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A slowly decaying singular tail makes the Frobenius term dominate

Example

Fix integers $k \geq 2$ and $p \geq 2$ . For every $n \geq k + p$ , let $A_{n} = diag (1, 2^{- 1 / 2}, \dots, n^{- 1 / 2})$ . In the real Gaussian range-finder bound, the second term involving ${(\sum_{j > k} σ_{j}^{2})}^{1 / 2}$ eventually exceeds the term involving $σ_{k + 1}$ by an arbitrarily large factor as $n \to \infty$ .

Facts & Assumptions

Given: Fixed $k, p$ as above and the family of finite diagonal matrices $A_{n}$ .

[F1]

The expected error bound has first term $(1 + \sqrt{k / (p - 1)}) σ_{k + 1}$ and second term $(e \sqrt{k + p} / p) (\sum_{j > k} σ_{j}^{2})^{1 / 2}$ (Expected spectral error of the real Gaussian range finder).

Verification

1.1givenF1algebra

The diagonal entries are decreasing, so they are the singular values of $A_{n}$ , and $σ_{k + 1} = (k + 1)^{- 1 / 2}$ while $\sum_{j > k} σ_{j}^{2} = \sum_{j = k + 1}^{n} 1 / j$ .

2.1step 1.1algebra∎

Since $x \mapsto 1 / x$ is decreasing, $\sum_{j = k + 1}^{n} 1 / j \geq \int_{k + 1}^{n + 1} d x / x = \log ((n + 1) / (k + 1))$ , which diverges with $n$ . The coefficient $e \sqrt{k + p} / p$ is positive and fixed, while the first-term coefficient and $σ_{k + 1}$ are fixed. Therefore the ratio of the second term to the first tends to infinity; choosing $n$ large makes the tail-Frobenius contribution dominate by any prescribed factor.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Halko, Martinsson, and Tropp, Theorem 10.6 and discussion after it (standard reference, not scraped)
Martinsson and Tropp, Theorem 11.5 and discussion after it (standard reference, not scraped)