Proofs from Markov systems

Non-negativity follows because each entry of $A\mathbf{x}$ is a sum of products of non-negative numbers. Let $\mathbf{1}$ be the all-ones vector. Since the columns of $A$ sum to $1$,

Therefore

The column-sum condition gives

so $1$ is an eigenvalue of $A^{\mathsf T}$. Since

$A$ and $A^{\mathsf T}$ have the same eigenvalues. Hence $1$ is an eigenvalue of $A$. The all-ones vector is generally an eigenvector of $A^{\mathsf T}$, not of $A$.

Let $v_1,\ldots,v_n$ be an eigenbasis with $Av_1=v_1$ and $Av_j=\lambda_jv_j$ for $j\ge2$. Expand the initial state as

Repeated application of $A$ gives

Since $|\lambda_j|<1$ for $j\ge2$, the sum tends to zero. Thus $\mathbf{x}(k)\to\alpha_1v_1$. For $j\ge2$, the column-sum identity $\mathbf1^{\mathsf T}A=\mathbf1^{\mathsf T}$ gives $\mathbf1^{\mathsf T}v_j=\lambda_j\mathbf1^{\mathsf T}v_j$, hence $\mathbf1^{\mathsf T}v_j=0$. Since the eigenvectors form a basis, $v_1$ must have non-zero component sum. The preserved initial sum therefore determines $\alpha_1v_1$ uniquely. If the initial state is a probability vector, every $\mathbf{x}(k)$ is one too, so its limit is non-negative and sums to $1$.