Numerical Linear Algebra
where the algebra stops being the arithmetic
Ladder

Moment matching — the ladder

3 distinct arguments against one idea, from the one that introduces it to the one that assumes the rest.
  1. 10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints4conditions bought8worst at a point5.3·10⁻¹⁶worst away from one8.4·10⁻⁴4 points, 8 conditionsand no bound in between

    Exact at the points that were named

    Balanced truncation asks for nothing and bounds everything, at a cost no large model can pay. The other kind of reduction asks for r numbers, costs r solves, is exact at every one of them — and bounds nothing anywhere else. That trade is the whole of large-scale model reduction.

    rung 1 · reduction
  2. 2345678110²10⁴10⁶10⁸10¹⁰vectors in the basisκ₂ of the basisas derivedsolves spread outone subspace, two spanning setseight moments at one point7.7·10⁹eight points, spread1growth per vector661/u4.5·10¹⁵the same subspaceand only one of them usable

    A basis that is the same subspace and not the same thing

    The interpolation conditions are conditions on a subspace, so any basis of it will do. The one a derivation writes down reaches a condition number of 7.7·10⁹ by its eighth vector, and the rate at which it gets there is set by a number the user chose with no information.

    rung 2 · reduction
  3. 01234567891011121310⁻¹110¹10²10³iterationinterpolation point σa fixed point that is a conditionorder4iterations13σ against −λ(Aᵣ)2.8·10⁻¹²H₂ error4.3·10⁻⁵it stopped movingand the condition holds there

    Interpolating at the model’s own poles

    One choice of interpolation points is not arbitrary — the mirrored poles of the model about to be built. It is a fixed point rather than a guess, and when it is reached it beats a method costing O(n³) — by 0.4 per cent, which is the honest size of the whole contest.

    rung 3 · reduction

All ladders