Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Matrix inverse — where it appears

The matrix undoing a linear map, which is the answer to a question about its entries and the wrong intermediate for a question about its action. Forming it costs three times a factorisation and loses digits, so a solve should go through the factors, and the essays that measure this find twelve decades between the two routes.

Named by 4 essays across 2 fields — each of them below, with the objects they name alongside it.

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 110¹10²10³10⁴10⁵10⁶10⁷10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸1/(1 − h), the leverage of the removed row‖R̄ᵀR̄ − (G − aaᵀ)‖ / ‖G − aaᵀ‖downdatedrefactorisedat h = 1 − 10⁻⁷κ of the downdated matrix4.3κ of the matrix downdated9.3·10⁶rotation's amplification344downdate residual3.5·10⁻¹⁰a hyperbolic rotation is not orthogonaland that is exactly what it is for

The observation that cannot be removed

Removing a rank-one term from a Cholesky factor needs a rotation that is not orthogonal, and the number under its square root is 1 − h, where h is the leverage of the row being removed. The algorithm's breakdown condition and the statistician's warning are the same quantity, arrived at from opposite ends, and neither field states it in the other's language.

leastsquares · Low-rank update 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 01234510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1correction steprelative errorLU route: η = 2.2·10⁻¹⁷LU route: forward 2.8·10⁻⁴forward errorbackward errorwhat a correction buysη before refinement4.5·10⁻⁵η after four steps2.8·10⁻¹⁷forward, unchanged7.1·10⁻⁴cost of a step, flops1800the residual is repairableand the accuracy floor is the problem's

The gap refinement can close

Multiplying by a computed inverse is not backward stable, and refinement at the working precision repairs it. That much is settled. The claim beside it — that the forward error does not move — was read at one conditioning and four corrections too late. Swept over ten, it moves at every one, and it lands on the LU route's own number after a single correction.

elimination · Inversion

Named alongside it

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

Condition numberForward errorBackward errorFlop countIterative refinementLeverageLow-rank updateLU factorisationCholeskyContraction factorGivens rotationHilbert matrix

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