Numerical Linear Algebra
where the algebra stops being the arithmetic
Series

Pencil — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 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.

    part 1 · spectra
  2. 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.

    part 2 · spectra
  3. -40-27-14-11225380123456789computed eigenvalueseedevery mark has a residual below 10⁻⁸a small residual, and no answercoefficients of det(A − λB)0worst residual over all seeds1.8·10⁻⁹spread of the answers66seeds drawn8the residual is small at every markand none of the marks means anything

    A problem with no answer

    If two matrices share a null vector then det(A − λB) is identically zero and every λ is an eigenvalue, which means none of them is. Perturb such a pencil by a ten-billionth and a solver returns six numbers with residuals below 10⁻⁹. Change the seed and it returns six different numbers, spread over forty-four, with residuals just as small.

    part 3 · spectra
  4. 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

    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.

    part 4 · spectra

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