Numerical Linear Algebra
where the algebra stops being the arithmetic
Field

When the problem arrives again

Every other field here solves one system and measures how wrong the answer is. Almost no computation is shaped like that: a solve is one step of an outer loop, and the vector it returns is an input rather than a deliverable. That changes what its accuracy is for. An outer iteration recomputes its residual from the matrix at every step, so whatever the inner solve got wrong is measured again rather than carried — and four quantities that looked like accuracy requirements turn out to be assets with a shelf life, priced here against a root that is known exactly.
nothing8.3%the right-hand side16.5%the matrix, slowly85.9%everything100.0%what can be reused: the answerwhat can be reused: the factorisationwhat can be reused: the factorisation, for a whilewhat can be reused: nothingshare of the cost of a sequence that shares nothingwhat 12 members cost in factorisationssame: factorisations1rhs: factorisations1drift: factorisations4independent: factorisations12what changes between the membersdecides what may be carried

The problem that arrives again

A hundred and thirty essays have solved a system once and measured how wrong the answer was. Almost no computation is shaped like that. A solve is one step of an outer loop, its answer is an input rather than a deliverable, and four quantities this collection has treated as accuracy requirements turn out to be assets with a shelf life.

 10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110⁻⁴10⁻³10⁻²10⁻¹1inner tolerance η, relative residual of the linear solvedistance from the root after the stepd² = 0.00138distance before the step, 0.03721093547231126the number by each point is the iterations it costeleven decades, one landing placedistance before the step0.037its square0.0014where η = 10⁻³ lands0.0025where η = 10⁻¹⁴ lands0.0025iterations for the first354iterations for the second1126the accuracy that is thrown awaymeasured against a root that is known

The accuracy that is thrown away

A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.

 0123456789101110⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³110³Newton steptolerance asked for, and iterations paiditerations paidtolerance asked forouter residualthe adaptive policy, step by stepNewton steps10inner iterations, total1009first step's cost1last step's cost271final outer residual3.4·10⁻¹¹the rule reads the last two residualsand asks for nothing it cannot use

A tolerance that reads its own residual

The cheapest constant forcing term costs 980 inner iterations and arrives with a hundred times the forward error of the dearest, which costs 9,358. A rule that sets each step's tolerance from the ratio of the last two residuals costs 1,009 and arrives with neither problem — and it is not a constant, so it does not appear on the curve the constants are compared on.

 024681012141618202210⁶10⁷10⁸members served by one factorisationmultiplications for the whole sequencedoes not convergethe contraction ruledrift 0.01 a memberevery member1.6·10⁷every 5 members9.3·10⁶contraction rule9.2·10⁶its factorisations4cliff at a period of20a factorisation has a shelf lifeand the cliff is past the optimum

A factorisation kept past its date

One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.

 10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹10112233445566relative drift ‖E‖ / ‖A₀‖preconditioned conjugate gradient iterationson the small eigenvalueson the large onessame drift, two placessmall end at 10⁻²38large end at 10⁻²9κ(M⁻¹A), small end36κ(M⁻¹A), large end1.3‖E‖/‖A₀‖, both0.01how much the matrix changedis not what the preconditioner cares about

Where the drift lands

The standing rule for when a preconditioner has gone stale is to rebuild it once the matrix has changed by more than some fraction of itself. Two drifts of exactly the same relative size cost 19 iterations and 5 on the same matrix, and the quantity that separates them is not in the rule at all — the perturbation is divided by the eigenvalue it lands on.

 10¹10²10³03691215what one rebuild is worth, in preconditioned iterationscheapest number of members between rebuildsa dense Cholesky here is worth 6.7 iterationsringed: where the growth rule beats every fixed periodcheapest period, by setup costsetup worth 51setup worth 102setup worth 203setup worth 506setup worth 1008setup worth 20012how long to keep itis a question about what it cost to build

What a rebuild is worth

One sequence, one drift, one preconditioner — and six different right answers, because the cheapest rebuild period depends on what a rebuild cost to build and on nothing else. The optimum walks from every member to every twelfth as the setup gets dearer, and the free rule that reads the iteration count beats it in the middle of that range and loses at both ends.

 10²10³10⁴10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²steps taken‖RᵀR − AᵀA‖ ⁄ ‖AᵀA‖the bound, linear in the steps√k · uevery step safe, the chain notdrift after the run3.9·10⁻¹⁴the bound there3.3·10⁻¹³√k · u there6.1·10⁻¹⁵worst single amplification2.7worst leverage met0.69refactorisations1backward stable onceand three thousand times is a different claim

Stable once, and three thousand times

A sliding window adds a row and removes one at every step and never looks at the data again. No single step of it amplifies by more than 2.72, no downdate fails, and after three thousand steps the triangular factor in memory is 3.9·10⁻¹⁴ from the matrix it is supposed to be a factor of — six hundred times growth from a per-step bound that says nothing about chains.

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