Papers
Topics
Authors
Recent
Detailed Answer
Quick Answer
Concise responses based on abstracts only
Detailed Answer
Well-researched responses based on abstracts and relevant paper content.
Custom Instructions Pro
Preferences or requirements that you'd like Emergent Mind to consider when generating responses
Gemini 2.5 Flash
Gemini 2.5 Flash 44 tok/s
Gemini 2.5 Pro 41 tok/s Pro
GPT-5 Medium 13 tok/s Pro
GPT-5 High 15 tok/s Pro
GPT-4o 86 tok/s Pro
Kimi K2 208 tok/s Pro
GPT OSS 120B 447 tok/s Pro
Claude Sonnet 4 36 tok/s Pro
2000 character limit reached

Stability of the Lanczos Method for Matrix Function Approximation (1708.07788v2)

Published 25 Aug 2017 in cs.DS, math.NA, and cs.NA

Abstract: The ubiquitous Lanczos method can approximate $f(A)x$ for any symmetric $n \times n$ matrix $A$, vector $x$, and function $f$. In exact arithmetic, the method's error after $k$ iterations is bounded by the error of the best degree-$k$ polynomial uniformly approximating $f(x)$ on the range $[\lambda_{min}(A), \lambda_{max}(A)]$. However, despite decades of work, it has been unclear if this powerful guarantee holds in finite precision. We resolve this problem, proving that when $\max_{x \in [\lambda_{min}, \lambda_{max}]}|f(x)| \le C$, Lanczos essentially matches the exact arithmetic guarantee if computations use roughly $\log(nC|A|)$ bits of precision. Our proof extends work of Druskin and Knizhnerman [DK91], leveraging the stability of the classic Chebyshev recurrence to bound the stability of any polynomial approximating $f(x)$. We also study the special case of $f(A) = A{-1}$, where stronger guarantees hold. In exact arithmetic Lanczos performs as well as the best polynomial approximating $1/x$ at each of $A$'s eigenvalues, rather than on the full eigenvalue range. In seminal work, Greenbaum gives an approach to extending this bound to finite precision: she proves that finite precision Lanczos and the related CG method match any polynomial approximating $1/x$ in a tiny range around each eigenvalue [Gre89]. For $A{-1}$, this bound appears stronger than ours. However, we exhibit matrices with condition number $\kappa$ where exact arithmetic Lanczos converges in $polylog(\kappa)$ iterations, but Greenbaum's bound predicts $\Omega(\kappa{1/5})$ iterations. It thus cannot offer significant improvement over the $O(\kappa{1/2})$ bound achievable via our result. Our analysis raises the question of if convergence in less than $poly(\kappa)$ iterations can be expected in finite precision, even for matrices with clustered, skewed, or otherwise favorable eigenvalue distributions.

Citations (44)

Summary

We haven't generated a summary for this paper yet.

List To Do Tasks Checklist Streamline Icon: https://streamlinehq.com

Collections

Sign up for free to add this paper to one or more collections.

Lightbulb On Streamline Icon: https://streamlinehq.com

Continue Learning

We haven't generated follow-up questions for this paper yet.

X Twitter Logo Streamline Icon: https://streamlinehq.com