Numerical Linear Algebra
where the algebra stops being the arithmetic
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The thread: Two routes to a number — page 7

Essays 145 to 151 of the 151 on this theme, in the same order.
σ14.87σ22.55σ31.08σ40.782σ58.04·10⁻¹⁷σ610⁻¹⁸σ710⁻¹⁸n = 7, and det(A − λB) has degree 4an integer, and a judgementdegree of det(A − λB), exactly4infinite eigenvalues, from the degree3singular values below the cut3largest gap in the spectrum∞a degree cannot be nearly threeand a singular value can be nearly zero Eigenvalues, singular values, rank

The largest gap is inside the null space

The rule recommended for counting a pencil's infinite eigenvalues is to cut at the largest gap in the singular values of B. On integer pencils, with no perturbation anywhere and an exact answer available from the characteristic polynomial, it returns the wrong count on nine of twenty-five — because the singular values that are mathematically zero come back spread over a hundred and forty orders of magnitude, and the largest ratio in the list is between two of them.

 0183654729010812610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹products with Asizeresidual boundtrue errorbounded memorybasis vectors kept8products with A140worst error in the k wanted7.1·10⁻¹⁵the bound is free and the error is notand the basis never grows Eigenvalues, singular values, rank

The same budget, spent five ways

A restarted method has one budget — products with A — and two ways to spend it, in many short cycles or a few long ones. At about a hundred and forty products the answer is the same to a factor of seven whichever split is chosen, and the residual bound the method reports spans ten orders of magnitude across the same five runs.

 00.7853981.57082.356193.1415900.250.50.751frequency θdamping |g(θ)|the oscillatory half →⅓ — the symmetric optimumno convectionwith convectionits imaginary parta modulus, not a valuesmoothing factor at this ω0.71best over every ω0.71the symmetric operator's, at ω = 2/30.33the imaginary part does not depend on ωso no ω removes it Iterating, instead of factorising

A smoother that stops being one

Weighted Jacobi's smoothing factor on the convection–diffusion operator is a function of the cell Péclet number and nothing else — identical to eight digits at five grid sizes at matched Pe. It is 0.3335 at Pe = 0.016, exactly 1/√2 at Pe = 1, and 5.2190 at Pe = 7.8, where the sweep amplifies the modes it exists to remove.

 10⁻¹100.20.40.60.8strength threshold θresidual reduction per cycleθ = εsemi-coarseningkept whole rowsfull coarseningone parameter, two methodsbest factor above ε0.053best factor below ε0.1the ratio across the switch231×31 anisotropic operatora switch, not a dial Iterating, instead of factorising

The switch does not know which side is better

The strength threshold moves the coarsening from full to semi at θ = ε exactly, at every anisotropy. Which of the two converges faster is a separate question with a separate answer, and it changes sign between ε = 0.33 and ε = 0.34 — where nothing whatever happens to the switch.

 10⁻³10⁻²10⁻¹1024681012size of the negative eigenvalue, −λproducts before the test firesharder to find, and milder8 spectra, n = 50products at the largest λ3products at the smallest10smallest share of λ recovered0.14largest0.34the one that hidesis the one that matters least Iterating, instead of factorising

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.

 22.32.62.910⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised832storage, second-order416Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations Iterating, instead of factorising

The answer that arrives when the space runs out

A second-order Krylov recurrence holds vectors of length n for a problem with 2n eigenvalues, so it is exact at n steps where the linearised route needs 2n. The machine-precision reading at forty-four vectors on a chain of forty is that exhaustion rather than convergence, and it arrives through a basis whose ‖QᵀQ − I‖ is above one.

 110¹00.30.60.91.2trust-region radius Δshare of the exact model decreasethe exact subproblem7 radii, n = 60share at the smallest radius0.95share at the largest0.3products, at most8radii stopped by the curvature4a few products against an eigendecompositionand most of the decrease Iterating, instead of factorising

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 λ_min = −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.

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