Stationary distribution — where it appears
Named by 10 essays across 2 fields — each of them below, with the objects they name alongside it.
A ranking that is an eigenvector
PageRank is the stationary vector of a walk that follows links with probability α and jumps at random otherwise. The iteration and the elimination agree to 4·10⁻¹⁷. What α is set to changes which pages come third, fourth and fifth.
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.
A ranking whose order is not determined
Of the fourteen adjacent comparisons in a top-fifteen, fourteen survive perturbing the arithmetic at the rounding level, eleven survive moving the teleportation parameter across its usual range, and three survive removing one link. The computation is the strongest part of the answer.
A Laplacian that is not symmetric
Point the edges and the matrix stops being symmetric. Its row sums are still exactly zero, so the null vector survives; everything built on the quadratic form does not, and the eigenvalues need a plane rather than a line. Asymmetry permits that and does not force it, which the smallest case here demonstrates by being asymmetric and real.
A conductance the arcs do not measure
Symmetrising a directed Laplacian with respect to its walk recovers everything the arrows took — a real spectrum, a sweep cut, a Cheeger inequality. What it does not recover is the quantity: the inequality bounds the probability that a step of the walk crosses the cut, which on one graph here is three times the weight of the arcs that do.
The well on the far side of the band
Gradual underflow was said to buy a predicate and not an answer, because a quantity that has decayed into the subnormal range is already lost. A quantity that passes through the band on its way somewhere else is not. The stationary distribution of a two-well chain, computed in half precision across a barrier whose top is two to the minus sixteen of the near well, keeps its far well's probability of 0.2454 to three digits with subnormals and returns exactly zero without them. The normwise backward error calls both answers exact.
The far well in a byte
Half precision's subnormals carried a two-well chain's far well ten octaves deeper than flush to zero, and the essay that measured it predicted that on the eight-bit formats the band would be worth almost nothing and starting high almost everything. Measured on nine chains, the band is worth what it always was, two or three octaves; starting at the top of the range is worth ten and fourteen; and at three significant bits the far well is lost on eight chains of nine with no exponent limit at all, because the precision fails before the range does.
Room for the step, not the value
Starting a recurrence at the top of an eight-bit format's range spends its headroom once. An exponent counter beside the stored value was predicted to spend it at every step, remove every range failure and leave only the precision. Rescaled on the value — back to one whenever it nears the bottom or the top — it does not: E4M3 still loses its far well on seven or more chains of nine, because one climbing step multiplies the value by forty and E4M3 clips silently at 448. Rescaled into a window read off the chain's own rates, it reproduces the unbounded exponent bit for bit on all 531 runs of every format, and what is left of E4M3's limit is nothing.
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.
The stall was in the block
On two dense clusters joined by a coupling ε, the power iteration needs about 7/ε steps and then stalls at about u/(4ε). Aggregation–disaggregation was predicted to need a number of steps independent of ε and to stall at rounding, and with a power step as its smoother it does: fourteen steps at every coupling from a tenth to 10⁻¹², every entry to a relative 10⁻¹⁵. The textbook smoother, block Gauss–Seidel, is faster and stalls at u/ε again — because each block solve forms one minus a self-probability, and a block whose rows sum to 1 − ε is singular to within ε. Solved without that subtraction, the same blocks finish in one to four steps at rounding, and get faster as the coupling weakens.
Named alongside it
The objects these essays reach for when they reach for this one.
Perron frobeniusExponent rangePagerankPower iterationSubnormal numbersTeleportationBackward errorComponentwise condition numberDangling nodeDirected graphFlush-to-zeroFp8