Consider the following probability problem:
Assume that if it rains some day, then the probability to rain the following day is $0.75$. Now if it doesn't rain some day, then the probability to rain the following day is $0.1$.
If we note $p_n$ the probability to rain in the $n$-th day and $q_n=1-p_n$, then we would have $$p_{n+1}=0.75p_n+0.1q_n$$ $$q_{n+1}=0.25p_n+0.9q_n$$ Thus we will get the following geometric recurrence relation $$U_{n+1}=\begin{pmatrix} p_{n+1} \\ q_{n+1} \end{pmatrix}=\begin{pmatrix} 0.75 & 0.1\\ 0.25 & 0.9 \end{pmatrix}\begin{pmatrix} p_{n} \\ q_{n} \end{pmatrix}=QU_n$$ How can we prove that in general if the transition matrix $Q$ contains only positive entries, then the sequence of probability state $(U_n)_n$ will converge to the steady states $U$ which solves the equation: $$U=QU.$$ In this case $$U=\begin{pmatrix} \frac{2}{7} \\ \frac{5}{7} \end{pmatrix}.$$
I heard that this has to do with the Perron-Frobenius Theorem.
Note: I know that we can solve this problem simply by solving the recurrence relation $$p_{n+1}=0.75p_n+0.1q_n=0.65p_n+0.1.$$ But I am just interested in the matricial aspect of the problem.