Concept

Worst-case analysis — where it appears

A bound over every input, which for several methods here is attained only by matrices somebody had to construct on purpose. Several bounds here are attained only on matrices somebody built to attain them, which is a fact about the bound rather than a reason to distrust it.

Named by 15 essays across 4 fields — each of them below, with the objects they name alongside it.

10²10⁴10⁶10⁸10¹⁰10¹²10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹κ of the sumrelative sizethe boundκ · umeasured spreadtwo curves and one constantspread ÷ κu, low0.25spread ÷ κu, high0.26bound ÷ spread, low7932decades swept10the spread is computablethe bound cannot see the order

A bound every answer satisfies

The classical bound on a summation error is correct, it covers all twenty-six answers one vector produced, and it is 7,932 times larger than the difference between them. A statement true of every ordering cannot say which ordering you got.

machine · Reduction order
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
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.

error · Conditioning
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.

error · Condition-estimation
10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹110²10⁴standard deviation of the noise σmedian growth factormargin 0.00564shootingWilkinsonwhat the growth rests onWilkinson, no noise5.5·10¹¹Wilkinson, σ = 10⁻¹⁴2shooting, σ = 10⁻⁸1.1·10⁴pivot margin0.0056thirty draws at every pointa tie breaks at any noise; a margin needs its own size

A worst case is as fragile as its margin

Wilkinson's matrix grows by 5.5·10¹¹ under partial pivoting, and adding Gaussian noise of 10⁻¹⁴ to every entry takes its median growth to exactly 2. The shooting matrix from a boundary-value problem grows by 1.1·10⁴, and noise ten million times larger leaves it untouched. The difference is what each worst case rests on. Wilkinson's rests on exact ties between candidate pivots, which any noise breaks. The shooting matrix's rests on a choice made by a margin of 5.6·10⁻³, and between 10⁻⁶ and 10⁻³ its median growth is that margin divided by the noise, times a constant between one half and four thirds.

elimination · Growth
0246810121416110¹10²10³10⁴10⁵10⁶length of the interval Tgrowth factorpartialrookcompleteh = 0.3, 102 unknownspartial, T = 15.01.3·10⁵e^(5T/6)/21.3·10⁵complete2largest κ8.3a boundary-value problem, not a constructionκ single-digit throughout

The growth a boundary-value problem supplies

Large growth under partial pivoting is usually said to need a matrix built for it. A two-point boundary-value problem solved by multiple shooting supplies one without being asked: its growth factor is e^(5T/6)/2 to four figures — 1.1·10⁴ at an interval of 12, 1.3·10⁵ at 15 — on a matrix whose condition number never exceeds 8.3, while rook and complete pivoting keep it below 2. And it is the finer shooting grid that grows: below a step of 0.3397 the choice partial pivoting makes turns on one entry against one, and above it there is no growth at all.

elimination · Growth
10⁻⁹10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²110¹10²10³10⁴standard deviation of the noise σmedian growth factorthe marginevery entrynonzeros onlyrelative, per entryone error in Estep 0.3, margin 0.00564, thirty drawsno noise1.1·10⁴every entry, σ = 10⁻⁵536nonzeros only, σ = 10⁻³1.1·10⁴one error in E, σ = 10⁻², draws kept21every curve is the same matrixonly what the noise touches changes

Noise the growth amplifies

The shooting matrix's growth of 1.1·10⁴ fell under noise as its pivot margin divided by the noise, and the explanation offered was noise reversing partial pivoting's choices. But noise of 10⁻⁵ is five hundred times smaller than that margin. Add the same noise only to the entries that are not zero and every draw keeps the whole growth up to 10⁻³. The dense noise was not reversing the comparisons by itself: it sat in the zeros, the elimination multiplied it by the growth already made, and a comparison flips when that product reaches about four tenths of the margin — at every noise level from 10⁻⁶ to 10⁻³.

elimination · Growth
10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³r½ recorded during one factorisationnoise that halves the growthT 6, h 0.3: 75T 9, h 0.3: 910T 12, h 0.3: 11000T 15, h 0.3: 1.3·10⁵T 12, h 0.25: 11000T 12, h 0.2: 11000T 9, h 0.15: 910measured ÷ predicted, shooting matricessmallest ratio0.92largest ratio2.4Wilkinson: recorded r½0Wilkinson: noise that halves it10⁻¹⁶one factorisation, one extra comparison a stepdashed: a factor of three either way

A margin the factorisation records

Partial pivoting finds the second-largest candidate in every column it scans, and throws it away. Keep it, divide the gap by the growth reached at that step, and take the smallest over the steps before half the final growth has arrived. That one number, recorded by the factorisation that is already running, predicts the dense noise that halves the growth to within a factor of 0.92 to 2.4 on shooting matrices whose growth runs from 75 to 1.3·10⁵ and whose fragility spans three and a half decades — and on Wilkinson's matrix it is exactly zero.

elimination · Growth
noise that halves the growthshooting, partial pivoting0.0018shooting, τ = 0.10.56Wilkinson, τ = 0.10.1800.10.20.30.40.50.60.70.80.9110⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1one minus the thresholdstored noise that halves the growthshooting, T = 12, h = 0.3Wilkinson, n = 24the same growth at every thresholdheld by a choice noise must push further to reverse

A threshold that holds the growth still

Partial pivoting's large growth on the shooting matrix rests on a margin of 5.6·10⁻³, and Wilkinson's rests on exact ties, which a perturbation at rounding breaks. A sparse code pivots with a threshold instead, taking the sparsest row among candidates within a factor τ of the largest. At every threshold from 1 to 0.1 both matrices grow by exactly as much as under partial pivoting. But the growth is now held by row counts, which no perturbation of the values can reverse: at τ = 0.1 it survives noise on the stored entries up to 0.56 on the shooting matrix, three hundred times more, and 0.18 on Wilkinson's, where partial pivoting's goes at 10⁻¹⁶. The margin that predicts it is the chosen candidate's height above the threshold line, and it needs no division by the growth.

elimination · Growth
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.

error · Condition-estimation
LAPACK's walk, exact83.0% · 5.2 productsLAPACK's walk, worst × 10037.7% · 5.2 productsblock of one, exact84.0% · 4.4 productsblock of one, worst × 10037.7% · 4.4 productsblock of two, exact96.5% · 8.5 productsblock of two, worst × 10059.6% · 8.5 productsblock of four, exact100.0% · 16.8 productsblock of four, worst × 100100.0% · 16.8 productseach bar a percentage — of the sample, or of the truthtwo random vectors close most of the tail

Two columns see what one walk cannot

The condition estimator every library ships walks from the all-ones vector, and a matrix whose largest column cancels against that vector hides from it: at n = 24 it reports five per cent of the truth. The block estimator behind MATLAB's condest walks with two vectors, the second random. On the same matrix at three sizes it is exact on every one of twenty seeds. On four hundred random 8 × 8 matrices it is exact on 96.5 per cent where the single walk is exact on 83.0, and its worst case, 0.596, is reached in the first fifty draws and not lowered by the next 1,550. The single walk's worst was still falling at 1,600. Four vectors are exact on all 400.

error · Condition-estimation
step 0.3rate per interval1.3largest r that catches it1.3110¹10²10³10⁴trigger sensitivity rgrowth factor1.021.324dashed: e to the five-sixths of the stepthe trigger must be finer than the growth

A trigger finer than the growth

Threshold pivoting cannot remove the shooting matrix's growth of 1.1·10⁴, because at τ = 1 it is partial pivoting. The proposal was a factorisation that notices growth on a step and searches as rook pivoting does after it, paying for the stricter search only where growth is made. But no step makes much: the largest one-step rise is exactly the growing mode's rate over one shooting interval, 1.284 at a step of 0.3, and 1.133 at 0.15. A trigger finer than that rate holds the growth at 2.00 for about two thirds of rook's extra search; one coarser never fires. And a trigger fine enough searches as rook on a quarter of an ordinary random matrix's steps.

elimination · Growth
μ = −0.1 and −0.01both hidden: steps a decade81the faster direction's race81the slower direction's race14710¹10²10³decades of the hidden directions removedstep the verdict settles0246810121416allboth directions hiddenonly the faster hiddenonly the slower hiddenμ = −0.1 aloneμ = −0.01 alonedashed: a single saddle of each curvatureone hyperplane hides nothing

One hyperplane hides nothing

Iterative refinement against a saddle-point matrix gives its verdict when the growing direction overtakes the minimum's slowest decay, and a right-hand side with that direction removed delays it by a fixed number of steps a decade. With two negative curvatures, the prediction was that hiding both is paid for by the faster, and that hiding only the faster leaves the slower one's race to run. The first half holds to half a per cent on three pairs, and to one and a half on two whose curvatures differ by a factor of two and of 1.2. The second does not: hiding either direction alone moves the verdict by a fixed handful of steps, the same at two decades as at sixteen, because the direction left in view is already growing. Only the intersection of the two hyperplanes hides anything — and it hides less than either curvature can alone, 1,127 steps at worst against 2,907, with random right-hand sides' slowest at 79 against 417.

constraint · Quasi-definite
each steering column half the hidden oneone random vector, 34 columns: share fooled0.77block of two, 34 columns: share fooled0.64block of four, 34 columns: share fooled0.390510152025303500.20.40.60.81steering columnsshare of seeds fooledone random vectorblock of twoblock of fourdashed: the single vector's share squared and to the fourthmore columns, more seeds, same factor

Columns that steer a random start

A condition estimator whose first vector is random cannot be built against, and concentrating the hidden column did not hide it: its own entries steer the first sign vector onto it. Give the steering to other columns — each equal on both rows of every pair the hidden column alternates over — and the random start is fooled after all. The prediction was a factor of one over the square root of the number of steering columns, on a share of seeds that falls as they lose to the hidden column, and the block of two fooled as often. The factor is whatever the builder chooses; the share, at a fixed factor of one half, rises with every column added, from 9 per cent with four to 77 with thirty-four. And the block of two is not fooled as often: it is fooled on the square of the single start's share, and the block of four on its fourth power — independent guesses, which is the one thing that still works.

error · Condition-estimation
exact share, points apartsize 8: random start less all-ones start0.025size 16: random start less all-ones start0.00570%80%90%100%share of matrices on which the estimate is exactsize 8size 16LAPACK's walk83.0% · 5.2 productsone walk from all ones84.0% · 4.4 productsone walk from random signs86.5% · 4.3 productsblock of two96.5% · 8.5 productsLAPACK's walk83.5% · 5.3 productsone walk from all ones83.5% · 4.4 productsone walk from random signs83.0% · 4.3 productsblock of two96.0% · 8.5 productsbars start at seventy per centthe first vector's sign pattern costs nothing

A first vector nobody can build against

The matrix built to fool a condition estimator is built against one vector, the all-ones vector its walk starts from, and the block estimator escaped it by adding a second, random one. Starting the single walk from random signs instead escapes it on every one of forty seeds at every size from 8 to 48, for the same 4.3 products, and loses nothing on random matrices — 86.5 per cent exact at size 8 against 84.0 from all ones, with tails that cross between sizes. The obvious way to build against a random start, a hidden column on few rows whose signs a random vector cancels half the time, fails on every seed: the hidden column writes itself into the walk's first product and turns the walk towards it. What the block of two's second vector buys is ten points of exact share, not the escape.

error · Condition-estimation

Named alongside it

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

Backward errorCondition numberGaussian eliminationGrowth factorPartial pivotingLower boundCondition-estimationCounterexampleMatrix normComplete pivotingRook pivotingSeeded generator

All concepts