On the inverse eigenvalue problem of symmetric nonnegative matrices

Abstract

In this paper, at first for a given set of real numbers with only one positive number, and in continue for a given set of real numbers in special conditions, we construct a symmetric nonnegative matrix such that the given set is its spectrum.

Introduction

The symmetric nonnegative inverse eigenvalue problem (SNIEP) asks for necessary and sufficient conditions on a list $\sigma =(\lambda _1,\lambda _2, \ldots ,\lambda _n)$ of real numbers in order that it be the spectrum of a symmetric nonnegative matrix A with spectrum $\sigma$, we will say that $\sigma$ is symmetrically realizable and that it is symmetric realization of $\sigma$.

Fielder [1] obtained some necessary and some sufficient conditions for a set of n real numbers $\sigma =\{\lambda _1,\lambda _2, \ldots ,\lambda _n\}$ that it to be the set of eigenvalues of $n \times n$ symmetric nonnegative matrix. Sufficient conditions for the SNIEP have been obtained in [2,3,4,5,6,7,8,9]. Soto and Julio [10] showed that most of these sufficient conditions can be obtained by the use of a result by Soto et al. [11]. They tried to find always a solution matrix.

Laffey and Helena S̆migoc [12] presented a method that by combination of two symmetric nonnegative matrices with prescribed eigenvalues, find a new symmetric matrix of union of eigenvalues of two matrices except Perron eigenvalues of one of these matrices.

One of the special and interesting cases of SNIEP is inverse eigenvalues of Euclidean distance matrix (EDM). For instance, T.L. Hayden, R. Reams and J. Wells have solved the inverse eigenvalue problem for Euclidean distance matrices of order $n = 3, 4, 5, 6$, and any n for which there exists a Hadamard matrix and also they solved this problem: If for $n \in {\mathbb {N}}$ there exists a Hadamard matrix of order n, then there is an $(n+1)\times (n+1)$ and an $(n + 2) \times (n + 2)$ distance matrix with eigenvalues which hold under special conditions for $n \leqslant 16$ [13]. In the paper [14], Jaklic̆ and Modic offered a method for constructing a symmetric nonnegative matrix with zero diagonal and eigenvalues $\lambda _i$, where $\sum _{i=1}^{n} \lambda _i =0$ and $\lambda _1> 0 >\lambda _2 \geqslant \cdots \geqslant \lambda _n$, then they survey the inverse eigenvalue problem for Euclidean distance matrices, which are a subclass of such matrices. Nazari and Mahdinasab solved this problem without using any Hadamard matrix [15].

Through this paper the following notation is used. The spectral radius of symmetric nonnegative matrix A denoted by $\lambda _1=\rho (A)$. There is a right and a left eigenvector associated with the Perron eigenvalue with nonnegative entries. In addition $s_k$ the kth power sum of the eigenvalues $\lambda _i$ and in the list $\sigma$, $\lambda _1$ is the Perron element.

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 $\sigma$ (Perron–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) [16, 17].

In this paper we use Theorem 2.1 from [18] and again prove the Theorem 2.4 from [1] and find a solution for SNIEP by a recursive method. The Theorem 2.1 is similar to the Lemma 5 of [2].

Construction

Theorem 2.1

[18] Let B be a $m\times m$ nonnegative matrix, $M_1=\{\mu _1,\mu _2,\ldots ,\mu _m\}$ be its eigenvalues and $\mu _1$ be Perron eigenvalue of B. Also assume that A is an $n\times n$ nonnegative matrix in following form

$$\begin{aligned} A=\left( \begin{array}{cc}A_1&{}\quad a\\ b^T &{}\quad \mu _1\end{array}\right) , \end{aligned}$$

where $A_1$ is an $(n-1)\times (n-1)$ matrix, a and b are arbitrary vectors in ${\mathbb {R}}^{n-1}$ and $M_2=\{\lambda _1,\lambda _2,\ldots ,\lambda _n\}$ is the set of eigenvalues of A. Then there exists an $(m+n-1)\times (m+n-1)$ nonnegative matrix such that $M=\{\mu _2,\ldots ,\mu _m,\lambda _1,\lambda _2,\ldots ,\lambda _n\}$ is its eigenvalues.

In above Theorem, this is very important that the Perron eigenvalue of matrix B exactly lies on the main diagonal of matrix A. In the following Theorem (Fiedler [1]) that we again prove by mathematical induction, we try to lie the Perron eigenvalue of matrix B on the main diagonal of matrix A. This solution of problem can be constructed by $C= \left( \begin{array}{cc} A_1 &{}\quad a s^*\\ sb^T &{}\quad B \end{array} \right)$, where s is normalized eigenvector corrsponding to the Perron eigenvalue of B and $s^*$ is transpose conjugate of vector s and $b^T$ is transpose of vector b.

Theorem 2.2

Let $\sigma =\{\lambda _1,\lambda _2, \ldots ,\lambda _n\}$ be a set of real numbers such that only $\lambda _1$ is a positive number and

$$\begin{aligned} \lambda _1+\lambda _2+\cdots +\lambda _n \ge 0, \end{aligned}$$

(2.1)

then there exists an $n \times n$ symmetric nonnegative matrix that $\sigma$ is its spectrum.

Proof

We provide proof by induction on n. Let $n=2$, then by Fiedler [1] the following symmetric nonnegative $2 \times 2$ matrix is solution

$$\begin{aligned} A=\left( \begin{array}{cc}0&{}\quad \sqrt{-\lambda _1\lambda _2}\\ \sqrt{-\lambda _1\lambda _2}&{}\quad \lambda _1+\lambda _2\end{array}\right) . \end{aligned}$$

(2.2)

For $n=3$, we set $\sigma _1=\{\lambda _1,\lambda _2\}$ then the symmetric nonnegative matrix (2.2) realizes $\sigma _1$. Now let $\sigma _2=\{\lambda _1+\lambda _2,\lambda _3\}$, then the following matrix realizes $\sigma _2$,

$$\begin{aligned} B=\left( \begin{array}{cc}0&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}\\ \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad \lambda _1+\lambda _2+\lambda _3\end{array}\right) . \end{aligned}$$

Since $\lambda _1+\lambda _2\ge |\lambda _3|$, then $\lambda _1+\lambda _2$ is the Perron eigenvalue of B. The normalized eigenvector corresponding to $\lambda _1+\lambda _2$ is

$$\begin{aligned} s=\left( \sqrt{\frac{-\lambda _3}{\lambda _1+\lambda _2-\lambda _3}}, \sqrt{\frac{\lambda _1+\lambda _2}{\lambda _1+\lambda _2-\lambda _3}}\right) ^T. \end{aligned}$$

Whereas $\lambda _1+\lambda _2$ set on the main diagonal of A and $a=\left( \sqrt{-\lambda _1\lambda _2}\right)$, so

$$\begin{aligned} as^T=\left( \begin{array}{cc} \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&\frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}} \end{array} \right) , \end{aligned}$$

then by Theorem 2.1 the symmetric nonnegative matrix

$$\begin{aligned} A &= \left( \begin{array}{cc} A_1 &{}\quad a s^T \\ sa^T &{}\quad B \end{array} \right) \nonumber \\&= \left( \begin{array}{ccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad 0&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}\\ \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad \lambda _1+\lambda _2+\lambda _3\ \end{array}\right) , \end{aligned}$$

(2.3)

has the eigenvalues $\sigma =\{\lambda _1,\lambda _2,\lambda _3\}$.

Now let $\sigma =\{\lambda _1,\lambda _2,\ldots ,\lambda _k,\lambda _{k+1}\}$ be a set of numbers such that $\lambda _1$ is the only positive element of $\sigma$ and $\lambda _1+\lambda _2+\cdots +\lambda _{k+1} \ge 0$. Assume that $\sigma _1=\{\lambda _1,\lambda _2,\ldots ,\lambda _k\}$ and $L_k =\lambda _1+\lambda _2+\cdots +\lambda _k$, for $k=1,2,\cdots$, then $L_k \ge |\lambda _{k+1}|$ therefore by hypothesis of induction and continuing the above process, we can construct the symmetric nonnegative matrix

$$\begin{aligned} A=\left( \begin{array}{cc}A_1&{}\quad a\\ a^T &{}\quad L_k \end{array}\right) , \end{aligned}$$

with the eigenvalues $\{\lambda _1,\lambda _2,\ldots ,\lambda _k \}$, where $A_1$ is a $(k-1) \times (k-1)$ symmetric nonnegative matrix and $a\in {\mathbb {R}}^{k-1}$ (recall that the (2,2) entry of symmetric matrix (2.2) and the (3,3) entry of symmetric matrix (2.3) are $L_2$ and $L_3$, respectively). On the other hand the matrix

$$\begin{aligned} B=\left( \begin{array}{cc}0&{}\quad \sqrt{-L_k\lambda _{k+1}}\\ \sqrt{-L_k\lambda _{k+1}}&{}\quad L_k+\lambda _{k+1}\end{array}\right) , \end{aligned}$$

has eigenvalues $\{L_k , \lambda _{k+1}\}$. It is clear that $L_k$ is the Perron eigenvalue of this matrix and its normalized eigenvector is $s=\left( \sqrt{\frac{-\lambda _{k+1}}{L_k -\lambda _{k+1}}}, \sqrt{\frac{L_k }{L_k -\lambda _{k+1}}}\right) ^T$. By noting that the matrices A and B satisfy the condition of Theorem 2.1, then by this theorem the following nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A_1&{}\quad as^T\\ sa^T &{}\quad B\end{array}\right) , \end{aligned}$$

has eigenvalues $\sigma _1=\{\lambda _1,\lambda _2,\ldots ,\lambda _k, \lambda _{k+1}\}$. $\square$

In continue we try to construct a symmetric nonnegative matrix for a given set of real numbers with nonnegative summation that holds in special conditions.

The case $n=2$

For $n=2$ we have only two disjoint cases that brings in following Theorem.

Theorem 3.1

Let $\sigma =\{\lambda _1,\lambda _2\}$ be a set of two real numbers such that $\lambda _1\ge |\lambda _2|$. Then $\sigma$ is the set of eigenvalues of a symmetric nonnegative matrix.

Proof

$\sigma$ has only one of following cases:

(a) If $\lambda _2\ge 0$, then $A={\mathrm {diag}}(\lambda _1,\lambda _2)$ is a solution of problem.

(b) If $\lambda _2<0$, then the matrix

$$\begin{aligned} A=\left( \begin{array}{cc}0&{}\quad \sqrt{-\lambda _1\lambda _2}\\ \sqrt{-\lambda _1\lambda _2}&{}\quad \lambda _1+\lambda _2\end{array}\right) , \end{aligned}$$

(3.1)

solves the problem. $\square$

The case $n=3$

For $n=3$ if necessary condition $(s_1=\lambda _1+ \lambda _2 + \lambda _3 \geqslant 0)$ is hold, then we can find solution for all cases of $\lambda _2$ and $\lambda _3$.

Theorem 4.1

Let $\sigma =\{\lambda _1,\lambda _2,\lambda _3\}$ be a set of real numbers. Then there exists a symmetric nonnegative matrix that realizes $\sigma$, if it satisfies in the following conditions:

$$\begin{aligned}&\lambda _1+\lambda _2+\lambda _3\ge 0, \end{aligned}$$

(4.1)

$$\begin{aligned}&\lambda _1\in {\mathbb {R}},\;\lambda _1\ge |\lambda _i|;\,\,i=2,3. \end{aligned}$$

(4.2)

Proof

By the above conditions we have the following cases:

(a) If $\lambda _2,\lambda _3\ge 0$, then the nonnegative matrix $C={\mathrm {diag}}(\lambda _1,\lambda _2,\lambda _3)$ is a solution of our problem.

(b) If $\lambda _2,\lambda _3<0$, then by Theorem 2.2 the following symmetric nonnegative matrix is solution of our problem

$$\begin{aligned} A= \left( \begin{array}{ccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad 0&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}\\ \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad \lambda _1+\lambda _2+\lambda _3\ \end{array}\right) . \end{aligned}$$

(4.3)

(c) If $\lambda _2<0$ and $\lambda _3\ge 0$, then the symmetric nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A_1&{}\quad a\\ a^T&{}\quad \lambda _3\end{array}\right) , \end{aligned}$$

(4.4)

is a solution of our problem, where $A_1$ is matrix (3.1) and a is zero vector with dimension of $2\times 1$. $\square$

The case $n=4$

For $n=4$ if holds only the necessary condition $s_1=\lambda _1+\lambda _2+\lambda _3+\lambda _4 \ge 0$ with Perron eigenvalue $\lambda _1$, then the SINEP problem has solution. We consider 5 different cases that brings in following Theorem.

Theorem 5.1

Let $\sigma =\{\lambda _1,\lambda _2,\lambda _3,\lambda _4\}$ be a set of real numbers with the following conditions

$$\begin{aligned}&\lambda _1+\lambda _2+\lambda _3+\lambda _4\ge 0, \end{aligned}$$

(5.1)

$$\begin{aligned}&\lambda _1\ge |\lambda _i|, \qquad i=2,3,4. \end{aligned}$$

(5.2)

Then there exists a symmetric nonnegative matrix that $\sigma$ is its spectrum.

Proof

In accordance with the above conditions we consider the following cases:

(a) If $\lambda _2,\lambda _3,\lambda _4\ge 0$, then the nonnegative matrix $C={\mathrm {diag}}(\lambda _1,\lambda _2,\lambda _3,\lambda _4)$ is a desired matrix.

(b) If $\lambda _2,\lambda _3,\lambda _4<0$, then by using Theorem 2.2, the following symmetric nonnegative matrix has spectrum $\sigma$:

$$\begin{aligned} C=\left( \begin{array}{cccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{-\lambda '\lambda \lambda _1\lambda _2}}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad 0&{}\quad \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{-\lambda _3\lambda \lambda '}}{\sqrt{(\lambda '-\lambda _4)}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad 0&{}\quad \sqrt{-\lambda '\lambda _4}\\ \frac{\sqrt{-\lambda '\lambda \lambda _1\lambda _2}}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{-\lambda _3\lambda \lambda '}}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad \sqrt{-\lambda '\lambda _4}&{}\quad \lambda '+\lambda _4 \end{array}\right) , \end{aligned}$$

(5.3)

where $\lambda =\lambda _1+\lambda _2$, $\lambda '=\lambda _1+\lambda _2+\lambda _3$.

(c) If $\lambda _2<0$ and $\lambda _3,\lambda _4\ge 0$, then the symmetric nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A_1&{}\quad a\\ a^T&{}\quad \lambda _4\end{array}\right) , \end{aligned}$$

(5.4)

is a solution of this problem, where $A_1$ is matrix (4.4) and a is zero vector with dimension of $3\times 1$.

(d) If $\lambda _2,\lambda _3\le 0,\lambda _4>0$ and at least for one of the eigenvalues $\lambda _2$ and $\lambda _3$, for example $\lambda _3$, we have $\lambda _3+\lambda _4\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cccc}0&{}\quad \sqrt{-\lambda _1\lambda _2}&{}\quad 0&{}\quad 0\\ \sqrt{-\lambda _1\lambda _2}&{}\quad \lambda _1+\lambda _2&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}\\ 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}&{}\quad \lambda _3+\lambda _4\end{array}\right) , \end{aligned}$$

is a solution of our problem.

(e) If $\lambda _2,\lambda _3\le 0,\lambda _4>0$ and we have $\lambda _2+\lambda _4\le 0,\lambda _3+\lambda _4\le 0$, then by Lowey and London [16] we consider $C=(c_{ij})$ in the following form

$$\begin{aligned} C=U\left( \begin{array}{cccc}\lambda _2&{}\quad &{}\quad &{}\quad \\ &{}\quad \lambda _3&{}\quad &{}\quad \\ {} &{}\quad &{}\quad \lambda _4&{}\quad \\ {} &{}\quad &{}\quad &{}\quad \lambda _1 \end{array}\right) U^T, \end{aligned}$$

(5.5)

where U is orthogonal matrix

$$\begin{aligned} U=\frac{1}{2}\left( \begin{array}{cccc}1&{}\quad 1&{}\quad 1&{}\quad 1\\ -1&{}\quad -1&{}\quad 1&{}\quad 1\\ 1&{}\quad -1&{}\quad -1&{}\quad 1\\ - 1&{}\quad 1&{}\quad -1&{}\quad 1\end{array}\right) . \end{aligned}$$

We have also the following relations:

$$\begin{aligned} c_{11} &= {} c_{22}=c_{33}=c_{44}=\frac{1}{4}(\lambda _1+\lambda _2+\lambda _3+\lambda _4)\ge 0\\ c_{12} &= {} c_{21}=c_{34}=c_{43}=\frac{1}{4}(\lambda _1-(\lambda _2+\lambda _3)+\lambda _4)\ge 0\\ c_{13} &= {} c_{31}=c_{24}=c_{42}=\frac{1}{4}((\lambda _1+\lambda _2)-(\lambda _3+\lambda _4))\ge 0\\ c_{14} &= {} c_{41}=c_{23}=c_{32}=\frac{1}{4}((\lambda _1+\lambda _3)-(\lambda _2+\lambda _4))\ge 0. \end{aligned}$$

So C is a symmetric nonnegative matrix and solves our problem.

The case $n=5$

Theorem 6.1

Let $\sigma =\{\lambda _1,\lambda _2,\lambda _3,\lambda _4,\lambda _5\}$ be a set of real numbers and satisfies in the following conditions

$$\begin{aligned}&\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5\ge 0, \end{aligned}$$

(6.1)

$$\begin{aligned}&\lambda _1\in {\mathbb {R}},\;\lambda _1\ge |\lambda _i|;i=2,3,4,5, \end{aligned}$$

(6.2)

$$\begin{aligned}&if\;(\lambda _2,\lambda _3<0,\lambda _4,\lambda _5\ge 0) \longrightarrow \;\lambda _1+\lambda _2+\lambda _3\ge 0, \end{aligned}$$

(6.3)

$$\begin{aligned}&if\;(\lambda _2,\lambda _3,\lambda _4<0,\lambda _5\ge 0) \longrightarrow \;\lambda _1+\lambda _2+\lambda _3+\lambda _4\ge 0. \end{aligned}$$

(6.4)

Then there exists a symmetric nonnegative matrix that realize $\sigma$.

Proof

We present the proof in following cases.

(a) If $\lambda _2,\lambda _3,\lambda _4,\lambda _5\ge 0$, then the symmetric nonnegative matrix $C={\mathrm {diag}}(\lambda _1,\lambda _2,\lambda _3,\lambda _4,\lambda _5)$ is solution.

(b) If $\lambda _2,\lambda _3,\lambda _4,\lambda _5<0$, then by Theorem 2.2, we construct the solution. Since the symmetric nonnegative matrix(5.3), with the spectrum $\sigma _1=\{\lambda _1,\lambda _2,\lambda _3,\lambda _4\}$ and nonnegative matrix

$$\begin{aligned} B=\left( \begin{array}{cc}0 &{}\quad \sqrt{-(\lambda _1+\lambda _2+\lambda _3+\lambda _4)\lambda _5}\\ \sqrt{-(\lambda _1+\lambda _2+\lambda _3+\lambda _4)\lambda _5}&{}\quad \lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5\end{array}\right) , \end{aligned}$$

with the spectrum $\sigma _2=\{\lambda _1+\lambda _2+\lambda _3+\lambda _4,\lambda _5\}$ have the conditions of matrices A and B of Theorem 2.1, respectively, the normalized eigenvector corresponding to the Perron eigenvalue of $\lambda =\lambda _1+\lambda _2+\lambda _3+\lambda _4$ of nonnegative matrix B has the form

$$\begin{aligned} s=\left( \sqrt{\frac{-\lambda _5}{\lambda _1+\lambda _2+\lambda _3+\lambda _4-\lambda _5}}, \sqrt{\frac{\lambda _1+\lambda _2+\lambda _3+\lambda _4}{\lambda _1+\lambda _2+\lambda _3+\lambda _4-\lambda _5}}\right) ^T, \end{aligned}$$

then by Theorem 2.2 the following matrix is solution

$$\begin{aligned} \tiny {C=\left( \begin{array}{ccccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{} \frac{\sqrt{\lambda '\lambda \lambda _1\lambda _2\lambda _5}}{z}&{}\quad \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda _1\lambda _2}}{z}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad 0&{}\quad \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{\lambda _3\lambda _5\lambda \lambda '}}{y}&{}\quad \frac{\sqrt{-\lambda _3\lambda \lambda '\lambda ''}}{y}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{} \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad 0&{}\quad \frac{\sqrt{\lambda _4\lambda _5\lambda '}}{\sqrt{\lambda ''-\lambda _5}}&{}\quad \frac{\sqrt{-\lambda _4\lambda '\lambda ''}}{\sqrt{\lambda ''-\lambda _5}}\\ \frac{\sqrt{\lambda '\lambda \lambda _1\lambda _2\lambda _5}}{z}&{}\quad \frac{\sqrt{\lambda _3\lambda _5\lambda \lambda '}}{y}&{}\quad \frac{\sqrt{\lambda _4\lambda _5\lambda '}}{\sqrt{\lambda ''-\lambda _5}}&{} 0&{}\quad \sqrt{-\lambda ''\lambda _5}\\ \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda _1\lambda _2}}{z}&{}\quad \frac{\sqrt{-\lambda _3\lambda \lambda '\lambda ''}}{y}&{}\quad \frac{\sqrt{-\lambda _4\lambda '\lambda ''}}{\sqrt{\lambda ''-\lambda _5}}&{}\quad \sqrt{-\lambda ''\lambda _5}&{} \lambda ''+\lambda _5 \end{array}\right) ,} \end{aligned}$$

(6.5)

where $\lambda =\lambda _1+\lambda _2$,$\lambda '=\lambda _1+\lambda _2+\lambda _3$, $\lambda ''=\lambda _1+\lambda _2+\lambda _3+\lambda _4$, $z=\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)(\lambda ''-\lambda _5)}$, $y=\sqrt{(\lambda '-\lambda _4)(\lambda ''-\lambda _5)}$.

(c) If $\lambda _2<0$ and $\lambda _3,\lambda _4,\lambda _5\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A&{}a\\ a^T&{}\lambda _5\end{array}\right) , \end{aligned}$$

(6.6)

is a solution of our problem, where A is matrix (5.4) and a is zero vectors with dimension of $4\times 1$.

(d) If $\lambda _2,\lambda _3<0$ and $\lambda _4,\lambda _5\ge 0$, then by (6.3) we have $\lambda _3+\lambda _4+\lambda _5\ge 0$, thus the nonnegative matrix

$$\begin{aligned} A=\left( \begin{array}{ccccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}} &{}\quad \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad 0&{}\quad 0\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}} &{}\quad 0&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad 0&{}\quad 0\\ \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}} &{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad \lambda _1+\lambda _2+\lambda _3&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \lambda _4&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _5 \end{array}\right) , \end{aligned}$$

(6.7)

is a solution of problem.

(e) If $\lambda _2,\lambda _3\le 0,\lambda _4,\lambda _5>0$ and at least for one of the eigenvalues $\lambda _2$ and $\lambda _3$, for example $\lambda _3$, we have $\lambda _3+\lambda _4\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{ccccc}0&{}\quad \sqrt{-\lambda _1\lambda _2}&{}\quad 0&{}\quad 0&{}\quad 0\\ \sqrt{-\lambda _1\lambda _2}&{}\quad \lambda _1+\lambda _2&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}&{}\quad 0\\ 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}&{}\quad \lambda _3+\lambda _4&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _5 \end{array}\right) , \end{aligned}$$

(6.8)

is a solution of problem.

(f) If $\lambda _2,\lambda _3,\lambda _4<0$ and $\lambda _5\ge 0$, then by (6.4) we have $\lambda _1+\lambda _2+\lambda _3+\lambda _4\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A&{}\quad a\\ a^T&{}\quad \lambda _5\end{array}\right) , \end{aligned}$$

(6.9)

is a solution of this problem, where A is matrix (5.3) and a is zero vector with dimension of $4\times 1$. $\square$

The case $n=6$

Theorem 7.1

Let $\sigma =\{\lambda _1,\lambda _2,\lambda _3,\lambda _4,\lambda _5,\lambda _6\}$ be a set of real numbers and satisfies in the following conditions,

$$\begin{aligned}&\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5+\lambda _6\ge 0, \end{aligned}$$

(7.1)

$$\begin{aligned}&\lambda _1\in {\mathbb {R}},\;\lambda _1\ge |\lambda _i|;i=2,3,4,5,6, \end{aligned}$$

(7.2)

$$\begin{aligned}&if\;(\lambda _2,\lambda _3<0,\lambda _4,\lambda _5,\lambda _6\ge 0)\longrightarrow \;\lambda _1+\lambda _2+\lambda _3\ge 0, \end{aligned}$$

(7.3)

$$\begin{aligned}&if\;(\lambda _2,\lambda _3,\lambda _4<0,\lambda _5,\lambda _6\ge 0)\longrightarrow \;\lambda _1+\lambda _2+\lambda _3+\lambda _4\ge 0. \end{aligned}$$

(7.4)

$$\begin{aligned}&if\;(\lambda _2,\lambda _3,\lambda _4,\lambda _5<0,\lambda _6\ge 0)\longrightarrow \;\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5\ge 0. \end{aligned}$$

(7.5)

Then there exists a symmetric nonnegative matrix that realizes $\sigma$.

Proof

We present the proof in following cases.

(a) If $\lambda _2,\lambda _3,\lambda _4,\lambda _5,\lambda _6\ge 0$, then the symmetric nonnegative matrix $C={\mathrm {diag}}(\lambda _1,\lambda _2,\lambda _3,\lambda _4,\lambda _5,\lambda _6)$ is solution.

(b) If $\lambda _2,\lambda _3,\lambda _4,\lambda _5,\lambda _6<0$, then by Theorem 2.2, we construct the solution since the symmetric nonnegative matrix(6.5), with the spectrum $\sigma _1=\{\lambda _1,\lambda _2,\lambda _3,\lambda _4,\lambda _5\}$ and nonnegative matrix

$$\begin{aligned} B=\left( \begin{array}{cc}0 &{}\quad \sqrt{-(\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5)\lambda _6}\\ \sqrt{-(\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5)\lambda _6}&{}\quad \lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5+\lambda _6\end{array}\right) , \end{aligned}$$

with the spectrum $\sigma _2=\{\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5,\lambda _6\}$ have the conditions of matrices A and B of theorem 2.1, respectively. The normalized eigenvector corresponding to Perron eigenvalue of $\lambda =\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5$ of nonnegative matrix B has the form of

$$\begin{aligned} s=\left( \sqrt{\frac{-\lambda _6}{\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5-\lambda _6}}, \sqrt{\frac{\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5}{\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5-\lambda _6}}\right) ^T, \end{aligned}$$

then by Theorem 2.2 the following matrix is solution

$$\begin{aligned} C=\left( \begin{array}{cc} C_1&\quad C_2 \end{array}\right) , \end{aligned}$$

where

$$\begin{aligned} C_1 &= {} \left( \begin{array}{ccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda -\lambda _3}}&{}\quad 0&{}\quad \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _4\lambda }}{\sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)}}&{}\quad \frac{\sqrt{\lambda _3\lambda _4\lambda }}{\sqrt{(\lambda '-\lambda _4)}}&{}\quad 0\\ \frac{\sqrt{\lambda '\lambda \lambda _1\lambda _2\lambda _5}}{x}&{}\quad \frac{\sqrt{\lambda _3\lambda _5\lambda \lambda '}}{\sqrt{(\lambda '-\lambda _4)(\lambda ''-\lambda _5)}}&{}\quad \frac{\sqrt{\lambda _4\lambda _5\lambda '}}{\sqrt{\lambda ''-\lambda _5}}\\ \frac{\sqrt{\lambda \lambda '\lambda ''\lambda _1\lambda _2\lambda _6}}{z}&{}\quad \frac{\sqrt{\lambda _3\lambda _6\lambda \lambda '\lambda ''}}{y}&{}\quad \frac{\sqrt{\lambda _4\lambda '\lambda ''}}{w}\\ \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda '''\lambda _1\lambda _2}}{z}&{}\quad \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda '''\lambda _3}}{y} \end{array}\right) ,\\ C_2 &= {} \left( \begin{array}{cccccc} \frac{\sqrt{\lambda '\lambda \lambda _1\lambda _2\lambda _5}}{x}&{}\quad \frac{\sqrt{\lambda \lambda '\lambda ''\lambda _1\lambda _2\lambda _6}}{z}&{}\quad \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda '''\lambda _1\lambda _2}}{z}\\ \frac{\sqrt{\lambda _3\lambda _5\lambda \lambda '}}{\sqrt{(\lambda '-\lambda _4)(\lambda ''-\lambda _5)}}&{}\quad \frac{\sqrt{\lambda _3\lambda _6\lambda \lambda '\lambda ''}}{y}&{}\quad \frac{\sqrt{-\lambda \lambda '\lambda ''\lambda '''\lambda _3}}{y}\\ \frac{\sqrt{\lambda _4\lambda _5\lambda '}}{\sqrt{\lambda ''-\lambda _5}}&{}\quad \frac{\sqrt{\lambda _4\lambda '\lambda ''}}{w}&{}\quad \frac{\sqrt{-\lambda _4\lambda '\lambda ''\lambda '''}}{w} \\ 0&{}\quad \frac{\sqrt{\lambda _5\lambda _6\lambda ''}}{\sqrt{\lambda ''-\lambda _6}}&{}\quad \frac{\sqrt{-\lambda _5\lambda ''\lambda '''}}{\sqrt{(\lambda ''-\lambda _4)}} \\ \frac{\sqrt{\lambda _5\lambda _6\lambda ''}}{\sqrt{\lambda ''-\lambda _6}}&{}\quad 0&{}\quad \sqrt{-\lambda '''\lambda _6}\\ \frac{\sqrt{-\lambda _5\lambda ''\lambda '''}}{\sqrt{(\lambda ''-\lambda _4)}}&{}\quad \sqrt{-\lambda '''\lambda _6}&{}\quad \lambda '''+\lambda _6 \end{array}\right) , \end{aligned}$$

and

$$\begin{aligned} \lambda &= {} \lambda _1+\lambda _2 ,\lambda '=\lambda _1+\lambda _2+\lambda _3,\\ \lambda '' &= {} \lambda _1+\lambda _2+\lambda _3+\lambda _4,\lambda '''=\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5,\\ x &= {} \sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)(\lambda ''-\lambda _5)} ,\\ z &= {} \sqrt{(\lambda -\lambda _3)(\lambda '-\lambda _4)(\lambda ''-\lambda _5)(\lambda '''-\lambda _6)},\\ y &= {} \sqrt{(\lambda '-\lambda _4)(\lambda ''-\lambda _5)(\lambda '''-\lambda _6)},\\ w &= {} \sqrt{(\lambda ''-\lambda _5)(\lambda '''-\lambda _6)}. \end{aligned}$$

(c) If $\lambda _2<0$ and $\lambda _3,\lambda _4,\lambda _5,\lambda _6\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A&{}\quad a\\ a^T&{}\quad \lambda _6\end{array}\right) , \end{aligned}$$

is a solution of this problem,where A is matrix (6.6) and a, b are zero vectors with dimension of $5\times 1$.

(d) If $\lambda _2,\lambda _3<0$ and $\lambda _4,\lambda _5,\lambda _6\ge 0$, then by (7.3) we have $\lambda _1+\lambda _2+\lambda _3\ge 0$ then the nonnegative matrix

$$\begin{aligned} A=\left( \begin{array}{cccccc} 0&{}\quad \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2- \lambda _3}}&{}\quad \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}} &{}\quad 0&{}\quad 0&{}\quad 0\\ \frac{\sqrt{\lambda _1\lambda _2\lambda _3}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad 0&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3} &{}\quad 0&{}\quad 0&{}\quad 0\\ \frac{\sqrt{-(\lambda _1+\lambda _2)\lambda _1\lambda _2}}{\sqrt{\lambda _1+\lambda _2-\lambda _3}}&{}\quad \sqrt{-(\lambda _1+\lambda _2)\lambda _3}&{}\quad \lambda _1+\lambda _2+\lambda _3&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \lambda _4&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _5&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _6 \end{array}\right) , \end{aligned}$$

is a solution of problem.

(e) If $\lambda _2,\lambda _3\le 0,\lambda _4,\lambda _5,\lambda _6>0$ and at least for one of the eigenvalues $\lambda _2$ and $\lambda _3$, for example $\lambda _3$, we have $\lambda _3+\lambda _4\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cccccc}0&{}\quad \sqrt{-\lambda _1\lambda _2}&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ \sqrt{-\lambda _1\lambda _2}&{}\quad \lambda _1+\lambda _2&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad \sqrt{-\lambda _3\lambda _4}&{}\quad \lambda _3+\lambda _4&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _5&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad \lambda _6\end{array}\right) , \end{aligned}$$

is a solution of problem.

(f) If $\lambda _2,\lambda _3,\lambda _4<0$ and $\lambda _5,\lambda _6\ge 0$, then by (7.4) we have $\lambda _1+\lambda _2+\lambda _3+\lambda _4\ge 0$, then the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A&{}\quad a\\ a^T&{}\quad \lambda _6\end{array}\right) , \end{aligned}$$

is a solution of this problem, where A is matrix (6.9) and a, b are zero vectors with dimension of $5\times 1$.

(g) If $\lambda _2,\lambda _3,\lambda _4,\lambda _5<0$ and $\lambda _6\le 0$, then by (7.5) we have $\lambda _1+\lambda _2+\lambda _3+\lambda _4+\lambda _5\ge 0$, so the nonnegative matrix

$$\begin{aligned} C=\left( \begin{array}{cc}A&{}\quad a\\ a^T&{}\quad \lambda _6\end{array}\right) , \end{aligned}$$

is a solution of our problem, where A is the matrix (6.5) and a, b are zero vector of dimension $5\times 1$. $\square$

Numerical example

Example 8.1

Assume given

$$\begin{aligned} \sigma =\{\lambda _1=7,\lambda _2=-4, \lambda _3=-4, \lambda _4=5, \lambda _5=a\ge 0\}, \end{aligned}$$

then we have $\lambda _3+\lambda _4\ge 0$, consequently by case (e) of Theorem 6.1 the following matrix realized $\sigma$

$$\begin{aligned} C=\left( \begin{array}{cccccc}0&{}\quad \sqrt{28}&{}\quad 0&{}\quad 0&{}\quad 0\\ \sqrt{28}&{}\quad 3&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad \sqrt{20}&{}\quad 0\\ 0&{}\quad 0&{}\quad \sqrt{20}&{}\quad 1&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad a\end{array}\right) . \end{aligned}$$

Example 8.2

Assume given

$$\begin{aligned} \sigma =\{\lambda _1=18,\lambda _2=-1, \lambda _3=-6, \lambda _4=-4, \lambda _5=-3, \lambda _6=-3\}, \end{aligned}$$

since $\lambda _2, \lambda _3, \lambda _4, \lambda _5, \lambda _6 \le 0$ and $\Sigma _{i=1}^6 \lambda _i\ge 0$ then by Theorem 2.2 we construct a solution for $\sigma$. At first it is easy to see that the symmetric matrix $A_1= \left[ \begin{array}{cc} 0&{}\quad 3\,\sqrt{2}\\ 3\,\sqrt{2}&{}\quad 17\end{array} \right]$ has eigenvalues $\left[ \begin{array}{c} 18\\ - 1\end{array} \right]$ and the symmetric matrix $B=\left[ \begin{array}{cc} 0&{}\quad \sqrt{102}\\ \sqrt{102 }&{}\quad 11\end{array} \right]$ has eigenvalues $\left[ \begin{array}{c} 17\\ - 6\end{array} \right]$, with Perron eigenvector $\left[ \begin{array}{c} 1/23\,\sqrt{138}\\ 1/23\, \sqrt{391}\end{array} \right]$, let $a= \left[ \begin{array}{c} 3\,\sqrt{2}\end{array} \right]$, then the $3\times 3$ symmetric matrix

$$\begin{aligned} C_1=\left( \begin{array}{cc}A_2&{}\quad as^*\\ sa^T &{}\quad B\end{array}\right) = \left[ \begin{array}{ccc} 0&{}\quad {\frac{6}{23}}\,\sqrt{69}&{}\quad {\frac{3}{ 23}}\,\sqrt{2}\sqrt{391}\\ {\frac{6}{23}}\,\sqrt{ 69}&{}\quad 0&{}\quad \sqrt{102}\\ {\frac{3}{23}}\,\sqrt{2}\sqrt{ 391}&{}\quad \sqrt{102}&{}\quad 11\end{array} \right] \end{aligned}$$

has eigenvalues $\left[ \begin{array}{c} 18\\ -1\\ -6\end{array} \right]$, and in next step $B=\left[ \begin{array}{cc} 0&{}\quad 2\,\sqrt{11}\\ 2\, \sqrt{11}&{}\quad 7\end{array} \right]$ with eigenvalues $\left[ \begin{array}{c} 11\\ - 4\end{array} \right]$ and $a= \left[ \begin{array}{c} {\frac{3}{23}}\,\sqrt{2}\sqrt{391} \\ \sqrt{102}\end{array} \right]$ and Perron eigenvector $s=\left[ \begin{array}{c} 2/15\,\sqrt{15}\\ 1/15\, \sqrt{165}\end{array} \right]$ then the $4\times 4$ symmetric matrix

$$\begin{aligned} \begin{array}{l} C=\left( \begin{array}{cc}A_2&{}\quad as^*\\ sa^T &{}\quad B\end{array}\right) =\\ \left[ \begin{array}{cccc} 0&{}\quad {\frac{3}{23}}\,\sqrt{2}\sqrt{138}&{}\quad { \frac{2}{115}}\,\sqrt{2}\sqrt{391}\sqrt{15}&{}\quad {\frac{1}{115}}\, \sqrt{2}\sqrt{391}\sqrt{165}\\ {\frac{3}{23}}\, \sqrt{2}\sqrt{138}&{}\quad 0&{}\quad 2/15\,\sqrt{102}\sqrt{15}&{}\quad 1/15\,\sqrt{102} \sqrt{165}\\ {\frac{2}{115}}\,\sqrt{2}\sqrt{391} \sqrt{15}&{}\quad 2/15\,\sqrt{102}\sqrt{15}&{}\quad 0&{}\quad 2\,\sqrt{11} \\ {\frac{1}{115}}\,\sqrt{2}\sqrt{391}\sqrt{165}&{}\quad 1/15\,\sqrt{102}\sqrt{165}&{}\quad 2\,\sqrt{11}&{}\quad 7\end{array} \right] \end{array} \end{aligned}$$

has eigenvalues $\left[ \begin{array}{c} 18\\ - 6\\ -4\\ - 1\end{array} \right]$. With continue this method in 3 cases and round the solution with 5 floating point, we have the following solution for $\sigma$:

$$\begin{aligned} \left[ \begin{array}{cccccc} 0.0&{}\quad 2.1668&{}\quad 1.8834&{}\quad 1.7108&{}\quad 1.7108&{}\quad 1.9755\\ 2.1668&{}\quad 0.0&{}\quad 5.2155&{}\quad 4.7371&{}\quad 4.7371&{}\quad 5.4702\\ 1.8834&{}\quad 5.2155&{}\quad 0.0&{}\quad 3.6332&{}\quad 3.6331&{}\quad 4.1953\\ 1.7108&{}\quad 4.7371&{}\quad 3.6332&{}\quad 0.0&{}\quad 3.0&{}\quad 3.4642 \\ 1.7108&{}\quad 4.7371&{}\quad 3.6331&{}\quad 3.0&{}\quad 0.0&{}\quad 3.4642 \\ 1.9755&{}\quad 5.4702&{}\quad 4.1953&{}\quad 3.4642&{}\quad 3.4642&{}\quad 1.0 \end{array} \right] . \end{aligned}$$

In the next example, we try to find a symmetric nonnegative matrix for a given set of real numbers that only has a positive number with zero summation, whose its spectrum is the same as the spectrum of a distance matrix(EDM).

Example 8.3

Assume given

$$\begin{aligned} \sigma =\{\lambda _1=15,\lambda _2=-1, \lambda _3=-2, \lambda _4=-3, \lambda _5=-4, \lambda _6=-5\}, \end{aligned}$$

since $\lambda _2, \lambda _3, \lambda _4, \lambda _5, \lambda _6 \le 0$ and $\Sigma _{i=1}^6 \lambda _i\ge 0$ then by Theorem 2.2 we construct a solution for $\sigma$. At first it is easy to see that the symmetric matrix $A_1= \left[ \begin{array}{cc} 0&{}\quad \sqrt{15}\\ \sqrt{15}&{}\quad 14\end{array} \right]$ has eigenvalues $\left[ \begin{array}{c} 15 \\ - 1\end{array} \right]$ and the symmetric matrix $B=\left[ \begin{array}{cc} 0&{}\quad 2 \, \sqrt{7}\\ 2 \, \sqrt{7} &{}\quad 12\end{array} \right]$ has eigenvalues $\left[ \begin{array}{c} 14\\ - 2\end{array} \right]$, with Perron eigenvector $\left[ \begin{array}{c} \frac{1}{16}\,\sqrt{32}\\ \frac{1}{16}\, \sqrt{224}\end{array} \right]$, then let $a= \left[ \begin{array}{c} 2 \, \sqrt{7}\end{array} \right]$, then the $3\times 3$ symmetric matrix

$$\begin{aligned} C_1=\left( \begin{array}{cc}A_2&{}\quad as^*\\ sa^T &{}\quad B\end{array}\right) = \left[ \begin{array}{ccc} 0&{}\quad {\frac{1}{2}}\,\sqrt{2}\, \sqrt{14}&{}\quad {\frac{8}{ 16}}\,\sqrt{7}\sqrt{14}\\ {\frac{1}{2}}\,\sqrt{2}\, \sqrt{14} &{}\quad 0&{}\quad 2 \, \sqrt{7}\\ {\frac{8}{ 16}}\,\sqrt{7}\sqrt{14}&{}\quad 2 \, \sqrt{7}&{}\quad 12\end{array} \right] \end{aligned}$$

has eigenvalues $\left[ \begin{array}{c} 15\\ - 1\\ -2\end{array} \right]$. With continue this method and round the solution with 5 floating point, we have

$$\begin{aligned} \left[ \begin{array}{cccccc} 0.0&{}\quad 1.36931 &{}\quad 1.62019 &{}\quad 1.79743 &{}\quad 1.90647 &{}\quad 1.90647 \\ 1.36931 &{}\quad 0.0&{}\quad 2.36643 &{}\quad 2.62532 &{}\quad 2.78457 &{}\quad 2.78457 \\ 1.62019 &{}\quad 2.36643 &{}\quad 0.0&{}\quad 3.32820 &{}\quad 3.53009 &{}\quad 3.53009 \\ 1.79743 &{}\quad 2.62532 &{}\quad 3.32820 &{}\quad 0.0&{}\quad 4.24264 &{}\quad 4.24264 \\ 1.90647 &{}\quad 2.78457 &{}\quad 3.53009 &{}\quad 4.24264 &{}\quad 0.0&{}\quad 5 \\ 1.90647 &{}\quad 2.78457 &{}\quad 3.53009 &{}\quad 4.24264 &{}\quad 5 &{}\quad 0.0 \end{array} \right] . \end{aligned}$$

(8.1)

But in Example 2.10. [15], if we choose

$$\begin{aligned} W'= \left[ \begin{array}{ccccc} 1 &{}\quad -2&{}\quad 3 &{}\quad 4&{}\quad -5\\ -1 &{}\quad -2&{}\quad 3 &{}\quad -4&{}\quad 5\\ 1&{}\quad 2&{}\quad 3&{}\quad -4 &{}\quad 5\\ -1&{}\quad 2&{}\quad -3 &{}\quad 4&{}\quad 5\\ 1 &{}\quad -2 &{}\quad -3&{}\quad 4&{}\quad -5\\ -1 &{}\quad 2&{}\quad -3 &{}\quad -4&{}\quad -5 \end{array} \right] , \end{aligned}$$

then

$$\begin{aligned} D= \left[ \begin{array}{cccccc} 0&{}\quad \frac{49}{15}&{}\quad \frac{18}{5}&{}\quad \frac{ 353}{120}&{}\quad \frac{159}{40}&{}\quad \frac{77}{30}\\ \frac{49}{15}&{}\quad 0 &{}\quad \frac{5}{3}&{}\quad \frac{121}{40}&{}\quad \frac{55}{24}&{}\quad \frac{17}{5}\\ \frac{18}{5}&{}\quad \frac{5}{3}&{}\quad 0 &{}\quad \frac{323}{120}&{}\quad \frac{21}{8} &{}\quad \frac{46}{15}\\ \frac{353}{120}&{}\quad \frac{121}{40}&{}\quad \frac{323}{120}&{}\quad 0 &{}\quad \frac{77}{30} &{}\quad \frac{35}{8} \\ \frac{159}{40}&{}\quad \frac{55}{24} &{}\quad \frac{21}{8} &{}\quad \frac{77}{30} &{}\quad 0 &{}\quad \frac{ 353}{120} \\ \frac{77}{30} &{}\quad \frac{17}{5} &{}\quad \frac{46}{15} &{}\quad \frac{35}{8} &{}\quad \frac{353}{120} &{}\quad 0 \end{array} \right] \end{aligned}$$

is an EDM which is not spherical with eigenvalues

$$\begin{aligned} \left[ \begin{array}{c} 15.06857044\\ -1.002439569\\ -2.001141510\\ -3.007884565\\ -4.010967084\\ -5.046137716\\ \end{array} \right] , \end{aligned}$$

and for

$$\begin{aligned}W'= \left[ \begin{array}{ccccc} 1 &{}\quad -2&{}\quad 3&{}\quad 4&{}\quad -5\\ -1&{}\quad -2 &{}\quad 3 &{}\quad -4 &{}\quad 5\\ 1 &{}\quad 2 &{}\quad 3&{}\quad 4&{}\quad 5\\ -1 &{}\quad 2 &{}\quad -3&{}\quad 4&{}\quad -5\\ 1 &{}\quad 2 &{}\quad -3&{}\quad -4&{}\quad -5\\ -1&{}\quad -2 &{}\quad -3 &{}\quad -4 &{}\quad 5\\ \end{array} \right] , \quad D= \left[ \begin{array}{cccccc} 0 &{}\quad \frac{67}{24} &{}\quad \frac{7}{2} &{}\quad \frac{5}{3} &{}\quad \frac{25}{8}&{}\quad \frac{19}{6} \\ \frac{67}{24} &{}\quad 0 &{}\quad \frac{73}{24} &{}\quad \frac{25}{8} &{}\quad \frac{13}{6} &{}\quad \frac{31}{8} \\ \frac{7}{2} &{}\quad \frac{73}{24} &{}\quad 0 &{}\quad \frac{19}{6} &{}\quad \frac{31}{8} &{}\quad \frac{13}{6} \\ \frac{5}{3} &{}\quad \frac{25}{8} &{}\quad \frac{19}{6} &{}\quad 0&{}\quad \frac{ 67}{24} &{}\quad \frac{7}{2}\\ \frac{25}{8} &{}\quad \frac{13}{6} &{}\quad \frac{31}{8} &{}\quad \frac{67}{24} &{}\quad 0 &{}\quad \frac{ 73}{24} \\ \frac{19}{6}&{}\quad \frac{31}{8} &{}\quad \frac{13}{6} &{}\quad \frac{7}{2} &{}\quad \frac{73}{24} &{}\quad 0 \end{array} \right] \end{aligned}$$

is a distance matrix with eigenvalues

$$\begin{aligned} \left[ \begin{array}{c} -3.\\ -2.\\ -1.\\ 15.01897846\\ -5.014117791\\ -4.004860665\\ \end{array} \right] , \end{aligned}$$

and D is similar to the matrix

$$\begin{aligned} \left[ \begin{array}{cccccc} -1&{}\quad 0 &{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad -2&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad -3&{}\quad 0&{}\quad 0&{}\quad 0\\ 0&{}\quad 0&{}\quad 0&{}\quad -4&{}\quad 0&{}\quad \frac{1}{8} \sqrt{6}\\ 0 &{}\quad 0 &{}\quad 0&{}\quad 0 &{}\quad -5 &{}\quad \frac{-1}{8}\sqrt{6}\sqrt{3} \\ 0 &{}\quad 0 &{}\quad 0&{}\quad \frac{1}{8} \sqrt{6} &{}\quad \frac{-1}{8}\sqrt{6}\sqrt{3}&{}\quad 15 \end{array} \right] . \end{aligned}$$

And by using Theorem 2.6, and Example 2.9 in [15], we have $\rho = \sqrt{(-\lambda _{n+1}-\lambda _{n+2})(\lambda _{1}+\lambda _{n+1})}= \sqrt{(-\lambda _{5}-\lambda _{6})(\lambda _{1}+\lambda _{5})} = 3 \sqrt{11}$, the spectrum of ${D_1} = \left( {\begin{array}{*{20}{c}} 0 &{}\quad {\frac{5}{2}} &{}\quad 2 &{}\quad {\frac{3}{2}} \\ {\frac{5}{2}} &{}\quad 0 &{}\quad {\frac{3}{2}} &{}\quad 2 \\ 2 &{}\quad {\frac{3}{2}} &{}\quad 0 &{}\quad {\frac{5}{2}} \\ {\frac{3}{2}} &{}\quad 2 &{}\quad {\frac{5}{2}} &{}\quad 0 \\ \end{array} } \right) ,$ is $\{6,-1,-2,-3\}$, and

$$\begin{aligned} D=\left[ \begin{array}{cccccc} 0 &{}\quad \frac{3}{2} &{}\quad \frac{5}{2} &{}\quad 2&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}\\ \frac{3}{2}&{}\quad 0&{}\quad 2 &{}\quad \frac{5}{2}&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}\\ \frac{5}{2} &{}\quad 2 &{}\quad 0 &{}\quad \frac{3}{2}&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}\\ 2&{}\quad \frac{5}{2}&{}\quad \frac{3}{2}&{}\quad 0&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}\\ \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}&{}\quad 0 &{}\quad 4\\ \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}&{}\quad \frac{3}{ 4}\sqrt{22}&{}\quad \frac{3}{4}\sqrt{22}&{}\quad 4 &{}\quad 0\\ \end{array} \right] , \end{aligned}$$

is an EDM and same matrix (8.1) has spectrum $\sigma$.

Conclusion

This is important problem that what is the set of eigenvalues is realized by symmetric nonnegative matrix. The necessary conditions said to us that for instance the set of eigenvalues $\{5, -5, 0.6, 0.5, 0.5, -1.5\}$ that although has nonnegative summation but it is easy to see that for large odd k has negative summation $s_k=\Sigma _{i=1}^{6}\lambda _i^k$ then for this example we have no solution. Therefore, we can say that if for every set of eigenvalues, that has nonnegative sumation and holds in condition (2) of necessary conditions, then the problem is solved.