Finding stationary Distribution I need to know how to find the stationary distribution for this matrix:
$$
Q= 
\begin{bmatrix}
-2 & 2  & 0 & 0 \\
1  & -2 & 1 & 0 \\
0 & 1 & -2 &1\\
0  & 0  & 2 & -2 \\
\end{bmatrix}$$
in three different ways.
I only know one, which is to replace the last column with $1$'s and then take the inverse. The last row will be the stationary distribution.
Thank you
Update: Sorry, I forgot a row. Thanks for the comments!
 A: Since you are asking for different proofs, here is a proof by exhibition: the unique stationary distribution $\pi$ is
$$
\pi=\left(\tfrac16,\tfrac13,\tfrac13,\tfrac16\right).
$$
And a proof by uniformization: assume that the state space is $S=\{1,2,3,4\}$, duplicate state $2$ into states $2$ and $6$, and duplicate state $3$ into states $3$ and $5$. Thus, the transition rates from state $1$ are now $+1$ to state $2$ and $+1$ to state $6$. Likewise, the transition rates from state $4$ are now $+1$ to state $3$ and $+1$ to state $5$. The transition rates from states $2$ and $3$ are unchanged and the transition rates from states $5$ and $6$ are those from states $3$ and $2$ respectively.
Phew! Almost done... The resulting Markov chain performs a standard random walk on the state space $\bar S=\{1,2,3,4,5,6\}$, best viewed as the discrete circle $\mathbb Z/6\mathbb Z$. Its  stationary distribution $\bar\pi$ is obviously uniform on $\bar S$ and $\pi$ is the projection of $\bar\pi$ on $S$. States $1$ and $4$ correspond to one state in $\bar S$ while states $2$ and $3$ correspond to two, hence the ratios $1:2:2:1$ in $\pi$.
And finally a proof by definition (probably the most direct in this context): solve $\pi Q=0$.
