The thread: Exact ground truth — page 4
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.
Eigenvalues, singular values, rankA 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.
Eigenvalues, singular values, rankSmall compared to what
This site's own singular value routine has carried a sentence since the month it was written — that one-sided Jacobi computes the small singular values to high relative accuracy and the standard method does not. It has never been measured here, because measuring it needs a σ that is known rather than computed. A bidiagonal matrix and a Sturm count in exact rationals supply one.
Eigenvalues, singular values, rankAccurate is not a property of a method
A bidiagonal matrix whose every entry is 1 or 4096 has singular values spanning thirty decades. On it, the method recommended for small singular values loses the small one by one and a half per cent, the sweep with the theorem behind it does not converge at all, and the shift the theorem is a warning about gets every value to 5·10⁻¹⁶. Nothing there contradicts the theory.
Eigenvalues, singular values, rankAn 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.
Eigenvalues, singular values, rankA threshold the matrix does not set
Two numbers come out of a relative-accuracy comparison and they belong to different things. The size of the matrix moves the constant of the routes that never fail, by a factor of 2.7 between n = 4 and n = 10; it does not move the point where the route through BᵀB stops returning an answer, which sits between ten and eleven decades of grading at every size drawn.
Iterating, instead of factorisingA different equation on every grid
Upwinding is the exact discretisation of a convection–diffusion problem with diffusion ε + h/2, entry for entry, at a relative difference of between 0 and 1.26·10⁻¹⁶ on every mesh from 15 points to 511. The equation it is exact for is chosen by the mesh and not by ε — the added diffusion is 0.01563 on a 31-point grid whether ε is 0.2 or 0.001.
Eigenvalues, singular values, rankThe error the method already knows
Summing the exponential's Taylor series throws away a known number of digits, and the number is on the machine while the sum is being formed. The largest term divided by the answer, times the unit roundoff, tracks the relative error that comes out — to within a factor of nine, across fourteen orders of magnitude of it — and nothing reports it.
Eigenvalues, singular values, rankThe 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.
Iterating, instead of factorisingA 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.
Iterating, instead of factorisingThe 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.