Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Mixing time — where it appears

How many steps a random walk takes to come within a stated distance of its stationary distribution. It is predicted by the second eigenvalue, and on the graphs measured here the prediction runs about five per cent high.

Named by 3 essays across one field — each of them below, with the objects they name alongside it.

0173451688510211910⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹1steps of the walkdistance from stationarya rate read as a timeλ₂ of the walk0.9steps measured122steps predicted131ratio0.93the dashed line is the eigenvaluethe curve is the walk

The rate is the second eigenvalue

A walk forgets where it started at a rate the graph's second eigenvalue names exactly. Across three orders of magnitude in the step count the prediction is five per cent high — and the published rate for PageRank is right for a reason nobody states, which is that a link graph is in pieces.

graph · Random walk 02244668811013215417610⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²iterationchange between iteratestwo rates, one curveα0.85λ₂(P)1predicted rate0.85measured0.84iterations176against the solve1.4·10⁻¹⁶the upper dashed line is αᵏthe curve is on the other one

A chain with no stationary vector

A page with no outgoing links loses forty per cent of the walker's probability in six hundred steps. A directed cycle never converges at all. And on a graph whose links only run one way, the entire rank of half the vertices is exactly one minus the teleportation parameter.

graph · Random walk 10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁶10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1coupling ε between the clustersworst relative entry errorGTHGaussian eliminationpower iterationthe stationary vector is computed against an exact rational solutiononly the elimination with no subtraction holds

No subtraction, and no waiting

The power iteration keeps a positive vector's small entries correct because it never subtracts, and it pays for that in steps. On a Markov chain made of two clusters joined by a coupling ε it needs about 7/ε of them — 6,900 at ε = 10⁻³, 69,000 at 10⁻⁴ — and then stalls at a fixed point of its own rounded arithmetic about u/(4ε) from the answer; at ε = 10⁻⁶ it is still 45 per cent wrong after 200,000. Gaussian elimination needs no waiting and subtracts: on the same chains its error grows like 1/ε, and on a chain whose entries fall to 10⁻¹⁸⁴ it returns the small ones as noise, eight of them negative. The Grassmann–Taksar–Heyman elimination does neither. Every entry of every chain comes back to a relative 10⁻¹⁵, in a third of n³ operations, because the stationary vector is well conditioned under the perturbations it makes.

graph · Perron frobenius

Named alongside it

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

PagerankPerron frobeniusPower iterationStationary distributionBipartiteCollatz wielandtComponentwise condition numberConductanceDangling nodeGaussian eliminationIrreducibilityNormalised laplacian

All concepts