Numerical Linear Algebra
where the algebra stops being the arithmetic
Theme

The thread: Where the algebra stops being the arithmetic — page 9

Essays 193 to 216 of the 271 on this theme, in the same order.
darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.3eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it Eigenvalues, singular values, rank

The eigenvalues that are not there

For a normal matrix the resolvent norm is exactly one over the distance to the nearest eigenvalue, so a picture of it carries nothing the spectrum did not. Move one entry above the diagonal and the region a perturbation of 10⁻⁸ can put an eigenvalue into stops being a disc and reaches out past the unit circle, while every eigenvalue stays at 0.8.

 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 Two errors, and whose fault they are

The fifth author

Four authors of a wrong answer have been named already 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.

 10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸10⁻⁷10⁻⁶|h|, the curvature of H along the weak directionσ at which the count is first wrong10⁻⁴10⁻³10⁻²10⁻¹110¹10²no interchangesLDLᵀ of Krank tolerancethrough Z: right to 10⁻¹⁶10 × 4, eight draws per curvatureh = -1·10⁻⁴: LDLᵀ first wrong at σ1.8·10⁻¹¹h = -0.01: LDLᵀ first wrong at σ1.8·10⁻¹⁰h = -1: LDLᵀ first wrong at σ1.4·10⁻⁹h = -100: LDLᵀ first wrong at σ1.8·10⁻⁸the pair's small eigenvalue is σ²/|h|the count loses it long before the rank does The matrix a constraint makes

A constraint the count stops seeing

Let one constraint drift towards being a combination of the others and the inertia of the saddle-point matrix keeps its promise only while σ²/|h| can be resolved — σ the constraint's smallest singular value, h the curvature along the direction it barely constrains. At h = −1 the count stops seeing the constraint at σ = 1.4·10⁻⁹, six decades before any rank test would drop it, and below that it reports a genuine minimum as a saddle on three to six draws in eight. No shift of H brings the constraint back: the correction loop shifts a problem that needed nothing by as much as 2,620. A perturbation of the constraint block does not bring it back either — it decides, at σ = √(|h|δ).

 0123401020304050z, the effective noise ratiotrials mirrored, %1 thin: 1 × 1 model2 thin: 2 × 2 model3 thin: 3 × 3 modelper cent mirrored at z = 0.7, 600 trialsn = 3, 1 thin, 20 points9.7n = 5, 1 thin, 20 points13n = 10, 1 thin, 40 points12n = 5, 2 thin, 20 points23n = 10, 2 thin, 40 points24n = 8, 3 thin, 40 points32points: measured in n dimensionslines: a k × k determinant, no n in it Orthogonality, measured

A mirror decided in the thin directions

In n dimensions the nearest orthogonal matrix to a noisy alignment is still sometimes a reflection, and the rate at which it is does not depend on n. Three, five and ten dimensions with one thin direction mirror alike; two thin directions mirror like each other in five dimensions and in ten. The rate is the chance that a k × k matrix built from the k thin directions has a negative determinant — 21.7 per cent at a noise ratio of 0.7 for k = 2, measured at 23.0 — and it is well above k independent coin flips. With two or more thin directions the determinant correction still fires, and it no longer rescues the rotation: the answer is eleven noise-widths off whether or not it was mirrored.

 edge of the numerical rangespectrum, -3.49zero37.9-0.986-3.496.59unstable one-sided models, of nineteen placements051015-3-2-1.5-1-0.500.511.522.53cond 1.4·10⁶cond 112t, the rescaling exponent — the k-th state is multiplied by γ to the power tk; t = 1 symmetrisesorder 1order 2order 3order 6the transfer function is the same at every tonly the inner product changes Reduction, and what a model is for

A certificate written in coordinates

Every one-sided reduced model of the convection–diffusion system was stable, because its poles cannot leave the numerical range and the range sat in the left half plane. Rescale the state by a diagonal matrix and nothing about the system changes — not its eigenvalues, not its transfer function, not one pole of any two-sided model — but the numerical range moves. At Péclet 40 a rescaling with condition number 260 pushes its edge to +6.6 and ten placements of nineteen return unstable one-sided models. Projecting in the inner product of a Lyapunov solution puts the certificate back in any coordinates.

 n = 512, ε = 10⁻⁸operations over stored numbers2cheapest leaf, arithmetic alone16cheapest leaf at 0 a block16097194291388485leaf sizeoperations per unknown48163264the arithmetic aloneand the blocks it reacheslarge dot: the cheapest leaf at this chargethe lower curve is the storage curve doubled Neither sparse nor dense

A second objective that is the first one doubled

Counting the operations a hierarchical product does was supposed to give the leaf size a second optimum, somewhere other than where the storage puts it. The count is 160,128, 139,904, 135,936, 155,136 and 216,064 against stored totals of 80,064, 69,952, 67,968, 77,568 and 108,032 — twice each, at every leaf, exactly. The two objectives cannot disagree, and what separates the leaf that stores least from the leaf that runs fastest is a charge of about 139 operations for reaching a block at all.

 00.250.50.75100.51xuexact & tunedupwindcentralone of these is exacttuned, worst nodal error2.4·10⁻¹⁷upwind, worst nodal error0.14central, points outside [0, 1]16exact at every nodeand only at the nodes Iterating, instead of factorising

The diffusion that makes the answer exact

Upwinding adds h/2 of artificial diffusion. Central differencing adds none. Add ε·ξ·Pe with ξ = coth(Pe) − 1/Pe and the computed solution is the exact one at every grid point, to 2.4·10⁻¹⁷ — at every Péclet number, on the problem it was derived from and on no other.

 a triangle and a square, joinedK₂,₃ with a pendant edge1234560123456indexeigenvaluethe same, and not the samevertices each6edges each7spanning trees12spectra differ by5.3·10⁻¹⁵highest degree, left4highest degree, right3one has a trianglethe other is bipartite The matrix that is a graph

The spectrum is not the graph

Two graphs on six vertices with the same Laplacian characteristic polynomial — as integer polynomials, not to fourteen digits. One contains a triangle; the other is bipartite. Every method in this field that reads only the spectrum is answering about the class.

 027548110813510⁻⁶10⁻⁴10⁻²110²10⁴10⁶power‖Aᵏ‖Kreiss constant 6760e · n · K‖Aᵏ‖ρᵏtwo routes to one peakspectral radius0.8peak of ‖Aᵏ‖2·10⁴Kreiss constant6757e · n · K1.1·10⁵everything here decays in the endand one of these curves says how much first Eigenvalues, singular values, rank

A spectral radius that grows first

ρ(A) below one guarantees that the powers of A go to zero and says nothing about what they do on the way. Here they rise by a factor of twenty thousand before turning over, and the peak is bracketed above and below by a constant computed from the resolvent norms outside the unit circle — two routes to one number, one through the plane and one through the powers.

 11.31.610⁻¹110¹10²10³log₁₀ ntropical root ÷ actual modulusexactsmallest moduluslargest modulusthree norms, two endslarge root, n = 40.96large root, n = 320.94small root, n = 45.4small root, n = 32230a maximum predicts a maximumand says nothing about a minimum The eigenvalue problem that is not linear

An estimate that does not move

The tropical roots are said to miss the bottom of a spectrum by a factor growing like n². The estimate does not get worse. It changes by four per cent between four masses and sixty-four while the modulus it names falls by a factor of a hundred and sixty, and an estimate that is constant in the variable the answer depends on is a different defect from an inaccurate one.

 110¹10²10³110¹10²10³the accumulating quantity, relative to its first valuethe error, relative to its first valuethe bounds: slope 1what all three do: slope ½three mechanisms, one exponenta left-to-right sum0.49a chain of rotations0.55a residual recurrence0.51every bound's slope1spread of the three0.067a bound is a sum of the roundingsand the roundings have signs The arithmetic underneath

Three walks and one bound

A left-to-right sum, a chain of three thousand rotations and a conjugate gradient residual recurrence share no arithmetic and no vocabulary. Each has a standard bound that is linear in whatever it accumulates against. All three come out at a half — 0.486, 0.554 and 0.507 — and nothing is rescaled.

 n = 512, leaf 16blocks, tightest to loosest592and at the loosest94mean rank, tightest3.3and at the loosest8.9052104156208260the separation constantstored per unknown0.20.30.40.50.951.4stored per unknownmean rank, compressed blocksboth curves move togetherwhich is what the other two knobs never do Neither sparse nor dense

A geometry setting that is a second accuracy

The tolerance raises every rank and moves no block; the leaf moves every block and no rank. The constant that decides how far apart two clusters must be before their block may be compressed moves both — 592 blocks at mean rank 3.25 at one end and 94 at mean rank 8.94 at the other — and the storage falls the whole way while the error against the dense matrix rises by a factor of 3.5, at a tolerance that never changed. It has no best setting, and it is not a third knob on the storage.

 a 28 × 28 saddle-point matrixnatural order, entries113sparsest order, entries72two-by-two pivots, loosest test0and at the strictest3023466992115the constant in the pivot testentries in the factor0.10.30.640.80.95133natural ordersparsest ordergrowth, naturalgrowth, sparsestnumbers above the points: two-by-two pivots takenboth curves favour the sparser order Sparsity, and what elimination costs

The freedom a symmetric factorisation does not have

Permuting rows and columns together leaves no column to choose, so the conflict between the sparsest pivot and the sound one should be worse rather than better. On a saddle-point matrix whose constraint rows have no diagonal entry at all, it is not there: taking the sparsest available pivot holds 70 entries against the natural order's 113 and a growth of 1.28 against 1.83 — better on both currencies at once, at every setting of the pivot test. The two-by-two blocks that make it legal cost 1.33 entries apiece.

 -12-10-8-6-4-2024681011.522.533.5leaf width minus (√M − 2)words ÷ pure recursionM = 64M = 144M = 256the edge is the same column at every memoryand the cheapest leaf sits on it Where the flop count stopped predicting the time

The leaf that sits on the edge

A recursive elimination's base case was found to be a block size in disguise, with a cliff where the blocked elimination's is, and the choice read as a trade: the processor wants wide leaves, the cache wants narrow ones. Measured at every width rather than at the halvings of 96, there is no trade inside the edge. Leaves exactly √M − 2 wide are the cheapest the recursion can have in words as well as calls — 0.81, 0.80 and 0.84 of the pure recursion's traffic at 64, 144 and 256 words — and one column wider moves 1.77 to 3.15 times it. And a matrix of 100 columns, which halves unevenly, meets the cliff in two steps rather than one.

 75 aggregates over 225 unknownsthe matrix chose thismean extent along y3mean extent along x1points adopted by pass two15no coordinate enters the methodand the shape follows the coupling Iterating, instead of factorising

Aggregating what the matrix calls strong

The depth phase measured every method it had on the 45°-rotated anisotropic operator — 0.784, 0.883, 0.844 — and diagnosed the failure as being in the discretisation rather than in the hierarchy. Smoothed aggregation is the standard answer to anisotropy. It returns 0.789.

 finite eigenvalues (degree of det Q)13at infinity (2n − degree)1at infinity, by the rank of M12n, if M were nonsingular14a degree, not a decisiondegree of det Q13at infinity1by the rank of M1singular-value gap∞the count is a degreeand the other route is a judgement The eigenvalue problem that is not linear

One mass removed, and one eigenvalue gone

A coordinate with no inertia reads like a coordinate that has been deleted, and a chain of eight masses with one of them removed would then be a chain of seven, with fourteen eigenvalues. It has fifteen. The massless coordinate is still there, still carrying a damper, and it contributes a first-order equation rather than none.

 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 Two errors, and whose fault they are

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.

 10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁴10⁻¹10¹²10²⁵10³⁸10⁵¹10⁶⁴δ, the gap between consecutive eigenvaluesrelative error in eᴬan answer with no correct digitsV f(Λ) V⁻¹scaling and squaring‖A(δ) − A₀‖exact eigenvalues throughoutκ(V) at the smallest δ3.3·10⁸²eigen route2.9·10⁶⁵scaling and squaring4.1·10⁻¹²distance to the limit7·10⁻¹²the eigenvalues are the diagonaland they are exact at every stop Eigenvalues, singular values, rank

A function of a matrix is not a function of its entries

Everybody learns that f(A) means diagonalise, apply f to the eigenvalues, undiagonalise. That is a definition, not a method. On a matrix seven picometres from a defective one — with exact eigenvalues and eigenvectors from a closed form — the definition returns an answer wrong by sixty-five orders of magnitude, and a method that never mentions an eigenvalue returns the right one.

 Re λIm λwhat survives the arrowsvertices18arcs18worst row sum0worst column sum0largest |Im λ|0.98asymmetry1the null vector is still exactand nothing else about the spectrum is real The matrix that is a graph

A Laplacian that is not symmetric

Point the edges and the matrix stops being symmetric. Its row sums are still exactly zero, so the null vector survives; everything built on the quadratic form does not, and the eigenvalues need a plane rather than a line. Asymmetry permits that and does not force it, which the smallest case here demonstrates by being asymmetric and real.

 02468101201change of units, by exponent10⁰10¹10²10³10⁴10⁶10¹⁰10¹⁶10¹⁹10²⁰10⁴⁰10¹⁵⁰as writtenafter scalinga range questionlargest finite value3.4·10³⁸predicted boundary γ1.8·10¹⁹last γ that forms10¹⁹stops where scaling fails0not a poor answerno answer at all The arithmetic underneath

The units that overflow before the answer does

A change of variable that is exact in the algebra requires γ² times a matrix to be a number the format can hold. In binary64 that is a bound nobody meets by accident. In binary32 it arrives at 10¹⁹ and in fp16 at 256, and past it there is no answer rather than a poor one.

 μ = −0.01, δ = 7saddle's growth a step0.0014minimum's fall a step0.014residual verdict, step6205010015020010⁻¹10¹refinement steprelative residual, relative errorsaddle, residualsaddle, errorminimum, residualminimum, errorthe pivot signs call both of these a minimumthe residual does not The matrix a constraint makes

The residual turns before the error doubles

Regularise a saddle-point matrix past the Hessian's most negative eigenvalue and its pivot signs certify every shallow saddle as a minimum; refinement against the unregularised matrix keeps the test, as a rate, and a saddle at a hundredth of the regularisation grows by only 1.0014 a step — 485 steps to double. That was read as hundreds of steps before the history says anything. The residual, which is what a solver actually has, says it at step 62: a fit of its logarithm over the last ten steps turns positive there and stays positive, while the minimum's is negative from step 10. Across five regularisations the verdict comes at an eighth of the doubling time.

 ties broken by 10⁻¹²one step ahead, n = 248.4·10⁶two steps ahead, n = 2424812162024110¹10²10³10⁴10⁵10⁶10⁷matrix size ngrowth factorgreedyone step aheadtwo steps aheadone step ahead lies exactly on greedythe damage is done a step before it shows Elimination, and the swap

One step ahead is one step short

Partial pivoting takes the largest entry in the column and, on Wilkinson's matrix, walks into growth of 2^(n−1) that a cyclic shift of the rows avoids entirely. A rule that chose instead the pivot whose elimination leaves the smallest trailing submatrix was expected to see the good order at the first step. It sees nothing there: every first pivot leaves a largest entry of exactly 2, and with the ties broken by 10⁻¹² it prefers the greedy row by 10⁻¹². It attains 2^(n−1) at every size. Looking two eliminations ahead, the greedy row scores 4 and every other row 2, and the growth is 2 at every size up to 24. On random matrices one step of look-ahead helps below n = 16 and is worse than greedy on more than half of them by n = 32.

 96 × 64, κ = 10⁸blocks of eight, final1.4·10⁻¹⁴their root-sum-square1.2·10⁻¹⁴one at a time, final2.9·10⁻¹⁴081624324048566410⁻¹⁵10⁻¹⁴10⁻¹³columns factorised‖QᵀQ − I‖sum of the blocks' ownone reflector at a timeblocks of eightroot-sum-squareeach block is one factor, each reflector anotherthe departure grows with the square root of the count Orthogonality, measured

Eight blocks and sixty-four reflections

One block of sixteen reflectors, assembled as I − Y T Yᵀ, departed from orthogonality five times as far as a single reflection, and the question was whether a factorisation of many blocks multiplies that factor. It does not. On 96 × 64 matrices eight blocks of eight end at 1.42·10⁻¹⁴ — within a fifth of the root-sum-square of their own departures, and less than half their sum — while the same factorisation taken one reflector at a time ends at 2.91·10⁻¹⁴. A block departs more than a reflector, and there are an eighth as many of them. And a nearly dependent column that swells the triangle's entries to 10²⁶ costs the product nothing.

 Cheeger's band, for the circulationcirculation · arcsvertices on the smaller sidewhich of the two the theorem is aboutλ₂ of 𝓛0.066λ₂/20.033√(2λ₂)0.36circulation Φ0.12arc conductance0.04their ratio3.1the inequality holdsabout a quantity nobody counts The matrix that is a graph

A conductance the arcs do not measure

Symmetrising a directed Laplacian with respect to its walk recovers everything the arrows took — a real spectrum, a sweep cut, a Cheeger inequality. What it does not recover is the quantity: the inequality bounds the probability that a step of the walk crosses the cut, which on one graph here is three times the weight of the arcs that do.

All themes