# Indices of diagonalizable and universal realizability of spectra††thanks:
Supported by Universidad Católica del Norte-VRIDT 036-2020, NUCLEO UCN
VRIDT-083-2020, Chile.

Charles R. Johnson${}^{a},$ Ana I. Juliob, Ricardo L. Sotob
aDepartment of Mathematics, College of William and Mary
Williamsburg, VA, USA
bDepartamento de Matemáticas, Universidad Católica del Norte
Antofagasta, Chile.
Casilla 1280, Antofagasta, Chile. Corresponding author<EMAIL_ADDRESS>(Charles
R. Johnson<EMAIL_ADDRESS>(Ana I. Julio<EMAIL_ADDRESS>(Ricardo L. Soto)

###### Abstract

A list $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ of complex numbers
(repeats allowed) is said to be realizable if it is the spectrum of an
entrywise nonnegative matrix $A$. $\Lambda$ is diagonalizably realizable if
the realizing matrix $A$ is diagonalizable. $\Lambda$ is said to be
universally realizable if it is  realizable for each possible Jordan canonical
form allowed by $\Lambda.$ Here, we study the connection between
diagonalizable realizability and universal realizability of spectra. In
particular, we establish  indices of realizability for diagonalizable and
universal realizability. We also define the merge of two spectra and we prove
a result that allow us to easily decide, in many cases, about the universal
realizability of spectra.

AMS classification: 15A18, 15A20, 15A29

Key words: Nonnegative matrices, Spectra diagonalizably realizable, Spectra
universal realizability, Jordan structure, Eigenvalues and eigenvectors

##  1 Introduction

A list $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ of complex numbers (with
repeats allowed) is said to be realizable if there is an $n$-by-$n$
nonnegative matrix $A$ with spectrum $\Lambda$. In this case $A$ is said to be
a realizing matrix. The problem of characterizing the realizability of lists
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ of complex numbers is called
the nonnegative inverse eigenvalue problem (NIEP). We say that $\Lambda$ is
diagonalizably realizable (DR) if it is the spectrum of a diagonalizable
nonnegative matrix. The problem of determining this kind of realizability is
called the diagonalizable realizability problem. $\Lambda$ is universally
realizable (UR) if it is realizable for every possible Jordan canonical form
(JCF) allowed by $\Lambda.$ The problem of determining the universal
realizability of spectra is called the universal realizability problem (URP).
The URP seeks to extend the question of determining the spectral properties
allowed by a nonnegative matrix, not only regarding the eigenvalues
themselves, but also from the point of view of the corresponding JCF. The URP
contains the NIEP and both problems are equivalent if the given complex
numbers $\lambda_{1},\ldots,\lambda_{n}$ are distinct. The NIEP has attracted
the interest of many linear algebraist researchers. The URP, on the other
hand, is a new and even more difficult problem. Both problems remain unsolved
for $n\geq 5.$ A complete solution, if any, is still far away from the present
state of the art. The URP is studied module the NIEP (see [15]). This means
that the methods applied to the NIEP are not only useful but also in many
cases necessary for the URP. Then, getting answers for some topics and open
questions about the URP means a positive impact on the progress towards a
solution.

The NIEP, as we know it today, begins with the works by Suleĭmanova [24] and
Perfect [17, 18]. The NIEP has been solved only for the cases $n=3$ by Loewy
and London [13], and for $n=4$ by Meehan [14] and also independently by Torre-
Mayo et al. [25]. The first known results on the URP are due to Minc [15, 16].
In [15] Minc proved that if $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ is
the spectrum of a diagonalizable positive matrix, then $\Lambda$ is UR. We
want to point out here that previously the URP was called nonnegative inverse
elementary divisors problem [15], and that in [4] the authors used for the
first time the concept and name universal realizability. There are spectra,
not positively realizable, that are known to be UR [3, 22, 23]. For instance,
spectra in the left half-plane, that is,
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ with $\lambda_{1}>0,$
$\mbox{Re}\lambda_{i}\leq 0,$ $i=2,\ldots,n,$ such as real Suleĭmanova spectra
[22] $\lambda_{1}>0>\lambda_{2}\geq\cdots\geq\lambda_{n}$; complex Suleĭmanova
spectra [23]

$\lambda_{1}>0,\text{ }\lambda_{i}\in\left\\{z\in\mathbb{C}:\mbox{Re}z\leq
0,\text{ }\left|\mbox{Re}z\right|\geq\left|\mbox{Im}z\right|\right\\},\text{
}i=2,\ldots,n;$

and Šmigoc spectra [3]

$\lambda_{1}>0,\text{ }\lambda_{i}\in\left\\{z\in\mathbb{C}:\mbox{Re}z\leq
0,\text{
}\left|\sqrt{3}\mbox{Re}z\right|\geq\left|\mbox{Im}z\right|\right\\},\text{
}i=2,\ldots,n,$

which contains the real and complex Suleimanova spectra. These spectra are
realizable if and only if they are UR if and only if
$\sum_{i=1}^{n}\lambda_{i}\geq 0.$ The good behavior of these kind of spectra
led to think that any left half-plane list was UR. Now, we know that this is
not true [9, 10].

In [12] Laffey and Šmigoc proved that a list
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ in the left half-plane is
realizable if and only if

$s_{1}=\sum_{i=1}^{n}\lambda_{i}\geq 0,\text{ \
}s_{2}=\sum_{i=1}^{n}\lambda_{i}^{2}\geq 0,\text{ \ }s_{1}^{2}\leq ns_{2}.$

The positivity and diagonalizability conditions in the result of Minc [15] are
essential for his proof. Minc says that “it is not known if the the theorem
holds for a general nonnegative matrix”, question that has been open for
almost 40 years. Recently, two extensions for nonnegative realizations have
been obtained: the first one by Collao, Salas, and Soto [2] shows that if
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ is the spectrum of a
diagonalizable nonnegative matrix with constant row sums and a positive row or
column, then $\Lambda$ is UR. The second extension by Johnson, Julio, and Soto
[7], shows that if $\Lambda$ is realizable by a diagonalizable ODP matrix,
that is, a diagonalizable nonnegative matrix having all off-diagonal entries
being positive (zeros on diagonal are permitted), then $\Lambda$ is also UR.
Observe that in both extensions, the set of realizing matrices contains the
set of positive realizing matrices. Moreover, the extension in [7] allows to
decide about the universal realizability of lists with
$\sum\limits_{i=1}^{n}\lambda_{i}=0$, which it is not possible from the Minc’s
result. These two extensions open a research line that allow to prove the
universal realizability of non-positively realizable spectra, thus
significantly extending the class of spectra that can be shown to be UR. Since
DR is a necessary condition for UR, all lists of complex numbers that are DR,
are natural candidates to be UR.

Despite the progress that has been made on the URP, there are still numerous
open questions. Two of them, until recently open, were: Is any left half-plane
list UR? In [10] the authors show that a realizable left half-plane list is
not necessarily UR. In fact, they prove that the spectrum

$\Lambda=\left\\{a,-\frac{a}{4}+\frac{\sqrt{5}a}{4}i,-\frac{a}{4}-\frac{\sqrt{5}a}{4}i,-\frac{a}{4}+\frac{\sqrt{5}a}{4}i,-\frac{a}{4}-\frac{\sqrt{5}a}{4}i\right\\}\text{
}a>0,$ (1)

is realizable, but not DR and therefore not UR The other open question was
whether DR realizable implies UR. In [9] the authors show that the spectra
$\\{\lambda_{1},\lambda_{1},\lambda_{2},\lambda_{2}\\}$ with
$\lambda_{1}>0>\lambda_{2}\geq-\lambda_{1},$ $\lambda_{1}+2\lambda_{2}<0,$
have no realization with a JCF having a Jordan block $J_{2}(\lambda_{2})$ of
size $2.$ Thus, for instance, $\Lambda=\\{1,1,-1,-1\\}$ is not UR although it
is DR. Since $\Lambda=\\{1,1,-1,-1\\}$ has a reducible realization, what may
be said about irreducible realizations?. In [10] it was also shown that
irreducible diagonalizable realizations are not necessarily UR. Then, it
remains to know under what conditions a DR left half-plane list of complex
numbers is UR. From extensions in [2, 7] we may say that DR implies UR if
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ is the spectrum of a
diagonalizable nonnegative matrix with constant row sums and a positive row or
column, or it is diagonalizably ODP realizable. The importance of the diagonal
JCF lies in the fact that we know how to join Jordan blocks to obtain larger
Jordan blocks. Then if we may obtain a diagonalizable realizing matrix for
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\},$ we may also obtain a
nonnegative matrix with spectrum $\Lambda$ for each possible JCF allowed by
$\Lambda.$ What about criteria which allow to decide about the universal
realizability of a list of complex numbers? In this work we also introduce an
operation that we name the merge of two spectra and a result which allow to
easily decide, in many cases, about the universal realizability of spectra. We
also establish indexes of Guo type [5] for diagonalizable and universal
realizability.

A matrix $A=[a_{ij}]$ of order $n$ is said to have constant row sums if all
its rows sum to the same constant $\alpha$. We denote by
$\mathcal{CS}_{\alpha}$ the set of all $n$-by-$n$ real matrices with constant
row sums equal to $\alpha$. It is clear that any matrix in
$\mathcal{CS}_{\alpha}$ has an eigenvector
$\mathbf{e}^{\textsuperscript{T}}=[1,\ldots,1]$ corresponding to the
eigenvalue $\alpha$. The relevance of the real matrices with constant row sums
is due to the fact that, the problem of finding a nonnegative matrix with
spectrum $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$, $\lambda_{1}$ being
the Perron eigenvalue, is equivalent to the problem of finding a nonnegative
matrix in $\mathcal{CS}_{\lambda_{1}},$ with spectrum $\Lambda$ (see [6]). We
denote by $E_{i,j}$ the matrix with $1$ in position $(i,j)$ and zeros
elsewhere and we define the matrix

$E(K)=\sum\limits_{i\in K}E_{i,i+1},\text{ \ }K\subset\\{1,2,\ldots,n\\}.$ (2)

The paper is organized as follows: In Section $2$, we introduce the
diagonalizable realizability index $g_{d}(\Lambda/\lambda_{1})$ and the
universal realizability index $g_{u}(\Lambda/\lambda_{1}).$ Then, we show that
a realizable list $\Lambda=\\{\mu,\lambda_{2},\ldots,\lambda_{n}\\}$ of
complex numbers is DR for all $\mu\geq g_{d}(\Lambda/\lambda_{1})$ and that
$\Lambda$ is UR for all $\mu\geq g_{u}(\Lambda/\lambda_{1}).$ In Section $3$,
we define the merge of two spectra and show that if
$\Gamma_{1}=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ and
$\Gamma_{2}=\\{\mu_{1},\ldots,\mu_{m}\\}$ have a diagonalizable ODP
realization, then
$\Gamma=\\{\lambda_{1}+\mu_{1},\lambda_{2},\ldots,\lambda_{n},\mu_{2},\ldots,\mu_{m}\\}$
has also a diagonalizable ODP realization and therefore $\Gamma$ is UR. This
result becomes a useful criterion to decide, in many cases, the universal
realizability of spectra.

##  2 Diagonalizable and universal realizability indices.

In what follows we will use the following results, Theorems 2.1 to 2.3 below.
Theorem 2.1, due to Brauer [1], is a perturbation result that shows how to
change one single eigenvalue of an $n$-by-$n$ matrix without changing any of
the remaining $(n-1)$ eigenvalues. Theorem 2.2, by Soto and Ccapa [22],
establishes the JCF of the Brauer perturbation
$A+\mathbf{eq}^{\textsuperscript{T}}.$ Theorem 2.3, by Šmigoc [20], shows how
to construct a matrix $C$ with a particular JCF from two given matrices $A$
and $B.$

###### Theorem 2.1

[1] Brauer. Let $A$ be an $n$-by-$n$ matrix with spectrum
$\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$. Let
$\mathbf{v}^{\textsuperscript{T}}=[v_{1},\ldots,v_{n}]$ be an eigenvector of
$A$ associated with the eigenvalue $\lambda_{k}$ and let $\mathbf{q}$ be any
$n$-dimensional vector. Then the matrix $A+\mathbf{vq}^{\textsuperscript{T}}$
has eigenvalues
$\lambda_{1},\ldots,\lambda_{k-1},\lambda_{k}+\mathbf{v}^{\textsuperscript{T}}\mathbf{q},\lambda_{k+1},\ldots,\lambda_{n}$.

###### Theorem 2.2

[22] Soto and Ccapa. Let
$\mathbf{q}^{\textsuperscript{T}}=[q_{1},\ldots,q_{n}]$ be an arbitrary
$n-$dimensional vector. Let $A\in\mathcal{CS}_{\lambda_{1}}$ with JCF

$J(A)=S^{-1}AS=diag\left\\{J_{1}(\lambda_{1}),J_{n_{2}}(\lambda_{2}),\ldots,J_{n_{k}}(\lambda_{k})\right\\}.$

If $\lambda_{1}+\sum_{i=1}^{n}q_{i}\neq\lambda_{i},$ $i=2,\ldots,n$, then the
matrix $A+\mathbf{eq}^{\textsuperscript{T}}$ has Jordan canonical form
$J(A)+\left(\sum_{i=1}^{n}q_{i}\right)E_{11}$. In particular, if
$\sum_{i=1}^{n}q_{i}=0$ then $A$ and $A+\mathbf{eq}^{\textsuperscript{T}}$ are
similar.

###### Theorem 2.3

[20] Šmigoc. Suppose $B$ is an $m$-by-$m$ matrix with JCF that contains at
least one $1$-by-$1$ Jordan block corresponding to the eigenvalue $c$:

$J(B)=\left[\begin{array}[]{cc}c&0\\\ 0&I(B)\end{array}\right].$

Let $\mathbf{t}$ and $\mathbf{s}$, respectively, be the left and the right
eigenvectors of $B$ associated with the $1$-by-$1$ Jordan block in the above
canonical form. Furthermore, we normalize vectors $\mathbf{t}$ and
$\mathbf{s}$ so that $\mathbf{t}^{\textsuperscript{T}}\mathbf{s}=1.$ Let
$J(A)$ be a JCF for an $n$-by-$n$ matrix

$A=\left[\begin{array}[]{cc}A_{1}&\mathbf{a}\\\
\mathbf{b^{\textsuperscript{T}}}&c\end{array}\right],$

where $A_{1}$ is an $(n-1)$-by-$(n-1)$ matrix and $\mathbf{a}$ and
$\mathbf{b}$ are vectors in $\mathbb{C}^{\textsuperscript{n-1}}.$ Then the
matrix

$C=\left[\begin{array}[]{cc}A_{1}&\mathbf{at}^{\textsuperscript{T}}\\\
\mathbf{sb}^{\textsuperscript{T}}&B\end{array}\right]$

has JCF

$J(C)=\left[\begin{array}[]{cc}J(A)&0\\\ 0&I(B)\end{array}\right].$

Let $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ and
$\Lambda/\lambda_{1}=\\{\lambda_{2},\ldots,\lambda_{n}\\}$ be self-conjugate
lists of complex numbers. Guo [5] proved that there is a minimal nonnegative
number $g_{r}(\Lambda/\lambda_{1})$ such that
$\Lambda_{\mu}=\\{\mu,\lambda_{2},\ldots,\lambda_{n}\\}$ is realizable for all
$\mu\geq g_{r}(\Lambda/\lambda_{1}).$ Moreover, Guo established that

$\max_{2\leq j\leq n}\left|\lambda_{j}\right|\leq
g_{r}(\Lambda/\lambda_{1})\leq 2n\max_{2\leq j\leq
n}\left|\lambda_{j}\right|.$ (3)

In [11] it was shown that the upper bound in (3) may be reduced to
$(n-1)\underset{2\leq j\leq n}{\max}\left|\lambda_{j}\right|,$ with $n\geq 5$
in the case when $\lambda_{k},$ $k=2,\ldots,n,$ are conjugates complex, and
that this bound is sharp. In [19] the authors show how to calculate the Guo
index $g_{r}(\Lambda/\lambda_{1})$ for circulant nonnegative matrices.
However, to compute this index becomes a prohibitive task for large $n.$ In
this section, we prove that there is also a minimal nonnegative number
$g_{d}(\Lambda/\lambda_{1})$, called diagonalizable realizability index, such
that $\Lambda_{\mu}=\\{\mu,\lambda_{2},\ldots,\lambda_{n}\\}$ is DR if
$\mu\geq g_{d}(\Lambda/\lambda_{1})$. Of course
$g_{d}(\Lambda/\lambda_{1})\geq g_{r}(\Lambda/\lambda_{1}).$ Equality occurs
if $\Lambda/\lambda_{1}$ has distinct elements and may occur otherwise. The
proof is similar to the proof in [11], but it considers all possible cases,
which it does not occur in [11].

###### Theorem 2.4

If $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ is realizable, then there is
a nonnegative real number $g_{d}(\Lambda/\lambda_{1})$ such that
$\Lambda_{\mu}=\\{\mu,\lambda_{2},\ldots,\lambda_{n}\\}$ is DR for every
$\mu\geq g_{d}(\Lambda/\lambda_{1}).$ Moreover

$g_{r}(\Lambda/\lambda_{1})\leq g_{d}(\Lambda/\lambda_{1})\leq(n-1)\max_{2\leq
j\leq n}\left|\lambda_{j}\right|.$ (4)

Proof. First, we exhibit a value $\mu$ such that $\Lambda_{\mu}$ is DR. Let
$A$ be a realizing matrix for $\Lambda.$ Without loss of generality we assume
that $A\in\mathcal{CS}_{\lambda_{1}}$, that is,
$A\mathbf{e}=\lambda_{1}\mathbf{e}$. If $A$ is diagonalizable we are done. If
not, we take $A=SJS^{-1},$ where $J$ is the JCF of $A$ and
$S\mathbf{e}_{1}=\mathbf{e}$. Now let $\widetilde{J}$ be the same as $J,$
except that any superdiagonal $1^{\prime}s$ are replaced with $0{\acute{}}s.$
So, $\widetilde{J}$ is diagonal with spectrum $\Lambda.$ Define
$\widetilde{A}=S\widetilde{J}S^{-1}.$ If $\widetilde{A}$ is nonnegative we are
done. If not, since $\widetilde{A}\in\mathcal{CS}_{\lambda_{1}}$ we apply
Brauer’s Theorem to produce a nonnegative matrix
$A^{\prime}=\widetilde{A}+\mathbf{eq}^{\textsuperscript{T}}$, where
$\mathbf{q}^{\textsuperscript{T}}=[q_{1},\ldots,q_{n}]$ is an appropriate
nonnegative vector. From Theorem 2.2 $A^{\prime}$ is diagonalizable with
spectrum
$\left\\{\lambda_{1}+\sum\limits_{i=1}^{n}q_{i},\lambda_{2},\ldots,\lambda_{n}\right\\}.$
Let $g_{d}(\Lambda/\lambda_{1})=\lambda_{1}+\sum\limits_{i=1}^{n}q_{i}$. Thus
we have established the existence of a value $g_{d}(\Lambda/\lambda_{1})$ such
that $\Lambda_{\mu}$ is DR for every $\mu\geq g_{d}(\Lambda/\lambda_{1}).$
Next, we show that $g_{d}(\Lambda/\lambda_{1})$ satisfies (4). The lower bound
is clear. Although similar to the proof of Theorem $3.2$ in [11], this is more
complete and consider all cases. In particular, the case in which $\mu_{j},$
$j=2,\ldots,n,$ are all conjugate complex. For the sake of completeness and
since the result is of independent interest, we set the proof here. Let
$m=\underset{2\leq j\leq n}{\max}\left|\lambda_{j}\right|$ and let
$\mu_{j}=\frac{\lambda_{j}}{(n-1)m},$ $j=2,\ldots,n.$ Then,
$\\{\mu_{2},\ldots,\mu_{n}\\}$ is a list of complex numbers such that $\
\left|\mu_{j}\right|\leq\frac{1}{n-1},$ $j=2,\ldots,n.$ Consider the initial
matrix

$B=\begin{bmatrix}0&0&0&\cdots&\cdots&\cdots&\cdots&\cdots&0\\\
-\mu_{2}&\mu_{2}&\ddots&&&&&&\vdots\\\
\vdots&\ddots&\ddots&\ddots&&&&&\vdots\\\
-\mu_{p}&\vdots&\ddots&\mu_{p}&0&&&&\vdots\\\
-x_{s}&y_{s}&&\ddots&x_{s}&-y_{s}&&&\vdots\\\
-x_{s}&-y_{s}&&&y_{s}&x_{s}&\ddots&&\vdots\\\
\vdots&\vdots&&&&\ddots&\ddots&\ddots&0\\\
-x_{t}&y_{t}&&&&&\ddots&x_{t}&-y_{t}\\\
-x_{t}&-y_{t}&\cdots&\cdots&\cdots&\cdots&0&y_{t}&x_{t}\end{bmatrix},$

where $\mu_{2},\ldots,\mu_{p}$ are real, $x_{j}=\mbox{Re}\mu_{j}$,
$y_{j}=\mbox{Im}\mu_{j},$ $p+1\leq j\leq\frac{n+p}{2}.$ Then
$B\in\mathcal{CS}_{0}$ has eigenvalues
$0,\mu_{2},\ldots,\mu_{p},\mu_{p+1},\ldots,\mu_{n}$ and is clear that $B$ is
diagonalizable.

  * •

If $\mbox{Re}\mu_{j}\leq 0,$ $j=2,\ldots,n,$ then all entries in the first
column of $B$ are nonnegative. Let

$\mathbf{q}^{\textsuperscript{T}}=\left[0,\frac{1}{n-1},\ldots,\frac{1}{n-1}\right].$

From Theorem 2.1 and Theorem 2.2, $A^{\prime}=B+\mathbf{\
eq}^{\textsuperscript{T}}$ is diagonalizable nonnegative with spectrum
$\\{1,\mu_{2},\ldots,\mu_{n}\\}$ and $A=(n-1)mA^{\prime}$ is diagonalizable
nonnegative with spectrum
$\left\\{(n-1)m,\lambda_{2},\ldots,\lambda_{n}\right\\}$.

  * •

If $\mbox{Re}\mu_{j}>0$ for some $j,$ $3\leq j\leq n$, then all the entries in
the $j^{\textsuperscript{st}}$ column of $B$ (or in the
$(j-1)^{\textsuperscript{st}}$ column of $B$ if $j$ corresponds to the second
column in the corresponding $2$-by-$2$ complex block) are nonnegative. Let

$\mathbf{q}^{\textsuperscript{T}}=\left[\frac{1}{n-1},\ldots,\frac{1}{n-1},0,\frac{1}{n-1},\ldots,\frac{1}{n-1}\right],$

with zero in the $j^{\textsuperscript{st}}$ position
($(j-1)^{\textsuperscript{st}}$ position). Then, again,
$A^{\prime}=B+\mathbf{eq}^{\textsuperscript{T}}$ is diagonalizable nonnegative
with spectrum $\\{1,\mu_{2},\ldots,\mu_{n}\\}$ and $A=(n-1)mA^{\prime}$ is
diagonalizable nonnegative with spectrum
$\\{(n-1)m,\lambda_{2},\ldots,\lambda_{n}\\}$. Observe that, from the
necessary and sufficient conditions by Loewy and London [13], the result still
holds for the special case $n=3$ with $\Lambda=\\{\lambda_{1},a+bi,a-bi\\}.$

  * •

If $\mu_{2}>0$ with $\mbox{Re}\mu_{j}<0,$ $j=3,\ldots,n,$ then we write the
$-\mbox{Re}\mu_{j}^{\prime}s,$ $3\leq j\leq n,$ along the second column of $B$
and the $\pm\mbox{Im}\mu_{j}^{\prime}s,$ $p+1\leq j\leq n,$ along the first
column of $B.$ Then again with

$\mathbf{q}^{\textsuperscript{T}}=\left[\frac{1}{n-1},0,\frac{1}{n-1},\ldots,\frac{1}{n-1}\right],$

we obtain, as before, the diagonalizable nonnegative matrix
$A=(n-1)mA^{\prime}$ with the required spectrum.

  * •

Now we consider the case in which $\mu_{j},$ $j=2,\ldots,n,$ are all conjugate
complex numbers, that is, $\mbox{Im}\mu_{j}\neq 0.$ Let the $2$-by-$2$
diagonal blocks

$\left[\begin{array}[]{cc}\mbox{Re}\mu_{j}&-\mbox{Im}\mu_{j}\\\
\mbox{Im}\mu_{j}&\mbox{Re}\mu_{j}\end{array}\right],\text{ \ }j=2k,\text{
}k=1,2,\ldots,\frac{n-1}{2}$

in the matrix $B.$ In this case we only can use the first column of $B$ to set
$-(\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j})$ in order that $B$
$\in\mathcal{CS}_{0}.$ Now we have the following cases:

    * –

If $\mbox{Re}\mu_{j}\geq 0$ for one or more indexes $j,$ then the
corresponding $j^{\textsuperscript{st}}$ columns in $B$ are nonnegative and
the proof follows as before. If
$\left|\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j}\right|>\frac{1}{n-1}$ for some $j$,
then we set the corresponding diagonal block

$\left[\begin{array}[]{cc}\mbox{Re}\mu_{j}&-\mbox{Im}\mu_{j}\\\
\mbox{Im}\mu_{j}&\mbox{Re}\mu_{j}\end{array}\right]$

on the last $2$-by-$2$ diagonal position in the matrix $B$ to distribute any
possible negative amount (from the reciprocal of
$\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j}$) through the last row under some
appropriate columns. Observe that there is at least one negative amount
($-\mbox{Im}\mu_{j}$) in each above $2$-by-$2$ block in $B.$ Thus,
$A^{\prime}$ is nonnegative with spectrum
$\left\\{\sum\limits_{i=1}^{n}q_{i},\mu_{2},\ldots,\mu_{n}\right\\},$
$\sum\limits_{i=1}^{n}q_{i}\leq 1$ and $A=(n-1)mA^{\prime}$ is nonnegative
with spectrum
$\left\\{(n-1)m\sum\limits_{i=1}^{n}q_{i},\lambda_{2},\ldots,\lambda_{n}\right\\},$
where $(n-1)m\sum\limits_{i=1}^{n}q_{i}\leq(n-1)m.$ In fact, for $n=5$ (the
minimum case) with $\mbox{Re}\mu_{i}\geq 0,$ we have

$\left[\begin{array}[]{ccccc}0&0&0&0&0\\\
-\mbox{Re}\mu_{i}+\mbox{Im}\mu_{i}&\mbox{Re}\mu_{i}&-\mbox{Im}\mu_{i}&0&0\\\
-\mbox{Re}\mu_{i}-\mbox{Im}\mu_{i}&\mbox{Im}\mu_{i}&\mbox{Re}\mu_{i}&0&0\\\
-\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j}&0&0&\mbox{Re}\mu_{j}&-\mbox{Im}\mu_{j}\\\
-\mbox{Re}\mu_{j}-\mbox{Im}\mu_{j}+\mbox{Im}\mu_{i}&0&-\mbox{Im}\mu_{i}&\mbox{Im}\mu_{j}&\mbox{Re}\mu_{j}\end{array}\right]$

Then

$\mathbf{q}^{T}={\Large[}\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j},0,\mbox{Im}\mu_{i},0,\mbox{Im}\mu_{j}{\Large]}$

and $\sum\limits_{i=1}^{5}q_{i}=\mbox{Re}\mu_{j}+2\mbox{Im}\mu_{j}+\mbox{
Im}\mu_{i}\leq\frac{4}{n-1}\leq 1.$ As it was said above, the validity of the
case $n=3$ is guaranteed from the necessary and sufficient conditions by Loewy
and London [13].

    * –

If $\mbox{Re}\mu_{j}<0$ with
$\left|\mbox{Re}\mu_{j}\right|\geq\left|\mbox{Im}\mu_{j}\right|,$
$j=2,\ldots,n,$ then the first column in $B$ is nonnegative and the proof
follows as before.

    * –

If $\mbox{Re}\mu_{j}<0$ with
$\left|\mbox{Re}\mu_{j}\right|<\left|\mbox{Im}\mu_{j}\right|$, $j=2,\ldots,n,$
then

$\left|\mbox{Re}\mu_{j}+\mbox{Im}\mu_{j}\right|<\left|\mbox{
Im}\mu_{j}\right|<\left|\mu_{j}\right|\leq\frac{1}{n-1},$

and the proof follows as before again.

The upper bound in (4) is also sharp for $g_{d}(\Lambda/\lambda_{1}),$ as it
is shown by a diagonalizable realization of the list

$\Lambda=\\{(n-1),\underset{(n-1)\text{times}}{\underbrace{-1,\ldots,-1}}\\}.$

The spectrum $\Lambda=\cup_{i=1}^{\frac{n}{2}}\\{\lambda_{i},-\lambda_{i}\\},$
with real $\lambda_{i},$ shows that the lower bound in (4) is also sharp for
$g_{d}(\Lambda/\lambda_{1}).$ We may also define, similar to
$g_{r}(\Lambda/\lambda_{1})$ and $g_{d}(\Lambda/\lambda_{1}),$ a universal
realizability index $g_{u}(\Lambda/\lambda_{1}).$ Then, the following result
is clear.

###### Corollary 2.1

From Theorem 2.4 we have that

$g_{d}(\Lambda/\lambda_{1})\leq g_{u}(\Lambda/\lambda_{1})\leq(n-1)\max_{2\leq
j\leq n}\left|\lambda_{j}\right|\text{.}$ (5)

Proof. The lower bound is clear. To prove the upper bound we consider the
matrix $A=(n-1)mA^{\prime},$ with
$A^{\prime}=(B+\mathbf{eq}^{T})\in\mathcal{CS}_{1}$, in the proof of Theorem
2.4, which is diagonalizable nonnegative with JCF $J(A)=S^{-1}AS$, where
$S\mathbf{e}_{1}=\mathbf{e}$. Then, for an appropriate matrix $E(K)$ as
defined in (2),

$J(A)+E(K)=S^{-1}AS+E(K)=S^{-1}(A+SE(K)S^{-1})S.$ (6)

Then we may reach any JCF allowed by the spectrum of $A.$ In fact,
$A+SE(K)S^{-1}$ has the desired JCF, $J(A)+E(K)$, although it is no
necessarily nonnegative. Since $SE(K)S^{-1}\in\mathcal{CS}_{0}$, then for a
convenient nonnegative vector
$\mathbf{r}^{\textsuperscript{T}}=[r_{1},\ldots,r_{n}]$ and $\epsilon>0$ small
enough, the matrix $M=\left(A+\epsilon SE(K)S^{-1}\right)+\mathbf{\
er}^{\textsuperscript{T}}$ is nonnegative with spectrum
$\\{(n-1)m,\lambda_{2},\ldots,\lambda_{n}\\}$ or spectrum
$\\{(n-1)m\sum\limits_{i=1}^{n}q_{i},\lambda_{2},\ldots,\lambda_{n}\\},$ with
$\sum\limits_{i=1}^{n}q_{i}\leq 1$ in the case $\mu_{j},$ $j=2,\ldots,n,$ are
conjugate complex numbers. Then, since
$(n-1)m\sum\limits_{i=1}^{n}q_{i}\leq(n-1)m,$ $\Lambda$ is UR and
$g_{u}(\Lambda/\lambda_{1})\leq(n-1)\underset{2\leq j\leq
n}{\max\left|\lambda_{j}\right|.}$

The following example illustrates Theorem 2.4 and Corollary 2.1. In particular
the case where $\mu_{j},$ $j=2,\ldots,n,$ are all conjugates complex numbers.

###### Example 2.1

Consider the list

${\huge\\{}0{\huge,}1+i\sqrt{15},1-i\sqrt{15},-\sqrt{15}+i,-\sqrt{15}-i{\huge\\}.}$

Then $n=5,$ $m=\underset{2\leq j\leq 5}{\max}\left|\lambda_{j}\right|=4$ and
$(n-1)m=16.$ In order to have $\sum\limits_{i=1}^{5}q_{i}\leq 1$ we need to
take the initial matrix $B$ as

$B=\frac{1}{16}\left[\begin{array}[]{ccccc}0&0&0&0&0\\\
1+\sqrt{15}&-\sqrt{15}&-1&0&0\\\ -1+\sqrt{15}&1&-\sqrt{15}&0&0\\\
-1+\sqrt{15}&0&0&1&-\sqrt{15}\\\
0&-1&-\sqrt{15}&\sqrt{15}&1\end{array}\right].$

Then, $A^{\prime}=B+\mathbf{eq}^{T}$ with

$\mathbf{q}^{T}=\frac{1}{16}\left[1,\sqrt{15},\sqrt{15},0,\sqrt{15}\right]$

has the spectrum

$\frac{1}{16}{\LARGE\\{}\sum\limits_{i=1}^{5}q_{i},1+i\sqrt{15},1-i\sqrt{15},-\sqrt{15}+i,-\sqrt{15}-i{\LARGE\\}}{\huge,}$

with $\sum\limits_{i=1}^{5}q_{i}\leq 1$ and
$(n-1)m\sum\limits_{i=1}^{5}q_{i}\leq(n-1)m.$

Since the list $\Lambda=\\{(n-1),-1,\ldots,-1\\}$ is UR, then the upper bound
in (4) is also sharp for $g_{u}(\Lambda/\lambda_{1}).$ Next, we have:

###### Theorem 2.5

Let $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ be a DR list of complex
numbers and let $g_{d}(\Lambda/\lambda_{1})$ be the diagonalizable
realizability index of $\Lambda$. Then for every
$\mu>g_{d}(\Lambda/\lambda_{1}),$
$\Lambda_{\mu}=\\{\mu,\lambda_{2},\ldots,\lambda_{n}\\}$ is UR.

Proof. The result is a straightforward consequence of the Minc’s result, and
the fact that if the Perron eigenvalue is increased for any $\epsilon>0,$ then
any diagonalizably realizable list becomes a diagonalizably positively
realizable list.

###### Remark 2.1

We have seen that $g_{d}(\Lambda/\lambda_{1})\geq g_{r}(\Lambda/\lambda_{1}).$
Of course, if a realizable list $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$
of complex numbers has all its elements distinct, then
$g_{d}(\Lambda/\lambda_{1})=g_{r}(\Lambda/\lambda_{1}).$ In order that
$g_{d}(\Lambda/\lambda_{1})>g_{r}(\Lambda/\lambda_{1}),$ $\Lambda$ must admit
repeats. This is necessary, but not sufficient. For instance, the list
$\Lambda=\\{6,1,1,-4,-4\\}$ is DR (see [21, Lemma $1$]), but
$g_{d}(\Lambda/\lambda_{1})=g_{r}(\Lambda/\lambda_{1})$. It is clear that
$g_{d}(\Lambda/\lambda_{1})>g_{r}(\Lambda/\lambda_{1})$ if and only if
$\Lambda$ is not DR.

###### Example 2.2

The list

$\Lambda=\\{7,5,1,1,-4,-4,-6\\},$

is symmetrically realizable by

$A=\left[\begin{array}[]{ccccccc}0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}\\\
\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}\\\
\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}\\\
\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}\\\
\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}\\\
\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&0&6\\\
\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&\frac{\sqrt{10}}{10}&6&0\end{array}\right].$

Then, since $A$ is diagonalizable ODP, $\Lambda$ is UR and
$g_{r}(\Lambda/\lambda_{1})=g_{d}(\Lambda/\lambda_{1})=g_{u}(\Lambda/\lambda_{1})=7.$

A realizable list $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ of complex
numbers is said to be Perron extreme, if the list
$\Lambda_{\epsilon}=\\{\lambda_{1}-\epsilon,\lambda_{2},\ldots,\lambda_{n}\\}$
is not realizable for every $\epsilon>0.$ If $\Lambda$ is not Perron extreme,
there is an $\epsilon>0$ such that $\Lambda_{\epsilon}$ is realizable. Then we
have the following result, which is equivalent to Theorem 2.5.

###### Corollary 2.2

Let $\Lambda=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ be a list not Perron
extreme of complex numbers with
$\Lambda_{\epsilon}=\\{\lambda_{1}-\epsilon,\lambda_{2},\ldots,\lambda_{n}\\},$
$\epsilon>0,$ being DR. Then $\Lambda$ is UR.

Proof. Let $B\in\mathcal{CS}_{\lambda_{1}-\epsilon}$ be a diagonalizable
nonnegative matrix with spectrum $\Lambda_{\epsilon}.$ Then

$A=B+\mathbf{eq}^{\textsuperscript{T}},\text{ with
}\mathbf{q}^{\textsuperscript{T}}=\left[\frac{\epsilon}{n},\ldots,\frac{\epsilon}{n}\right]$

is positive with spectrum $\Lambda.$ Moreover, from Theorem 2.2, $A$ is
diagonalizable. Hence, $\Lambda$ is UR.

###### Example 2.3

The list $\Lambda=\\{8,2,2,-3,-4,-4\\}$ is not Perron extreme. Since
$\Lambda^{\prime}=\\{7,2,2,-3,-4,-4\\}$ is DR by

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

then $A=B+\mathbf{eq}^{\textsuperscript{T}},$ where
$\mathbf{q}^{\textsuperscript{T}}=\left[\begin{array}[]{cccccc}\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}&\frac{1}{6}\end{array}\right]$
is diagonalizable positive with spectrum $\Lambda.$ Therefore $\Lambda$ is UR.

###### Remark 2.2

We know that DR does not necessarily imply UR. We also know of two important
extensions [2, 7] of Minc’s result [15]. As far as we know, these extensions
are the most general universal realizability criteria for a solution to URP.
Then, for $\Lambda=\\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\\}$
diagonalizably realizable we may say that DR implies UR if:
$i)$ The realizing matrix for $\Lambda$ is ODP (positive matrices are ODP), or
$ii)$ $\Lambda$ is the spectrum of a nonnegative matrix
$A\in\mathcal{CS}_{\lambda_{1}}$ with a positive row or column.
There are spectra, however, which are UR, without to have necessarily a
realizing matrix of some above classes. For instance:
$iii)$
$\Lambda=\\{\lambda_{1},\lambda_{2},\ldots,\lambda_{\frac{n}{2}},-\lambda_{\frac{n}{2}},\ldots,-\lambda_{2},-\lambda_{1}\\}.$
Is clear that $\Lambda$ has a diagonalizable realization and if it has
$2\times 2$ blocks repeated, we may obtain any coarser JCF.
$iv)$ $\Lambda=\\{\lambda_{1},\lambda_{2},-\lambda_{3},\ldots,-\lambda_{n}\\}$
with $\lambda_{j}>0,$ $j=1,2,\ldots,n.$ $\lambda_{1}>\lambda_{2},$
$\lambda_{1}+\lambda_{2}-\sum\limits_{j=3}^{n}\lambda_{j}=0,$ It was proved in
[4] that $\Lambda$ is UR.

##  3 The merge of two spectra

In this section we define the merge of two spectra in the following way:

###### Definition 3.1

Let $\Gamma_{1}=\\{\lambda_{1},\lambda_{2},\ldots,\lambda_{n}\\}$ and
$\Gamma_{2}=\\{\mu_{1},\mu_{2},\ldots,\mu_{m}\\}$ be lists of complex numbers.
The merge $\Gamma_{1}$ with $\Gamma_{2}$ is

$\Gamma=\\{\lambda_{1}+\mu_{1},\lambda_{2},\ldots,\lambda_{n},\mu_{2},\ldots,\mu_{m}\\}.$

Then, we have:

###### Theorem 3.1

Suppose that $A$ and $B$ are $n$-by-$n$ and $m$-by-$m$ diagonalizable ODP
matrices with spectrum $\Gamma_{1}=\\{\lambda_{1},\ldots,\lambda_{n}\\}$ and
$\Gamma_{2}=\\{\mu_{1},\ldots,\mu_{m}\\},$ respectively. Then, there is an
$(n+m-1)$-by-$(n+m-1)$ diagonalizable ODP matrix $C$ with spectrum
$\Gamma=\\{\lambda_{1}+\mu_{1},\lambda_{2},\ldots,\lambda_{n},\mu_{2},\ldots,\mu_{m}\\}.$
Hence, $\Gamma$ is UR.

Proof. Since $A$ is diagonalizable ODP with spectrum $\Gamma_{1},$ it is
irreducible. Then, there is a positive vector
$\mathbf{v}^{\textsuperscript{T}}=[v_{1},\ldots,v_{n}]$ such that
$A\mathbf{v}=\lambda_{1}\mathbf{v}$. Thus, if
$D=diag\\{v_{1},\ldots,v_{n}\\},$ then
$\tilde{A}=D^{-1}AD\in\mathcal{CS}_{\lambda_{1}}$ is diagonalizable ODP. If
$d_{1},\ldots,d_{n}$ are the diagonal entries of $\tilde{A}$, then from
Theorem 2.1,
$A_{1}=\tilde{A}+\mathbf{e}[0,0,\ldots,\mu_{1}]=\begin{bmatrix}A_{11}&\mathbf{a}\\\
\mathbf{b^{\textsuperscript{T}}}&d_{n}+\mu_{1}\end{bmatrix}$ has spectrum
$\\{\lambda_{1}+\mu_{1},\lambda_{2},\ldots,\lambda_{n}\\}$ and diagonal
entries $d_{1},d_{2},\ldots,d_{n}+\mu_{1}$. It is clear, from Theorem 2.2,
that $A_{1}$ is a diagonalizable ODP matrix.
Since $B$ is diagonalizable ODP with spectrum $\Gamma_{2}$ then, as before,
there is a diagonalizable ODP matrix $\tilde{B}\in\mathcal{CS}_{\mu_{1}}$ with
spectrum $\Gamma_{2}$. Then, the matrix
$B_{1}=\tilde{B}+\mathbf{e}[d_{n},0\ldots,0]$ is diagonalizable ODP with
spectrum $\\{\mu_{1}+d_{n},\mu_{2},\ldots,\mu_{m}\\}$. Finally, by applying
the Šmigoc’s glue technique, Theorem 2.3, there is a matrix

$C=\begin{bmatrix}A_{11}&\mathbf{at^{\textsuperscript{T}}}\\\
\mathbf{sb^{\textsuperscript{T}}}&B_{1}\end{bmatrix}$

with $A_{11}$ the $(n-1)$-by-$(n-1)$ matrix in $A_{1}$ and with
$\mathbf{s},\mathbf{t}$ such that $\mathbf{t^{\textsuperscript{T}}s}=1$, being
the right and left eigenvectors of $B_{1}$, respectively. Now, $C$ has the
spectrum
$\\{\lambda_{1}+\mu_{1},\lambda_{2},\ldots,\lambda_{n},\mu_{2},\ldots,\mu_{m}\\},$
with Jordan canonical form $J(C)=J(A_{1})\oplus I(B_{1})$ with
$J(B_{1})=\begin{bmatrix}\mu_{1}+d_{n}&\\\ &I(B_{1})\end{bmatrix}$. Since
$\mathbf{a},\mathbf{b},\mathbf{s}$ and $\mathbf{t}$ are positive vectors, it
is clear that $C$ is a diagonalizable ODP matrix and therefore $\Gamma$ is UR.

In many cases, Theorem 3.1 may be a useful tool to decide about the universal
realizability of a list of complex numbers, as the following example shows.

###### Example 3.1

Is

$\Lambda=\\{13,1,1,-3,-4,-4,1+3i,1-3i\\},$

universally realizable? To answer the question, consider the lists

$\displaystyle\Lambda_{1}$ $\displaystyle=$
$\displaystyle\\{7,-3,1+3i,1-3i\\}.$ $\displaystyle\Lambda_{2}$
$\displaystyle=$ $\displaystyle\\{6,1,1,-4,-4\\}.$

From [8], $\Lambda_{1}$ has the normal nonnegative realization

$A_{1}=\left[\begin{array}[]{cccc}3&2-\sqrt{3}&\frac{\sqrt{6}+2\sqrt{2}}{2}&\frac{\sqrt{6}+2\sqrt{2}}{2}\\\
2+\sqrt{3}&3&\frac{2\sqrt{2}-\sqrt{6}}{2}&\frac{2\sqrt{2}-\sqrt{6}}{2}\\\
\frac{2\sqrt{2}-\sqrt{6}}{2}&\frac{\sqrt{6}+2\sqrt{2}}{2}&0&3\\\
\frac{2\sqrt{2}-\sqrt{6}}{2}&\frac{\sqrt{6}+2\sqrt{2}}{2}&3&0\end{array}\right],$

which is diagonalizable ODP. From [21, Lemma $1$], $\Lambda_{2}$ has the
symmetric ODP realization

$A_{2}=\left[\begin{array}[]{ccccc}0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}\\\
\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}\\\
\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}\\\
\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0&\frac{3+\sqrt{5}}{2}\\\
\frac{3+\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3-\sqrt{5}}{2}&\frac{3+\sqrt{5}}{2}&0\end{array}\right].$

Then, from Theorem 3.1, the merge $\Lambda_{1}$ with $\Lambda_{2},$ that is,

$\Lambda=\\{13,1,1,-3,-4,-4,1+3i,1-3i\\}$

has a diagonalizable ODP realization. Hence, $\Lambda$ is UR.

###### Remark 3.1

It is known that lists of Suleĭmanova type, real or complex, are UR. Now, this
result can be also proved via Theorem 3.1, in the case of real Suleĭmanova
spectra, and of complex Suleĭmanova spectra with
$\left|\mbox{Re}\lambda_{i}\right|>\left|\mbox{Im}\lambda_{i}\right|.$ In
fact, in both cases, we may obtain diagonalizable ODP realizations.

Declaration of Competing Interest

There is no competing interest

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