Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Modified Gram–Schmidt — where it appears

Orthogonalising a vector against each earlier one in turn, using the partially corrected vector each time. It is the same algebra as the classical version and returns an orthogonality of 10⁻¹⁵ where that returns 1.

Named by 6 essays across 2 fields — each of them below, with the objects they name alongside it.

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 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 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 a generic tall matrixHouseholder: aimed over median8b appended: aimed over median11column MGS: aimed over median1110⁻⁴10⁻³10⁻²10⁻¹1error ÷ κ²uρthe boundHouseholderb appendedcolumn MGSdots: random directions · bar: aimed · ring: the fixed directionaiming closes a factor of seven

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.

orthogonality · Gram–Schmidt 051015202530354010⁻³10⁻¹10¹GMRES steprelative residualthe two curves are the same curvea number re-derived, not carriedworst reported/actual factor1.8at step40‖VᵀV − I‖ of the basis1.4reported at the last step0.041actual at the last step0.074the same family of methodsand only one of them lies

The number that is re-derived

GMRES prints a residual it never computes from its answer either. On the matrix that sends a conjugate gradient recurrence 7.3·10¹⁰ wrong, and on two others chosen to be worse, its number is never more than a factor of 2.86 out — while the basis it is computed from has lost orthogonality entirely. The disease is not iterative methods, and it is not floating point.

iterative · Residual gap 33.310⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised2080storage, second-order1040Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations

A Krylov space for a problem that is not linear

A quadratic eigenvalue problem has no matrix to build a Krylov space out of. The recurrence that builds one anyway stores half as many numbers, returns twice as many Ritz values — and stops being a basis at twenty vectors while the answer it gives keeps improving.

iterative · Krylov

Named alongside it

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

Householder reflectionOrthogonalityCondition squaringExact ground truthGram–SchmidtLeast-squaresLoss of orthogonalityBackward errorKrylov subspaceResidualArnoldiArnoldi iteration

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