Numerical Linear Algebra
where the algebra stops being the arithmetic
Field

When the index is a tuple

A discretisation on a d-dimensional grid has n^d unknowns, so its matrix has n²ᵈ entries — a number with no physical meaning. What such a matrix usually is instead is a sum of d Kronecker products of n × n matrices, and nothing has been approximated: assembling it was the mistake. Then the same word is carried one step further, to arrays with three or more indices, and every theorem this site has relied on about rank stops being true. The best approximation may not exist, the rank depends on the field the entries are read over, and there is no decomposition that is orthogonal and diagonal at once. What survives is not the definition but the algorithm: take the SVDs of the reshapes, and existence, computability and a quasi-optimality factor all come back.
03672108144180216024681012eigenvalues in orderλthe closed formmarks: the assembled matrix, decomposeda spectrum nobody computedrows of the matrix216numbers that describe it108λ smallest0.59λ largest11worst |computed − exact|7.1·10⁻¹³the matrix is never neededand neither is its decomposition

An index that is a pair

A discretisation on a two-dimensional grid of n points a side has n² unknowns and a matrix with n⁴ entries — 10⁸ at n = 100. What that matrix is instead is two Kronecker products of an n × n matrix, which is 2n² numbers, and nothing has been approximated: assembling it was the mistake.

 024681012141610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹index of the singular value at the cutσ ⁄ σ₁eight digits10^-4: 5 Kronecker terms10^-8: 7 Kronecker terms10^-12: 8 Kronecker termsnot closed, and nearly closedrank at the cut8a Kronecker product's1terms at 10⁻⁴5terms at 10⁻⁸7terms a decade0.5the inverse leaves the formatby half a term a decade

A solve that is d decompositions

A Kronecker sum is closed under nothing useful — its inverse is not a Kronecker sum and no factorisation of it is one. What it has instead is eigenvectors that are Kronecker products, so a solve with 1,728 unknowns takes one decomposition of a 12 × 12 matrix and nothing else.

 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.

 10¹10²10³10⁴00.050.10.150.20.250.30.350.40.450.50.550.60.650.70.750.80.850.90.951drawsshare with real rank twoπ/4 = 0.78539815,705 of 19,953 have rank twoa probability with a closed formdraws2·10⁴rank two1.6·10⁴share0.79π/40.79standard errors out0.59two typical ranksand the split is π/4

A rank that is not a property of the tensor

The same eight real numbers have rank three over the reals and rank two over the complexes, and a random 2 × 2 × 2 tensor has rank two with probability exactly π/4. Neither sentence has an analogue for matrices, where the rank is one number and a random matrix has the largest one.

 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.

 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.

 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.

 110¹10²10³10⁴10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹sweeprelative error, and the largest term's sizerising: the swamp's largest rank-one termfalling, slowly: its errorfalling, once: a fit with an answera plateau with a rising floorswamp error0.0014swamp term10term growth2.2benign error9.7·10⁻¹⁵benign term growth1the error alone cannot tellthe size of the terms can

An iteration that walks out of the set

Every sweep of alternating least squares is the exact minimiser of its own subproblem, so the objective can only fall. What it cannot do is converge, when the target's nearest rank-r point is not in the rank-r set — and a plateau at a small residual looks identical to slow convergence unless the size of the terms is plotted beside it.

 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.

 15913172110¹10⁶10¹¹10¹⁶10²¹10²⁶number of indicesnumbers storedthe tensor: n^dthe core: r^d + dnrthe train: (d − 2)nr² + 2nrthe curse, movedentries at d = 2010²⁶core1.1·10¹²train5920core ⁄ train1.9·10⁸entries ⁄ core9.5·10¹³the definition is repairedthe size is not

A compression of 10¹⁴ that still does not fit

A Tucker core of a twenty-index array at rank four is 1.1·10¹² numbers against the tensor's 1.05·10²⁶ — a compression by a factor of 9.5·10¹³ that is still nearly nine terabytes. The ratio is not the verdict. The verdict is a ceiling, and the ceiling is a number of indices.

 110¹10²10⁴10⁶10⁸10¹⁰10¹²n, points along one axismultiplicationsa dense factorisationthrough the eigenbasis6 decompositions of an n × nunknowns1.6·10⁴multiplications9.4·10⁵dense factorisation2.5·10¹²fitted exponent7‖Ax − b‖ ⁄ ‖b‖1.8·10⁻¹⁵nothing of size n^dis ever factorised

Five indices are cheaper than two

The same 4,096 unknowns cost 1.049·10⁶ multiplications indexed as a 64 × 64 grid and 1.966·10⁵ indexed as six axes of four. The dense factorisation that ignores the indexing costs 4.581·10¹⁰ at every one of them, and the residual improves in the same direction as the cost.

 -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.

 01210⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹terms fitted1 − the largest cosine between two fitted terms3 — the tensor's rank45below this line, two terms are the same termthe residual says nothing3 terms, collinearity0.524 terms, collinearity15 terms, collinearity13 terms, error2.1·10⁻¹⁴4 terms, error10⁻¹³4 terms, sweeps76one term too manyand two of them cancel

One term too many

A tensor built from three rank-one terms, fitted with four, reaches a relative error of 1.0·10⁻¹³ in seventy-six sweeps. Two of the four terms come back with a cosine of 0.99927 between them, subtracting each other, and the smallest weight is a fifth of the largest rather than a rounding. Nothing in the residual says any of it.

 10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1λ, the ridge on every subproblemerror, and the largest rank-one termthe terms a swamp was growingthe error it costs a fit that has an answerthe error it costs one that has nota bound, at its own priceterms, no ridge6.7terms, heaviest1.2sweeps, no ridge3000sweeps, heaviest69benign error at λ0.089terms still the right terms1the plateau is boundedand the answer is biased by exactly λ

The repair that costs exactly itself

A ridge on every subproblem is the standard cure for a swamp and it works — the terms stop at 1.61 instead of climbing past 6.7, and a run that never finished finishes in 798 sweeps. On a tensor that does have an answer the error it costs is the ridge itself, to within a factor of two, at every setting from 10⁻¹⁰ to 10⁻¹. And the terms it returns are still the right terms.

 01210¹10²10³the three kinds of runsweep at which the test firesno answer to reachbadly conditionedordinaryhollow: never fired, drawn where the run endedwhat it fires on, and whenthreshold0.25boundary, fired6boundary, median sweep22collinear, fired6collinear, median sweep759ordinary, fired0the value does not separate themand the sweep does

The test that is a deadline

The quantity that separates a boundary from slow convergence fires on every run that has nothing to reach, at sweep 21 of four thousand, and on no ordinary fit. It also fires on every badly conditioned fit that does converge — at sweep 758. No threshold between 0.10 and 0.40 separates those two, and the sweep it fires at separates them by a factor of thirty-six.

 at rank threelowest across tensors1highest across tensors1at rank fourlowest across tensors0.056highest across tensors0.25234500.20.40.60.81rank fittedagreement among the starts that reach itsix planted tensorsdashed: the true rankeach line one tensor, fitted at ranks two to fivefive starting points at every rank

The rank a sweep can vouch for

To decide a tensor's rank, fit it at one rank after another and stop where something changes. Three things can be read off each rank's fits — the error, how nearly parallel the terms are, and whether fits from different starting points agree — and over six planted rank-three tensors at four noise levels they decide it 24, 12 and 15 times out of 24; the ratio of consecutive errors, which needs nothing, decides it 24 times. The error is right every time and has to be told the noise level. The collinearity is a coin toss. Agreement among starts is never wrong and, under noise, cannot decide on nine of the twenty-four — and the rank past the answer, where every decision is made, costs ten times the answer.

 no noisesweeps to stop49final error2.5·10⁻¹⁴1% noisesweeps to stop6000largest term, ÷ data0.67110¹10²10³10⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1sweeprelative change in the error, per sweepthe stopping test1% noiseno noise: stops at 10⁻¹⁴the clean run stops; the noisy one crawlsits terms stay the size they were

A fit that has an answer and cannot stop

Fit a noisy rank-three tensor with four terms and the prediction was that the spare term would find real structure in the noise and stop pairing off with the others. It does not: the largest cosine between two fitted terms has a median of 0.977 at 1% noise against 0.975 without noise. What changes is the solver. Without noise the four-term fit stops in fifty sweeps; with noise it never stops: its error keeps moving by a part in a million a sweep for six thousand sweeps, while on most tensors its terms stand still. The extra term takes exactly its share of the noise, and the error is settled by sweep 150.

 start 1, errorreaches2.8·10⁻¹³stopped by δ = 0.010.0099stopped by δ = 0.0010.0098stopped by δ = 10⁻⁴2.6·10⁻¹³110¹10²10³10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²sweeprelative errorδ = 0.01δ = 0.001δ = 10⁻⁴δ = 10⁻⁵a plateau is a still errorand a still error is what the test asks for

A still error is not a settled one

A CP fit with one term too many on a noisy tensor never meets the usual stopping test, and its error has settled by sweep 150. A test that asks only whether the error has moved by less than δ of itself over ten sweeps stops those fits by sweep 25, and over twenty-four noisy tensors it names the same rank as the usual test every time, at δ = 10⁻² for 30,250 sweeps against 311,458. But δ = 10⁻² is the tolerance a rank decision states, and on a slow fit that converges it stops five starts of six on a plateau ten orders above the answer and names the wrong rank for four tensors of six. A tenth of the decision's tolerance keeps every decision on both families.

 share of rank nn = 2 (π/4 = 0.785)0.79n = 30.5n = 80.004234567891010⁻⁴10⁻³10⁻²10⁻¹1n, the size of each sliceshare with rank n0 of 4000dashed: e to the minus 0.087 n squaredthe lower rank stops being typical in practice

The rank that stops being typical

A random 2 × 2 × 2 tensor has rank two with probability π/4 and rank three otherwise, and the sentence has no analogue for matrices. It is the first of a family. An n × n × 2 tensor is a pencil of two slices, and it has rank n exactly when the pencil's eigenvalues are all real, n + 1 otherwise. Over draws, the share of rank n is 0.786 at n = 2, 0.500 at 3, 0.264 at 4, 0.039 at 6, 0.004 at 8 and none of 4,000 at 10 — falling like e^(−0.087n²) — because the mean number of real eigenvalues grows only like the square root of n, to Edelman, Kostlan and Shub's closed form within two per cent. Both ranks stay typical in theory; in practice the lower one disappears.

 fit ÷ distance, less oneleast excess over the distance1.8·10⁻⁴greatest0.01510⁻²10⁻¹110⁻²10⁻¹1distance to the double-eigenvalue surfaceerror the fit settles atdashed: equalitytwo routes to one number

A fit with no answer to find

Half of all random 3 × 3 × 2 tensors, and most larger ones, have no rank-three decomposition, because their pencil has a complex pair. A rank-n fit to one of them does not wander and does not stall. Two starts settle at the same error to five digits, and that error is the distance from the tensor to the surface where its pencil has a double eigenvalue — found with no fitting at all, and matched to within one and a half per cent. Meanwhile the fit's terms grow without limit, like the square root of the sweep count, while the fitted pencil's two closest eigenvalues close on each other at exactly the rate the terms grow. The error has an answer; the decomposition does not.

 term size ÷ normτ 0.01, smallest term ratio3.8τ 0.003, smallest term ratio6.9τ 0.001, smallest term ratio1210⁻³10⁻²10⁻¹110¹stop when the error is within τ of the distancelargest term ÷ tensor's normmedianthree times the normgrey: one line per drawone per cent is not early enough

A stop that knows the distance

A rank-two fit to a random 2 × 2 × 2 tensor of rank three settles at the tensor's distance to the boundary of the rank-two set while its terms grow without limit, and the distance can be computed without fitting. So a fit can be stopped when its error is within a stated fraction of it, and the prediction was that at one per cent the terms would still be within three times the tensor's norm, because one per cent is reached early. On 23 random tensors the terms at the one-per-cent stop are 3.8 to 12.9 times the norm — none within three — and the reason is the law the earlier essay found: the excess falls as the inverse square of the term size, so the size at the stop is the square root of a constant over τ times the distance, and a tensor close to the boundary pays twice, in larger terms and in sweeps. The two closest tensors never reach one per cent in 20,000 sweeps. The stall test a code would use stops in the same range by accident. And one of the 24 computed distances was wrong, which the fit itself exposed.

 per unit of perturbationtrue move, θ = 10⁻³200ALS off by, θ = 10⁻³200ALS off by, θ = 17.9·10⁻⁴10⁻³10⁻²10⁻¹110¹10²10³angle θ between two columns of the third factorchange ÷ perturbation110⁻¹10⁻²10⁻³true decomposition's moveALS's distance from itKruskal: unique at every θ hereθ closes to the leftunique, and not determined

Unique, and not determined

Kruskal's condition says a CP decomposition is unique, and it says it as a yes or a no. Turn two columns of one factor of a 4 × 4 × 4 rank-three tensor toward each other and the condition keeps saying yes at every angle above zero, by one to spare. What it does not say is how well the unique decomposition is determined. The Jacobian of the map from factors to tensor loses its smallest singular value in proportion to the angle, and the true decomposition of a tensor perturbed by one part in 10⁸ moves by 1.6 parts at θ = 0.1 and by 200 at θ = 10⁻³. Alternating least squares, started at the exact answer, finds the moved decomposition while the angle is large, slows by ten times or more on the way to θ = 0.1, and below θ = 0.03 never arrives: at θ = 10⁻³ its own stopping test fires after 11 to 65 sweeps with its terms 200 to 420 perturbations from the answer and its residual within a per cent of the answer's.

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