Kronecker product — where it appears
Named by 11 essays across 3 fields — each of them below, with the objects they name alongside it.
An index that is a pair
A discretisation on a two-dimensional grid of n points a side has n² unknowns and a matrix with n⁴ entries — 10⁸ at n = 100. What that matrix is instead is two Kronecker products of an n × n matrix, which is 2n² numbers, and nothing has been approximated: assembling it was the mistake.
A solve that is d decompositions
A Kronecker sum is closed under nothing useful — its inverse is not a Kronecker sum and no factorisation of it is one. What it has instead is eigenvectors that are Kronecker products, so a solve with 1,728 unknowns takes one decomposition of a 12 × 12 matrix and nothing else.
An equation whose unknown is a matrix
AX + XB = C is linear in X, so it has a coefficient matrix, and writing it down is the obvious thing to do. At n = 100 that matrix has a hundred million entries for a problem with ten thousand unknowns, and the algorithm everybody uses instead never forms it. Its conditioning is not the eigenvalue gap either, which is the number a reader is invited to consult.
Five indices are cheaper than two
The same 4,096 unknowns cost 1.049·10⁶ multiplications indexed as a 64 × 64 grid and 1.966·10⁵ indexed as six axes of four. The dense factorisation that ignores the indexing costs 4.581·10¹⁰ at every one of them, and the residual improves in the same direction as the cost.
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 elimination the matrix does not need
The Kronecker form of AX + XB = C is dismissed with a hundred million entries and (2/3)n⁶ operations. Both price an elimination, and after the reduction both routes take, the matrix has exactly n³ nonzeros, none of them above the block diagonal, and nothing left to eliminate.
The staircase a separable kernel builds
A block-circulant preconditioner took 18, 21, 22, 21, 19 and 18 steps on a 10 × 10 Toeplitz-block-Toeplitz system as the correlation rose from 0.5 to 0.98 and the unpreconditioned count rose from 43 to 178 — the parameter that makes the problem hard was the one the solver did not notice. That kernel factorises. The isotropic kernel with the same correlation along each axis does not, and on it the preconditioned count climbs 17, 22, 27, 29, 33, 36, while the preconditioner buys a factor of 1.4 where it bought ten. The reason is the spectrum's shape: a product of two one-dimensional spectra is a staircase of ten treads, and a kernel that is not a product gives a ramp of thirty-eight.
The symbol that builds the staircase
A block-circulant preconditioner left the isotropic kernel's spectrum a ramp of thirty-eight clusters where the separable kernel's was a staircase of ten, and the proposal was a circulant sampled from the isotropic kernel's own symbol, to gather the ramp into treads of its own. It gathers the separable kernel instead: six clusters at every grid from 4 to 10 and every correlation, sixty-four of a hundred eigenvalues at exactly one, and eight conjugate-gradient steps at every side from 6 to 16 while Chan's circulant climbs from 18 to 28. On the isotropic kernel it shortens the ramp from 38 clusters to 30 and the step count by about a sixth, and both still grow with the grid. The treads come from a Kronecker product, not from a small difference: on both kernels the matrix and its symbol circulant differ in dozens of directions.
Six steps were six eigenvalues
A circulant sampled from a separable kernel's symbol held conjugate gradients at eight steps on every grid, because the preconditioned spectrum was six values. Add a fraction ε of the isotropic kernel and the prediction was that the six treads would widen, the count would stay near eight while they stayed under one per cent, and then climb towards the isotropic count. A ten-thousandth of isotropy leaves six clusters at the one-per-cent rule and takes the count from 8 to 13–15; a millionth takes it to 11. The eight steps were finite termination on six distinct eigenvalues, not convergence on six clusters, and once the eigenvalues are distinct what a tread costs is set by its width against the solver's tolerance: at a tolerance of 10⁻⁴ a ten-thousandth of isotropy costs two steps. The widths grow in proportion to ε, and summed tread by tread they account for the climb to within four steps. Past a tenth the mixture is harder than either kernel alone — 37 steps on a 16 × 16 grid against 35 for the isotropic kernel — while its condition number is lower than both.
A backward error the answer does not feel
Both ways of solving AX + XB = C leave a residual at the rounding level, and for a linear system that would be backward stability. For the equation it is not: the smallest perturbation of A, B and C that makes the computed X exact can be far larger. On a family built to make it so — X increasingly ill conditioned, C increasingly small beside AX — the structured backward error of Bartels and Stewart's solution climbs to 2.1·10⁵ times its residual, and the Kronecker system's, whose residual is four times smaller, to 2.0·10⁶, within the bound Higham's μ sets. And the error of the answer does not move. Against the exact X it stays where the size of C puts it, 6·10⁻¹⁵ to 1·10⁻⁹, across eight decades of κ(X). The perturbation that makes the backward error large has to sit on C, along directions X is insensitive to, and so the backward error rises and the answer never feels it.
A doubling that arrives two steps late
A pivot rule that looks one elimination ahead chooses Wilkinson's growth row for row, because the doubling it sets up arrives one step later; looking two ahead escapes it. The argument said a matrix whose damage arrives two steps late would defeat the two-step rule the same way. It exists, and it is two Wilkinson matrices interleaved: W ⊗ I₂. Every look-ahead of depth one or two reproduces the greedy rule's growth on it exactly, 4 to 64 from n = 6 to 14, and depth three gives 2. Three copies defeat depth three and yield to four. The growth each depth is held to is 2^(n/d − 1): a fixed depth stretches the exponential and never removes it, at a cost that rises by a factor of about n for every step it looks.
Named alongside it
The objects these essays reach for when they reach for this one.
Condition numberSeparabilityPreconditioningCirculant preconditionerClustered spectrumConjugate gradientsFlop countKronecker sumToeplitz matrixDiscrete laplacianExact ground truthMatrix equation