Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Flop count — where it appears

The number of arithmetic operations an algorithm performs, which stopped deciding which of two implementations is faster about thirty years ago.

Named by 9 essays across 6 fields — each of them below, with the objects they name alongside it.

10²10³10⁴10⁵matrix size ncountoperations, bothwords, unblockedwords, blocked (b = 6)the answer does not move‖PA − LU‖/‖A‖, unblocked2.8·10⁻¹⁶‖PA − LU‖/‖A‖, blocked2.8·10⁻¹⁶difference between them0the dashed curve is both orderings' operation countthe solid pair is what they cost

The same arithmetic at a different price

A blocked and an unblocked elimination perform 72,568 operations each — the same operations, associated differently — choose the same pivots, and return a factorisation identical to the last bit: ‖PA − LU‖/‖A‖ = 4.487946226420872·10⁻¹⁶ in both. One of them moves 41,332 words between fast and slow memory and the other moves 19,476.

cost · blocking 110¹10⁴10⁵block size bwords movedthe count: √(M/3) = 5measured best: b = 8words movedat the best block1.6·10⁴at b = 13.9·10⁴at b = 243.9·10⁴derived from M with no measurement, and scannedthe two agree

A block size is a property of the machine

Three lines of counting say the best block size is √(M/3). Scanned over every integer at five fast memories, the measured optimum is √M − 2 — exactly, at all five. The count has the right scaling and the wrong constant, low by a factor of 1.56, and the wrong form: the answer is affine in √M rather than proportional to it.

cost · blocking 10¹10³10⁵10⁷10⁹10¹¹10¹³10¹⁵10⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1κ₂(A), the matrix that was updatedrelative forward errorSherman–Morrisondirect solve of A + uvᵀone answer, two routesκ of the answer's matrix1κ of the matrix replaced10·10¹³update formula's error2.5·10⁻⁴direct solve's error1.1·10⁻¹⁶the question's condition number is 1at every point on this axis

A correction cheaper than the problem

Sherman and Morrison's formula updates a solved system for a rank-one change to the matrix, at 4n² operations instead of (2/3)n³. It is exact algebra. On a problem whose updated matrix is the identity — condition number one, the easiest system there is — it returns a forward error of 2.5·10⁻⁴ where a direct solve returns 10⁻¹⁶.

leastsquares · low rank update 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.

orthogonality · polar decomposition 01234567810⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹μ, the entry above the diagonalsep, and the eigenvalue gapmin |λᵢ + μⱼ|sep(A, B)the spectra never moveeigenvalue gap, throughout2sep at μ = 02sep at μ = 89.5·10⁻⁴amplification there1056solvability is the eigenvaluesand conditioning is not

An equation whose unknown is a matrix

AX + XB = C is linear in X, so it has a coefficient matrix, and writing it down is the obvious thing to do. At n = 100 that matrix has a hundred million entries for a problem with ten thousand unknowns, and the algorithm everybody uses instead never forms it. Its conditioning is not the eigenvalue gap either, which is the number a reader is invited to consult.

structure · matrix equation 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

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.

elimination · determinant 10²10⁴10⁶10⁸10¹⁰10¹²10¹⁴10⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1κ₂(A)relative errorforward, A⁻¹bforward, A\bbackward, A⁻¹bbackward, A\bat κ = 10¹⁴η, LU solve2.2·10⁻¹⁷η, via the inverse4.5·10⁻⁵forward, LU solve2.8·10⁻⁴forward, via the inverse0.015one factorisation, two ways to use itand one of them forfeits the backward error

The inverse that is never formed

x = A⁻¹b is how the solution of a linear system is written and it is not how it is computed. The usual reason given is cost — three times the arithmetic. The real reason is that one of the two routes is backward stable and the other is not, and at κ = 10¹⁴ they differ by twelve orders of magnitude in the number that says whose fault a wrong answer is.

elimination · inversion 0246810121410⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸squarings, srelative error against the closed formbest: s = 2the approximant is out of rangethe squarings amplifyboth sides costerror at s = 06·10⁻¹²best, at s = 26.9·10⁻¹⁶error at s = 153.2·10⁻¹²‖A‖ / 2^s at the best1.1the squarings are exact in the algebraand each one doubles the rounding

The series that has to be squared back

The Taylor series for the matrix exponential is not wrong — every term is computed correctly — and on Moler and Van Loan's two-by-two its largest term is 5.4 million times the answer it sums to. The method that replaces it scales the matrix down and squares the result back, and both halves of that sentence cost: too few squarings and the approximant is out of range, too many and each one doubles the rounding.

spectra · matrix function 0481216202410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹matrix–vector products, mrelative error in e^Abforming e^A: 3·10⁻¹⁶crosses at m = 18the vector, not the matrixsteps to the dense answer18dimension100Krylov megaflops0.36dense megaflops2the exponential that is computedis 18×18

The vector was what was wanted

Nobody who computes a matrix exponential wants the matrix. They want e^{At}b — one vector, the state of a system at a later time. Twenty matrix–vector products get it to sixteen digits on a hundred-by-hundred problem, without ever forming a hundred-by-hundred exponential, and the exponential that does get computed is twenty by twenty.

spectra · matrix function

Named alongside it

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

Condition numberLU factorisationBackward errorBlocked algorithmCacheCancellationData movementExact ground truthForward errorLoop orderMatrix exponentialMatrix-free

All concepts