Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Bit length — where it appears

How many binary digits an integer occupies, which in exact arithmetic is the unit every cost is measured in. An operation on a forty-thousand-bit integer is not one operation, so a count of arithmetic operations does not price an exact algorithm and a count weighted by operand length does.

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

3456789101112110¹10²10³10⁴10⁵10⁶nwidest intermediate, in bitsrationals, not reducedfraction-free · rationals reduced · the answerone answer, three widthsn12answer40Hadamard bound52fraction-free40reduced rationals39unreduced1.4·10⁶the error is zero on every curvethe cost is the length of the numbers

An answer with no error in it

An integer matrix eliminated over the rationals rounds nothing, so the forward error is zero, the residual is the zero vector, and the identity this site is built on has no terms left. The cost does not vanish with the error. It moves into the length of the numbers, where three correct routes differ by four orders of magnitude.

exact · Exact cost 1-2-6-7-474-26-12-5-8-524-394151435A, the matrix as given1-2-6-7-401840552700112217207003943011710017279153after 2 fraction-free steps3 × 3 minors25 divisions, all exactthe intermediates are minorswhich is why the divisions come out whole

Every intermediate is a minor

Fraction-free elimination divides by the previous pivot at every step and the division is always exact. Not usually, not for these entries — always, because the number being divided is a determinant with that pivot as a factor, which is a theorem and is checked here against the minors themselves.

exact · Fraction-free 3456789101110²10³nbitsthe answerthe question · det Abits in, bits outn11question484answer729det A33widest part37the answer is the floorand no route writes it down more cheaply

The answer is longer than the question

An exact solution of an integer system is a vector of fractions, each of them a ratio of two determinants. So the output carries 2n long integers where the input carried n² short ones, and no algorithm can write it down more cheaply — the length of the answer is a floor under every exact solver rather than a property of one.

exact · Exact cost 34567891011110¹10²nbitsthe budget and what it buysrandom, bound47random, actual33primes needed2Hadamard n = 8, bound13Hadamard n = 8, actual13the count is decided by a theorembefore any arithmetic happens

How many primes the answer needs

Work modulo a word-sized prime and no intermediate can exceed twenty-six bits, whatever the matrix does. The catch is that the answer must be reassembled from several such computations, and the number of them has to be fixed before the first one runs — by a theorem about how large a determinant can be, not by trying more until it settles.

exact · Modular lift 2^18modulus, as a power of twoa lattice with one short vectornumerator355denominator1132·max(n, d)², bits18first recovered at18moduli tried42below the bound there are two answersand the algorithm cannot prefer one

A fraction recovered from one remainder

A solution over the rationals can be computed modulo a prime power and then recovered — the residue determines the fraction uniquely, but only once the modulus is twice the square of the fraction's longer part. Below that there is no partial credit: the algorithm returns a different fraction with the same residue, and it is a perfectly good one.

exact · Modular lift 10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²110⁻⁷10⁻⁵10⁻³10⁻¹10¹relative error in the datarelative error in the answerthe answer's errorthe data's errornothing here is the arithmetic'sdata error10⁻¹²answer error10·10⁻⁴amplification10·10⁸rounding committed0residual0the residual is exactly zeroand the answer is wrong anyway

An exact answer to a measured problem

The residual is the zero vector, nothing was rounded at any step, and the answer is wrong in its first digit. Data accurate to fourteen places, an exact solve of the system it defines, and an error of 10⁻⁵ — because conditioning was never a statement about arithmetic and removing the arithmetic error removes none of it.

exact · Exact cost

Named alongside it

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

Exact arithmeticFraction-free eliminationHadamard boundDeterminantBackward errorChinese remainder theoremCondition numberCramers ruleExact ground truthModular arithmeticForward errorHilbert matrix

All concepts