How to prove the irreducible non-negative matrix has positive geometric sum?

Question

Here is a question:
A matrix $A=(A_{ij})_{n\times n}$ is called irreducible iff for every pairs $(i,j)$ we have $(A^k)_{ij}>0$ for some $1\leq k\leq n-1$.

How to prove that matrix $A$ is irreducible $\Leftrightarrow$ $I+A+A^2+\cdots+A^{n-1}>0$ ?

I searched many books but I did not find the proof.

Answer 1

Hint: For a list of non-negative numbers $a_1,\dots,a_n$, it holds that $a_k > 0$ for at least one value of $k$ if and only if $a_1 + \cdots + a_n > 0$.

The $i,j$ entry of the sum is given by $$ [I]_{ij}+[A]_{i,j}+[A^2]_{i,j}+\cdots+[A^{n-1}]_{i,j} $$