Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Infinite eigenvalue — where it appears

An eigenvalue of a pencil whose B-component is zero, which is an algebraic constraint of the model rather than a numerical failure.

Named by 2 essays across 2 fields — each of them below, with the objects they name alongside it.

significand bits10³10⁴10⁵10⁶10⁷10⁸10⁹10¹⁰κ(A)87892634456329625911109435761199127514162024every matrix positive definiteruns producing a false certificate14of runs in total72never above, in significand bits12first κ at eight bits10⁵the comparison was correctand what it proved was not true

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.

error · deliberate zero λ = 0λ = ∞λ = ∞-0.37690.5654.3836.4282 eigenvalues here, and it is one placecounted exactly, in rationalsfinite eigenvalues4at infinity2degree of det(A − λB)4worst residual, either kind5.5·10⁻¹⁴an eigenvalue is a ratioand a ratio has a direction, not a size

An eigenvalue with no value

If the second matrix of a pencil is singular then some of the eigenvalues are infinite, and that is not a degeneracy — it is the algebraic constraints of the model, one per constraint. What survives is a pair of numbers rather than one, and on the line those pairs live on, infinity is an ordinary point with an ordinary residual.

spectra · pencil

Named alongside it

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

Numerical rankBackward errorCertificateCounterexampleDescriptor systemDeterminantExact arithmeticGeneralised eigenvalue problemHalf precisionInvariant subspaceKrylov subspaceMachine epsilon

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