Question on transition probability matrices Question: $P$ is the transition matrix of a finite state space Markov chain. Which of the following statements are necessarily true?
$1.$ If $P$ is irreducible, then $P^2$ is irreducible.
$2.$ If $P$ is not irreducible then $P^2$ is not irreducible.
$3.$ If $P$ is irreducible then $\frac{I+P}{2} $ is irreducible.
$4.$ If $P$ is irreducible and aperiodic, then $P^3$ is irreducible.
Mt Attempt:
For $(1)$, I found a counterexample, take $P= \left[ \begin{matrix} 0&1 \\ 1&0    \end{matrix} \right]$. Here, $P$ is irreducible but $P^2=I$, which is reducible. I tried some examples for $(2),(3)$ and $(4)$ but they could not find any counterexample for these, I guess that last three options are correct, however, I was unable to prove them.

Kindly help me to prove the last three options. Is there any theorem regarding these results? Can these results be generalized for $P^n$? Thank you so much for your effort(s)!

 A: It's not actually a proof, but maybe it helps.


*If $P$ is not irreducible, there are some states $i$ and $j$, such that you can't get from $i$ to $j$. If $P^2$ is not irreducible, there are some states $i$ and $j$, such that you can't get from $i$ to $j$ with an even number of steps.
Since you aren't able to get from $i$ to $j$ with any number of steps, especially you can't get from $i$ to $j$ with an even number of steps.


*If $P$ is irreducible, then you can get from any state to any other. Further observe that in the matrix $\frac{I+P}2$ the same entries are non-zero as before, since from $p_{ij}>0$ follows $\frac{p_{ij}}{2}>0$. It means we can still get from any state to any other, but the probabilities change. The underlying graph is almost the same, except that we add loops for every node.


*If $P$ is irreducible, then you can get from any state $i$ to any other state $j$. If $P^3$ is irreducible, then you can get from any state $i$ to any other state $j$, such that the number of steps from $i$ to $j$ is divisible by 3. I am not sure about this, but I guess it is true. Sure you need $P$ to be aperiodic, otherwise $\begin{bmatrix} 0 &1&0\\0&0&1\\1&0&0\end{bmatrix}$ is a counterexample.
