# Properties of eigenvalues of general nonnegative matrices

I am aware, that an answer to this question can be found via Perron-Frobenius theory or something very similar, but unfortunately I am far from being an expert in the field and I am unable to find the answer in literature by myself.

My question is: does any of these statements hold?:

• all eigenvalues of a nonnegative (not necessarily irreducible) matrix are a $n$-th complex root of some real number for some $n \in \mathbb{N}$
• all leading eigenvalues of a nonnegative matrix are a $n$-th complex root of some real number for some $n \in \mathbb{N}$

If my knowledge is correct, the second statement should hold (because of any nonnegative matrix can be decomposed to a triangular block matrix with irreducible diagonal blocks). But I have no idea, if the first statement holds, and if so, why. Moreover, if the matrix is a nonnegative integer matrix, does it somehow simplify it's spectral properties?

I am interested in the problem because I am trying out to analyze a behavior of solutions of certain linear systems of recurrences, and any of these properties would significantly simplify the analysis.

Any ideas and references to literature would be extremely helpful. Thank you in advance.

• An irreducible non-negative matrix has a unique positive real eigenvalue, $\rho$ say. If $\lambda$ is another eigenvalue then $\lambda \le \rho$; if equality holds then $\lambda/\rho$ is a $k$-root of unity. So some version of your second statement might be true, but I am not sure what a "leading eigenvalue" is. Your first statement is false, for example the eigenvalues of a path on $n$ vertices are of the form $2\cos(\pi r/(n+1))$ for $r=1,\ldots,n$. – Chris Godsil Jun 17 '12 at 17:13
• @Chris. No the positive real eigenvalue need not be unique. But the maximal one is simple. – Denis Serre Jun 18 '12 at 8:17
• A reference for the second statement: for an irreducible non-negative matrix, the eigenvalues of maximal modulus are simple and form a regular $m$-agon. In addition, the matrix has a circulant form. See my book Matrices, GTM 216, Springer-Verlag. – Denis Serre Jun 18 '12 at 8:20

The first of these is certainly false. Here's one way to see it (probably not the most elegant). The matrix $A$ given by $$A=\begin{pmatrix} 0&0&0&1 \cr 1&0&0&1 \cr 0&1&0&1 \cr 0&0&1&1 \end{pmatrix}$$ is the companion matrix to the polynomial $\lambda^4-\lambda^3-\lambda^2-\lambda-1$.
The roots to this polynomial are (using mathematica) -0.7748, 1.9276 and a pair of complex roots. Mathematica also verifies that this is an irreducible polynomial. I am sure that the pair of complex roots are not roots of any real number, but don't know how to prove it. However it doesn't matter. The polynomial depends continuously on the roots. This means that perturbing the roots, as long as the complex roots remain a conjugate pair, one obtains a polynomial with first coefficient 1 and remaining coefficients negative and real. So you can just perturb the roots so as to make the argument not a rational multiple of $\pi$ and obtain a polynomial of this type. Now you can convert the polynomial back into a matrix whose eigenvalues are exactly the roots of the polynomial. Since the non-leading terms of the polynomial were negative, the entries of the matrix will be non-negative (and positive in the same places as in the matrix $A$).