One closed class makes the normalized stationary system nonsingular

If $U=\mathrm{\varnothing }$, $K=K_C$ is irreducible and row-stochastic. Otherwise, contract the strongly connected components of the finite positive-transition graph to form its component graph. This graph is acyclic, since a directed cycle would make its components mutually reachable and hence a single component. Every component therefore reaches a sink by following outgoing edges; the process terminates because the component graph is finite and acyclic. By [F2], sinks are exactly closed classes; C is the unique sink class. From each state in U, follow a component path to C. Each component edge is realized by a positive transition, and strong connectivity supplies finite positive paths from the current state to the source of each such transition and from its endpoint to the next transition's source. Concatenating these paths gives a positive path to C; before its first entry into C it stays in U. Order the states as $C,U$. Closedness makes the block form $K=\left(\begin{array}{cc}{K}_C& 0\\ R& T\end{array}\right)$, where T is row-substochastic; the path property gives the escape hypothesis in [F3].

Let $x$ satisfy $Px=x$ and put $w=x^{\mathsf{T}}$, so $wK=w$. In the block form of step 1.1, $w_U=w_{U}T$. If $U\mathrm{\ne }\mathrm{\varnothing }$, iterating $w_U=w_{U}T$ gives $w_U=w_U{T}^L$, and [F3] gives $T^L\mathbf{1}\le (1-\epsilon )\mathbf{1}$. Hence $\mathrm{\parallel }{w}_U{\mathrm{\parallel }}_1=\mathrm{\parallel }{w}_U{T}^L{\mathrm{\parallel }}_1\le {\sum }_{i\in U}\mathrm{\mid }(w_U{)}_i\mathrm{\mid }{\sum }_{j\in U}(T^L{)}_{ij}\le (1-\epsilon )\mathrm{\parallel }{w}_U{\mathrm{\parallel }}_1$, so $w_U=0$. With no transient states this conclusion is immediate.

The remaining block satisfies $w_C{K}_C=w_C$. The class $C$ is strongly connected, so [F4] gives $w_C=s{\pi }_C$ for some real $s$ and the unique invariant probability ${\pi }_C$. Thus every real fixed vector of $P$ is a scalar multiple of ${\pi }_C^{\mathsf{T}}$, and $\ker (I-P)$ is one-dimensional.

Fix any row index $r$, and let $A_r$ be $I-P$ with row $r$ replaced by ${\mathbf{1}}^{\mathsf{T}}$. If $A_{r}z=0$, then $(I-P)z$ is zero outside coordinate $r$. But ${\mathbf{1}}^{\mathsf{T}}(I-P)=0$ because P is column-stochastic, so its remaining coordinate is also zero. Hence $z=c{\pi }_C^{\mathsf{T}}$ by step 3.1; the replacement row gives $0={\mathbf{1}}^{\mathsf{T}}z=c$, so $z=0$. Therefore $A_r$ is nonsingular.

The vector ${\pi }_C^{\mathsf{T}}$ satisfies $(I-P){\pi }_C^{\mathsf{T}}=0$ and has coordinate sum one. Any real solution of the normalized system is fixed by P and has the same normalization, so step 3.1 makes it ${\pi }_C^{\mathsf{T}}$. Thus it is the unique solution, including the one-state boundary case.