I have that a sequence of random matrices, $M\_n$, converges almost surely to a diagonal matrix, $D$, with finite real entries on its diagonal. During convergence, the off-diagonals are not necessarily zero. How can I go about showing that the smallest eigenvector of $M\_n$ converges almost surely to the smallest eigenvector of $D$?


**Edit:** If it helps, at least one diagonal entry of $D$ is zero and each $M\_n$ is symmetric.



Almost sure convergence of smallest eigenvector of diagonal matrix
The set of symmetric matrices that have multiple eigenvalues has Lebesgue measure 0. If the probability measure you use on the space of matrices is absolutely continuous w.r.t. the Lebesgue measure, then the probability that a random matrix has multiple eigenvalues is zero. I assume that this is the case. Assume that the matrices are $d\times d$.


$\newcommand{\bP}{\mathbb{P}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eO}{\mathscr{O}}$ Suppose that your random matrices are defined on the probability space $(\Omega,\eO,\bP)$. Let $N\in\eO$ such that $\bP[N]=0$ and for any $\omega\in\Omega\setminus N$ all the matrices $M\_n(\omega)$ have simple eigenvalues and


$$\lim\_{n\to\infty} M\_n(\omega)=D. $$


Denote by $\lambda\_1(\omega)$ the smallest eigenvalue of $M\_n(\omega)$ and by $\phi\_n(\omega)$ an eigenvector of Euclidean norm $1$ corresponding to this eigenvalue. (There are two choices for $\phi\_n(\omega)$ since $\lambda\_1(\omega)$ is simple.) Then for any $\phi\in\bR^d$ of Euclidean norm $1$ we have


$$ \lambda\_1(\omega)=\langle\; M\_n(\omega)\phi\_n(\omega),\phi\_n(\omega)\;\rangle\leq \langle\; M\_n(\omega)\phi,\phi\;\rangle. $$


If we let $n\to \infty$ along a subsequence $n\_k$ such that $\phi\_{n\_k}(\omega)$ converges to some $\phi\_\infty(\omega)$ of norm $1$ we deduce


$$ \langle\; D\phi\_\infty(\omega),\phi\_\infty(\omega)\;\rangle \leq \langle\; D\phi,\phi\; \rangle,\;\;\forall \phi\in\bR^d,\;\;\Vert\phi\Vert=1. $$


This proves that any limit vector $\phi\_\infty(\omega)$ of the sequence $(\; \phi\_n(\omega)\;)$ is an eigenvector of norm $1$ of $D$ corresponding to the lowest eigenvalue. That is the best that you can hope.


