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.

  • $\begingroup$ 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 $k\ge1$? $\endgroup$ – Cm7F7Bb Jun 6 '18 at 14:04
  • $\begingroup$ 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.) $\endgroup$ – Willie Wong Jun 6 '18 at 14:34
  • $\begingroup$ Intuitively, it would seem that the faster the operators converge, the faster should the eigenvectors converge. Would you think that it would be possible to say anything about the convergence rate of the eigenvectors given the convergence rate of the operators? $\endgroup$ – Cm7F7Bb Jun 6 '18 at 14:50
  • $\begingroup$ 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 $\inf \lambda_k = 0$. $\endgroup$ – Willie Wong Jun 6 '18 at 16:41

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