An irreducible finite class has a unique invariant vector

By [F3], choose an invariant probability row vector $\pi$. At least one coordinate ${\pi }_j$ is positive because its finite nonnegative coordinates sum to one.

Fix any $i\in C$. By irreducibility there is a finite positive-probability path from $j$ to $i$, so for some $r\ge 0$ the matrix power satisfies $(K^r{)}_{ji}>0$. Invariance implies $\pi K^r=\pi$, hence ${\pi }_i={\sum }_k{\pi }_k(K^r{)}_{ki}\ge {\pi }_j(K^r{)}_{ji}>0$. Thus every coordinate of $\pi$ is positive.

For any real invariant row vector $w$, define $f_i=w_i\mathrm{/}{\pi }_i$. The matrix $R$ with entries $R_{ik}={\pi }_k{K}_{ki}\mathrm{/}{\pi }_i$ is well-defined by step 2.1, is nonnegative, and has row sums ${\sum }_k{R}_{ik}={\sum }_k{\pi }_k{K}_{ki}\mathrm{/}{\pi }_i=1$ by invariance of $\pi$. Also $f_i={\sum }_k{R}_{ik}{f}_k$, by invariance of $w$. The directed graph of $R$ reverses that of $K$, so it is irreducible.

Since $C$ is finite, choose an index $i$ at which $f_i$ attains its maximum $M$. The identity $M=f_i={\sum }_k{R}_{ik}{f}_k$ is a convex combination of numbers at most $M$, so every $k$ with $R_{ik}>0$ has $f_k=M$. Repeating along directed paths of the irreducible matrix $R$ gives $f_k=M$ for every $k\in C$. Therefore $w=M\pi$.

Taking $w$ to be any invariant probability shows it equals $\pi$, because its coordinate sum is one; taking arbitrary real invariant $w$ gives the asserted one-dimensional fixed space. For $n=1$ the same argument has a single coordinate and is immediate. The proof uses finite paths, finite maxima, and finite sums only.