Numerical Linear Algebra
where the algebra stops being the arithmetic
Theme

The thread: Whose fault is it — page 2

Essays 25 to 48 of the 58 on this theme, in the same order.
the matrix105 entriescorner first — sparsest227 entries, growth 1.9·10¹¹largest first — safe242 entries, growth 1.19the middle factor is the smaller one, and its answer has no correct digitsboth factorisations reproduce the matrix‖PA − LU‖/‖A‖, sparsest3.8·10⁻¹⁷‖PA − LU‖/‖A‖, pivoted5.4·10⁻¹⁷forward error, sparsest3·10⁻⁵forward error, pivoted4.8·10⁻¹⁶red marks are entries elimination createdthe fill argument and the stability argument disagree Sparsity, and what elimination costs

Structure and stability stop being separable

The sparsest variable to eliminate on this matrix has a diagonal entry of 10⁻¹². Eliminating it produces the smaller factor, reproduces the matrix to 3.8·10⁻¹⁷ — better than pivoting does — and returns an answer wrong in the fifth digit.

 0481216202428321rank keptrelative errormedianthe answer movesspread at rank 81.8spread at rank 241best median error0.14widest where the method is worstand the bound does not say so Methods that were designed apart

An answer that changes with the seed

A randomised rank-k solve is a truncation computed in a random subspace, and it reaches the same floor as the deterministic ones. What it does not do is return the same answer twice — a factor of 1.84 across four seeds at rank 8, and 1.02 at the rank where the method is best.

 10²020406080100120140160size niterationsno preconditionerwrapped (Strang)averaged (T. Chan)both are circulant approximations‖C − T‖/‖T‖, averaged0.13‖C − T‖/‖T‖, wrapped0.25smallest eigenvalue, wrapped, n = 16-0.65one of them is positive definiteand it is the one that is nearer Structure, and the solver that cannot see it

The circulant that cannot be indefinite

The previous essay found a preconditioner taking 117 steps against an unpreconditioned 59, because its smallest eigenvalue was −0.173. Average the two diagonals instead of choosing between them and the count is 7, 8, 9, 10, 10 across a factor of sixteen in size.

 rounds on the critical pathHouseholder sweep48reduction tree4Cholesky QR4words sentHouseholder sweep1170reduction tree1170Cholesky QR2160two counts, two rankingsrounds, sweep ÷ tree12words, Cholesky ÷ tree1.8arithmetic, tree ÷ sweep1.5the rounds separate the threeand the words do not Where the flop count stopped predicting the time

The message and the word

Three factorisations of one matrix on sixteen processors: 48 communication rounds, 4, and 4. The words sent are 1,170, 1,170 and 2,160 — so the method with the fewest rounds sends the most words, and the count that separates the three is the one no operation count can see.

 forward error, relative to a solution of exactly (1, 1)no pivoting · as given1 0 interchangesno pivoting · rows scaled1 0 interchangespartial · as given0 1 interchangepartial · rows scaled1 0 interchangesscaled partial · as given0 1 interchangescaled partial · rows scaled0 1 interchangecomplete · as given0 1 interchangecomplete · rows scaled0 1 interchangethe same problem twicepartial, as given10⁻¹⁸partial, rows scaled1its relative residual10⁻¹⁷complete, rows scaled10⁻¹⁸the two systems have the same solutionand one pivot rule cannot see it Elimination, and the swap

The pivot that reads the units

Partial pivoting compares the entries of a column and takes the largest. Those entries carry units, so the comparison depends on them — and there is a row scaling, on the standard two-by-two that pivoting exists to fix, which makes partial pivoting perform the identical catastrophic elimination it was introduced to prevent, with no interchange at all.

 1112131415193111.365129.731148.096166.461probes takenrunning estimate of the tracenormal±1one probe, no errorthe exact trace99±1 variance, this matrix0±1 variance, rotated57normal variance545the same spectrum in a general basiscosts the ±1 probe its whole advantage Randomised, and the guarantee that changes kind

Counting what cannot be looked at

The trace is n additions and one of the most expensive quantities in the subject to estimate, because the matrices whose trace is wanted are never stored. Hutchinson's estimator is unbiased with one line of algebra — and its variance depends on which random vector is used, by a factor that is a property of the matrix, and on a diagonal matrix one choice is exact from the first probe and the other is not.

 00.250.50.75110⁻¹110¹share of the noise placed in the matrixleast-squares error ÷ total least-squares errorequally accuratetotal leastsquares aheadordinary leastsquares aheadthe model, not the methodadvantage, all noise in b0.28advantage, all noise in A2.5seeds at each share40the same total noise at every pointand only where it sits changes Least squares, and the road not to take

When the matrix is wrong too

Every least-squares problem on this site has assumed A is exact and b is not, and moved b onto the column space of A. Where both were measured, the smallest correction that makes the system consistent moves the matrix as well — and on the problems where that answer is more accurate, it has the larger residual, by construction rather than by luck.

 10¹10²10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰110⁴10⁸10¹²10¹⁶condition number κ(A)× short of a double solvebf16fp16fp32bf16fp16fp32the reference is soundreference solve, worst backward error10⁻¹⁶bfloat16 threshold κ256fp32 threshold κ1.7·10⁷eight refinement steps, residual always in doubleflat at 1 means it reached double The arithmetic underneath

Where the hardware went

bfloat16 carries eight mantissa bits, which puts its refinement threshold at a condition number of 256. That is not an exotic matrix. It is an ordinary one, and past it the method still improves the answer by a factor of four hundred while getting nowhere near a usable one.

 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 Sparsity, and what elimination costs

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.

 10²10³10⁴10⁵10⁶10⁷10⁸10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²condition number‖QᵀQ − I‖one passsweeptwicetreewhat the second pass removesfitted slope, one pass2fitted slope, two passes0.96rounds, two passes6rounds, the sweep24one pass squares the condition numberand two do not Where the flop count stopped predicting the time

Doing it twice

Cholesky QR squares the condition number — a fitted slope of 1.95 in κ against the Householder sweep's 1.00. Run the identical routine a second time on the Q it returned and the slope is 0.93, the orthogonality is at or below the sweep's at every κ, and the price is one more all-reduce.

 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 Two errors, and whose fault they are

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.

 10¹10³10⁵10⁷10⁹10¹¹10¹³10¹⁵10⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1κ₂(A), the matrix that was updatedrelative forward errorSherman–Morrisondirect solve of A + uvᵀone answer, two routesκ of the answer's matrix1κ of the matrix replaced10·10¹³update formula's error2.5·10⁻⁴direct solve's error1.1·10⁻¹⁶the question's condition number is 1at every point on this axis Least squares, and the road not to take

A correction cheaper than the problem

Sherman and Morrison's formula updates a solved system for a rank-one change to the matrix, at 4n² operations instead of (2/3)n³. It is exact algebra. On a problem whose updated matrix is the identity — condition number one, the easiest system there is — it returns a forward error of 2.5·10⁻⁴ where a direct solve returns 10⁻¹⁶.

 10²11.251.51.7522.25processorstraffic saved, as a factorthe √4 the law promisesbreak-evenmeasureda limit is not a sizesaving at p = 640.88saving at p = 5761.4what the law promises2memory, as a factor4a loss at sixty-four processorsand 72% of the law at five hundred Where the flop count stopped predicting the time

Memory bought with messages

Holding four copies of the data instead of one is supposed to cut a matrix multiplication's communication by √4. Measured on a machine of 64 processors it costs 14% more traffic; at 576 it saves 44%, which is 72% of what the law promises. The memory is exactly four times, and that part is not asymptotic.

 D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero Elimination, and the swap

When symmetry is not enough

The matrix [[0, 1], [1, 0]] is symmetric, nonsingular and perfectly conditioned, and there is no diagonal entry to pivot on. Every factorisation restricted to symmetric interchanges and one-by-one pivots fails on it, at any depth of searching, because every entry it could search is zero. The repair is to take two variables at once.

 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 Two errors, and whose fault they are

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.

 110¹10²10³10⁴10⁵10⁶10⁷10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸1/(1 − h), the leverage of the removed row‖R̄ᵀR̄ − (G − aaᵀ)‖ / ‖G − aaᵀ‖downdatedrefactorisedat h = 1 − 10⁻⁷κ of the downdated matrix4.3κ of the matrix downdated9.3·10⁶rotation's amplification344downdate residual3.5·10⁻¹⁰a hyperbolic rotation is not orthogonaland that is exactly what it is for Least squares, and the road not to take

The observation that cannot be removed

Removing a rank-one term from a Cholesky factor needs a rotation that is not orthogonal, and the number under its square root is 1 − h, where h is the leverage of the row being removed. The algorithm's breakdown condition and the statistician's warning are the same quantity, arrived at from opposite ends, and neither field states it in the other's language.

 01234567810⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹μ, the entry above the diagonalsep, and the eigenvalue gapmin |λᵢ + μⱼ|sep(A, B)the spectra never moveeigenvalue gap, throughout2sep at μ = 02sep at μ = 89.5·10⁻⁴amplification there1056solvability is the eigenvaluesand conditioning is not Structure, and the solver that cannot see it

An equation whose unknown is a matrix

AX + XB = C is linear in X, so it has a coefficient matrix, and writing it down is the obvious thing to do. At n = 100 that matrix has a hundred million entries for a problem with ten thousand unknowns, and the algorithm everybody uses instead never forms it. Its conditioning is not the eigenvalue gap either, which is the number a reader is invited to consult.

 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 Two errors, and whose fault they are

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.

 10⁻²10⁻¹110¹10²10³10⁴10⁻⁴10⁻³10⁻²10⁻¹1backward error, in units of ushare of systems above ituCramereliminationCramer, control24-bit arithmeticworst Cramer, in u395worst elimination, in u1.2control, worst Cramer4.9κ of the worst system3.4·10⁶one derivation, two computationsand only one of them is stable Elimination, and the swap

A rule that is correct and unusable

Cramer's rule gives every component of the solution in closed form, in terms of determinants, and it is a theorem. On two-by-two systems whose rows are nearly parallel it returns an answer with a backward error of 458 units of roundoff where elimination returns 1.3 — on a matrix whose condition number is 32,000 and which elimination solved perfectly.

 10¹10¹.³10¹.⁶10¹.⁹⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰¹10².²00.250.50.751grid points nresidual reduction per stepJacobiGauss–SeidelV-cycleV-cycle spread, 8× in size0.003Jacobi at n = 1270.99work exponent, fitted0.079the dashed curve is cos(πh), Jacobi's closed formthe flat line is the whole method Iterating, instead of factorising

A rate that does not notice the size

The V-cycle reduces the residual by a factor of ten a cycle at fifteen points and at a hundred and twenty-seven. Jacobi on the same four problems goes from 0.981 to 0.9978, climbing towards one. One of those is a constant and the other is an exponent, and that is the whole distinction the field turns on.

 110¹10²10³110¹10²10³off-diagonal entry ccondition number of the eigenvalue√(1 + c²)decoupled: 1measuredthree routes, one number‖A − ZTZᵀ‖/‖A‖1.7·10⁻¹⁵closed form100computed 1/|yᵀx|100worst measured movement46four eigenvalues, two conditioning numbersthe symmetric case has one, and it is 1 Eigenvalues, singular values, rank

A condition number for one eigenvalue

In the symmetric case every eigenvalue has condition number exactly one. In this four-by-four matrix two of them have condition number 100.005 and the other two have exactly 1, and the number belongs to the eigenvalue rather than to the matrix.

 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)σ_min1/κ = σ_min/σ_max0.1·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 Two errors, and whose fault they are

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.

 10²10⁴10⁶10⁸10¹⁰10¹²10¹⁴10⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1κ₂(A)relative errorforward, A⁻¹bforward, A\bbackward, A⁻¹bbackward, A\bat κ = 10¹⁴η, LU solve2.2·10⁻¹⁷η, via the inverse4.5·10⁻⁵forward, LU solve2.8·10⁻⁴forward, via the inverse0.015one factorisation, two ways to use itand one of them forfeits the backward error Elimination, and the swap

The inverse that is never formed

x = A⁻¹b is how the solution of a linear system is written and it is not how it is computed. The usual reason given is cost — three times the arithmetic. The real reason is that one of the two routes is backward stable and the other is not, and at κ = 10¹⁴ they differ by twelve orders of magnitude in the number that says whose fault a wrong answer is.

 10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻⁸10⁻⁵10⁻²gap between the two eigenvalueshow far it movedthe eigenvectorsthe eigenvaluestheir plane‖E‖ / gapone perturbation, three answerseigenvalue shift, spread over the sweep1plane angle, spread over the sweep1eigenvector angle, spread1.6·10⁵the dashed line is Davis–Kahan's ‖E‖/gaptwo of the three never noticed Eigenvalues, singular values, rank

The gap decides the eigenvector

A symmetric matrix's eigenvalues move by at most the size of the perturbation, whatever the spectrum looks like. Its eigenvectors are governed by a completely different quantity — the distance to the neighbouring eigenvalue — and at a gap of 10⁻⁹ the same perturbation turns them through 27°.

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