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 n^{2d} 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‖A_n − A‖ and the largest term's norm‖A_n − A‖the larger of its two termsan infimum that is not attainedn1024‖A_n − 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 − T_tt‖ ⁄ ‖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.

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