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\begin{document}
\title{M-Matrices}
\author{Stanislaw H. \.{Z}ak \thanks{School of Electrical and Computer Engineering, Purdue University, West Lafayette, IN 47907}}
\maketitle


In this write-up, we summarize some basic properties of the so called M-matrices that are useful in the design of neural associative memories. For more information on the subject, we refer to Berman and Plemmons~\cite{berman94}.
\begin{defn}
A real $n \times n$ matrix $\mb{\bm$A$}$ is an M-matrix if all its off-diagonal elements are non-positive, that is, $a_{ij} \le 0$, $i \ne j$, and all its principal minors are positive.
\end{defn}
Note that if $\mb{\bm$A$}$ is an M-matrix, then its diagonal elements are positive because its principal minors are positive.

In establishing the needed properties of the M-matrices, we use the following technical lemma due to Fiedler and Pt\'{a}k~\cite{fiedler62}.
\begin{lemma}
\label{lemma:lu}
The following properties of an arbitrary matrix $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ are equivalent:
\begin{itemize}
\item[(i)] The leading principal minors of $\mb{\bm$A$}$ are positive,
\item[(ii)] There exists a lower triangular matrix $\mb{\bm$L$} \in \mb{\Bb R}^{n \times n}$ and an upper triangular matrix $\mb{\bm$U$} \in \mb{\Bb R}^{n \times n}$ both with positive diagonal elements such that $\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$}$
\end{itemize}
\end{lemma}
{\bf Proof}\quad We first prove the implication (i)$ \Rightarrow $(ii). The proof is by induction. Obviously the implication holds for $n=1$. Suppose the statement holds for all real matrices of order $n-1$. Represent $\b A$ as:
\[
\mb{\bm$A$} = \left[\begin{array}{cc}
\mb{\bm$A$}_{n-1} & \mb{\bm$a$}\\
\b b^T & a_{nn}
\end{array} \right],
\]
where $\mb{\bm$A$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$, $\mb{\bm$a$} \in \mb{\Bb R}^{(n-1) \times 1}$, $\b b^T \in \mb{\Bb R}^{1 \times (n-1)}$, and $a_{nn} \in \mb{\Bb R}$. From the inductive hypothesis, the matrix $\mb{\bm$A$}_{n-1}$ has positive leading principal minors, $\mb{\bm$A$}_{n-1}=\mb{\bm$L$}_{n-1}\mb{\bm$U$}_{n-1}$, where $\mb{\bm$L$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$ is a lower triangular matrix with positive diagonal elements, and $\mb{\bm$U$}_{n-1} \in \mb{\Bb R}^{(n-1) \times (n-1)}$ is an upper triangular matrix with positive diagonal elements. Observe that
\[
\left[\begin{array}{cc}
\mb{\bm$I$}_{n-1} & \mb{\bm$0$}\\
-\b b^T \mb{\bm$A$}_{n-1}^{-1} & 1
\end{array} \right] \mb{\bm$A$} = \left[\begin{array}{cc}
\mb{\bm$A$}_{n-1} & \mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right],
\]
where it follows from the assumptions that $\det \mb{\bm$A$}_{n-1}>0$, and $\det \mb{\bm$A$}>0$. Hence,
\[
0< \det \mb{\bm$A$}=\left( a_{nn} - \b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$} \right) \det \mb{\bm$A$}_{n-1}.
\]
Thus, we have
\[
a_{nn} - \b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}>0.
\]
Let
\[
\mb{\bm$L$}=\left[\begin{array}{cc}
\mb{\bm$L$}_{n-1} & \mb{\bm$0$}\\
\b b^T \mb{\bm$U$}_{n-1}^{-1} & 1
\end{array} \right],
\]
and
\[
\mb{\bm$U$}=\left[\begin{array}{cc}
\mb{\bm$U$}_{n-1} & \mb{\bm$L$}_{n-1}^{-1}\mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \b b^T \mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right].
\]
One can check that
\[
\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$},
\]
and that the matrices $\mb{\bm$L$}$ and $\mb{\bm$U$}$ satisfy~(ii).

We now prove the implication (i)$ \Leftarrow $(ii). From the generalized Cauchy's theorem it follows that the leading principal minors of $\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$}$ are just the products of the corresponding leading principal minors of $\mb{\bm$L$}$ and $\mb{\bm$U$}$. This is because the matrix $\mb{\bm$L$}$ is lower triangular while the matrix $\mb{\bm$U$}$ is upper triangular. Because the diagonal elements of $\b L$ and $\b U$ are positive, the principal minors of $\b A$ must be positive. This completes our proof of the lemma.\\
\hspace*{5.5in} $\Box$

The following theorem, also due to Fiedler and Pt\'{a}k~\cite{fiedler62}, establishes some properties of the M-matrices in terms of their LU-decomposition.
\begin{thm}
\label{thm:m_matlu}
Let $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ be an M-matrix. Then, there is a lower triangular matrix $\mb{\bm$L$} \in \mb{\Bb R}^{n \times n}$ and an upper triangular matrix $\mb{\bm$U$} \in \mb{\Bb R}^{n \times n}$ both with positive diagonal elements. Their off-diagonal elements are non-positive. Furthermore, both $\mb{\bm$L$}^{-1}$ and $\mb{\bm$U$}^{-1}$ are non-negative which implies that $\mb{\bm$A$}^{-1}=\mb{\bm$U$}^{-1}\mb{\bm$L$}^{-1}$ is non-negative.
\end{thm}
{\bf Proof}\quad We prove this theorem by induction. The theorem holds for $n=1$. Suppose the theorem holds for matrices of order $n-1$. Consider the factorization from the proof of Lemma~\ref{lemma:lu},
\[
\mb{\bm$A$}=\mb{\bm$L$}\mb{\bm$U$} = \left[\begin{array}{cc}
\mb{\bm$L$}_{n-1} & \mb{\bm$0$}\\
\b b^T\mb{\bm$U$}_{n-1}^{-1} & 1
\end{array} \right]\left[\begin{array}{cc}
\mb{\bm$U$}_{n-1} & \mb{\bm$L$}_{n-1}^{-1}\mb{\bm$a$}\\
\mb{\bm$0$} & a_{nn} - \b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}
\end{array}\right],
\]
where $\mb{\bm$L$}_{n-1}$, $\mb{\bm$U$}_{n-1}$, $\mb{\bm$L$}_{n-1}^{-1}$, $\mb{\bm$U$}_{n-1}^{-1}$ and $\mb{\bm$A$}_{n-1}^{-1}=\mb{\bm$U$}_{n-1}^{-1}\mb{\bm$L$}_{n-1}^{-1}$ satisfy the inductive hypotheses. We see that $l_{ij} \le 0$ for $i>j$ and $u_{kl} \le 0$ for $k<l$ which means that the off-diagonal elements of $\mb{\bm$L$}$ and $\mb{\bm$U$}$ are non-positive. Their diagonal elements are positive. The inverse of $\mb{\bm$L$}$ is
\[
\mb{\bm$L$}^{-1}=\left[\begin{array}{cc}
\mb{\bm$L$}_{n-1}^{-1} & \mb{\bm$0$}\\
-\b b^T\mb{\bm$L$}_{n-1}^{-1} & 1
\end{array}\right].
\]
From the inductive assumption $\mb{\bm$L$}_{n-1}^{-1}$ and $\mb{\bm$A$}_{n-1}^{-1}$ are non-negative and hence $\mb{\bm$L$}_{n-1}^{-1}$ is non-negative. The inverse of $\mb{\bm$U$}$ is
\[
\mb{\bm$U$}^{-1} = \left[\begin{array}{cc}
\mb{\bm$U$}_{n-1}^{-1} & -\mb{\bm$A$}_{n-1}^{-1} \frac{\mb{\bm$a$}}{ a_{nn} - \b b^T \mb{\bm$A$}_{n-1}^{-1} \mb{\bm$a$}}\\
\mb{\bm$0$} & \left( a_{nn}-\b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$} \right) ^{-1}
\end{array} \right].
\]
From the inductive assumptions and the fact that $a_{nn}-\b b^T\mb{\bm$A$}_{n-1}^{-1}\mb{\bm$a$}>0$, the matrix $\mb{\bm$U$}^{-1}$ is non-negative, and the proof is complete.\\
\hspace*{5.5in} $\Box$

We now consider a more general class of matrices called Z-matrices.
\begin{defn}
A real $n \times n$ matrix whose off-diagonal elements are non-positive is called a Z-matrix.
\end{defn}
Note that M-matrices are Z-matrices.

In the following, we use the notation $\mb{\bm$B$}>0$, to mean that all the elements of $\mb{\bm$B$}$ are positive, where $\mb{\bm$B$}$ can be a matrix or a vector. Respectively, $\mb{\bm$B$} \ge 0$ means that all the elements of $\mb{\bm$B$}$ are non-negative. We are now ready to state and prove the following theorem concerning properties of Z-matrices. The theorem can be found in Fiedler and Pt\'{a}k~\cite{fiedler62}.
\begin{thm}
\label{thm:z_mat}
Let $\mb{\bm$A$} \in \mb{\Bb R}^{n \times n}$ be a Z-matrix. Then, the following conditions are equivalent:
\begin{itemize}
\item[(i)] There exists a vector $\mb{\bm$x$} \ge 0$ such that $\mb{\bm$Ax$}>0$;
\item[(ii)] There exists a vector $\mb{\bm$y$}>0$ such that $\mb{\bm$Ay$}>0$;
\item[(iii)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal elements such that $\b A\b D\b e>0$, where $\b e=\left[\begin{array}{ccc}
1 & \cdots & 1
\end{array}\right]^T \in \mb{\Bb R}^n$;
\item[(iv)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal elements such that the matrix $\b W=\b A\b D$ has a dominant positive main diagonal, that is, for $i=1,2,\ldots,n$, we have $w_{ii}>\sum_{{j=1}\atop{j \ne i}}^n w_{ij}$.
\end{itemize}
\end{thm}
{\bf Proof}\quad We prove this theorem using the following scheme:
\[
\mb{(i)} \Rightarrow \mb{(ii)} \Rightarrow \mb{(iii)} \Rightarrow \mb{(iv)} \Rightarrow \mb{(i)}.
\]
We begin with the implication (i)$ \Rightarrow $(ii). Thus suppose that there exists a vector $\mb{\bm$x$} \ge 0$ such that $\mb{\bm$Ax$} > 0$. Let $\mb{\bm$e$}=\left[\begin{array}{ccc}
1 & \cdots & 1
\end{array}\right]^T \in \mb{\Bb R}^n$, $\epsilon > 0$, and $\mb{\bm$y$} = \mb{\bm$x$} + \epsilon \mb{\bm$e$} > 0$. Then,
\[
\mb{\bm$Ay$} = \mb{\bm$A(x$} + \epsilon \mb{\bm$e$}) = \mb{\bm$Ax$} + \epsilon \mb{\bm$Ae$} > 0
\]
for a sufficiently small $\epsilon > 0$.

We next prove that (ii)$ \Rightarrow $(iii). For this let $\mb{\bm$y$}=\left[\begin{array}{ccc}
y_1 & \cdots & y_n
\end{array}\right]^T > 0$ be such that $\mb{\bm$Ay$} > 0$. Let
\[
\b D = \left[\begin{array}{ccccc}
y_1 & 0 & 0 & \cdots & 0\\
0 & y_2 & 0 & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
0 & 0 & 0 & \cdots & y_n
\end{array} \right].
\]
It then follows that $\b A\b D\b e = \b A\b y > 0$.

To prove (iii)$ \Rightarrow $(iv), let $\mb{\bm$D$} > 0$ be a diagonal matrix with positive diagonal elements such that $\mb{\bm$ADe$} > 0$. We denote
\[
\b A\b D=\b W = \left[\begin{array}{cccc}
w_{11} & -w_{12} & \cdots & -w_{1n}\\
-w_{21} & w_{22} & \cdots & -w_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-w_{n1} & -w_{n2} & \cdots & w_{nn}
\end{array} \right],
\]
where $w_{ij} \ge 0$ for $i \ne j$. Note that $\mb{\bm$ADe$}=\mb{\bm$We$}>0$ which is equivalent to
\[
w_{ii} > \sum_{{j=1}\atop{j \ne i}}^{n} w_{ij}, \qquad i=1,2,\ldots ,n.
\]
Since $w_{ij} \ge 0$, we have
\[
w_{ii} > \sum_{{j=1}\atop{j \ne i}}^{n} |w_{ij}|.
\]
The above means that $\mb{\bm$W$}$ is a row diagonal dominant matrix with positive main diagonal. Note $a_{ii} > 0$ since $0 < w_{ii} = a_{ii} d_i$ and $d_i > 0$.

To complete the proof of the theorem, we show that (iv)$ \Rightarrow $(i). For this let $\mb{\bm$D$}$ be a diagonal matrix with positive diagonal elements such that $\mb{\bm$W$} = \b A\b D$ has a positive dominant main diagonal. Then,
\[
0 < \b W\b e = \b A\b D\b e = \b A\b x,
\]
where
\[
\mb{\bm$x$}=\left[\begin{array}{ccc}
d_1 & \cdots & 0\\
\vdots & \ddots & \vdots\\
0 & \cdots & d_n
\end{array} \right] \left[\begin{array}{c}
1\\
\vdots\\
1
\end{array} \right]=\left[\begin{array}{c}
d_1\\
\vdots\\
d_n
\end{array}\right] > 0.
\]
Hence $(i)$ is satisfied.\\
\hspace*{5.5in} $\Box$

To proceed further, we need the following two technical results.
\begin{lemma}
\label{lemma:hadamard}
An $n \times n$ matrix $\mb{\bm$A$}$ with complex coefficients such that
\[
|a_{ii}| > \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}|\quad\mb{for}\quad i=1,2,\ldots,n,
\]
is nonsingular, that is, $\det \mb{\bm$A$} \ne 0$.
\end{lemma}
{\bf Proof}\quad We prove the above lemma by contraposition. Suppose that $\det \mb{\bm$A$} = 0$. Then, there exists a nontrivial solution $\mb{\bm$x$} \ne \mb{\bm$0$}$ to $\mb{\bm$Ax=0$}$, that is, for $k=1,2,\ldots,n$,
\[
\sum_{j=1}^{n} a_{kj} x_j = 0.
\]
Let
\[
x_i = \max_{1 \le k \le n} |x_k|.
\]
Rearranging $\sum_{j=1}^{n} a_{ij} x_j = 0$ yields
\[
-a_{ii}x_i=a_{i1}x_1 + \cdots + a_{i,i-1}x_{i-1} + a_{i,i+1}x_{i+1} + \cdots + a_{in}x_n.
\]
Taking the absolute values of both sides of the above equation gives
\[
|a_{ii}||x_i|=|a_{i1}x_1 + \cdots + a_{i,i-1}x_{i-1} + a_{i,i+1}x_{i+1} + \cdots + a_{in}x_n| \le \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}||x_j|.
\]
Thus,
\[
|a_{ii}| \le \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}| \frac{|x_j|}{|x_i|} \le  \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}|
\]
which contradicts the assumption that $|a_{ii}| > \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}|$.\\
\hspace*{5.5in} $\Box$

The above lemma says that if a square matrix is row diagonal dominant then it is nonsingular. Following the arguments of the proof of the lemma, we can show that if a square, complex, matrix is column diagonal dominant then it is also nonsingular. We state this result in the following corollary.
\begin{corollary}
A complex $n \times n$ matrix $\mb{\bm$A$}$ such that 
for $j=1,2,\ldots,n$,
\[
|a_{jj}| > \sum_{{k=1}\atop{k \ne j}}^{n} |a_{kj}|
\]
is nonsingular. 
\end{corollary}

We now use Lemma~\ref{lemma:hadamard} to prove the following theorem due to Gershgorin.
\begin{thm}
\label{thm:gershgorin}
Given a complex $n \times n$ matrix $\b A$. Let $D_i$, $i=1,2,\ldots,n$, denote the $i$-th closed disk in the complex plane centered at $a_{ii}$ whose radius is $r_i=\sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}|$. Then the eigenvalues of $\mb{\bm$A$}$ are located in the union $\bigcup_{i=1}^{n}D_i$. 
\end{thm}
{\bf Proof}\quad The eigenvalues of $\mb{\bm$A$}$ are the roots of the characteristic equation
\[
\det [\lambda \mb{\bm$I$} - \mb{\bm$A$}] = \det \left[\begin{array}{cccc}
\lambda - a_{11} & -a_{12} & \cdots & -a_{1n}\\
-a_{21} & \lambda - a_{22} & \cdots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots \\
-a_{n1} & -a_{n2} & \cdots & \lambda -a_{nn}
\end{array} \right]=0.
\]
By Lemma~\ref{lemma:hadamard}, the characteristic polynomial is nonzero if for all $i=1,2,\ldots,n$,
\[
|\lambda -a_{ii}| > \sum_{{j=1}\atop{j \ne i}}^{n} |a_{ij}|,
\]
that is, for all $i=1,2,\ldots,n$
\[
|\lambda -a_{ii}| > r_i.
\]
Thus the eigenvalues of $\mb{\bm$A$}$ must be in the set $\bigcup_{i=1}^{n} D_i$.\\
\hspace*{5.5in} $\Box$

In the following theorem, due to Fiedler and Pt\'{a}k~\cite{fiedler62}, we collect a number of equivalent conditions characterizing the M-matrices.
\begin{thm}
\label{thm:m_matsum}
Let $\mb{\bm$A$}$ be a Z-matrix. Then, the following conditions are: equivalent
\begin{itemize}
\item[(i)] There exists a vector $\mb{\bm$x$} \ge 0$ such that $\b A\b x > 0$;
\item[(ii)] There exists a diagonal matrix $\mb{\bm$D$}$ with positive diagonal such that the matrix $\b W=\b A\b D$ has a positive main diagonal and is row diagonal dominant;
\item[(iii)] Eigenvalues of $\mb{\bm$W$}$ have positive real parts and principal minors of $\mb{\bm$W$}$ are positive;
\item[(iv)] All principal minors of $\mb{\bm$A$}$ are positive, that is, $\mb{\bm$A$}$ is an M-matrix;
\item[(v)] All leading principal minors of $\mb{\bm$A$}$ are positive;
\item[(vi)] There exists a lower triangular matrix $\mb{\bm$L$}$ and an upper triangular matrix $\mb{\bm$U$}$ both with positive diagonals such that $\mb{\bm$A=LU$}$. Furthermore, $l_{ij} \le 0$ for $i>j$, and $u_{kl} \le 0$ for $k<l$;
\item[(vii)] $\mb{\bm$A$}^{-1}$ exists and $\mb{\bm$A$}^{-1} \ge 0$.
\end{itemize}
\end{thm}
{\bf Proof}\quad The implication (i)$ \Rightarrow $(ii) follows from Theorem~\ref{thm:z_mat}. To prove (ii)$ \Rightarrow $(iii), we represent the matrix $\b W=\b A\b D$ as follows
\[
\b W=\b A\b D=\left[\begin{array}{cccc}
w_{11} & -w_{12} & \cdots & -w_{1n}\\
-w_{21} & w_{22} & \cdots & -w_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-w_{n1} & -w_{n2} & \cdots & w_{nn}
\end{array} \right],
\]
where $w_{ij} \ge 0$ for $i \ne j$, and $w_{ii} > 0$, as follows from Theorem~\ref{thm:z_mat}. Furthermore,
\[
w_{ii} > \sum_{{j=1}\atop{j \ne i}}^{n}w_{ij} \quad \mbox{for} \quad i=1,2,\ldots,n.
\]
The above implies that each principal minor matrix of $\mb{\bm$W$}$ is also row diagonal dominant. By the Theorem of Gershgorin, the eigenvalues of $\mb{\bm$W$}$ as well as its principal minor matrices are located on the disks with centers at $w_{ii}$ and radii $r_i < w_{ii}$. Hence they must have real positive parts.

We shall now show that (iii)$ \Rightarrow $(iv) holds. First note that since the determinant of a square matrix is the product of its eigenvalues and the product of complex eigenvalues of a real matrix is positive, it follows that all principal minors of the matrix $\mb{\bm$W$}$ are positive. Next, the matrix $\mb{\bm$W$}$ has the form
\[
\b W=\b A\b D=\left[\begin{array}{cccc}
a_{11} & -a_{12} & \cdots & -a_{1n}\\
-a_{21} & a_{22} & \cdots & -a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
-a_{n1} & -a_{n2} & \cdots & a_{nn}
\end{array} \right]
\left[\begin{array}{cccc}
d_{11} & \cdots & {} & 0\\
{} & d_{22} & {} & {}\\
\vdots & {} & \ddots & \vdots\\
0 & \cdots & {} & d_{nn}
\end{array} \right],
\]
where $a_{ij} \ge 0$ for $i \ne j$, $a_{ii} > 0$, and $d_{ii} > 0$. Thus if the principal minors of $\mb{\bm$W$}$ are positive, the principal minors of $\mb{\bm$A$}$ must be positive, and vice versa. This is because
\[
\b W\left(\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right)=
\b A\left(\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right)
\left[\begin{array}{cccc}
d_{i_1 i_1} & \ldots & {} & 0\\
{} & d_{i_2 i_2} & {}& {}\\
\vdots & {} & \ddots & \vdots\\
0 & \ldots & {} & d_{i_k i_k}
\end{array}\right],
\]
where
\[
\b W\left(\begin{array}{cccc}
i_1 & i_2 & \ldots & i_k\\
i_1 & i_2 & \ldots & i_k
\end{array} \right)
\]
is a matrix formed using $i_1,i_2,\ldots,i_k$ columns and $i_1,i_2,\ldots,i_k$ rows of the matrix $\mb{\bm$W$}$.

The implication (iv)$ \Rightarrow $(v) follows from the observation that the leading principal minors are also principal minors.

The implications (v)$ \Rightarrow \mb{(vi)} \Rightarrow $(vii) follow from Theorem~\ref{thm:m_matlu}.

Finally, to complete the proof, we show that (vii)$ \Rightarrow $(i). For this note that if $\mb{\bm$A$}^{-1}$ exists and $\mb{\bm$A$}^{-1} \ge \mb{\bm$0$}$, then
\[
\mb{\bm$x=A$}^{-1}\mb{\bm$e$} \ge \mb{\bm$0$},
\]
where $\mb{\bm$e$}=\left[\begin{array}{cccc}
1 & 1 & \cdots & 1
\end{array}\right]^T \in \mb{\Bb R}^n$. Hence
\[
\b A\b x=\b e > \mb{\bm$0$},
\]
which implies that (i) is satisfied.\\
\hspace*{5.5in} $\Box$

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