Tropical roots — where it appears
Named by 5 essays across one field — each of them below, with the objects they name alongside it.
The scaling that buys ten orders
Two lines computed from three norms, a change of variable that is exact in both directions, and the whole of the loss the previous essay measured comes back — flat, at every stop, because after scaling every stop is the same problem.
An estimate that does not move
The tropical roots are said to miss the bottom of a spectrum by a factor growing like n². The estimate does not get worse. It changes by four per cent between four masses and sixty-four while the modulus it names falls by a factor of a hundred and sixty, and an estimate that is constant in the variable the answer depends on is a different defect from an inaccurate one.
Two groups need two reductions
When a quadratic eigenproblem's tropical roots are far apart its eigenvalues fall into two groups, and the remedy offered is a second scaling — one per tropical root, each run keeping the group its root predicts. Raise the damping on an eight-mass chain until the roots are 10²³ apart and measure every eigenvalue: no scaling rescues the small group. Its backward error grows as the unit roundoff times the separation under Fan–Lin–Van Dooren's scaling, under both tropical scalings and under none, to 10⁻⁵ at a separation of 4·10¹¹. What rescues it is the other reduction of the same pencil — inverting the constant term instead of the leading one — which keeps the small group at rounding at every separation and loses the large one instead. Both reductions, each keeping its own group, give all sixteen eigenvalues to 10⁻¹¹ at worst. On the way, the closed form that served as the exact answer turned out to lose the small roots to cancellation, from the same separation.
The middle group needs no reduction of its own
A quadratic whose two groups of eigenvalues sit far apart needs two reductions, one per end. A cubic can have three groups, and the middle one is inverted by neither reduction, so the worry was that it would need a third route — a shift-and-invert at the middle tropical root. Build order-four cubics with three groups up to eight decades apart and measure every eigenvalue: the leading reduction keeps the middle group at 10⁻¹¹ or better at every separation, the trailing reduction loses it to 5·10⁻⁹ by eight decades, and the reversed polynomial's leading reduction, which inverts the same constant coefficient as the trailing one, keeps it at 2·10⁻¹⁴. The middle group's loss belongs to the pencil, not to the end. Two reductions still give all twelve eigenvalues to 2·10⁻¹¹. And the tropical scaling at the small root, which rescues the small group when every coefficient shares one diagonal form, does nothing for it once the coefficients are perturbed out of that form.
An interior root needs its own scaling
A cubic matrix polynomial with three groups of eigenvalues needed only two reductions: the leading one kept the middle group at every separation. A quartic has two interior groups, and the prediction was that the leading reduction would keep both, so that two reductions would still do. At eight decades between groups it keeps neither — both interior groups are wrong in the second digit — and no unscaled route keeps the second group at all. Two reductions still suffice, but each must be scaled at an interior tropical root: the trailing one at the second root for the lower two groups, the leading one at the third for the upper two. And the second companion pencil, predicted to keep the middle groups as the reversed polynomial does, loses them as the first pencil does.
Named alongside it
The objects these essays reach for when they reach for this one.
Matrix polynomialScalingAsymptotic analysisBackward errorExact ground truthFrobenius normLinearisationQuadratic eigenvalue problemA-priori boundCompanion formCondition number