What conditions on matrix $A$ such that there exists a $k$ such that $A^k$ is a strictly positive matrix You have matrix a of size $n × n$.
Matrix $A$ meets the following two conditions:

*

*for any numbers $i, j, a_{ij} \geq 0$


*Matrix $b$ is strictly positive, if for any numbers $i, j$ $(1 \leq i, j \leq n)$ the inequality $b_{ij} > 0$ holds.
What are the condition such that there exists a $k$ such that $A^k$ is a strictly positive matrix?
It has something to do with the reachibility of the elements but I cannot see how and why this is considered.
 A: Only the first one is a condition for $A$, the second sentence is a definition.
Assign a directed graph $D$ to $A$ in the following way. The vertices are $\{1,2,\ldots,n\}$. There is a directed egde from $i$ to $j$ iff $a_{ij}>0$.
Note that the $ij$-entry of $A^k$ is positive iff there is a walk of length $k$ from $i$ to $j$ in the assigned digraph $D$. So the requirement that $A^k$ be positive for some $k\in \mathbb N$ translates to the following combinatorial property: there is a $k\in \mathbb N$ such that there is a walk of length $k$ between any (ordered) pair of vertices in $D$. So obviously, the graph $D$ has to be strongly connected, but that is not enough.
Consider the oriented 3-cycle: $1 \rightarrow 2 \rightarrow 3\rightarrow 1$: then starting from 1, you only reenter 1 in every step whose residue modulo 3 is 0, you only enter 2 in every step whose residue modulo 3 is 1, and you only enter 3 in every step whose residue modulo 3 is 2. So there is no appropriate $k$ that settles all three at once. This suggests the idea of aperiodicity. The period of a vertex in a digraph is the greatest common divisor of all positive integers that occur as a length of a closed walk starting and ending in that vertex. For example, in the oriented 3-cycle, every vertex has period 3, and that is a clear obstruction.
It is not hard to show that in a strongly connected graph, every vertex has the same period. Clearly, if this common period is not 1, then the condition you are investigating fails. It is easy to prove that if the common period is 1, then the condition holds.
So to sum up, the equivalent condition is as follows: the assigned digraph $D$ is strongly connected and aperiodic (that is, the common period of the vertices is 1).
