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Let $B=A^T$ and choose any column probability vector $q$, for example the first coordinate unit vector. Because $A$ has nonnegative entries and each row sums to one, $B$ maps the probability simplexinto itself: $Bv\ge 0$ and $1^{T}Bv=(A1{)}^{T}v=1^{T}v=1$.

The iterates $B^{j}q$ need not converge, so take their arithmetic averages,A telescoping sum givesBoth vectors on the right before subtraction have L1 norm one, so $\Vert B{x}_N-x_N{\Vert }_1\le 2/N\to 0$.

The probability simplex is closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. It follows that a subsequence $x_{N_j}$ converges to some $x\in \Delta$. Passing to the limit in the displayed identity, using continuity of the linear map $B$, gives $A^{T}x=Bx=x$, with $x\ge 0$ and ${\sum }_i{x}_i=1$.

This proves stationary distribution existence for a finite Markov chain without any Markov chain irreducibility or aperiodicity assumption. Those stronger hypotheses concern uniqueness and convergence, not the existence required here.

A probability vector lists the probabilities of the outcomes in a finite sample space. Its entries are nonnegative and sum to one, so a column probability vector is a point of a probability simplex. Row and column conventions describe the same probability distribution, but they determine whether a stochastic matrix acts on the right or its transpose acts on the left.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.