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

Question

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?

Answer 1

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

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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.

Answer 2

It is a matrix such that $P^4= P^2$ and some additional relations. We see that

$$ 0 = P^4 - P^2 = P^2(P^2 - I) = P^2(P-I)(P+I)$$

so the minimal polynomial $\mu_P(x)$ divides $x^2(x-1)(x+1)$.

However, the minimal polynomial is not any of $$x, x^2, x-1, x+1, \underbrace{x(x-1)}_{P^2 \ne P}, \underbrace{x(x+1)}_{P^2 \ne -P}, \underbrace{x^2(x-1)}_{P^3 \ne P^2}, \underbrace{x^2(x-1)}_{P^3 \ne -P^2}, \underbrace{(x-1)(x+1)}_{P^2 \ne I}, \underbrace{x(x-1)(x+1)}_{P^3 \ne P}$$ so it has to be precisely $x^2(x-1)(x+1)$.

The possibilities for the Jordan normal form of $P$ are:

$$\pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0}, \pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & 1}, \pmatrix{0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 & -1}$$

meaning $5 \times 5$ matrices have the above properties if and only if they are similar to one of the three matrices above.

Answer 3

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.

Answer 4

The entries are all between 0 and 1 and in each row the entries add up to 1. They are called stochastic matrices. They have 1 as an eigen value with $(1,1,\ldots,1)^T$ as corresponding eigenvector. This property is inherited by all the powers of that matrix.

Answer 5

You can draw a graph for which your matrix is the matrix of probabilities of state change (some weights on the edges). Paint states A, D and E as white, and B and C as black. Then any possible move changes the color.

Since P^n is the probablility matrix of moving from X to Y in exacly n steps (or sum of products of weights spotted by any n-step path) then the oscillation occurs because your graph is bipartite. You can produce oscillation with other period if you wish this way.