Let the transition matrix for a stochastic process be
$$P=\begin{bmatrix} 1 & 0.7 & 0 & 0 & 0 \\ 0 & 0 & 0.5 & 0 & 0 \\ 0 & 0.3 & 0 & 0.65 & 0 \\ 0 & 0 & 0.5 & 0 & 0 \\ 0 & 0 & 0 & 0.35 & 1 \end{bmatrix}$$
and let $\mathbf{x}_0$ be the initial state. Let $s_1$ be the state with the vector
$$\mathbf{x}_0=\begin{bmatrix}1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
The communication class (I don't know the English word) $K_1 = \{s_1\}$ is a closed class, with means that if $s_1 \leadsto s_i$ then $s_i \in K_1$.
The formula for calculating the probabilty that an initial state $s_j$ will end up in $K_1$, sooner or later, is
$$x_j^{K_1} = 1, s_j \in K_1$$ $$x_j^{K_1} = \sum_{i=1}^n{p_{ij}x_i}, s_j \notin K_1$$
where $p_{ij}$ is the entry in row $i$ and column $j$ in P.
Here comes my problem. I get the following result: if the initial state is $s_2$ then the probability that the process will end in $K_1$ is $0$!
Because I solve the system $P\mathbf{x} = I_5\mathbf{x}$ which is $(P-I_5)=\mathbf{0}$. The solution is the vector
$$\begin{bmatrix}x_1 \\ 0 \\ 0 \\0 \\ x_5 \end{bmatrix}$$
which is
$$\begin{bmatrix}1 \\ 0 \\ 0 \\0 \\ 0 \end{bmatrix}$$
in relation to $K_1$.
This I cannot understand. The probability that $s_2$ could end up in $K_1$ can't be zero?
