If $P^r$ has all positive entries, then so does $P^n$

Question

Let $P$ be the transition probability matrix of a Markov Chain. Argue that it for some positive integer r, $P^r$ has all positive entries, then so does $P^n$, for all integers $n\geq r$

I know that $$p_{ij}=P(X_{n+1}=j|X_n=i)$$ $$p(i,j)^r>0\space\forall i,j$$ If $r>0$ and $s>0$ such that $r+s=n$ $$p(i,j)^n=\sum_kp(i,k)^rp(k,j)^s>0\Rightarrow p(i,j)^n>0$$

Is that right? I'm bad when it comes to prove things.

Answer 1

We want to see that $$\forall\; (i,j):\;p(i,j)^r>0 \Rightarrow \forall(i,j)\;:\;p(i,j)^{r+1}>0$$

If $\sum_k p(k,j) = 0$ then $p (l,j) = 0$ for every $l$ therefore $p(k,j)^r = \sum_{l}p(k,l)^{r-1}p(l,j) = 0$ which is a contradiction.

Therefore $ \exists\, k^*: p(k^*,j) > 0 $ and we conclude that $$ p_(i,j)^n \geq p(i,k^*)^rp(k^*,j)>0$$

Note: Use induction to get the general case.