A characterization of trace zero bisymmetric nonnegative $5 \times 5$ matrices

TL;DR

This paper characterizes the eigenvalues of traceless bisymmetric nonnegative 5x5 matrices, extending prior results on symmetric nonnegative matrices by establishing necessary and sufficient conditions.

Contribution

It provides a complete characterization of eigenvalues for traceless bisymmetric nonnegative 5x5 matrices, generalizing previous symmetric matrix results.

Findings

Necessary and sufficient conditions for eigenvalues of traceless bisymmetric nonnegative matrices.

Extension of Spector's conditions to bisymmetric matrices.

Eigenvalue characterization for 5x5 matrices with trace zero.

Abstract

Let $\lambda_1 \geq \lambda_2 \geq \lambda_3 \geq \lambda_4 \geq \lambda_5 \geq -\lambda_1$ be real numbers such that $\sum_{i=1}^5 \lambda_i =0$. In \cite{oren}, O. Spector prove that a necessary and sufficient condition for $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5$ to be the eigenvalues of a symmetric nonnegative $5 \times 5$ matrix is "$\lambda_2+\lambda_5<0$ and $\sum_{i=1}^5 \lambda_{i}^{3} \geq 0"$. In this article, we show that this condition is also a necessary and sufficient condition for $\lambda_1, \lambda_2, \lambda_3, \lambda_4, \lambda_5$ to be the spectrum of a traceless bisymmetric nonnegative $5 \times 5$ matrix.

Equations

$$L_1=\left[\begin{array}{ccccc}\frac{{\lambda }_1+{\lambda }_5}{2}& & & & \frac{{\lambda }_1-{\lambda }_5}{2}\\ & \frac{{\lambda }_2+{\lambda }_4}{2}& & \frac{{\lambda }_2-{\lambda }_4}{2}& \\ & & {\lambda }_3& & \\ & \frac{{\lambda }_2-{\lambda }_4}{2}& & \frac{{\lambda }_2+{\lambda }_4}{2}& \\ \frac{{\lambda }_1-{\lambda }_5}{2}& & & & \frac{{\lambda }_1+{\lambda }_5}{2}\end{array}\right].$$

$$L_2=\left[\begin{array}{ccccc}0& a& b& a& -{\lambda }_5\\ a& 0& c& -{\lambda }_4& a\\ b& c& {\sum }_{i=1}^5{\lambda }_i& c& b\\ a& -{\lambda }_4& c& 0& a\\ -{\lambda }_5& a& b& a& 0\end{array}\right],$$

$$a=\frac{1}{2}\sqrt{\frac{({\lambda }_1+{\lambda }_5)({\lambda }_2+{\lambda }_5)({\lambda }_3+{\lambda }_4)}{{\lambda }_1+{\lambda }_2-{\lambda }_3+{\lambda }_5}},$$

$$b=\sqrt{\frac{-({\lambda }_1+{\lambda }_5)({\lambda }_2+{\lambda }_5)({\lambda }_1+{\lambda }_2+{\lambda }_4+{\lambda }_5)}{2({\lambda }_1+{\lambda }_2-{\lambda }_3+{\lambda }_5)}},$$

$$c=\sqrt{\frac{-({\lambda }_3+{\lambda }_4)({\lambda }_1+{\lambda }_2+{\lambda }_4+{\lambda }_5)}{2}}.$$

$$L_3=\left[\begin{array}{ccccc}\frac{{\lambda }_2+{\lambda }_5}{2}& 0& 0& 0& \frac{{\lambda }_2-{\lambda }_5}{2}\\ 0& 0& d& -{\lambda }_4& 0\\ 0& d& {\lambda }_1+{\lambda }_3+{\lambda }_4& d& 0\\ 0& -{\lambda }_4& d& 0& 0\\ \frac{{\lambda }_2-{\lambda }_5}{2}& 0& 0& 0& \frac{{\lambda }_2+{\lambda }_5}{2}\end{array}\right],$$

$$L_4=\left[\begin{array}{ccccc}\frac{{\lambda }_3+{\lambda }_4}{2}& 0& 0& 0& \frac{{\lambda }_3-{\lambda }_4}{2}\\ 0& 0& g& -{\lambda }_5& 0\\ 0& g& {\lambda }_1+{\lambda }_2+{\lambda }_5& g& 0\\ 0& -{\lambda }_5& g& 0& 0\\ \frac{{\lambda }_3-{\lambda }_4}{2}& 0& 0& 0& \frac{{\lambda }_3+{\lambda }_4}{2}\end{array}\right],$$

$$Q=\frac{1}{2}\left[\begin{array}{ccc}X+Y& \sqrt{2}x& (X-Y)J\\ \sqrt{2}{x}^T& p& \sqrt{2}{x}^{T}J\\ J(X-Y)& \sqrt{2}Jx& J(X+Y)J\end{array}\right]$$

$${\lambda }_2{\lambda }_4-{\lambda }_3{\lambda }_5-{\lambda }_3-{\lambda }_5\ge 0\text{ and }{\lambda }_2{\lambda }_4-{\lambda }_3{\lambda }_5-{\lambda }_3-{\lambda }_5\ge \frac{{\lambda }_3{\lambda }_5}{{\lambda }_2+{\lambda }_4}.$$

$$-{\lambda }_5+{\lambda }_2{\lambda }_4-{\lambda }_3-{\lambda }_3{\lambda }_5\ge -{\lambda }_2+{\lambda }_3{\lambda }_4-{\lambda }_5-{\lambda }_3{\lambda }_5={\lambda }_3({\lambda }_4-{\lambda }_5)-({\lambda }_2+{\lambda }_5)\ge 0.$$

$$0\le 1+\sum _{i=2}^5{\lambda }_i^3=1+{\lambda }_2^3+{\lambda }_3^3+{\lambda }_4^3-(1+{\lambda }_2+{\lambda }_3+{\lambda }_4{)}^3.$$

$$A-JC=\left[\begin{array}{cc}{\lambda }_2+{\lambda }_4& \sqrt{-{\lambda }_2{\lambda }_4}\\ \sqrt{-{\lambda }_2{\lambda }_4}& 0\end{array}\right]\text{ and }A+JC=\left[\begin{array}{cc}-({\lambda }_2+{\lambda }_4)& \sqrt{-{\lambda }_2{\lambda }_4}\\ \sqrt{-{\lambda }_2{\lambda }_4}& 0\end{array}\right].$$

$$U_{a,b}=\left[\begin{array}{ccc}0& a& b\\ a& -({\lambda }_2+{\lambda }_4)& \sqrt{-{\lambda }_2{\lambda }_4}\\ b& \sqrt{-{\lambda }_2{\lambda }_4}& 0\end{array}\right],$$

$$a^2+b^2={\lambda }_2{\lambda }_4-{\lambda }_3{\lambda }_5-{\lambda }_3-{\lambda }_5\text{ and }-({\lambda }_2+{\lambda }_4)b^2-2ab\sqrt{-{\lambda }_2{\lambda }_4}=-{\lambda }_3{\lambda }_5.$$

$$r=\sqrt{{\lambda }_2{\lambda }_4-{\lambda }_3{\lambda }_5-{\lambda }_3-{\lambda }_5}$$

$$b=\frac{2a\sqrt{-{\lambda }_2{\lambda }_4}\pm \sqrt{-4a^2{\lambda }_2{\lambda }_4+4({\lambda }_2+{\lambda }_4){\lambda }_3{\lambda }_5}}{-2({\lambda }_2+{\lambda }_4)}.$$

$$f(a)=\frac{2a\sqrt{-{\lambda }_2{\lambda }_4}+\sqrt{-4a^2{\lambda }_2{\lambda }_4+4({\lambda }_2+{\lambda }_4){\lambda }_3{\lambda }_5}}{-2({\lambda }_2+{\lambda }_4)}$$

$${\lambda }_2+{\lambda }_5\le 0\text{ and }\sum _{i=1}^5{\lambda }_i^3\ge 0.$$

$$A-JC=\left[\begin{array}{cc}{\lambda }_2+{\lambda }_4& \sqrt{-{\lambda }_2{\lambda }_4}\\ \sqrt{-{\lambda }_2{\lambda }_4}& 0\end{array}\right],A+JC=\left[\begin{array}{cc}1+{\lambda }_3+{\lambda }_5& \sqrt{-{\lambda }_2{\lambda }_4}\\ \sqrt{-{\lambda }_2{\lambda }_4}& 0\end{array}\right]$$

$$a^2+b^2={\lambda }_2{\lambda }_4-{\lambda }_3{\lambda }_5-{\lambda }_3-{\lambda }_5\text{ and }(1+{\lambda }_3+{\lambda }_5)b^2-2ab\sqrt{-{\lambda }_2{\lambda }_4}=-{\lambda }_3{\lambda }_5.$$

$$Q=\left[\begin{array}{ccccc}\frac{1+{\lambda }_2+{\lambda }_3+{\lambda }_4+{\lambda }_5}{2}& \sqrt{-{\lambda }_2{\lambda }_4}& \frac{a_0}{\sqrt{2}}& 0& \frac{1+{\lambda }_3+{\lambda }_5-{\lambda }_2-{\lambda }_4}{2}\\ \sqrt{-{\lambda }_2{\lambda }_4}& 0& \frac{b_0}{\sqrt{2}}& 0& 0\\ \frac{a_0}{\sqrt{2}}& \frac{b_0}{\sqrt{2}}& 0& \frac{b_0}{\sqrt{2}}& \frac{a_0}{\sqrt{2}}\\ 0& 0& \frac{b_0}{\sqrt{2}}& 0& \sqrt{-{\lambda }_2{\lambda }_4}\\ \frac{1+{\lambda }_3+{\lambda }_5-{\lambda }_2-{\lambda }_4}{2}& 0& \frac{a_0}{\sqrt{2}}& \sqrt{-{\lambda }_2{\lambda }_4}& \frac{1+{\lambda }_2+{\lambda }_3+{\lambda }_4+{\lambda }_5}{2}\end{array}\right]$$

$$A-JC=\left[\begin{array}{cc}-0.4& \sqrt{0.21}\\ \sqrt{0.21}& 0\end{array}\right]\text{ and }A+JC=\left[\begin{array}{cc}0.4& \sqrt{0.21}\\ \sqrt{0.21}& 0\end{array}\right]$$

$$a^2+b^2=0.55\text{ and }0.4b^2-2\sqrt{0.21}ab=\mathrm{0.16.}$$

$$Q=\left[\begin{array}{ccc}A& \frac{x}{2}& JCJ\\ \frac{x^T}{2}& 0& \frac{x^T}{2}J\\ C& J\frac{x}{2}& JAJ\end{array}\right]=\left[\begin{array}{ccccc}0& \sqrt{0.21}& \frac{a_0}{\sqrt{2}}& 0& 0.4\\ \sqrt{0.21}& 0& \frac{b_0}{\sqrt{2}}& 0& 0\\ \frac{a_0}{\sqrt{2}}& \frac{b_0}{\sqrt{2}}& 0& \frac{b_0}{\sqrt{2}}& \frac{a_0}{\sqrt{2}}\\ 0& 0& \frac{b_0}{\sqrt{2}}& 0& \sqrt{0.21}\\ 0.4& 0& \frac{a_0}{\sqrt{2}}& \sqrt{0.21}& 0\end{array}\right],$$

Full text

1. Introduction

The problem of finding necessary and sufficient conditions for $n$ complex numbers to be the spectra of a nonnegative matrix is known as the nonnegative inverse eigenvalues problem (NIEP). This problem was first studied by Suleimanova in [15] and then several authors had been extensively studied this problem. Several results for this problem can be found in [1, 5, 9, 11, 12, 13, 15, 7, 16]. If we consider this problem only for the list of $n$ real numbers, then the problem will be called the real nonnegative inverse eigenvalue problem (RNIEP).

The problem of finding the necessary and sufficient conditions on the list of $n$ real numbers to be a spectrum of a symmetric (bisymmetric) nonnegative matrix is called the symmetric (bisymmetric) nonnegative inverse eigenvalue problem (SNIEP (BNIEP)). In [4], A. I. Julio and R. L. Soto give sufficient conditions for the BNIEP. This article discusses the BNIEP for $5\times 5$ traceless matrices.

Recall that a square matrix $A$ is symmetric, persymmetric or centrosymmetric if $A=A^{T}$, $AJ=JA^{T}$ or $AJ=JA$, respectively, where $J$ is the reverse identity matrix, i.e., J=\begin{bmatrix}0&&1\\ &\reflectbox{\ddots}&\\ 1&&0\end{bmatrix} (note that $J^{2}=J$). A square matrix possessing any two of these conditions possesses the third, and we call it bisymmetric. Many results on bisymmetric matrices were discussed by A. Cantoni and P. Butler in [2].

The NIEP, RNIEP, SNIEP and BNIEP have been unsolved for general $n$. The NIEP for $n\leq 3$ and the RNIEP for $n\leq 4$ have been solved by Loewy and London [7]. In [16], Wuwen showed that the RNIEP and the SNIEP are equivalent when $n\leq 4$. The fact that the RNIEP and the SNIEP are different for $n>4$ was proved by Johnson et al. in [3]. The NIEP for $4\times 4$ traceless matrices was solved by Reams in [10] and $5\times 5$ traceless matrices was solved by Laffey and Meehan in [6]. In [14], O. Spector showed that a necessary and sufficient condition for a list of $5$ real numbers $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}$ such that $\lambda_{5}\geq-\lambda_{1}$ and $\sum_{i=1}^{5}\lambda_{i}=0$ to be the eigenvalues of a $5\times 5$ traceless symmetric nonnegative matrix is $\lambda_{2}+\lambda_{5}\leq 0$ and $\sum_{i=1}^{5}\lambda_{i}^{3}\geq 0$. In this article, we prove that the above condition is also a necessary and sufficient condition of the BNIEP for traceless $5\times 5$ matrices. We organize our paper as follows. In Section $2$, we give sufficient conditions of the BNIEP for $5\times 5$ matrices. This result is Proposition 2.1. Then in Section $3$, we characterize all bisymmetric nonnegative traceless $5\times 5$ matrices. Our main result of this characterization is Theorem 3.2.

2. Sufficient conditions of the BNIEP for $5\times 5$ matrices

Let $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}$ be real numbers. If $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the eigenvalues of a $5\times 5$ nonnegative matrix, then $\sum_{i=1}^{5}\lambda_{i}\geq 0$ and by the Perron-Frobenius Theorem, $\lambda_{1}\geq|\lambda_{5}|$. Since we consider the BNIEP for $5\times 5$ matrices, we from now on assume that $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}\geq-\lambda_{1}$ and $\sum_{i=1}^{5}\lambda_{i}\geq 0$.

If $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}\geq 0$ or $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq 0>\lambda_{5}$ then a $5\times 5$ nonnegative bisymmetric matrix with the eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the matrix

[TABLE]

If $\lambda_{1}\geq 0>\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}$ then a $5\times 5$ nonnegative bisymmetric matrix with the eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the matrix

[TABLE]

where

[TABLE]

[TABLE]

and

[TABLE]

Now we consider the case $\lambda_{1}\geq\lambda_{2}\geq 0>\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}$. If $\lambda_{1}+\lambda_{3}+\lambda_{4}\geq 0$ and $\lambda_{2}+\lambda_{5}<0$ then the matrix $L_{2}$ in the previous case is nonnegative and hence it is our desired matrix. If $\lambda_{2}+\lambda_{5}\geq 0$ and $\lambda_{1}+\lambda_{3}+\lambda_{4}\geq 0$ then a $5\times 5$ nonnegative bisymmetric matrix with the eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the matrix

[TABLE]

where $d=\sqrt{\dfrac{-(\lambda_{3}+\lambda_{4})(\lambda_{1}+\lambda_{4})}{2}}$. Finally, if $\lambda_{2}+\lambda_{5}\geq 0$ and $\lambda_{1}+\lambda_{3}+\lambda_{4}<0$ then $\sum_{i=1}^{5}\lambda_{i}<\lambda_{2}+\lambda_{5}$ contradicting a necessary condition of the result (Lemma $1$ in [8]) by R. Loewy and J. J. McDonald and hence in this case there is no nonnegative bisymmetric matrix with the eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$.

Next, we consider the case $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq 0>\lambda_{4}\geq\lambda_{5}.$ If $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}\geq 0$ and $\lambda_{2}+\lambda_{5}\geq 0$, then $\lambda_{2}+\lambda_{4}\geq 0$ and the matrix $L_{1}$ is our desired matrix. If $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}\geq 0$, $\lambda_{2}+\lambda_{5}<0$ and $\lambda_{3}+\lambda_{4}\geq 0,$ then the matrix

[TABLE]

where $g=\sqrt{\dfrac{-(\lambda_{2}+\lambda_{5})(\lambda_{1}+\lambda_{5})}{2}}$,  will be our desired matrix. If $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}\geq 0$, $\lambda_{2}+\lambda_{5}<0$ and $\lambda_{3}+\lambda_{4}<0$, then again the matrix $L_{2}$ can be used. We do not have a complete solution for the case when $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}<0$. However, some partial result (Corollary 3.4) will be given at the end of Section $3$.

Summing up what we have done so far, we get the following result.

Proposition 2.1**.**

Let $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}\geq-\lambda_{1}$ be real numbers such that $\sum_{i=1}^{5}\lambda_{i}\geq 0$. If $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ satisfies one of the following conditions

• (1)

$\lambda_{4}\geq 0$* or $\lambda_{2}\leq 0$,*

• (2)

$\lambda_{2}\geq 0>\lambda_{3}$* and $\sum_{i=1}^{5}\lambda_{i}\geq\lambda_{2}+\lambda_{5}$,*

• (3)

$\lambda_{3}\geq 0>\lambda_{4}$* and $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}\geq 0$,*

then there is a $5\times 5$ nonnegative bisymmetric matrix with the eigenvalues $\lambda_{1}$, $\lambda_{2}$, $\lambda_{3}$, $\lambda_{4}$, $\lambda_{5}.$

Proposition 2.1 will be employed to obtain our main result, Theorem 3.3 in Section $3$.

3. A necessary and sufficient condition of the BNIEP for $5\times 5$ traceless matrices

In this section, we give a characterization of bisymmetric nonnegative traceless $5\times 5$ matrices according to the characterization of symmetric nonnegative traceless $5\times 5$ matrices given by O. Spector in [14]. To obtain our main result, we use Proposition 2.1 in the previous section along with a useful result of Cantoni and Butler in [2] which gives a standard form of a bisymmetric matrix. This result of Cantoni and Butler (Lemma 2(ii) and Lemma 3(b) in [2]) states that a $(2m+1)\times(2m+1)$ matrix $Q$ is bisymmetric if and only if $Q$ is of the form $\begin{bmatrix}A&x&JCJ\\ x^{T}&p&x^{T}J\\ C&Jx&JAJ\end{bmatrix},$ where $A$ and $C$ are $m\times m$ matrices, $A=A^{T}$, $JC^{T}J=C$, $x$ is an $m\times 1$ matrix and $p$ is a real number, and in this case the eigenvalue of $Q$ can be partitioned into the eigenvalue of $A-JC$ and the eigenvalues of $\begin{bmatrix}p&\sqrt{2}x^{T}\\ \sqrt{2}x&A+JC\end{bmatrix}$.

So, if $p\in\mathbb{R}$, $x$ is an $m\times 1$ matrix and $X$, $Y$ are $m\times m$ symmetric matrices, then $A=\dfrac{X+Y}{2}$ and $JC=\dfrac{X-Y}{2}$ are symmetric and hence the matrix

[TABLE]

is a bisymmetric matrix with the eigenvalues obtained from the matrices $\begin{bmatrix}p&x^{T}\\ x&X\end{bmatrix}$ and $Y$. In terms of $A$ and $C$, the matrix $Q$ is of the form $\begin{bmatrix}A&\dfrac{x}{\sqrt{2}}&JCJ\\ \dfrac{x^{T}}{\sqrt{2}}&p&\dfrac{x^{T}J}{\sqrt{2}}\\ C&\dfrac{Jx}{\sqrt{2}}&JAJ\end{bmatrix}$. Note that, $Q$ is a nonnegative matrix if and only if $p\geq 0$ and $x,X+Y$ and $X-Y$ are nonnegative matrices. So $Q$ is nonnegative if we have the condition that $\begin{bmatrix}p&x^{T}\\ x&A+JC\end{bmatrix}$, $A$ and $C$ are nonnegative. Therefore, to construct a $(2m+1)\times(2m+1)$ bisymmetric nonnegative matrix from $m\times m$ symmetric matrices $X$ and $Y$, $p\in\mathbb{R}$ and an $m\times 1$ matrix $x$, we need to find matrices $A$ and $C$ such that $X=\dfrac{A+JC}{2}$, $Y=\dfrac{A-JC}{2}$ and $\begin{bmatrix}p&x^{T}\\ x&A+JC\end{bmatrix}$, $A$ and $C$ are nonnegative. We will use this method in the proof of Theorem 3.2

Before we proof our main theorem, we first need the following auxiliary lemma.

Lemma 3.1**.**

Let $1\geq\lambda_{2}\geq\lambda_{3}>0>\lambda_{4}\geq\lambda_{5}\geq-1$ be real numbers such that $1+\sum_{i=2}^{5}\lambda_{i}=0$ and  $1+\sum_{i=2}^{5}\lambda_{i}^{3}\geq 0.$ If  $1+\lambda_{2}+\lambda_{4}+\lambda_{5}<0$, then

[TABLE]

Proof.

Suppose that $1+\lambda_{2}+\lambda_{4}+\lambda_{5}<0$. Since $1\geq\lambda_{2}\geq\lambda_{3}>0>\lambda_{4}\geq\lambda_{5}\geq-1$ and $1+\lambda_{2}+\lambda_{4}+\lambda_{5}<0$, we have $\lambda_{2}+\lambda_{5}<0.$ Since $1>\lambda_{2}\geq\lambda_{3}$ and $-\lambda_{5}>\lambda_{4}\geq-1$, by the rearrangement inequality, we have $-\lambda_{5}+\lambda_{2}\lambda_{4}-\lambda_{3}\geq-\lambda_{2}+\lambda_{3}\lambda_{4}-\lambda_{5}.$ So,

[TABLE]

Therefore, we have $\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}\geq 0.$

Since $\lambda_{5}=-(1+\lambda_{2}+\lambda_{3}+\lambda_{4})$,

[TABLE]

Simplifying the right-hand side of the previous inequality, we obtain

$0\leq\lambda_{2}\lambda_{5}-\lambda_{2}^{2}\lambda_{4}+\lambda_{2}\lambda_{3}\lambda_{5}+\lambda_{2}\lambda_{3}+\lambda_{2}\lambda_{5}-\lambda_{2}\lambda_{4}^{2}+\lambda_{3}\lambda_{4}\lambda_{5}+\lambda_{3}\lambda_{4}+\lambda_{4}\lambda_{5}.$

Therefore,

$\lambda_{2}\lambda_{5}\geq\lambda_{2}^{2}\lambda_{4}-\lambda_{2}\lambda_{3}\lambda_{5}-\lambda_{2}\lambda_{3}-\lambda_{2}\lambda_{5}+\lambda_{2}\lambda_{4}^{2}-\lambda_{3}\lambda_{4}\lambda_{5}-\lambda_{3}\lambda_{4}-\lambda_{4}\lambda_{5}$

$=(\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5})(\lambda_{2}+\lambda_{4}).$

Since  $\lambda_{2}+\lambda_{4}<0$, we have $\dfrac{\lambda_{3}\lambda_{5}}{\lambda_{2}+\lambda_{4}}$  $\leq\ \lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}$. ∎

Now, we are ready to prove the following result.

Theorem 3.2**.**

Let $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}>0>\lambda_{4}\geq\lambda_{5}\geq-\lambda_{1}$ be real numbers such that $\sum_{i=1}^{5}\lambda_{i}=0$ and $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}<0.$ Then $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the eigenvalue of a $5\times 5$ nonnegative bisymmetric matrix if and only if  $\sum_{i=1}^{5}\lambda_{i}^{3}\ \geq 0.$

Proof.

Normalizing by the Perron root, without loss of generality, we assume that $\lambda_{1}=1$.

If there is a $5\times 5$ nonnegative bisymmetric matrix $Q$ with the eigenvalues $1,\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$, then $Q^{3}$ is a nonnegative diagonalizable matrix and hence $\displaystyle 1+\sum_{i=2}^{5}\lambda_{i}^{3}=$ Tr($Q^{3}$) $\geq 0$. So $\displaystyle 1+\sum_{i=2}^{5}\lambda_{i}^{3}\geq 0$ is a necessary condition.

Assume that $\displaystyle 1+\sum_{i=2}^{5}\lambda_{i}^{3}\geq 0.$ Then, by Lemma 3.1, we have $\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}\geq 0$ and $\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}\geq\dfrac{\lambda_{3}\lambda_{5}}{\lambda_{2}+\lambda_{4}}.$ Let $A$ and $C$ be nonnegative matrices such that

[TABLE]

Then $A=\begin{bmatrix}0&\sqrt{-\lambda_{2}\lambda_{4}}\\ \sqrt{-\lambda_{2}\lambda_{4}}&0\end{bmatrix}$ and $C=\begin{bmatrix}0&0\\ -(\lambda_{2}+\lambda_{4})&0\end{bmatrix}$. Note that $A-JC$ has  $\lambda_{2}$ and $\lambda_{4}$  as its eigenvalues. Now, let

[TABLE]

where $a,b$ are nonnegative real numbers. We want to show that there are real numbers $a$ and $b$ such that the matrix $U_{a,b}$ has the eigenvalues $1,\lambda_{3},\lambda_{5}$.

Suppose that $1,\lambda_{3},\lambda_{5}$ are the eigenvalues of $U_{a,b}.$ Then by comparing the coefficients of $(x-1)(x-\lambda_{3})(x-\lambda_{5})$ and the characteristic polynomial of $U_{a,b}$ we have the system of equations

[TABLE]

We observe that the first equation represent a circle with radius

[TABLE]

and the second equation gives the hyperbola with $b$-intercept at $b=\pm\sqrt{\dfrac{\lambda_{3}\lambda_{5}}{\lambda_{2}+\lambda_{4}}}$ in $ab$-coordinates. By Lemma 3.1, $r^{2}\geq\dfrac{\lambda_{3}\lambda_{5}}{\lambda_{2}+\lambda_{4}}$. So, the hyperbola intersects $b$-axis inside the circle. Using the quadratic formula to the equation $-(\lambda_{2}+\lambda_{4})b^{2}-2ab\sqrt{-\lambda_{2}\lambda_{4}}=-\lambda_{3}\lambda_{5}$ we have

[TABLE]

We see that the function $f$ defined by

[TABLE]

for $a\geq 0$, is increasing and $f(a)\geq 0$ for $a\geq 0$. Therefore the hyperbola and the circle are intersect at some point in the first quadrant. So, there are nonnegative real numbers $a_{0}$ and $b_{0}$ satisfying our system and hence the matrix $U_{a_{0},b_{0}}$ has the eigenvalue $1,\lambda_{3},\lambda_{5}$. Finally, we define the matrix $Q=\begin{bmatrix}A&\dfrac{x}{\sqrt{2}}&JCJ\\ \dfrac{x^{T}}{\sqrt{2}}&0&\dfrac{x^{T}J}{\sqrt{2}}\\ C&\dfrac{Jx}{\sqrt{2}}&JAJ\end{bmatrix}$ where $x^{T}=\begin{bmatrix}a_{0}&b_{0}\end{bmatrix}.$ Then $Q$ is our desired matrix. ∎

Now, we state and prove our main result for this section.

Theorem 3.3**.**

Let $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq\lambda_{4}\geq\lambda_{5}\geq-\lambda_{1}$ be real numbers such that $\sum_{i=1}^{5}\lambda_{i}=0$. Then $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ is the eigenvalue of a $5\times 5$ nonnegative bisymmetric matrix if and only if

[TABLE]

Proof.

We have seen that $\sum_{i=1}^{5}\lambda_{i}^{3}\geq 0$ is a necessary condition. In [8], R. Loewy and J. J. McDonald proved that for any $5\times 5$ nonnegative symmetric matrix $A$ with the eigenvalues $\alpha_{1}\geq\alpha_{2}\geq\alpha_{3}\geq\alpha_{4}\geq\alpha_{5}$, Tr($A$) $\geq\alpha_{2}+\alpha_{5}$. So, for any traceless bisymmetric matrix $Q$ with the eigenvalues $\alpha_{1}\geq\alpha_{2}\geq\alpha_{3}\geq\alpha_{4}\geq\alpha_{5}$, we have $\alpha_{2}+\alpha_{5}\leq$ Tr($Q$) $=0$. So, we prove the direct part of our assertion.

Conversely, assume that $\lambda_{2}+\lambda_{5}<0$ and $\sum_{i=1}^{5}\lambda_{i}^{3}\geq 0$.

Case 1: If $\lambda_{5}\geq 0$, then all $\lambda_{i}$’s are [math] and the zero $5\times 5$ matrix is our desired matrix.

Case 2: If $\lambda_{4}\geq 0>\lambda_{5}$ or $\lambda_{2}\geq 0>\lambda_{3}$ or $\lambda_{1}\geq 0>\lambda_{2}$, then we can apply Propositon 2.1.

Case 3: Suppose that $\lambda_{3}\geq 0>\lambda_{4}$. If $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}\geq 0,$ then we can use Proposition 2.1. If $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}<0,$ then $\lambda_{3}\neq 0$ and hence Theorem 3.2 can be applied. ∎

Finally, we give a partial result concerning the case that we do not have a complete answer, that is the case of finding a $5\times 5$ bisymmetric nonnegative matrix having the eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}$ satisfying the condition $\lambda_{1}\geq\lambda_{2}\geq\lambda_{3}\geq 0>\lambda_{4}\geq\lambda_{5}\geq-\lambda_{1}$, $\sum_{i=1}^{5}\lambda_{i}>0$ and $\lambda_{1}+\lambda_{2}+\lambda_{4}+\lambda_{5}<0.$ Note that, in this condition, we have $\lambda_{1}\neq 0$ and $\lambda_{3}\neq 0.$ Therefore, without loss of generality, we may assume that $\lambda_{1}=1$ in this condition.

Corollary 3.4**.**

Let $1\geq\lambda_{2}\geq\lambda_{3}>0>\lambda_{4}\geq\lambda_{5}\geq-1$ be real numbers such that $1+\sum_{i=2}^{5}\lambda_{i}>0$ and  $1+\lambda_{2}+\lambda_{4}+\lambda_{5}<0$. If  $\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}\geq\dfrac{-\lambda_{3}\lambda_{5}}{1+\lambda_{3}+\lambda_{5}},$ then there is a $5\times 5$ nonnegative bisymmetric matrix with the eigenvalues $1,\lambda_{2},\lambda_{3},\lambda_{4},\lambda_{5}.$

Proof.

Again, by setting

[TABLE]

and $U_{a,b}=\begin{bmatrix}0&a&b\\ a&1+\lambda_{3}+\lambda_{5}&\sqrt{-\lambda_{2}\lambda_{4}}\\ b&\sqrt{-\lambda_{2}\lambda_{4}}&0\\ \end{bmatrix}$ and then comparing the coefficient of characteristic polynomials of $U_{a,b}$ and the polynomials $(x-1)(x-\lambda_{3})(x-\lambda_{5})$, we have the system

[TABLE]

Using the idea in the proof of Theorem 3.2, the inequality condition $\lambda_{2}\lambda_{4}-\lambda_{3}\lambda_{5}-\lambda_{3}-\lambda_{5}\geq\dfrac{-\lambda_{3}\lambda_{5}}{1+\lambda_{3}+\lambda_{5}}$ give us a nonnegative solution $(a_{0},b_{0})$ of the above system. Then the matrix

[TABLE]

is our desired matrix. ∎

Example 3.5**.**

The set of real numbers $\{1,0.3,0.2,-0.7,-0.8\}$ satisfies the condition of Theorem 3.2. Using the construction in the proof of Theorem 3.2, we let $A$ and $C$ be two matrices such that

[TABLE]

and let $U_{a,b}=\begin{bmatrix}0&a&b\\ a&0.4&\sqrt{0.21}\\ b&\sqrt{0.21}&0\end{bmatrix}$ be such that $U_{a,b}$ has the eigenvalues $1$, $0.2$, $-0.8$. That is, we want to find nonnegative real numbers $a$ and $b$ satisfying the system

[TABLE]

One nonnegative solution $(a_{0},b_{0})$ of this system is given by $a_{0}=\sqrt{\dfrac{51-3\sqrt{273}}{200}}$ and $b_{0}=\sqrt{\dfrac{59+3\sqrt{273}}{200}}$. Therefore the matrix

[TABLE]

where $x^{T}=\begin{bmatrix}a_{0}&b_{0}\end{bmatrix}$, is a nonnegative bisymmetric matrix with the eigenvalues $1,0.3,0.2,-0.7,-0.8$.