Numerical Linear Algebra
where the algebra stops being the arithmetic
Field

Orthogonality, measured

Orthogonal is not an adjective, it is a number: ‖QᵀQ − I‖. Two algorithms that are the same algebra written in a different order return 10⁻¹⁵ and 1 for it on the same matrix — and the one that fails still reconstructs the matrix perfectly, which is why nothing warns you.
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.

 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.

 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.

 how far is it from A to an orthogonal matrix?smaller is nearer · the polar factor minimises this in every unitarily invariant normpolar factor U1.8554QR, signs fixed2.1265QR as returned3.8226200 drawn at randomκ = 10polar factor1.9QR, signs fixed2.1QR as returned3.8best of 200 random2.7‖A − QR‖ is the same either wayand ‖A − Q‖ is not

The nearest orthogonal matrix

Every field that has to clean up a drifted rotation reaches for QR, and QR does not answer the question. The nearest orthogonal matrix is the orthogonal factor of the polar decomposition — nearer by about a tenth, and, more to the point, the same matrix whatever order the columns were written in. QR's answer changes completely.

 159131721252910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps7Newton–Schulz, steps28Newton at step 657scaled Newton at step 64.4·10⁻¹³a Newton step needs an inverseand a Schulz step needs two products

An iteration that only multiplies

Newton's iteration for the polar factor needs an inverse every step. Newton–Schulz needs only matrix products — nothing that reads an entry, nothing that pivots — and it converges if and only if every singular value is below √3. At 1.73205 it converges and at 1.73206 it returns an orthogonal matrix that is not the answer, with a residual of 5·10⁻¹⁶ and nothing to say so.

 κ(A) = 10⁶ throughout · κ(H) = 100 · the answer is the same answer for every basisorthonormal — κ(Z)1κ(ZᵀHZ)25.6relative error1.07·10⁻¹⁵first m basic — κ(Z)1.99·10⁸κ(ZᵀHZ)3.8·10¹⁶relative error0.0518pivoted basic — κ(Z)2.06κ(ZᵀHZ)31.9relative error6.71·10⁻¹⁶what the choice costsdensity, orthonormal1density, fundamental0.5κ(ZᵀHZ) ÷ κ(Z)², naive0.96error, pivoted choice6.7·10⁻¹⁶every one of them is a basisand one of them loses fourteen digits

The basis nobody chose on purpose

A method that eliminates a constraint has to pick a basis for its null space, and every basis is correct. Their condition numbers are eight orders apart, the reduced problem inherits the square, and the choice is usually made by a one-line rule nobody thought of as a numerical decision.

 polar: U₁P − U₂QR: Q₁P − Q₂-3.3·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶-1.9·10⁻¹⁶5.6·10⁻¹⁷-1.7·10⁻¹⁶6.7·10⁻¹⁶-8.9·10⁻¹⁶2.8·10⁻¹⁶-5.6·10⁻¹⁶-3.6·10⁻¹⁶6.1·10⁻¹⁶-8.3·10⁻¹⁶8.3·10⁻¹⁶-7.2·10⁻¹⁶-10·10⁻¹⁶2.2·10⁻¹⁶4.4·10⁻¹⁶1.7·10⁻¹⁶-4.4·10⁻¹⁶-5.6·10⁻¹⁷5.6·10⁻¹⁷0-7.2·10⁻¹⁶-4.4·10⁻¹⁶3.9·10⁻¹⁶5.6·10⁻¹⁷2.8·10⁻¹⁶-1.1·10⁻¹⁶-3.3·10⁻¹⁶2.2·10⁻¹⁶1.1·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶-2.8·10⁻¹⁶5.6·10⁻¹⁶0.67-0.35-0.23-0.620.19-0.72-0.770.170.20.290.190.93-0.180.55-0.0290.06-0.460.11-0.17-0.61-0.230.38-0.55-0.380.750.870.661.1-0.36-0.530.42-0.670.180.180.52-0.3Frobenius norms, columns reordered‖U₁P − U₂‖2.8·10⁻¹⁵‖Q₁P − Q₂‖3‖Q‖, for scale2.4κ of the matrix100Frobenius distance between the two answers, on one scale0 to 4polar2.8·10⁻¹⁵QR3.048‖Q‖ = 2.449the column space did not moveand one of the two answers did

A test with no answer in it

A caller with no reference answer can still ask whether a routine answered the right question: reverse the columns, run it again, compare. The polar factor's two answers agree to 10⁻¹⁵ at every conditioning drawn; a QR's differ by 2.353 on matrices whose own norm is 2.449. The test has a floor, and the floor is measurable too.

 110¹10²01020304050noise ÷ thicknesstrials mirrored, %a coin: 50%t = 10⁻²t = 10⁻³t = 10⁻⁴per cent mirroredσ/t = 3, mean of three7.7σ/t = 10, mean of three34σ/t = 100, mean of three4720 points, 400 trials a stop, one seed per thicknessthe ratio decides, not the thinness

A rotation that comes back mirrored

Align twenty noisy points and the nearest orthogonal matrix to the answer is a reflection in 7.7 per cent of trials at noise three times the set's thickness and a third of them at ten — at thicknesses of 10⁻², 10⁻³ and 10⁻⁴ alike. The determinant fix is never a small correction. It moves the answer by exactly 2, it costs exactly 4σ₃ of residual, and it leaves the rotation's error at half the noise however thin the set becomes.

 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.

 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.

 110¹10²10³10⁴10⁵10⁶10¹10²10³10⁴10⁵spread of resistances, largest ÷ smallest possiblecondition number, unit diagonalleast-resistance treebreadth-first treegreatest-resistance treenode equationsspread resistances make the loops easyand the nodes hard

Spread resistances make the loops easy

Scaled to a unit diagonal, the loop equations on the least-resistance tree get easier as a network's resistances spread — from 120 to 5.44 over six decades — and stop depending on the grid's size, while the node equations of the same flow get harder, from 538 to 4.6·10⁴. The spread that ruins the range-space formulation rescues the null-space one, though the loops' density means the work saved is a factor of two, not the factor of nine the iteration counts suggest.

 groundarcs of the treearcs off the treethe loop worst servedleast-resistanceκ(Z)11κ(ZᵀHZ)143κ, unit diagonal9.3nonzeros in Z426longest loop, arcs26worst path ÷ own arc0.97every tree gives a basis of whole numbersthe resistances decide which one to want

The tree the resistances choose

On a network every basic set is a spanning tree and every null-space basis is a set of loops with entries 0 and ±1, so no tree can make Z badly conditioned. The tree with the best-conditioned Z still gives loop equations 4.8 times worse than the tree of least resistance: the basis has to be chosen against the Hessian, and pivoting finds it only when it pivots on the resistances too.

 8×8, κ = 10⁸direction at ε = 10⁻², ‖QᵀQ − I‖1.5·10⁻¹⁵scalar at ε = 10⁻², ‖QᵀQ − I‖0.065direction at ε = 10⁻², residual0.00510⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²relative perturbation εdeparture‖QᵀQ − I‖, scalar‖A − QR‖/‖A‖, scalar‖A − QR‖/‖A‖, direction‖QᵀQ − I‖, directionflat line: a different reflection is still a reflectionsloped line: a non-reflection is not

One number that has to be right

Householder's orthogonality was called structural: a reflection is built from a unit vector, so rounding the vector names a different reflection rather than a broken one. Tested by breaking it, the claim is narrower and sharper. Perturb every component of the reflector by a relative 10⁻², and ‖QᵀQ − I‖ stays at 1.5·10⁻¹⁵ while the factorisation moves to 5·10⁻³. Perturb the one stored scalar by the same amount and ‖QᵀQ − I‖ is 6.5·10⁻². The structure is one degree of freedom, and the departure is four times its relative error.

 what a block storesnumbers in T, block of 11numbers in T, block of 16136‖QᵀQ − I‖ there3.9·10⁻¹⁵10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶block size‖QᵀQ − I‖124816T perturbed by 10⁻⁶T as computed1, 3, 10, 36, 136 computed numbers in the triangleand a factor of five in what they produce

A triangle where the scalar was

Every level-3 QR assembles a block of reflectors into Q = I − Y T Yᵀ, and T is computed by a recurrence whose inputs are its own previous columns. A block of sixteen carries 136 computed numbers where sixteen separate reflections carry sixteen. The orthogonality it produces is 3.9·10⁻¹⁵ against the single reflector's 7.8·10⁻¹⁶ — a factor of five for a hundred and thirty-six times as many things that have to be right.

 10⁻¹101020304050z, the effective noise over thicknesstrials mirrored, %5 points, equal20 points, equal80 points, equal● five precise, 1/σ²■ five weighted ×100per cent mirrored at z = 0.45, 1,000 trialstwenty points1.8eighty points1.3five points9.5five precise, weighted 1/σ²7.9equal noise, five weighted ×1006.8both weighted schemes have five effective pointstwenty and eighty points share one curve

Five precise points are five points

Weighting each sighting by its reliability is the standard form of an attitude or registration fit, and it changes how often the nearest orthogonal matrix comes back as a mirror. Measured, the rate is a function of two numbers: the weighted noise over thickness, and the effective count (Σw)²/Σw². Five points with a tenth of the noise, weighted by 1/σ², carry the information of 515 equal points and mirror like five — 7.9 per cent at a noise ratio where twenty points mirror 1.8 and five mirror 9.5. The √m the earlier measurement left unchecked is right, and it counts what carries the thin direction.

 0123401020304050z, the effective noise ratiotrials mirrored, %1 thin: 1 × 1 model2 thin: 2 × 2 model3 thin: 3 × 3 modelper cent mirrored at z = 0.7, 600 trialsn = 3, 1 thin, 20 points9.7n = 5, 1 thin, 20 points13n = 10, 1 thin, 40 points12n = 5, 2 thin, 20 points23n = 10, 2 thin, 40 points24n = 8, 3 thin, 40 points32points: measured in n dimensionslines: a k × k determinant, no n in it

A mirror decided in the thin directions

In n dimensions the nearest orthogonal matrix to a noisy alignment is still sometimes a reflection, and the rate at which it is does not depend on n. Three, five and ten dimensions with one thin direction mirror alike; two thin directions mirror like each other in five dimensions and in ten. The rate is the chance that a k × k matrix built from the k thin directions has a negative determinant — 21.7 per cent at a noise ratio of 0.7 for k = 2, measured at 23.0 — and it is well above k independent coin flips. With two or more thin directions the determinant correction still fires, and it no longer rescues the rotation: the answer is eleven noise-widths off whether or not it was mirrored.

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