Numerical Linear Algebra
where the algebra stops being the arithmetic
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The thread: Precision as the knob

The decompositions here are written against an arithmetic context rather than against plus and times, so the same code runs at any significand width from sixteen bits to fifty-three. Where a figure has a slider, the slider is usually the mantissa — and every frame it can reach was generated, and asserted, at build time.
[½, 1)[1, 2)[2, 4)0.5124gap 0.125gap 0.25 — twice as wide8 values per octavespacing doubles at each power of two The arithmetic underneath

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.

 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 Least squares, and the road not to take

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.

 [ 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 Elimination, and the swap

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.

 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 Orthogonality, measured

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.

 0369121501122334455digits asked for, −log₁₀ εcolumns keptwhat the geometry promiseswhat the matrix costsa rank is a number of digitscolumns a decade0.55bound, a decade3.3rank at 10⁻⁸5bound at 10⁻⁸28q0.5the shape is rightand the constant is not Neither sparse nor dense

A rank that is a number of digits

Ask a kernel block for two digits and it costs two columns; ask for fourteen and it costs nine. The curve is a straight line at 0.55 columns a decade, and the bound the geometry gives is a straight line too — at 3.32, which is the same shape and six times the price.

 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 Eigenvalues, singular values, rank

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.

 21018263442500102030405060708090100110120130140150160170180190200210220230240250260270280290300310320330340significand bitsiterationspreconditioner roundedarithmetic roundedno preconditionersame bits, different casualtyerror, 3-bit preconditioner8.8·10⁻¹³error, 3-bit arithmetic0.16‖A − LLᵀ‖/‖A‖ of the factor0.083a direction may be roundeda measurement may not Methods that were designed apart

The part of a solver that may be rounded

A preconditioner computed and applied with a three-bit significand still returns thirteen correct digits — it costs seventeen extra iterations and nothing else. Round the working arithmetic instead and the step count barely moves while the answer loses exactly the digits the format dropped.

 0246810⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹log₁₀ γ, the change of unitsforward erroras writtenafter scalingone change of variableunscaled, worst0.0013scaled, worst1.7·10⁻¹³orders recovered10scaled coefficient spread4.5the answer was never the problemthe units were The eigenvalue problem that is not linear

The scaling that buys ten orders

Two lines computed from three norms, a change of variable that is exact in both directions, and the whole of the loss the previous essay measured comes back — flat, at every stop, because after scaling every stop is the same problem.

 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 Two errors, and whose fault they are

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.

 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 arithmetic underneath

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.

 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 The arithmetic underneath

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.

 10¹10³10⁵10⁷10⁹10¹¹110¹10²10³10⁴condition number of the matrixgrowth factorbound 2^11partial pivotingCholeskyno pivot to gain fromCholesky growth, every κ1Cholesky interchanges0partial pivoting, worst9the bound, 2^112048both eliminations reach the same growthand only one of them had to swap to get there Elimination, and the swap

A factorisation with nothing to pivot for

Cholesky's growth factor is not bounded by one. It is equal to one, at every size and every condition number, and the two-line reason is why the algorithm needs no pivoting at all — not "usually gets away without it". Its only failure is the square root of a non-positive number, which is exactly the test for definiteness, and in floating point that test moves with the precision.

 10¹10²10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰110⁴10⁸10¹²10¹⁶condition number κ(A)× short of a double solvebf16fp16fp32bf16fp16fp32the reference is soundreference solve, worst backward error10⁻¹⁶bfloat16 threshold κ256fp32 threshold κ1.7·10⁷eight refinement steps, residual always in doubleflat at 1 means it reached double The arithmetic underneath

Where the hardware went

bfloat16 carries eight mantissa bits, which puts its refinement threshold at a condition number of 256. That is not an exotic matrix. It is an ordinary one, and past it the method still improves the answer by a factor of four hundred while getting nowhere near a usable one.

 3456789101100.250.50.751bits in the exponent fieldeach curve as a fraction of its own maximumbfloat16fp16rangeprecisionto 617 decadesto 3.9 digitswhat the split buysbfloat16: largest number3.4·10³⁸fp16: largest number6.6·10⁴a bit of exponent doubles the rangea bit of significand adds a third of a digit The arithmetic underneath

The other half of a format

fp16 and tf32 have the same eleven significand bits and their largest numbers are 65,504 and 3.4·10³⁸. For two phases this site simulated the significand alone, so it was obliged to report them as the same format — which is a claim, and a false one.

 10⁻²10⁻¹110¹10²10³10⁴10⁻⁴10⁻³10⁻²10⁻¹1backward error, in units of ushare of systems above ituCramereliminationCramer, control24-bit arithmeticworst Cramer, in u395worst elimination, in u1.2control, worst Cramer4.9κ of the worst system3.4·10⁶one derivation, two computationsand only one of them is stable Elimination, and the swap

A rule that is correct and unusable

Cramer's rule gives every component of the solution in closed form, in terms of determinants, and it is a theorem. On two-by-two systems whose rows are nearly parallel it returns an answer with a backward error of 458 units of roundoff where elimination returns 1.3 — on a matrix whose condition number is 32,000 and which elimination solved perfectly.

 -9-7-5-3-1110⁵10⁶10⁷log₁₀ ε of the preconditionermultiplications for the whole solveno preconditioner: 41 stepsiteration count, rescaledmultiplications spenttwo curves, opposite waysκ21steps, no preconditioner41cheapest ε0.5its rank1tightest ⁄ cheapest2.4the count is what is printedand the cost is what is spent Neither sparse nor dense

The accuracy worth paying for

Used as a preconditioner, a hierarchical representation gets better at every accuracy — the iteration count falls monotonically all the way to the tightest tolerance. The total work does not. Its minimum sits at a rank-one preconditioner on an easy problem and six decades further along on a hard one.

 10⁻³10⁻¹10¹10³01magnitude448NaN — no ∞0.0156 — the smallest normaldrawn from the format's own rulespositive finite values126largest finite value448worst round-trip error0the subnormals are the evenly spaced ticks at the lefteverything a byte can be The arithmetic underneath

Eight bits, and a format that breaks the rules

E4M3 reuses the exponent code IEEE reserves for infinities, so it reaches 448 where the same bits under IEEE's rules would reach 240 — and has no infinity left to signal an overflow with. The same computation is a NaN on one conforming device and 448 on another.

 10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²ε the blocks were compressed at‖A − LLᵀ‖ ⁄ ‖A‖the factorisationthe knob, on a factorisationtruncations34residual at 10⁻⁴2.2·10⁻⁵residual at 10⁻¹⁰9.9·10⁻¹²slope1worst ratio to the representation1.8ten decadesand a slope of one Neither sparse nor dense

A knob calibrated in residuals

A formatted Cholesky has two numbers in it and only one of them is an accuracy. Across twelve trees — three sizes by four leaf sizes — the leaf moves the truncation count from 0 to 258 and moves the ranks of the blocks not at all, while the residual follows the tolerance at slopes between 1.022 and 1.046 and sits at about a tenth of it throughout.

 significand bits10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰κ(A)87892634456329625911109435761199127514162024every matrix positive definiteruns producing a false certificate14of runs in total72never above, in significand bits12first κ at eight bits10⁵the comparison was correctand what it proved was not true Two errors, and whose fault they are

Deciding that a zero has arrived

The previous tolerances were offers — accept this much error, save this much work. A detection threshold is not an offer, because both directions are failures. One matrix here has three genuinely near-invariant subspaces, and the constant somebody typed decides which of them the recurrence stops at; at eight significand bits the same kind of constant produces a proof of something false.

 02040608010010⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸truncations performed by the factorisationrelative errorthe representation's own error‖A − LLᵀ‖ ⁄ ‖A‖no decomposition without its residualresidual, 0 truncations2.1·10⁻¹⁰residual, 98 truncations1.1·10⁻⁹representation, deepest1.4·10⁻⁹residual ⁄ representation0.81levels, deepest5a hundred approximate stepsand an exact-looking factorisation Neither sparse nor dense

The count that is not the budget

A Cholesky performed inside a low-rank format truncates 0, 2, 10, 34 and 98 times as the leaf falls from 128 to 8, and those five integers are the same at every accuracy from 10⁻¹² to 10⁻². Across all ten decades the factorisation's residual stays below the representation's own error at a ratio between 0.81 and 1.00 — with two entries that read 1.83 and 1.78, and neither of them is accumulation.

 10⁻²10⁻¹10¹10²drift between consecutive membersinner steps over the whole runthe previous answer — a warm startthe parabola through the last threethe line through the last twodegree zero is a choicemembers per factorisation4warm, at drift 0.02160linear, at 0.0212quadratic, at 0.0226the factor13drifts where warm fails2one more term in the extrapolationand a factor of thirteen When the problem arrives again

A warm start is degree zero

The previous answer used as the next member's starting point costs 160 inner steps over twenty members. The line through the last two answers costs 12 — a factor of thirteen, for three vector operations and no extra storage. The parabola through the last three costs 26, which is worse than the line and better than the point.

 half a stepthe 32 values of one block, in the order they arrivestep 0.0625one scale, thirty-two valuesoctaves inside the block1.9entries rounded to zero0worst error over its bound131 levels either side of zerothe largest entry chose the step The arithmetic underneath

One exponent for thirty-two numbers

Share the exponent across a block and the cost per value drops from eight bits to 6.25, and the accuracy improves — up to about three octaves of spread inside a block. Past that a single outlier deletes the thirty-one values beside it, and the 2-norm barely notices.

 -13-11-9-7-5-3-102468log₁₀ of the accuracy asked forlargest train rank0.30 of rank per decadea cost that is typed inentries216stored at 10⁻²90stored at 10⁻¹²288rank per decade0.3worst error ⁄ tolerance0.83the storage is chosena constant of rank a decade When the index is a tuple

The digit that costs more than the tensor

Ask a three-index reciprocal tensor on six points a side for seven digits and its train is 288 numbers against 216 entries. The break-even rank is n − 1 at all four grids measured, and a train that reaches it fits with exactly n numbers to spare.

 10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹relative compression error, chosen‖x − x*‖ ⁄ ‖x*‖κ × the chosen backward errorthe error in the answerboth factors known firstκ21chosen at 10⁻⁸1.4·10⁻⁹predicted forward2.9·10⁻⁸measured forward1.6·10⁻⁹bound ⁄ measured26the amplifier, used forwardsfor once Neither sparse nor dense

The knob that moved two things

Decide how many digits the answer needs, divide by the condition number, and compress to that. It is the one rule licensed in advance here, and its two factors are not the independent inputs it reads as: the partition's leaf moves neither of them and moves the answer by nearly a factor of three, and the only knob here that raises κ halves the ranks while it does so.

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