Spectrally Simple Spectra without Irreducible Nonnegative Realizations

Introduction and main results

For an integer $n \ge1$, let $I_n$ denote the $n \times n$ identity matrix. The 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 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 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 irreducibly realizable if it has an irreducible nonnegative realization.

For a realizable spectrum $\sigma$, its Perron root is $\rho(\sigma) = \max_{\lambda\in\sigma} |\lambda|$, which, by the Perron–Frobenius theorem, is a nonnegative real member of $\sigma$. Its spectral circle is $\{z \in\mathbb{C} : |z| = \rho(\sigma)\}$. Following Johnson, Marijuán, and Pisonero [1], we call $\sigma$ 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 [1 p. 3] 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 [1 Sections 8–10]. In their analysis of nonreal spectra of order four, they also conjectured that a particular family has no irreducible realizations [1 p. 12]. Our main result proves the latter conjecture and disproves the former.

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.

An eigenvector inequality

For a nonnegative $4 \times4$ matrix $A=(a_{ij})$, put

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

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

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 (4).

Proof of the main result

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

Proof of Theorem 1.1. The definition of $s$ and (1.1) give

The nonnegative matrix

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,

and $0 < b < a$, so $\sigma$ is spectrally simple.

Suppose now that $A=(a_{ij})$ is an irreducible nonnegative realization. Put

and write $l=(l_1,l_2,l_3,l_4)$. Write the characteristic polynomial as

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

Indeed, expanding the determinant gives $\sum_{i<j}(l_i l_j-a_{ij}a_{ji})$ 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 $Q\ge0$, (8) yields

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 {(xref)}(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 {(xref)}(3.3) gives

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

This contradiction proves that $\sigma$ is not irreducibly realizable.

Proof of Corollary 1.2. Take $(a,b,c,d)=(1,1/2,0,1/\sqrt{3})$. Then $s=1$, and every hypothesis of Theorem 1.1 holds. The construction in {(xref)}(3.2) gives the explicit realization

Theorem 1.1 excludes every irreducible nonnegative realization.

Proof. Corollary 1.2 gives a realizable, spectrally simple spectrum with no irreducible nonnegative realization.

Johnson, Marijuán, and Pisonero [1 Section 10, p. 12] 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

They conjectured that $\sigma=\{a,b,c\pm di\}$ has no irreducible nonnegative realization.

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

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.