Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Method of weighting — where it appears

Imposing an equality constraint by stacking it onto a least-squares problem with a large weight instead of enforcing it exactly. The answer approaches the constrained one as the reciprocal of the square of the weight, and how large a weight the arithmetic permits is a property of the solver.

Named by 5 essays across 2 fields — each of them below, with the objects they name alongside it.

0246810121410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹10⁴10⁷10¹⁰log₁₀ τ — the weight on the constraintrelative error against the exact answerτ = 1/√uGram–Schmidtnormal equationsHouseholder QRthe ceiling is the method'sHouseholder, τ = 10¹⁴4.8·10⁻¹⁵normal equations14Gram–Schmidt4.9·10¹⁰1/√u6.7·10⁷a constraint is a weight at infinityand the solver decides how far infinity is

A constraint is a weight at infinity

Stack an equality constraint on top of a least-squares problem with a large weight and the answer approaches the constrained one like 1/τ². The limit is takeable to any accuracy — and how far it can be taken is a property of the solver, not of the problem. One of them stops at the square root of the precision, and one of them does not stop.

leastsquares · Constrained least squares 2468101210⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²log₁₀ τ‖Bx − d‖ / ‖d‖sketched with the objectiveweighted, not sketchedkept out of the sketchone power instead of twokept out — feasibility1.7·10⁻¹⁶sketched at τ = 10⁸1.7·10⁻⁸unsketched at τ = 10⁸1.7·10⁻¹⁶objective ÷ optimum1.2a sketch preserves a normand a constraint is not one

The half of a problem a sketch may touch

A sketch guarantees that a norm is preserved to within a factor. An equality constraint is a statement that a quantity is zero, and no multiplicative guarantee says anything about zero. Sketch a constrained problem written as a weighted one and the constraint is not destroyed — it is demoted, from a violation of 1/τ² to one of ε/τ, exactly half the exponent.

randomised · Sketching 10¹10³10⁵10⁷10⁹10¹¹10¹³10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹κ(B), the conditioning of the constraint blockrelative error, and feasibilitydashes: the method of weightingthe saddle-point routethe null-space routealong the bottom: how nearly every answer satisfies the constraintsfeasible and wrongκ(B)4.6·10¹²κ(B)·u0.001null space1.2·10⁻⁴saddle point0.0082weighting2.6feasibility10⁻¹⁵every constraint is satisfiedand the answer has no digits

Feasible and wrong

A third constraint that nearly repeats the first takes the best route's answer from 2.96·10⁻¹⁵ to 1.16·10⁻⁴, and the other two routes to no correct digit at all. Every one of those answers satisfies every constraint to 10⁻¹⁵. The quantity a caller checks after a constrained solve is the one quantity here that says nothing.

leastsquares · Constrained least squares heavy row first1 − h at 4³⁰7.4·10⁻¹⁸κ(A) at 4³⁰3.1·10⁸complement's error2.1·10⁻¹⁶110³10⁶10⁹10¹²10¹⁵10¹⁸10⁻¹⁷10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹weight on the rowrelative error of 1 − h1 − ‖R⁻ᵀa‖²1 − ‖row of Q₁‖²‖row of Q₂‖²κ(A)·uthe rows are the same in every positiononly the order the factorisation meets them changes

The weight the factor met first

The route to one minus a leverage through the orthogonal factor was said to lose a digit for every decade of the condition number, whatever else it does. Put a weight on one row and it does not. With the heavy row first, the complement keeps every digit at κ(A) = 2.5·10⁹ while both subtractions return nothing. With the same row last it loses digits as the row's scale grows. And two heavy rows that leave κ(A) at 3.1 still lose six digits when the light rows come first. The law was about the order the factor met the rows, and the condition number had been standing in for it.

leastsquares · Leverage heavy row firstresidual at 4³⁰-2.8·10⁻¹⁰complement's error9.1·10⁻¹⁶110³10⁶10⁹10¹²10¹⁵10¹⁸10⁻¹⁷10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹weight on the rowrelative error of its residualb − Axb − Ax, exactlyQ₂Q₂ᵀbu·wthe same rows, the same data, in every positionthe solution misfits the heavy row by its own rounding

The residual the solution cannot hold

Sorting a weighted fit's rows heaviest first gave every digit of one minus the heavy row's leverage back. It gives nothing back to the heavy row's residual, if that residual is computed the way every textbook computes it — as the datum minus the fitted value. The fitted value is a double, and a double cannot resolve a misfit smaller than its own last digit times the weight: at a weight of 4²⁴ the residual formed from the solution is wrong in its second digit in every order, and forming the subtraction exactly changes nothing. Taken from the same orthogonal factor as the divisor, the residual keeps fifteen digits, and so does Cook's distance at 3.4·10¹⁷.

leastsquares · Leverage

Named alongside it

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

Exact ground truthCondition numberQR factorisationEquality-constrained least squaresRow orderingSaddle-point systemsCatastrophic cancellationHouseholder reflectionLeast squaresLeverageResidualUnit roundoff

All concepts