I'm trying to understand this solution to a question related finding the steady state matrix $s$ for a regular markov chain.
Specifically I'm having trouble understanding how my textbook got
$$ \left( \begin{matrix}s1\\s_2 \end{matrix}\right) = \left( \begin{matrix}2r \\r \end{matrix}\right) $$
If the transition matrix $T$ is $\left( \begin{array}{ccc} \frac{1}{2} & 1 \\ \frac{1}{2} & 0 \end{array} \right)$ and $Ts=s$, $(T-I)s=0$
The textbook reduced the augmented matrix $[(T-I)|0]$ and somehow got $ \left( \begin{matrix}s1\\s_2 \end{matrix}\right) = \left( \begin{matrix}2r \\r \end{matrix}\right) $
I don't understand why this works?
In order to find the steady-state vector $s = \begin{pmatrix} s_1 \\ s_2 \end{pmatrix}$ you need to solve a simple matrix equation \begin{equation} (T - I)s = 0. \end{equation} But \begin{equation} T - I = \begin{pmatrix} -\frac{1}{2} & 1 \\ \frac{1}{2} & -1 \end{pmatrix} \end{equation} and, thus, you have a linear system \begin{equation} \begin{cases} -s_1/2 + s_2 = 0; \\ s_1/2 - s_2 = 0. \end{cases} \end{equation} Hence, we obtain $s_1 = 2s_2$.
So, if you denote $r = s_2$ then vector $s = \begin{pmatrix} 2r \\ r \end{pmatrix}$ is a solution of the matrix equation for every real number $r$.
And finally, vector $s$ must satisfy the normalization constraint: \begin{equation} s_1 + s_2 = 1. \end{equation} Therefore, $2r + r = 1$, that is $r = 1/3$ and the steady-state vector is the following: $$s = \begin{pmatrix} 2/3 \\ 1/3 \end{pmatrix}.$$