Numerical Linear Algebra
where the algebra stops being the arithmetic
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The thread: Assertions that reject — page 7

Essays 145 to 150 of the 150 on this theme, in the same order.
01020304050607010⁻⁶10⁻³110³10⁶10⁹k‖Aᵏ/k!‖‖e^A‖ = 2.6largest term 1.4·10⁷what the series throws awaylargest term1.4·10⁷‖e^A‖2.6digits cancelled away5.4·10⁶error after the sum5.2·10⁻⁹every term is computed correctlyand the sum has lost seven digits Eigenvalues, singular values, rank

The error the method already knows

Summing the exponential's Taylor series throws away a known number of digits, and the number is on the machine while the sum is being formed. The largest term divided by the answer, times the unit roundoff, tracks the relative error that comes out — to within a factor of nine, across fourteen orders of magnitude of it — and nothing reports it.

 012345678910⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹rank budgetresidual the run stalls atb with no structure at allb constant: the answer is a traintwo ladders, one truncationstructured, rank 10.12structured, rank 45.5·10⁻⁶structured, rank 85.9·10⁻¹⁴unstructured, rank 10.97unstructured, rank 80.47the floor is not the truncation'sit is the answer's Iterating, instead of factorising

A run that is over at step five

A conjugate gradient whose every iterate is cut to a rank budget reaches the floor that budget allows at step 5, 36, 42 or 59, and then does nothing for the rest of the run. Four times the iterations move the floor by a factor of 1.8, and past the answer's own rank they move it the wrong way.

 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.

 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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