The Lanczos Algorithm Under Few Iterations: Concentration and Location of the Output

We study the Lanczos algorithm where the initial vector is sampled uniformly from $\mathbb{S}^{n-1}$. Let $A$ be an $n \times n$ Hermitian matrix. We show that when run for few iterations, the output of Lanczos on $A$ is almost deterministic. More precisely, we show that for any $\varepsilon \in (0, 1)$ there exists $c >0$ depending only on $\varepsilon$ and a certain global property of the spectrum of $A$ (in particular, not depending on $n$) such that when Lanczos is run for at most $c \log n$ iterations, the output Jacobi coefficients deviate from their medians by $t$ with probability at most $\exp(-n^\varepsilon t^2)$ for $t<\Vert A \Vert$. We directly obtain a similar result for the Ritz values and vectors. Our techniques also yield asymptotic results: Suppose one runs Lanczos on a sequence of Hermitian matrices $A_n \in M_n(\mathbb{C})$ whose spectral distributions converge in Kolmogorov distance with rate $O(n^{-\varepsilon})$ to a density $μ$ for some $\varepsilon > 0$. Then we show that for large enough $n$, and for $k=O(\sqrt{\log n})$, the Jacobi coefficients output after $k$ iterations concentrate around those for $μ$. The asymptotic setting is relevant since Lanczos is often used to approximate the spectral density of an infinite-dimensional operator by way of the Jacobi coefficients; our result provides some theoretical justification for this approach. In a different direction, we show that Lanczos fails with high probability to identify outliers of the spectrum when run for at most $c' \log n$ iterations, where again $c'$ depends only on the same global property of the spectrum of $A$. Classical results imply that the bound $c' \log n$ is tight up to a constant factor.

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