other eigenvalue of $A$ is $\overline{z_{A}}.$

Let ${\cal C}^{(1)}$ and ${\cal C}^{(-1)}$ denote two subclasses of $\cal C$
defined by:

${\cal C}^{(1)}:=\\{A\in{\cal C}:\alpha_{0}+\alpha_{1}+\alpha_{2}=1\\}$ (2.4)

and

${\cal C}^{(-1)}:=\\{A\in{\cal C}:\alpha_{0}+\alpha_{1}+\alpha_{2}=-1\\}$
(2.5)

It is shown in [8, Theorem 4.2] that if $A\in{\cal C}^{(1)}$, then $A$ is
uniquely determined by $z_{A}.$ More precisely, any $A$ of the form 2.3,
satisfies $A\in{\cal C}^{(1)}$ iff
$z_{A}=\alpha_{0}+\alpha_{1}\omega+\alpha_{2}\omega^{2},$ whenever
$\alpha_{0}+\alpha_{1}+\alpha_{2}=1.$ In particular, in the same result, it is
shown that $A\in{\cal C}^{(1)}$ is an $M$-matrix iff $z_{A}=a+ib$ satisfies
the inequality

$a\geq 1+\sqrt{3}|b|.$

Now, let us turn our attention to matrices in $\cal C$ which have a negative
eigenvalue. For an $N_{0}$-matrix, here is a result that was proved in [28,
Theorem 3.1]. $A\in{\cal C}^{(-1)}$ is uniquely determined by $z_{A}.$ In
particular, $A\in{\cal C}^{(-1)}$ is an $N_{0}$-matrix iff $z_{A}=a+ib$
satisfies the inequalities

$a\geq\rm max\\{\frac{1}{2},-1+\sqrt{3}|b|\\}$ and $(a-1)^{2}+b^{2}\geq 1.$

We obtain a version of these results for the classes $F_{0}$-matrices, inverse
$F_{0}$-matrices, inverse $N_{0}$-matrices and inverse $M$-matrices, in
Theorem 4.17.

A number of results stated here for the specific subclasses, have
generalizations to the larger class of nonsingular $Z$-matrices. We refer the
reader to the works [9], [20], [22]. For some recent results see [21] and for
a short survey, refer to [29].

We close this section by giving an overview of the new results of this
article. In Theorem 4.1, we show that the group inverse of a reducible
$F_{0}$-matrix, is nonpositive. For irreducible $F_{0}$-matrices of order
$3\times 3$, we obtain the same conclusion in Theorem 4.3. While an analogoue
of Theorem 4.1 is shown to be false for the Moore-Penrose inverse, once again,
affirmative answers are obtained for matrices of order $3\times 3$. These are
proved in Theorem 4.8 and Theorem 4.9. Some specific results relevant to the
linear complementarity problem are collected in Theorem 4.13. Results on
$3\times 3$ circulant matrices pertinent to the matrix classes considered
here, are presented in Theorem 4.17. Finally, certain well known invertible
matrices are shown to be inverse $F_{0}$-matrices in Theorem 4.21.

##  3 Preliminaries

Let $A\in\mathbf{M}_{m,n}\,(\mathbb{R})$. Recall that the Moore-Penrose
inverse of $A$ is the unique matrix $X\in\mathbf{M}_{n,m}\,(\mathbb{R})$ that
satisfies the equations $AXA=A,XAX=X,(AX)^{*}=AX$ and $(XA)^{*}=XA.$ We denote
the Moore-Penrose inverse of $A$ by $A^{{\dagger}}$. The group inverse of a
matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is the unique matrix
$X\in\mathbf{M}_{n}\,(\mathbb{R})$, if it exists, that satisfies the equations
$AXA=A,XAX=X$ and $AX=XA$. If it exists, then the group inverse is denoted by
$A^{\\#}$. A necessary and sufficient condition for the group inverse of a
matrix $A$ to exist is that $R(A^{2})=R(A)$, which is equivalent to the
condition $N(A^{2})=N(A)$. Another characterization is that $A^{\\#}$ exists
iff $R(A)$ and $N(A)$ are complementary subspaces of $\mathbb{R}^{n}$. It is
easy to show that a nilpotent matrix does not have group inverse, whereas the
group inverse of an idempotent matrix (exists and) is itself.

Next, we recall the full rank factorization. Let
$A\in\mathbf{M}_{m,n}\,(\mathbb{R})$ with $\rm rk(A)=r>0.$ We say that $A$ has
a full rank factorization, if there exist
$F\in\mathbf{M}_{m,r}\,(\mathbb{R}),G\in\mathbf{M}_{r,n}\,(\mathbb{R})$ such
that $\rm rk(F)=\rm rk(G)=r$ and $A=FG$. It is well known that any nonzero
matrix has a full rank factorization. Let $A=FG$ be a full rank factorization
of $A$. Then
$A^{{\dagger}}=G^{{\dagger}}F^{{\dagger}}=G^{*}(GG^{*})^{-1}(F^{*}F)^{-1}F^{*}.$
Again, let $A=FG$ be a full rank factorization of
$A\in\mathbf{M}_{n}\,(\mathbb{R}).$ Then, a necessary and sufficient condition
for $A^{\\#}$ to exist is that $GF$ is invertible. In such a case, one also
has the formula $A^{\\#}=F(GF)^{-2}G.$ For further details on these
generalized inverses, we refer the reader to [1].

We will have occasion to discuss relationships between a matrix class and the
linear complementarity problem. Let us recall this notion. Let
$A\in\mathbf{M}_{n}\,(\mathbb{R})$ and $q\in\mathbb{R}^{n}.$ The Linear
Complementarity Problem LCP$(A,q)$ is to determine if there exists
$x\in\mathbb{R}^{n}$ such that

$x\geq 0,\leavevmode\nobreak\ y:=Ax+q\geq 0\leavevmode\nobreak\
\textit{and}\leavevmode\nobreak\ x^{T}y=0.$

A vector $x$ that satisfies the first two conditions is called a feasible
vector, and the set of all feasible vectors for LCP$(A,q)$ is denoted by
FEA$(A,q)$. A vector $x$ that satisfies all the three conditions is called a
(complementary) solution, and the set of all solutions for LCP$(A,q)$ is
denoted by SOL$(A,q)$.

A matrix $A$ is called a $Q$-matrix if for any $q\in\mathbb{R}^{n}$, the
linear complementarity problem LCP$(A,q)$ has at least one solution. A matrix
$A$ is called a $Q_{0}$-matrix if whenever a feasible vector exists for
LCP$(A,q)$, there exists at least one complementary solution. That is, for any
$q\in\mathbb{R}^{n}$ such that FEA$(A,q)\neq\emptyset$, we must have
SOL$(A,q)\neq\emptyset$.

A matrix $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is said to be an $R_{0}$-matrix,
if the LCP$(A,0)$ has zero as the only solution. This is equivalent to the
implication:

$x\geq 0,Ax\geq 0\leavevmode\nobreak\ {\textit{a}nd}\leavevmode\nobreak\
x^{T}Ax=0\Longrightarrow x=0.$

Matrix $A$ is called semimontone if

$0\neq x\geq 0\Longrightarrow\leavevmode\nobreak\
{\textit{t}here\leavevmode\nobreak\ exists}\leavevmode\nobreak\
k\leavevmode\nobreak\ {\textit{s}uch\leavevmode\nobreak\
that}\leavevmode\nobreak\ x_{k}>0\leavevmode\nobreak\
{\textit{a}nd}\leavevmode\nobreak\ (Ax)_{k}\geq 0.$

##  4 Main Results

###  4.1 Nonpositivity/$Z$-property of generalized inverses of
$F_{0}$-matrices

As mentioned earlier, if $M\in\mathbf{M}_{n+1}\,(\mathbb{R})$ is an
$F_{0}$-matrix, then, $M$ may be written as

$M=\begin{pmatrix}A&b\\\ c^{T}&d\end{pmatrix},$

where $A\in\mathbf{M}_{n}\,(\mathbb{R})$ is an $N_{0}$-matrix,
$b,c\in\mathbb{R}^{n}$ are nonpositive and $d\in\mathbb{R}$ is nonnegative. In
general, $F_{0}$-matrices may be singular and possess some interesting
nonnegativity/nonpositivity properties. One of the objectives of this survey
is to prove some of those. An illustrative example gives a hint at what we are
interested in. Let $A=-\begin{pmatrix}0&1\\\
1&0\end{pmatrix},\leavevmode\nobreak\ b=c^{T}=(-1,0)^{T}$ and $d=0.$ Then $A$
is an $N_{0}$-matrix, $b,c$ are nonpositive vectors and

$M=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 0&-1&-1\\\
-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0&\leavevmode\nobreak\
\leavevmode\nobreak\ 0\\\ -1&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}$

is a singular $F_{0}$-matrix. It is easy to verify that $M^{\\#}$ exists and
that $M^{\\#}=\frac{1}{2}M.$ Thus, $M^{\\#}$ is a $Z$-matrix. Further, it is
important to observe in this simple example the fact that one may also have
all the diagonal entries, of the group inverse, to be zero.

Motivated by this example, we may ask the question: Does Theorem 2.5 have a
generalization for singular matrices, which however, are group invertible? We
show that the answer is in the affirmative. This is presented in Theorem 4.1.

Let us make some preliminary observations. Let $M$ be as above. It is easy to
observe that, for $M$ to be singular, it is necessary and sufficient that the
number $d$ satisfies $d=c^{T}A^{-1}b.$ Due to the nonpositivity of the off-
diagonal entries and the matrix $A^{-1},$ it follows that $d\leq 0.$ This, in
conjunction with the fact that all diagonal entries of an $F_{0}$-matrix are
nonnegative, implies that $d=0.$ Thus, a singular $F_{0}$-matrix is of the
form

$M=\begin{pmatrix}A&b\\\ c^{T}&0\end{pmatrix},$

with the additional assumptions on $b,c$ and $A$ as above. In our first
result, we show that the group inverse of $M$ exists. This is rather
interesting and is a consequence of the nonnegativity/nonpositivity
assumptions on the relevant matrix/vectors. We also show that the group
inverse of $M$ is a $Z$-matrix, if it is reducible. Thus, we have an extension
of Theorem 2.5 for the case of singular matrices. We also prove some results
relating to the LCP.

We will make use of the fact that if $A$ is an invertible matrix, then a
necessary and sufficient condition for the (rank one perturbed) matrix
$A+uv^{T}$ to be invertible (given by the Sherman-Morrison-Woodbury formula),
is the statement that $1+v^{T}A^{-1}u\neq 0.$ In such a case, we also have the
explicit expression for the inverse given by:

$(A+uv^{T})^{-1}=A^{-1}-\frac{1}{1+v^{T}A^{-1}u}A^{-1}uv^{T}A^{-1}.$

######  Theorem 4.1.

Let $M$ be an $F_{0}$ matrix, with the representation given as above. Let $M$
be singular. We then have the following:
(1) $\rm rk(M)=n.$
(2) $M^{\\#}$ exists.
(3) Let $M$ be reducible. Then $M^{\\#}\leq 0$ (so that $M^{\\#}$ is a
$Z$-matrix).

###### Proof.

(1) This is easy to see, as $M$ is singular, while the leading maximal
principal submatrix $A$, is nonsingular.
(2) We have

$M=\begin{pmatrix}A&b\\\ c^{T}&0\end{pmatrix}=FG=:\begin{pmatrix}A\\\
c^{T}\end{pmatrix}\begin{pmatrix}I&A^{-1}b.\end{pmatrix}$

Then

$GF=\begin{pmatrix}I&A^{-1}b\end{pmatrix}\begin{pmatrix}A\\\
c^{T}\end{pmatrix}=A+gc^{T},$

where $g:=A^{-1}b.$ Note that

$1+c^{T}A^{-1}g=1+c^{T}A^{-2}b>0,$

since $A^{-2}=(A^{-1})^{2}\geq 0$ and $b,c$ are nonpositive vectors. Thus, by
the Sherman-Morrison-Woodbury formula, $GF$ is invertible and so $M^{\\#}$
exists.
(3) Let $M$ be reducible. Since $A$ is an $N_{0}$-matrix, it is irreducible
(as mentioned earlier) and so it follows that $c=0$. From the full rank
factorization given earlier, we now have $GF=A$ (which is invertible) so that

$M^{\\#}=F(GF)^{-2}G=FA^{-2}G=\begin{pmatrix}A\\\
0\end{pmatrix}A^{-2}\begin{pmatrix}I&A^{-1}b\end{pmatrix}=\begin{pmatrix}A^{-1}&A^{-2}b\\\
0&0\end{pmatrix}.$

Now, $A^{-1}$ is a nonpositive matrix, $A^{-2}$ is a nonnegative matrix and
$b$ is a nonpositive vector and so $M^{\\#}\leq 0$. In particular, $M^{\\#}$
is a $Z$-matrix. ∎

In relation to item (3), one may ask as to what conclusion one could draw,
when the matrix is irreducible. A number of randomly generated numerical
examples have all turned out to show that the group inverse of a singular
irreducible matrix is also nonpositive. However, a general proof seems
elusive. Nevertheless, we are able to prove this result for irreducible
matrices of order $3\times 3$.

First, in the next auxiliary result, we identify $N_{0}$-matrices of order
$2\times 2.$ The proof is trivial. We use this to show that all
$F_{0}$-matrices of order $3\times 3$ (reducible or irreducible) have the
property that their group inverses are nonpositive.

######  Lemma 4.2.

Let $A\in\mathbb{R}^{2\times 2},$ be an $N_{0}$-matrix. Then $A$ is of one of
the four forms given by:

$A=\begin{pmatrix}0&\beta\\\ \gamma&\delta\end{pmatrix},\leavevmode\nobreak\
{\textit{w}ith}\leavevmode\nobreak\ \delta>0,\leavevmode\nobreak\
{\textit{a}nd}\leavevmode\nobreak\ \beta,\gamma<0.$ (4.6)
$A=\begin{pmatrix}\alpha&\beta\\\ \gamma&0\end{pmatrix},\leavevmode\nobreak\
{\textit{w}ith}\leavevmode\nobreak\ \alpha>0\leavevmode\nobreak\
{\textit{and}\leavevmode\nobreak\ }\beta,\gamma<0.$ (4.7)
$A=\begin{pmatrix}0&\beta\\\ \gamma&0\end{pmatrix},\leavevmode\nobreak\
{\textit{with}}\leavevmode\nobreak\ \beta,\gamma<0.$ (4.8)
$A=\begin{pmatrix}\alpha&\beta\\\
\gamma&\delta\end{pmatrix},\leavevmode\nobreak\
{\textit{with}}\leavevmode\nobreak\ \alpha,\delta>0,\leavevmode\nobreak\
\beta,\gamma<0\leavevmode\nobreak\ {\textit{and}}\leavevmode\nobreak\
\alpha\delta<\beta\gamma.$ (4.9)

######  Theorem 4.3.

Let $M\in\mathbb{R}^{3\times 3}$ be a singular irreducible $F_{0}$-matrix.
Then $M^{\\#}\leq 0.$

###### Proof.

Let $M=\begin{pmatrix}A&b\\\ c^{T}&0\end{pmatrix},$ where $A$ is an
$N_{0}$-matrix, and $b,c$ are nonpositive column vectors. First, we assume
that $b,c$ have at least one nonzero component. Otherwise, $M$ (or $M^{T}$) is
reducible and this has been considered in (3) of Theorem 4.1.

We make use of Lemma 4.2 for the four forms that the matrix $A$ can take. Let
us begin with $A$ as in 4.9. Then
$A^{-1}=\frac{1}{\alpha\delta-\gamma\beta}\begin{pmatrix}\leavevmode\nobreak\
\delta&-\beta\\\ -\gamma&\leavevmode\nobreak\ \alpha\end{pmatrix}$ and so, the
condition that $c^{T}A^{-1}b=0$ translates to

$c_{1}\delta b_{1}-c_{1}\beta b_{2}-c_{2}\gamma b_{1}+c_{2}\alpha b_{2}=0.$

Since each term on the left hand side is nonnegative, each must equal zero.
So, either $c=0$ or $b=0.$ Thus, once again, $M$ is reducible.

Next, let $A$ have form 4.8. Then $A^{-1}=\begin{pmatrix}0&\frac{1}{\gamma}\\\
\frac{1}{\beta}&0\end{pmatrix}$ and so,

$0=c^{T}A^{-1}b=\frac{1}{\gamma}c_{1}b_{2}+\frac{1}{\beta}b_{1}c_{2}.$

The right hand side is a sum of nonnegative terms and so each equals zero. So,
$c_{1}b_{2}=0=b_{1}c_{2}.$ This results in the following two possibilities,
employing irreducibility.

$M=\begin{pmatrix}0&\beta&b_{1}\\\ \gamma&0&0\\\ c_{1}&0&0\end{pmatrix}$ and
$M=\begin{pmatrix}0&\beta&0\\\ \gamma&0&b_{2}\\\ 0&c_{2}&0\end{pmatrix}.$

In the first case, one has

$M^{\\#}=\frac{1}{\beta\gamma+b_{1}c_{1}}\begin{pmatrix}\gamma&\beta&b_{1}\\\
\gamma&0&0\\\ c_{1}&0&0\end{pmatrix}\leq 0$

and in the latter, we have

$M^{\\#}=\frac{1}{\beta\gamma+b_{2}c_{2}}M\leq 0.$

Finally, we note that a matrix $A$ given as in 4.7 is permutationally
equivalent to 4.6 and so, it suffices to consider only a matrix of type 4.6.
Then $A^{-1}=-\frac{1}{\beta\gamma}\begin{pmatrix}\leavevmode\nobreak\
\leavevmode\nobreak\ \delta&-\beta\\\ -\gamma&\leavevmode\nobreak\
\leavevmode\nobreak\ 0\end{pmatrix}$ and so, again using $c^{T}A^{-1}b=0$, we
have

$\delta c_{1}b_{1}-\beta c_{1}b_{2}-\gamma b_{1}c_{2}=0.$

As before, each term on the right hand side equals zero. Taking into account
the irreducibility of the matrix, we have:

$M=\begin{pmatrix}0&\beta&0\\\ \gamma&\delta&b_{2}\\\ 0&c_{2}&0\end{pmatrix}.$

It is easy to verify that for this case,

$M^{\\#}=\frac{1}{(\beta\gamma+b_{2}c_{2})^{2}}\begin{pmatrix}-\beta\gamma\delta&\beta(\beta\gamma+b_{2}c_{2})&-\beta\gamma\delta\\\
\leavevmode\nobreak\ \gamma(\beta\gamma+b_{2}c_{2})&0&\leavevmode\nobreak\
b_{2}(\beta\gamma+b_{2}c_{2})\\\ -\gamma\delta
c_{2}&c_{2}(\beta\gamma+b_{2}c_{2})&\delta b_{2}c_{2}\end{pmatrix}\leq 0,$

completing the proof. ∎

Next, we turn our attention to the case of the Moore-Penrose inverse. A
natural question in relation to item (3) of Theorem 4.3, is whether the Moore-
Penrose inverse of $M$ is a $Z$-matrix. We show that this is not the case,
even when $M$ is irreducible.

######  Example 4.4.

Let

$M=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 2&-4&-1&-1\\\
-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 4&-1&-4\\\
-2&-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 1&-6\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ 0&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0&\leavevmode\nobreak\
\leavevmode\nobreak\ 0\end{pmatrix}.$

Here $A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 2&-4&-1\\\
-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 4&-1\\\
-2&-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 1\end{pmatrix},$ is a
$Z$-matrix, $A^{-1}$ is irreducible and $A^{-1}\leq 0.$ So, $A$ is an
$N_{0}$-matrix. All the diagonal entries and all the $2\times 2$ principal
minors of $M$ are nonnegative. Hence, $M$ is an $F_{0}$-matrix. Now,

$M^{\dagger}=\frac{1}{2609}\begin{pmatrix}\leavevmode\nobreak\
\leavevmode\nobreak\ 538&\leavevmode\nobreak\ \leavevmode\nobreak\
245&-327&0\\\ -149&\leavevmode\nobreak\ \leavevmode\nobreak\ 257&-220&0\\\
-692&-577&\leavevmode\nobreak\ \leavevmode\nobreak\ 440&0\\\
-245&-291&-176&0\end{pmatrix}$

has a positive off-diagonal entry and so $M^{\dagger}$ is not a $Z$-matrix.

Since, in general, $M^{{\dagger}}$ is not a nonpositive matrix, one may ask
whether $M^{{\dagger}}$ is nonpositive on $R(M).$ In the next example, we show
that even this not true. Let us make this precise.

A matrix $G\in\mathbf{M}_{n}\,(\mathbb{R})$ is said to be row-monotone if

$Gx\geq 0,\leavevmode\nobreak\ x\in R(G^{T})\Longrightarrow x\geq 0.$

Then, $G$ is row-monotone iff ([18, Lemma 2.4], [3])

$x\in R(G)\cap\mathbb{R}^{n}_{+}\Longrightarrow G^{{\dagger}}x\geq 0.$

This generalizes the requirement of the nonnegativitiy of $G^{{\dagger}}$, to
$G^{{\dagger}}$ being nonnegative on its range space $R(G)$. Specifically, if
$G^{{\dagger}}\geq 0,$ then $G$ is row-monotone, while the converse is false.

######  Example 4.5.

Let $M$ be as in Example 4.4. The vector $(37,14,30,-12)^{T}$ spans the null
space $N(M).$ So, if $(x_{1},x_{2},x_{3},x_{4})^{T}=x\in R(M^{T}),$ then we
have

$37x_{1}+14x_{2}+30x_{3}-12x_{4}=0.$

Thus, the vector $u:=(6,3,-6,7)^{T}\in R(M^{T})$, has both positive and
negative coordinates, and $Mu=-(1,22,66,0)^{T}\leq 0$. Thus, $-M$ is not row-
monotone.

Let us also record the following ”monotonicity type” affirmative result,
implied by the inverse nonpositivity of the leading principal submatrix $A$.

######  Theorem 4.6.

Let $M=\begin{pmatrix}A&b\\\
c^{T}&0\end{pmatrix}\in\mathbb{R}^{(n+1)\times(n+1)}$. Then the following
implication holds:

$Mx\leq 0,\leavevmode\nobreak\ x_{n+1}=0\Longrightarrow x\geq 0.$

###### Proof.

Let $x=(w,0)^{T}$ be such that $0\geq Mx=\begin{pmatrix}A&b\\\
c^{T}&0\end{pmatrix}\begin{pmatrix}w\\\ 0\end{pmatrix}=\begin{pmatrix}Aw\\\
c^{T}w\end{pmatrix}.$ Then $Aw\leq 0,$ which together with $A^{-1}\leq 0$
implies that $w\geq 0.$ Thus $x\geq 0.$ ∎

Let us illustrate the above result for the matrix of Example 4.4.

######  Example 4.7.

Let $M$ be as in Example 4.4 and $x=(x_{1},x_{2},x_{3},0)^{T}$. Then $Mx\leq
0$ unpacks into $2x_{1}-4x_{2}-x_{3}\leq 0,-2x_{1}+4x_{2}-x_{3}\leq 0$ and
$-2x_{1}-2x_{2}+x_{3}\leq 0.$ Adding the first and the second inequality
results in $x_{3}\geq 0,$ the sum of the first and the third yield $x_{2}\geq
0$ while summing the second and the third give $-4x_{1}+2x_{2}\leq 0,$ so that
$x_{1}\geq 0.$ We have shown that $x\geq 0.$

Interestingly, for the case of $3\times 3$ matrices, as in the case of the
group inverse (which was even nonpositive), we show that the Moore-Penrose
inverse of $M$ is a $Z$-matrix. We treat the cases of reducible and
irreducible matrices, separately.

######  Theorem 4.8.

Let $M\in\mathbb{R}^{3\times 3}$ be a singular reducible $F_{0}$-matrix. Then
$M^{{\dagger}}$ is a $Z$-matrix.

###### Proof.

Since the transpose of a $Z$-matrix is again a $Z$-matrix, there is no loss of
generality in assuming that $M=\begin{pmatrix}A&b\\\ 0&0\end{pmatrix},$ where
$A$ is an $N_{0}$-matrix and $b$ is a nonpositive column vector. By using the
Greville’s method ([1], [26]) (which appears to be more suitable than the full
rank factorization, in this case) it may be shown that

$M^{{\dagger}}=\frac{1}{1+{\parallel
g\parallel}^{2}}\begin{pmatrix}({1+{\parallel
g\parallel}^{2}})A^{-1}-gg^{T}A^{-1}&0\\\ g^{T}A^{-1}&0\end{pmatrix},$

where $g:=A^{-1}b\geq 0.$ Note that since $g^{T}A^{-1}\leq 0$ the bottom-left
row vector is nonpositive. We show that the top left (leading principal)
submatrix viz.,

$(({1+{\parallel g\parallel}^{2}})I-gg^{T})A^{-1},$

is a $Z$-matrix, for all the four possible forms that the matrix $A$ can
assume. This means that we must prove that

$\langle(({1+{\parallel g\parallel}^{2}})I-gg^{T})A^{-1}e^{1},e^{2}\rangle$

and

$\langle(({1+{\parallel g\parallel}^{2}})I-gg^{T})A^{-1}e^{2},e^{1}\rangle$

are nonpositive. Note that for any two vectors $u,v$, we have

$\langle(({1+{\parallel
g\parallel}^{2}})I-gg^{T})A^{-1}u,v\rangle=(1+{\parallel
g\parallel}^{2})\langle A^{-1}u,v\rangle-(g^{T}A^{-1}u)\langle g,v\rangle,$

where we have deliberately used both the dot product vectors as well as the
inner product notation, for convenience. Thus, we need to prove that the
numbers

$(1+{\parallel g\parallel}^{2})\langle
A^{-1}e^{1},e^{2}\rangle-(g^{T}A^{-1}e^{1})\langle g,e^{2}\rangle$ (4.10)

and

$(1+{\parallel g\parallel}^{2})\langle
A^{-1}e^{2},e^{1}\rangle-(g^{T}A^{-1}e^{2})\langle g,e^{1}\rangle$ (4.11)

are both nonpositive.

Let $A=\begin{pmatrix}\alpha&\beta\\\ \gamma&\delta\end{pmatrix}$, where
$\alpha,\delta$ are nonnegative, $\beta,\gamma$ are nonpositive, and
$b^{T}=(b_{1},b_{2})$ with $b_{1},b_{2}\leq 0.$ We have $A^{-1}=\frac{1}{\rm
detA}\begin{pmatrix}\delta&-\beta\\\ -\gamma&\alpha\end{pmatrix}$ and so

$A^{-1}e^{1}=\frac{1}{\rm detA}\begin{pmatrix}\delta\\\
-\gamma\end{pmatrix}\leavevmode\nobreak\ \textit{and}\leavevmode\nobreak\
A^{-1}e^{2}=\frac{1}{\rm detA}\begin{pmatrix}-\beta\\\ \alpha\end{pmatrix}.$

Thus,

$g^{T}=A^{-1}b=\frac{1}{{\rm det}A}(\delta b_{1}-\beta b_{2},\alpha
b_{2}-\gamma b_{1}).$

It may be verified that

$(1+{\parallel g\parallel}^{2})\langle
A^{-1}e^{2},e^{1}\rangle=\frac{-\beta}{({\rm det}A)^{3}}\\{(\alpha
b_{2}-\gamma b_{1})^{2}+(\beta b_{2}-\delta b_{1})^{2}+({\rm det}A)^{2}\\},$

Now,

$\displaystyle(1+{\parallel g\parallel}^{2})\langle
A^{-1}e^{2},e^{1}\rangle-(g^{T}A^{-1}e^{2})\langle g,e^{1}\rangle$
$\displaystyle=$ $\displaystyle\frac{(\alpha b_{2}-\gamma
b_{1})(\beta\gamma-\alpha\delta)}{({\rm det}A)^{3}}-\frac{\beta}{{\rm det}A}$
$\displaystyle=$ $\displaystyle-\frac{b_{1}(\alpha b_{2}-\gamma b_{1})}{({\rm
det}A)^{2}}-\frac{\beta}{{\rm det}A}\leq 0.$

Similarly, one may verify that

$\displaystyle(1+{\parallel g\parallel}^{2})\langle
A^{-1}e^{1},e^{2}\rangle-(g^{T}A^{-1}e^{1})\langle g,e^{2}\rangle$
$\displaystyle=$ $\displaystyle\frac{b_{2}(\beta b_{2}-\delta
b_{1})(\alpha\delta-\beta\gamma)}{({\rm det}A)^{3}}-\frac{\gamma}{{\rm det}A}$
$\displaystyle=$ $\displaystyle\frac{b_{2}(\beta b_{2}-\delta b_{1})}{({\rm
det}A)^{2}}-\frac{\gamma}{{\rm det}A}\leq 0.$

∎

######  Theorem 4.9.

Let $M\in\mathbb{R}^{3\times 3}$ be a singular irreducible $F_{0}$-matrix.
Then $M^{{\dagger}}\leq 0.$

###### Proof.

Let $M=\begin{pmatrix}A&b\\\ c^{T}&0\end{pmatrix},$ where $A$ is an
$N_{0}$-matrix, and $b,c$ are nonpositive column vectors. We show that there
is a full rank factorization of $M$ given by $M=FG$, where $F^{{\dagger}}\leq
0$ and $G^{{\dagger}}\geq 0$. The proof would then follow, since
$M^{{\dagger}}=G^{{\dagger}}F^{{\dagger}}$.

Again, we consider the three forms that the matrix $A$ can take. Assume that
$A$ is given as in 4.8. Then, as argued earlier, there are only two
possibilities. In the first case, we have

$M=\begin{pmatrix}0&\beta&b_{1}\\\ \gamma&0&0\\\ c_{1}&0&0\end{pmatrix}$

and $M$ has the full rank factorization $M=FG$, where

$F=\begin{pmatrix}0&\beta\\\ \gamma&0\\\ c_{1}&0\end{pmatrix}$ and
$G=\begin{pmatrix}1&0&0\\\ 0&1&\frac{b_{1}}{\beta}\end{pmatrix}.$

It may be verified that

$F^{{\dagger}}=\frac{1}{\beta^{2}(\gamma^{2}+c_{1}^{2})}\begin{pmatrix}0&\beta^{2}\gamma&\beta^{2}c_{1}\\\
\beta(\gamma^{2}+c_{1}^{2})&0&0\end{pmatrix}\leq 0$

whereas

$G^{{\dagger}}=\frac{1}{\beta^{2}+b_{1}^{2}}\begin{pmatrix}\beta^{2}+b_{1}^{2}&0\\\
0&\beta^{2}\\\ 0&\beta b_{1}\end{pmatrix}\geq 0.$

The other possibility for $M$ is given as

$M=\begin{pmatrix}0&\beta&0\\\ \gamma&0&b_{2}\\\ 0&c_{2}&0\end{pmatrix}.$

Here, $M$ has the full rank factorization $M=FG$, where

$F=\begin{pmatrix}0&\beta\\\ \gamma&0\\\ 0&c_{2}\end{pmatrix}$ and
$G=\begin{pmatrix}1&0&\frac{b_{2}}{\gamma}\\\ 0&1&0\end{pmatrix}.$

Then

$F^{{\dagger}}=\frac{1}{\gamma(\beta^{2}+c_{2}^{2})}\begin{pmatrix}0&\beta^{2}+c_{2}^{2}&0\\\
\beta\gamma&0&c_{2}\gamma\end{pmatrix}\leq 0$

and

$G^{{\dagger}}=\frac{1}{\gamma^{2}+b_{2}^{2}}\begin{pmatrix}\gamma^{2}&0\\\
0&\gamma^{2}+b_{2}^{2}\\\ \gamma b_{2}&0\end{pmatrix}\geq 0.$

Next, assume that $A$ is as in 4.6. Here,

$M=\begin{pmatrix}0&\beta&0\\\ \gamma&\delta&b_{2}\\\ 0&c_{2}&0\end{pmatrix}$

and a full rank factorization $M=FG$ is given by

$F=\begin{pmatrix}0&\beta\\\ \gamma&\delta\\\ 0&c_{2}\end{pmatrix},$

whereas $G$ is the same matrix as in the previous case. So, $G^{{\dagger}}\geq
0$, while

$F^{\dagger}=\begin{pmatrix}\frac{-\delta\beta}{\gamma(\delta^{2}+c_{2}^{2})}&\frac{1}{\gamma}&\frac{-\delta
c_{2}}{\gamma(\delta^{2}+c_{2}^{2})}\\\
\frac{\beta}{\beta^{2}+c_{2}^{2}}&0&\frac{c_{2}}{\beta^{2}+c_{2}^{2}}\end{pmatrix}\leq
0.$

This completes the proof. ∎

We illustrate Theorem 4.8 and Theorem 4.9, by examples.

######  Example 4.10.

Let $M=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 1&-2&-1\\\
-3&\leavevmode\nobreak\ \leavevmode\nobreak\ 1&-2\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ 0&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}.$ Then $M$ is
reducible and is an $F_{0}$-matrix. We have

$M^{\dagger}=\begin{pmatrix}\leavevmode\nobreak\
\frac{1}{15}&-\frac{1}{5}&0\\\ -\frac{1}{3}&\leavevmode\nobreak\
\leavevmode\nobreak\ 0&0\\\ -\frac{4}{15}&-\frac{1}{5}&0\end{pmatrix},$

showing that, while the Moore-Penrose inverse is a $Z$-matrix, it is not
necessarily nonpositive.

######  Example 4.11.

Let $M=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 0&-1&-1\\\
-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 0&\leavevmode\nobreak\
\leavevmode\nobreak\ 0\\\ -1&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}.$ Then $M$ is an
irrreducible $F_{0}$-matrix and its Moore-Penrose inverse is given by

$M^{\dagger}=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\
0&-\frac{2}{5}&-\frac{1}{5}\\\ -\frac{1}{2}&\leavevmode\nobreak\
\leavevmode\nobreak\ 0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\\\
-\frac{1}{2}&\leavevmode\nobreak\ \leavevmode\nobreak\ 0&\leavevmode\nobreak\
\leavevmode\nobreak\ 0\end{pmatrix}\leq 0.$

###  4.2 Complementarity properties

In this section, we prove some consequences for $F_{0}$-matrices in the
context of the linear complementarity problem.

Let us recall a result that presents a necessary condition for a matrix to
belong to $Q_{0}\setminus Q.$ Note that if a matrix has a nonpositive row,
then it cannot be a $Q$-matrix. By taking a vector $q$ whose last coordinate
is negative, it is immediate to observe that LCP$(A,q)$ does not have a
solution. Next, we show that such

######  Lemma 4.12.

Let $A$ be a $Q_{0}$-matrix which is not a $Q$-matrix. Then there exists a
vector $0\neq y\geq 0$ such that $A^{T}y\leq 0.$

######  Theorem 4.13.

Let $M$ be a singular $F_{0}$-matrix, given by $M=\begin{pmatrix}A&b\\\
c^{T}&0\end{pmatrix}.$ Then
(1) $M$ is an $R_{0}$-matrix iff $b\neq 0.$
(2) $M$ is not a semimonotone matrix.
(3) If $-M$ is a $Q_{0}$-matrix, then $c=0.$

###### Proof.

(1) Let $b\neq 0.$ Assume that $x:=(y,\lambda)^{T}$ is chosen such that $x\geq
0$ and $Mx\geq 0$. Then $y\geq 0,\lambda\geq 0$ satisfy the inequalities

$Ay+\lambda b\geq 0\leavevmode\nobreak\ \textit{and}\leavevmode\nobreak\
c^{T}y+\lambda d\geq 0.$

The first inequality, upon premultiplying by $A^{-1}\leq 0$, yields $y+\lambda
A^{-1}b\leq 0,$ so that $y\leq-\lambda A^{-1}b\leq 0$, since $A^{-1}b\geq 0.$
Thus $y=0$ and since $0\neq b\leq 0,$ (by the simplified first inequality), we
then have $\lambda=0$. Thus $x=0,$ proving that $M$ is an $R_{0}$-matrix.

If $b=0,$ then any vector of the form $x:=(0,\lambda)^{T}$ for $\lambda>0$ is
a nonnegative nonzero vector that satisfies $Mx=0$. Hence $M$ is not an
$R_{0}$-matrix.

(2) Let $y\in\mathbb{R}^{n}$ be chosen such that $y<0$ and $x:=A^{-1}y.$ Then
$x>0$ and $(Ax)_{k}<0$ for every $k$. If we set
$z:=(x,0)^{T}\in\mathbb{R}^{n+1},$ then $z\geq 0$ and $Mz=(Ax,u^{T}x)^{T}.$ If
$z_{k}>0,$ then $1\leq k\leq n$ and for every such $k$, we have
$(Mz)_{k}=(Ax)_{k}<0,$ proving that $M$ is not semimonotone.
(3) Let $-M$ be a $Q_{0}$-matrix. Then by Lemma 4.12, there exists a vector
$y\neq 0,$ $y=(w,\lambda)\geq 0$ such that $-M^{T}y\leq 0.$ Expanding this, we
obtain $b^{T}w\geq 0$ and

$\displaystyle A^{T}w+\lambda c\geq 0.$ (4.12)

Now, premultiplying 4.12 by $(A^{T})^{-1}$ yields

$w+\lambda(A^{-1})^{T}c\leq 0,$

so that

$w\leq-\lambda(A^{-1})^{T}c\leq 0.$

So, $w=0$. Now, 4.12 gives $\lambda c\geq 0$ and so $\lambda c=0$ (since
$\lambda$ is nonnegative and $c\leq 0$). If $c$ is a nonzero vector, then we
must have $\lambda=0.$ Hence $y=0,$ contradicting $y\neq 0.$ Therefore, $c=0.$
∎

######  Remark 4.14.

In the proof of (1), the complementarity condition $x^{T}Mx=0$ has not been
used. This means that one has

$x\geq 0\leavevmode\nobreak\ \textit{and}\leavevmode\nobreak\ Mx\geq
0\Longrightarrow x=0.$

It follows that $-M$ has a left poverse, that is there exist nonnegative
matrices $U,V$ such that $I+V+UM=0.$ We refer the reader to Theorem 8.6,
Remark 8.7, Example 8.22 and the relevant citation in [19] for more details.

Next, we show that the converse in (3) of the above result is not true.

######  Example 4.15.

Let

$M=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 0&-1&-1\\\
-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0&-1\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ 0&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}.$

Then $M$ is a singular $F_{0}$-matrix. If $q=(-1,1,1)^{T},$ then
$x^{0}:=(1,1,0)^{T}$ satisfies $-Mx^{0}+q\geq 0$ and so $x^{0}$ is feasible
for LCP$(-M,q).$ If possible, let $u=(u_{1},u_{2},u_{3})^{T}$ be a solution
for LCP$(-M,q).$ Then $v:=-Mu+q\geq 0$ and $u^{T}v=0.$ Since $q_{3}$ is
positive, the complementarity constraint implies that $u_{3}=0$. The first two
inequalities are $v_{1}=u_{2}-1\geq 0$ and $v_{2}=u_{1}+1\geq 0$ Now, since
$u_{2}\geq 1$, from the second complementarity condition, (which is
$u_{2}v_{2}=0$), it follows that $u_{1}+1=0$. This contradicts the
nonnegativity of $u_{1},$ showing that $-M$ is not a $Q_{0}$-matrix.

However, we have the following result.

######  Lemma 4.16.

Let $M$ be a singular reducible $F_{0}$-matrix. Let $q=(p,q_{n})^{T},$ where
$p\in\mathbb{R}^{n}$ such that $p\leq 0$ and $q_{n}\geq 0$. Then LCP$(-M,q)$
has a solution.

###### Proof.

We have $M=\begin{pmatrix}A&b\\\ 0&0\end{pmatrix},$ where $A$ is an
$N_{0}$-matrix and $b\leq 0.$ Define $u:=A^{-1}p\geq 0.$ If $v:=(u,0)^{T},$
then $v\geq 0$ and so

$y:=-Mv+q=\begin{pmatrix}-A&-b\\\ \leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}\begin{pmatrix}u\\\
0\end{pmatrix}+\begin{pmatrix}p\\\ q_{n}\end{pmatrix}=\begin{pmatrix}-Au+p\\\
q_{n}\end{pmatrix}=\begin{pmatrix}0\\\ q_{n}\end{pmatrix}\geq 0.$

Also, $v^{T}y=0.$ Hence, $v\in$ SOL$(-M,q).$ ∎

###  4.3 Results for circulant matrices

Here, we present characterizations for $3\times 3$ circulant matrices to
belong to some of the pertinent matrix classes, motivated by the results
mentioned earlier.

######  Theorem 4.17.

Let $A$ be a $3\times 3$ circulant matrix and $z_{A}=a+ib$, both be as defined
earlier. The following hold:
(i) Let $A\in{\cal C}^{(-1)}$. Then
(1) $A$ is an $F_{0}$-matrix iff

$a>-1+|b|\sqrt{3},\leavevmode\nobreak\ a\geq\frac{1}{2}$ and
$(a-1)^{2}+b^{2}<1.$

(2) $A$ is an inverse $N_{0}$-matrix iff

$(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\geq 1,\leavevmode\nobreak\
(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\geq 1$ and $a\leq\frac{1}{2}.$

(3) $A$ is an inverse $F_{0}$-matrix iff

$(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\geq 1,\leavevmode\nobreak\
(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\geq 1$ and $a>\frac{1}{2}.$

(ii) Let $A\in C^{(1)}$. Then $A$ is an inverse $M$-matrix iff

$(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\leq 1$ and
$(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\leq 1.$

###### Proof.

(i) Recall that a matrix $A$ is of the form 2.3 iff

$A=\begin{pmatrix}\alpha_{0}&\alpha_{1}&\alpha_{2}\\\
\alpha_{2}&\alpha_{0}&\alpha_{1}\\\
\alpha_{1}&\alpha_{2}&\alpha_{0}\end{pmatrix}.$

Then, $A\in{\cal C}^{(-1)}$ iff
$z_{A}=\alpha_{0}+\alpha_{1}\omega+\alpha_{2}\omega^{2},$ where
$\alpha_{0}+\alpha_{1}+\alpha_{2}=-1.$ We have $a+ib=z_{A}.$ Thus,
$a=\alpha_{0}-\frac{1}{2}(\alpha_{1}+\alpha_{2})$, which simplifies to
$a=-1-\frac{3}{2}(\alpha_{1}+\alpha_{2})$. Also,
$b=\frac{\sqrt{3}}{2}(\alpha_{1}-\alpha_{2}).$ Hence,

$\displaystyle\alpha_{1}=-\frac{a+1}{3}+\frac{b}{\sqrt{3}}$ (4.13)
$\displaystyle\alpha_{2}=-\frac{a+1}{3}-\frac{b}{\sqrt{3}}.$ (4.14)

(1) For $A$ to be an $F_{0}$-matrix, the following inequalities are necessary
and sufficient:

$\alpha_{0}\geq 0,\leavevmode\nobreak\ \alpha_{1}<0,\leavevmode\nobreak\
\alpha_{2}<0$ and $\alpha_{0}^{2}<\alpha_{1}\alpha_{2}.$

The nonpositivity of $\alpha_{1}$ and $\alpha_{2}$ yield the inequality
$a>-1+|b|\sqrt{3}.$ The nonnegativity of $\alpha_{0}$ is equivalent to
$a\geq\frac{1}{2}.$ The inequality $\alpha_{0}^{2}<\alpha_{1}\alpha_{2}$
reduces to

$\displaystyle(\alpha_{1}+\alpha_{2})^{2}+2(\alpha_{1}+\alpha_{2})-\alpha_{1}\alpha_{2}+1<0.$
(4.15)

This is the requirement $(a-1)^{2}+b^{2}<1.$

(2) The matrix $A$ is an inverse $N_{0}$-matrix iff

$\alpha_{2}^{2}-\alpha_{0}\alpha_{1}\geq 0$
$\alpha_{1}^{2}-\alpha_{0}\alpha_{2}\geq 0$ and $\alpha_{0}\leq 0.$

The nonnegativity requirement on the numbers
$\alpha_{2}^{2}-\alpha_{0}\alpha_{1}$ and
$\alpha_{1}^{2}-\alpha_{0}\alpha_{2}$ yield the inequalities

$(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\geq 1$ and
$(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\geq 1,$

respectively. The nonpositivity of $\alpha_{0}$ is equivalent to
$a\leq\frac{1}{2}.$

(3) The matrix $A$ is an inverse $F_{0}$-matrix iff

$\alpha_{2}^{2}-\alpha_{0}\alpha_{1}\geq 0,$
$\alpha_{1}^{2}-\alpha_{0}\alpha_{2}\geq 0$ and $\alpha_{0}>0.$

The nonnegativity requirements $\alpha_{2}^{2}-\alpha_{0}\alpha_{1}\geq 0$ and
$\alpha_{1}^{2}-\alpha_{0}\alpha_{2}\geq 0$ reduce to

$(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\geq 1$ and
$(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\geq 1,$ respectively.

The condition $\alpha_{0}>0$ is equivalent to $a>\frac{1}{2}.$

(ii) For a matrix $A$ of the form 4.3 to be in ${\cal C}^{(1)}$ it is
necessary and sufficient that
$z_{A}=\alpha_{0}+\alpha_{1}\omega+\alpha_{2}\omega^{2},$ where
$\alpha_{0}+\alpha_{1}+\alpha_{2}=1.$ For this matrix class, we have
$a=\alpha_{0}-\frac{1}{2}(\alpha_{1}+\alpha_{2})$, which simplifies to
$a=1-\frac{3}{2}(\alpha_{1}+\alpha_{2})$. Also,
$b=\frac{\sqrt{3}}{2}(\alpha_{1}-\alpha_{2}).$ Thus,

$\displaystyle\alpha_{1}=\frac{1}{3}(1-a)+\frac{b}{\sqrt{3}}$ (4.16)
$\displaystyle\alpha_{2}=\frac{1}{3}(1-a)-\frac{b}{\sqrt{3}}.$ (4.17)

Now, $A$ is an inverse $M$-matrix iff

$\alpha_{2}^{2}-\alpha_{0}\alpha_{1}\leq 0$ and
$\alpha_{1}^{2}-\alpha_{0}\alpha_{2}\leq 0.$

The nonpositivity of $\alpha_{2}^{2}-\alpha_{0}\alpha_{1}$ is the same as the
inequality $(a-\frac{1}{2})^{2}+(b-\frac{\sqrt{3}}{2})^{2}\leq 1,$ while the
nonpositivity of $\alpha_{1}^{2}-\alpha_{0}\alpha_{2}$ is equivalent to
$(a-\frac{1}{2})^{2}+(b+\frac{\sqrt{3}}{2})^{2}\leq 1.$ ∎

###  4.4 Inverse $F_{0}$-matrices

In the last part of this work, we present a condition for an invertible matrix
to be an inverse $F_{0}$-matrix, in Theorem 4.21. Even though this result
follows as a particular case of [20, Theorem 3.12], our proof is more
elementary.

Let us recall that a matrix $M$ is said to be totally nonpositive if all its
minors, of any order, are nonpositive. For a matrix $A$, we denote by
$A(i:j)$, the submatrix of $A$ obtained by deleting the row indexed by $i$ and
the column indexed by $j.$

######  Theorem 4.18.

Let $M\in\mathbb{R}^{(n+1)\times(n+1)}$ satisfy the conditions that it has at
least one positive diagonal entry and all of whose minors of order at least
two, are nonpositive. Then $M$ is an inverse $F_{0}$-matrix iff
$\rm{det}(M)<0$ and $\rm{det}M(i:j)=0$ whenever the indices $i,j$ satisfy
$i+j=2k,$ $i\neq j$, for any positive integer $k$.

###### Proof.

Suppose that $M$ is an inverse $F_{0}$-matrix, so that ${\rm det}M<0$. Set
$M^{-1}=(m_{ij})$. Then

$m_{ij}=(-1)^{i+j}{\rm det}M(i:j)/{\rm det}M.$

Since $m_{ij}\leq 0$ whenever $i\neq j$, and since the determinants of the
submatrices $M(i:j)$ are nonpositive for all $i,j$, we conclude that ${\rm
det}M(i:j)=0$ for any $i\neq j$ such that $i+j$ is even.

For the converse, assume that $\rm{det}(M)<0$ and $\rm{det}M(i:j)=0$ whenever
$i,j$ are such that $i\neq j$ and $i+j=2k$, for some positive integer $k$.
This means that, for $i\neq j$, if $i+j$ is even, then $m_{ij}=0$. Also, when
$i+j$ is odd, then $m_{ij}\leq 0$. Therefore, $M^{-1}$ is a $Z$-matrix. The
rest of the proof follows, by appealing to Theorem 2.6. ∎

We need an intermediate result, which in turn, uses another proposition. This
is stated next.

######  Theorem 4.19.

[12, Theorem 2.3] If $M\in\mathbb{R}^{n\times n}$ is a matrix of type $D,$ and
$a_{n}<0,$ then all minors of $M$ are nonpositive.

######  Theorem 4.20.

Let $M\in\mathbb{R}^{(n+1)\times(n+1)}$ be a matrix of type $D,$ where
$a_{n+1}>0$ and $a_{n}<0.$ Then all minors of $M$ of order at least two, are
nonpositive.

###### Proof.

The matrix $M$ is given by

$D_{n+1}=\begin{pmatrix}a_{1}&a_{1}&a_{1}&....&a_{1}&a_{1}\\\
a_{1}&a_{2}&a_{2}&....&a_{2}&a_{2}\\\ \vdots&&\vdots&...&\vdots&\vdots\\\
a_{1}&a_{2}&a_{3}&....&a_{n}&a_{n}\\\
a_{1}&a_{2}&a_{3}&....&a_{n}&a_{n+1}\end{pmatrix},$

which we rewrite using block matrices as

$M=\begin{pmatrix}D_{n}&b\\\ b^{T}&a_{n+1}\end{pmatrix},$

where $b=(a_{1},a_{2},...,a_{n})^{T}$ has all its entries negative. Since $M$
is invertible and $M/D_{n}\geq 0$, we have ${\rm det}(M)={\rm
det}(A)(a_{n+1}-b^{T}A^{-1}b)<0$. Consider $M(i:j)$ (which is a submatrix of
$D_{n+1}$). To compute the determinant of $M(i:j,)$ we consider the following
cases:

$\underline{Case\leavevmode\nobreak\ 1}:$ $i=j=n+1.$ Then $M(i:j)=D_{n},$ and
so by Theorem 4.19, ${\rm det}(M(i:j))<0.$

$\underline{Case\leavevmode\nobreak\ 2}:$ $M(i:j)$ contains both the last row
and the last column of $M$. The same argument used earlier to show that ${\rm
det}(M)<0$, applies. We get ${\rm det}M(i:j)\leq 0$.

$\underline{Case\leavevmode\nobreak\ 3}:$ $1\leq i\leq n,j=n+1.$ Here, either
$M(i:j)=D_{n}$ or $M(i:j)$ is a submatrix of $D_{n+1}$ with two identical
rows. So, either ${\rm det}M(i:j)={\rm det}(D_{n})<0$ or ${\rm det}M(i:j)=0.$

$\underline{Case\leavevmode\nobreak\ 4}:$ This is the same as the previous
case, with the roles of $i$ and $j$ interchanged. Hence, once again, either
${\rm det}M(i:j)<0$ or ${\rm det}M(i:j)=0.$

An entirely similar argument as above may be used to show that every minor of
order at least two, is nonpositive. This completes the proof. ∎

As a consequence of this and Theorem 4.18 and 4.20, we have the result that
was stated earlier.

######  Theorem 4.21.

Let $M\in\mathbb{R}^{(n+1)\times(n+1)}$ be a type $D$ matrix, with $a_{n+1}>0$
and $a_{n}<0.$ Then $M^{-1}$ is a tridiagonal $F_{0}$-matrix.

###### Proof.

First, note that the diagonal entry $a_{n+1}>0$. Also, by Theorem 4.20, we
have $\rm det(M)<0.$ Thus, the hypotheses of Theorem 4.18 are satisfied. To
conclude that $M$ is an inverse $F_{0}$-matrix, we only need to prove that
${\rm det}M(i:j)=0$ for all $i,j$ with $i\neq j$ such that $i+j=2k$, where $k$
is a positive integer. This condition, however follows from the fact that
$M^{-1}$ is a tridiagonal matrix [30, Theorem 8]. ∎

We illustrate Theorem 4.21, by an example.

######  Example 4.22.

Consider the matrix

$M=\begin{pmatrix}-3&-3&-3&-3\\\ -3&-2&-2&-2\\\ -3&-2&-1&-1\\\
-3&-2&-1&\leavevmode\nobreak\ 1\end{pmatrix}.$

Then $M$ is a type $D$ matrix, satisfying the conditions of Theorem 4.21. It
may verified that $M^{-1}$ is the tridiagonal matrix given by

$M^{-1}=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\
\frac{2}{3}&-1&\leavevmode\nobreak\ \leavevmode\nobreak\
0&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\\\ -1&\leavevmode\nobreak\
\leavevmode\nobreak\ 2&-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\\\
\leavevmode\nobreak\ \leavevmode\nobreak\ 0&-1&\leavevmode\nobreak\
\leavevmode\nobreak\ \frac{3}{2}&-\frac{1}{2}\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ 0&\leavevmode\nobreak\ \leavevmode\nobreak\
0&-\frac{1}{2}&\leavevmode\nobreak\ \leavevmode\nobreak\
\frac{1}{2}\end{pmatrix}=\begin{pmatrix}A&b\\\ c^{T}&d\end{pmatrix},$

where $A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\
\frac{2}{3}&-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\\\
-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 2&-1\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ 0&-1&\leavevmode\nobreak\ \leavevmode\nobreak\
\frac{3}{2}\end{pmatrix}$ is a $N_{0}$-matrix, $b=c=(0,0,-\frac{1}{2})^{T}\leq
0,$ and $d=\frac{1}{2}\neq c^{T}A^{-1}b.$ Hence, $M^{-1}$ is an
$F_{0}$-matrix.

##  5 Concluding remarks

We have looked at the problem of extending results known for a class of
invertible $Z$-matrices to the case, when the matrices are singular. This
study has opened up the problem of seeking generalizations involving the
bigger class of all $Z$-matrices. Let us just point to a couple of interesting
questions that were not investigated here. If
$A=\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\ 3&-2&-2\\\
-2&-1&-1\\\ -2&-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix},$
then $A$ is not an $F_{0}$-matrix (since it has a negative diagonal) and
$A^{-1}=\frac{1}{7}\begin{pmatrix}\leavevmode\nobreak\ \leavevmode\nobreak\
1&-2&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\\\ -2&\leavevmode\nobreak\
\leavevmode\nobreak\ 4&-1\\\ \leavevmode\nobreak\ \leavevmode\nobreak\
0&-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 1\end{pmatrix}.$ The comparison
matrix of $A$ is given by ${\cal M}_{A}=\begin{pmatrix}\leavevmode\nobreak\
\leavevmode\nobreak\ 3&-2&-2\\\ -2&\leavevmode\nobreak\ \leavevmode\nobreak\
1&-1\\\ -2&-1&\leavevmode\nobreak\ \leavevmode\nobreak\ 0\end{pmatrix}.$
${\cal M}_{A}$ is nonsingular and ${{\cal
M}_{A}}^{-1}=\frac{1}{15}\begin{pmatrix}\leavevmode\nobreak\
\leavevmode\nobreak\ 1&-2&-4\\\ -2&\leavevmode\nobreak\ \leavevmode\nobreak\
4&-7\\\ -4&-7&\leavevmode\nobreak\ \leavevmode\nobreak\ 1\end{pmatrix}.$ We
then have

${{\cal M}_{A}}^{-1}\leq|A^{-1}|,$

an inequality that goes in the reverse direction to that of Ostrowski (Theorem
2.1)! This motivates the question: Is this true, in general for the class of
$Z$-matrices, which have at least one negative diagonal entry? Since the
comparison matrix of an $F_{0}$-matrix is itself, the above inequality holds
trivially for $F_{0}$-matrices, since for such matrices, we have

${{\cal M}_{A}}^{-1}={A}^{-1}\leq|A^{-1}|.$

The second question is whether the result of Fan (and its converse, namely
that $A-I$ is an invertible $M$-matrix iff both $A$ and $I-A^{-1}$ are
invertible $M$-matrices), holds good for $F_{0}$-matrices? More generally, is
it true for the class of $Z$-matrices?

Acknowledgements:

This work was done when the second author was visiting the Department of
Mathematics, University of California, Santa Barbara during the Winter and
Spring of 2023. The author records his thanks to Dr. Maria Isabel for the
excellent hospitality that was extended to him. The first author acknowledges
funding received from the Prime Minister’s Research Fellowship (PMRF),
Ministry of Education, Government of India, for carrying out this work.

## References

  * [1] A. Ben-Israel and T.N.E. Greville, Generalized Inverses - Theory and Applications, CMS Books in Mathematics vol. 15. Springer-Verlag, New York, 2003.
  * [2] A. Berman and R.J. Plemmons, Nonnegative Matrices in the Mathematical Sciences, Classics in Applied Mathematics, SIAM, Philadelphia, PA, 1994.
  * [3] A. Berman and R.J. Plemmons, Monotonicity and the generalized inverse, SIAM J. Appl. Math., 22 (1972) 155-161.
  * [4] M. Chatterjee and K.C. Sivakumar, Inequalities for group invertible H-matrices, Linear Algebra Appl., 576 (2019) 158-180.
  * [5] M. Chatterjee and K.C. Sivakumar, Intervals of $H$-matrices and inverse $M$-matrices, Linear Algebra Appl., 614 (2021) 24-43.
  * [6] Y. Chen, Notes on $F_{0}$-matrices, Linear Algebra Appl., 142, (1990) 167-172.
  * [7] K. Fan, Some matrix inequalities, Abh. Math. Se. Univ. Hamburg 29, (1966) 185-196.
  * [8] M. Fiedler, T.L. Markham, and M. Neumann, Classes of products of $M$-matrices and inverse $M$-matrices, Linear Algebra Appl., 52/53 (1983) 265-287.
  * [9] M. Fiedler, T.L. Markham, A classification of matrices of class $Z$, Linear Algebra Appl., 173 (1992), 115-124.
  * [10] C.R. Johnson, Inverse $M$-matrices, Linear Algebra Appl., 47 (1982) 195-216.
  * [11] G.A. Johnson, A generalization of $N$-matrices, Linear Algebra Appl., 48, (1982) 201-217.
  * [12] G.A. Johnson, Inverse $N_{0}$-matrices, Linear Algebra Appl., 64, (1985) 215-222.
  * [13] C.R. Johnson, R.L. Smith, Inverse $M$-matrices II, Linear Algebra Appl., 435 (2011) 195-216.
  * [14] C.R. Johnson, R.L. Smith and M.J. Tsatsomeros, Matrix Positivity, Cambridge University Press, Cambridge, 2020.
  * [15] A. Kalauch, S. Lavanya and K.C. Sivakumar, Singular irreducible $M$-matrices revisited, Linear Algebra Appl., 565 (2019) 47-64.
  * [16] A. Kalauch, S. Lavanya and K.C. Sivakumar, Matrices whose group inverses are $M$-matrices, Linear Algebra Appl., 614 (2021) 44-67.
  * [17] T.L. Markham, Nonnegative matrices whose inverses are $M$-matrices, Proc. Amer. Math. Soc. 36(2) (1972) 326-330.
  * [18] D. Mishra and K.C. Sivakumar, Generalizations of matrix monotonicity and their relationships with certain subclasses of proper splittings, Linear Algebra Appl., 436 (2012) 2604-2614.
  * [19] G.S.R. Murthy, K.C. Sivakumar and P. Sushmitha, T. Parthasarathy’s contributions to complementarity problems: a survey, Ann. Oper. Res., 287 (2020) 867-894.
  * [20] R. Nabben, $Z$-matrices and inverse $Z$-matrices, Linear Algebra Appl., 256 (1997), 31-48.
  * [21] R.Nabben, On relationships between several classes of Z-matrices, M-matrices and nonnegative matrices, Linear Algebra Appl., 421 (2007), 417-439.
  * [22] R. Nabben ahd R.S. Varga, On classes of inverse $Z$-matrices, Linear Algebra Appl., 223/224 (1995), 521-552.
  * [23] A. Neumaier, Interval methods for systems of equations, Encyclopedia of Mathematics and its Applications, 37 Cambridge University Press, Cambridge, 1990.
  * [24] A. Ostrowski, Über die determinanten mit überwiegender Hauptdiagonale (German), Comment. Math. Helv., 10 (1937) 69-96.
  * [25] S. Mondal, K.C. Sivakumar and M.J. Tsatsomeros, New results on $M$-matrices, $H$-matrices and their inverse classes, Electron. J. Linear Algebra, 38 (2022) 729-744.
  * [26] K.C. Sivakumar, Proof by verification of the Greville/Udwadia/Kalaba formula for the Moore-Penrose inverse of a matrix, J. Optim. Theory Appl., 131 (2006) 307-311.
  * [27] R.L. Smith, On the spectrum of $N_{0}$-matrices, Linear Algebra Appl., 83, (1986) 129-134.
  * [28] R.L. Smith, Some notes on $Z$-matrices, Linear Algebra Appl., 106, (1988) 219-231.
  * [29] J.L. Stuart, Special families of matrices - a talk in honor of Miroslav Fiedler, Linear Algebra Appl., 439, (2013) 830-835.
  * [30] J.L. Stuart, Nested matrices and inverse M-matrices, Czechoslovak Math. J., 65 (2015) 537-544.