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Suppose $A$ is an $n\times n$ matrix over the set of non-negative integers. Is there a necessary condition for $A$ so that it would have at least one real non-zero eigenvalue?

Ignore what's written below:

I'm just wondering, given an $n\times n$ matrix $A$ over the set of non-negative integers, is there a necessary condition for $A$ to have at least one real, non-zero eigenvalue?

Edit: If we know that $A$ doesn't have a row that is completely zero, would this suffice?

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  • $\begingroup$ $A$ has at least one real, non-zero eigenvalue iff its characteristic polynomial has at least one real, non-zero root. $\endgroup$ – Rob Arthan Aug 4 '13 at 23:49
  • $\begingroup$ @RobArthan sure, but what if the entries of $A$ are unknown? So I was hoping for something that only relies on a particular property of $A$ (if there is such...) $\endgroup$ – Eric Aug 4 '13 at 23:54
  • $\begingroup$ You ask first for a necessary condition. Then the edit suggests a possible sufficient condition. Which are you looking for? $\endgroup$ – OR. Aug 5 '13 at 0:00
  • $\begingroup$ @RGB Hmm, good point. let me edit it again. I think I've made up my mind on what my question would be like. $\endgroup$ – Eric Aug 5 '13 at 0:06
  • $\begingroup$ What do you mean by a "particular property"? The conjecture you added in your edit is wrong - rotations in the plane have no non-zero eigenvalues, but are non-singular. $\endgroup$ – Rob Arthan Aug 5 '13 at 0:07
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By the Perron-Frobenius theorem and its extension to the nonnegative matrices, a matrix with nonnegative elements either has only zeroes as the eigenvalues or it has to have at least one real non-zero eigenvalue, because it has one that is - in absolute value - greater than or equal to all the others (so it cannot be zero).

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