TITLE: Convergence of a sequence of eigenvectors (nonnegative matrix)
QUESTION [9 upvotes]: Let $A$ be a $n\times n$ matrix with coefficients in $ [0,1] $. Let $ B $ be the matrix filled up only with the value $ \frac{1}{2} $:
$$B = \begin{pmatrix}
\frac{1}{2} & \dots  & \frac{1}{2} \\ 
\vdots & \ddots  & \vdots \\ 
\frac{1}{2} & \dots  & \frac{1}{2} 
\end{pmatrix}\,.$$
For all $ t \in ]0,1] $ let $ A(t) = tB + (1-t)A $. The matrix $ A(t) $ is primitive for any fixed $ t $. Hence, from the Perron-Frobenius theorem, we have that $ \rho(t) $ (i.e. the spectral radius of the matrix $ A(t) $) is a simple eigenvalue, with the relative eigenvector that can be taken positive (i.e. every component is strictly positive). Let me call it $ x(t) $, choosing it such that $ \|x(t)\|_1 = \rho(t) $ (i.e. the sum of all the components is equal to the spectral radius of the matrix). In this way, I obtain the following properties:
\begin{matrix}
A(t)x(t) = \rho(t)x(t) \\
\|x(t)\|_1 = \rho(t)\\ 
x(t) > 0
\end{matrix}
My question is: does there exist $ \lim_{t \to 0^+}x(t) $?
I have read that spectral radius is continuous with respect to any matrix norm, and then I would have: $ \lim_{t \to 0^+}A(t) = A \Rightarrow \lim_{t \to 0^+}\rho(t) = \rho(0) $. Is it correct?
Anyway, passing to the sequences with $ t = \frac{1}{n} $, I did not succeed in proving that the generated sequence $ x_n = x(\frac{1}{n}) $ is a Cauchy sequence (this fact would imply that $ x_n $ is convergent, because of the sequentially compactness of $ [0,1]^n $).
I think that the better way in order to prove the existence of the limit above is to prove the monotonicity of the components of $ x_n $ using the monotonicity of the coefficients $ a_{ij}(t) $ of $ A(t) $. It is only an idea but I do not know if it works.
I really thank you in advance.

REPLY [3 votes]: For $4$ days, I have been stuck essentially at the place where Surb arrived. I detail the basic reasoning
Note that, when $A(t)$ is continuous, the spectral radius $\rho(t)$ of $A(t)$, is always continuous on $[0,1]$.
When $t>0$: $A(t)$ is an analytic function of $t$ and a positive matrix;  moreover $\rho(t)>0$ is a simple eigenvalue s.t. $A(t)x(t)=\rho(t)x(t)$ where $x(t)$ is a positive vector satisfying $||x(t)||=1$ for some norm. Then, according to
https://www.math.upenn.edu/~kazdan/504/eigenv.pdf
$\rho(t)$ and $x(t)$ are analytic functions on $(0,1]$.
$\textbf{Proposition 1.}$ If $A$ is symmetric, or more generally, $A$ is normal and $(A-A^T)B=B(A-A^T)$, then $x(t)$ converges.
$\textbf{Proof.}$ Since $A(t)$ is normal over $[0,1]$, $\rho(t)$ and $x(t)$ are analytic over $[0,1]$. cf. Theorem(A) in
https://arxiv.org/pdf/1111.4475v2.pdf
$\square$
$\textbf{Proposition 2.}$ The cluster points of $x(t)$ are some non-negative vectors in $\ker(A-\rho(A)I)$.
$\textbf{Proof}.$ Consider a sequence $(t_p)$ that tends to $0$ s.t. $x(t_p)\rightarrow x\;(\geq 0)$; then $Ax=\rho(A)x$. $\square$
$\textbf{Remark.}$ It "remains " to show that all these subsequences tend to the same limit.
$\textbf{Corollary.}$ If $dim(\ker(A-\rho(A)I))=1$, then $x(t)$ converges.
EDIT. 
$\textbf {Proposition 3.}$ $\lim x(t)$ (if it exists) is not a continuous function of $A$.
$\textbf{Proof.}$  Let $A=\begin{pmatrix}1&a&0\\0&4/5&1/10\\0&2/5&4/5\end{pmatrix}$ where $\rho(A)=1$.  We use $||.||_{\infty}$.
If $a=0$, then  $dim(\ker(A-I))=2$ and $x(t)\rightarrow [2/3,1/2,1]^T$.
If $a\not= 0$, then $dim(\ker(A-I))=1$ and $x(t)\rightarrow [1,0,0]^T$.  $\square$
EDIT 2. About the  sasquires' post and the following reference that user1551 gave (indirectly) 
[1] A note on perturbations of stochastic matrices, Huppert, Willems, Journal of Algebra (2000). -in free access-
https://reader.elsevier.com/reader/sd/pii/S0021869300985442?token=A1062A5C61A49B95037719BF2E60DBE3462EA3D00AD1980BC8CFC9C479C5C1250AC05536F459134264505903FCFFA812
One has $2$ questions. i) Does $x(t)$ converge ?  ii) If yes, towards what ?
i) For almost stochastic matrices, the authors of [1] claim yes in Theorem 3-2 (a). Yet, I am not convinced by the end of the proof of (a); could someone read this proof and say what he/she thinks ?
ii) sasquires claims, in his post, that the following is true (I think that too)
$\textbf{Conjecture.}$ Let $A\in M_n$ be non-negative where $\rho(A)>0$ is semi simple and (do we need that ?) is the unique eigenvalue with modulus $\rho(A)$. IF $x(t)$ CONVERGES, then the limit can be explicitly calculated as follows
Let $(v_i)_{i\leq k},(u_i)_{i\leq k}$ be bases of $\ker(A-\rho(A) I)$ and $\ker(A^T-\rho(A)I)$ that satisfy $u_i^Tv_i=1$ and, when $i\not= j$, $u_i^Tv_j=0$.
-Of course, one must prove that such bases exist-
$R$ is the $k\times k$ matrix $R_{i,j}=u_i^T(B-A)v_j$; $c$ is the eigenvector associated to $\max( spectrum(R))$; finally -up to a factor- $\lim x(t)=\sum_i c_iv_i$.
sasquires, it's your idea; then, it's your job to write a flawless proof.
$\textbf{Remark.}$ If $x(t)$ converges and the above conjecture is true, then 
$\textbf{Lemma.}$ Let $a,b>0$ and let $y(t)$ be the eigenvector associated to $(aA,bB)$. Then $\lim y(t)=\lim x(t)$.
$\textbf{Proof.}$ We may assume that $b=1$. Then the new matrix $R$ is $S_{i,j}=u_i^T(B-aA)v_j=u_i^TBv_j-au_i^TAv_j$ and $S=[u_i^TBv_j]-a\rho(A)I$. Finally we may replace $R$ with the matrix $[u_i^TBv_j]$. $\square$
In particular, we may replace $B$ with the matrix $[1/n]$, that is, a stochastic matrix and $A$ with a matrix satisfying $\rho(A)=1$.