Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 2 above the diagonal

One function in the pseudo library, called 7 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 9 claims it put to the test while drawing them, and where it stands against the rule this site is named for.

At its defaults it draws σₘᵢₙ(zi − a) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 2 above the diagonal. A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.33, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

resolvent-map is one function in lib/figures/pseudo.js — pseudospectra — where the eigenvalues would be, and what the powers do first. Everything below came out of it during this build, at arguments taken from the essays rather than invented for this page. A figure here is the figure a reader meets in an essay, and if the generator changes, this page changes with it.

At its defaults

Drawn even though every essay passes arguments — which on this site is every essay, at 100% of placements since the standard pass. A default nothing exercises is a trap for the next essay to call this with none, and this is the page where a default that has drifted from the figures around it becomes visible.

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 2 above the diagonalA square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.33, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.3eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.33, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

m: 3

The arguments are the ones A spectral radius that grows first passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 3 above the diagonalA square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.54, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.5eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.54, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

m: 2

The arguments are the ones The eigenvalues that are not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 2 above the diagonalA square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.33, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.3eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.33, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

m: 0

The arguments are the ones The eigenvalues that are not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

σₘᵢₙ(zI − A) over the complex plane, for a normal 6×6 matrix with every eigenvalue at 0.8A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. The dark bands are circles centred on the eigenvalue at 0.8, because for a normal matrix σₘᵢₙ is exactly the distance to the nearest eigenvalue and the picture says nothing the spectrum did not. The 10⁻³ level is a disc of radius 10⁻³ around the eigenvalue, which is smaller than one cell of this grid.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level0eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. The dark bands are circles centred on the eigenvalue at 0.8, because for a normal matrix σₘᵢₙ is exactly the distance to the nearest eigenvalue and the picture says nothing the spectrum did not. The 10⁻³ level is a disc of radius 10⁻³ around the eigenvalue, which is smaller than one cell of this grid.

m: 1

The arguments are the ones The eigenvalues that are not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 1 above the diagonalA square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.12, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.1eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.12, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

m: 4

The arguments are the ones The eigenvalues that are not there passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

σₘᵢₙ(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 4 above the diagonalA square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.69, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.7eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it

A square of the complex plane shaded by how small σₘᵢₙ(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.69, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.

What it checked while drawing

Every figure above checked its own claims on the way to being drawn, and a claim that failed would have stopped the picture rather than shipped a wrong one. Those checks used to leave no trace at all: a passing one returned true and the only evidence the figure had checked anything was that nothing crashed. The list below is what they actually said, collected by running this generator with an observer installed — not a description of what it is believed to check.

9 distinct claims across 6 sets of arguments, grouped below by shape — because most of them are one sentence with a different number in it, and how many separate times that sentence was put to the test is the informative part.

a grid the figure has room for and can afford

a size the grid of small SVDs can afford

a superdiagonal between none and the largest drawn

because σₘᵢₙ is exactly the distance to the nearest eigenvalue

matmul shapes agree

on a normal matrix the 10⁻³ level does not leave the unit disc

the 10⁻³ pseudospectrum reaches outside the unit circle

the spectral radius is 0.8 whatever the superdiagonal is

well beyond every eigenvalue

Against the rule

It draws a decomposition and prints its residual. It calls spectrum, spectralRadius, pseudospectrumGrid, spectrum, and every figure above carries the badge — which residualcheck verifies by looking for it in the emitted SVG rather than by finding the call that builds one. A badge that is constructed and then left out of the body is the failure that check exists for.

Across the library: the rule bites on 217 of 397 generators — 199 print a residual and 18 are exempt with a published reason; 180 factorise nothing. Read from lib/residual-rule.js, which is the same body the gate enforces from, and the gate's last check fails the build if this page and it disagree about any generator.

Where it is called

Changing this generator changes every figure on this list. That is what makes the list worth publishing rather than keeping in a check script.

Eigenvalues, singular values, rank

A spectral radius that grows first

ρ(A) below one guarantees that the powers of A go to zero and says nothing about what they do on the way. Here they rise by a factor of twenty thousand before turning over, and the peak is bracketed above and below by a constant computed from the resolvent norms outside the unit circle — two routes to one number, one through the plane and one through the powers.

 Eigenvalues, singular values, rank

The eigenvalues that are not there

For a normal matrix the resolvent norm is exactly one over the distance to the nearest eigenvalue, so a picture of it carries nothing the spectrum did not. Move one entry above the diagonal and the region a perturbation of 10⁻⁸ can put an eigenvalue into stops being a disc and reaches out past the unit circle, while every eigenvalue stays at 0.8.

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