Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
$A$ has at least one real, non-zero eigenvalue iff its characteristic polynomial has at least one real, non-zero root. –  Rob Arthan Aug 4 '13 at 23:49
    
@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...) –  Eric Aug 4 '13 at 23:54
    
You ask first for a necessary condition. Then the edit suggests a possible sufficient condition. Which are you looking for? –  ABC Aug 5 '13 at 0:00
    
@RGB Hmm, good point. let me edit it again. I think I've made up my mind on what my question would be like. –  Eric Aug 5 '13 at 0:06
    
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. –  Rob Arthan Aug 5 '13 at 0:07
show 2 more comments

1 Answer 1

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.