Numerical Linear Algebra
where the algebra stops being the arithmetic
Theme

The thread: Two routes to a number — page 2

Essays 25 to 48 of the 59 on this theme, in the same order.
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.

 how far is it from A to an orthogonal matrix?smaller is nearer · the polar factor minimises this in every unitarily invariant normpolar factor U1.8554QR, signs fixed2.1265QR as returned3.8226200 drawn at randomκ = 10polar factor1.9QR, signs fixed2.1QR as returned3.8best of 200 random2.7‖A − QR‖ is the same either wayand ‖A − Q‖ is not Orthogonality, measured

The nearest orthogonal matrix

Every field that has to clean up a drifted rotation reaches for QR, and QR does not answer the question. The nearest orthogonal matrix is the orthogonal factor of the polar decomposition — nearer by about a tenth, and, more to the point, the same matrix whatever order the columns were written in. QR's answer changes completely.

 015030045060075090010⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹iteration‖r‖ / ‖b‖JacobiGauss–SeidelSOR ω=1.78closed form vs measuredρ Jacobi, exact0.99ρ measured0.99ρ Gauss–Seidel, exact0.981D Laplacian, n = 24ω optimal = 1.777 Iterating, instead of factorising

A rate that is known in advance

On the model problem, Jacobi contracts by cos(π/(n+1)) per step, Gauss–Seidel by its square, and optimally relaxed SOR by a number given in closed form. Three rates, all known before anything runs, and all measurable against what runs.

 -0.16-0.63-0.493.43-0.025-0.632.3-0.69-0.95-1.70.076-0.49-0.694.3-1.6-1.41.93.4-0.95-1.65.40.0141.63-1.7-1.40.0144.60.055-0.0250.0761.91.60.0553.1A, symmetric→-0.164.600004.65.92.500002.51-1.20000-1.24.90.3800000.385.80.2400000.242H = QᵀAQ, tridiagonal‖A − QHQᵀ‖/‖A‖1.1·10⁻¹⁵below the subdiagonal0worst eigenvalue movement7.1·10⁻¹⁵a similarity, so the spectrum is untouched — and every later step is O(n²) rather than O(n³)one reduction, then every iteration is cheapthe eigenvalues did not move Eigenvalues, singular values, rank

The form that makes it affordable

One Householder reduction, done once, turns every subsequent iteration of the eigenvalue algorithm from cubic to quadratic cost. It changes no answer at all, which is why it is easy to describe as an optimisation and wrong to.

 10²020406080100120unknownsiterations2D, no preconditioner2D, preconditioned1D, preconditionedone construction, two dimensions2D steps at 16 unknowns102D steps at 100 unknowns211D steps at 100 unknowns10the same averaging, the same transformand a count that no longer stops growing Structure, and the solver that cannot see it

Two dimensions, and the cluster that thins

The same kernel, the same averaging, the same transform — applied along two axes instead of one. In one dimension the preconditioned step count is 7, 10, 10, 10; on square grids with the same unknown counts it is 10, 18, 20, 21, and the share of the spectrum near one falls from 56% to 17%.

 1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 20no penaltypenalty insidewhat stopping is worthbest without a penalty0.14and at step 40161best with one0.14and at step 400.14the same floor, reached twiceand only one run stays on it Methods that were designed apart

The step that stops mattering

Regularise the problem the iteration has built rather than the problem it was given, and the error curve stops turning. The unregularised run ends 1,127 times above its own best; the same run with a penalty inside it ends 1.000000000003 times above.

 10²10⁴10⁶10⁸10¹⁰110²10⁴10⁶10⁸10¹⁰condition number of the matrixcondition number seen by the iterationκ(A), unchangedκ(AR⁻¹)a bound with no κ(A) in itκ(AR⁻¹), every κ(A)2.2κ(SU), the other route2.2κ(A), across the sweep10⁸κ(AR⁻¹), across the sweep1the sketch never sees the spectrumand the spectrum cancels out of the answer Randomised, and the guarantee that changes kind

The sketch that is not the answer

Sketch-and-solve throws away the original problem and keeps the small one's answer, which is why its answer moves with the seed. Use the same sketch as a preconditioner instead and the condition number the iteration sees is the same number at every κ from a hundred to ten billion — identically the same, to nine digits, because the spectrum cancels out of it.

 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.

 00.250.50.751-1-0.500.51mode frequency θ / πdamping factor per sweeppredicted±0.333measureda coarse grid seestwo routes to one factorsmoothing factor, scanned0.33smoothing factor, closed form0.33worst mode disagreement4.4·10⁻¹⁶63 interior points, one sweepthe left-hand end is what the coarse grid is for Iterating, instead of factorising

The error smoothing cannot reach

One weighted Jacobi sweep multiplies every mode of the error by a number, and the number is a sine. Half the modes are cut by three or better, and the other half come back at 0.999 — which is not a failure of the method but the fact the whole of multigrid is built on.

 the bound211no pivoting127τ = 0.00198τ = 0.003100τ = 0.01106τ = 0.03111τ = 0.1116τ = 0.3117τ = 1138entries in Ua bound, and its slackthe bound, from the graph alone211the worst that occurs138loose by1.5no arithmetic was done to compute the bound — only the pattern of AᵀA and its elimination graphGeorge and Ng: U fits inside chol(AᵀA), whatever the swapsprovable, cheap, and loose Sparsity, and what elimination costs

What the symbolic phase can only bound

Without pivoting, the fill can be computed from the graph and the count is exact — 233 predicted, 233 measured. With pivoting it is 233 predicted and 242 measured, and what survives is a bound that is right at every threshold and loose by 1.7 times at the largest grid drawn.

 27121722273237424710⁻²10⁻¹110¹subspace size kvalueno answer below 8GCV's traceλ from GCVλ from the residuala denominator that is not the problem'strace at k = 41trace at k = 4827λ range across the run2.2subspace before an answer8the divisor moves by twenty-sevenand the answer does not move Methods that were designed apart

A parameter chosen on a smaller problem

Inside a hybrid method the regularisation parameter is chosen on a 25×24 problem rather than a 64×64 one. The rule that reads a residual transfers exactly; the rule that reads a trace is biased by exactly two grid steps at twenty-four steps and one at forty, at every noise level from 10% to 0.1%.

 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.

 15731pointsfinest gridone unknown — the recursion bottoms out in a divisionthe coarse operator, two ways‖RA_hP − A_2h‖/‖A_2h‖10⁻¹⁸unknowns / finest grid1.7cycle cost, in fine sweeps12each coarse point reaches three fine ones½, 1, ½ — and the restriction is its transpose Iterating, instead of factorising

The same problem on a coarser grid

Restriction, the coarse operator and interpolation are three matrices with nine distinct entries between them. Two of the three are each other's transpose, and their product with the fine operator is the coarse discretisation exactly — not approximately, entry for entry, at every level.

 0.46-1.11.5-0.470.2-0.181.90.690.42-0.160.028-0.0160-1.50.160.880.099-0.1100.680.762.30.930.0480-0.250.110.952.11.500001.52.3after reflector 2the highlighted entries are the bulge — the only thing that is not Hessenbergentries below the subdiagonal3reflectors used2reflectors in a whole step5the shifts are never formed — only their sum and productand both of those are real Eigenvalues, singular values, rank

Two shifts that are never formed

The double shift is defined as a factorisation of (A − μI)(A − μ̄I), which nobody computes. What is computed is the first column of that product — three numbers — and the bulge those three numbers create, pushed down the subdiagonal by n − 2 reflectors until it falls off the bottom.

 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.

 10⁻⁷10⁻⁵10⁻³10⁻¹⁰10⁻⁸10⁻⁶magnitudespacing to the next numberthe smallest normalgradualflush to zerosmallest normal6.1·10⁻⁵smallest subnormal6·10⁻⁸octaves of subnormals10pairs that lie under FTZ10the spacing stops halving and stays putwhich is what makes x − y = 0 mean x = y The arithmetic underneath

The numbers below the smallest one

Below the smallest normal number the spacing stops halving and stays put, all the way to zero. That is what gradual underflow is, and the thing it buys is the sentence every algorithm assumes without being told — x minus y is zero only when x equals y.

 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.

 -0.0833-0.167-0.0833-0.1671-0.167-0.0833-0.167-0.0833R · A · P, normalised to a unit centrestored entries per rowlevel 0 · 31×314.87/rowlevel 1 · 15×158.22/rowlevel 2 · 7×77.37/rowlevel 3 · 3×35.44/rowlevel 4 · 1×11.00/rowstill a stencil, still annihilates a constantentries in an interior row9weight outside the 3×30row sum0the isotropic model problemnine, at every level below the first Iterating, instead of factorising

The coarse problem is a different problem

In one dimension the Galerkin coarse operator is the coarse discretisation, entry for entry — this site asserted it. In two dimensions a five-point operator produces a nine-point coarse one, so the recursion solves a different discretisation at every level below the first, and converges at 0.20 a cycle regardless.

 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°.

 θx (frequency across x)θy0π/2π0π/2πthe coarse grid'sunder 0.2under 0.40under 0.60under 0.80under 0.95under 1.01damping per sweeptwo routessmoothing factor, scanned1closed form130×30 frequency cellsthe marker is the mode nothing removes Iterating, instead of factorising

A direction the smoother cannot see

Give the Laplacian a strong direction and multigrid stops working — from 0.2016 a cycle to 0.9565 — with every component unchanged and the condition number identical to twelve digits. The problem did not get harder. The link between the method's two halves broke.

 10²10³10⁴10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴terms summedrelative error+∞, −∞ 1.01zero 1.00stochastic 0.50nearest 0.47√n against n, fittednearest, fitted exponent0.47stochastic, fitted exponent0.5toward +∞, fitted exponent1twelve seeds averaged at each sizethe slope is the bias, not the precision The arithmetic underneath

The direction the error leans

The size of one rounding error is set by the precision. How ten thousand of them combine is set by something else entirely — the rounding mode — and the fitted exponents are 0.47 for round-to-nearest and 1.01 for round-toward-infinity, on identical data at identical precision.

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