Numerical Linear Algebra
where the algebra stops being the arithmetic
Field

Two errors, and whose fault they are

A good algorithm returns the exact answer to a nearby problem. So a wrong answer has two possible authors, and they can be measured separately: the backward error says how well the algorithm did, the condition number says how much the problem amplifies it, and only their product is what anyone sees.
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.

 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.

 110¹10²10³10⁴10⁵10⁶10⁷00.250.50.751amplification of the input perturbationfraction of directions at or belowκ = 10·10⁵worst found 7.6·10⁵6×6, 200 directionsmedian reaches 0.29 of κ

The condition number is an amplifier

κ is usually introduced as a definition and then quoted. It is a measurement: perturb the input by a known amount, look at how much the output moves, and the largest ratio you can find is the number.

 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.

 0246810110²10⁴10⁶10⁸10¹⁰10¹²spread of the row units (decades)condition numberκ∞(DA)cond(DA)Hilbert κ∞Hilbert condone system, two numbersκ∞ at no spread9.8κ∞ at 10 decades1.9·10¹⁰cond, either end7Hilbert, equilibrated1.3·10¹⁰the solution is the same at every spreadand one of these curves knows it

The units the matrix is measured in

One linear system, written twice. The rows of the second are the rows of the first in different units, the solution is identical to the last bit, and the condition number has moved by eight orders of magnitude. One of those two numbers is a fact about the problem and the other is a fact about the notation.

 10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1relative size of the entrywise perturbationrelative forward errorκ∞ · εcond(A,x) · εmeasuredboth bounds holdκ∞(A)1.9·10⁸cond(A, x)4.8ratio of the bounds4·10⁷both curves above the data are boundsand only one of them is a measurement

A condition number scaling cannot move

Skeel's componentwise condition number is invariant under any row scaling — exactly, before any norm is taken, because two diagonal factors cancel entry by entry. It is never larger than the normwise one and can be arbitrarily smaller, and the ratio between them is a diagnostic for which kind of ill-conditioning a matrix has.

 the estimator maximises this quantity over the columns it visitscolumn 1 ‹visited›12column 2 ‹the answer›114column 311.4column 411.4column 511.4column 611.4column 711.4column 811.4column 911.4column 1011.4column 1111.4column 1211.4estimate 12.0a walk that stopped earlythe estimate returned12the true 1-norm114columns visited1products with the matrix5the walk's own stopping test firedand every column it could see was smaller

An estimate that can be fooled

Nobody computes a condition number, because forming an inverse costs more than the solve did. Every library estimates it instead, from four or five products with a factorisation already in hand. The estimate is exactly right on four random matrices out of five — and there is a matrix, three distinct entries wide, on which it returns a twentieth of the truth.

 does this matrix look nearly singular?green: the test agrees with the truth · red: it does not · the bar under each number is its magnitude, over sixty-two decades|det A||det A|^(1/n)σₘᵢₙ1/κ = σₘᵢₙ/σₘₐₓ0.1·I at n = 40perfectly conditioned10⁻⁴⁰0.10.11κ = 10¹⁰, |det| = 1nearly singular1110·10⁻⁶10·10⁻¹¹Hilbert at n = 8nearly singular2.7·10⁻³³8.5·10⁻⁵1.1·10⁻¹⁰6.6·10⁻¹¹the two counterexamplesκ of the scaled identity1its determinant10⁻⁴⁰κ of the normalised matrix10¹⁰its determinant1det(cA) = cⁿ det(A)so a determinant carries the units n times over

The number that decides nothing

The determinant is the first scalar anybody attaches to a matrix and the last one worth consulting. A tenth of the identity has a determinant of 10⁻⁶⁰ and a condition number of exactly one. The Hilbert matrix's determinant stops being right at n = 13 and stops being a number at n = 29, and nothing in between reports either.

 015304560759010512010⁻²10⁻¹110¹steprelative sizeleast error: 20discrepancy stop: 7errorresidualthe knob is an integerleast error, at step20error there0.14error at step 1206the residual falls at every stepthe error turns and keeps rising

The zero you are allowed to write

A deflation criterion sets a subdiagonal entry to zero because it is small. A drop tolerance discards an entry of a factor because it is small. A truncation discards a singular value because it is small. Three fields, three vocabularies, no shared arithmetic — and plotted as work saved against error accepted, one curve.

 significand bits10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰κ(A)87892634456329625911109435761199127514162024every matrix positive definiteruns producing a false certificate14of runs in total72never above, in significand bits12first κ at eight bits10⁵the comparison was correctand what it proved was not true

Deciding that a zero has arrived

The previous tolerances were offers — accept this much error, save this much work. A detection threshold is not an offer, because both directions are failures. One matrix here has three genuinely near-invariant subspaces, and the constant somebody typed decides which of them the recurrence stops at; at eight significand bits the same kind of constant produces a proof of something false.

 10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³relative compression error, the representation‖b − Ax‖ ⁄ ‖A‖‖x‖, the solveequala backward error, chosenslope0.99representation at 10⁻⁸1.2·10⁻⁹backward error there1.3·10⁻¹⁰representation at 10⁻¹²1.2·10⁻¹³backward error there2.5·10⁻¹⁴the compression is not an approximationit is a perturbation of the problem

An accuracy that is a backward error

Every backward error on this site is something an algorithm produced and somebody then measured. This one is a line in the program. Solving with a compressed matrix gives a residual that is the compression's own error, at a slope of 1.000 over ten decades, so the knob that sets the storage sets the backward error directly.

 110¹10²10⁻¹²10⁻⁸10⁻⁴110⁴ncondition number, and residual reachedmarks above: κ of the step · dashes: its closed formmiddle: a fit from a random startbelow: the same fit started at the answera decomposition that is ill-conditionedκ at n = 1283.3·10⁴its closed form3.3·10⁴cosine of the terms1from a random start0.012from the answer1.4·10⁻¹⁰the answer existsand cannot be found

A tensor that cannot be decomposed

Every member of a certain sequence is exactly a sum of two rank-one terms, and both terms are written down in closed form. A three-hundred-sweep fit from a random start does not find them, and stalls at the same one per cent however far the sequence goes — while a fit started at the answer loses digits exactly as 2n² says it should.

 -12-10-8-6-4-2010⁻¹⁷10⁻¹³10⁻⁹10⁻⁵10⁻¹10³10⁷10¹¹10¹⁵log₁₀ μcondition number, and relative errorκ₂, augmentedcomponentwise, condensedcomponentwise, augmentederror, augmentedone matrix, two numbersκ₂ at μ = 10⁻¹²3·10¹³componentwise, same matrix13their ratio2.3·10¹²measured error9.4·10⁻¹⁶every library prints the top lineand the error obeys the third

Two condition numbers of one matrix

κ₂ is a worst case over perturbations of a given norm, and a normwise perturbation may put its whole budget on the smallest entry. The componentwise number is a worst case over perturbations proportional to the entries, which is what a backward-stable factorisation actually makes. On one matrix they are 3·10¹³ and 13.3, and the error obeys the second.

 57911131517192110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²degreerelative error of the worst rootmeasuredκ × ua polynomial is its coefficientsworst root, measured0.0076predicted, κ × u0.0023root condition number10¹³largest coefficient1.4·10¹⁹the roots are integersand the coefficients are not the roots

The roots are not the coefficients

A polynomial whose roots are the integers one to twenty, expanded exactly, handed to the routine every library uses. The computed roots are wrong in the third digit, the computation is backward stable for the matrix it factorised, and above degree eighteen the coefficients are not double-precision numbers at all.

 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.

 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.

 10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²-1-0.500.51relative backward error acceptedfraction of the work not doneQR deflation criterionincomplete Cholesky droplow-rank truncationthree fields, no shared arithmeticslope, deflation criterion0.039slope, drop tolerance0.051slope, rank truncation0.23widest apart, as a ratio5.7a tolerance is an offerand the three offers are one offer

A tolerance is priced by the problem

Three tolerances from three fields sit on one pair of axes and agree to within a factor of 5.74. That factor is the ratio of the two curves that cannot move. Change the only problem in the comparison and the third curve's fitted slope swings from 0.188 to 0.040 while the printed spread does not shift by a digit.

 each bar is a percentage — of the sample, or of the true condition numberexactly right86.0%inside 10%91.0%inside a factor of 291.0%worst in the sample, ×10035.7%the constructed matrix, ×1007.7%usually exactexact share0.86worst of the sample0.36the constructed matrix0.077a routine that is right most of the timeand never wrong in the safe direction

The tail a sample never reaches

Hager's estimator is exactly right on four random matrices in five, and that share is stable — between 80.5 and 87.5 per cent across nine sizes. The worst underestimate is not stable at all: it falls every time more matrices are drawn, from 0.746 at sixty to 0.377 at four hundred, and the matrix built to defeat the estimator sits five times below anything four hundred draws found.

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