Doubly stochastic matrix proof

Proof:

We first must note that $\pi_j$ is the unique solution to $\pi_j=\sum \limits_{i=0} \pi_i P_{ij}$ and $\sum \limits_{i=0}\pi_i=1$.

Let's use $\pi_i=1$. From the double stochastic nature of the matrix, we have $$\pi_j=\sum_{i=0}^M \pi_iP_{ij}=\sum_{i=0}^M P_{ij}=1$$ Hence, $\pi_i=1$ is a valid solution to the first set of equations, and to make it a solution to the second we must normalize it by dividing by $M+1$.

Then by uniqueness as mentioned above, $\pi_j=\dfrac{1}{M+1}$. $$ \blacksquare$$

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Note : To understand this proof, one must recall the definition of a stationary distribution.

A vector $\mathbf{\pi}$ is called a stationary distribution vector of a Markov process if the elements of $\mathbf{\pi}$ satisfy: $$ \mathbf{\pi} = \mathbf{\pi} \cdot \mathbf{P}, \sum_{i \in S} \pi_{i} = 1 \text{ , and } \pi_{i} > 0\text{ }\forall \text{ } i \in S $$ Note that a stationary distribution may not exist, and may not be unique.