The collection

Every essay — page 6

Essays 121 to 144 of 507, in the same order.

Regularisation, and the answer that is chosen

Some problems do not determine their own answer. The data is consistent with a range of solutions that differ by orders of magnitude, no arithmetic can choose between them, and something outside the data has to. That choice is the computation rather than a preliminary to it — and every published rule for making it is a heuristic scored, here, against a truth that exists only because the problem was constructed.

10141822263034384210⁻¹10⁻⁰.⁵110⁰.⁵grid points nrelative error against the continuous signalsampled kernel, hatsintegrated, hatsintegrated, spline0.1% noise, no λsampled kernel, hats, best0.13integrated, hats, best0.12integrated, spline, best0.12truncation of 64 points0.12same nodal unknowns, same databetter at representing the noise too

A better discretisation is a weaker filter

A coarse grid's error has two sources — how well the discrete operator approximates the integral, and how well the grid's function represents the answer — and the grid essay could not separate them. Changed one at a time they separate: integrating the kernel against the hat functions takes a fifth off the 12-point error, reading the answer as a cubic spline takes 15 per cent more, and both roughly double the condition number on every grid. At 0.1% noise the spline discretisation's unregularised solve on 26 points reaches the best truncation of a 64-point grid to 0.3%. At 1% it is worse than the crude grid.

8 figures · Regularisation, essay 7
122028364452606876849210010⁻¹110¹grid points nrelative error against the continuous signalbest with no λ: n = 26no λeach grid's best λTikhonov, 64 points0.1% noise, 16 drawsbest grid with no λ26its error0.1396 points with its λ0.12best λ on fine grids0.0056each point is a median over the same drawsλ belongs to the problem, not to the grid

Where the grid hands over to λ

An unregularised solve on a coarse grid comes within a tenth of the best Tikhonov answer on a fine one, and the pair of a grid and a λ was left unmeasured. Measured, the two do not trade. On every grid up to the best unregularised one no λ helps at all. On every grid of 40 points and more the best λ is the same to within a quarter of a decade — 3.2·10⁻² at 1% noise per sample, 10⁻³ at 0.01% — and the 96-point grid with it beats the best coarse grid by 7, 9 and 13 per cent. The grids between the two, given their own λ, land between them.

8 figures · Regularisation, essay 6
where each λ sitsoracle's share0.0037corner's share0.1corner's cost25and what a target costsbest fixed target0.0035its cost1estimate ÷ truth at the oracle0.9210⁻⁸10⁻⁶10⁻⁴10⁻²110⁻³10⁻²10⁻¹110¹10²10³10⁴λnoise part ÷ signal parttarget 0.0035oracle λ: share 0.0037corner: share 0.102solid: the share on this draw · dashed: the share a rule can estimatethe rule is a level set of the dashed curve

A rule that has to be told how good its answer will be

The L-curve's corner reads a noise share of 0.10 to 0.21 across fifteen pairings of signal and penalty, and the share the best λ sits at runs from 0.0037 to 0.43 — a factor of a hundred and fifteen. A rule aimed at the right share is within a few per cent of the oracle on every one of them. The right share is about a third to four-fifths of the relative error that λ will achieve, which is the number the answer was wanted for.

7 figures · Parameter choice, essay 7
what each reachesbest pair0.0034‖x‖ alone0.0048‖L₁x‖ alone0.0035λ₁, on ‖x‖λ₂, on ‖L₁x‖ — log₁₀0-6-5-4-3-2-100-6-5-4-3-2-10outlined row and column: one penalty switched offred cell: the best pair, 0.00343darker is worse, on a logarithmic scalethe optimum sits on or beside an edge

A second penalty is not a second parameter

Penalise ‖x‖ and ‖L₁x‖ at once and there are two λ to choose. Over ten draws on five signals the best pair beats the better single penalty by between 0.00% and 3.1%, and one of its two parameters is exactly zero on 30 to 70 per cent of draws. Choosing the wrong one of the two costs up to 54%. The surface is a choice between two curves with a knob nobody needs.

6 figures · Parameter choice, essay 8
1220283644526068768492100101214161820grid points nrelative error against the continuous signal, per centsampled kernel, hatsintegrated, hatsintegrated, splinespline, no λ1% noise per sample, median of sixteen drawssampled kernel, hats: 16 points0.18integrated, hats: 16 points0.16integrated, spline: 16 points0.16sampled kernel, hats: 96 points0.14integrated, hats: 96 points0.14integrated, spline: 96 points0.14spline, no λ, its best grid0.15dotted: the 96-point answerevery point is a grid with its own λ

The grid on which the discretisation stops mattering

Without regularisation, integrating the blur's kernel against cubic splines beat sampling it at 0.1% noise and lost to it at 1%. Give each discretisation its own best λ on every grid and the difference shrinks to nothing where grids are fine — 0.17, 0.28 and 0.20 per cent apart on 96 points at the three noise levels, with every discretisation choosing the same λ — and stays at 17 to 18 per cent on 16 points. The choice between them is a choice of how coarse a grid can be: at 0.1% noise the integrated discretisations reach the fine-grid answer on 26 points and the sampled one needs 40.

6 figures · Regularisation, essay 8
on 96 pointsno breakpoint0.12two jumps0.05doubled knots0.13one box0.013the box on fewer points16 points0.0526 points0.02248 points0.00710203040506070809010010⁻²10⁻¹.⁵10⁻¹grid pointsrelative error, best λno breakpointtwo jumpsdoubled knotsone boxevery reading spends the same n unknownsand is given its own best λ on each grid

A corner the penalty can afford

Every smooth reading of the deconvolution's grid needed about forty points and then stopped improving, and the step was the suspect. Give the step one coefficient of its own and forty-eight points reach an error of 0.0070 at 0.1% noise, against 0.118 for the best smooth reading on ninety-six — the step was most of the error. But the same step given two coefficients recovers half as well, and given a doubled node at each edge it recovers worse than no breakpoint at all, while representing the signal to 0.07%. What decides is what the penalty is charged for the corner, and whether the data can say where it is.

6 figures · Regularisation, essay 9
worst draw on any grid, % over the oraclediscrepancy13GCV4·10⁶L-curve291median on 96 points, % overdiscrepancy3.4GCV0.18L-curve282030405060708090100110¹grid pointserror ÷ the oracle's, same drawdiscrepancy, mediandiscrepancy, worstGCV, medianGCV, worstL-curve, mediansolid: the median draw; dashed: the worst of sixteenthe axis stops at twenty times the oracle

The data count their dimensions, not the step's

Every grid in the deconvolution essays was chosen with the answer in hand, and so was every λ. From the data alone, the discrepancy principle's worst draw is within 16 per cent of the oracle on every grid from 16 points to 96; generalised cross-validation is better on the median draw and, on grids of thirty points and more, has draws thousands of times worse. And the data can say how many dimensions they carry — about 20, 25 and 29 at three noise levels, one number once the grid exceeds it — but not how many more the step needs: the grid that count chooses is 14 to 19 per cent worse than forty points at the lower two.

6 figures · Regularisation, essay 10
worst draw on any grid, ÷ the oracleGCV4·10⁴rightmost minimum1.6discrepancy1.1draws more than twice the oracleGCV10rightmost minimum02030405060708090100110¹10²10³10⁴grid pointsworst error ÷ the oracle'sGCV, worst drawrightmost minimumdiscrepancythe worst of sixteen draws on each gridthe axis stops at ten thousand times the oracle

The minimum on the right

Generalised cross-validation's worst draws on a fine grid were all one mistake: a second dip in its function at λ near zero, deeper than the real minimum. The proposed repair was a residual threshold, one number, refusing any λ whose residual falls too far below the real minimum's. Measured over 528 draws, a threshold of one half still lets two hundredfold misses through; only the extreme value, which is no threshold at all but the rule 'take the rightmost local minimum', removes all eleven. It costs nothing on the coarse grids where the dip is the right answer, and on the collection's own problem over five thousand draws it turns 243 tenfold misses into 17.

7 figures · Regularisation, essay 11
square systemfloor minima21interior dips28four samples per unknownfloor minima0interior dips4051015202530samples per unknowndraws of 24011.251.524minimum at the floorinterior dipGCV over 10× the oracleover 100×horizontal axis doubles at each gridlinethe floor goes at once and the dip does not

More samples take the floor and leave the dip

GCV's catastrophic misses on fine grids were blamed on squareness: on an n × n system the residual and n − t both reach zero as λ does, and their ratio can dip there. With more samples than unknowns neither reaches zero, and the prediction was that the dip would be gone by construction. Half of it is. The minimum at the floor of the scale, 21 draws in 240 on square systems, is gone at every ratio. The interior dip is not — 28, 20, 13, 8 and 4 draws at one to four samples per unknown — and at four per unknown one draw still misses the oracle by 877 times.

5 figures · Regularisation, essay 12
1×1.1×1.5×2×3×stepoffsetsmoothoscillatingspikeswrong penalty ÷ right oneone penalty ÷ best pairbest pair ÷ best triplewrong penalty ÷ right oneone penalty ÷ best pairbest pair ÷ best triplewrong penalty ÷ right oneone penalty ÷ best pairbest pair ÷ best triplewrong penalty ÷ right oneone penalty ÷ best pairbest pair ÷ best triplewrong penalty ÷ right oneone penalty ÷ best pairbest pair ÷ best tripleeach tick one drawthe choice is worth a factor, the third penalty nothing

A third penalty on a flat floor

Two penalties at once were worth three per cent at most, and the explanation offered was that one penalty already does the work — which predicts that a third buys less still and that the best choice sits on a face of the parameter cube. The third buys a median of exactly nothing on all five signals and 0.66% on its best draw. But on twelve draws of forty the best triple does use all three, and on every one of them the nearest point with a penalty switched off is within that same 0.66%. The minimum is not on a face; it is on a floor so flat that where it lands is noise. Choosing the right single penalty is worth a factor of two.

6 figures · Parameter choice, essay 9
45 interior dipsρ at the dip, true noise, median0.84ρ at the dip, estimated, median0.980.50.7511.251.51.7520.50.7511.251.51.752ρ at the dip, true noiseρ̂ at the dip, estimated noisedashed: the estimate equal to the truththe estimate pins every dip at one

An estimate that shares the dip's luck

GCV's interior dips form where the residual per remaining degree of freedom is low, and the proposal was a guard that estimates the noise from the least-squares residual and refuses a minimum whose residual is implausibly small. Over 960 draws holding 45 dips it refuses none, and picks plain GCV's minimum on every draw. At the dip ρ is 0.84 with the true noise and 0.98 with the estimate, because the estimate is made from the residual at the dip's own end of the scale and inherits its luck. Told the true noise instead, the guard can refuse 14 dips only by refusing 78 real minima. Handed the same estimate, the discrepancy principle misses by 570,000 times.

5 figures · Regularisation, essay 13
worst of 960, over the oraclediscrepancy, least-squares estimate, worst5.7·10⁵GCV with the held-out guard, worst1.8·10⁴discrepancy, held-out estimate, worst18rightmost minimum, worst3.7110¹10²10³10⁴10⁵10⁶samples per unknownerror ÷ oracle's, worst draw1.251.524discrepancy, least-squares estimateGCV with the held-out guarddiscrepancy, held-out estimaterightmost minimum960 draws, four ratiosan estimate that shares nothing with the fit

An estimate that shares nothing with the fit

GCV's guard failed because the noise estimate it read came from the residual its dips live in, and the proposal was to estimate the noise from samples held out of the fit instead. Over the same 960 draws the held-out guard refuses none of the 45 dips at the 1% threshold — fewer than the guard told the true noise, which refuses three — because a quarter of the samples gives its threshold too few degrees of freedom to refuse anything. Loosened to 10%, it refuses sixteen dips and 170 real minima. But the discrepancy principle, which missed by 570,000 times when told the least-squares estimate, never misses by more than eighteen told the held-out one, at every ratio and not only where the held-out set is the larger. The held-out estimate is the worse estimate, biased upward by up to half. What makes it the better input is that it is independent of the residual it is compared with.

5 figures · Regularisation, essay 14

Eigenvalues, singular values, rank

A symmetric matrix gives up its eigenvalues without complaint. An unsymmetric one can move them by the eighth root of a perturbation, so rounding error alone shifts them by a hundredth. And rank is not a property a floating-point matrix has — it is a decision about a gap, and the gap is worth printing beside it. The algorithm that libraries actually run lives here too, and it needs a shift before it is an algorithm at all.

10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹size of the perturbation ‖δA‖how far the eigenvalues moveJordan block, ε^(1/8)symmetric, ≤ ‖δA‖rounding error alone moves it to 10⁻²six seeds per symmetric point; Jordan is closed formsymmetry beats precision

Symmetry is worth more than precision

A symmetric matrix gives up its eigenvalues to full accuracy however ill-conditioned it is. An unsymmetric one can move them by the eighth root of a perturbation, so the rounding involved in merely storing the matrix shifts the spectrum by a hundredth.

6 figures · Eigen conditioning, essay 1
1234567891010⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1indexsingular valuecutoff, σ₁ · 10⁻¹⁰numerical rank 10gap 8.2·10⁶an opiniontrue rank 410×10, built with 4 nonzero valuesrank is a decision

Rank is a decision

A floating-point matrix does not have a rank. It has a spectrum of singular values, and somewhere in that spectrum is a place where the values stop being signal and start being noise. Deciding where is a judgement, and the evidence for it is a gap.

9 figures · Rank, essay 2
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

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.

6 figures · SVD, essay 2
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

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.

7 figures · The QR algorithm, essay 1
-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

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.

7 figures · The QR algorithm, essay 2
3000000120000-21000000-0.5-1.500001.5-0.5000000-2T = ZᵀAZthe highlighted boxes each hold one conjugate pair, and no real rotation removes themthe form, and that it is one‖A − ZTZᵀ‖/‖A‖1.8·10⁻¹⁵‖ZᵀZ − I‖2.5·10⁻¹⁵worst eigenvalue error2.7·10⁻¹⁵surviving subdiagonal26×6, spectrum chosen before the matrix was builtquasi-triangular is as far as the reals go

The form a real matrix can reach

A real matrix with complex eigenvalues has no real triangular form, and the reason is one line — a real triangular matrix has a real diagonal, and a similarity does not move the spectrum. What it has instead is triangular except for one two-by-two block per conjugate pair, and the count is decided by the matrix rather than by where the iteration stopped.

5 figures · Real schur, essay 1
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

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.

8 figures · Francis, essay 1
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

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.

6 figures · Eigen conditioning, essay 2
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

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°.

6 figures · Eigen conditioning, essay 3
the invariant planesolid: beforedashed: aftersame perturbation, two questionsthe vectors turned, radians0.029the plane turned, radians7.6·10⁻⁸what left the plane5.6·10⁻⁸drawn in the unperturbed plane's own basisa radius is not determined; the circle is

The plane survives what its vectors do not

At a gap of 10⁻⁹ a perturbation of 10⁻⁶ turns the two eigenvectors through half a radian and turns the plane they span through 7.6·10⁻⁸ — a ratio of six million. Ask for the subspace instead of the vectors and a hopeless computation becomes a well-conditioned one, with no change to the arithmetic.

6 figures · Invariant subspace, essay 1
1357910⁻⁵10⁻⁴10⁻³10⁻²10⁻¹1eigenvalue λamplificationkeptdiscardeda filter on the starting vectormeasured against the polynomial3.9·10⁻¹¹worst kept direction1best discarded direction0.8the roots are the discarded Ritz valuesand the vertical lines are where they sit

Restarting is a filter

A restart throws away the Ritz values it does not want and begins again from a new starting vector. Written in the eigenbasis, that vector's components have been multiplied by a polynomial with its roots at the discarded values — measured component by component, and agreeing with the polynomial to rounding.

8 figures · Lanczos, essay 2
12345610⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹Ritz value, largest firstdistance from 10one vectora block of twoa space, not a ratecopies found, one vector1copies found, block of two2Krylov dimension, one vector16the second copy is not in the spaceat any number of steps

An eigenvalue one vector cannot see

A matrix with an exactly doubled eigenvalue at 10. Twelve Lanczos steps find it once; twenty-four find it once, on a Krylov space of dimension 23 in a 24-dimensional problem. A block of two vectors finds it twice. This is not slow convergence — the second copy is not in the space.

7 figures · Invariant subspace, essay 2