Hadamard bound — where it appears
Named by 8 essays across one field — each of them below, with the objects they name alongside it.
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.
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.
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.
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.
A prime that divides the answer
A modular elimination reports a singular matrix and is telling the truth — over the field with p elements the matrix is singular. Over the rationals it is not. Nothing in the residue distinguishes the two cases, no quantity is small enough to be suspicious, and the wrong answer is a correct computation of a different question.
A bound on every intermediate at once
Fraction-free elimination's intermediates are minors of the original, which is a theorem about exactness. It is also a bound: Hadamard's inequality applies to every minor, so one inequality bounds the whole run before it starts. The bound on the k-th step is the one on (k+1)×(k+1) minors, not the one on the whole matrix — and on a 10×10 with entries in ±6 the difference is ten bits, with the run reaching 2.7 bits a step against the bound's 3.4.
Three orders and one last entry
Over the integers there is no stability to pivot for, so a fraction-free elimination swaps rows only when the pivot is zero. Choosing a pivot for length instead does change the sequence of minors — the smallest-nonzero rule makes seven exchanges where the natural order makes none and keeps the profile two bits lower through the middle. It cannot change the peak. The last entry of the elimination is the determinant, and the determinant does not know what order it was computed in.
The pivot is in every product
A fraction-free elimination's intermediates are minors, and Hadamard bounds a minor by the lengths of the rows it is made of — so the rule that picks the smallest pivot entry looked like a proxy for a rule that picks the shortest row. Measured, the two keep the bit-length profile equally low: 0.969 and 0.966 of the natural order's area. They part on what the arithmetic costs. Counting every multiplication and division at the product of its operands' lengths, the smallest-pivot rule costs 0.70 of the natural order and the shortest-row rule 0.84, because the pivot multiplies every entry of the step and divides every entry of the next. Three per cent of area is thirty per cent of arithmetic.
Named alongside it
The objects these essays reach for when they reach for this one.
Exact arithmeticBit lengthDeterminantFraction-free eliminationExact ground truthMinorSylvesters identityChinese remainder theoremModular arithmeticPermutationBackward errorCondition number