I have some difficulty in proving the following proposition about non-negative matrix which means that every element of it is non-negative.
Suppose $A$ is a non-negative matrix. If there exist a $k\geq 1$ such that $A$ is a positive matrix, then $\rho (A)$ is a single eigenvalue.
According to the Perron-Frobenius theorem, $\rho(A^k)$ is a single eigenvalue of $A^k$. $\rho(A^k) = \rho(A)^k$.
We only need to prove that $\rho(A)$ is a solution of $\det(\lambda-A)=0$ mutiplicity one. Or some equivelent forms of single eigenvalue.
The statement is trivial if you've been taught Perron-Frobenius theorem for irreducible matrices (since $A^k>0$, $A$ must be irreducible and hence $\rho(A)$ is simple, by PFT). But you sound like you've only been taught PFT for positive matrices.
In this case, let $\rho(A),\lambda_2,\ldots,\lambda_n$ be the eigenvalues of $A$ over $\mathbb C$. Then the eigenvalues of $A^k$ are $\rho(A^k),\lambda_2^k,\ldots,\lambda_n^k$ respectively. Since $A^k$ is positive, PFT says that $\rho(A^k)$ is a simple eigenvalue of $A^k$. Thus $\lambda_2^k,\ldots,\lambda_n^k\ne\rho(A^k)=\rho(A)^k$. In turn, $\lambda_2,\ldots,\lambda_n\ne\rho(A)$ and hence $\rho(A)$ is a simple eigenvalue of $A$.