Why does spectral radius $\rho\left(A\right) > 0$ for a non-negative matrix $A \in M_n$, i.e., $A \geq 0$ with $A^k > 0$ for some positive integer $k$?
I guess it is related to Perron-Frobenius theorem? right? if yes, do we have a short and simple proof for non-experts like me?
Thank you very much
$A$ is nonnegative, so it has a nonnegative eigenvector $v$ for the eigenvalue $\rho(A)$ and in turn, $A^kv=\rho(A)^kv$. Since $A^k>0$ and $v$ is nonnegative but nonzero, the product $A^kv$ is positive. Therefore the equality $A^kv=\rho(A)^kv$ implies that $\rho(A)>0$ and $v>0$.
As pointed out in a comment, this result is also regarded as part of Perron-Frobenius theorem for primitive matrices.