(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Let A be an operator from V to V(finite). If the matrix of A is positive, i.e. all entries are strictly positive, and A^k=A for some integer k>1 what is the rank of A?

2. Relevant equations

3. The attempt at a solution

I think that rank(A)=1 and i believe that i can prove in the case where k=2.

Proof: if A^2=A then the eigenvalues of A are 0 and 1, which means that the perron root of A is 1. Perron-frobenius theory then states the the eigenspace of 1 has geometric mult.=algebraic mult. = 1. Furthermore since A(Ax)=(Ax) for all vectors x in V, the range is a subspace of the eigenspace corresponding to the eigenvalue 1. And since Ax=x is a subspace of the range, the range of A = the eigenspace corresponding to the eigenvalue 1, which has geom. mult = dim(range space) = rank(A)=1

Im have trouble finding a general proof. Any help is greatly appreciated. I know I'm new here but I promise to look around and see if I can lend a hand to anyone :)

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# Homework Help: A proof needed, problem in Perron-frobenius theory

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