How Does a Non-Negative Matrix Ensure a Positive Eigenvector?

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Homework Statement


If A≥0 and Ak>0 for some k≥1, show that A has a positive eigenvector.



Homework Equations





The Attempt at a Solution


A is nxn

Well from a previous problem we know that the spectral radius ρ(A)>0

We also know that if A≥0, then ρ(A) is an eigenvalue of A and there is a non negative vector x, x=/=0 such that Ax=ρ(A)x

Kinda stuck
 
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What do you mean with a "nonnegative vector" or "positive vector"?
 
non negative vector means all the entries in that vector is greater than zero,

if the vector is positive all entries in that vector is positive

i.e. if x≥0 all components of x are greater than or equal to zero

similarly if a matrix A≥0

all [aij]≥0

positive just means everything is greater than 0
 
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BrainHurts said:

Homework Statement


If A≥0 and Ak>0 for some k≥1, show that A has a positive eigenvector.



Homework Equations





The Attempt at a Solution


A is nxn

Well from a previous problem we know that the spectral radius ρ(A)>0

We also know that if A≥0, then ρ(A) is an eigenvalue of A and there is a non negative vector x, x=/=0 such that Ax=ρ(A)x

Kinda stuck

Google Perron-Frobenius theorem.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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