What kind of matrix is this and why does this happen?

So I was studying Markov chains and I came across this matrix \begin{align*}P=\left( \begin{array}{ccccc} 0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\ \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ \end{array} \right).\end{align*}

I noticed (by brute force) that \begin{align*}P^2=\left( \begin{array}{ccccc} \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ \frac{3}{8} & 0 & 0 & \frac{1}{4}& \frac{1}{2}\\ \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\ \end{array} \right),\end{align*} and \begin{align*}P^3=\left( \begin{array}{ccccc} 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\ \frac{7}{16} & 0 & 0 & \frac{1}{4} & \frac{5}{16}\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ \end{array} \right).\end{align*}

In fact; using a computer I found that every even power takes the form of the $P^2$ matrix and every odd power takes the form of the $P^3$ matrix.

I just wanted to know why that oscillation occurs? Is there a special name for the kind of matrix that $P$ is for it to exhibit that kind of behaviour?

Solution 1:

This is a periodic Markov chain (with period $2$). Otherwise, there's not much that's terribly unusual about it.

User "Iwillnotexist Idonotexist" raised an important point in the comments:

Well, something that can be noted for periodic Markov chains is that by definition they cannot be ergodic, another very important property of MCs that you may encounter soon. Roughly speaking, an ergodic MC that runs long enough "forgets" everything about its initial state. If the MC is periodic, then clearly you must remember some information about the contents of the initial state, because you're stuck in a loop of states that you keep on coming back to and aren't forgetting.

Solution 2:

Notice that your matrix $$\begin{align*}P=\left( \begin{array}{ccccc} 0 & \frac{1}{4} & \frac{3}{4} & 0 & 0\\ \frac{1}{4} & 0 & 0 & \frac{1}{4} & \frac{1}{2}\\ \frac{1}{2} & 0 & 0 & \frac{1}{4}& \frac{1}{4}\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ 0 & \frac{1}{4} & \frac{3}{4} & 0& 0\\ \end{array} \right).\end{align*}$$

has a special block decomposition as

$$ P=\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}$$

Where A,B,C,D are non zero submatrices.

To find $P^2$ we use the new form to get

$$ P^2 = \begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}\begin {pmatrix} 0&A &0\\B&0&C\\0&D&0\end{pmatrix}=\begin {pmatrix} AB&0 &AC\\0&BA+CD&0\\DB&0&DC\end{pmatrix} $$

As you know we can find powers of P by multiplying the new form of P as many times as we wish.

The alternating repeating form of powers of P is due to the block decomposition form of P.