I am dealing with a topological question concerning the mapping that maps a positive stochastic matrix onto its invariant distribution. I am asking myself if such a mapping is continuous (or measurable) with respect to suitable topologies ($\sigma$-fields). Let me describe the problem some more precisely.
Let $n>1$ and $S(n,+)$ be the space of (element-wise) positive (row-)stochastic matrices and let $A\in S(n,+)$.
Due to the Perron-Frobenius theorem, $\rho(A)=$1 is the unique eigenvalue on the spectral radius of $A$ with corresponding (right) eigenvector $v=v_A>0$, where $v$ is normalized, i.e. it is (row-)stochastic and hence can be viewed to be the unique $A$-invariant distribution. Furthermore, $B:=B_A:=\lim\limits_{n\rightarrow \infty}A^n=\lim\limits_{n\rightarrow \infty}\frac{1}{n}\sum\limits_{k=1}^n A^k\in S(n,+)$, where the rows of $B$ are all equal and are all stochastic (left) eigenvectors of $A$ and $BA=AB=B$ as well as $B^2=B$. Thus, $B$ is a projection, called the Perron-projection and can be written as $B=vw^t$, where $v,w$ are such that $Av=v$, $w^tA=w^t$ and $w^tv=1$.
I am wondering if the mapping $A\mapsto B_A$ or $A\mapsto v_A$, respectively, is continuous with respect to the topologies of coordinate-wise convergence.
In other words, do minor changes in the positive stochastic matrix only give rise to minor changes in the corresponding one-dimensional Perron-eigenspaces?
How can this problem best be regarded and approached? In particular in situations where it is not efficient to solve the identity $w^t A=w^t$ explicitly.
If it was a known problem, please let me know where it is described or solved, respectively.
Thanks in advance and regards
EDIT 1. Your vector $v$ is $[1,\cdots,1]^T$; then, $A\rightarrow B_A$ and $A\rightarrow w$ are essentially the same functions.
Let $I$ be an interval of $\mathbb{R}$ and $t\in I\rightarrow A(t)$ be $C^{\infty}$ with values in the positive stochastic matrices. For every $t\in I$, the eigenvalue $1$ is simple and the associated left-eigenvector is defined up to a factor. Therefore, $(*)$ there is a $C^{\infty}$ function $t\in I\rightarrow w(t)\in (0,+\infty)^n$ s.t., for every $t\in I$, $w(t)^TA(t)=w(t)^T$ (note that $(*)$ is a global result).
The proof is easy. Let $t_0\in I$; here $\lambda=1\in \mathbb{Q}$ and there is a dimension $n-1$ submatrix $B(t)$ of $A(t)-\lambda I$ s.t. $\det(B(t_0))\not=0$. While $\det(B(t))\not= 0$, $w(t)$ (for instance $=(w_1(t),\cdots,w_{n-1}(t),1)$) is a rational fraction in the $(a_{ij}(t))$ and in $\lambda$, then in the $(a_{ij}(t))$. More generally, if $A(t)$ and one of its simple eigenvalue $\lambda(t)$ are $C^{\infty}$ (resp. holomorphic), then an eigenvector associated to $\lambda(t)$ is locally given by a $C^{\infty}$ (resp. holomorphic) function.
Beware, $(*)$ is no longer valid if $A(t)$ is only non-negative stochastic.
In fact, you ask about the function $A\rightarrow w$. Here we have only local results. Assume that $A\in M_n(\mathbb{C})$ and consider a small varaiation of $A$; then the spectrum of $A$ varies; in general, there does not exist a natural ordering of the eigenvalues of $A$, except when the eigenvalues $(\lambda_i)$ are distinct. Hopefully, when the eigenvalues are distinct, using the induced ordering, Kato proved the local existence of holomorphic functions $A\rightarrow \lambda_i$ and $A\rightarrow w_i$ where $w_i$ is an eigenvector associated to $\lambda_i$.
Yet, if a special eigenvalue $\lambda$ can be isolated s.t. $A\rightarrow \lambda$ is continuous and simple, then we can prove (in a similar way) the existence of holomorphic functions associated to $\lambda$.
The set of positive real matrices, $M^+$, is open in $M_n(\mathbb{R})$.
Proposition. Let $A_0\in M^+$. There are analytic functions $A\rightarrow \rho(A)$, $A\rightarrow v\in (0,\infty)^n$ (s.t. $Av=\rho(A)v$) that are defined in a neighborhood of $A_0$.
EDIT 2. @ K.O.T. Let $w^T=[z,1],A-I=\begin{pmatrix}B&c\\l&\alpha\end{pmatrix}$ where $\det(A-I)=0$, that is $lB^{-1}c=\alpha$ and $B$ is invertible. Then $w^T(A-I)=0$ can be written $zB+l=0,zc+\alpha=0$. Finally $z=-lB^{-1}$ and then $zc+\alpha=0$ is automatically satisfied. $l$ is a part of $A$ and the entries of $B^{-1}$ are rational fractions in the $(a_{i,j})$; that implies that the $(z_j)$ are rational fractions in the $(a_{i,j})$.