Numerical Linear Algebra
where the algebra stops being the arithmetic
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 11 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 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 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–SchmidtKrylov subspaceOrthogonalityCondition numberHouseholder reflectionReorthogonalisationResidualConjugate gradientsFinite terminationBackward stabilityBlock methodsCholesky QR

All concepts