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Low-order Hilbert matrices already have large condition numbers

Example

The Hilbert matrices $H_{n}$ with entries $(H_{n})_{i j} = 1 / (i + j + 1)$ for $0 \leq i, j < n$ are already severely ill conditioned at order two and three:

$κ_{\infty} (H_{2}) = 27, κ_{\infty} (H_{3}) = 748.$

The point is not asymptotic growth but that the ill conditioning appears at the smallest possible orders.

Facts & Assumptions

Given: The Hilbert matrices $H_{2} = (\begin{array}{cc} 1 & 1 / 2 \\ 1 / 2 & 1 / 3 \end{array})$ and $H_{3} = (\begin{array}{ccc} 1 & 1 / 2 & 1 / 3 \\ 1 / 2 & 1 / 3 & 1 / 4 \\ 1 / 3 & 1 / 4 & 1 / 5 \end{array})$ .

[L1]

$κ_{p} (A) = ∥ A ∥_{p} ∥ A^{- 1} ∥_{p}$ (The condition number kappa_p(A) = ||A||_p ||A^{-1}||_p of a nonsingular linear system).

[L2]

The induced $\infty$ -norm is the maximum row sum (The induced 1-norm is the maximum column sum and the induced infinity-norm is the maximum row sum).

Verification

technique · direct

1.1

The inverse of $H_{2}$ is $H_{2}^{- 1} = (\begin{array}{cc} 4 & - 6 \\ - 6 & 12 \end{array})$ : multiplying, $(H_{2} H_{2}^{- 1})_{00} = 4 - 3 = 1$ , $(H_{2} H_{2}^{- 1})_{01} = - 6 + 6 = 0$ , $(H_{2} H_{2}^{- 1})_{10} = 2 - 2 = 0$ and $(H_{2} H_{2}^{- 1})_{11} = - 3 + 4 = 1$ .

algebra

1.2

By [L2], $∥ H_{2} ∥_{\infty} = 1 + 1 / 2 = 3 / 2$ and $∥ H_{2}^{- 1} ∥_{\infty} = \max {4 + 6, 6 + 12} = 18$ , so [L1] gives $κ_{\infty} (H_{2}) = (3 / 2) \cdot 18 = 27$ .

L1L2algebra

1.3

The inverse of $H_{3}$ is $H_{3}^{- 1} = (\begin{array}{ccc} 9 & - 36 & 30 \\ - 36 & 192 & - 180 \\ 30 & - 180 & 180 \end{array}),$ as a direct multiplication verifies: the first column against the rows of $H_{3}$ gives $9 - 18 + 10 = 1$ , $9 / 2 - 12 + 15 / 2 = 0$ , $3 - 9 + 6 = 0$ , and the other six entries similarly reduce to the identity matrix.

algebra

1.4

By [L2], $∥ H_{3} ∥_{\infty} = 1 + 1 / 2 + 1 / 3 = 11 / 6$ , and the row sums of $H_{3}^{- 1}$ are $9 + 36 + 30 = 75$ , $36 + 192 + 180 = 408$ and $30 + 180 + 180 = 390$ , so $∥ H_{3}^{- 1} ∥_{\infty} = 408$ ; [L1] gives $κ_{\infty} (H_{3}) = (11 / 6) \cdot 408 = 748$ .

L1L2algebra

2.1

Steps 1.2 and 1.4 exhibit condition numbers $27$ and $748$ at orders $2$ and $3$ : the Hilbert family is badly conditioned from the start.

step 1.2step 1.4∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

L. N. Trefethen and D. Bau III, Numerical Linear Algebra, Lecture 12 (standard reference, not scraped)