Understanding the Long Term Behavior of a Stochastic Matrix using Eigenpairs

Question

I have a problem with understanding a stochastic matrix and a situation concerning the long term transition of a state

Suppose that a particular town has three states: Commercially used (25%), Industrially used (20%), and Residentially used (55%). Then suppose that we have the following transition probabilities given in matrix form

$$A=\left[\begin{matrix} 0.7 & 0.1 & 0 \\ 0.2 & 0.9 & 0.2 \\ 0.1 & 0 & 0.8 \\ \end{matrix}\right]$$

Here the first column is from Commercial transitioning to (in order) Commercial, Industrial, and Residential. In other words, after the first iteration, 70% of commercially allocated land will remain commercially allocated, 20% will transition to industrially allocated land, and 0.1 will transition to residential. Second column is industrial, and last is residential. So it makes sense that the largest probabilistic entries are those where no transition takes place.

So it's easy to see how the transition of land occurs over intervals of time, through Matrix multiplication. But the problem that I am tasked to solve is the long term behaviour of this matrix on the original vector $\mathbf{x}=\left[\begin{matrix} 0.25 & 0.2 & 0.55 \\ \end{matrix}\right]^T$. The textbook I am using (Kreyszig's Advanced Engineering Mathematics) says that we can determine the long term probabiliy by finding the eigenvector associated with the eigenvalue of 1. (All stochastic matrices have an eigenvalue of 1 or less than 1, from what I read). How does this process work? Why does finding the eigenvalue of 1 and thus finding the corresponding eigenvector give us the long term behavior of $A$ on any vector $\mathbf{x}$?

I thought by finding the eigenvalue of 1, and the corresponding eigenvector of $\mathbf{x}=\left[\begin{matrix} t & t & t \\ \end{matrix}\right]^T$ for some parameter $t$, we would be finding all initial state vectors that are preserved by the matrix multplication and scaled appropriately. Here, we would need a specific value of $t$ as $\frac{1}{9}$ so that the long term state would become $\mathbf{x}=\left[\begin{matrix} \frac{2}{9} & \frac{1}{3} & \frac{1}{9} \\ \end{matrix}\right]^T$

Since the eigenvector has direction, and since the original land state vector is not a scalar multiple of the long term vector, why did we need to find eigenvalues and vectors? Very confused as to the mechanics here.

EDIT: Here is the problem and solution in the text I am using. You can see that it appears very 'hand-wavy' and to me, really doesn't explain why this process produces long term states and has no bearing on the original land state;

Answer 1

First observe that the chain is irreducible.Which means that there is one class of communication and is recurrent (i.e all the states communicate to each other).

Your t.p.m is column wise stochastic.And your long-term distribution is $\pi(x) = \{\frac{2}{9},\frac{6}{9},\frac{1}{9}\}$.The distribution vector must sum to 1.

The need of eigenvalues and eigenvectors now. The transition probabilities of higher order satisfy the Chapman - Kolmogorov equations :

$$p^{m+n}(x,y)=\sum_{z \in \mathbb{X}} p^{(m)}(x,z)p^{n}(z,y)$$ for all $m,n \in \mathbb{N}_{0} \quad ,x,y \in \mathbb{X}$.

Taking into account these equations which are the Law of Total Probability for the event $\{X_{m+n}=y\}$ in relation with the condition of the chain the time $m$.

Define the power $$P^{n} =\{p^{(n)}(x,y) \}_{x,y \in \mathbb{X}}$$ of a stochastic matrix $P$ for $n \in \mathbb{N}_{0}$ and using the Fubinni - Tonelli theorem we have:

$P^{0}=I$ and in general $$P^{n}=P^{n-1}P $$ for $n \in \mathbb{N}$. By this all you have to do it send $n$ to large number but you must calculate each time this product.Avoiding this calculations you can decompose the tpm matrix.

Diagonilizing the matrix $P$ we will obtain the eigenvalue decomposition: $$Q\Lambda Q^{-1}$$.In the eigenvalue matrix the $\lambda$'s will be $$\Lambda = \begin{bmatrix} 1 & 0& 0 \\ 0 & \lambda_{1}^{n}& 0 \\ 0 & 0& \lambda_{2}^{n} \\ \end{bmatrix}$$ Sending $n \to \infty$ and reconstruct the tpm $P$ you will obtain the long term distribution of the chain.

Edit

In Python you can confirm:

import numpy as np
from scipy.linalg import eig 
transition_mat = np.matrix([
  [.7, .2, .1],\
  [.1, .9,0],\
  [0, .2,.8]])

S, U = eig(transition_mat.T)
stationary = np.array(U[:, np.where(np.abs(S - 1.) < 1e-8)[0][0]].flat)
stationary = stationary / np.sum(stationary)

print(stationary.real)

[0.22222222 0.66666667 0.11111111]