Certificate — where it appears
Named by 20 essays across 6 fields — each of them below, with the objects they name alongside it.
A minimum the Hessian cannot see
A Hessian with four negative eigenvalues can sit at a constrained minimum, and a Cholesky of it stops at the third row. One symmetric indefinite factorisation of the saddle-point matrix settles the question anyway — ten positive pivots and four negative — without a basis for the null space ever being formed. The count is exact in the algebra and blind in floating point, in a band that grows like κ(A)²; the route through the null space is blind in one that grows like κ(A).
The active set before the digits
An interior-point method takes fifteen iterations on a quadratic programme with forty constraints, and its iterate has eight correct digits at the eleventh. Take the constraints its diagonal calls active at the first iterate, solve the equality problem they define once, and check the answer against the conditions for optimality. It passes, to thirteen digits. The step's matrix had a condition number of 43 at that iterate, and 7·10¹⁵ at the last.
An eigenvector that must not change sign
Perron's theorem says the leading eigenvector of a connected nonnegative matrix is strictly positive. On a clique with a long tail, four of its thirty-six entries come back negative — and beside them is the one two-sided bound on this site that is proved rather than estimated.
Deciding that a zero has arrived
The previous tolerances were offers — accept this much error, save this much work. A detection threshold is not an offer, because both directions are failures. One matrix here has three genuinely near-invariant subspaces, and the constant somebody typed decides which of them the recurrence stops at; at eight significand bits the same kind of constant produces a proof of something false.
A shift that certifies a saddle
On a constrained problem whose Hessian has four negative eigenvalues, a saddle-point matrix is quasi-definite only once H + δI is positive definite — past δ = 5.08 here. Its pivot signs then count the curvature of ZᵀHZ + δI rather than of ZᵀHZ, so a saddle with a negative curvature of −1 is certified a minimum from δ = 1.05 on, and every saddle shallower than δ goes the same way. Iterative refinement against the unregularised matrix keeps the second-order test the count gave up: it contracts on the minimum at δ/(μ + δ), 0.980 a step at δ = 5, and on the saddle it grows at exactly 1.25.
Two repairs for one symptom
Rescale a quadratic programme's constraint rows over six decades and the crossover that certified its answer at iterate 1.5 first certifies at 69.8, with two of six programmes never certifying at all. Normalising the rows removes the spread completely — the same numbers at 10¹, 10² and 10³ either way. Starting the method at the magnitudes the rows imply repairs the iteration count completely and the identification only halfway. They are two repairs and they fix different halves.
A test with no tolerance in it
An interior-point method's own stopping test is a tolerance on μ, and at the tightest it can be set to it stops after 15 iterations with 1.6·10⁻¹². A crossover from the iterate at 1.5 returns a point whose error is 1.8·10⁻¹⁴ — ten times sooner and a hundred times better, from an iterate carrying one correct digit. One attempt costs an eighth of a step, and the guess's own margin says which iterate to spend it on.
The shift that stops at the first right count
A nonconvex solver that finds the wrong inertia adds δI to H and tries again, and the δ it settles on is used as though it measured the curvature it corrects. It does not. On a well-conditioned constraint it is the schedule's number — 1.8·10⁴ times the need at a curvature of 5.6·10⁻⁹, between one and 7.3 times above 10⁻² — and on an ill-conditioned one the loop stops wherever the count first reads right: 63 of 152 saddles at κ(A) = 10⁸, the deepest with curvature 56. Refining the shift by bisection removes the first error and adds to the second.
A constraint the count stops seeing
Let one constraint drift towards being a combination of the others and the inertia of the saddle-point matrix keeps its promise only while σ²/|h| can be resolved — σ the constraint's smallest singular value, h the curvature along the direction it barely constrains. At h = −1 the count stops seeing the constraint at σ = 1.4·10⁻⁹, six decades before any rank test would drop it, and below that it reports a genuine minimum as a saddle on three to six draws in eight. No shift of H brings the constraint back: the correction loop shifts a problem that needed nothing by as much as 2,620. A perturbation of the constraint block does not bring it back either — it decides, at σ = √(|h|δ).
The room a relation has to stand out
A lattice search for an integer relation returns its shortest vector, and the proposal was to return the gap to the next one as well, so a caller could tell a relation from an accident. Measured, the gap is a certificate with a budget: the digits the numbers really have, shared among all but one of them, less the size of the relation. A found relation's gap sits half a digit under that budget, accidents stay near zero, and a one-digit gap vouches for 81 of 96 relations among three numbers and for 1 of 35 among six. The test numbers the proposal came from turned out to have relations of their own.
Two precisions guard the other edge
Run a lattice relation search at N digits and again at N/10, and accept its answer only if both runs return the same vector. Among six measured numbers, where the gap between the shortest and next vector vouches for one found relation in 35, the two runs agree on 34. Past a double's sixteen digits they never once agree on an accident, 0 of 561. They do agree on 62 accidents at fifteen digits or fewer — approximate relations that really are the shortest vector there — and the gap, which cannot see a relation among six numbers, can see those.
A loop that asks the null space why
An inertia-correction loop sees only an integer, and two different faults produce the same wrong one: curvature that needs a shift, and a constraint too weak for the count to see. One QR of the constraint matrix on a wrong count tells them apart — it shifts none of the 22 weak-constraint minima the ordinary loop shifted by up to 2,621 — and its reduced eigenvalue gives the shift a saddle needs in one step, twice the need exactly, where the schedule overshoots by up to 17,783 times. But at κ(A) = 10⁸ the loop still certifies 57 saddles of 152, because a false certificate is a count that read right, and a check made only on wrong counts never sees it. Asking every time leaves five, all shallower than 2·10⁻⁸.
Two machines, one certificate
Nothing a solver returns says which of its answers you got. Four things could be reported instead — the summation condition number, the partition count, an exactly accumulated residual and a directed-rounding interval — and each costs about one pass over data the routine already has in hand.
The residual turns before the error doubles
Regularise a saddle-point matrix past the Hessian's most negative eigenvalue and its pivot signs certify every shallow saddle as a minimum; refinement against the unregularised matrix keeps the test, as a rate, and a saddle at a hundredth of the regularisation grows by only 1.0014 a step — 485 steps to double. That was read as hundreds of steps before the history says anything. The residual, which is what a solver actually has, says it at step 62: a fit of its logarithm over the last ten steps turns positive there and stays positive, while the minimum's is negative from step 10. Across five regularisations the verdict comes at an eighth of the doubling time.
The digits between the two searches
A lattice relation search run at N digits and again at N/10 never agrees with itself on a relation among the rounding, but it does agree on 62 approximate relations at fifteen digits or fewer — accidents that really are the shortest vector there, and that dropping a digit was said to be unable to dislodge, since a combination that cancels to D digits cancels to D − 1. It cancels, but it stops being the shortest. Two digits apart the accepted accidents fall to 17 and three digits apart to 5, while the relations kept fall from 711 of 755 to 652 and 602 — every one of the losses at fifteen digits or fewer. Past a double's sixteen digits the separation costs nothing and buys nothing.
The right-hand side that hides the saddle
Refinement against an unregularised saddle-point matrix gives the second-order verdict the pivot signs cannot, from step 62 on the shallowest saddle — for one right-hand side. The verdict depends on how much of the growing direction the right-hand side contains, and it can contain none. Each decade removed delays the verdict by a fixed number of steps, 147 on the shallowest saddle, set by the growth against the minimum's slowest contraction rather than by the growth alone, which predicted 1,611. Rounding stops the hiding at about 2,700 steps — but by then refinement has solved the system to a residual of 2.6 times ten to the minus thirteen, and any stopping test has already accepted it. A random kick to the starting point costs nothing and finds it.
One hyperplane hides nothing
Iterative refinement against a saddle-point matrix gives its verdict when the growing direction overtakes the minimum's slowest decay, and a right-hand side with that direction removed delays it by a fixed number of steps a decade. With two negative curvatures, the prediction was that hiding both is paid for by the faster, and that hiding only the faster leaves the slower one's race to run. The first half holds to half a per cent on three pairs, and to one and a half on two whose curvatures differ by a factor of two and of 1.2. The second does not: hiding either direction alone moves the verdict by a fixed handful of steps, the same at two decades as at sixteen, because the direction left in view is already growing. Only the intersection of the two hyperplanes hides anything — and it hides less than either curvature can alone, 1,127 steps at worst against 2,907, with random right-hand sides' slowest at 79 against 417.
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.
The certificate that arrives soonest is worth least
The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λₘᵢₙ = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.
Named alongside it
The objects these essays reach for when they reach for this one.
Indefinite matrixSaddle-point systemsReduced hessianInertiaCondition numberLDLᵀ factorisationNegative curvatureQuasi-definite matrixIterative refinementKrylov subspaceStopping criterionActive set