Almost sure convergence of eigenvectors under nice assumptions I have that a sequence of symmetric, real, positive semi-definite random matrices, $M_n$, converges almost surely to a real-valued positive semi-definite diagonal matrix, $D$, with at least one diagonal entry equal to zero. How can I go about showing that the smallest eigenvector of $M_n$ converges almost surely to the smallest eigenvector of $D$, or that $v_i$ converges almost surely to $0$ if $D_{i,i}\neq 0$, where $v$ is the smallest eigenvector? Note that each $M_n$ is not necessarily diagonal.
 A: By smallest eigenvector of $M_n$, I assume that you consider an eigenvector associated to a smallest eigenvalue of $M_n$. Anyway, we consider only eigenvectors with norm $1$. Let $S_p$ be the set of symmetric real matrices of dimension $p$.
EDIT. Clearly you are not interested in my first answer; I rewrite my post and then, I leave the party. More I read your question, less it seems to me clear. If I correctly understand:
One has for a.e. $t\in \Omega$, $M_n(t)$ converges to $D(t)$.  Let $\lambda_n(t)$ be the smallest eigenvalue of $M_n(t)$; we assume that the $M_n[i,j]$ follow a probability law with a $C^{\infty}$ density function. 
It is known that if $(n,t)\rightarrow M_n(t)$ is continuous in $(+\infty,0)$, then $(t,n)\rightarrow \lambda_n(t)$ too; thus $\lambda_n$ converges a.s. to $0$. Unfortunately a unit eigenvector associated to $\lambda_n(t)$ is not necessarily continuous and you have no results about its convergence. 
Yet, if $D$ has a sole zero on the diagonal and if the $M_n[i,j]$ are $C^{\infty}$, then 
the convergence of the previous eigenvector works.
