Convergence for Perron eigenvectors

Question

Let $M_n$ be a sequence of matrices that converges to $M$, component-wise: $(M_n)_{ij}\to M_{ij}$ as $n\to\infty$ for each pair $i,j$. Suppose that $\lambda_n,v_n$ is an eigenpair for $M_n$ each $n$, and that $\lambda,v$ is an eigenpair for $M$.

A general question of interest is: when does $M_n\to M$ and $\lambda_n\to \lambda$ imply that $v_n\to v$?

My specific situation is that: all matrices are nonnegative (each matrix component is nonnegative), irreducible (strongly connected directed graph), aperiodic (gcd 1 for all directed paths $i$ to $j$) and hence primitive (unique max mod eigenvalue) and that the eigenpairs are all Perron eigenpairs. (See below for additional background.)

Now suppose that $\lambda_n,v_n$ is the Perron eigenpair for $M_n$ and similarly for $\lambda,v$. Assume that $M_n\to M$ and $\lambda_n\to \lambda$ as $n\to\infty$. Is it possible to prove that $v_n\to v$? Edit: I also meant to mention that each eigenvector sums to one, i.e. $\sum_j v_j=1$.

I don't write all of my own attempts here, but they are all variations on basic analysis methods, triangle inequalities, $\epsilon/3$ stuff etc. I though maybe taking a subsequence: since $(v_n)_j$ (the sequence for the $j^{th}$ component of $v_n$) is a bounded sequence, it has a convergent subsequence. But the subsequence of indices could be different for each $j$.

Maybe someone can come up with a counterexample?

Additional background:

"Irreducible" means that if you draw a directed edge between indexes $i\to j$ when $M_{ij}>0$, and $i\leftarrow j$ when $M_{ji}>0$, and that if we draw that directed graph for the entire matrix we can find a directed path from each index to every other index.

This graph is called "aperiodic" if the greatest common divisor for the number of steps for all possible paths between indices $i$ and $j$ is one---i.e. that you don't have a situation where any path from $i$ to $j$ is, say, always a multiple of $2$ (giving period 2).

"Primitive" means that there is a unique eigenvalue of maximum modulus which is algebraically simple, i.e. every other eigenvalue will have a strictly less absolute value. We call the unique max mod eigenvalue and its eigenvector the Perron eigenpair. This vector is also strictly positive.

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EDIT: (solution attempt)

We have $M_nv_n=\lambda_nv_n$ for each $n$, $\lambda_n\to\lambda$, and that $M_n\to M$ and that $v$ is the only eigenvector (up to scalar multiple) for eigenvalue $\lambda$. I am also assuming that the eigenvectors are strictly positive and that the components of each $v_n$ and $v$ sum to one, e.g. $\sum_j v_j=1$.

We have: $$\begin{aligned} (\lambda I-M)v_n&=(\lambda_n I-M - \lambda_n I +\lambda I)v_n\\ &=(M_n I-M)v_n - (\lambda_n -\lambda)v_n \end{aligned}$$ Hence taking the $j^{th}$ component: $$\left|[(\lambda I-M)v_n]_j\right| \leq \left|[(M_n-M)v_n]_j\right| + \left| \lambda_n-\lambda\right| |[v_n]_j|.$$

Now we can choose $N$ so large such that for any $n>N$ we have each term on the right arbitrarily small. Thus the left side can be made arbitrarily small for all $j$ simultaneously (note that $v_n$ has bounded components since they are "normalized" to sum to one and nonnegative). This means that $(\lambda I-M)v_n$ can be made arbitrarily close to the zero vector. This is only possible if $v_n$ gets arbitrarily close to $v$, since $(\lambda I-M)x$ can only be close to zero if $x$ is close to $v$ (because $v$ is the only eigenvector that goes with eigenvalue $\lambda$ for $M$--- the only one whose components sum to one that is).

Answer 1

More generally, suppose $M$ is a matrix with a simple eigenvalue $\lambda$ (i.e. generalized eigenspace one-dimensional) and $M_n$ a sequence of matrices converging to $M$. The eigenprojection for $M$ on $\lambda$ is $P(M) = \frac{1}{2\pi i} \oint_C (zI - M)^{-1}\; dz$, where $C$ is a positively oriented closed contour with $\lambda$ inside and all other eigenvalues for $M$ outside. If $n$ is sufficiently large, $P(M_n) = \frac{1}{2\pi i} \oint_C (zI-M_n)^{-1}\; dz$ is also a one-dimensional projection, and $P(M_n) \to P(M)$ as $n \to \infty$. In fact, $P(M_n)$ is the eigenprojection for a simple eigenvalue $\lambda_n$ of $M_n$ with $\lambda_n \to \lambda$ as $n \to \infty$, and the corresponding eigenvector $v_n$ (which we can take as $P(M_n) v$ where $v$ is an eigenvector of $M$ for $\lambda$) converges to $v$ as $n \to \infty$.

Answer 2

The unit sphere of an Euclidean space is compact so there is a subsequence $(v_{n_i})$ converging to some vector $w$ of norm $1$. Taking limits in the equation $M_{n_i}v_{n_i}=\lambda_{n_i}v_{n_i}$ we get $Mw=\lambda w$. Since the eigen space coresponding to $\lambda $ is one dimensional we get $w=v$.

Now use the fact that $v_n \to v$ iff every subsequence of $(v_n)$ has a further subsquence which converges to $v$.