MINRES — where it appears
Named by 8 essays across 2 fields — each of them below, with the objects they name alongside it.
Three eigenvalues, and two are the golden ratio
Precondition a saddle-point system by the block diagonal of its own two definite pieces and the preconditioned matrix has exactly three distinct eigenvalues — 1, and the two roots of λ² − λ − 1. A minimal polynomial of degree three means three steps, at every conditioning, and the preconditioner nobody can afford turns out to be the statement the affordable ones are measured against.
A preconditioner that need not know the constraint
Keep the constraint block exactly and replace the objective block by anything positive definite on the null space. The preconditioned matrix then has 2m eigenvalues at exactly one, and its remaining n − m are the generalised eigenvalues of a pencil in which the constraint does not appear. Sweep its condition number over six decades and they do not move in six digits.
One eigenvalue and two steps
Put the off-diagonal block back into a block-diagonal saddle-point preconditioner and every eigenvalue of the preconditioned matrix becomes exactly one. GMRES still needs two steps, because the matrix is the identity plus a nilpotent part of norm 54, and a computed eigenvalue at one comes back as a ring of radius 8·10⁻⁸ — the square root of the rounding, not the rounding. With an approximate Schur complement the triangular form leaves one copy of each value where the diagonal form leaves two, and the step count halves.
Where the augmentation puts the cost
Add γAᵀA to the objective block of a saddle-point system and its Schur complement tends to I/γ, so the cheapest possible approximation becomes the right one and the golden-ratio spectrum arrives — within 7.6·10⁻⁶ at γ = 10⁶. MINRES falls from 21 steps to 6. The inner solve with the augmented block rises from 14 conjugate gradient steps to 43, their product does not fall at all, and the answer loses seven and a half digits on the way.
An augmentation read in the smallest eigenvalue
An augmented Lagrangian preconditioner's least work sat at γ = 1, 10⁴ and 1 on three systems, and two of the three optima sat where the smallest generalised eigenvalue was about 0.06 — which suggested a rule written in ν rather than γ. On twenty-four systems whose norms are all one, the least-work γ spans seven decades and ν at the optimum spans a factor of thirty, so there is no one ν. There is still a rule: take the smallest γ at which ν reaches 0.03, and the median system pays 7 per cent over its least work and the worst 29, where the best single γ pays 61. Its price is digits — twelve times the best forward error on the median system, 4,561 times on the worst.
A square root the residual does not pay
With the exact Schur complement, the triangular saddle-point preconditioner puts every eigenvalue at one in Jordan blocks of size two, and GMRES finishes in two steps. Solve the first block only to a relative accuracy τ and the blocks split: their eigenvalues leave one like √τ — 0.42 to 0.62 times √τ on eighteen systems across six decades, each pair exactly where a five-by-five eigenproblem puts it — while the rest move like τ. The question left open was whether that square root reaches the iteration. It does not. GMRES's residual after two steps grows like τ, with a slope of 0.86 to 1.27 on every system, because its polynomial has a double root at one and cancels each split pair to second order; every further pair of steps divides the residual by about the same factor; and a rule written in τ alone predicts the step count exactly on 179 of 234 runs and within one on 228.
The division that cannot be done
Conjugate gradients divides by pᵀAp at every step, and on a matrix that is not positive definite that number can be zero or negative. The guard against it has been here from the first essay and described it as a failure. In the method that made conjugate gradients famous it is the single most valuable object the iteration can produce, and it costs six matrix–vector products.
A proof that does not ask how large the matrix is
Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.
Named alongside it
The objects these essays reach for when they reach for this one.
Block preconditionerSaddle-point systemsSchur complementConjugate gradientsKrylov subspacePreconditioningExact ground truthIndefinite matrixMinimal polynomialAugmented lagrangianCertificateCholesky