Exact ground truth — where it appears
Named by 94 essays across 17 fields — each of them below, with the objects they name alongside it.
The problem that arrives again
A hundred and thirty essays have solved a system once and measured how wrong the answer was. Almost no computation is shaped like that. A solve is one step of an outer loop, its answer is an input rather than a deliverable, and four quantities treated here as accuracy requirements turn out to be assets with a shelf life.
The zero that is not a missing entry
A constrained minimisation produces a matrix with a zero block, and the zero is a theorem rather than a sparsity pattern. No pivot order makes it positive definite, no precision changes that, and Cholesky does not fail somewhere on it — it fails at the first constraint row, on a number the problem already contained.
A matrix that depends on its own eigenvalue
A damped structure does not produce Ax = λx. It produces (λ²M + λC + K)x = 0, where the matrix whose null vector is wanted is a function of the number being solved for — so there is nothing to factorise, an n × n problem has 2n answers, and the eigenvectors cannot be a basis.
A model that is a rational function
A state matrix has a hundred thousand rows and the thing anyone wants from it is a function of one complex variable. The number that says how much of that size was ever the complexity is a rank — and the rank a derivation writes down cannot be computed, while one built from samples alone can.
An answer with no error in it
An integer matrix eliminated over the rationals rounds nothing, so the forward error is zero, the residual is the zero vector, and the identity this site is built on has no terms left. The cost does not vanish with the error. It moves into the length of the numbers, where three correct routes differ by four orders of magnitude.
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.
A matrix with no numbers in it
A graph arrives as vertices and edges. Two different matrices can be built from it, they answer different questions, and one of them has a null vector that is exact — the only object in these essays whose kernel is known before anything runs.
The accuracy that is thrown away
A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.
The bound that is known in advance
Almost every error in numerical linear algebra is measured after the fact. Balanced truncation has one that is computable before the reduced model exists, in a norm of a function rather than of a residual — and on ordinary problems it is not an upper bound that is loose. It is attained.
A backward-stable answer to a problem nobody asked
One quadratic eigenvalue problem, in nine systems of units, with a change of variable that is exact in both directions. The residual the solver prints stays at the rounding level at every stop. The answer loses eleven orders of magnitude, and the two facts are consistent.
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.
Three eigenvalues, and two are the golden ratio
Precondition a saddle-point system by the block diagonal of its own two definite pieces and the preconditioned matrix has exactly three distinct eigenvalues — 1, and the two roots of λ² − λ − 1. A minimal polynomial of degree three means three steps, at every conditioning, and the preconditioner nobody can afford turns out to be the statement the affordable ones are measured against.
The scaling that buys ten orders
Two lines computed from three norms, a change of variable that is exact in both directions, and the whole of the loss the previous essay measured comes back — flat, at every stop, because after scaling every stop is the same problem.
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.
A factorisation kept past its date
One Cholesky factor can serve five members of a drifting sequence and save 44 per cent of the work. Kept for twenty it does not lose accuracy — it stops converging altogether. The optimum and the cliff are four members apart, both move with the drift, and a rule written in a ratio the iteration has already computed finds them without being told what the drift is.
A condition number sent to infinity
An interior-point method manufactures an ill-conditioned matrix on every iteration, deliberately, because the separating of a diagonal is how it discovers which constraints are active. Written one way the answer keeps fifteen digits at a condition number of 3·10¹⁵. Written the other way — the way almost every code writes it — it has none left.
Every eigenvalue real, and a test that says so
A quadratic eigenvalue problem has no reason to have real eigenvalues. One class does, as a property rather than an outcome, and the proof is a Cholesky that completes. The boundary of the class has a closed form, and at the boundary the arithmetic loses half its digits with nothing ill conditioned anywhere.
The units the matrix is measured in
One linear system, written twice. The rows of the second are the rows of the first in different units, the solution is identical to the last bit, and the condition number has moved by eight orders of magnitude. One of those two numbers is a fact about the problem and the other is a fact about the notation.
Stable once, and three thousand times
A sliding window adds a row and removes one at every step and never looks at the data again. No single step of it amplifies by more than 2.72, no downdate fails, and after three thousand steps the triangular factor in memory is 3.9·10⁻¹⁴ from the matrix it is supposed to be a factor of — six hundred times growth from a per-step bound that says nothing about chains.
A constraint is a weight at infinity
Stack an equality constraint on top of a least-squares problem with a large weight and the answer approaches the constrained one like 1/τ². The limit is takeable to any accuracy — and how far it can be taken is a property of the solver, not of the problem. One of them stops at the square root of the precision, and one of them does not stop.
A problem with infinitely many eigenvalues
Let the matrix depend on λ through something that is not a polynomial and three things stop being true at once. There is no linearisation, there is no characteristic polynomial, and "compute the spectrum" is not a request that can be granted — the only finite question is how many eigenvalues are inside this circle.
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.
The problem the solver was actually given
A linearisation is exact — it has the polynomial's eigenvalues, with their multiplicities, and the whole loss is arithmetic. A nonlinear eigenvalue problem does not offer that. Every algorithm replaces the function first, and the term that replacement contributes is committed before any number is rounded and appears in no residual.
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.
The half of a problem a sketch may touch
A sketch guarantees that a norm is preserved to within a factor. An equality constraint is a statement that a quantity is zero, and no multiplicative guarantee says anything about zero. Sketch a constrained problem written as a weighted one and the constraint is not destroyed — it is demoted, from a violation of 1/τ² to one of ε/τ, exactly half the exponent.
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.
An exact answer to a measured problem
The residual is the zero vector, nothing was rounded at any step, and the answer is wrong in its first digit. Data accurate to fourteen places, an exact solve of the system it defines, and an error of 10⁻⁵ — because conditioning was never a statement about arithmetic and removing the arithmetic error removes none of it.
The right-hand side as one more column
Modified Gram–Schmidt's Q is 4.3·10⁻⁹ from orthogonal at κ = 10⁸, and a least-squares solve that multiplies b by it is wrong by 0.13. Hand the same routine b as an extra column instead and the answer is right to 2.7·10⁻¹⁰ — closer than Householder's 4.0·10⁻⁹. Classical Gram–Schmidt gains nothing from the same trick, to the last bit.
Where the augmentation puts the cost
Add γAᵀA to the objective block of a saddle-point system and its Schur complement tends to I/γ, so the cheapest possible approximation becomes the right one and the golden-ratio spectrum arrives — within 7.6·10⁻⁶ at γ = 10⁶. MINRES falls from 21 steps to 6. The inner solve with the augmented block rises from 14 conjugate gradient steps to 43, their product does not fall at all, and the answer loses seven and a half digits on the way.
The repair the drift did not need
A sliding window's carried Cholesky factor drifts 3.9·10⁻¹⁴ from its data, and multiplying by κ(AᵀA) predicts eight lost digits in the coefficients, a stream conditioned at 10¹² losing the answer, and a periodic refresh of the factor as the default repair. Measured against coefficients computed exactly in rationals, all three come out differently. On a stream made ill-conditioned by scaling, the conditioning never reaches the coefficients. On a collinear stream, a freshly recomputed factor is as wrong as the drifted one. And one correction from the window's own rows reaches Householder's accuracy for a fraction of a refresh's cost.
The eigenvalues that are answers to nothing
A rational approximant of degree five turns a six-by-six problem into a thirty-six-by-thirty-six one, and thirty-six numbers come back. Six are the answer. The rest are exact eigenvalues of the approximant, lying where the function it approximates is not a real number at all.
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.
A distance computed by a solve
Effective resistance is the one quantity in this field with no combinatorial route to it — it is defined by a linear system. On a small unweighted graph the answer is a ratio of two integers, so for once the error is known rather than estimated, and every resistance in a graph has to add up to a number fixed in advance.
A ceiling with a knob on it
A contour method returns at most as many eigenvalues as its probe block has columns, and the object that comes back does not distinguish that from having found everything. One line of the derivation multiplies the ceiling by a number the caller chooses, and it costs no extra solves at all.
A tensor that cannot be decomposed
Every member of a certain sequence is exactly a sum of two rank-one terms, and both terms are written down in closed form. A three-hundred-sweep fit from a random start does not find them, and stalls at the same one per cent however far the sequence goes — while a fit started at the answer loses digits exactly as 2n² says it should.
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.
The definition asks for more of what defeats it
The rank of a p × p Hankel matrix of Markov parameters resolves a degree of p, and it needs 2p parameters to do it. Those parameters grow like the norm of the state matrix raised to their index, so the count that buys resolution is the same count that buys dynamic range. One model, four run lengths, and a spread that runs from 10¹¹ to 10¹⁰².
Two condition numbers of one matrix
κ₂ is a worst case over perturbations of a given norm, and a normwise perturbation may put its whole budget on the smallest entry. The componentwise number is a worst case over perturbations proportional to the entries, which is what a backward-stable factorisation actually makes. On one matrix they are 3·10¹³ and 13.3, and the error obeys the second.
Feasible and wrong
A third constraint that nearly repeats the first takes the best route's answer from 2.96·10⁻¹⁵ to 1.16·10⁻⁴, and the other two routes to no correct digit at all. Every one of those answers satisfies every constraint to 10⁻¹⁵. The quantity a caller checks after a constrained solve is the one quantity here that says nothing.
Two repairs for one symptom
Rescale a quadratic programme's constraint rows over six decades and the crossover that certified its answer at iterate 1.5 first certifies at 69.8, with two of six programmes never certifying at all. Normalising the rows removes the spread completely — the same numbers at 10¹, 10² and 10³ either way. Starting the method at the magnitudes the rows imply repairs the iteration count completely and the identification only halfway. They are two repairs and they fix different halves.
Which of the choices is doing the work
Elimination makes n − 1 decisions and they are not worth the same. On 8×8 standard normal matrices, removing the first pivot search and leaving the other six multiplies the median growth factor by 1.624; removing the last multiplies it by 1.000. The cost falls monotonically along the run, and the worst single matrix in the sweep grows by 2,366 when one early decision goes — so the median is the wrong statistic and the tail is where pivoting earns its reputation.
One number that has to be right
Householder's orthogonality was called structural: a reflection is built from a unit vector, so rounding the vector names a different reflection rather than a broken one. Tested by breaking it, the claim is narrower and sharper. Perturb every component of the reflector by a relative 10⁻², and ‖QᵀQ − I‖ stays at 1.5·10⁻¹⁵ while the factorisation moves to 5·10⁻³. Perturb the one stored scalar by the same amount and ‖QᵀQ − I‖ is 6.5·10⁻². The structure is one degree of freedom, and the departure is four times its relative error.
Two minima that are one minimum
The order that decides the memory found the operation count behaving like the square of the fill, which leaves room for an order with slightly more fill but a shorter heaviest column to do less arithmetic. Searched exactly over every elimination order on forty graphs, that order does not exist: one order attains both minima on thirty-nine of forty, and on the fortieth the least-fill order's arithmetic is 1.0099 times the least. Minimum degree attains both on the same thirty-three graphs and neither on the same seven.
A guess worth two per cent
The previous Newton step looks like a free guess at the next one, and it is worth nothing. Started from it unscaled, the inner solve costs 4 to 61 per cent more than starting from zero, because the guess is 15 to 209 times too large. Scaled by the ratio of the two residual norms it is the right size and halves the starting residual — which buys a constant handful of inner iterations, not a share, because conjugate gradients costs the logarithm of its tolerance.
The split nobody is in a position to choose
Hutch++ spends two thirds of its budget on a sketch and a third on probes, and the third is published as a constant. Swept across six rates of spectral decay at a fixed budget of 48 products, the best share is 0.45 on the fastest and 0.00 on the slowest — sketch nothing at all — and the published third costs between 1.09 and 4.11 times the best error. The decay that decides it is readable from the sketch's own singular values, for products the estimator was going to spend anyway.
The roots are not the coefficients
A polynomial whose roots are the integers one to twenty, expanded exactly, handed to the routine every library uses. The computed roots are wrong in the third digit, the computation is backward stable for the matrix it factorised, and above degree eighteen the coefficients are not double-precision numbers at all.
The condition number that does not know
Two constrained fits with the same size, the same number of constraints and the same κ(A) to twelve figures. One returns 4.7·10⁻¹⁶ and the other 3.0·10⁻⁴. What separates them is the conditioning of A restricted to the constraint's null space — 1.00 against 10¹² — which every solver computes on the way and none reports.
The order the greedy rule cannot choose
Wilkinson's matrix is the standard demonstration that partial pivoting's growth bound of 2^(n−1) is attained. It is attained by the row order the greedy rule picks, and not by the matrix: a single cyclic shift of the rows gives growth 2 at every size, with no multiplier above one. At n = 7 that is 64 against 2. And perturbing one entry by 10⁻¹² leaves the good order exactly where it was while putting every tie-break of the greedy rule back on 64.
A triangle where the scalar was
Every level-3 QR assembles a block of reflectors into Q = I − Y T Yᵀ, and T is computed by a recurrence whose inputs are its own previous columns. A block of sixteen carries 136 computed numbers where sixteen separate reflections carry sixteen. The orthogonality it produces is 3.9·10⁻¹⁵ against the single reflector's 7.8·10⁻¹⁶ — a factor of five for a hundred and thirty-six times as many things that have to be right.
A rule that reads only its own probes
A trace estimator is a mean of independent samples, so its own standard error is estimable from the samples and a stopping rule needs nothing the estimator does not already have. Over forty draws it is calibrated in the middle and not at the edge: at a target relative standard error of 1% the median error reached is 5.3·10⁻³ and the worst of forty is 2.9·10⁻² — three times the target. And the cost of the target is the estimator's own square root: tightening it from 3% to 1% takes the median probe count from 75 to 696.
One line that buys a quarter of the run
The adaptive forcing rule has a floor on it that no published statement of the rule carries: do not solve a step to an accuracy the outer loop will not use. Removing it costs 9 to 27 per cent of the whole inner run. Keeping it costs between 23 and 2,600 times the forward error — accuracy the residual test never asked for and both runs satisfy the test either way. The line is a trade between a residual and an error, and which of the two the caller meant decides whether it is a saving.
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.
A test with no tolerance in it
An interior-point method's own stopping test is a tolerance on μ, and at the tightest it can be set to it stops after 15 iterations with 1.6·10⁻¹². A crossover from the iterate at 1.5 returns a point whose error is 1.8·10⁻¹⁴ — ten times sooner and a hundred times better, from an iterate carrying one correct digit. One attempt costs an eighth of a step, and the guess's own margin says which iterate to spend it on.
Three errors and one number
This site's identity has two factors and a division of blame between them. Two fields have now added a third party and a fourth, and only one of the four is a property of anything — the others are decisions, made before the arithmetic, reported by nothing.
The reference was a method
The optimality conditions of a constrained fit contain AᵀA, so solving them is the road that squares the problem wearing a block structure. At κ(A) = 10¹¹ the route that never forms a cross-product returns 1.89·10⁻⁹ and the route that does returns 4.64·10⁻⁴ — and forming AᵀA and then solving it in exact rationals returns 3.45·10⁻⁴, so nearly all of the loss happens before any elimination begins.
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.
The rank a certificate charges
A randomised range finder can choose its own rank: grow the basis a column at a time and stop when ten fresh probes all come back short. With the published safety factor it never stopped early in any draw measured, and on a matrix whose singular values fall by 0.8 a step it stopped at rank 30 for a tolerance the best rank-11 approximation already meets. The nineteen extra columns are three separate prices — four for building the basis from random vectors, five because a probe reads more than the spectral norm, and ten for the constant — and the spectrum decides which of them dominates.
A straight path has nothing for a parabola to fit
The line through the last two answers beat the parabola through the last three on a drifting sequence, and the reason offered was that the parabola amplifies the stored answers' error. Tightening the solve tolerance from 10⁻⁶ to 10⁻¹⁴ should have reversed that, and it does not: the line needs 7 to 18 inner steps over twenty members at every tolerance and the parabola 14 to 23. The sequence's roots move along a straight line, so the line is exact and there is nothing else to fit. Bend the path by a part in ten thousand and the parabola wins below 10⁻⁹, by 105 steps to 75 at 10⁻¹⁴.
Where the box is cut
A branch-and-bound with an interval operator settles a whole square — two roots proved unique, forty-two regions proved empty, nothing left undecided, in 87 evaluations. Move the roots so one lands on the first bisection and it proves nothing at all, at any depth. Cutting at 0.485 instead of 0.5 finds both, in a quarter of the work.
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.
A class a longer chain takes away
Symmetry survives a bigger problem. Hyperbolicity does not. The damping that certifies a chain of seven masses is refused by a chain of eight, the damping the class demands grows like the length without bound, and the certificate has to be earned again at every size — which costs one Cholesky, and the alternative is a proof quietly inherited from a smaller problem.
The degree the history chooses
A sequence of solves can start each member from the line through its last two answers or the parabola through its last three, and which is better depends on how much its path bends — which a code does not know. Over thirty runs of bend and tolerance, always taking the line costs 456 inner steps more than the better choice; always taking the parabola costs 33. A free rule reading the stored answers closes that to 11. A rule that evaluates the residual at both starts picks the better one on 29 runs of 30, and pays 135 steps for the evaluations.
The spectrum is not the graph
Two graphs on six vertices with the same Laplacian characteristic polynomial — as integer polynomials, not to fourteen digits. One contains a triangle; the other is bipartite. Every method in this field that reads only the spectrum is answering about the class.
An estimate that does not move
The tropical roots are said to miss the bottom of a spectrum by a factor growing like n². The estimate does not get worse. It changes by four per cent between four masses and sixty-four while the modulus it names falls by a factor of a hundred and sixty, and an estimate that is constant in the variable the answer depends on is a different defect from an inaccurate one.
Three walks and one bound
A left-to-right sum, a chain of three thousand rotations and a conjugate gradient residual recurrence share no arithmetic and no vocabulary. Each has a standard bound that is linear in whatever it accumulates against. All three come out at a half — 0.486, 0.554 and 0.507 — and nothing is rescaled.
The miss a normal table already priced
A trace estimator that stops when its own standard error reaches a target misses the target on about a third of draws, and the essay that measured it read the loose targets as the worst calibrated. Over 400 draws the loose target is the better covered — 76% at 10% against 64% at 3% — because the warm-up stops most of its runs with probes to spare. Where the criterion decides, the misses are a normal distribution's: a margin of c on the standard error buys what a normal table says, 94.8% at 1.96, and costs c² in probes, 3.86 times.
The weight the factor met first
The route to one minus a leverage through the orthogonal factor was said to lose a digit for every decade of the condition number, whatever else it does. Put a weight on one row and it does not. With the heavy row first, the complement keeps every digit at κ(A) = 2.5·10⁹ while both subtractions return nothing. With the same row last it loses digits as the row's scale grows. And two heavy rows that leave κ(A) at 3.1 still lose six digits when the light rows come first. The law was about the order the factor met the rows, and the condition number had been standing in for it.
The answer the last window left
A sliding window that corrects its least-squares answer at every step could start each correction from the previous step's corrected answer instead of from a fresh solve: the two windows share all but one row. On a stream with any noise in it, that start is three orders worse. The window's exact answer moves by 0.79 of itself in one step at κ(A) = 3·10⁶, a fresh seminormal solve is wrong by only 1.8·10⁻⁴, and one correction contracts either start by the same factor — so the fresh start ends at 2.2·10⁻⁸ and the carried one at 3.7·10⁻⁵. Only on data that agree exactly does carrying win.
What the appended block inherits
Modified Gram–Schmidt on [A b] solves least squares as well as Householder, although its Q is not orthogonal. A block code appends b as one more block. Block modified Gram–Schmidt inherits the rescue at every placement of the ill-conditioning: at κ = 10⁸ the appended block gives 6.9·10⁻¹⁰ where the same Q through Qᵀb gives 8.9·10⁻³. Block classical Gram–Schmidt gets the same wrong answer both ways, to the last bit. And the variant whose Q is orthogonal to 10⁻¹⁵ — two passes with Cholesky QR inside — is a hundred thousand times worse than Householder when the ill-conditioning is inside the blocks, because its R is wrong.
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.
A function of a matrix is not a function of its entries
Everybody learns that f(A) means diagonalise, apply f to the eigenvalues, undiagonalise. That is a definition, not a method. On a matrix seven picometres from a defective one — with exact eigenvalues and eigenvectors from a closed form — the definition returns an answer wrong by sixty-five orders of magnitude, and a method that never mentions an eigenvalue returns the right one.
The residual the solution cannot hold
Sorting a weighted fit's rows heaviest first gave every digit of one minus the heavy row's leverage back. It gives nothing back to the heavy row's residual, if that residual is computed the way every textbook computes it — as the datum minus the fitted value. The fitted value is a double, and a double cannot resolve a misfit smaller than its own last digit times the weight: at a weight of 4²⁴ the residual formed from the solution is wrong in its second digit in every order, and forming the subtraction exactly changes nothing. Taken from the same orthogonal factor as the divisor, the residual keeps fifteen digits, and so does Cook's distance at 3.4·10¹⁷.
A near-tie is a factor of four
A beam over contraction orders spends its width by level — wide late was the best schedule, finding the exhaustive order on 37 of 60 seven-tensor networks for 78 pairings priced. The alternative was to widen only where the ranking is a near-tie. Measured, near-ties buy nothing: keeping every candidate within a quarter of the best changes nothing at all, because the pairing the exhaustive order wants is almost never tied with the ranking's first choice. It sits a factor of two to four further down. Widen to keep everything within a factor of three and the beam finds the exhaustive order on 46 networks for 150 pairings, and at nine tensors has a median of 1.007 where every schedule tried before had 1.056 or worse below 1,380 pairings.
One step ahead is one step short
Partial pivoting takes the largest entry in the column and, on Wilkinson's matrix, walks into growth of 2^(n−1) that a cyclic shift of the rows avoids entirely. A rule that chose instead the pivot whose elimination leaves the smallest trailing submatrix was expected to see the good order at the first step. It sees nothing there: every first pivot leaves a largest entry of exactly 2, and with the ties broken by 10⁻¹² it prefers the greedy row by 10⁻¹². It attains 2^(n−1) at every size. Looking two eliminations ahead, the greedy row scores 4 and every other row 2, and the growth is 2 at every size up to 24. On random matrices one step of look-ahead helps below n = 16 and is worse than greedy on more than half of them by n = 32.
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.
The series that has to be squared back
The Taylor series for the matrix exponential is not wrong — every term is computed correctly — and on Moler and Van Loan's two-by-two its largest term is 5.4 million times the answer it sums to. The method that replaces it scales the matrix down and squares the result back, and both halves of that sentence cost: too few squarings and the approximant is out of range, too many and each one doubles the rounding.
A multiplier is a force
A third constraint nearly parallel to the first made the multipliers of a constrained fit rise in exact proportion to κ(B), which looked like the conditioning measured a second, dearer way. It was not. Give the third constraint a datum that asks for nothing new and, at the same κ(B) = 4.6·10¹², the multipliers are eighteen thousand times smaller; give it a strain δ and they are 0.0133 δ/ε², a force on a lever of length ε. What they measure is what the constraint asks. What they do not measure is the error of the best route, which sits at the same level whether the constraint asks for nothing or for a displacement of 3·10⁹.
An augmentation read in the smallest eigenvalue
An augmented Lagrangian preconditioner's least work sat at γ = 1, 10⁴ and 1 on three systems, and two of the three optima sat where the smallest generalised eigenvalue was about 0.06 — which suggested a rule written in ν rather than γ. On twenty-four systems whose norms are all one, the least-work γ spans seven decades and ν at the optimum spans a factor of thirty, so there is no one ν. There is still a rule: take the smallest γ at which ν reaches 0.03, and the median system pays 7 per cent over its least work and the worst 29, where the best single γ pays 61. Its price is digits — twelve times the best forward error on the median system, 4,561 times on the worst.
The residual the appended block cannot remove
Appending b as one more block made block modified Gram–Schmidt solve least squares as well as Householder, ten million times better than the same Q through Qᵀb at κ = 10⁸ — on problems with no residual. Give b a component outside the range and every stable route's error rises with it, while Qᵀb's, already at κ²u, does not move. The appended block's advantage then falls as one over the residual: 5,400 at a relative residual of 10⁻⁴, 54 at a per cent, none at one. It never falls behind Householder by more than a factor of four. What the residual decides is whether the extra block is worth its synchronisations, and the answer is yes up to a residual of about a per cent.
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.
The vector was what was wanted
Nobody who computes a matrix exponential wants the matrix. They want eᴬᵗb — one vector, the state of a system at a later time. Twenty matrix–vector products get it to sixteen digits on a hundred-by-hundred problem, without ever forming a hundred-by-hundred exponential, and the exponential that does get computed is twenty by twenty.
Where a contour's budget should go
A contour method's ceiling is the number of probes times the number of moments, and the moments are free in solves while the probes are not. Four ways of reaching one ceiling come out four orders apart, the ordering is not monotone, and what separates the best two is not the usual draw but the unlucky one.
The step the two rows owe
A sliding least-squares window can start each step's correction from a fresh solve or from the answer it already has. The answer it has is three orders worse on noisy data, because the exact answer moves by most of itself in a step. The proposal was a start that moves too: the previous answer plus the change the entering and leaving rows imply, two triangular solves from the factor the window keeps. Its start lands exactly where one correction of the carried answer lands — the update is that correction, computed from two rows instead of twenty-four — and one correction after it ends 2.9 to 440 times below the fresh start at κ(A) = 3·10⁶, and level with the carried answer when the data agree exactly.
Balanced is not symmetric
The Sylvester–Kac matrix is made of small integers, so a double holds it exactly, and its eigenvalues are the integers from −(n − 1) to n − 1 in steps of two. Every digit an eigensolver loses on it is therefore the solver's own, and it loses them at exactly the rate first-order perturbation theory predicts: the median error is half the prediction across 1,568 eigenvalues. Balancing, the preprocessing libraries apply for this kind of matrix, divides every condition number by about sixty and leaves their growth untouched, and at order 112 the unbalanced solver returns eighteen complex eigenvalues for a spectrum of integers.
A finer invariant that hears less
The Laplacian spectrum fixes a graph's number of spanning trees and not the group those trees form, so the Smith normal form of the grounded Laplacian is a strictly finer integer invariant — and on the six-vertex pair the spectrum cannot separate, it does: ℤ₁₂ against ℤ₂ ⊕ ℤ₆. Over all 1,022 Laplacian-cospectral pairs of connected graphs on eight vertices it separates 435. The adjacency polynomial, a second spectrum rather than a finer one, separates all of them.
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.
The circle between two eigenvalues
A contour count's error is not approximately governed by the nearest eigenvalue; it is exactly one closed-form term per eigenvalue, and summing those terms reproduces the quadrature to a millionth at 1,720 radius and point pairs. Three things follow. The rate is the ratio of the two moduli the circle sits between, not a distance, so two circles 0.2 from their nearest eigenvalue converge three times apart. Ten digits cost about 21 points divided by log₁₀ of that ratio, 71 points with twelve eigenvalues inside and 3,476 with six. And the best circle is not halfway: at the geometric mean of two eigenvalues on one ray their two terms are equal and opposite, and ten digits cost 27 points where the midpoint needs 84.
Each spectrum hears the other's pairs
Pairs of graphs with the same Laplacian spectrum share a spanning-tree count, so the critical group could only tell them apart by how that count factors — and did on two pairs in five. Turn the census round, to the 733 pairs of eight-vertex graphs with the same adjacency spectrum, and the tree count is free to differ. It does on 717 of them, by a median of six per cent. The critical group separates exactly those 717 and not one more: on all sixteen pairs whose counts agree, including a non-cyclic one, the two graphs have the same group. And the Laplacian spectrum separates all 733, as the adjacency spectrum separated every Laplacian pair.
The scale that only moved a pivot
Multiply the constraint rows of a saddle-point system until its multipliers are the size of its solution, and the extra error the route was blamed for — 4.6·10⁻⁴ against the null-space route's 1.8·10⁻⁸ — falls to 3.3·10⁻⁸. The prediction holds and its reason does not. A scale of ten does what a scale of 6·10⁵ does; hold the elimination's row order fixed and nine decades of scale move the error by less than a factor of five. What the scale changed was which row partial pivoting took at the second step, and taking the constraint rows first does the same job with no scale at all.
The worst residual belongs to the route
Every stable least-squares route's error rises with the residual at about a fiftieth of the bound κ²uρ, and the question left open was whether a residual aimed at the weak directions closes the gap. It can be aimed exactly: the map from residual to error is, on these problems, one direction and rounding. Aimed, Householder's constant rises from a median of 0.014 to 0.11 — √(m − n) = 6.9 times a typical direction, at 32, 64 and 128 rows to three figures — and stops a factor of nine short of the bound. The direction is each route's own: aimed at one, another route draws a fifth of its worst. And aimed at the appended block, the one-per-cent rule that made it worth its extra block falls to half a per cent, with the block twelve times behind Householder on the same data.
Two groups need two reductions
When a quadratic eigenproblem's tropical roots are far apart its eigenvalues fall into two groups, and the remedy offered is a second scaling — one per tropical root, each run keeping the group its root predicts. Raise the damping on an eight-mass chain until the roots are 10²³ apart and measure every eigenvalue: no scaling rescues the small group. Its backward error grows as the unit roundoff times the separation under Fan–Lin–Van Dooren's scaling, under both tropical scalings and under none, to 10⁻⁵ at a separation of 4·10¹¹. What rescues it is the other reduction of the same pencil — inverting the constant term instead of the leading one — which keeps the small group at rounding at every separation and loses the large one instead. Both reductions, each keeping its own group, give all sixteen eigenvalues to 10⁻¹¹ at worst. On the way, the closed form that served as the exact answer turned out to lose the small roots to cancellation, from the same separation.
An eigenvalue count that cannot be slightly wrong
Every spectral computation here returns floats with errors in them. Counting eigenvalues below a shift by the signs of an unpivoted elimination returns an integer, and an integer cannot be 6.9999999997 — so the answer is exactly right, or wrong by a whole eigenvalue, and where the second happens is a band of measurable width.
A different equation on every grid
Upwinding is the exact discretisation of a convection–diffusion problem with diffusion ε + h/2, entry for entry, at a relative difference of between 0 and 1.26·10⁻¹⁶ on every mesh from 15 points to 511. The equation it is exact for is chosen by the mesh and not by ε — the added diffusion is 0.01563 on a 31-point grid whether ε is 0.2 or 0.001.
The error the method already knows
Summing the exponential's Taylor series throws away a known number of digits, and the number is on the machine while the sum is being formed. The largest term divided by the answer, times the unit roundoff, tracks the relative error that comes out — to within a factor of nine, across fourteen orders of magnitude of it — and nothing reports it.
Named alongside it
The objects these essays reach for when they reach for this one.
Condition numberBackward errorFlop countForward errorSaddle-point systemsResidualLinearisationNewton iterationQR factorisationCholesky factorisationCondition squaringDeterminant