Symbol — where it appears
Named by 9 essays across one field — each of them below, with the objects they name alongside it.
A limit the matrix never reaches
Szegő's theorem gives a Toeplitz family's condition number in closed form — ((1+ρ)/(1−ρ))², which is 81 at ρ = 0.8. The 8×8 section reaches 52% of it, the 128×128 reaches 98.9%, and none of them ever arrives. A statement about a family is not a statement about the matrix in front of you.
Two dimensions, and the cluster that thins
The same kernel, the same averaging, the same transform — applied along two axes instead of one. In one dimension the preconditioned step count is 7, 10, 10, 10; on square grids with the same unknown counts it is 10, 18, 20, 21, and the share of the spectrum near one falls from 56% to 17%.
Four orders of conditioning, and four steps
On a 10×10 grid the two-dimensional kernel's condition number runs from 62 at ρ = 0.5 to 818,561 at ρ = 0.98. The preconditioned step count over the same range runs 18, 21, 21, 22, 21, 19, 18, and the count of eigenvalues the preconditioner actually brings within half a unit of one does not move at all — it is 9, 11, 13, 17 at every correlation the figure will draw.
The circulant the problem did not contain
A matrix that differs from a circulant in two corner entries can be solved through the circulant, by a transform and a two-by-two correction, and the cost claim is exact. The accuracy claim is not. On tridiag(−1, 2 + σ, −1), whose condition number stops at 1,712, the correction is wrong by 1.2·10⁻⁴ at σ = 10⁻⁸ while elimination is right to 1.1·10⁻¹⁴ — because the periodic neighbour is singular at σ = 0 and the two-by-two system inherits that. Solved by Cramer's rule, as here, the two amplifications multiply; a later measurement found that a pivoted solve of the same two-by-two system removes the second.
Two near-zeros cost less than one
Solve a well-conditioned tridiagonal matrix through a nearly singular wrap and the correction's accuracy is not set by how singular the wrap is. At κ = 4·10⁷ one wrap returns the answer to 8.8·10⁻¹¹ — better than κ·u — and another, at κ = 3.8·10⁷, returns it to 3.3·10⁻⁶. The difference is how many of its samples sit near the symbol's zeros. A real wrap lands on a conjugate pair, a rank-two correction absorbs the pair exactly, and its two-by-two system has condition number 1.00 — which mattered because that system was solved by Cramer's rule; solved with pivoting, the single landing costs what the pair costs.
The correction lost to its own two-by-two solve
Solving a band matrix through a circulant and a small correction was measured losing the answer to 10⁻⁴ where elimination kept it to 10⁻¹⁴, and a wrap that landed one sample on a zero was measured costing four orders more than one that landed a pair. Both measurements solved the two-by-two correction system by Cramer's rule. Solved with a row interchange, the corrected solve on the same matrix loses 2.9·10⁻¹⁰ — the cancellation, and nothing multiplied onto it — and the single landing costs what the pair costs. The loss was in the determinant.
A zero no twist can step around
Solved through its circulant wrap with the small capacitance system factorised stably, a banded matrix was found to lose only what the cancellation in the correction costs, and a twist of half a step kept that cancellation near one. That holds only while the cancellation is large. With it kept small, the corrected solve's error is a quarter of the wrap's condition number times u on every case measured — and on a symbol with a zero of fourth order, the square of the Laplacian, no twist can keep the wrap well conditioned, because the nearest sample any twist can reach sees (π/n)⁴. At n = 64 the corrected solve is 34 times less accurate than elimination, where the Laplacian's is four.
One step past the zero
The squared Laplacian, solved through its twisted circulant and a rank-four correction, lost 6 to 34 times more accuracy than elimination, because its symbol's fourth-order zero keeps the wrap as ill conditioned as the matrix whatever the twist. One step of iterative refinement — a residual formed with the band itself, a second solve by the same route — takes it to 0.59, 0.68, 0.73 and 1.40 times elimination's error at n = 16, 32, 64 and 128, and the second and third steps go no lower. The loss the zero imposed was the solver's, and a solver's loss is what refinement removes.
Where one step stops being enough
One step of iterative refinement took the squared Laplacian's circulant-wrap solve to elimination's accuracy at every size, and the account was that each step multiplies the error by the wrap's condition number times the rounding, so one step suffices while that product is small. Raised to higher powers, the band tests the account and half of it holds: each step does contract by about κ(wrap)·u. The other half fails. The fifth power at sixteen points needs three steps with κ(wrap)·u near 10⁻⁶, where the third power at 128 points needs one with thirty times more, because the first solve starts up to a thousand times further from the answer than κ(wrap)·u says.
Named alongside it
The objects these essays reach for when they reach for this one.
Condition numberCirculant matrixCapacitance matrixIterative refinementLow-rank updateToeplitz matrixAsymptotic analysisBackward errorSherman morrison woodburyWoodbury identityCancellationCirculant preconditioner