Concept

Loss of orthogonality — where it appears

The drift of a computed basis away from orthogonality, which in a Lanczos recurrence collapses rather than decaying and starts when the first value converges. It begins at the moment the first eigenvalue converges, which is a mechanism rather than a drift, and it makes a converged value reappear as a duplicate.

Named by 15 essays across 4 fields — each of them below, with the objects they name alongside it.

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.

orthogonality · Orthogonality
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.

orthogonality · Gram–Schmidt
05101520253010⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹iteration‖RᵀR − I‖ and ‖r‖/‖b‖step n‖RᵀR − I‖residual30×30, run for exactly n stepsexact arithmetic would end here

An orthogonalisation nobody calls one

Conjugate gradients are derived as a minimisation and behave as an orthogonalisation, which is why the finite-termination property in every textbook is not a property the method has in floating point.

iterative · Krylov
26101418222630343810⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹stepsizeleast error: 20‖QᵀQ − I‖filter disagreementan identity with an expiry datedisagreement at step 1210⁻¹³disagreement at step 400.014least error at step20exact while the basis is orthogonaland false where the method is best

An expiry date the noise does not move

The polynomial description of conjugate gradients leaves the level of rounding at step 17 or 18 on this operator, at every noise level from 10% to 0.1%. The step worth stopping at moves from 3 to 44 across the same range. They coincide at about 1% noise, which is where the coincidence was first read, and it is a fact about the noise rather than about the method.

combination · Iterative regularisation
10¹10³10⁵10⁷10⁹10¹¹10¹³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹condition number κ(A)‖QᵀQ − I‖classical once, Householdermodified once, Householderclassical twice, Householderclassical twice, Cholesky QRκ²uκuat κ = 10⁸classical once, Householder0.0042modified once, Householder5.7·10⁻⁹classical twice, Householder3.1·10⁻¹⁵classical twice, Cholesky QR10⁻¹⁵64×16 in blocks of 4, three seedsHouseholder inside does not help between

A stable block is not a stable basis

Block Gram–Schmidt orthogonalises twice over — between blocks, and inside each one. Householder inside the blocks does not stop the classical between-block step losing orthogonality like κ², 4.2·10⁻³ at κ = 4.3·10⁷, and a second pass does not stop Cholesky QR inside the blocks breaking down at κ = 10⁸. Each level fails only on ill-conditioning placed at its own level, and one variant holds 3·10⁻¹⁵ on every placement.

orthogonality · Gram–Schmidt
10²10⁴10⁶10⁸10¹⁰10¹²10¹⁴10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹condition number κ(A)‖x̂ − x‖ / ‖x‖modified, through Qᵀbclassical, either route‖QᵀQ − I‖, modifiedHouseholdermodified, on [A b]κ²uκuforward error at κ = 10⁸Householder4·10⁻⁹modified, through Qᵀb0.13modified, on [A b]2.7·10⁻¹⁰classical840×8, three seeds, against an exact rational solveQ is not orthogonal; x is right

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.

orthogonality · Gram–Schmidt
three seeds, medianblock MGS, Qᵀb, κ 10⁸0.23block MGS, b as a block1.1·10⁻⁹Householder2.8·10⁻⁹10²10⁴10⁶10⁸10¹⁰10¹²10⁻¹⁶10⁻¹²10⁻⁸10⁻⁴110⁴10⁸κ(A)relative error of xblock MGS, Qᵀbblock CGS, either routeblock MGS, b as a blockHouseholderthe same Q in both block MGS routesonly the order in which b meets it differs

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.

orthogonality · Gram–Schmidt
‖QᵀQ − I‖ of the implied Qκ = 10⁸, Cholesky0.37κ = 10⁸, sweep8.5·10⁻⁹κ = 10¹⁰, Cholesky1.3κ = 10¹⁰, sweep3·10⁻⁷κ = 3·10¹⁰, Choleskyrefusedκ = 3·10¹⁰, sweep6.7·10⁻⁶κ = 10¹², Cholesky1.5κ = 10¹², sweep1.4·10⁻⁴the safe run is the one that failsrefusals in the range1wholly non-orthogonal returns2the pivot it refused on-6.5·10⁻¹⁷one of these outcomes is safeand it is the refusal

The licence is not the boundary

Cholesky QR is licensed by κ²u ≪ 1, which reaches equality at κ = 9.5·10⁷ in double precision. At 10⁸ the factor it returns is already 0.37 away from orthogonal, and it goes on returning factors as far as 10¹³ — refusing at scattered condition numbers in between, at different ones for eight columns and for six.

machine · Algorithm selection
κ = 10⁸, acrossQᵀb ÷ appended, ρ = 01.3·10⁷Qᵀb ÷ appended, ρ = 10.6710⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1relative residual ρforward error010⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1block MGS, Qᵀbblock MGS, appendedHouseholderdotted: κ squared times u times ρthe residual brings every route to the same term

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.

orthogonality · Gram–Schmidt
median departureone reflector at a time, formed7.5·10⁻¹⁵one reflector at a time, applied3.4·10⁻¹⁵blocks of 64, applied3.7·10⁻¹⁵10⁻¹⁵10⁻¹⁴block sizedeparture from orthogonality12481632642·10⁻¹⁵4·10⁻¹⁵6·10⁻¹⁵8·10⁻¹⁵Q formedapplied through the blocksopen dots: each of the five seedsblocking helps the formed matrix and not the operator

The factor nobody forms

A blocked Householder factorisation's orthogonal factor, multiplied out, departs from orthogonality half as far in blocks of sixteen as one reflector at a time, and that was read as blocking buying a factor of two. Libraries do not multiply it out. Applied to vectors through its stored blocks — which is how every caller uses it — the same factor departs by 3.1 to 4.0·10⁻¹⁵ at every block size from one to sixty-four, and stops growing after about twenty reflectors instead of adding them up. The factor of two was the price of forming the product, and a factor that is never formed never pays it.

orthogonality · Householder
11.5234blur 1.5, noise 1%, answerblur 1.5, noise 1%, 1.3×blur 2.5, noise 1%, answerblur 2.5, noise 1%, 1.3×blur 4, noise 1%, answerblur 4, noise 1%, 1.3×blur 1.5, noise 0.1%, answerblur 1.5, noise 0.1%, 1.3×blur 2.5, noise 0.1%, answerblur 2.5, noise 0.1%, 1.3×blur 4, noise 0.1%, answerblur 4, noise 0.1%, 1.3×the power m, median of twelve drawsthe iteration's, from its Ritz valuesthe operator's singular values above the same cornerthe blur width's closed form, same cornershaded: within 0.1 of the iteration's mthe operator does not set the power

A power the spectrum does not set

Conjugate gradients' filter past the answer is a ramp raised to a power m between 1.27 and 1.50, read off its own Ritz values, and the power was predicted to be the blur's: computable from the decay of the singular values to within 0.1, so the iteration's filter could be built without running it. It is not. Summed over the singular values above the iteration's own corner, the decay gives anything from 0.76 to 4.3, because the Ritz values are not those singular values — fewer of them early in the run, each standing for a cluster, and more of them late, the leading ones found again and again. The power is the spacing between the smallest Ritz value and its two neighbours. And nothing depends on it: placed by GCV's first minimum, the power ramp at every m from 1 to 3 lands within a tenth of its best on 63 to 66 draws of 72.

combination · Iterative regularisation
024681012141618024681012141618distinct eigenvalues in the spectrumstep the recurrence stops atthe step is m, not nn = 30 throughoutspectra drawn8every one breaking at m8worst residual at the breakdown5.6·10⁻¹⁶smallest gain over the step before3.5·10¹⁰an invariant subspace contains the answerand its dimension is what the method costs

The zero that means it is finished

Every Krylov method ends by dividing by a number the previous step produced, and when that number is zero the recurrence stops. In Arnoldi the stop is the answer — the subspace has closed, the solution is inside it, and the residual is at the unit roundoff. The literature calls it a lucky breakdown, and the adjective is doing real work.

iterative · Breakdown
10203040506010¹10²squared singular values above the cornerRitz values heldshort recurrencereorthogonaliseddashed: one Ritz value per singular valuewithout copies, still not one each

The copies cost steps, not the filter

Past the answer, conjugate gradients holds more Ritz values than singular values above its corner, because rounding lets it find the leading ones again and again. Those copies were a suspect for why the filter's power could not be read off the spectrum. Reorthogonalising every basis vector removes them, and the run reaches the same iterate — the same best error to a per cent — in up to four times fewer steps. The power barely moves, the operator's sum still misses it on 62 draws of 72, and without copies there are still fewer Ritz values than singular values above the corner: the clusters, not the copies, broke the prediction.

combination · Iterative regularisation
012345678910⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹rank budgetresidual the run stalls atb with no structure at allb constant: the answer is a traintwo ladders, one truncationstructured, rank 10.12structured, rank 45.5·10⁻⁶structured, rank 85.9·10⁻¹⁴unstructured, rank 10.97unstructured, rank 80.47the floor is not the truncation'sit is the answer's

A run that is over at step five

A conjugate gradient whose every iterate is cut to a rank budget reaches the floor that budget allows at step 5, 36, 42 or 59, and then does nothing for the rest of the run. Four times the iterations move the floor by a factor of 1.8, and past the answer's own rank they move it the wrong way.

iterative · Low-rank iteration
22.32.62.910⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised832storage, second-order416Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations

The answer that arrives when the space runs out

A second-order Krylov recurrence holds vectors of length n for a problem with 2n eigenvalues, so it is exact at n steps where the linearised route needs 2n. The machine-precision reading at forty-four vectors on a chain of forty is that exhaustion rather than convergence, and it arrives through a basis whose ‖QᵀQ − I‖ is above one.

iterative · Krylov

Named alongside it

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

Gram–SchmidtHouseholder reflectionOrthogonalityKrylov subspaceCondition numberConjugate gradientsReorthogonalisationResidualRitz valuesBlock methodsExact ground truthFilter factors

All concepts