Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Approximation before linearisation — where it appears

Replacing the function in a nonlinear eigenvalue problem by one that can be linearised, before any eigenvalue is computed. The error it contributes is chosen rather than incurred, is not amplified by any condition number, and appears in no residual against the object the solver factorised.

Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.

23456710⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²poles in the approximantresiduals and errorforward error‖T(λ)x‖‖T̃(λ)x‖one extra evaluationagainst the approximant1.5·10⁻¹²against the problem asked1.9·10⁻⁶forward error4.8·10⁻⁵‖g − r‖ there8.5·10⁻⁵the free residual is flatand the answer is not

The problem the solver was actually given

A linearisation is exact — it has the polynomial's eigenvalues, with their multiplicities, and the whole loss is arithmetic. A nonlinear eigenvalue problem does not offer that. Every algorithm replaces the function first, and the term that replacement contributes is committed before any number is rounded and appears in no residual.

polynomial · Approximation before linearisation 110¹10⁻⁷10⁻⁵10⁻³distance from the branch point, λ + c|g − r|the eigenvaluesdegree 98 poleschosen, not computedreach0.06left end, from the cut0.2rational, worst8.5·10⁻⁴polynomial, worst0.0068linearisation size, both54committed before the solveand invisible to it

An error committed before the arithmetic

Before a nonlinear eigenvalue problem is solved, somebody says where they think the eigenvalues are. That sentence sets the accuracy of everything that follows by five orders, costs nothing to say, and cannot be revised once the approximation built on it is in hand.

polynomial · Approximation before linearisation 263850627410⁻⁵10⁻³10⁻¹rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.06largest size drawn78rational there2·10⁻⁵polynomial there0.0015the ratio73no winner on an easy targetand two orders on a hard one

Two approximants and one matrix size

A polynomial approximant linearises to nd rows and a rational one to n(m+1), so the fair contest fixes the matrix and varies the basis. On an easy target set the two are indistinguishable and the ordering flips with the noise; on one that reaches a branch point the rational pulls away by two orders.

polynomial · Approximant choice -10123456789-101λbranch point at −0.4the closed formwhich of these is an answereigenvalues returned36wanted6past the branch point6complex24worst against the closed form7.7·10⁻⁴all of them exactfor a problem nobody asked

The eigenvalues that are answers to nothing

A rational approximant of degree five turns a six-by-six problem into a thirty-six-by-thirty-six one, and thirty-six numbers come back. Six are the answer. The rest are exact eigenvalues of the approximant, lying where the function it approximates is not a real number at all.

polynomial · Spurious spectrum 12345678910111210⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹degree of the approximantworst relative error on the target setchosen by the residualdegree reached12error there10⁻¹²even support, same degree1.1·10⁻⁷advantage702nearest support point0.096clustering ratio1.5filled: points chosen by the erroropen: points spread evenly

The points the algorithm chose

A rational approximant whose support points are picked by its own residual clusters geometrically at a branch point nobody named — recovering by measurement the rule a hand-built approximant is given. At degree ten it is seven hundred and fifty times more accurate than the same form with its points spread evenly.

polynomial · Adaptive interpolation the problem you posedA = H10b = A·(1, 2, …, 10)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.3·10⁻¹⁷the answer you wantedx = (1, 2, …, 10), exactlythe answer you gotx̂, wrong by 2.7·10⁻⁴ relativebackward error 2.3·10⁻¹⁷forward error 2.7·10⁻⁴κ = 1.6·10¹³κ · η = 3.6·10⁻⁴, and the measured forward error is 2.7·10⁻⁴.The algorithm is not at fault. The problem is.H10, LU with partial pivotingresidual and error differ

Three errors and one number

This site's identity has two factors and a division of blame between them. Two fields have now added a third party and a fourth, and only one of the four is a property of anything — the others are decisions, made before the arithmetic, reported by nothing.

error · Backward error the problem you posedA = H12b = A·(1, 2, …, 12)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 1.8·10⁻¹⁷the answer you wantedx = (1, 2, …, 12), exactlythe answer you gotx̂, wrong by 0.02 relativebackward error 1.8·10⁻¹⁷forward error 0.02κ = 1.8·10¹⁶κ · η = 0.33, and the measured forward error is 0.02.The algorithm is not at fault. The problem is.H12, LU with partial pivotingresidual and error differ

The fifth author

Four authors of a wrong answer have been named on this site and each is a statement about one computation. The fifth is not: it is what separates two computations that are both correct, it is a backward error of measurable size, and no residual, bound or condition number contains it.

error · Backward error

Named alongside it

The objects these essays reach for when they reach for this one.

Condition numberLinearisationRational approximationBranch pointResidualBackward errorExact ground truthForward errorNonlinear eigenvalue problemSpurious spectrumTarget setAdaptive interpolation

All concepts