Numerical Linear Algebra
where the algebra stops being the arithmetic
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The thread: Measured, not asserted — page 13

Essays 289 to 293 of the 293 on this theme, in the same order.
10⁻¹100.20.40.60.8strength threshold θresidual reduction per cycleθ = εsemi-coarseningkept whole rowsfull coarseningone parameter, two methodsbest factor above ε0.053best factor below ε0.1the ratio across the switch231×31 anisotropic operatora switch, not a dial Iterating, instead of factorising

The switch does not know which side is better

The strength threshold moves the coarsening from full to semi at θ = ε exactly, at every anisotropy. Which of the two converges faster is a separate question with a separate answer, and it changes sign between ε = 0.33 and ε = 0.34 — where nothing whatever happens to the switch.

 10⁻³10⁻²10⁻¹1024681012size of the negative eigenvalue, −λproducts before the test firesharder to find, and milder8 spectra, n = 50products at the largest λ3products at the smallest10smallest share of λ recovered0.14largest0.34the one that hidesis the one that matters least Iterating, instead of factorising

A proof that does not ask how large the matrix is

Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.

 10³10⁶10⁹10¹²10¹⁵10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1largest iterate on the way, ‖x‖difference between the two residualsthe bound, linear in ‖x‖what it does, slope 0.46a bound of one, a walk of a halffitted slope0.46the bound's slope1smallest gap measured3.9·10⁻¹⁵largest gap measured1.5·10⁻¹⁰share of the bound, near end9.1·10⁻⁴share of the bound, far end3.4·10⁻⁹the bound is rightand loose by a square root Iterating, instead of factorising

A walk needs a length

The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.

 22.32.62.910⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised832storage, second-order416Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations Iterating, instead of factorising

The answer that arrives when the space runs out

A second-order Krylov recurrence holds vectors of length n for a problem with 2n eigenvalues, so it is exact at n steps where the linearised route needs 2n. The machine-precision reading at forty-four vectors on a chain of forty is that exhaustion rather than convergence, and it arrives through a basis whose ‖QᵀQ − I‖ is above one.

 110¹00.30.60.91.2trust-region radius Δshare of the exact model decreasethe exact subproblem7 radii, n = 60share at the smallest radius0.95share at the largest0.3products, at most8radii stopped by the curvature4a few products against an eigendecompositionand most of the decrease Iterating, instead of factorising

The certificate that arrives soonest is worth least

The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λ_min = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.

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