# Convergence rate of eigenvectors

Let us suppose that $A,A_1,A_2,\ldots$ are non-negative definite self-adjoint bounded linear operators in $L(\mathbb H)$, where $\mathbb H$ is a separable Hilbert space. $(v_j)_{j\ge1}$ and $(\lambda_j)_{j\ge1}$ are the eigenvectors and the eigenvalues of $A$ such that $\lambda_1>\lambda_2>\ldots$ and $\sum\lambda_j<\infty$. For $n\ge1$, $(v_{nj})_{j\ge1}$ and $(\lambda_{nj})_{j\ge1}$ are the eigenvectors and the eigenvalues of $A_n$.

Suppose that $\|A-A_n\|_{op}=o(a_n)$ as $n\to\infty$, where $\|\cdot\|_{op}$ is the operator norm and $(a_n)_{n\ge1}$ is a sequence such that $a_n\to0$ as $n\to\infty$. Can we say anything about the convergence rate of $\|v_k-v_{nk}\|$ as $n\to\infty$ with $k\ge1$? Is it true that $\|v_k-v_{nk}\|\le C\|A-A_n\|_{op}$ for $k\ge1$, where $C$ is some positive constant? Do we need stronger assumptions to say anything about the convergence rate of the eigenvectors?

It seems that if $\|A-A_n\|_{op}\to0$ as $n\to\infty$, then $\lambda_{nk}\to\lambda_k$ and $\|v_{nk}-v_k\|\to0$ as $n\to\infty$ for $k\ge1$ (see this paper by Joachim Weidmann). I am interested in the convergence rate of the eigenvectors given the convergence rate of the operators. Maybe some additional assumptions are needed to establish the convergence rate of eigenvectors (for example, a stronger convergence of the operators).

References are very welcome. Any help is much appreciated!

Consider the case of $\ell_2$ as you Hilbert space and let $A$ be the operator with the standard basis $(e_i)_{i = 1}^\infty$ as the eigenvectors and eigenvalues $\lambda_i = i^{-2}$.
Let $A_n$ be the operator that is identical to $A$, except for the action on the subspace spanned by $e_{2n - 1}$ and $e_{2n}$. On this subspace $A$ acts like $$\begin{pmatrix} (2n-1)^{-2} & \\ & (2n)^{-2} \end{pmatrix}$$ We choose $A_n$ to act instead like $$\frac12 \begin{pmatrix} (2n-1)^{-2} + (2n)^{-2} & (2n-1)^{-2} - (2n)^{-2}\\ (2n-1)^{-2} - (2n)^{-2} & (2n-1)^{-2} + (2n)^{-2} \end{pmatrix}$$ so that the corresponding normalized eigenvectors are $\frac{1}{\sqrt{2}} ( e_{2n-1} \pm e_{2n})$.
Then no-matter how you number the eigenvectors, $$\sup_{k} \| v_{nk} - v_k \| \geq \sqrt{2 - \sqrt{2}}$$ does not go toward zero, but $A_n \to A$ in operator norm since their difference is of size $n^{-2}$. So a uniform decay rate of the eigenvectors cannot be expected.
• Thanks for the answer (+1)! So the supremum over all $k$ does not necessarily go to $0$ as $n\to\infty$. If the operators converge in the operator norm, can we deduce that $\|v_{nk}-v_k\|\to0$ as $n\to\infty$ for each $k\ge1$? – Cm7F7Bb Jun 6 '18 at 14:04
• Provided you make sure that the corresponding eigenvalues $\lambda_{nk} \to \lambda_k$, and that you take care of issues regarding multiplicity (eigenspace being more than 1 dimensional) and sign ($v_{nk}$ versus $- v_{nk}$). (Which is why Weidmann stated the result in his paper the way he did.) – Willie Wong Jun 6 '18 at 14:34
• Your intuition is not 100% correct. The operator norm difference tells you, roughly, how far $A_n v_k$ is from being an eigenvector with eigenvalue $\lambda_k$, or, it says something about how close $\lambda_k^{-1} A_n v_k$ is to a normalized eigenvector. In the case where you have a lower bound on $\lambda_k$ (for example, strictly positive definite operators in finite dimensions), this means you can translate (roughly) the bound on $\|A_n - A\|$ to a bound $(\inf \lambda_k)^{-1} \|A_n - A\|$ on the difference of eigenvectors. But in your case $\inf \lambda_k = 0$. – Willie Wong Jun 6 '18 at 16:41