Numerical Linear Algebra
where the algebra stops being the arithmetic
Depth

Ladders

A field says what an essay is about. A ladder says what else there is to say about it — the distinct arguments that stand against one idea, from the one that introduces it to the one that assumes all the others.
the problem you posedA = H10b = A·(1, 2, …, 10)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.3·10⁻¹⁷the answer you wantedx = (1, 2, …, 10), exactlythe answer you gotx̂, wrong by 2.7·10⁻⁴ relativebackward error 2.3·10⁻¹⁷forward error 2.7·10⁻⁴κ = 1.6·10¹³κ · η = 3.6·10⁻⁴, and the measured forward error is 2.7·10⁻⁴.The algorithm is not at fault. The problem is.H10, LU with partial pivotingresidual and error differ

The exact answer to a nearby problem

A good algorithm does not give an approximate answer to your problem. It gives the exact answer to a problem very close to yours — and once that is the definition, a wrong result has two possible authors and they can be measured apart.

2 rungs · error the matrix, lower triangle408 entriesits Cholesky factor1739 entries · 1331 created→‖A − LLᵀ‖/‖A‖1.4·10⁻¹⁶fill, symbolic1331fill, numeric1331n = 144 · density 3.2% · bandwidth 12same matrix, renumberedthe answer is identical to rounding

The factor is not sparse

A sparse matrix has a factor that is not sparse, and the gap between them is the entire reason iterative methods exist. The entries elimination creates can be counted before any arithmetic runs, from the graph alone.

2 rungs · sparsity 0408012016020024010⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹iteration‖e‖_A / ‖e₀‖_Ameasuredκ bound119 steps40×40, spectrum spread evenly in logbound permits 1417

The rate the condition number predicts

Conjugate gradients converge at a rate governed by the square root of the condition number. That is a bound rather than an estimate, it is provable, and it is loose enough that provisioning iterations from it wastes nine out of ten.

2 rungs · iterative 012345610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹refinement step‖x − x*‖ / ‖x*‖a full double-precision solveresidual in24-bitresidual indoubleone argument apartκ·u of the factorisation6·10⁻⁴double residual, final3.2·10⁻¹³same-precision, final1.3·10⁻⁴30×30, κ = 10⁴, same factors in both runsidentical cost

Buying the accuracy back

Factorise in single precision, then correct the answer using residuals computed in double, and the result is what a full double-precision solve would have given. Compute those residuals in single instead and the identical algorithm, at identical cost, recovers nothing.

2 rungs · arithmetic 0357010514017521024528010⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹iteration|subdiagonal entry|no shiftRayleighWilkinsontwo routes to one raterate, from the spectrum0.9rate, measured0.9iterations, none / Wilkinson45symmetric 4×4, spectrum 8, 4, 2, 1.8the dashed line is the prediction

The algorithm the libraries actually run

Factorise, multiply the factors back in the other order, repeat. That description is complete and correct and produces something nobody would use — on a matrix with eigenvalues +1 and −1 it does not converge at all, and the subdiagonal entry does not move by so much as a rounding error.

2 rungs · spectra 04812162010⁻¹10⁻⁰.⁵1target rank k‖A − A_k‖₂published boundrandomisedσ_{k+1}, optimalhow far apart the three areworst seed spread1.6bound / median at k = 125.9median / optimum at k = 121.960×60, 6 seeds, oversampling p = 5band is best to worst

A bound that holds with probability

Every other guarantee on this site is deterministic. The randomised low-rank approximation offers one that holds with a probability, the seed changes the answer, and the honest figure is a band rather than a line.

2 rungs · randomised 10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹xrelative error of the computed value(1 − cos x)/x², as written2 sin²(x/2)/x²no digits left at alldouble precision throughoutone function, two spellings

Cancellation takes the answer, not a digit

Subtracting two nearly equal numbers is exact. That is what makes it dangerous — the subtraction introduces no error at all, it exposes error the operands were already carrying, and the exposure can consume every significant figure at once.

1 rung · arithmetic 110¹10²10³10⁴10⁵10⁶10⁷00.250.50.751amplification of the input perturbationfraction of directions at or belowκ = 10·10⁵worst found 7.6·10⁵6×6, 200 directionsmedian reaches 0.29 of κ

The condition number is an amplifier

κ is usually introduced as a definition and then quoted. It is a measurement: perturb the input by a known amount, look at how much the output moves, and the largest ratio you can find is the number.

1 rung · error 10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹size of the perturbation ‖δA‖how far the eigenvalues moveJordan block, ε^(1/8)symmetric, ≤ ‖δA‖rounding error alone moves it to 10⁻²six seeds per symmetric point; Jordan is closed formsymmetry beats precision

Symmetry is worth more than precision

A symmetric matrix gives up its eigenvalues to full accuracy however ill-conditioned it is. An unsymmetric one can move them by the eighth root of a perturbation, so the rounding involved in merely storing the matrix shifts the spectrum by a hundredth.

1 rung · spectra 21-13-3-121-212-443-12as givenrows in the order 1 2 3 443-1201.251.252.502.51.5-30-0.5-0.52after step 1pivot 443-1202.51.5-3000.5400-0.21.4after step 2pivot 2.543-1202.51.5-3000.540003after step 3pivot 0.5‖PA − LU‖/‖A‖0largest multiplier0.75row order 4 3 2 1the pivot is chosen

Elimination is a sequence of choices

Gaussian elimination is taught as a procedure with no decisions in it. There is one decision at every step — which row to use — and every stability property the algorithm has comes from making it well.

1 rung · elimination H13 x = b, b formed exactly so that x = (1, 2, …, 13)12345678910111213exact1.00002.00003.00133.97865.18864.993010.46720.048921.2657-2.575319.21478.906013.5113computed6.6 correct digits4.8 correct digits3.4 correct digits2.3 correct digits1.4 correct digit0.8 correct digitno correct digitsno correct digitsno correct digitsno correct digitsno correct digits0.6 correct digit1.4 correct digitbackward error2.2·10⁻¹⁷κ = 1.7·10¹⁸The algorithm solved a neighbouring problem perfectly. That problem's answer is this one.right-hand side built in BigInt rationalsthe truth is known

An answer that is known

Almost every demonstration of numerical error estimates the error by computing the same thing more carefully. The Hilbert matrix does not need that: its inverse is a closed form in integers, so the true answer is available exactly and the error is measured rather than approximated.

1 rung · error 10⁻⁴10⁻³10⁻²10⁻¹110¹10²10³10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹relative change in the coefficients, along the worst directionrelative increase in the residualcoefficients doubled39% change, fit unmoved in the sixth digit308×: the third digit movesκ(A) = 3.6·10⁶. Exact arithmetic would pick one point on this floor. It would not raise it.24 points, degree 9, monomial basisthe data leaves them free

The valley with no bottom

A degree-nine fit's coefficients can be moved by a third of their own size before the residual changes in the sixth significant figure. The arithmetic did not lose those digits. The data never contained them.

1 rung · leastsquares [½, 1)[1, 2)[2, 4)0.5124gap 0.125gap 0.25 — twice as wide8 values per octavespacing doubles at each power of two

What a float can hold

The representable numbers are not a fine fuzz spread evenly over the line. They are evenly spaced inside each power-of-two interval and twice as far apart in the next one up, and almost everything else in this subject is a consequence of that one fact.

1 rung · arithmetic 02468101210⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1iteration‖r‖ / ‖b‖Laplaciancyclic shiftno progress at allevery eigenvalue of the shift is on the unit circleand it predicts nothing

The spectrum that predicts nothing

For a symmetric matrix the eigenvalues govern how fast an iteration converges. Drop symmetry and they stop governing anything — there is a matrix whose eigenvalues are as evenly spread as eigenvalues can be, on which GMRES makes no progress at all until the last possible step.

1 rung · iterative for each previous column i, subtract the projection of column j onto qᵢclassicalr[i][j] = qᵢ · a[j] ↑ the ORIGINAL columnv = v − r[i][j] · qᵢthe three worst |qᵢ · qⱼ|:columns 7 and 8: 1columns 6 and 8: 0.13columns 6 and 7: 0.13modifiedr[i][j] = qᵢ · v ↑ what is LEFT of itv = v − r[i][j] · qᵢthe three worst |qᵢ · qⱼ|:columns 1 and 8: 4.4·10⁻⁷columns 2 and 8: 2.7·10⁻⁷columns 3 and 8: 2.4·10⁻⁸The two R factors agree to 1.2·10⁻⁶ relative. The two Q factors do not.the 8×8 Hilbert matrixone word, eight orders

Two Gram–Schmidts

One argument changes. Classical Gram–Schmidt projects the original column onto each previous direction; modified projects what is left of it. In exact arithmetic the coefficients are identical. In floating point they differ by eight orders of magnitude in the thing that matters.

1 rung · orthogonality 0816243240110²10⁴10⁶10⁸10¹⁰10¹²10¹⁴matrix size ngrowth factor max|u| / max|a|the 2ⁿ⁻¹ boundworst of 30 randommedian randomWilkinson's matrix sits on the bound30 Gaussian matrices per sizeat n = 40: bound 5.5·10¹¹, worst 4.8

The bound that is never attained

Partial pivoting's stability guarantee permits the entries to double at every step — a factor of 5.5·10¹¹ at n = 40. The measured growth on random matrices of that size is about three. The gap is eleven orders of magnitude, and the guarantee is still worth having.

1 rung · elimination the mirrorx, length 4.000Hx = (-4.000, 0)v = x − αe₁safe sign: α = −‖x‖, so v is formed from a sum and nothing cancelsunsafe sign: α = +‖x‖ gives ‖v‖ only 35.1% of ‖x‖ + ‖x‖ — the digits go‖HᵀH − I‖5·10⁻¹⁶‖Hx‖ − ‖x‖8.9·10⁻¹⁶second component2.2·10⁻¹⁶built from a unit vectororthogonality is structural

A reflection cannot stop being one

Householder QR holds orthogonality at 10⁻¹⁵ whatever the condition number of the matrix, and Gram–Schmidt does not. The reason is not that it is more careful. It is that its Q is built from unit vectors, and rounding a unit vector gives a different reflection rather than a broken one.

1 rung · orthogonality everything Ax can reachb = (1.1, 0.4, 1.5)Ax, the closest reachable pointr = b − Ax‖Aᵀr‖ / (‖A‖‖r‖)1.7·10⁻¹⁶‖b‖² − ‖Ax‖² − ‖r‖²1.3·10⁻¹⁵‖r‖1.3200 random nearby points of the plane were tried; none is closer.a 3×2 system, Householder QRperpendicularity is checked

The projection and the right angle

The least-squares solution is the one whose residual is perpendicular to everything the columns can reach. That is not a mnemonic — it is an equation, Aᵀr = 0, and the computed answer satisfies it to 10⁻¹⁶.

1 rung · leastsquares 10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹10¹ε in Läuchli's matrix (smaller ε, larger κ)relative error in the coefficientsAᵀA exactly singularnormal equationsQRκ from 1.7·10⁸ to 17the cliff is at √u = 2.4·10⁻⁴

The road that squares the problem

The normal equations are the first method every course teaches and the method no library uses. Forming AᵀA squares the condition number, and below ε = √u it does not degrade — it produces a matrix that is exactly singular, from data that was perfectly usable.

1 rung · leastsquares natural1739reverse Cuthill–McKee1354minimum degree1026nested dissection1413matrix: 408 entries · dense factor: 10440bandwidth 12 · 4.26× the matrixbandwidth 12 · 3.32× the matrixbandwidth 123 · 2.51× the matrixbandwidth 108 · 3.46× the matrixn = 144, five-point stencilevery ordering fills in; none avoids it

The order decides the memory

Four elimination orderings on one matrix give factors of 1,739, 1,354, 1,413 and 1,026 entries. All four factorisations are exact, all four return the same answer, and the one with the better asymptotics is not the one that wins.

1 rung · sparsity A = H8 · κ = 1.5·10¹⁰ · both factorisations reconstruct A to 6·10⁻¹⁷the diagonal is 1 in both — every column is a unit vector either way1.000000000001.000000000001.000000000001.000000000001.00000.002-0.002000001.0000.125-0.13300000.0020.1251.000-1.0000000-0.002-0.133-1.0001.000classical Gram–Schmidt1.000000000001.000000000001.000000000001.000000000001.000000000001.000000000001.000000000001.000Householderclassical ‖QᵀQ − I‖1.4Householder ‖QᵀQ − I‖1.4·10⁻¹⁵largest off-diagonal 1 against 3.1·10⁻¹⁶length is not angle

Orthogonal is a number

"Q is orthogonal" is a claim about a measurable quantity, ‖QᵀQ − I‖, and on the eight-by-eight Hilbert matrix two standard algorithms return 10⁻¹⁵ and 1 for it. The one that returns 1 still reconstructs the matrix perfectly, which is why nothing warns you.

1 rung · orthogonality [ ε 1 ; 1 1 ] x = [ 1 ; 2 ], exact answer (1.000000, 1.000000)with partial pivoting1101U after elimination1.0000001.000000computed xbackward error 0forward error 0without10⁻¹⁷10-1·10¹⁷U after elimination0.0000001.000000computed xbackward error 0.25forward error 0.71no error is raisedgrowth 10¹⁷

The swap that is not optional

Run elimination without a row interchange on a matrix that needs one and nothing announces a failure. There is no division by zero, no warning, and an answer of the right shape. It is simply wrong, and how wrong depends on a number you did not look at.

1 rung · elimination 051015202530354010⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹iteration‖r‖ / ‖b‖plain CGIC(0) CGwhat the preconditioner didκ(A)48κ(L⁻¹AL⁻ᵀ)5.1‖A − LLᵀ‖/‖A‖0.0832D Laplacian, n = 100√κ ratio predicts 3.07×

Changing the condition number on purpose

Preconditioning is usually introduced as a trick that makes an iteration converge faster. It is not a trick. It is solving a different system with the same solution and a condition number chosen rather than inherited, and the new condition number is computable.

1 rung · iterative 1234567891010⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1indexsingular valuecutoff, σ₁ · 10⁻¹⁰numerical rank 10gap 8.2·10⁶an opiniontrue rank 410×10, built with 4 nonzero valuesrank is a decision

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.

1 rung · spectra 10²10².³10².⁵⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁶10⁻¹10⁻⁰.⁵1rows in the sketchworst relative distortiondimension 64dimension 2565 seeds per point, band is best to worstthe dimension does not appear

The dimension does not appear

A random projection preserves the lengths of a set of vectors to within a distortion that depends on how many vectors there are and not on how many coordinates each one has. That is the fact the whole field rests on, and it is genuinely surprising.

1 rung · randomised 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

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.

1 rung · iterative 10¹10²10³10⁴10⁵10⁶10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹number of terms addedrelative error against the exact sumin orderin a treecompensatedbinary32 · terms are 1/icompensated: 3·10⁻⁸

The order they are added in

Addition is associative in the algebra and is not associative in the arithmetic. The same million numbers, added in a different order, give answers that differ in the third significant figure — and the fix is not a wider float, it is a different order.

1 rung · arithmetic 12345678910⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1rank k of the approximation‖A − Aₖ‖measured, 2-normσₖ₊₁, from theorymeasured, Frobeniusthe first two agreeto 4.3·10⁻⁹worst |‖A−Aₖ‖₂ − σₖ₊₁| / σₖ₊₁4.3·10⁻⁹worst Frobenius discrepancy4.3·10⁻⁹κ = 10⁹; 30 random rank-3 matrices, none closerthe error is σₖ₊₁

The best approximation there is

The error of the best rank-k approximation is not bounded by the next singular value. It is equal to it. That is an unusually sharp theorem, and it makes the theorem itself usable as an independent check on the computation.

1 rung · spectra

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