Concept

Determinant — where it appears

The signed volume factor of a linear map, which is a product over the whole spectrum and therefore carries the units n times over. It underflows long before it is uninformative, and the questions it is asked — is this singular, is this definite — all have better answers elsewhere.

Named by 26 essays across 9 fields — each of them below, with the objects they name alongside it.

1-2-6-7-474-26-12-5-8-524-394151435A, the matrix as given1-2-6-7-401840552700112217207003943011710017279153after 2 fraction-free steps3 × 3 minors25 divisions, all exactthe intermediates are minorswhich is why the divisions come out whole

Every intermediate is a minor

Fraction-free elimination divides by the previous pivot at every step and the division is always exact. Not usually, not for these entries — always, because the number being divided is a determinant with that pivot as a factor, which is a theorem and is checked here against the minors themselves.

exact · Fraction-free
3456789101110²10³nbitsthe answerthe question · det Abits in, bits outn11question484answer729det A33widest part37the answer is the floorand no route writes it down more cheaply

The answer is longer than the question

An exact solution of an integer system is a vector of fractions, each of them a ratio of two determinants. So the output carries 2n long integers where the input carried n² short ones, and no algorithm can write it down more cheaply — the length of the answer is a floor under every exact solver rather than a property of one.

exact · Exact cost
34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

How many primes the answer needs

Work modulo a word-sized prime and no intermediate can exceed twenty-six bits, whatever the matrix does. The catch is that the answer must be reassembled from several such computations, and the number of them has to be fixed before the first one runs — by a theorem about how large a determinant can be, not by trying more until it settles.

exact · Modular lift
det mod p, as a fraction of p3713a prime that divides the answerdet A3·10⁴primes swept25unlucky5rate0.2det, in bits15singular mod p is not singularand one residue cannot tell them apart

A prime that divides the answer

A modular elimination reports a singular matrix and is telling the truth — over the field with p elements the matrix is singular. Over the rationals it is not. Nothing in the residue distinguishes the two cases, no quantity is small enough to be suspicious, and the wrong answer is a correct computation of a different question.

exact · Modular lift
10⁻²10⁻¹110¹10²10³10⁴10⁻⁴10⁻³10⁻²10⁻¹1backward error, in units of ushare of systems above ituCramereliminationCramer, control24-bit arithmeticworst Cramer, in u395worst elimination, in u1.2control, worst Cramer4.9κ of the worst system3.4·10⁶one derivation, two computationsand only one of them is stable

A rule that is correct and unusable

Cramer's rule gives every component of the solution in closed form, in terms of determinants, and it is a theorem. On two-by-two systems whose rows are nearly parallel it returns an answer with a backward error of 458 units of roundoff where elimination returns 1.3 — on a matrix whose condition number is 32,000 and which elimination solved perfectly.

elimination · Determinant
2022242628303210⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²k, where the entries are near 2ᵏrelative error of the determinantas writtenfused: exactproducts need 54 bitsone rounding, the whole answertrue determinant1naive, k = 300fused, k = 301first wrong at k27sizes returning 06both forms conformand the source does not say which

One multiply the compiler removed

A determinant whose value is exactly 1, computed as exactly 0 by the expression that is written down, and exactly 1 by the same expression with the multiply and the add fused. Both forms conform to IEEE-754, both are legal compilations of the same source, and nothing in the program says which one you have.

machine · Fma contraction
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
does this matrix look nearly singular?green: the test agrees with the truth · red: it does not · the bar under each number is its magnitude, over sixty-two decades|det A||det A|^(1/n)σₘᵢₙ1/κ = σₘᵢₙ/σₘₐₓ0.1·I at n = 40perfectly conditioned10⁻⁴⁰0.10.11κ = 10¹⁰, |det| = 1nearly singular1110·10⁻⁶10·10⁻¹¹Hilbert at n = 8nearly singular2.7·10⁻³³8.5·10⁻⁵1.1·10⁻¹⁰6.6·10⁻¹¹the two counterexamplesκ of the scaled identity1its determinant10⁻⁴⁰κ of the normalised matrix10¹⁰its determinant1det(cA) = cⁿ det(A)so a determinant carries the units n times over

The number that decides nothing

The determinant is the first scalar anybody attaches to a matrix and the last one worth consulting. A tenth of the identity has a determinant of 10⁻⁶⁰ and a condition number of exactly one. The Hilbert matrix's determinant stops being right at n = 13 and stops being a number at n = 29, and nothing in between reports either.

error · Determinant
s10 bitss20 bitss30 bitss40 bitss514 bitsinvariant factors, in bitstwo routes, and a conserved quantitydet A-2.3·10⁴Π invariants2.3·10⁴SNF widest15HNF widest24det of the transform-1an algorithm and a definitionagreeing as integers

What a determinant does not determine

Two integer matrices can have the same determinant, the same rank and the same size, and define genuinely different maps. What separates them is a list of integers each dividing the next — computed here twice, once by unimodular elimination and once from the gcds of every minor, which share no algorithm at all.

exact · Normal forms
the same lattice, twicewhat the reduction may not changedet, before1det, after1defect, before7.1defect, after1LLL steps1the determinant is the invariantand the defect is what is being reduced

A basis that describes its lattice badly

The same set of points has infinitely many bases, they are all correct, and they are not equally useful. One measurement separates them — the product of the vectors' lengths over the lattice determinant — and the determinant is the invariant the reduction may not change, which is what makes the reduction checkable.

exact · Lattice reduction
38434853586368110¹10²10³10⁴e, where the skew is 2ᵉ + 12345orthogonality defect of the basis returneda double holds k exactlyan orthogonal basisthe invariant sees none of itfloat, at 2^521float, at 2^531.4float, at 2^701.2·10⁴residue at 2^701.2·10⁴exact, everywhere1determinant, always1every step was unimodularand the answer is 12,345 times worse

The knob and the rounding

The Lovász parameter is the number a lattice reduction is specified by, and moving it from 0.50 to 0.99 strengthens the proved bound from 16.00 to 1.83, costs 91 per cent more steps, and returns a basis with the same orthogonality defect. The one rounding nobody writes down decides everything: above 2⁵³ the reduction returns a basis 12,345 times worse than it should, with the determinant invariant equal to one throughout.

exact · Lattice reduction
10×10, ±6determinant, bits27Hadamard's bound36median slack10024680510152025303540elimination stepbitsHadamard, on the minorsthe intermediatesevery intermediate is a minor of the originalso a bound on the minors bounds the run

A bound on every intermediate at once

Fraction-free elimination's intermediates are minors of the original, which is a theorem about exactness. It is also a bound: Hadamard's inequality applies to every minor, so one inequality bounds the whole run before it starts. The bound on the k-th step is the one on (k+1)×(k+1) minors, not the one on the whole matrix — and on a 10×10 with entries in ±6 the difference is ten bits, with the run reaching 2.7 bits a step against the bound's 3.4.

exact · Fraction-free
20 matricespeak, every rule29the determinant's own length29area, smallest ÷ natural0.9602468051015202530elimination stepbits, longest entrythe determinant: 29 bitsswap only on a zerolargest pivotsmallest nonzero pivotthree orders, three sequences of minorsand one last entry they all arrive at

Three orders and one last entry

Over the integers there is no stability to pivot for, so a fraction-free elimination swaps rows only when the pivot is zero. Choosing a pivot for length instead does change the sequence of minors — the smallest-nonzero rule makes seven exchanges where the natural order makes none and keeps the profile two bits lower through the middle. It cannot change the peak. The last entry of the elimination is the determinant, and the determinant does not know what order it was computed in.

exact · Fraction-free
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
57911131517192110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹vertices of the complete graphrelative sizethe algorithm is not the problemsizes counted18worst relative error9.4·10⁻¹⁵first wrong at n =16digits in the answer17spacing of doubles there8largest integer a double holds9·10¹⁵the elimination stays at the rounding levelthe format runs out

A count that comes out of a determinant

The number of spanning trees of a graph is the determinant of its grounded Laplacian, so it is a whole number known in advance. The elimination that computes it is backward stable at every size — and from sixteen vertices the answer is wrong, because the count has seventeen digits and a binary64 has sixteen.

graph · Graph elimination
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
4 numbers, coefficients to 30found relations72with a gap of a digit or more54accidents56accidents with that gap05101520253001234decimal digits the numbers are scaled togap, in digitsexact relation foundnot a relationthe room a relation hashorizontal: a gap of one digitdashed vertical: the digits a double has

The room a relation has to stand out

A lattice search for an integer relation returns its shortest vector, and the proposal was to return the gap to the next one as well, so a caller could tell a relation from an accident. Measured, the gap is a certificate with a budget: the digits the numbers really have, shared among all but one of them, less the size of the relation. A found relation's gap sits half a digit under that budget, accidents stay near zero, and a one-digit gap vouches for 81 of 96 relations among three numbers and for 1 of 35 among six. The test numbers the proposal came from turned out to have relations of their own.

exact · Lattice reduction
at σ = 10⁻⁸Cramer's rule1.2·10⁻⁴pivoted2.9·10⁻¹⁰elimination on T1.1·10⁻¹⁴cancellation × u1.7·10⁻¹⁰-10-9-8-7-6-5-4-3-210⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1σ, as a power of tenforward errorCramer's rulepivotedcancellation × uelimination on Tleft: σ small, the wrap nearly singularthe pivoted solve follows the cancellation

The correction lost to its own two-by-two solve

Solving a band matrix through a circulant and a small correction was measured losing the answer to 10⁻⁴ where elimination kept it to 10⁻¹⁴, and a wrap that landed one sample on a zero was measured costing four orders more than one that landed a pair. Both measurements solved the two-by-two correction system by Cramer's rule. Solved with a row interchange, the corrected solve on the same matrix loses 2.9·10⁻¹⁰ — the cancellation, and nothing multiplied onto it — and the single landing costs what the pair costs. The loss was in the determinant.

structure · Circulant
6 numbers, coefficients to 30relations, both agree34relations, they differ1accidents, both agree5accidents, they differ88510152025300123decimal digits the numbers are scaled togap, in digitsrelation, both agreerelation, they differaccident, both agreeaccident, they differhorizontal: a gap of one digitdashed vertical: the digits a double has

Two precisions guard the other edge

Run a lattice relation search at N digits and again at N/10, and accept its answer only if both runs return the same vector. Among six measured numbers, where the gap between the shortest and next vector vouches for one found relation in 35, the two runs agree on 34. Past a double's sixteen digits they never once agree on an accident, 0 of 561. They do agree on 62 accidents at fifteen digits or fewer — approximate relations that really are the shortest vector there — and the gap, which cannot see a relation among six numbers, can see those.

exact · Lattice reduction
finite eigenvalues (degree of det Q)13at infinity (2n − degree)1at infinity, by the rank of M12n, if M were nonsingular14a degree, not a decisiondegree of det Q13at infinity1by the rank of M1singular-value gap∞the count is a degreeand the other route is a judgement

One mass removed, and one eigenvalue gone

A coordinate with no inertia reads like a coordinate that has been deleted, and a chain of eight masses with one of them removed would then be a chain of seven, with fourteen eigenvalues. It has fifteen. The massless coordinate is still there, still carrying a damper, and it contributes a first-order equation rather than none.

polynomial · Polynomial eigenvalue
largest pivot, area1.037largest pivot, cost1.293smallest pivot, area0.969smallest pivot, cost0.704shortest row, original, area0.966shortest row, original, cost0.841shortest row, current, area0.964shortest row, current, cost0.760over the natural order: below one is cheaperthe rules agree on area and part on cost

The pivot is in every product

A fraction-free elimination's intermediates are minors, and Hadamard bounds a minor by the lengths of the rows it is made of — so the rule that picks the smallest pivot entry looked like a proxy for a rule that picks the shortest row. Measured, the two keep the bit-length profile equally low: 0.969 and 0.966 of the natural order's area. They part on what the arithmetic costs. Counting every multiplication and division at the product of its operands' lengths, the smallest-pivot rule costs 0.70 of the natural order and the shortest-row rule 0.84, because the pivot multiplies every entry of the step and divides every entry of the next. Three per cent of area is thirty per cent of arithmetic.

exact · Fraction-free
1.001.251.501.752.002.252.50cost ÷ the least any row order reachesnatural order1.733 · worst 2.41 · least on 0smallest pivot1.199 · worst 1.61 · least on 0cheapest step1.187 · worst 1.57 · least on 1two pivots ahead1.025 · worst 1.87 · least on 8three pivots ahead1.000 · worst 1.27 · least on 13smallest, then next1.217 · worst 1.89 · least on 0two ahead, estimated1.160 · worst 1.87 · least on 2bar: median · whisker: worst of twenty-fourthree pivots ahead finds the least on thirteen

What reading the next pivot buys

In a fraction-free elimination every pivot is paid for twice — it multiplies every entry of its own step and divides every entry of the next — and the rule that picks the smallest pivot entry left a median 20 per cent above the least arithmetic any row order reaches. A rule that charges two steps ahead brings the median matrix to within 2.5 per cent, and on one matrix of twenty-four costs 1.87 times the least, worse than the smallest pivot ever does. Charging three steps ahead finds the least of all 40,320 orders on thirteen matrices and is never more than 27 per cent above it. And none of it pays: choosing that way costs forty to two hundred and sixty times the elimination it chooses.

exact · Fraction-free
invertible sharerationals, n = 201𝔽₂, n = 200.29𝔽₃, n = 200.56𝔽₅, n = 200.74𝔽₇, n = 200.83246810121416182000.20.40.60.81size nshare invertibleover rationalsover 𝔽₂over 𝔽₃over 𝔽₅over 𝔽₇dashed: a matrix uniform over the fieldthe fields part company as n grows

The field decides it, usually

A matrix whose rank depends on the field it is read over was built, the first time, from its invariant factors outward, because random integer matrices never seemed to show the effect. Random 0/1 matrices show it at almost every size that is not tiny. At twenty rows, 99.8% of them are invertible over the rationals, 29% modulo two, 56% modulo three — and 71% of the ones the rationals call invertible are singular modulo two. Modulo two they obey, corank by corank, the law for uniformly random matrices over that field; modulo three and five, which their entries cannot fill, they converge to that field's law anyway.

exact · Exact rank
λ = 0λ = ∞λ = ∞-0.37690.5654.3836.4282 eigenvalues here, and it is one placecounted exactly, in rationalsfinite eigenvalues4at infinity2degree of det(A − λB)4worst residual, either kind5.5·10⁻¹⁴an eigenvalue is a ratioand a ratio has a direction, not a size

An eigenvalue with no value

If the second matrix of a pencil is singular then some of the eigenvalues are infinite, and that is not a degeneracy — it is the algebraic constraints of the model, one per constraint. What survives is a pair of numbers rather than one, and on the line those pairs live on, infinity is an ordinary point with an ordinary residual.

spectra · Pencil

A problem with no answer

If two matrices share a null vector then det(A − λB) is identically zero and every λ is an eigenvalue, which means none of them is. Perturb such a pencil by a ten-billionth and a solver returns six numbers with residuals below 10⁻⁹. Change the seed and it returns six different numbers, spread over forty-four, with residuals just as small.

spectra · Pencil

The largest gap is inside the null space

The rule recommended for counting a pencil's infinite eigenvalues is to cut at the largest gap in the singular values of B. On integer pencils, with no perturbation anywhere and an exact answer available from the characteristic polynomial, it returns the wrong count on nine of twenty-five — because the singular values that are mathematically zero come back spread over a hundred and forty orders of magnitude, and the largest ratio in the list is between two of them.

spectra · Pencil

Named alongside it

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

Exact arithmeticExact ground truthBit lengthFraction-free eliminationHadamard boundBackward errorMinorCondition numberLattice reductionOrthogonalityOrthogonality defectSingular values

All concepts