What's the period of this matrix?

Question

Consider the matrix $$ A = \begin{pmatrix} 0.1 & 0.3 & 0.4 & 0.2 \\ 0.2 & 0.4 & 0.0 & 0.4 \\ 0.0 & 0.3 & 0.5 & 0.2 \\ 0.5 & 0.3 & 0.2 & 0.0 \end{pmatrix}. $$ Note since $$ A^2 = \begin{pmatrix} 0.17 & 0.33 & 0.28 & 0.22 \\ 0.30 & 0.34 & 0.16 & 0.20 \\ 0.16 & 0.33 & 0.29 & 0.22 \\ 0.11 & 0.33 & 0.30 & 0.26 \end{pmatrix}, $$ the matrix $A$ is irreducible, as described in Definition 3 here. The confusing question now is, what's the period? Again according to the description linked above, the period of indices 1,2 and 3 is 1, since $(A)_{ii} > 0$ for $i = 1,2,3$. For index $4$, however, the period should be $2$ since $(A)_{44} = 0$ but $(A^2)_{44} > 0$. The link then says, "When A is irreducible, the period of every index is the same..." but I don't see how this is true in this case. Is the period 1 or 2? If it's 1, how does that not conflict with index 4's period or 2? If it's 2, that conflicts with the rest of the indices.

Answer 1

Since all entries of $A^2$ are strictly positive, so are all entries of $A^n$ for all $n > 2$. The period of index $4$ is the gcd of all $n$ such that $(A^n)_{44} > 0$, so that is $1$, not $2$.