Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Eigenvalues — where it appears

The numbers λ for which A − λI is singular, which describe a normal matrix completely and a non-normal one incompletely. They describe a normal matrix completely, and for a non-normal one they are a statement about the limit rather than about anything a computation does first.

Named by 9 essays across 4 fields — each of them below, with the objects they name alongside it.

024681012141610⁻¹1index kλthe transformthe eigensolverC = F* Λ F is a factorisationworst relative disagreement2.6·10⁻¹⁵‖Cx − b‖/‖b‖ from the transform solve4.7·10⁻¹⁶imaginary part of a real spectrum1.2·10⁻¹⁶n = 16, and the whole matrix is 16 numberseigenvectors known in advance

The matrix that is one row

A circulant of size 16 is sixteen numbers, has no zero entry anywhere, and hands over its entire spectrum in closed form — the discrete Fourier transform of its first column, exactly. An eigensolver spends a sweep of Jacobi rotations over 256 entries arriving at the same answer, and agrees to 1.2·10⁻¹⁵.

structure · Circulant 01234567810⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹μ, the entry above the diagonalsep, and the eigenvalue gapmin |λᵢ + μⱼ|sep(A, B)the spectra never moveeigenvalue gap, throughout2sep at μ = 02sep at μ = 89.5·10⁻⁴amplification there1056solvability is the eigenvaluesand conditioning is not

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.

structure · Matrix equation 10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁴10⁻¹10¹²10²⁵10³⁸10⁵¹10⁶⁴δ, the gap between consecutive eigenvaluesrelative error in eᴬan answer with no correct digitsV f(Λ) V⁻¹scaling and squaring‖A(δ) − A₀‖exact eigenvalues throughoutκ(V) at the smallest δ3.3·10⁸²eigen route2.9·10⁶⁵scaling and squaring4.1·10⁻¹²distance to the limit7·10⁻¹²the eigenvalues are the diagonaland they are exact at every stop

A function of a matrix is not a function of its entries

Everybody learns that f(A) means diagonalise, apply f to the eigenvalues, undiagonalise. That is a definition, not a method. On a matrix seven picometres from a defective one — with exact eigenvalues and eigenvectors from a closed form — the definition returns an answer wrong by sixty-five orders of magnitude, and a method that never mentions an eigenvalue returns the right one.

spectra · Matrix function share of rank nn = 2 (π/4 = 0.785)0.79n = 30.5n = 80.004234567891010⁻⁴10⁻³10⁻²10⁻¹1n, the size of each sliceshare with rank n0 of 4000dashed: e to the minus 0.087 n squaredthe lower rank stops being typical in practice

The rank that stops being typical

A random 2 × 2 × 2 tensor has rank two with probability π/4 and rank three otherwise, and the sentence has no analogue for matrices. It is the first of a family. An n × n × 2 tensor is a pencil of two slices, and it has rank n exactly when the pencil's eigenvalues are all real, n + 1 otherwise. Over draws, the share of rank n is 0.786 at n = 2, 0.500 at 3, 0.264 at 4, 0.039 at 6, 0.004 at 8 and none of 4,000 at 10 — falling like e^(−0.087n²) — because the mean number of real eigenvalues grows only like the square root of n, to Edelman, Kostlan and Shub's closed form within two per cent. Both ranks stay typical in theory; in practice the lower one disappears.

tensor · Tensor rank 10¹10⁴10⁷10¹⁰10¹³10¹⁶10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹κ(B)relative error, and asymmetryvia B⁻¹Avia Choleskyasymmetry of B⁻¹Au · κ(B)against a spectrum known exactlyslope, via B⁻¹A0.92slope, via Cholesky0.98worst ratio between them2.3asymmetry of B⁻¹A1.1the symmetry claim is trueand it is not about the accuracy

Two matrices and one problem

Ax = λBx is what a finite element model, a structural vibration and a constrained optimisation actually produce, and it is not the one-matrix problem with a change of variables. Everybody is told not to form B⁻¹A because it is not symmetric. That is true, the departure from symmetry is about one, and it is not what decides the accuracy.

spectra · Pencil 012345678-7-5-3-11357conjugate gradient steppᵀAp ⁄ pᵀpλₘᵢₙ = -0.1positive: a step existsnegative: a certificate existsone matrix, two questionsstep it turns at6quotient there-0.027share of λₘᵢₙ recovered0.27λₘᵢₙ, by construction-0.1MINRES steps on the same system37the division that cannot be doneis the answer to a different question

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.

iterative · Breakdown λ = 105 timesλ = 9.55 timesλ = 95 timesλ = 8.55 timesλ = 2.952 timesλ = 2.92 timeseigenvalues that arrived more than once — the matrix has 40 distinct onesa spectrum with the wrong multiplicitiesextra copies, no reorthogonalisation25extra copies, full reorthogonalisation0worst relative error among the copies1.9·10⁻⁸steps taken of 80 asked for, full40no arithmetic error was madeevery one of these is right to eight digits

An eigenvalue that arrives twice

A matrix with forty distinct eigenvalues, handed to Lanczos for eighty steps, returns twenty-five extra copies of thirteen of them — the largest arriving five times. Every copy is accurate to 1.9·10⁻⁸ relative. No arithmetic error was made, nothing overflowed, and a caller counting eigenvalues gets the wrong multiplicity from a computation in which no individual number is wrong.

spectra · Lanczos 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

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.

iterative · Breakdown 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

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.

iterative · Trust-region

Named alongside it

The objects these essays reach for when they reach for this one.

Condition numberKrylov subspaceCertificateCholeskyConjugate gradientsIndefinite matrixNegative curvatureFlop countGeneralised eigenvalue problemMatrix-freeMatrix pencilMINRES

All concepts