Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

unfolding-ranks

One function in the tensor library, called 19 times across 10 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 2 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws the three unfoldings of a 10 × 11 × 12 smooth tensor, and the singular values of each. A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 2.37, 2.37, 2.37, each normalised to its own mode below.

unfolding-ranks is one function in lib/figures/tensor.js — an index that is a tuple — a matrix that is d small ones, and the inverse that is nearly one. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

The three unfoldings of a 10 × 11 × 12 smooth tensor, and the singular values of eachA tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 2.37, 2.37, 2.37, each normalised to its own mode below.02468101210⁻¹⁶10⁻¹²10⁻⁸10⁻⁴1index of the singular valueσ ⁄ σ₁the tolerance the ranks are read atmode 1 · rank 8mode 2 · rank 8mode 3 · rank 8smooth: three matrices, one arrayentries1320mode-1 rank8mode-2 rank8mode-3 rank8‖T‖2.4three ranksand none of them is the tensor's

A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 2.37, 2.37, 2.37, each normalised to its own mode below.

family: "smooth"

The arguments are the ones A decomposition made only of SVDs passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The three unfoldings of a 10 × 11 × 12 smooth tensor, and the singular values of eachA tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 2.37, 2.37, 2.37, each normalised to its own mode below.02468101210⁻¹⁶10⁻¹²10⁻⁸10⁻⁴1index of the singular valueσ ⁄ σ₁the tolerance the ranks are read atmode 1 · rank 8mode 2 · rank 8mode 3 · rank 8smooth: three matrices, one arrayentries1320mode-1 rank8mode-2 rank8mode-3 rank8‖T‖2.4three ranksand none of them is the tensor's

A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 2.37, 2.37, 2.37, each normalised to its own mode below.

family: "hilbert"

The arguments are the ones A decomposition made only of SVDs passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The three unfoldings of a 10 × 11 × 12 hilbert tensor, and the singular values of eachA tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 3.2, 3.2, 3.2, each normalised to its own mode below.02468101210⁻¹⁶10⁻¹²10⁻⁸10⁻⁴1index of the singular valueσ ⁄ σ₁the tolerance the ranks are read atmode 1 · rank 8mode 2 · rank 8mode 3 · rank 8hilbert: three matrices, one arrayentries1320mode-1 rank8mode-2 rank8mode-3 rank8‖T‖3.3three ranksand none of them is the tensor's

A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 8, 8, 8 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 3.2, 3.2, 3.2, each normalised to its own mode below.

family: "noise"

The arguments are the ones A decomposition made only of SVDs passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The three unfoldings of a 10 × 11 × 12 noise tensor, and the singular values of eachA tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 10, 11, 12 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 14.2, 13.6, 13.4, each normalised to its own mode below.02468101210⁻¹⁶10⁻¹²10⁻⁸10⁻⁴1index of the singular valueσ ⁄ σ₁the tolerance the ranks are read atmode 1 · rank 10mode 2 · rank 11mode 3 · rank 12noise: three matrices, one arrayentries1320mode-1 rank10mode-2 rank11mode-3 rank12‖T‖36three ranksand none of them is the tensor's

A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 10, 11, 12 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 14.2, 13.6, 13.4, each normalised to its own mode below.

family: "wave"

The arguments are the ones A rank that is not a property of the tensor passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The three unfoldings of a 10 × 11 × 12 wave tensor, and the singular values of eachA tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 2, 2, 2 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 18.7, 19.3, 18.7, each normalised to its own mode below.02468101210⁻¹⁶10⁻¹²10⁻⁸10⁻⁴1index of the singular valueσ ⁄ σ₁the tolerance the ranks are read atmode 1 · rank 2mode 2 · rank 2mode 3 · rank 2wave: three matrices, one arrayentries1320mode-1 rank2mode-2 rank2mode-3 rank2‖T‖26three ranksand none of them is the tensor's

A tensor has one matrix per index — put that index on the rows and every other index down the columns — and each of those matrices has an ordinary rank. Here they are 2, 2, 2 at a relative tolerance of 10⁻⁸, from a tensor of 1320 entries whose modes are of different lengths. Nothing requires the three numbers to agree, and nothing requires any of them to be the tensor's own rank: they are three different matrices built from one array. The leading singular values are 18.7, 19.3, 18.7, each normalised to its own mode below.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

2 distinct claims across 5 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a family this site defines

every unfolding has something in it

Against the rule

It draws a decomposition and prints its residual. It calls modeSvd, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 146 of 287 generators — 131 print a residual and 15 are exempt with a published reason; 141 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

When the index is a tuple

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.

 When the index is a tuple

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.

 When the index is a tuple

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.

 When the index is a tuple

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.

 When the index is a tuple

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.

 When the index is a tuple

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.

 Eigenvalues, singular values, rank

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.

 Randomised, and the guarantee that changes kind

Sketching what is never unfolded

A range finder multiplies its matrix by a few random vectors. For a mode-k unfolding those vectors have n^{d−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.

 When the index is a tuple

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.

 When the index is a tuple

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.

The whole library · All essays · What must fail