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The thread: Measured, not asserted — page 2

Essays 25 to 48 of the 110 on this theme, in the same order.
the mirrorx, length 4.000Hx = (-4.000, 0)v = x − αe₁safe sign: α = −‖x‖, so v is formed from a sum and nothing cancelsunsafe sign: α = +‖x‖ gives ‖v‖ only 35.1% of ‖x‖ + ‖x‖ — the digits go‖HᵀH − I‖5·10⁻¹⁶‖Hx‖ − ‖x‖8.9·10⁻¹⁶second component2.2·10⁻¹⁶built from a unit vectororthogonality is structural Orthogonality, measured

A reflection cannot stop being one

Householder QR holds orthogonality at 10⁻¹⁵ whatever the condition number of the matrix, and Gram–Schmidt does not. The reason is not that it is more careful. It is that its Q is built from unit vectors, and rounding a unit vector gives a different reflection rather than a broken one.

12345678910⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1rank k of the approximation‖A − Aₖ‖measured, 2-normσₖ₊₁, from theorymeasured, Frobeniusthe first two agreeto 4.3·10⁻⁹worst |‖A−Aₖ‖₂ − σₖ₊₁| / σₖ₊₁4.3·10⁻⁹worst Frobenius discrepancy4.3·10⁻⁹κ = 10⁹; 30 random rank-3 matrices, none closerthe error is σₖ₊₁ Eigenvalues, singular values, rank

The best approximation there is

The error of the best rank-k approximation is not bounded by the next singular value. It is equal to it. That is an unusually sharp theorem, and it makes the theorem itself usable as an independent check on the computation.

the matrix43 entriestip eliminated first253 entriestip eliminated last43 entries‖A − LLᵀ‖/‖A‖, tip first1.4·10⁻¹⁶‖A − LLᵀ‖/‖A‖, tip last0dense factor is n(n+1)/2 = 253 · sparse factor is 2n − 1 = 43one row swapped to the endnothing numerical chose between them Sparsity, and what elimination costs

Two ends of the same arrow

One matrix, one row moved from the front of the elimination order to the back, and the factor goes from completely dense to no fill at all. Both factorisations are exact to rounding, and nothing numerical chose between them.

04812162010⁻¹10⁻⁰.⁵1target rank k‖A − A_k‖₂published boundrandomisedσ_{k+1}, optimalhow far apart the three areworst seed spread1.1bound / median at k = 1211median / optimum at k = 12160×60, 6 seeds, oversampling p = 5band is best to worst Randomised, and the guarantee that changes kind

Randomisation does not create structure

On a matrix whose singular values are all equal, a rank-ten randomised approximation has error 1.0 — and so does the optimal deterministic one. Neither achieved anything, and only one of them is usually sold with the implication that it might.

10⁻⁴10⁻³10⁻²10⁻¹110¹10²10³10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹relative change in the coefficients, along the worst directionrelative increase in the residualcoefficients doubled39% change, fit unmoved in the sixth digit308×: the third digit movesκ(A) = 3.6·10⁶. Exact arithmetic would pick one point on this floor. It would not raise it.24 points, degree 9, monomial basisthe data leaves them free Least squares, and the road not to take

The valley with no bottom

A degree-nine fit's coefficients can be moved by a third of their own size before the residual changes in the sixth significant figure. The arithmetic did not lose those digits. The data never contained them.

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 Two errors, and whose fault they are

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.

051015202530354010⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹iteration‖r‖ / ‖b‖plain CGIC(0) CGwhat the preconditioner didκ(A)48κ(L⁻¹AL⁻ᵀ)5.1‖A − LLᵀ‖/‖A‖0.0832D Laplacian, n = 100√κ ratio predicts 3.07× Iterating, instead of factorising

Changing the condition number on purpose

Preconditioning is usually introduced as a trick that makes an iteration converge faster. It is not a trick. It is solving a different system with the same solution and a condition number chosen rather than inherited, and the new condition number is computable.

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 Elimination, and the swap

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.

10²110¹10²size niterations to 10⁻¹⁰λ_min(C) changes signno preconditionerStrang's circulantthe preconditioner's own spectrumλ_min(C) at n = 16-0.4λ_min(C) at n = 32-0.14λ_min(C) at n = 640.016λ_min(C) at n = 1280.051λ_min(C) at n = 2560.053left of the line the repair costs stepsright of it, the count stops counting n Structure, and the solver that cannot see it

A preconditioner that changes sign

Strang's circulant preconditioner takes Toeplitz conjugate gradients from 179 steps to 10 at n = 256. At n = 64 on the same family it takes 66 steps to 109 — worse than doing nothing. Between those rows the preconditioner's smallest eigenvalue crosses zero, and nothing in the published account of the method mentions that it can be negative.

classical Gram–Schmidt4.62·10⁻¹⁰modified Gram–Schmidt1.49·10⁻¹²Householder, one sweep2.03·10⁻¹⁴reduction tree, 16 leaves1.48·10⁻¹⁵departure from orthogonality, logarithmicthe tree, at four depths‖AᵀA − RᵀR‖/‖AᵀA‖, depth 13.4·10⁻¹⁵‖AᵀA − RᵀR‖/‖AᵀA‖, depth 24.3·10⁻¹⁵‖AᵀA − RᵀR‖/‖AᵀA‖, depth 31.7·10⁻¹⁵‖AᵀA − RᵀR‖/‖AᵀA‖, depth 41.5·10⁻¹⁵the same algebra, four timestwo of them are products of reflections Where the flop count stopped predicting the time

A reduction that changes the order

A tall-skinny QR computed as a tree of independent block factorisations touches a 512×12 matrix once instead of twelve times, computes a completely different sequence of roundings from the sweep it replaces, and returns ‖AᵀA − RᵀR‖/‖AᵀA‖ = 1.65·10⁻¹⁵ against the sweep's 9.95·10⁻¹⁵. On the same matrix classical Gram–Schmidt returns 4.6·10⁻¹⁰.

10⁻²10⁻¹110¹10²10³10⁴‖Ax − b‖‖x‖the oraclediscrepancyL-curvegeneralisedscored against a truth none hasoracle, relative error0.11discrepancy principle, as a multiple1.1L-curve corner, as a multiple2.3generalised cross-validation, as a multiple1the oracle needs the exact answer and is not a methodit is the reference the others are scored on Regularisation, and the answer that is chosen

Choosing without knowing

Three published rules for choosing a regularisation parameter, scored against an oracle that requires the exact answer and is therefore not a method. Generalised cross-validation lands on the oracle's λ exactly; the discrepancy principle costs 6%; the L-curve costs 129%. And told a noise level ten times too small, the discrepancy principle's error goes from 0.112 to 10,449.

21018263442500102030405060708090100110120130140150160170180190200210220230240250260270280290300310320330340significand bitsiterationspreconditioner roundedarithmetic roundedno preconditionersame bits, different casualtyerror, 3-bit preconditioner8.8·10⁻¹³error, 3-bit arithmetic0.16‖A − LLᵀ‖/‖A‖ of the factor0.083a direction may be roundeda measurement may not Methods that were designed apart

The part of a solver that may be rounded

A preconditioner computed and applied with a three-bit significand still returns thirteen correct digits — it costs seventeen extra iterations and nothing else. Round the working arithmetic instead and the step count barely moves while the answer loses exactly the digits the format dropped.

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 Two errors, and whose fault they are

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.

02468101210⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1iteration‖r‖ / ‖b‖Laplaciancyclic shiftno progress at allevery eigenvalue of the shift is on the unit circleand it predicts nothing Iterating, instead of factorising

The spectrum that predicts nothing

For a symmetric matrix the eigenvalues govern how fast an iteration converges. Drop symmetry and they stop governing anything — there is a matrix whose eigenvalues are as evenly spread as eigenvalues can be, on which GMRES makes no progress at all until the last possible step.

0357010514017521024528010⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹iteration|subdiagonal entry|no shiftRayleighWilkinsontwo routes to one raterate, from the spectrum0.9rate, measured0.9iterations, none / Wilkinson45symmetric 4×4, spectrum 8, 4, 2, 1.8the dashed line is the prediction Eigenvalues, singular values, rank

The algorithm the libraries actually run

Factorise, multiply the factors back in the other order, repeat. That description is complete and correct and produces something nobody would use — on a matrix with eigenvalues +1 and −1 it does not converge at all, and the subdiagonal entry does not move by so much as a rounding error.

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 The arithmetic underneath

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.

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.

036912151821242700.250.50.7511.25index kweight appliedonebidiagonalArnoldiis it a function of σmisfit, even fit1.6·10⁻¹²misfit, general fit0.063‖A − Aᵀ‖/‖A‖0.086one method's weights do not notice the operatorand the other's stop being a function of σ Regularisation, and the answer that is chosen

The basis decides what a filter is

The vocabulary of regularisation is spectral — a method keeps a component or discards it, and the weights are a function of the singular value. Row-normalising a symmetric blur so that it preserves a constant makes it 8.6% asymmetric, and that is enough to move GMRES's weights from 7·10⁻¹⁴ off a function of σ to 4.4·10⁻².

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.

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