Question about Markov Process I've studied Markov Process with 2x2 matrices.
Using the linear algebra and calculus procedures is clear to me how a Markov chain works.
However, i'm still not able to grasp the intuitive and immediate meaning of a Markov chain.
Why intuitively, for $n\rightarrow +\infty $, the state of the system is independent of both initial state and the states reached during the process ?
 A: It does depends on the initial state. Consider the Markov chain which consist of two disjoint state, that is $$\begin{bmatrix} 1 & 0 \\  0 & 1\end{bmatrix}$$
Then if you start in state $1$, you will stuck in state $1$ forever as it is an absorbing state. Similarly if you start in state $2$. 
Those example that you may have encounter might have the property that they only contain a unique stationary distribution, hence no matter where you begins, it ends up there.
A: Suppose that your process with matrix $P$ has a unique stationary distribution $\mathbf\pi_\infty$. This vector is an eigenvector of $1$, since $\mathbf\pi_\infty P=\mathbf\pi_\infty$, and the other eigenvalue $\lambda$ of $P$ has absolute value less than one. Every state vector $\mathbf\pi$ can be decomposed into the sum of the stationary distribution $\mathbf\pi_\infty$ and an eigenvector $\mathbf v$ of $\lambda$. We then have $$\mathbf\pi P^n = \mathbf\pi_\infty P^n+\mathbf v P^n = \mathbf\pi_\infty+\lambda^n\mathbf v.$$ Since $|\lambda\lt1|$, the second term will get smaller and smaller with increasing $n$ and vanish in the limit: intuitively, the influence on the process of the difference between the initial state and the stationary distribution wanes over time.  
Graphically, the situation looks something like this:

Every state vector lies on the line $x+y=1$. With each iteration, the component of the state vector parallel to this line gets shorter and shorter, bringing the state vector closer and closer to $\mathbf\pi_\infty$.
