Numerical Linear Algebra
where the algebra stops being the arithmetic
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The thread: Exact ground truth — page 2

Essays 25 to 48 of the 157 on this theme, in the same order.
34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens Exact arithmetic, and what it costs instead

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.

 024681012141618202210⁶10⁷10⁸members served by one factorisationmultiplications for the whole sequencedoes not convergethe contraction ruledrift 0.01 a memberevery member1.6·10⁷every 5 members9.3·10⁶contraction rule9.2·10⁶its factorisations4cliff at a period of20a factorisation has a shelf lifeand the cliff is past the optimum When the problem arrives again

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.

 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 Exact arithmetic, and what it costs instead

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.

 10¹10³10⁵10⁷10⁹10¹¹110¹10²10³10⁴condition number of the matrixgrowth factorbound 2^11partial pivotingCholeskyno pivot to gain fromCholesky growth, every κ1Cholesky interchanges0partial pivoting, worst9the bound, 2^112048both eliminations reach the same growthand only one of them had to swap to get there Elimination, and the swap

A factorisation with nothing to pivot for

Cholesky's growth factor is not bounded by one. It is equal to one, at every size and every condition number, and the two-line reason is why the algorithm needs no pivoting at all — not "usually gets away without it". Its only failure is the square root of a non-positive number, which is exactly the test for definiteness, and in floating point that test moves with the precision.

 015030045060075090010⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹iteration‖r‖ / ‖b‖JacobiGauss–SeidelSOR ω=1.78closed form vs measuredρ Jacobi, exact0.99ρ measured0.99ρ Gauss–Seidel, exact0.981D Laplacian, n = 24ω optimal = 1.777 Iterating, instead of factorising

A rate that is known in advance

On the model problem, Jacobi contracts by cos(π/(n+1)) per step, Gauss–Seidel by its square, and optimally relaxed SOR by a number given in closed form. Three rates, all known before anything runs, and all measurable against what runs.

 192123252729313335373941434500.10.20.30.40.50.60.7position jweight on the true signal at jthe blur's row, 5.89 widethe kernel, 2.82 widecomputed from A and λ alonekernel width at half height2.8components kept, Σfₖ28width × Σfₖ / n1.2deepest negative lobe-0.075no data and no truth went into this curvethe answer is the truth seen through it Regularisation, and the answer that is chosen

A second blur, narrower than the first

A regularised answer is not the truth with the noise taken out. It is the truth seen through a second blur, V F Vᵀ, which depends on the operator and λ and on nothing that was measured. At the best λ for 0.1% noise its rows are 2.82 points wide against the instrument's 5.89, they dip to −0.075 on either side, and their width times the number of components kept stays between 1.10n and 1.27n across seven decades of λ. Two spikes four points apart come back as two; three apart, as one.

 1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 20no penaltypenalty insidewhat stopping is worthbest without a penalty0.14and at step 40161best with one0.14and at step 400.14the same floor, reached twiceand only one run stays on it Methods that were designed apart

The step that stops mattering

Regularise the problem the iteration has built rather than the problem it was given, and the error curve stops turning. The unregularised run ends 1,127 times above its own best; the same run with a penalty inside it ends 1.000000000003 times above.

 3000000120000-21000000-0.5-1.500001.5-0.5000000-2T = ZᵀAZthe highlighted boxes each hold one conjugate pair, and no real rotation removes themthe form, and that it is one‖A − ZTZᵀ‖/‖A‖1.8·10⁻¹⁵‖ZᵀZ − I‖2.5·10⁻¹⁵worst eigenvalue error2.7·10⁻¹⁵surviving subdiagonal26×6, spectrum chosen before the matrix was builtquasi-triangular is as far as the reals go Eigenvalues, singular values, rank

The form a real matrix can reach

A real matrix with complex eigenvalues has no real triangular form, and the reason is one line — a real triangular matrix has a real diagonal, and a similarity does not move the spectrum. What it has instead is triangular except for one two-by-two block per conjugate pair, and the count is decided by the matrix rather than by where the iteration stopped.

 2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one Exact arithmetic, and what it costs instead

A fraction recovered from one remainder

A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

 00.250.50.751-1-0.500.51mode frequency θ / πdamping factor per sweeppredictedmeasured±0.333a coarse grid seestwo routes to one factorsmoothing factor, scanned0.33smoothing factor, closed form0.33worst mode disagreement4.4·10⁻¹⁶63 interior points, one sweepthe left-hand end is what the coarse grid is for Iterating, instead of factorising

The error smoothing cannot reach

One weighted Jacobi sweep multiplies every mode of the error by a number, and the number is a sine. Half the modes are cut by three or better, and the other half come back at 0.999 — which is not a failure of the method but the fact the whole of multigrid is built on.

 10²10³10⁴10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²steps taken‖RᵀR − AᵀA‖ ⁄ ‖AᵀA‖the bound, linear in the steps√k · uevery step safe, the chain notdrift after the run3.9·10⁻¹⁴the bound there3.3·10⁻¹³√k · u there6.1·10⁻¹⁵worst single amplification2.7worst leverage met0.69refreshes1backward stable onceand three thousand times is a different claim When the problem arrives again

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.

 0246810121410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹10⁴10⁷10¹⁰log₁₀ τ — the weight on the constraintrelative error against the exact answerτ = 1/√uGram–Schmidtnormal equationsHouseholder QRthe ceiling is the method'sHouseholder, τ = 10¹⁴4.8·10⁻¹⁵normal equations14Gram–Schmidt4.9·10¹⁰1/√u6.7·10⁷a constraint is a weight at infinityand the solver decides how far infinity is Least squares, and the road not to take

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.

 ℚ6𝔽24𝔽35𝔽56𝔽76𝔽116𝔽136𝔽1016𝔽655376rank, by the ring the entries are read inthe rank of one matrixover ℚ6over 𝔽24over 𝔽35over 𝔽56over 𝔽76nothing is rounded hereand the answer still is not a property of the matrix Exact arithmetic, and what it costs instead

The rank depends on the ring

A floating-point rank is a decision about a threshold. Remove the arithmetic error entirely and the threshold goes away — and the answer still is not a property of the array of numbers, because one integer matrix has rank six over the rationals, five modulo three and four modulo two, with nothing rounded and nothing decided.

 -5-3-11350eigenvalueHZᵀHZK4 negative — Cholesky of H stops at row 30 negative — a minimum on the constraint(10, 4, 0) = In(ZᵀHZ) + (4, 4, 0)one factorisation, no Zpositive, LDLᵀ of K10negative, LDLᵀ of K4negative in H4negative in ZᵀHZ0the count follows the reduced Hessiannot the Hessian The matrix a constraint makes

A minimum the Hessian cannot see

A Hessian with four negative eigenvalues can sit at a constrained minimum, and a Cholesky of it stops at the third row. One symmetric indefinite factorisation of the saddle-point matrix settles the question anyway — ten positive pivots and four negative — without a basis for the null space ever being formed. The count is exact in the algebra and blind in floating point, in a band that grows like κ(A)²; the route through the null space is blind in one that grows like κ(A).

 the smallest perturbation of any kind — 1.89·10⁻¹⁷the smallest Toeplitz one — 2.65·10⁻¹³both exact for the same x̂smallest of any kind1.9·10⁻¹⁷smallest Toeplitz one2.7·10⁻¹³the price of the constraint1.4·10⁴diagonal defect, unconstrained0.97an exact answer to a nearby problemof a kind nobody posed Structure, and the solver that cannot see it

A nearby problem of the wrong kind

A good algorithm returns the exact answer to a nearby problem. A hundred and eighteen essays have measured the distance and not one has asked what the nearby problem looks like. On a Toeplitz system it is a rank-one matrix that is constant along none of its diagonals — and the smallest one that is Toeplitz is two and a half million times larger.

 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 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.

 15731pointsfinest gridone unknown — the recursion bottoms out in a divisionthe coarse operator, two ways‖RAₕP − A₂ₕ‖/‖A₂ₕ‖10⁻¹⁸unknowns / finest grid1.7cycle cost, in fine sweeps12each coarse point reaches three fine ones½, 1, ½ — and the restriction is its transpose Iterating, instead of factorising

The same problem on a coarser grid

Restriction, the coarse operator and interpolation are three matrices with nine distinct entries between them. Two of the three are each other's transpose, and their product with the fine operator is the coarse discretisation exactly — not approximately, entry for entry, at every level.

 1234567891010⁻¹⁹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹kλₖ₊₁ ÷ λ₁, and the boundZₖ²the Gramianpredicted from two numbersstates30κ of the spectrum389λ11 ÷ λ₁2.5·10⁻⁸the bound there5.2·10⁻⁴the cliff everything rests onand the reason for it Reduction, and what a model is for

Why a Gramian can be truncated at all

Every method in this field rests on one fact nobody states the reason for — the eigenvalues of a Gramian fall off a cliff. The equation defining it has a rank-one right-hand side and no low-rank structure anywhere — and the answer's decay is a rational approximation problem with a closed-form rate.

 110¹10²10³110¹10²10³off-diagonal entry ccondition number of the eigenvalue√(1 + c²)decoupled: 1measuredthree routes, one number‖A − ZTZᵀ‖/‖A‖1.7·10⁻¹⁵closed form100computed 1/|yᵀx|100worst measured movement46four eigenvalues, two conditioning numbersthe symmetric case has one, and it is 1 Eigenvalues, singular values, rank

A condition number for one eigenvalue

In the symmetric case every eigenvalue has condition number exactly one. In this four-by-four matrix two of them have condition number 100.005 and the other two have exactly 1, and the number belongs to the eigenvalue rather than to the matrix.

 10¹10¹.³10¹.⁶10¹.⁹⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰⁰¹10².²00.250.50.751grid points nresidual reduction per stepJacobiGauss–SeidelV-cycleV-cycle spread, 8× in size0.003Jacobi at n = 1270.99work exponent, fitted0.079the dashed curve is cos(πh), Jacobi's closed formthe flat line is the whole method Iterating, instead of factorising

A rate that does not notice the size

The V-cycle reduces the residual by a factor of ten a cycle at fifteen points and at a hundred and twenty-seven. Jacobi on the same four problems goes from 0.981 to 0.9978, climbing towards one. One of those is a constant and the other is an exponent, and that is the whole distinction the field turns on.

 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 Two errors, and whose fault they are

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.

 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 Exact arithmetic, and what it costs instead

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.

 -12-11-10-9-8024681012log₁₀ of how nearly dependent the columns areverdicts that disagreed, of 40724103which one is correctmatrices tested200verdicts disagreed26fused was right9unfused was right17lower part: thefused build was righta sign has no last digitso a verdict has nowhere to hide The answer that depends on the machine

A matrix that is definite on one machine

Two hundred Gram matrices, two conforming builds, and twenty-six of them get different answers to "is this positive definite". The exact verdict, from determinants in BigInt rationals, says the fused build is right nine times and the other one seventeen.

 -0.0833-0.167-0.0833-0.1671-0.167-0.0833-0.167-0.0833R · A · P, normalised to a unit centrestored entries per rowlevel 0 · 31×314.87/rowlevel 1 · 15×158.22/rowlevel 2 · 7×77.37/rowlevel 3 · 3×35.44/rowlevel 4 · 1×11.00/rowstill a stencil, still annihilates a constantentries in an interior row9weight outside the 3×30row sum0the isotropic model problemnine, at every level below the first Iterating, instead of factorising

The coarse problem is a different problem

In one dimension the Galerkin coarse operator is the coarse discretisation, entry for entry — this site asserted it. In two dimensions a five-point operator produces a nine-point coarse one, so the recursion solves a different discretisation at every level below the first, and converges at 0.20 a cycle regardless.

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