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The following question came up today during a discussion:

Suppose $A$ is an $n \times n$ real matrix. Is there some way to tell whether there exists an integer $q > 0$ such that $A^q$ is elementwise nonnegative? If so, can we compute this exponent $q$ quickly?

Thanks for your insights.

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3 Answers 3

Here is one case: Suppose $A$ has a unique eigenvalue $\lambda$ of greatest absolute value that has algebraic multiplicity 1, with left and right eigenvectors $u^T$ and $v$ having all entries nonzero, normalized so $u^T v = 1$. Since $A$ is a real matrix, its complex eigenvalues come in complex-conjugate pairs, so $\lambda$ must be real. Then $A^q = \lambda^q v u^T + o(|\lambda|^q)$ as $q \to \infty$. If all entries of $u^T$ and $v$ have the same sign, then all entries of $A^q$ are positive for all sufficiently large $q$ (if $\lambda > 0$) or all sufficiently large even $q$ (if $\lambda < 0$). If some entries of $u^T$ or $v$ have different signs, there will be entries of $A^q$ with different signs for all sufficiently large $q$, and therefore for all positive integers $q$ (if the elements of $A^q$ all have the same sign, so do the elements of $A^{kq}$ for all positive integers $k$).

EDIT: Here's a partial converse. By the Perron-Frobenius theorem, if $A^q$ has all its entries strictly positive, then $A^q$ has a positive eigenvalue $\mu$ which is greater in absolute value than all other eigenvalues, and is simple, with left and right eigenvectors $u^T$ and $v$ having all entries strictly positive. Since the eigenvalues of $A^q$ are the $q$'th powers of eigenvalues of $A$, there must be one eigenvalue $\lambda$ of $A$ with $\lambda^q = \mu$, also having left and right eigenvectors $u^T$ and $v$. Since $A$ is real and $\mu$ is a simple eigenvalue, $\lambda$ must be real, and we are in the situation of the previous paragraph.

Matters can be somewhat more complicated if $A^q$ is nonnegative but never all strictly positive.

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Thanks for your answer and insights Robert. –  Suvrit Oct 18 '11 at 18:24
I believe that the following is still open: given an $n\times n$ integer matrix $A$ and $1\leq i,j\leq n$, is it decidable whether there is some $q\geq 1$ for which $(A^q)_{ij}=0$? See citeseerx.ist.psu.edu/viewdoc/summary?doi= This suggests that the question posed here might also not be known to be decidable, but I could be wrong about this. –  Richard Stanley Oct 18 '11 at 19:33

The paper "On an inverse problem for nonnegative and eventually nonnegative matrices" gives necessary and sufficient conditions on the spectrum of eventually nonnegative matrices. This is not a full answer to your question.

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Thanks for the pointer! –  Suvrit Oct 18 '11 at 18:25

This paper is fairly interesting, and has reasonably extensive references:


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Thanks for that pointer Igor. –  Suvrit Oct 18 '11 at 18:24

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