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\begin{document}

\title{Inverse eigenvalue problem of bisymmetric nonnegative matrices}

\author[Nazari, Aslami, Nezami]{A.M. Nazari$^1$, P. Aslami, A. Nezami}
\address{$^{1}$ Department of Mathematics, Arak University, P. O. Box 38156-8-8349, Arak, Iran..}

\email{a-nazari@araku.ac.ir}

 \subjclass[2010]{47A55; 39B52; 34K20; 39B82.}

 \keywords{ Bisymmetric matrices; Perron eigenvalue; Spectrum of
matrix }
 \begin{abstract}
This paper considers an inverse eigenvalue problem for bisymmetric nonnegative matrices. We first discuss the specified structure of the bisymmetric matrices. Then for a given set of real numbers of order maximum five with special conditions, we construct a nonnegative bisymmetric matrix such that the given set is its spectrum. Finally, we solve the problem for arbitrary order $n$ in the special case of the spectrum. 
\end{abstract}

\maketitle

\section{Introduction}
Bisymmetric matrices have been widely discussed since 1939, and
are very useful
 in communication theory, engineering and statistics [1].
In fact, symmetric Toeplitz matrices and
 persymmetric Hankel matrices are
  two useful examples of bisymmetric matrices.
The  bisymmetric nonnegative  inverse eigenvalue problem is the
problem of finding necessary and sufficient conditions for a list
of $n$ real numbers to be the spectrum of an $n\times n$
 bisymmetric nonnegative  matrix. If there exists an
 $n\times n$ bisymmetric nonnegative  matrix $A$ with
  spectrum  $\sigma$, we say that σ is realizable
  and that $A$ realizes $\sigma$. 


The nonnegative inverse eigenvalue problem
  is very difficult
 and it is solved only for $n=3$  by Loewy and London
  and for matrices with trace $0$ of order $n=4$
  by Reams  and for $n=5$ in some special cases
  by Nazari and Sherafat in 2012 \cite{1M}. Recently Nazari et.al solve symmetric nonnegative inverse eigenvalue problem (SNIEP) with one positive eigenvalue and nonnegative summation \cite{Mashayekhi}. 

Through this paper the following notation is used. The spectral
radius of nonnegative matrix  $A$ denoted by  $\rho(A)$. For nonnegative matrices the largest eigenvalue is called Perron eigenvalue and denoted by $\lambda_1$ and we have  $\lambda_1=\rho(A)$, so   there is
a right and a
left eigenvector associated with the Perron eigenvalue
 with nonnegative entries.

Some necessary conditions  on the list of real number $\sigma=\{\lambda_1,\lambda_2, \ldots,\lambda_n\}$
to be the spectrum of a  nonnegative matrix are listed
below.\\
 (1) The Perron eigenvalue $\max \{|\lambda_i |; \lambda_i \in \sigma \}$
 belongs to $\sigma$ (PerronÂ–-Frobenius theorem). \\
(2) $s_k=\sum_{i=1}^n \lambda_i^k \ge 0.$\\
(3)$s_k^m \le n^{m-1}s_{km}$ for $k,m = 1, 2, \ldots$ (JLL
inequality)\cite{4M, 8}.

 This paper
is organized as follows. First, we discuss the specified
properties and structure of bisymmetric matrices 
 in section 2, and in the next section  find a solution for BSNIEP for order 2,3,5 and finally, we solve BSNIEP for a special  given spectrum of arbitrary order $n$.

\section{THE PROPERTIES OF BISYMMETRIC MATRICES }
A matrix for which the values on each line parallel to the main diagonal are constant, is called a Toeplitz matrix and   Hankel matrix, is a square matrix in which each ascending skew-diagonal from left to right is constant.

Let $A = (a_{ij})$ be an $n \times n$ matrix. $A$ is  persymmetric if for all $i,j$ we have 
$$
a_{ij}=A_{n-j+1,n-i+1}.
$$
This can be equivalently expressed as $AJ_n = J_nA^T$ where $J_n$ is the exchange matrix, i.e.$J_n=(e_{n},e_{n-1},...,e_{1})$ and we 
denote by $e_{i}$ the $i$th, $(i=1,...,n)$ column of  identity matrix $I_{n}$.  Also it is clear that
$$
J_{n}=J_{n}^T, \qquad J_{n}J_{n}^T=I_{n}.
$$
 If a symmetric matrix is rotated by $90$ degrees, it becomes a persymmetric matrix. Symmetric persymmetric matrices are sometimes called bisymmetric matrices \cite{Gloub}. 
\begin{definition}
{\rm A real $n \times n$ matrix $A=(a_{i,j} )$ is called a
bisymmetric matrix if its elements satisfy the properties
$$
a_{i,j} =a_{j,i}, \qquad a_{i,j} =a_{n-j+1,n-i+1}.
 $$

The set of all $ n \times n$  bisymmetric real matrices is denoted by
$BSR^{n\times n} $.}
\end{definition}
Clearly, a bisymmetric matrix is a square matrix that
 is symmetric about both of its main diagonals.
 
 If $A$ is a real bisymmetric matrix with distinct eigenvalues, then the matrices that commute with $A$ must be bisymmetric \cite{Yasuda}.
The inverse of bisymmetric matrices can be represented by recurrence formulas \cite{Wang}.
\begin{lemma} A matrix $A \in BSR^{n \times n}$  if and only if $A^T=A $
and $S_{n}AS_{n}=A$.
\end{lemma}
Noting that $BSR^{n\times n}\subset SR^{n\times n}$, all
eigenvalues of a bisymmetric matrix are real numbers, where $SR^{n\times n}$ denote the set of symmetric matrices of dimnesion  $n$.
\begin{definition}{\rm
 Given $A\in R^{n\times n}$, if $n-k$  is even, then the $k$-square
  central principal submatrix of $A$, denoted as  $A_{c}\left( k\right) $,
 is a $k$-square submatrix obtained by deleting the  first and last
   $\frac{n-k}{2}$  rows and columns of $A$,  that is
$$
A_{c}\left( k\right)=\left( 0\,I_{k}\, 0
\right)A{\left(0\,I_{k}\,0 \right)}^T, \qquad   0\in R^{(k)\times
(\frac{n-k}{2})}
$$
} central principal submatrices preserves the
bisymmetric structure of the given matrix.
\end{definition}
The product of two bisymmetric matrices is a centrosymmetric matrix.

\section{CONSTRUCTION}
\subsection{CASE n=2}
\begin{theorem}
 Let $\sigma=\left\lbrace \lambda_{1},\lambda_{2}\right\rbrace $ be
 a set of two real numbers such that $\lambda_{1}\geq|\lambda_{2}|$. Then $\sigma$  is the set of eigenvalues of a bisymmetric
nonnegative matrix.
\end{theorem}

\begin{proof}
we consider desired matrix is $A=\left[
\begin{matrix}a&b\\b&a \end{matrix}\right] $ by get roots of
characteristic polynomial of $A$, we have

$${\rm det}( A-\lambda I)=0 \Longrightarrow
\lbrace\lambda_{1}=a-b,\,\,\lambda_{2}=a+b\rbrace,$$
so $a=\frac
{\lambda_{1}+\lambda_{2}}{2}$, $b=\frac{\lambda_{1}-\lambda_{2}}{2}$
then the matrix
$$
A=\left[ \begin{matrix}\frac
{\lambda_{1}+\lambda_{2}}{2}&\frac{\lambda_{1}-
\lambda_{2}}{2}\\\frac{\lambda_{1}-\lambda_{2}}{2}&\frac
{\lambda_{1}+\lambda_{2}}{2}\end{matrix}\right],
$$
solves the problem.
\end{proof}
\subsection{CASE n=3}
\begin{theorem}\label{t3}
 Let $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3}\right\rbrace $ 
 be a set of  real numbers such that
\begin{enumerate}
\item
 $\lambda_{1}+\lambda_{2}+\lambda_{3}\geq 0,$
 
 \item
$\label{2} \lambda_{1}\in R, \,\,
\lambda_{1}\geq|\lambda_{i}|;i=2,3.,$
\end{enumerate}

Then there exist a bisymmetric nonnegative matrix that realize
$\sigma$
\end{theorem}
\begin{proof}
If $\lambda_1$ is Perron eigenvalues of real set  $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3}\right\rbrace $ with nonnegative $\lambda_2$ and this set  is the spectrum of $3 \times 3 $ nonnegative bisymmetric( also  $3\times 3 $ centrosymmetric, since all $3\times 3$ bisymmetric matrix  is centrosymmetric matrix)  matrix then the following  trivial solution solve the problem:
 $$
 \left[\begin{array}{ccc}
 \frac{\lambda_1+\lambda_3}{2} & 0 & \frac{\lambda_1-\lambda_3}{2}\\
 0 & \lambda_2 & 0\\
 \frac{\lambda_1-\lambda_3}{2} & 0 & \frac{\lambda_1+\lambda_3}{2}
 \end{array}
 \right],
 $$
 otherwise   in $\sigma$ we have $\lambda_1>0 \geq \lambda_3 \geq \lambda_2$ then if assume that the nonnegative matrix solution has the following form:
 \[
C= \left[ \begin {array}{ccc} a&b&c\\\noalign{\medskip}b&a&b
\\\noalign{\medskip}c&b&a\end {array} \right],
 \]
 then its charactristic polynomial is:
 \begin{eqnarray}\label{31}
 \begin{array}{l}
 P(\lambda)={\lambda}^{3}-3\,{\lambda}^{2}a- \left( {c}^{2}+2\,{b}^{2}-3\,{a}^{2}
 \right) \lambda+ \\a{c}^{2}-2\,c{b}^{2}+2\,a{b}^{2}-{a}^{3}=
 (\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3).
 \end{array}
 \end{eqnarray}
 If in \eqref{31} we find $a,b,$ and $c$ based on $\lambda_1 , \lambda_2$ and $\lambda_3$ then we have 
 $$
 \begin{array}{l}
 a =\frac{\lambda_1+\lambda_2+\lambda_3}{3},\\
 c=-2/3\,\lambda_{{3}}+1/3\,\lambda_{{2}}+1/3\,\lambda_{{1}},\\
 b=1/4\,\sqrt {2\,{\lambda_{{1}}}^{2}-4\,\lambda_{{2}}\lambda_{{1}}+2\,{
\lambda_{{2}}}^{2}-2\,{c}^{2}},
\end{array}
 $$
 so the $3\times 3$ nonnegative bisymmetric matrix is:
%\begin{small}
 \[
 \left[ \begin {array}{ccc} \frac{\lambda_1+\lambda_2+\lambda_3}{3}&b&\frac{\lambda_1-2\lambda_2+\lambda_3}{3}\\\noalign{\medskip}b&\frac{\lambda_1+\lambda_2+\lambda_3}{3}&b\\\noalign{\medskip}\frac{\lambda_1-2\lambda_2+\lambda_3}{3}&b&\frac{\lambda_1+\lambda_2+\lambda_3}{3}\end {array} \right],
 \] 
%\end{small} 
 and it is easy to see that this matrix is nonnegative bisymmetric and has spectrum $\sigma$. 
 \end{proof}
   Suppose  $\sigma \in \mathbb{Q}$, the following Theorem shows the conditions under which we can have a $3 \times 3$ bisymmetric nonnegative matrix from  rational numbers.
\begin{theorem}\label{t32}
 Let $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3}\right\rbrace $ 
 be a set of  real numbers such that
\begin{enumerate}
\item
 $\lambda_{1}+\lambda_{2}+\lambda_{3}\geq 0,$
 
 \item
$\label{2} \lambda_{1}\in R, \,\,
\lambda_{1}\geq|\lambda_{i}|;i=2,3.,$

\item \label{t4}
$\lambda_1+2\lambda_3 \geq 0,$

\item\label{t5}
$\lambda_1+\lambda_3 \geq \pm 3\lambda_2.$
\end{enumerate}

Then there exist a bisymmetric nonnegative matrix that realize
$\sigma$
\end{theorem}
\begin{proof}
We consider desired matrix as
$$A=\left[
\begin{matrix}a&b&c\\b&a+c-b&b\\c&b&a\end{matrix}\right],
$$
 and  by finding the roots of charactristic polynomial of matrix $A$ we have:
$$
{\rm det}( A-\lambda I)=0 \Longrightarrow
\lbrace\lambda_{1}=a+b+c,\lambda_{2}=a-c,\lambda_{3}=a-2b+c\rbrace
$$
so $a=\frac{2\lambda_{1}+3\lambda_{2}+\lambda_{3}}{6}$,$b=\frac{\lambda_{1}-\lambda_{3}}{3}$ , $c=\frac{2\lambda_{1}-3\lambda_{2}+\lambda_{3}}{6}$ ,$ a+c-b=\frac{2\lambda_{1}+4\lambda_{3}}{6}$  then the matrix\\
$$
A=\left[ \begin{matrix}\frac{2\lambda_{1}+3\lambda_{2}+
\lambda_{3}}{6}&\frac{\lambda_{1}-\lambda_{3}}{3}&\frac{2\lambda_{1}-3\lambda_{2
}+\lambda_{3}}{6}\\
\frac{\lambda_{1}-\lambda_{3}}{3}&\frac{2\lambda_{1}+
4\lambda_{3}}{6}&\frac{\lambda_{1}-\lambda_{3}}{3}\\
\frac{2\lambda_{1}-3\lambda_{2}+
\lambda_{3}}{6}&\frac{\lambda_{1}-\lambda_{3}}{3}&\frac{2\lambda_{1}+
3\lambda_{2}+\lambda_{3}}{6}\end{matrix}\right],
$$
is nonnegative bisymmetric matrix and solves the problem.
\end{proof}
\begin{example}
For $\sigma=\{5,-1,-3\}$ find a bisymmetric nonnegative matrix that realizes spectrum $\sigma$.

By Theorem \eqref{t3} the following bisymmetric nonnegative matrix is solution:
\[
\left[ \begin {array}{ccc} 1/3&2/3\,\sqrt {14}&4/3
\\\noalign{\medskip}2/3\,\sqrt {14}&1/3&2/3\,\sqrt {14}
\\\noalign{\medskip}4/3&2/3\,\sqrt {14}&1/3\end {array} \right].
\]
and we see that for this spectrum we cannot find a nonnegative bisymmetric matrix according to Theorem \eqref{t32} because the \eqref{t4} or \eqref{t5} condition of the Theorem \eqref{t32} will not always exist. But if we increase the Perron eigenvalue's to 6, the matrix of the following nonnegative bisymmetric matrix of the set of rational numbers will be the answer:
\[
\left[ \begin {array}{ccc} 1&3&2\\\noalign{\medskip}3&0&3
\\\noalign{\medskip}2&3&1\end {array} \right].
\]
\end{example}
\subsection{CASE n=4}
If $\lambda_1$ is Perron eigenvalues of  $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3}, \lambda_4\right\rbrace $ with nonnegative $\lambda_2$ and  $\lambda_2 \geq \lambda_4$ is the spectrum of $4 \times 4 $ nonnegative bisymmetric matrix then the following  trivial solution solve the problem:
$$
 \left[\begin{array}{cccc}
 \frac{\lambda_1+\lambda_3}{2} &0 & 0  & \frac{\lambda_1-\lambda_3}{2}\\
0  &  \frac{\lambda_2+\lambda_4}{2} & \frac{\lambda_2-\lambda_4}{2} & 0 \\
0  &  \frac{\lambda_2-\lambda_4}{2} & \frac{\lambda_2+\lambda_4}{2} & 0 \\
 \frac{\lambda_1-\lambda_3}{2} & 0 & 0& \frac{\lambda_1+\lambda_3}{2}
 \end{array}
 \right],
$$
otherwise we study some special cases in the following Theorem: 
\begin{theorem}
 Let $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}\right\rbrace $  be a set of  real numbers with following condition 
 \begin{enumerate}
 \item
$
\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}\geq 0,
$
\item
$ \lambda_{1}\in R, \,
\lambda_{1}\geq|\lambda_{i}|;i=2,3,4,  \lambda_{3}=\lambda_{4},
$
\item
$\lambda_1+\lambda_2 \geq \pm 2\lambda_3.
$
\end{enumerate}

Then there exist a bisymmetric nonnegative matrix that realizes
$\sigma$
\end{theorem}
\begin{proof}
In the first case, at first we assume that $\lambda_3=\lambda_4=0$, then  it easy to see that the following circulant nonnegative bisymmetric matrix has eigenvalues $\{\lambda_1,\lambda_2,0,0\}$:
\[
 \left[ \begin {array}{cccc} \frac{\lambda_1+\lambda_2}{4}&\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}&\frac{\lambda_1-\lambda_2}{4}\\\noalign{\medskip}\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}
&\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}\\\noalign{\medskip}\frac{\lambda_1+\lambda_2}{4}&\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}&\frac{\lambda_1-\lambda_2}{4}
\\\noalign{\medskip}\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}&\frac{\lambda_1-\lambda_2}{4}&\frac{\lambda_1+\lambda_2}{4}
\end {array} \right].
\]
Now  we consider the answer matrix as a $4 \times 4$ circulant matrix. It is clear that this matrix is a type of Teoplitz matrix.
$$
A=\left[ \begin{matrix}a&b&c&b\\b&a&b&c\\c&b&a&b\\b&c&b&a
\end{matrix}\right],
$$
 We  find the characteristic polynomials of this matrix and then determine its roots.
$$
{\rm det}( A-\lambda I)=0 \Longrightarrow
 \lbrace\lambda_{1}=a+2b+c,\lambda_{2}=a-2b+c,
 \lambda_{3}=a-c,\lambda_{4}=a-c\rbrace
 $$
so $a=\frac{\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}}{4}$, $b=\frac{\lambda_{1}+\lambda_{4}-\lambda_{2}-\lambda_{3}}{4}$, $c=\frac{\lambda_{1}+\lambda_{2}+\lambda_{4}-3\lambda_{3}}{4}$, 
then the following bisymmetric nonnegative matrix:\\
$$
 A=\left[\begin{matrix}\frac{\lambda_{1}+\lambda_{2}+
 \lambda_{3}+\lambda_{4}}{4}& \frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{2}+\lambda_{4}-3\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}\\
 \frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+\lambda_{2}+
 \lambda_{3}+\lambda_{4}}{4}&\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{2}+\lambda_{4}-3\lambda_{3}}{4}\\
 \frac{\lambda_{1}+\lambda_{2}+
 \lambda_{4}-3\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{2}+\lambda_{3}+\lambda_{4}}{4}&\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}\\\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{2}+\lambda_{4}-3\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{4}-\lambda_{2}-\lambda_{3}}{4}&\frac{\lambda_{1}+
 \lambda_{2}+\lambda_{3}+\lambda_{4}}{4}\end{matrix}\right],
 $$
is the solution, with Perron eigenvalue $\lambda_{1}$.\\
\end{proof}
\begin{theorem}
 Let $\sigma=\left\lbrace \lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4}\right\rbrace $  be a set of  real numbers with following condition 
 \begin{enumerate}
 \item
$
\lambda_{1}+\lambda_{2}+\lambda_{3}+\lambda_{4}\geq 0,
$
\item
$ \lambda_{1}\in R, \,
\lambda_{1}\geq|\lambda_{i}|;i=2,3,4,  \lambda_{3}=\lambda_{4},
$
\item
$\lambda_1+\lambda_2 \geq \pm (\lambda_3 +\lambda_4).
$
\end{enumerate}
Then there exist a bisymmetric nonnegative matrix that realizes
$\sigma$
\end{theorem}
\begin{proof}
We consider the nonnagative bisymmetric matrix as following Hankel form:
\[
 \left[ \begin {array}{cccc} a&b&c&b\\\noalign{\medskip}b&c&b&c
\\\noalign{\medskip}c&b&c&b\\\noalign{\medskip}b&c&b&a\end {array}
 \right].
\]
The charactristic polynomial of above matrix  is obtained as :
\[
\begin{array}{l}
P(\lambda)={\lambda}^{4}+ \left( -2\,a-2\,c \right) {\lambda}^{3}+ \left( -4\,{b}
^{2}-{c}^{2}+4\,ca+{a}^{2} \right) {\lambda}^{2}+ \\
\left( -4\,c{b}^{2}+
4\,a{b}^{2}+2\,{c}^{3}-2\,c{a}^{2} \right) \lambda- 
{a}^{2}{b}^{2}+2\,a
{b}^{2}c-{b}^{2}{c}^{2}+{c}^{4}-2\,{c}^{3}a+{a}^{2}{c}^{2},
\end{array}
\]
 we find the roots of above polynomial  and denote them by $\lambda_1,\lambda_2,\lambda_2$ and $\lambda_4$, so we have 
\begin{eqnarray}\label{41}
\left\{ \begin{array}{l}
\lambda_1=b+1/2\,a+1/2\,c+1/2\,\sqrt {4\,{b}^{2}+8\,bc+{a}^{2}-2\,ca+5\,{c}^{2}},\\
\lambda_2=b+1/2\,a+1/2\,c-1/2\,\sqrt {4\,{b}^{2}+8\,bc+{a}^{2}-2\,ca+5\,{c}^{2}},\\
\lambda_3=-b+1/2\,a+1/2\,c+1/2\,\sqrt {4\,{b}^{2}-8\,bc+{a}^{2}-2\,ca+5\,{c}^{2}},\\
\lambda_4=-b+1/2\,a+1/2\,c-1/2\,\sqrt {4\,{b}^{2}-8\,bc+{a}^{2}-2\,ca+5\,{c}^{2}}.
\end{array}
\right.
\end{eqnarray}
Now from \eqref{41} we find $a,b$ and $c$  and then we will provide the bisymmetric nonnegative matrix in this case. To do this, from the first two equations and then from the  last two equations of \eqref{41} respectively we have:
\begin{eqnarray}\label{42}
\begin{array}{l}
\lambda_1+\lambda_2=2b+a+c, \\
\lambda_3+\lambda_4=-2b+a+c,
\end{array}
\end{eqnarray}
then 
\begin{eqnarray}\label{44}
b=\frac{(\lambda_1+\lambda_2)-(\lambda_3+\lambda_4)}{4},
\end{eqnarray}

also from \eqref{41} we have
\begin{eqnarray}\label{43}
\begin{array}{l}
\lambda_1-\lambda_2=\sqrt {(2\,{b}+2\,c)^2+({a}-\,c)^{2}},\\
\lambda_3 -\lambda_4=\sqrt {(2\,{b}-2\,c)^2+({a}-\,c)^{2}},
\end{array}
\end{eqnarray}
so 
\[
c=\frac{(\lambda_1-\lambda_2)^2-(\lambda_3-\lambda_4)^2}{16b}=\frac{(\lambda_1-\lambda_2)^2-(\lambda_3-\lambda_4)^2}{4[(\lambda_1+\lambda_2)-(\lambda_3+\lambda_4)]}.
\]
By \eqref{43}we have $\lambda_1-\lambda_2 \geq \lambda_3 -\lambda_4$ then by hypthesis we have $b , c \geq 0$. Now, by combinig and  simplifying the  relations  \eqref{42} and \eqref{44}, we can obtain $a$. So
\begin{eqnarray} \label{45}
a=\frac{(\lambda_1+\lambda_2)^2-(\lambda_3+\lambda_4)^2+4\lambda_1\lambda_2-4\lambda_3\lambda_4}{4[(\lambda_1+\lambda_2)-(\lambda_3+\lambda_4)]},
\end{eqnarray}
By placing $a, b$ and $c$ in Hankel matrix, the desired matrix will be obtained.
\end{proof}
\begin{example}
Assume given  $$\sigma=\left\{ 9/2+1/2\,\sqrt {65},9/2-1
/2\,\sqrt {65}, 7/2+1/2\,\sqrt {65}
, 7/2-1/2\,\sqrt {65} \right\},
$$
by above Theorem we see that $\lambda_1+\lambda_2 \geq  (\lambda_3 +\lambda_4).
$ Then we have $a=8,b=\frac{1}{2}$ and $c=0$ and so  the following Hankel matrix is bisymmetric matrix and has eigenvalues $\sigma$:
\[
\left[ \begin {array}{cccc} 8&1/2&0&1/2\\\noalign{\medskip}1/2&0&1/2&0
\\\noalign{\medskip}0&1/2&0&1/2\\\noalign{\medskip}1/2&0&1/2&8
\end {array} \right] .
\]

\end{example}
\subsection{CASE n=5}
In this subsection at first we try to get a extention of problem that related to above subsection.
\begin{theorem}
Let $\sigma_1=\{\lambda_1,\lambda_2,\lambda_3\}$ be the spectrum of nonnegative bisymmetric matrix
 \[ 
\left[ \begin {array}{ccc} a&b&c\\\noalign{\medskip}b&d&b
\\\noalign{\medskip}c&b&a\end {array}
 \right],
\]
and $\lambda_4$ and $\lambda_5$ are two real numbers such that  $\lambda_4 \geq 0$ and $\lambda_4+\lambda_5 \geq 0$, then $\sigma=\{\lambda_1,\lambda_2,\lambda_3,\lambda_4, \lambda_5\}$ is realized by a nonnegative bisymmetric $5\times 5$ matrix. 
\end{theorem}
\begin{proof}
It is tivial that the following matrix 
 \[ 
\left[ \begin {array}{ccccc} \frac{\lambda_4+\lambda_5}{2}& 0& 0 &0 &\frac{\lambda_4-\lambda_5}{2} \\ 0&  a&b&c&0 \\\noalign{\medskip}0&b&d&b&0
\\\noalign{\medskip}0&c&b&a&0 \\
\frac{\lambda_4-\lambda_5}{2} &0&0&0&\frac{\lambda_4+\lambda_5}{2}\end {array}
 \right],
\]
is nonnegative and bisymmeytric and has spectrum $\sigma$. 
\end{proof}
\begin{theorem}
Let $\sigma=\{\lambda_1,\lambda_2,\lambda_3,\lambda_4\}$ be the spectrum of nonnegative bisymmetric matrix 
\[ 
\left[ \begin {array}{cccc} a&b&c&d\\\noalign{\medskip}b&c&b&c
\\\noalign{\medskip}c&b&c&b\\\noalign{\medskip}d&c&b&a\end {array}
 \right],
\]
and $\lambda_5 \geq 0$, then the following nonnegative bisymmetric matrix is realized the spectrum $\sigma=\{\lambda_1,\lambda_2,\lambda_3,\lambda_4, \lambda_5\}$ 
\[ 
\left[ \begin {array}{ccccc} a&b&0&c&d\\\noalign{\medskip}b&c&0&b&c 
\\\noalign{\medskip} 0& 0 & \lambda_5 & 0 & 0
\\\noalign{\medskip}c&b&0&c&b\\\noalign{\medskip}d&c&0&b&a\end {array}
 \right].
\]
\end{theorem}
\section{SPECAL CASES OF PROBLEM}
\begin{theorem}\label{s1}
Let's given $\sigma=\{\lambda_1,\lambda_2, \ldots,\lambda_n\} $  with the following conditions
\begin{enumerate}
\item
$\lambda_1\geq |\lambda_{i} |$, and $\sum_{i=1}^n\lambda_i\geq 0,$
\item
$\lambda_i= \lambda_j, \,\,\, i,j=2,3,\cdots, n$
\end{enumerate}
then the following nonnegative bisymmatric  matrix is realzied spectrum  $\sigma$
$$
C= \left[ \begin {array}{ccccc} \frac{\lambda_1+(n-1)\lambda_2}{n}&\frac{\lambda_1-\lambda_2}{n-1}&\cdots&\frac{\lambda_1-\lambda_2}{n-1}&\frac{\lambda_1-\lambda_2}{n-1}\\\noalign{\medskip}\frac{\lambda_1-\lambda_2}{n-1}& \frac{\lambda_1+(n-1)\lambda_2}{n}&\cdots&\frac{\lambda_1-\lambda_2}{n-1}&\frac{\lambda_1-\lambda_2}{n-1}
\\\noalign{\medskip}\frac{\lambda_1-\lambda_2}{n-1}&\frac{\lambda_1-\lambda_2}{n-1}& \cdots&\frac{\lambda_1-\lambda_2}{n-1}&\frac{\lambda_1-\lambda_2}{n-1}\\\noalign{\medskip}\vdots& \vdots & \vdots& \vdots &\vdots
\\\noalign{\medskip}\frac{\lambda_1-\lambda_2}{n-1}&\frac{\lambda_1-\lambda_2}{n-1}&\cdots&\frac{\lambda_1-\lambda_2}{n-1}& \frac{\lambda_1+(n-1)\lambda_2}{n}\end {array} \right].
$$
\begin{proof}
We select the matrices $A$ and $L$ as the following:
\[
A=\left[ \begin {array}{ccccccc} \lambda_2 &0&0&0&\cdots&0&\frac{\lambda_i}{n-1}\\\noalign{\medskip}0&\lambda_2&0&0&\cdots&0&2\frac{\lambda_1}{n-1}
\\\noalign{\medskip}0&0&\lambda_2&0&\cdots&0&3\frac{\lambda_1}{n-1}\\\noalign{\medskip}
\vdots& \vdots & \vdots& \vdots & \vdots &\vdots &\vdots\\\noalign{\medskip}
0&0&0& 0&\cdots&\lambda_2&(n-1)\frac{\lambda_1}{n-1}
\\\noalign{\medskip}0&0&0&0&\cdots&0&\lambda_1\end {array} \right],
\]
and 
\[
L=\left[ \begin {array}{cccccc} 1&0&0&0&\cdots&0\\\noalign{\medskip}1&1&0&0&\cdots&0
\\\noalign{\medskip}1&1&1&0&\cdots&0\\\noalign{\medskip}1&1&1&1&\cdots&0
\\\noalign{\medskip} \vdots& \vdots & \vdots& \vdots & \vdots &\vdots 
\\\noalign{\medskip}1&1&1&1&\cdots&1\end {array} \right],
\]
It is easy to see that we have  $C=L^{-1}AL.$
\end{proof}
\end{theorem}
\begin{example}
Let given $\sigma=\{70,-15,-15,-15,-15\}$ by Theorem \eqref{s1} we find a bisymmetric matrix such that  $\sigma$ is its spectrum.

We assume that 
\[
A=\left[ \begin {array}{ccccc} -15&0&0&0&17\\ \noalign{\medskip}0&-15&0
&0&34\\ \noalign{\medskip}0&0&-15&0&51\\ \noalign{\medskip}0&0&0&-15&
68\\ \noalign{\medskip}0&0&0&0&70\end {array} \right], 
\qquad
L=\left[ \begin {array}{ccccc} 1&0&0&0&0\\ \noalign{\medskip}1&1&0&0&0
\\ \noalign{\medskip}1&1&1&0&0\\ \noalign{\medskip}1&1&1&1&0
\\ \noalign{\medskip}1&1&1&1&1\end {array} \right],
\]
then we have 
\[
L^{-1}= \left[ \begin {array}{ccccc} 1&0&0&0&0\\ \noalign{\medskip}-1&1&0&0&0
\\ \noalign{\medskip}0&-1&1&0&0\\ \noalign{\medskip}0&0&-1&1&0
\\ \noalign{\medskip}0&0&0&-1&1\end {array} \right].
\]
So the soultion matrix is:
\[
C=L^{-1}AL= \left[ \begin {array}{ccccc} 2&17&17&17&17\\ \noalign{\medskip}17&2&
17&17&17\\ \noalign{\medskip}17&17&2&17&17\\ \noalign{\medskip}17&17&
17&2&17\\ \noalign{\medskip}17&17&17&17&2\end {array} \right].
\]
\end{example}
\bibliographystyle{amsplain}
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% Please cite your relevant papers but at most total 5 papers/books
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\end{document}

