Concept

Singular value decomposition — where it appears

The factorisation UΣVᵀ into two orthonormal bases and a diagonal, which answers every question about rank, norms and best approximation and costs the most. It answers rank, norms and best approximation exactly, and it costs several times an LU, which is why the cheaper factorisations exist at all.

Named by 13 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
04812162010⁻¹10⁻⁰.⁵1target rank k‖A − Aₖ‖₂published boundrandomisedσₖ₊₁, optimalhow far apart the three areworst seed spread1.6bound / median at k = 125.9median / optimum at k = 121.960×60, 6 seeds, oversampling p = 5band is best to worst

A bound that holds with probability

Every other guarantee in this collection is deterministic. The randomised low-rank approximation offers one that holds with a probability, the seed changes the answer, and the honest figure is a band rather than a line.

randomised · Randomised
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
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.

spectra · SVD
04812162010⁻¹10⁻⁰.⁵1target rank k‖A − Aₖ‖₂published boundrandomisedσₖ₊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

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.

randomised · Randomised
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

When the matrix is wrong too

Every least-squares problem here 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.

leastsquares · Total least-squares
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
polar: U₁P − U₂QR: Q₁P − Q₂-3.3·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶-1.9·10⁻¹⁶5.6·10⁻¹⁷-1.7·10⁻¹⁶6.7·10⁻¹⁶-8.9·10⁻¹⁶2.8·10⁻¹⁶-5.6·10⁻¹⁶-3.6·10⁻¹⁶6.1·10⁻¹⁶-8.3·10⁻¹⁶8.3·10⁻¹⁶-7.2·10⁻¹⁶-10·10⁻¹⁶2.2·10⁻¹⁶4.4·10⁻¹⁶1.7·10⁻¹⁶-4.4·10⁻¹⁶-5.6·10⁻¹⁷5.6·10⁻¹⁷0-7.2·10⁻¹⁶-4.4·10⁻¹⁶3.9·10⁻¹⁶5.6·10⁻¹⁷2.8·10⁻¹⁶-1.1·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶1.1·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶0.67-0.35-0.23-0.620.19-0.72-0.770.170.20.290.190.93-0.180.55-0.0290.06-0.460.11-0.17-0.61-0.230.38-0.55-0.380.750.870.661.1-0.36-0.530.42-0.670.180.180.52-0.3Frobenius norms, columns reordered‖U₁P − U₂‖2.8·10⁻¹⁵‖Q₁P − Q₂‖3‖Q‖, for scale2.4κ of the matrix100Frobenius distance between the two answers, on one scale0 to 4polar2.8·10⁻¹⁵QR3.048‖Q‖ = 2.449the column space did not moveand one of the two answers did

A test with no answer in it

A caller with no reference answer can still ask whether a routine answered the right question: reverse the columns, run it again, compare. The polar factor's two answers agree to 10⁻¹⁵ at every conditioning drawn; a QR's differ by 2.353 on matrices whose own norm is 2.449. The test has a floor, and the floor is measurable too.

orthogonality · Polar decomposition
110¹10²01020304050noise ÷ thicknesstrials mirrored, %a coin: 50%t = 10⁻²t = 10⁻³t = 10⁻⁴per cent mirroredσ/t = 3, mean of three7.7σ/t = 10, mean of three34σ/t = 100, mean of three4720 points, 400 trials a stop, one seed per thicknessthe ratio decides, not the thinness

A rotation that comes back mirrored

Align twenty noisy points and the nearest orthogonal matrix to the answer is a reflection in 7.7 per cent of trials at noise three times the set's thickness and a third of them at ten — at thicknesses of 10⁻², 10⁻³ and 10⁻⁴ alike. The determinant fix is never a small correction. It moves the answer by exactly 2, it costs exactly 4σ₃ of residual, and it leaves the rotation's error at half the noise however thin the set becomes.

orthogonality · Polar decomposition
the upper pair is distance from the truth; the lower pair is ‖Ax − b‖least squares · error0.08523total least squares · error0.037least squares · ‖Ax − b‖3.965total least squares · ‖Ax − b‖4.078two orderingserror ratio (ls ÷ tls)2.3residual ratio (tls ÷ ls)1seeds40no vector makes the residual smallernot even the one the problem was built from

The two numbers a caller has

Choosing between the two least-squares methods is a statement about where the noise is, and the two quantities a caller can compute are both blind to it. The residual separates the answers by 0.14 per cent where their accuracies differ by 14, and κ(A) falls from 3.54 to 2.46 across a sweep in which the error rises by a factor of sixty-two.

leastsquares · Total least-squares
1110¹10²10³10⁴10⁵the noise level it is told ÷ the true oneerror ÷ the oracle's0.40.50.71.523twice the oraclecliff, ρ = 0.69told the truthworst of 400middle 80%median draweach draw against its own oraclecliff, median ρ0.69told the truth, median1.1told a half, median3201shaded: the middle 80% of drawsbelow the cliff the error has doubled

Thirty-two coefficients instead of a noise level

The discrepancy principle has to be told the noise, and told too little it does not degrade — it falls off a cliff, at 0.80 of the truth when the noise is 10% and at 0.58 when it is 0.001%, exactly where the understatement forces the filter past its best truncation. The missing number is in the data. The root mean square of the last thirty-two coefficients never sends the rule over the cliff at or below 1% noise in four hundred draws, where eight coefficients with the same median do so thirty-five times.

regularisation · Parameter choice
10⁻¹101020304050z, the effective noise over thicknesstrials mirrored, %5 points, equal20 points, equal80 points, equal● five precise, 1/σ²■ five weighted ×100per cent mirrored at z = 0.45, 1,000 trialstwenty points1.8eighty points1.3five points9.5five precise, weighted 1/σ²7.9equal noise, five weighted ×1006.8both weighted schemes have five effective pointstwenty and eighty points share one curve

Five precise points are five points

Weighting each sighting by its reliability is the standard form of an attitude or registration fit, and it changes how often the nearest orthogonal matrix comes back as a mirror. Measured, the rate is a function of two numbers: the weighted noise over thickness, and the effective count (Σw)²/Σw². Five points with a tenth of the noise, weighted by 1/σ², carry the information of 515 equal points and mirror like five — 7.9 per cent at a noise ratio where twenty points mirror 1.8 and five mirror 9.5. The √m the earlier measurement left unchecked is right, and it counts what carries the thin direction.

orthogonality · Polar decomposition
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

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.

orthogonality · Polar decomposition

Named alongside it

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

Condition numberOrthogonalityFilter factorsIll-posed problemPolar decompositionRegularisationSingular valuesDeterminantEckart–YoungFrobenius normLow-rank approximationResidual

All concepts