The collection

Every essay — page 15

Essays 337 to 346 of 346, in the same order.

The matrix a constraint makes

Every difficult matrix in the other fields was difficult for a reason the arithmetic supplied: a Hilbert matrix arrives ill conditioned, a Wilkinson matrix grows under elimination, a kernel matrix is dense. Ask a problem to minimise something subject to a constraint and the matrix that results is difficult for a reason the algebra supplies. Its zero block is the second derivative of a Lagrangian with respect to its own multipliers, so no pivot order removes it and no precision changes that; its inertia is known before anything runs; and Cholesky does not fail somewhere on it, it fails at the first constraint row, on a number the problem already contained. Then the field's second surprise, which is the opposite one: an interior-point method drives the condition number of this matrix to 10¹⁵ deliberately, and the answer keeps fifteen digits — because the number that describes the error is not the one every library prints.

00.3670080.7340171.101031.468031.835040eigenvalue of P⁻¹Kwritten down, then computeddistinct3at 16φ computed1.6off the closed form2.9·10⁻¹⁴1 − φ1φthe preconditioner's effect is a theoremand the golden ratio is in 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.

7 figures · Block preconditioning rung 1
012345610⁻³10⁻²10⁻¹110¹log₁₀ κ(A)eigenvalue of P⁻¹Kthe constraint is not in itnontrivial6at one8movement, six decades7.1·10⁻⁶drift at one6·10⁻⁵2m at onethe marks are a pencil that never saw Aand the lines are the preconditioned matrix

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.

8 figures · Block preconditioning rung 2
-14-12-10-8-6-4-2010⁻¹⁷10⁻¹³10⁻⁹10⁻⁵10⁻¹10³10⁷10¹¹10¹⁵log₁₀ μ — the barrier parametercondition number, and relative errorκ₂, condensedκ₂, augmentederror, condensederror, augmentedagainst a BigInt answerκ₂ augmented, μ = 10⁻¹⁴3·10¹⁵its relative error10⁻¹⁵κ₂ condensed2.4·10¹⁶its relative error0.31the same step, written two waysand only one of them is solvable

A condition number sent to infinity

An interior-point method manufactures an ill-conditioned matrix on every iteration, deliberately, because the separating of a diagonal is how it discovers which constraints are active. Written one way the answer keeps fifteen digits at a condition number of 3·10¹⁵. Written the other way — the way almost every code writes it — it has none left.

8 figures · Interior-point conditioning rung 1
-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.69worst growth6.4·10⁵growth × γ0.64the ordering is free to chooseand not free of consequence

The regularisation that legalises every order

Perturb a saddle-point matrix's two blocks in opposite directions and it acquires a factorisation with a diagonal D under every symmetric permutation — not under a good one, under all of them. Five hundred random orderings, five hundred successes, and a growth factor that spans six orders across them.

7 figures · Quasi-definite rung 1
-5-3-11350eigenvalueHZᵀHZK4 negative — Cholesky of H stops at row 30 negative — a minimum on the constraint(10, 4, 0) = In(ZᵀHZ) + (4, 4, 0)one factorisation, no Zpositive, LDLᵀ of K10negative, LDLᵀ of K4negative in H4negative in ZᵀHZ0the count follows the reduced Hessiannot the Hessian

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).

7 figures · Saddle-point systems rung 3
-1-0.500.511.50eigenvalue of P⁻¹Ktriangular [[H, Aᵀ], [0, −Ŝ]] — 1 valuediagonal blkdiag(H, Ŝ) — 3 valuessteps to a residual of 10⁻¹⁰GMRES, triangular2MINRES, diagonal3‖P⁻¹K − I‖54computed |λ − 1| at c = 18.1·10⁻⁸one copy of each value against twoand the counts follow

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.

7 figures · Block preconditioning rung 3
0246810121410⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹interior-point iterationrelative error, and μcertified from iterate 1μthe iterate's errorthe crossover's errorone solve, checkediterations15first certified iterate1iterate error there0.22crossover error there4·10⁻¹⁴the active set arrives long before the digitsand a crossover collects them at once

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.

8 figures · Interior-point conditioning rung 2
-1-0.500.511.520eigenvalue of P⁻¹K1 − φ1φS never formedγ1smallest ν0.056furthest from φ, 1 − φ0.57MINRES steps11an exact Schur approximation, bought by changing Hthe golden ratio without S

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.

7 figures · Block preconditioning rung 4
01234567910δ added to Hpositive pivots−λmin(H) = 4.67|μ| = 1minimumsaddlewhat the signs are countingminimum, true count10saddle, true count9saddle read as minimum from1.1guarantee needs δ past5.1the signs count ZᵀHZ + δInot the curvature the problem has

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.

6 figures · Quasi-definite rung 3
-18-15-12-9-6-300log₁₀ ‖PKPᵀ − LDLᵀ‖ / ‖K‖share of orderingsbestworstexistence and stabilityfactorise1unregularised0.67worst growth5·10⁷growth × γ0.5the ordering is free to chooseand not free of consequence

The perturbation that does the work

A saddle-point matrix made quasi-definite is perturbed in both blocks, and the laws measured for it moved both together. Moved apart, the laws all belong to one block. The zero block's perturbation γ decides whether every ordering factorises, sets the worst ordering's growth at 0.51/γ, and costs the answer 1,451 per unit — the reciprocal of the smallest eigenvalue of AH⁻¹Aᵀ to three figures. The perturbation of H moves none of the first two and costs 19 per unit. Refinement removes each block's perturbation at the rate its own Schur complement sets, so γ's limit sits fifty times nearer than δ's.

7 figures · Quasi-definite rung 2