Concept

Truncated SVD — where it appears

Keeping the first K singular components of a solution and dropping the rest, which is the sharpest of the standard weightings. It needs the singular value decomposition, which is the most expensive factorisation on this site, and it is the one weighting with no parameter between kept and discarded.

Named by 24 essays across 5 fields — each of them below, with the objects they name alongside it.

081624324048566400.250.50.751index kfilter factor fₖno regularisation: fₖ = 1truncationTikhonovthe same sum, three weightsTikhonov, relative error0.11truncation, relative error0.11no filter at all5.5·10⁸both filters are one expression with a different weightfₖ = 1 is the catastrophe

When the answer is a choice

A backward-stable least-squares solve of this problem returns an answer whose relative error is 5.5·10⁸. Nothing went wrong. The singular values decay exponentially with no gap anywhere in them, the data does not determine the answer, and something outside the data has to choose — which is the computation rather than a preliminary to it.

regularisation · Regularisation
051015202530354010⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹singular value, in orderσ ⁄ σ₁eight digitsan independent draw per entry1 ⁄ r, falling to the floora cliff, and a control1/r rank at 10⁻⁸5log r rank at 10⁻⁸5noise rank at 10⁻⁸96σ₂ ⁄ σ₁0.024σ₆ ⁄ σ₁2.8·10⁻⁹the block has full rankand five useful columns

A block nobody can call sparse

A 96 × 96 block of a kernel matrix has ninety-six nonzero singular values and five that matter. It has no zero entries, it is not described by fewer numbers than it contains, and neither of the two ways this collection already knows to make a large matrix affordable applies to it.

hierarchy · Off-diagonal rank
00.250.50.75110⁻¹110¹10²10³fraction of the method's own rangerelative errorfloor 0.141truncation KTikhonov λCGLS steprandomised rankfour methods, one floortruncation K0.14Tikhonov λ0.14CGLS step0.14randomised rank0.14four knobs from four fieldsand one obstruction underneath them

Four knobs and one floor

A truncation, a Tikhonov parameter, a step count and a randomised rank, on one problem with an answer that is known. Their best errors are 0.1445, 0.1406, 0.1426 and 0.1449 — a spread of 3% across four methods that share no arithmetic.

combination · Parameter choice
081624324048566410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹index kmagnitudethe floor: k = 32best truncation: k = 28σₖ|uₖᵀb| exact|uₖᵀb| with noisetwo different indicesthe crossing, from the data alone32the truncation that is actually best28relative error there0.11the exact coefficients never flattenthe noisy ones stop at ‖e‖/√n

Where the answer stops being in the data

The Picard condition finds the index where a noisy right-hand side stops carrying signal, from the data alone, with no knowledge of the answer. It lands at 32 where the truncation that actually minimises the error is 28 — and at 45 where the best is 38. It overshoots at every stop from 10% noise to 0.0001%, and it overshoots for a reason. The best truncation walks up the spectrum in a straight line, six or seven indices a decade; the crossing climbs in jumps of 11, 0, 8, 5 and 1.

regularisation · Regularisation
0369121501122334455digits asked for, −log₁₀ εcolumns keptwhat the geometry promiseswhat the matrix costsa rank is a number of digitscolumns a decade0.55bound, a decade3.3rank at 10⁻⁸5bound at 10⁻⁸28q0.5the shape is rightand the constant is not

A rank that is a number of digits

Ask a kernel block for two digits and it costs two columns; ask for fourteen and it costs nine. The curve is a straight line at 0.55 columns a decade, and the bound the geometry gives is a straight line too — at 3.32, which is the same shape and six times the price.

hierarchy · Off-diagonal rank
110¹10²10³10⁻³10⁻²10⁻¹110¹10²10³n‖Aₙ − A‖ and the largest term's norm‖Aₙ − A‖the larger of its two termsan infimum that is not attainedn1024‖Aₙ − A‖0.0017largest term1024their product1.7√31.7the distance goes to zeroand nothing reaches it

A nearest point that is not there

Eckart and Young guarantee that a matrix has a best rank-k approximation and that the truncated SVD is it. For three indices the guarantee is false in the strongest available way — there are tensors whose distance to the rank-two set is zero and which no rank-two tensor equals.

tensor · Tensor rank
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

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.

combination · Randomised
024681010⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹rank kept in every moderelative errordashes above: √(Σ tail²), the upper bounddashes below: max tail, a floor under the bestsolid: what the projection returnssmooth: pinned to the upper boundrank 10 error1.1·10⁻¹¹its upper bound1.1·10⁻¹¹the lower bound6.3·10⁻¹²error ⁄ bound1error ⁄ lower1.7inside the boundand sitting on it

A decomposition made only of SVDs

Everything the definition of tensor rank loses comes back if the SVD's algorithm is carried across instead of its definition — take the leading left singular subspace of every unfolding and project onto all of them. It exists, it costs d matrix decompositions, and its error is within √d of the best there is.

tensor · Multilinear rank
192123252729313335373941434500.10.20.30.40.50.60.7position jweight on the true signal at jthe blur's row, 5.89 widethe kernel, 2.82 widecomputed from A and λ alonekernel width at half height2.8components kept, Σfₖ28width × Σfₖ / n1.2deepest negative lobe-0.075no data and no truth went into this curvethe answer is the truth seen through it

A second blur, narrower than the first

A regularised answer is not the truth with the noise taken out. It is the truth seen through a second blur, V F Vᵀ, which depends on the operator and λ and on nothing that was measured. At the best λ for 0.1% noise its rows are 2.82 points wide against the instrument's 5.89, they dip to −0.075 on either side, and their width times the number of components kept stays between 1.10n and 1.27n across seven decades of λ. Two spikes four points apart come back as two; three apart, as one.

regularisation · Regularisation
smooth97.7%hilbert94.5%wave26.6%noise0.4%share of the core's energy on its 6 superdiagonal entrieskeeping only them: 0.152 against 7.94·10⁻⁶keeping only them: 0.235 against 1.77·10⁻⁶keeping only them: 0.857 against 1.34·10⁻¹⁵keeping only them: 0.998 against 0.842orthogonal, and not diagonalsmooth on-diagonal0.98hilbert on-diagonal0.94wave on-diagonal0.27noise on-diagonal0.0044worst slice pair3.5·10⁻¹⁶the slices are orthogonalthe core is not diagonal

The orthogonality that cannot be diagonal

A matrix decomposition hands over orthonormal factors and a diagonal middle at once. For three indices the two come apart, and there is no arrangement that has both — so the question stops being which decomposition to use and becomes which of the two properties the computation needs.

tensor · Multilinear rank
123456710¹10²10³10⁴10⁵number of indicesnumbersentries: 6^dstored: 4n(d − 1)exponential against linearentries at d = 64.7·10⁴numbers stored120ratio389slope against d24‖T − Tₜₜ‖ ⁄ ‖T‖1.4·10⁻¹⁵one line is n^dthe other is a constant per index

The format that does not notice the dimension

A Tucker core is r^d numbers, so the format that repaired the definition still cannot go past five indices. Cutting between the indices rather than across them gives d − 1 ranks instead of d, storage linear in the number of indices, and a family whose ranks are two everywhere by an addition formula.

tensor · Tensor train
12345678910¹10³10⁵10⁷number of indicesrandom numbers drawna dense Gaussian sketchdashes: the tensor's own entriesa Khatri–Rao sketcha random matrix nobody can afforddense at d = 81.5·10⁸structured4032the tensor's entries1.7·10⁷dense ⁄ structured3.7·10⁴crossing at d2the sketch outgrows its tensorand the structured one does not

Sketching what is never unfolded

A range finder multiplies its matrix by a few random vectors. For a mode-k unfolding those vectors have nᵈ⁻¹ entries, so the random object is the size of the tensor divided by n — and by six indices it is larger than the tensor it is sketching.

randomised · Sketching
1216202428323640444810⁻¹110¹10²grid points nrelative error against the continuous signalbest truncation, 0.117best grid: n = 26best K = 28with noisenoise-freeno λ anywhere in this curvebest grid, error0.13best truncation, error0.12κ on the best grid61grid where it has doubled34every point is an unregularised solvethe grid chose the truncation

The grid was the first filter

A continuous deconvolution discretised on n points and solved with no regularisation at all is not unregularised. Its error against the continuous signal is least at 24, 26 and 34 points for noise of 1%, 0.1% and 0.01% per sample — beside best truncations of 24, 28 and 32 components on a 64-point grid — and within 4 to 16 per cent of their error. The grid's own filter factors sum to n exactly and fall through a half at k = n. Choosing the grid was choosing a truncation, before anybody chose a λ.

regularisation · Regularisation
6 × 6 × 6, rank 3 · 9 ≥ 81.00002 × 2 × 2, rank 3 · 6 < 80.02076 × 6 matrix, rank 3 · no condition0.1089worst agreement between two runs about the factors, 0 to 1every run fits to 3·10⁻¹³every run fits to 1.2·10⁻¹¹every run factorises to 2.4·10⁻¹⁵ and none agrees with anotherthe only thing that improvestensor, Kruskal holds1tensor, Kruskal fails0.021matrix0.11worst residual1.2·10⁻¹¹three successful fitsone recoverable answer

A factorisation that is unique for once

A rank-r factorisation of a matrix is never unique — AB is (AM)(M⁻¹B) for any invertible M, so no factor means anything on its own. For three indices a checkable condition on the factors' k-ranks makes the decomposition unique up to permuting and scaling the terms, and it holds generically.

tensor · Uniqueness
0816243211.11.21.31.4terms added, each followed by a truncationerror ⁄ best rank-k erroroptimalterms with nothing in commona subspace that driftsthe rounding nobody should have feareddrifting, worst excess1independent, worst excess1a linear bound would say32energy discarded, first1.8·10⁻⁷energy discarded, last0.03thirty-two roundingsand four per cent

The rounding that was not the problem

A rank-k block plus a rank-k block is a rank-2k block, exactly, so every arithmetic in this format truncates after every addition. A Cholesky performed inside it does ninety-eight of those and its residual is 1.14·10⁻⁹ against a representation error of 1.40·10⁻⁹ — the roundings cost nothing measurable.

hierarchy · Recompression
10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²ε the blocks were compressed at‖A − LLᵀ‖ ⁄ ‖A‖the factorisationthe knob, on a factorisationtruncations34residual at 10⁻⁴2.2·10⁻⁵residual at 10⁻¹⁰9.9·10⁻¹²slope1worst ratio to the representation1.8ten decadesand a slope of one

A knob calibrated in residuals

A formatted Cholesky has two numbers in it and only one of them is an accuracy. Across twelve trees — three sizes by four leaf sizes — the leaf moves the truncation count from 0 to 258 and moves the ranks of the blocks not at all, while the residual follows the tolerance at slopes between 1.022 and 1.046 and sits at about a tenth of it throughout.

hierarchy · Recompression
12345610⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹digits of measurement accuracybest relative errorTikhonovtruncationCGLS stepslope 0.89, optimalslope 2/3, Tikhonov's limithow fast the error fallsTikhonov slope0.7truncation slope0.94CGLS step slope0.89optimal 2ν/(2ν + 1)0.89Tikhonov's limit0.67better data, smaller errorat a rate the method may not be able to keep

The method that cannot use a smooth answer

On the collection's own signal four regularisers reach the same floor to a few per cent. Score them instead against answers of increasing smoothness and one stops improving. Across six decades of noise Tikhonov's error falls with a fitted slope of 0.70 whether the answer is twice or four times as smooth, while truncation's rises to 0.94 — and at 10⁻⁶ noise Tikhonov's best is 38 times truncation's.

combination · Iterative regularisation
02040608010010⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations2.1·10⁻¹⁰residual, 98 truncations1.1·10⁻⁹representation, deepest1.4·10⁻⁹residual ⁄ representation0.81levels, deepest5a hundred approximate stepsand an exact-looking factorisation

The count that is not the budget

A Cholesky performed inside a low-rank format truncates 0, 2, 10, 34 and 98 times as the leaf falls from 128 to 8, and those five integers are the same at every accuracy from 10⁻¹² to 10⁻². Across all ten decades the factorisation's residual stays below the representation's own error at a ratio between 0.81 and 1.00 — with two entries that read 1.83 and 1.78, and neither of them is accumulation.

hierarchy · Recompression
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.

regularisation · Regularisation
-13-11-9-7-5-3-102468log₁₀ of the accuracy asked forlargest train rank0.30 of rank per decadea cost that is typed inentries216stored at 10⁻²90stored at 10⁻¹²288rank per decade0.3worst error ⁄ tolerance0.83the storage is chosena constant of rank a decade

The digit that costs more than the tensor

Ask a three-index reciprocal tensor on six points a side for seven digits and its train is 288 numbers against 216 entries. The break-even rank is n − 1 at all four grids measured, and a train that reaches it fits with exactly n numbers to spare.

tensor · Tensor train
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.

regularisation · Regularisation
each method's best, dimensionconjugate gradients24Tikhonov23truncated SVD21and the error thereconjugate gradients0.14Tikhonov0.14truncated SVD0.140510152025301effective dimensionrelative errorconjugate gradientsTikhonovtruncated SVDdashed vertical: the iteration's bestthe iteration drawn while its dimension still rises

One arc, and what each filter pays to be on it

Conjugate gradients and Tikhonov stop at the same effective dimension, and that could have meant two curves crossing once or one curve. It is one curve over a stretch — on six problems, Tikhonov and truncation reach the iteration's error at the iteration's dimension to within 7.2 per cent from 0.7 of the answer to its top — and the two separate on either side. But the curve is shared by a trade, not by an identical answer: at the same dimension Tikhonov carries 22 to 42 per cent more noise than the iteration and up to five per cent less bias.

combination · Iterative regularisation
sum with overshoot clippedwidest departure, 1.1 to 1.30.17widest departure, 0.3 to 0.50.690.30.50.70.91.11.30.7511.251.51.7522.252.5clipped sum ÷ the answer'serror ÷ the iteration'sTikhonovtruncated SVDshaded column: 0.7 to 1.0 of the answerthe axis decides the band above the top

The overshoot was the lead

Past its best, conjugate gradients' error rose more slowly than Tikhonov's at the same effective dimension — on the narrow blur at 0.1% noise Tikhonov was 2.35 times worse at 1.3 of the answer — and the reading was that the iteration spends its dimension where the data has content. Count admitted directions instead of summing factors, so that a factor of 1.81 counts once, and the lead is gone: 0.96 on that problem, and within 0.17 of one on all six from 1.1 to 1.3. Tikhonov now carries less noise than the iteration there. Below half the answer, where the three filters also disagree, the count changes nothing.

combination · Iterative regularisation
p = 2widest departure, 0.3 to 0.50.11Tikhonov's, same stretch0.690.30.50.70.91.11.30.811.21.41.61.8clipped sum ÷ the answer'serror ÷ the iteration'sp = 2Tikhonovshaded column: the first one to five stepssharpening closes the band below half

A tail from Tikhonov and a corner from truncation

Below half the answer, conjugate gradients beat Tikhonov at a matched count of directions by up to 69 per cent, and the proposed measurement was the sharpness p of a roll-off between Tikhonov and truncation that matches the iteration there. Any p from 2 to 4 closes the gap to a tenth; truncation, the family's limit, reopens it to a third. But no p describes the iteration. Its filter has Tikhonov's slope exactly in its tail and a local sharpness of 2 to 5 on its shoulder, and a single fitted p is a compromise that drifts from 2.3 at the first step to 1.7 at the answer.

combination · Iterative regularisation

Named alongside it

The objects these essays reach for when they reach for this one.

Filter factorsLow-rank approximationIll-posed problemDeconvolutionTikhonov regularisationEckart–YoungRegularisationUnfoldingIterative regularisationCondition numberConjugate gradientsHigher-order SVD

All concepts