Intuitive explanation of the stationary distribution of Markov chains I'm trying to understand (intuitively) the concept of a stationary distribution as it is defined on pp. 10 of these lecture notes.
This concept is defined there in the following way: Consider a Markov chain that is described by an $n \times n $ transition matrix $P$ (this implies, we consider a time-homogenuous, finite-state Markov chain - but this fact is not important for this question).
A vector $\pi$ is then called stationary distribution, of $\pi =P \pi$. 
What does this intuitively even mean? I can understand the formal definition, but I can'T visualize it.
 A: Firstly, the stationary distribution is a solution of equation $\pi =\pi P$, not the equation $\pi =P \pi$!  
Read the paragraph preceding the definition: 

In other words, a Markov chain is stationary if and only if the distribution of $X_n$ does not change with $n$. Since, by (3), the distribution $\mu_n$ at time $n$ is related to the distribution $\mu_0$ at time $0$ by $\mu_n = \mu_0 P^n$, we see that the Markov chain is stationary if and only if the
  distribution $\mu_0$ is chosen so that $\mu_0 = \mu_0P$.

Nothing to be added here. If $\mu_0=\pi$ is the initial distribution of MC, and it is stationary, then after one step we get the same distribution $$\mu_1=\mu_0P=\pi P=\pi=\mu_0.$$
Let us consider simple example. Transition matrix for a chain with two states is 
$$
P=\begin{pmatrix}\frac13 & \frac23\\ \frac12 & \frac12\end{pmatrix}
$$
Let us check whether the distribution $\pi$ which assigns probability $\frac37$ to the first state and $\frac47$ to the second state is stationary:
$$
\pi P=\begin{pmatrix}\frac37\ , &\frac47\end{pmatrix}\begin{pmatrix}\frac13 & \frac23\\ \frac12 & \frac12\end{pmatrix}=\begin{pmatrix}\frac37\ , & \frac47\end{pmatrix}=\pi
$$
It means that if the chain starts at time $0$ from any of two states with probabilities $\frac37$ and $\frac47$, then at time $1$ it will be at first state with probability 
$$
\begin{align}
P(X_1=1) &= P(X_0=1)P(X_1=1\mid X_0=1)+P(X_0=2)P(X_1=1\mid X_0=2)\\ &= \frac37\cdot\frac13+\frac47\cdot \frac12 = \frac37=P(X_0=1)
\end{align}$$
And the same for $X_1=2$.
Please note that the equation $\pi =P \pi$ has the solution which is not the stationary distribution. Any vector $\pi$ with equal coordinates solves this equation.
