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\hypersetup{pdftitle={Spectrally Simple Spectra without Irreducible Nonnegative Realizations},pdfauthor={Yair Lavi}}
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\title{Spectrally Simple Spectra without Irreducible Nonnegative Realizations}
\author{Yair Lavi}
\date{9 October 2026}
\begin{document}
\maketitle
\begin{abstract}
We give a counterexample to a conjecture of Johnson, Marijuán,
and Pisonero: we show that the spectrum $\{1,1/2,\pm i/\sqrt3\}$ is realizable
and spectrally simple, but has no irreducible nonnegative realization. More
generally, we prove their nonexistence conjecture for an order-four boundary
family.
\end{abstract}
\noindent\textbf{Keywords.} Nonnegative inverse eigenvalue problem; irreducible realization;
Perron root.

\noindent\textbf{2020 Mathematics Subject Classification.} 15A29, 15B48.


\section{Introduction and main results}

For an integer $n\ge1$, let $I_n$ denote the $n\times n$ identity matrix. The
\textbf{spectrum} of a real $n\times n$ matrix $A$ is the list of its
eigenvalues, counted with algebraic multiplicity.
A list $\sigma$ of $n$ complex numbers is \textbf{realizable} if it is the spectrum
of an entrywise nonnegative real matrix. We write $A\ge0$ when every entry of
$A$ is nonnegative. Such a matrix $A=(a_{ij})$ is \textbf{irreducible} if, for every
nonempty proper subset $S$ of $\{1,\ldots,n\}$, some $i\in S$ and $j\notin S$
satisfy $a_{ij}>0$. A spectrum is \textbf{irreducibly realizable} if it has an
irreducible nonnegative realization.

For a realizable spectrum $\sigma$, its \textbf{Perron root} is
$\rho(\sigma)=\max_{\lambda\in\sigma}|\lambda|$, which, by the
Perron--Frobenius theorem, is a nonnegative real member of $\sigma$.
Its \textbf{spectral circle} is
$\{z\in\mathbb C:|z|=\rho(\sigma)\}$. Following Johnson, Marijuán, and
Pisonero \cite{ref1}, we call $\sigma$ \textbf{spectrally simple} if its Perron root occurs
once and every other member has strictly smaller modulus. For a matrix $A$,
write $\rho(A)$ for the maximum modulus of its eigenvalues.

Johnson, Marijuán, and Pisonero \cite[p.~3]{ref1} conjectured that every realizable,
spectrally simple spectrum is irreducibly realizable. They proved the
assertion in orders below four and for real spectra of order four
\cite[Sections 8--10]{ref1}. In their analysis of nonreal spectra of order four,
they also conjectured that a particular family has no irreducible
realizations \cite[p.~12]{ref1}. Our main result proves the latter conjecture and
disproves the former.

\begin{theorem}
Let $a,b,c,d\in\mathbb R$ satisfy $a>b$, $a>0$, and $d>0$.
Put $s=a+2c$, and suppose
\[
s>0,\qquad a-c=\sqrt3d,\qquad b>\frac{s}{3}.
\tag{1.1}
\]
Then $\sigma=\{a,b,c+di,c-di\}$ is realizable and spectrally simple, but is
not irreducibly realizable.
\end{theorem}

\begin{corollary}
The spectrum
\[
\left\{1,\frac12,\frac{i}{\sqrt3},-\frac{i}{\sqrt3}\right\}
\tag{1.2}
\]
is realizable and spectrally simple, but has no irreducible nonnegative
realization.
\end{corollary}

The proof uses an eigenvector for the smaller real eigenvalue $b$. Its
positive and negative coordinates force a strict lower bound on the sum of products
$a_{ij}a_{ji}$. The coefficient of $x^2$ in the characteristic polynomial
gives an incompatible upper bound. The argument includes eigenvectors with
zero coordinates.

\section{An eigenvector inequality}

For a nonnegative $4\times4$ matrix $A=(a_{ij})$, put
\[
l_i=a_{ii},\qquad q_{ij}=a_{ij}a_{ji}\quad(i<j),\qquad
Q=\sum_{i<j}q_{ij}.
\tag{2.1}
\]

\begin{lemma}
Let $A\ge0$ be an irreducible $4\times4$ matrix, and let
$b<\rho(A)$ be a real eigenvalue. Suppose $0\le l_i\le b$ for every $i$,
with at most one $l_i=b$. Put $r_i=b-l_i$. After simultaneously permuting
the rows and columns,
\[
Q>r_1r_2+r_3r_4.
\tag{2.2}
\]
\end{lemma}

\begin{proof}
Choose a real eigenvector $x$ with $Ax=bx$. The
Perron--Frobenius theorem gives a real vector $w$ whose entries are all
positive and which satisfies $w^{\mathsf T}A=\rho(A)w^{\mathsf T}$. Since
$(\rho(A)-b)w^{\mathsf T}x=0$ and $b<\rho(A)$, we have
$w^{\mathsf T}x=0$. Because $w$ has strictly positive entries, $x$ has both
positive and negative coordinates.

Suppose first that no coordinate is zero. A sign cannot occur just once:
if $x_i$ were the sole positive coordinate, the equation
\[
r_ix_i=\sum_{j\ne i}a_{ij}x_j\le0
\]
would force $r_i=0$ and $a_{ij}=0$ for all $j\ne i$, contrary to
irreducibility. Replacing $x$ by $-x$ treats a sole negative coordinate.
We may therefore arrange $x_1,x_2>0$ and $x_3,x_4<0$. Write
$u_i=x_i$ for $i=1,2$ and $v_i=-x_i$ for $i=3,4$. The first two rows give
\[
\begin{aligned}
a_{12}u_2&=r_1u_1+a_{13}v_3+a_{14}v_4,\\
a_{21}u_1&=r_2u_2+a_{23}v_3+a_{24}v_4.
\end{aligned}
\tag{2.3}
\]
Both left sides are positive: if one vanished, its row would have no positive
off-diagonal entry. Irreducibility supplies a positive entry from $\{1,2\}$
to $\{3,4\}$, so at least one of $a_{13},a_{14},a_{23},a_{24}$ is positive.
Since $v_3,v_4>0$, the sums
\[
a_{13}v_3+a_{14}v_4,\qquad a_{23}v_3+a_{24}v_4
\]
in (2.3) are nonnegative, and at least one is positive. Consequently,
\[
a_{12}u_2\ge r_1u_1,\qquad a_{21}u_1\ge r_2u_2,
\]
with at least one strict inequality. Both left sides are positive, as
established above. Multiplying gives
\[
q_{12}u_1u_2=(a_{12}u_2)(a_{21}u_1)>r_1r_2u_1u_2.
\]
If $r_1r_2=0$, the strict inequality follows directly from positivity of the
left side. Dividing by $u_1u_2>0$ yields $q_{12}>r_1r_2$.

The same argument, using a positive entry from $\{3,4\}$ to $\{1,2\}$,
gives $q_{34}>r_3r_4$. All remaining $q_{ij}$ are nonnegative, so (2.2)
follows from
\[
Q\ge q_{12}+q_{34}>r_1r_2+r_3r_4.
\]

If $x$ has two zero coordinates, its nonzero coordinates have opposite signs.
After relabelling, suppose $x_1>0$, $x_2<0$, and $x_3=x_4=0$. The first two
row equations become
\[
r_1x_1=a_{12}x_2,\qquad r_2x_2=a_{21}x_1.
\]
In the first equation, the left side is nonnegative and the right side is
nonpositive; in the second, the signs are reversed. Thus both equations have
both sides zero. Since $x_1,x_2\ne0$, we obtain $r_1=r_2=0$, hence
$l_1=l_2=b$, contradicting the assumption that at most one diagonal entry
equals $b$. More than two zero coordinates cannot give both signs.

It remains to consider exactly one zero. After relabelling and, if necessary,
replacing $x$ by $-x$, assume $x_1,x_2>0$, $x_3<0$, and $x_4=0$. The third
row gives $r_3=0$ and $a_{31}=a_{32}=0$. Irreducibility forces $a_{34}>0$.
Applied to the subset $\{3,4\}$, it also forces $a_{41}>0$ or $a_{42}>0$.
The fourth row equation
\[
a_{41}x_1+a_{42}x_2+a_{43}x_3=0
\]
therefore implies $a_{43}>0$. Hence $q_{34}>0=r_3r_4$. The first two row
equations, multiplied as above, give $q_{12}\ge r_1r_2$. This again proves
(2.2).
\end{proof}

\section{Proof of the main result}

For a list $z=(z_1,z_2,z_3,z_4)$, define
$e_2(z)=\sum_{i<j}z_iz_j$.

\begin{proof}[Proof of Theorem 1.1]
The definition of $s$ and (1.1) give
\[
c=\frac{s-a}{2},\qquad d=\frac{3a-s}{2\sqrt3},\qquad 0<s<3a.
\tag{3.1}
\]
The nonnegative matrix
\[
B=\frac13
\begin{pmatrix}
s&3a-s&0\\
0&s&3a-s\\
3a-s&0&s
\end{pmatrix}
\tag{3.2}
\]
has characteristic polynomial $(x-s/3)^3-(a-s/3)^3$ and eigenvalues
$a,c+di,c-di$. Adjoin a first row and column to $B$, placing $b$ in the
top-left entry and zero in every other new entry. The resulting nonnegative
matrix has characteristic polynomial $(x-b)\det(xI_3-B)$ and realizes
$\sigma$. Moreover,
\[
c^2+d^2=a^2-s\left(a-\frac{s}{3}\right)<a^2,
\]
and $0<b<a$, so $\sigma$ is spectrally simple.

Suppose now that $A=(a_{ij})$ is an irreducible nonnegative realization.
Put
\[
l_i=a_{ii},\qquad q_{ij}=a_{ij}a_{ji}\quad(i<j),\qquad
Q=\sum_{i<j}q_{ij},
\]
and write $l=(l_1,l_2,l_3,l_4)$. Write the characteristic polynomial as
\[
\det(xI_4-A)=x^4+k_1x^3+k_2x^2+k_3x+k_4.
\]
The first two characteristic coefficients are $k_1$, the coefficient of
$x^3$, and $k_2$, the coefficient of $x^2$. Computing them from the matrix
entries and from the prescribed eigenvalues gives
\[
k_1=-\sum_i l_i=-(b+s),\qquad
k_2=e_2(l)-Q=e_2(\sigma)=s\left(b+\frac{s}{3}\right).
\tag{3.3}
\]
Indeed, expanding the determinant gives
$\sum_{i<j}(l_il_j-a_{ij}a_{ji})$ as the coefficient of $x^2$.

Let $m=\max_i l_i$. The other three diagonal entries have sum $b+s-m$;
their pairwise products sum to at most $(b+s-m)^2/3$, with equality only
when those entries are equal. Since $Q\ge0$, (3.3) yields
\[
s\left(b+\frac{s}{3}\right)\le e_2(l)
\le m(b+s-m)+\frac{(b+s-m)^2}{3}.
\tag{3.4}
\]
The last expression, as a function of $m$, is strictly decreasing for
$m>(b+s)/4$. Since $b>s/3$, we have $b>(b+s)/4$, and its value at $m=b$
is $s(b+s/3)$. Consequently $m\le b$. If $m=b$, equality holds throughout
(3.4), so $Q=0$ and the other three diagonal entries are $s/3<b$.
Thus $0\le l_i\le b$, with at most one $l_i=b$.

Lemma 2.1 applies, because $b<a=\rho(A)$. Put $r=(r_1,r_2,r_3,r_4)$, where
$r_i=b-l_i$, and label the indices as in that lemma. Expanding
$e_2(l)=e_2(b-r_1,\ldots,b-r_4)$ in (3.3) gives
\[
\sum_i r_i=3b-s,\qquad
Q=e_2(r)-3b^2+2bs-\frac{s^2}{3}.
\tag{3.5}
\]
Set $X=r_1+r_2$ and $Z=r_3+r_4$. Since $X+Z=3b-s$, we have
$XZ\le(3b-s)^2/4$. It follows that
\[
\begin{aligned}
0<Q-r_1r_2-r_3r_4
&=XZ-3b^2+2bs-\frac{s^2}{3}\\
&\le\frac{(3b-s)^2}{4}-3b^2+2bs-\frac{s^2}{3}\\
&=-\frac{(3b-s)^2}{12}<0.
\end{aligned}
\tag{3.6}
\]
This contradiction proves that $\sigma$ is not irreducibly realizable.
\end{proof}

\begin{proof}[Proof of Corollary 1.2]
Take
$(a,b,c,d)=(1,1/2,0,1/\sqrt3)$. Then $s=1$, and every hypothesis of
Theorem 1.1 holds. The construction in (3.2) gives the explicit realization
\[
\begin{pmatrix}
1/2&0&0&0\\
0&1/3&2/3&0\\
0&0&1/3&2/3\\
0&2/3&0&1/3
\end{pmatrix}.
\]
Theorem 1.1 excludes every irreducible nonnegative realization.
\end{proof}

\begin{corollary}
The conjecture of Johnson, Marijuán, and Pisonero
\cite[p.~3]{ref1} that every realizable, spectrally simple spectrum is irreducibly
realizable is false.
\end{corollary}

\begin{proof}
Corollary 1.2 gives a realizable, spectrally simple spectrum with
no irreducible nonnegative realization.
\end{proof}

Johnson, Marijuán, and Pisonero \cite[Section 10, p.~12]{ref1} also proposed the
following family conjecture. Let $a,b,c,d\in\mathbb R$ satisfy $a>b$, $a>0$,
and $d>0$. Suppose $\{a,c\pm di\}$ is realizable and
\[
a+2c>0,\qquad a-c=\sqrt3d,\qquad b>\frac{a+2c}{3}.
\]
They conjectured that $\sigma=\{a,b,c\pm di\}$ has no irreducible
nonnegative realization.

\begin{corollary}
The family conjecture of Johnson, Marijuán, and Pisonero
stated above is true.
\end{corollary}

\begin{proof}
Put $s=a+2c$. The parameters satisfy every hypothesis of
Theorem 1.1, which excludes every irreducible nonnegative realization of
$\sigma$.
\end{proof}

\section*{Acknowledgments}

The proof of this conjecture was carried out by GPT-5.6-sol, GPT-6 Astra,
Claude Fable 5, and Claude Fable 5.1, under the guidance of the author. The author has reviewed the
resulting proof arguments. Responsibility for the final text rests with the
author.

\begin{thebibliography}{1}
\small

\bibitem{ref1}
C. R. Johnson, C. Marijuán, and M. Pisonero, \emph{The INIEP: Irreducible and
   Positive Realizations}, preprint (2026).
   \href{https://arxiv.org/abs/2605.23552v1}{arXiv:2605.23552v1}.
\end{thebibliography}
\end{document}