Permutative nonnegative matrices with prescribed spectrum

TL;DR

This paper extends previous results on permutative matrices with prescribed spectra to more general cases, providing new insights and answering an open question in the field of linear algebra.

Contribution

It generalizes existing theorems on permutative matrices and addresses a previously unresolved question about their spectral properties.

Findings

Extended the class of spectra for which permutative matrices can be constructed.

Provided a negative answer to an open question about spectral realization.

Enhanced understanding of the structure of permutative matrices with prescribed spectra.

Abstract

An n x n permutative matrix is a matrix in which every row is a permutation of the first row. In this paper the result given by Paparella in [Electron. J. Linear Algebra 31 (2016) 306-312] is extended to a more general lists of real and complex numbers, and a negative answer to a question posed by him is given.

Equations

$$P=\left[\begin{array}[]{c}\mathbf{x}^{T}\\ (P_{2}\mathbf{x})^{T}\\ \vdots\\ (P_{n}\mathbf{x})^{T}\end{array}\right].$$

$$C=\left[\begin{array}[]{ccccc}\lambda_{1}-\alpha&0&\ddots&\cdots&0\\ \lambda_{1}-\alpha-\lambda_{2}&\lambda_{2}&\ddots&\ddots&\vdots\\ \lambda_{1}-\alpha-\lambda_{3}&\ddots&\lambda_{3}&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&0\\ \lambda_{1}-\alpha-\lambda_{n}&0&\cdots&0&\lambda_{n}\end{array}\right]\in\mathcal{CS}_{\lambda_{1}-\alpha}.$$

$$A=\left[\begin{array}[]{ccccc}a_{11}&a_{12}&a_{13}&\cdots&a_{1n}\\ a_{11}-\lambda_{2}&a_{12}+\lambda_{2}&a_{13}&\cdots&a_{1n}\\ a_{11}-\lambda_{3}&a_{12}&a_{13}+\lambda_{3}&\cdots&a_{1n}\\ \vdots&\vdots&a_{13}&\ddots&\vdots\\ a_{11}-\lambda_{n}&a_{12}&\cdots&\cdots&a_{1n}+\lambda_{n}\end{array}\right]$$

$$a_{11}=\frac{1}{n}\sum _{k=1}^n{\lambda }_k,a_{11}-{\lambda }_k\ge 0,k=2,\dots ,n,$$

$${\Lambda }_0=\{{\lambda }_{01},{\lambda }_{02},\dots ,{\lambda }_{0r}\},{\Lambda }_1=\{{\lambda }_{11},{\lambda }_{12},\dots ,{\lambda }_{1p}\},$$

$$A=\left[\ \begin{array}[]{cccc}P_{1}&&&\\ &P_{1}&&\\ &&\ddots&\\ &&&P_{1}\end{array}\right],$$

$$C=B-\Omega,\text{ \ }X=\left[\ \begin{array}[]{cccc}\mathbf{x}&0&\cdots&0\\ 0&\mathbf{x}&\ddots&\vdots\\ \vdots&\ddots&\mathbf{\ddots}&0\\ 0&\ldots&0&\mathbf{x}\end{array}\right],$$

$$\Lambda$$

$${\Lambda }_0$$

$$P_{1}=\left[\begin{array}[]{ccc}0&1&3\\ 1&0&3\\ 3&1&0\end{array}\right],\text{ \ }B=\left[\begin{array}[]{cccc}4&0&2&4\\ 0&4&2&4\\ 2&0&4&4\\ 4&0&2&4\end{array}\right],$$

$$A=\left[\begin{array}[]{cccc}P_{1}&&&\\ &P_{1}&&\\ &&P_{1}&\\ &&&P_{1}\end{array}\right],\text{ \ }X=\left[\begin{array}[]{cccc}\mathbf{x}&0&0&0\\ 0&\mathbf{x}&0&0\\ 0&0&\mathbf{x}&0\\ 0&0&0&\mathbf{x}\end{array}\right],\text{ \ with \ }\mathbf{x=}\frac{1}{\sqrt{3}}\mathbf{e.}$$

$$P$$

$$\displaystyle\left[\begin{array}[]{ccc}P_{1}&&\\ &P_{1}&\\ &&P_{1}\end{array}\right]+\left[\begin{array}[]{ccc}\mathbf{x}&0&0\\ 0&\mathbf{x}&0\\ 0&0&\mathbf{x}\end{array}\right]\left[\begin{array}[]{ccc}0&c&c\\ c&0&c\\ c&c&0\end{array}\right]\left[\begin{array}[]{ccc}\mathbf{x}^{T}&0&0\\ 0&\mathbf{x}^{T}&0\\ 0&0&\mathbf{x}^{T}\end{array}\right]$$

$$i)\text{ }\left[\begin{array}[]{ccc}a&b&c\\ b&c&a\\ c&a&b\end{array}\right],\text{ }ii)\text{ }\left[\begin{array}[]{ccc}a&a&b\\ a&b&a\\ b&a&a\end{array}\right],\text{ }iii)\text{ }\left[\begin{array}[]{ccc}a&b&b\\ b&a&b\\ b&b&a\end{array}\right],$$

$$ii)a$$

$$iii)a$$

$${\Lambda }_0=\{8,6,3,3\},{\Lambda }_1=\{-5\},{\Gamma }_1=\{5,-5\}.$$

$$P_{1}=\left[\begin{array}[]{cc}0&5\\ 5&0\end{array}\right],\text{ \ }B=\left[\begin{array}[]{cccc}5&2&\frac{1}{2}&\frac{1}{2}\\ 2&5&\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}&5&2\\ \frac{1}{2}&\frac{1}{2}&2&5\end{array}\right],$$

$$A$$

$$A=\left[\begin{array}[]{ccc}3&0&2\\ 0&6&0\\ 2&0&3\end{array}\right]\text{ has the spectrum }\Lambda.$$

$$A=\left[\begin{array}[]{ccc}\frac{11}{2}&\frac{1}{2}&0\\ \frac{1}{2}&\frac{11}{2}&0\\ 0&0&1\end{array}\right].$$

$$\begin{array}[]{ccc}\Lambda=&\overline{\Lambda},\text{ \ }\lambda_{1}\geq\left|\lambda_{j}\right|,\text{ \ }j=2,3,\text{ \ }\lambda_{1}+\lambda_{2}+\lambda_{3}&\geq 0\\ &(\lambda_{1}+\lambda_{2}+\lambda_{3})^{2}\leq 3(\lambda_{1}^{2}+\lambda_{2}^{2}+\lambda_{3}^{2}).&\end{array}$$

$${\Lambda }_0=\{3,2,1\},{\Gamma }_1=\{2,-1+i,-1-i\}.$$

$$P_{1}=\left[\begin{array}[]{ccc}0&1-\frac{\sqrt{3}}{3}&1+\frac{\sqrt{3}}{3}\\ 1+\frac{\sqrt{3}}{3}&0&1-\frac{\sqrt{3}}{3}\\ 1-\frac{\sqrt{3}}{3}&1+\frac{\sqrt{3}}{3}&0\end{array}\right],\text{ \ }B=\left[\begin{array}[]{ccc}2&0&1\\ 0&2&1\\ 0&1&2\end{array}\right]$$

$$P=\left[\begin{array}[]{ccc}P_{1}&&\\ &P_{1}&\\ &&P_{1}\end{array}\right]+XCX^{T}=\left[\begin{array}[]{ccc}P_{1}&0&\frac{1}{3}\mathbf{ee}^{T}\\ 0&P_{1}&\frac{1}{3}\mathbf{ee}^{T}\\ 0&\frac{1}{3}\mathbf{ee}^{T}&P_{1}\end{array}\right]$$

$$C=\left[\begin{array}[]{ccccc}c_{0}&c_{1}&c_{2}&\cdots&c_{n-1}\\ c_{n-1}&c_{0}&c_{1}&\ddots&\vdots\\ \ddots&c_{n-1}&\ddots&\ddots&c_{2}\\ c_{2}&\ddots&\ddots&\ddots&c_{1}\\ c_{1}&c_{2}&\cdots&c_{n-1}&c_{0}\end{array}\right],$$

$${\lambda }_1$$

$${\lambda }_j$$

$$\omega$$

$${\lambda }_{n-j+2}=\overline{{\lambda }_j},j=2,3,\dots ,\left[\frac{n+1}{2}\right].$$

$$F$$

$$1$$

$${\mathbf{v}}_j$$

Full text

AMS classification:   15A18.

Key words: permutative matrices, nonnegative matrices.

1 Introduction

The *nonnegative inverse eigenvalue problem *(NIEP) is the problem of characterizing all posible spectra of entrywise nonnegative matrices. This problem remain unsolved. A complete solution is known only for $n\leq 4.$ A list $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ of complex numbers is said to be realizable if $\Lambda$ is the spectrum of an $n\times n$ nonnegative matrix $A$. In this case $A$  is said to be a realizing matrix. From the Perron-Frobenius theory, we have that if $\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ is the spectrum of an $n\times n$ nonnegative matrix $A$ then $\rho(A)=\max\limits_{1\leq i\leq n}\left|\lambda_{i}\right|$ is an eigenvalue of $A.$ This eigenvalue is called the Perron eigenvalue of $A$ and we shall assume, in this paper, that $\rho(A)=\lambda_{1}.$ A matrix $A=[a_{ij}]$ is said to have *constant row sums if all its rows sum up to the same constant, say $\alpha,$ i.e., $\mathop{\displaystyle\sum}\limits_{j=1}^{n}a_{ij}=\alpha,$ $i=1,\ldots,n.$ The set of all matrices with constant row sums equal to $\alpha$ will be denoted by $\mathcal{CS}_{\alpha}.$ It is clear that any matrix in *$\mathcal{CS}_{\alpha}$ has the eigenvector $\mathbf{e}=(1,1,\ldots,1)^{T}$ corresponding to the eigenvalue $\alpha.$ We shall denote by $\mathbf{e}_{k}$ the n-dimensional vector with one in the $kth$ position and zeros elsewhere. The real matrices with constant row sums are important because it is known that the problem of finding a nonnegative matrix with spectrum $\Lambda=\{\lambda_{1},\ldots,\lambda_{n}\}$ is equivalent to the problem of finding a nonnegative matrix in * *$\mathcal{CS}_{\lambda_{1}}$ with spectrum $\Lambda.$

The following definition, given in [5], is due to Charles Johnson:

Definition 1.1

Let $\mathbf{x}\in\mathbb{C}^{n}$ and let $P_{2},\ldots,P_{n}$ be $n\times n$ permutation matrices. A permutative matrix is any matrix of the form

[TABLE]

It is clear that $P\in\mathcal{CS}_{S},$ where $S$ is the sum of the entries of the vector $\mathbf{x}.$ In [5], the author prove that a list $\Lambda=\{\lambda_{1},\ldots,\lambda_{n}\}$ of real numbers of Suleimanova type [12], that is $\lambda_{1}>0\geq\lambda_{2}\geq\ldots\geq\lambda_{n},$ is realizable by a permutative nonnegative matrix. The author in [5] also pose the question: can all relizable lists of real numbers be realized by a permutative nonnegative matrix? The following result was announced by Suleimanova [12] and proved by Perfect [6].

Theorem 1.1

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a list of real numbers with $\lambda_{i}<0,$ $i=2,\ldots,n.$ Then $\Lambda$ is the spectrum of an $n\times n$ nonnegative matrix if and only if $\mathop{\displaystyle\sum}\limits_{i=1}^{n}\lambda_{i}\geq 0.$

The following two results, which we set here for completeness, have been shown to be very useful, not only to derive sufficient conditions for the realizability of the *NIEP, *but for constructing a realizing matrix as well. The first result, due to Brauer [1], shows how to modify one single eigenvalue of a matrix, via a rank-one perturbation, without changing any of its remaining eigenvalues (see [6, 8, 11] and the references therein to see how Brauer result has been applied to NIEP). The second result, due to Rado and introduced by Perfect in [7] is an extension of Brauer result and it shows how to change $r$ eigenvalues of an $n\times n$ matrix ($r<n),$ via a perturbation of rank $r,$ without changing any of its remaining $n-r$ eigenvalues (see [7, 9] to see how Rado result has been applied to NIEP). Both results will be employed here to obtain conditions for lists of real and complex numbers to be the spectrum of a permutative nonnegative matrix.

Theorem 1.2

Brauer [1] Let $A$ be an $n\times n$ arbitrary matrix with eigenvalues $\lambda_{1},\lambda_{2},\ldots,\lambda_{n}.$ Let $\mathbf{v}=(v_{1},\ldots,v_{n})^{T}$ be an eigenvector of $A$ corresponding to the eigenvalue $\lambda_{k}$ and let $\mathbf{q}$ be any $n-$dimensional vector. Then the matrix $A+\mathbf{vq}^{T}$ has eigenvalues $\lambda_{1},\ldots\lambda_{k-1},\lambda_{k}+\mathbf{v}^{T}\mathbf{q,}\lambda_{k+1},\ldots,\lambda_{n}.$

Theorem 1.3

Rado [7] Let $A$ be an $n\times n$ arbitrary matrix with spectrum $\Lambda=\{\lambda_{1},\ldots,\lambda_{n}\}.$ Let $X=\left[\mathbf{x}_{1}\mid\cdots\mid\mathbf{x}_{r}\right]$ be such that $rank(X)=r$ and $A\mathbf{x}_{i}=\lambda_{i}\mathbf{x}_{i},$ $i=1,\ldots,r,$ $r\leq n.$ Let $C$ be an $r\times n$ arbitrary matrix. Then $A+XC$ has eigenvalues $\mu_{1},\ldots,\mu_{r},\lambda_{r+1},\ldots\lambda_{n},$ where $\mu_{1},\ldots,\mu_{r}$ are eigenvalues of the matrix $\Omega+CX$ with $\Omega=diag\{\lambda_{1},\ldots,\lambda_{r}\}.$

A simple proof of Theorem 1.1 was given in [8] by applying Brauer result. The following result in [10], is a symmetric version of the Rado result, which we shall use to obtain some of the results in this paper:

Theorem 1.4

[10]** Let $A$ be an $n\times n$ real symmetric matrix with spectrum $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\},$ and for some $r\leq n,$ let $\{\mathbf{x}_{1},\mathbf{x}_{2},\ldots,\mathbf{x}_{r}\}$ be an orthonormal set of eigenvectors of $A$ spanning the invariant subspace associated with $\lambda_{1},\lambda_{2},\ldots,\lambda_{r}.$ Let $X$ be the $n\times r$ matrix with $i-th$ column $\mathbf{x}_{i},$ let $\Omega=diag\{\lambda_{1},\ldots,\lambda_{r}\},$ and let $C$ be any $r\times r$ symmetric matrix. Then the symmetric matrix $A+XCX^{T}$ has eigenvalues $\mu_{1},\ldots,\mu_{r},\lambda_{r+1},\ldots,\lambda_{n},$ where $\mu_{1},\ldots,\mu_{r}$ are eigenvalues of the matrix $\Omega+C.$

In this paper we give very simple and short proofs to show that both, a list of real numbers of Suleimanova type and a list of complex numbers of Suleimanova type, that is, $\mathop{\mathrm{R}e}\lambda_{k}\leq 0,$ $\left|\mathop{\mathrm{R}e}\lambda_{k}\right|\geq\left|\mathop{\mathrm{I}m}\lambda_{k}\right|,$ $k=2,\ldots,n,$ are realizable by a permutative nonnegative matrix. We use theorems 1.3 and 1.4 to obtain sufficient conditions for more general lists to be the spectrum of a permutative nonnegative matrix and the spectrum of a symmetric permutative nonnegative matrix. Our results generate an algorithmic procedure to compute a realizing matrix. The paper is organized as follows: In Section $2$ we show that a list of real numbers of Suleimanova type is always the spectrum of a permutative nonnegative matrix, and we give sufficient conditions for the problem to have a solution in the case of more general lists of real numbers. We also explore on the existence and construction of symmetric permutative nonnegative matrices with prescribed spectrum. We show that the question in [5] has a negative answer, that is, there are realizable lists of real numbers which are not the spectrum of a permutative nonnegative matrix. In section $3$ we consider the case of realizable lists of complex numbers $\Lambda=\{\lambda_{1},\ldots,\lambda_{n}\}$ of Suleimanova type, with the condition $\lambda_{n-j+2}=\overline{\lambda_{j}},$  $j=2,3,\ldots,\left[\frac{n+1}{2}\right],$ and we show that they are also realizable by permutative nonnegative matrices. We also give some examples to illustrate the results.

2 Permutative matrices with prescribed real spectrum

In this section we give a short simple proof of Theorem $3.3$ in [5], and we prove sufficient conditions for the existence of a (symmetric) permutative nonnegative matrix with prescribed real spectrum. We also give a response to the question in [5].

Theorem 2.1

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a a list of real numbers with $\lambda_{1}>0,$ $\lambda_{i}<0,$  $i=2,\ldots,n.$ Then $\Lambda$ is the spectrum of an $n\times n$ permutative nonnegative matrix if and only if $\mathop{\displaystyle\sum}\limits_{i=1}^{n}\lambda_{i}\geq 0.$

Proof. The need is clear. Suppose that $\alpha=\mathop{\displaystyle\sum}\limits_{i=1}^{n}\lambda_{i}\geq 0.$ Then we take the list $\Lambda_{\alpha}=\{\lambda_{1}-\alpha,\lambda_{2},\ldots,\lambda_{n}\}$ and consider the initial matrix

[TABLE]

From the Brauer result, for $\mathbf{q}=\left[\alpha-\lambda_{1},-\lambda_{2},-\lambda_{3},\ldots,-\lambda_{n}\right],$ we have that $B=C+\mathbf{eq}^{T}$ is a permutative nonnegative matrix with spectrum $\Lambda_{\alpha},$ and $B\in\mathcal{CS}_{\lambda_{1}-\alpha}.$ Now, $A=B+\mathbf{er}^{T},$ where $\mathbf{r}=\left[\frac{\alpha}{n},\frac{\alpha}{n},\ldots,\frac{\alpha}{n}\right],$ is the desired permutative nonnegative matrix with spectrum $\Lambda$ (of course we may also take $A=C+\mathbf{e(q}^{T}+\mathbf{r}^{T})$).

Now we give sufficient conditions for more general lists of real numbers:

Lemma 2.1

The matrix

[TABLE]

has eigenvalues $\lambda_{1}=\mathop{\displaystyle\sum}\limits_{j=1}^{n}a_{1j},\lambda_{2},\ldots,\lambda_{n}.$

Proof. Since $A$ has constant row sums equal to $\mathop{\displaystyle\sum}\limits_{j=1}^{n}a_{1j},$ then $\lambda_{1}=\mathop{\displaystyle\sum}\limits_{j=1}^{n}a_{1j}.$ Moreover, it is clear that $\det(\lambda I-A)=0$ for $\lambda=\lambda_{i},$ $i=2,\ldots,n.$

Theorem 2.2

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a list of real numbers and let $a_{11},a_{12},\ldots,a_{1n}$ be real nonnegative numbers. If

[TABLE]

then the matrix $A$ in (1) is permutative nonnegative. If $\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k}>0$ and $a_{11}-\lambda_{k}>0,$ then $A$ in (1) becomes permutative positive.

Proof. It is enough to take $a_{1k}=a_{11}-\lambda_{k},$ $k=2,3,\ldots,n.$ Then $\mathop{\displaystyle\sum}\limits_{k=1}^{n}a_{1k}=na_{11}-\mathop{\displaystyle\sum}\limits_{k=2}^{n}\lambda_{k}=\lambda_{1}$ and $a_{11}=\frac{1}{n}\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k}.$ Thus the $k^{th}$ row of $A,$ $k=2,\ldots,n,$ is a permutation of the first row and $A$ is an $n\times n$ permutative nonnegative matrix with spectrum $\Lambda.$ It is clear that if $\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k}>0$ and $a_{11}-\lambda_{k}>0$ then $A$ is positive.

Corollary 2.1

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a list of real numbers with $\lambda_{k}<0,$ $k=2,\ldots,n,$ and $\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k}\geq 0.$ Then $\Lambda$ is the spectrum of a permutative nonnegative (positive) matrix.

Proof. It is enough to take $a_{1k}=a_{11}-\lambda_{k},$ with $a_{11}=\frac{1}{n}\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k},$ $k=2,3,\ldots,n.$ Then $A$ in (1) is permutative nonnegative. If $\mathop{\displaystyle\sum}\limits_{k=1}^{n}\lambda_{k}>0$ then $A$ becomes positive.

Theorem 2.3

*Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a list of real numbers. Suppose that:

$i)$  There exist a partition $\Lambda=\Lambda_{0}\cup\underset{r\text{ times}}{\underbrace{\Lambda_{1}\cup\cdots\cup\Lambda_{1}}},$ where*

[TABLE]

*such that $\Gamma_{1}=\{\lambda,\lambda_{11},\lambda_{12},\ldots,\lambda_{1p}\},$ $0\leq\lambda\leq\lambda_{1},$ is the spectrum of a $(p+1)\times(p+1)$ permutative nonnegative matrix.

$ii)$ There exists an $r\times r$ permutative nonnegative matrix with spectrum $\Lambda_{0}$ and diagonal entries $\lambda,\lambda,\ldots,\lambda$ $(r$ times$).$

Then, there exists a permutative nonnegative matrix $P$ with spectrum $\Lambda.$*

Proof. From $i)$ let $P_{1}$ be a $(p+1)\times(p+1)$ permutative nonnegative matrix with spectrum $\Gamma_{1}$ and let

[TABLE]

with $r$ blocks $P_{1}.$ Then $P_{1}\mathbf{x}=\lambda\mathbf{x}$ with $\mathbf{x}=\frac{1}{\sqrt{p+1}}\mathbf{e}$ (that is $\left\|\mathbf{x}\right\|=1$), where $\lambda$ and $\mathbf{x}$ are the Perron eigenvalue and the Perron eigenvector of $P_{1}$, respectively.

From $ii)$ let $B$ be the $r\times r$ permutative nonnegative matrix with spectrum $\Lambda_{0}$ and diagonal entries $\lambda,\lambda,\ldots,\lambda$ $(r$ times$).$ Let $\Omega$ be the $r\times r$ diagonal matrix $\Omega=diag\{\lambda,\lambda,\ldots,\lambda\}.$ Then for

[TABLE]

where $X$ is the $r(p+1)\times r$ matrix of eigenvectors of $A,$ it follows that $XCX^{T}$ is a permutative nonnegative matrix, and from Theorem 1.3, with $C=CX^{T},$ $P=A+XCX^{T}$ is a permutative nonnegative matrix with spectrum $\Lambda.$ Observe that $P$ is also an $r\times r$ block permutative nonnegative matrix.

Example 2.1

Let $\Lambda=\{10,4,2,0,-1,-1,-1,-1,-3,-3,-3,-3\}.$ We take the partition

[TABLE]

Then we look for a permutative nonnegative matrix $P_{1}$ with spectrum $\Gamma_{1},$ and a permutative nonnegative matrix $B$ with spectrum $\Lambda_{0}$ and diagonal entries $4,4,4,4.$ These matrices are

[TABLE]

obtained from Theorem 1.2 and Lemma 2.1, respectively. Let

[TABLE]

Then for $C=B-diag\{4,4,4,4\}$ we have that

[TABLE]

is a permutative nonnegative matrix with spectrum $\Lambda.$ Observe that $P$ is also a $4\times 4$ block permutative nonnegative matrix with permutative blocks.

Remark 2.1

If in the proof of Theorem 2.3 the matrices $P_{1}$ and $B$ can be choosen as symmetric permutative nonnegative, then $A+XCX^{T}$ becomes symmetric permutative nonnegative. In fact, if $r=3,$ for instance, we have that

[TABLE]

is symmetric permutative nonnegative.

In particular, for $n=3,$ we have the following pattern of symmetric permutative nonnegative matrices $B:$

[TABLE]

with eigenvalues of the form $\lambda_{1},\lambda_{2},-\lambda_{2},$  and  $\lambda_{1},\lambda_{2},\lambda_{2}.$

Conditions for cases $ii)$ and $iii)$ are

[TABLE]

However, except for the case $iii),$ $C=B-diagB$ need not to be permutative. Consider the following example:

Example 2.2

Let $\Lambda=\{8,6,3,3,-5,-5,-5,-5\}$ with the partition

[TABLE]

We compute the matrices

[TABLE]

with spectrum $\Gamma_{1}$ and spectrum $\Lambda_{0}$ and diagonal entries $5,5,5,5,$ respectively. Then

[TABLE]

is symmetric permutative nonnegative with spectrum $\Lambda.$

We finish this section by given a negative answer the question in [5], that is, we show that there are lists of real numbers which are the spectrum of a nonnegative matrix, but not the spectrum of a permutative nonnegative matrix.

Lemma 2.2

There is no permutative nonnegative matrix with spectrum $\Lambda=\{6,5,1\}.$

Proof. It is clear that $\Lambda=\{6,5,1\}$ is realizable (trivially by $A=diag\{6,5,1\}$): For instance, the matrix

[TABLE]

Suppose $P$ is a permutative nonnegative matrix with spectrum $\Lambda$ and first row $\left(a,b,c\right).$ Then $a+b+c=6,$ $P\in\mathcal{CS}_{6},$ $tr(P)=12.$ We have three cases:

$i)$  The entries of the main diagonal of $P$ are of the form $x,x,x.$ Then $x=4$ and the Perfect necessary and sufficient conditions for the existence of a nonnegative matrix with prescibed spectrum $\Lambda$ and diagonal entries $4,4,4$ are not satisfied [7, Theorem $4,$ condition $(4)$]. Therefore, there is no permutative nonnegative matrix with spectrum $\Lambda.$

$ii)$ The entries of the main diagonal of $P$ are of the form $x,x,y.$ Then $2x+y=12,$ and from de Perfect conditions [7, Theorem $4,$ condition $(4)$], $x\geq 5,$ $y\leq 2$ or $y\geq 5,$ $x\leq\frac{7}{2},$ which contradicts $x+y+z=6,$ except for $x=6,$ $y=z=0.$ In this last case however, the conditions in [7, Theorem $4$] are not satisfied either.

$iii)$ The entries of the main diagonal of $P$ are of the form $x,y,z.$ Then $x+y+z=12$ contradicts $x+y+z=6.$

Thus, $\Lambda$ cannot be the spectrum of a permutative nonnegative matrix with spectrum $\Lambda.$

Observe, however that $\Lambda=\{6,5,1\}$ is the spectrum of the direct sum of permutative nonnegative matrices

[TABLE]

After this paper was submitted, R. Loewy [4] showed that a realizable list of real numbers need not to be the spectrum of a permutative nonnegative matrix nor the spectrum of a direct sum of permutative nonnegative matrices.

3 Permutative matrices with precribed complex spectrum

In this section we show that certain lists of complex numbers are realizable by permutative nonnegative matrices. First we recall the result of Loewy and London [3], which solves the NIEP for $n=3:$

Theorem 3.1

Let $\Lambda=\{\lambda_{1},\lambda_{2},\lambda_{3}\}$ be a list of complex numbers. Then $\Lambda$ is the spectrum of a nonnegative matrix if and only if

[TABLE]

Then we have the following

Corollary 3.1

Every realizable list $\Lambda=\{\lambda_{1},a+bi,a-bi\}$ of complex numbers is in particular realizable by a permutative nonnegative matrix.

Proof. The realizing matrix in the proof of Theorem 3.1 is circulant. Since circulant matrices are permutative the result follows.

Corollary 3.2

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots\lambda_{n}\}$ be a list of complex numbers. If there exists a partition $\Lambda=\Lambda_{1}\cup\cdots\cup\Lambda_{t},$ where $\Lambda_{i}=\{\lambda_{i1},\lambda_{i2},\lambda_{i3}\},$ $i=1,2,\ldots,t,$ satisfies conditions of Theorem 3.1, then $\Lambda$ is the spectrum of a direct sum of permutative nonnegative matrices with spectrum $\Lambda.$

Proof. The proof is immediate from Corollary 3.1.

Remark 3.1

Theorem 2.3 can also be applied to a list of complex numbers, as the following example shows:

Example 3.1

Let $\Lambda=\{3,2,1,-1\pm i,-1\pm i,-1\pm i\}$ with

[TABLE]

The matrices

[TABLE]

are permutative with spectrum $\Gamma_{1}$ and $\Lambda_{0},$ respectively. Moreover $B$ has the required diagonal entries. Then

[TABLE]

is permutative nonnegative with spectrum $\Lambda.$

Next we recall that an $n\times n$ circulant matrix is a matrix of the form

[TABLE]

and it is uniquely determined by the entries of its first row, which we denoted by $\mathbf{c}=\left(c_{0},c_{1},\ldots,c_{n-1}\right).$ It is clear that $C$ is also permutative. Let $\mathbf{\lambda}=\left(\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\right)$ with

[TABLE]

being the eigenvalues of the circulant matrix $C=circ\left(c_{0},c_{1},\ldots,c_{n-1}\right).$ Then

[TABLE]

Let

[TABLE]

Then

[TABLE]

The following result shows that a list of complex Suleimanova type, with the property (8), is realizable by a permutative nonnegative matrix.

Theorem 3.2

Let $\Lambda=\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\}$ be a list of complex numbers with

[TABLE]

satisfying $\lambda_{n-j+2}=\overline{\lambda_{j}},$  $j=2,3,\ldots,\left[\frac{n+1}{2}\right].$ Then $\Lambda$ is the spectrum of a permutative nonnegative matrix if and only if $\mathop{\displaystyle\sum}\limits_{j=1}^{n}\lambda_{j}\geq 0.$

Proof. The condition is necessary. Suppose $\mathop{\displaystyle\sum}\limits_{j=1}^{n}\lambda_{j}\geq 0.$ Let $\mu=-\mathop{\displaystyle\sum}\limits_{j=2}^{n}\lambda_{j}.$ Then the list $\Lambda_{\mu}=\{\mu,\lambda_{2},\ldots,\lambda_{n}\}$ is realizable. From (9) we have that

[TABLE]

$k=0,1,\ldots,2m$  if $n=2m+1,$ and

[TABLE]

$k=0,1,\ldots,2m+1$  if $n=2m+2.$ Let $n=2m+1.$ Then

[TABLE]

Since $\mathop{\mathrm{R}e}\lambda_{j}\leq 0,$ then ${\LARGE(}\cos\frac{2k(j-1)\pi}{2m+1}-1{\LARGE)}\mathop{\mathrm{R}e}\lambda_{j}\geq 0,$ $j=2,\ldots,m+1.$ Moreover Im$\lambda_{j}\sin\frac{2k(j-1)\pi}{2m+1}\geq 0,$ $j=2,\ldots,m+1,$ $k=0,1,\ldots,2m.$ Then

[TABLE]

and $C=circ\left(c_{0},c_{1},\ldots,c_{n-1}\right)$ is a circulant nonnegative matrix with spectrum $\Lambda_{\mu}$, which is also permutative nonnegative. The proof is similar for $n=2m+2.$ If $\mathop{\displaystyle\sum}\limits_{j=1}^{n}\lambda_{j}=\alpha>0,$ then $\alpha=\lambda_{1}-\mu,$ and the matrix $C^{\prime}=C+\frac{\alpha}{n}\mathbf{ee}^{T}$ is circulant nonnegative (permutative nonnegative) with spectrum $\Lambda.$