A proof needed, problem in Perron-frobenius theory

[Jakesaccount]

Homework Statement

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?

Homework Equations

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 :)

 

[Jakesaccount]

Allright so I solved this myself. Here is the solution if anyone is interested:

Let y be an eigenvalue to A. Then y^n=y which implies y^n-y=0 so y(y^(n-1)-1)=0.
This implies either that y=0 or that y=e^(2pi(k-1)i/(n-1) ) for k=1,...,n-2. This means that all y which are not 0 has abs(y)=1 and that 1 is an eigenvalue. But this means that y=1 is the perron root of A and so by the perron theorem all other eigenvalues must be smaller in modulus, which implies that all other eigenvalues are zero.

But y=0 has the algebraic multiplicity 1 and by the perron theorem so does y=1. Since V can be written as the direct sum of the generalized eigenspaces this means that range(A) = the eigenspace corresponding to y=1 which has dimension 1 by the perron theorem. So rank(A)=1 QED.