Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Backward error — where it appears

8 essays name this object, across 4 fields. What follows is each of them, and the objects they name alongside it.
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

The exact answer to a nearby problem

A good algorithm does not give an approximate answer to your problem. It gives the exact answer to a problem very close to yours — and once that is the definition, a wrong result has two possible authors and they can be measured apart.

error · backward error 21-13-3-121-212-443-12as givenrows in the order 1 2 3 443-1201.251.252.502.51.5-30-0.5-0.52after step 1pivot 443-1202.51.5-3000.5400-0.21.4after step 2pivot 2.543-1202.51.5-3000.540003after step 3pivot 0.5‖PA − LU‖/‖A‖0largest multiplier0.75row order 4 3 2 1the pivot is chosen

Elimination is a sequence of choices

Gaussian elimination is taught as a procedure with no decisions in it. There is one decision at every step — which row to use — and every stability property the algorithm has comes from making it well.

elimination · elimination 110²10⁴10⁶10⁸10¹⁰10¹²10¹⁴10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹condition number κ(A)relative errorforward errorbackward errorpredicted: κ · u8×8, 20 seeds per κ; dashed is the worstthe problem worsens, not the method

A small residual is not a small error

Substituting the answer back and finding that it fits is the most natural check there is, and it verifies the wrong thing. A residual of 10⁻¹⁷ is entirely compatible with an answer whose second digit is wrong.

error · backward error [ ε 1 ; 1 1 ] x = [ 1 ; 2 ], exact answer (1.000000, 1.000000)with partial pivoting1101U after elimination1.0000001.000000computed xbackward error 0forward error 0without10⁻¹⁷10-1·10¹⁷U after elimination0.0000001.000000computed xbackward error 0.25forward error 0.71no error is raisedgrowth 10¹⁷

The swap that is not optional

Run elimination without a row interchange on a matrix that needs one and nothing announces a failure. There is no division by zero, no warning, and an answer of the right shape. It is simply wrong, and how wrong depends on a number you did not look at.

elimination · pivoting 0816243240110²10⁴10⁶10⁸10¹⁰10¹²10¹⁴matrix size ngrowth factor max|u| / max|a|the 2ⁿ⁻¹ boundworst of 30 randommedian randomWilkinson's matrix sits on the bound30 Gaussian matrices per sizeat n = 40: bound 5.5·10¹¹, worst 4.8

The bound that is never attained

Partial pivoting's stability guarantee permits the entries to double at every step — a factor of 5.5·10¹¹ at n = 40. The measured growth on random matrices of that size is about three. The gap is eleven orders of magnitude, and the guarantee is still worth having.

elimination · growth H13 x = b, b formed exactly so that x = (1, 2, …, 13)12345678910111213exact1.00002.00003.00133.97865.18864.993010.46720.048921.2657-2.575319.21478.906013.5113computed6.6 correct digits4.8 correct digits3.4 correct digits2.3 correct digits1.4 correct digit0.8 correct digitno correct digitsno correct digitsno correct digitsno correct digitsno correct digits0.6 correct digit1.4 correct digitbackward error2.2·10⁻¹⁷κ = 1.7·10¹⁸The algorithm solved a neighbouring problem perfectly. That problem's answer is this one.right-hand side built in BigInt rationalsthe truth is known

An answer that is known

Almost every demonstration of numerical error estimates the error by computing the same thing more carefully. The Hilbert matrix does not need that: its inverse is a closed form in integers, so the true answer is available exactly and the error is measured rather than approximated.

error · exact ground truth 012345610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹refinement step‖x − x*‖ / ‖x*‖a full double-precision solveresidual in24-bitresidual indoubleone argument apartκ·u of the factorisation6·10⁻⁴double residual, final3.2·10⁻¹³same-precision, final1.3·10⁻⁴30×30, κ = 10⁴, same factors in both runsidentical cost

Buying the accuracy back

Factorise in single precision, then correct the answer using residuals computed in double, and the result is what a full double-precision solve would have given. Compute those residuals in single instead and the identical algorithm, at identical cost, recovers nothing.

arithmetic · mixed precision 10⁻³10⁻²10⁻¹1110¹10²10³10⁴pivot threshold τgrowth factor · entries in L+U, ÷ entries in Agrowthfillthe library default‖PA − LU‖/‖A‖ at τ = 0.12.9·10⁻¹⁶growth at τ = 0.138entries in L+U at τ = 0.1372one knob, two measurements, opposite directionsand the default is most of both

A threshold between fill and growth

One number decides how small a pivot an elimination will accept. At 0.001 the factor holds 172 entries and the matrix grows by 1,330; at 1 it holds 260 and grows by 1.2. The libraries ship 0.1, and the measurement says why.

sparsity · sparse pivoting

Named alongside it

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

ResidualGaussian eliminationGrowth factorBackward stabilityForward errorPartial pivotingCondition numberHilbert matrixIterative refinementMultipliersPermutationPerturbation

All concepts