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.

  • $\begingroup$ You need to make some assumptions on the nature of randomness of the matrices $M_n$ that I gather are finite dimensional. If you assume that the entries of $M_n$ are continuous random variables then the result is true because in this case, with probability one, the spectrum of $M_n$ is simple. $\endgroup$ Dec 21, 2015 at 15:44
  • $\begingroup$ @LiviuNicolaescu "because in this case, with probability one, the spectrum of $M_n$ is simple." Can you elaborate on why this implies the result? $\endgroup$
    – PThomasCS
    Dec 21, 2015 at 16:42
  • $\begingroup$ I think the idea is that if the spectrum is simple, then the eigenvectors and eigenvalues vary continuously. Hence if the limiting diagonal matrix has almost surely got distinct eigenvalues, then the eigenspaces of the limiting matrix are one-dimensional and are spanned by elements of the standard basis. By continuity, this is also true for nearby matrices. $\endgroup$ Dec 22, 2015 at 1:26

1 Answer 1


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.

  • $\begingroup$ Thank you for this answer. I have a follow-up question: In your proof, you used the fact that $M_n$ has distinct eigenvalues with probability $1$. We do not need to assume this for the limiting matrix $D$ right? For example, a sequence of Wishart matrices with $n$ degrees of freedom converges to identity, but the proof would hold (and we can show that the smallest eigenvalue converges to $1$), right? $\endgroup$
    – Student88
    Jan 8, 2023 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.