Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Symmetric indefinite — where it appears

A symmetric matrix with eigenvalues of both signs, for which no factorisation restricted to symmetric interchanges and 1×1 pivots need exist. It is the case where the symmetric elimination has to abandon 1×1 pivots, and the smallest example of it is a saddle-point system.

Named by 4 essays across 3 fields — each of them below, with the objects they name alongside it.

-1-0.582271-0.1645420.2531860.6709151.088641.506370eigenvalue4 negative10 positivecounted before it was formedpositive10negative4at zero0innermost ratio39the zero block is a theoremand so is the count either side of it

The zero that is not a missing entry

A constrained minimisation produces a matrix with a zero block, and the zero is a theorem rather than a sparsity pattern. No pivot order makes it positive definite, no precision changes that, and Cholesky does not fail somewhere on it — it fails at the first constraint row, on a number the problem already contained.

constraint · saddle point system 10¹10³10⁵10⁷10⁹10¹¹110¹10²10³10⁴condition number of the matrixgrowth factorbound 2^11partial pivotingCholeskyno pivot to gain fromCholesky growth, every κ1Cholesky interchanges0partial pivoting, worst9the bound, 2^112048both eliminations reach the same growthand only one of them had to swap to get there

A factorisation with nothing to pivot for

Cholesky's growth factor is not bounded by one. It is equal to one, at every size and every condition number, and the two-line reason is why the algorithm needs no pivoting at all — not "usually gets away without it". Its only failure is the square root of a non-positive number, which is exactly the test for definiteness, and in floating point that test moves with the precision.

elimination · cholesky D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

When symmetry is not enough

The matrix [[0, 1], [1, 0]] is symmetric, nonsingular and perfectly conditioned, and there is no diagonal entry to pivot on. Every factorisation restricted to symmetric interchanges and one-by-one pivots fails on it, at any depth of searching, because every entry it could search is zero. The repair is to take two variables at once.

elimination · cholesky -4-2.35371-0.7074290.9388572.585144.231435.8777102468101214shift σν(σ)an answer that is an integershifts2000disagreements0eigenvalues14steps14the marks are a Jacobi decompositionand the staircase never saw one

An eigenvalue count that cannot be slightly wrong

Every spectral computation on this site returns floats with errors in them. Counting eigenvalues below a shift by the signs of an unpivoted elimination returns an integer, and an integer cannot be 6.9999999997 — so the answer is exactly right, or wrong by a whole eigenvalue, and where the second happens is a band of measurable width.

spectra · inertia

Named alongside it

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

CholeskyLDLᵀ factorisationSaddle-point systemsCondition numberConstrained minimisationExact ground truthGrowth factorIndefinite matrixInertiaPartial pivotingBisectionBunch–Kaufman

All concepts