Generator

pencil-projective

One function in the pencil library, called 6 times across 1 essay. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 8 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 the eigenvalues of a 6×6 pencil with 2 algebraic constraints, on the projective line. λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 4 finite eigenvalues — -0.3769, 0.565, 4.383, 6.428 — sit inside the arc, and the 2 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 5.53·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

pencil-projective is one function in lib/figures/pencil.js — pencils — two matrices, an eigenvalue with no value, and a problem with no answer. 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.

The eigenvalues of a 6×6 pencil with 2 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 4 finite eigenvalues — -0.3769, 0.565, 4.383, 6.428 — sit inside the arc, and the 2 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 5.53·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.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

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 4 finite eigenvalues — -0.3769, 0.565, 4.383, 6.428 — sit inside the arc, and the 2 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 5.53·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

k: 2

The arguments are the ones An eigenvalue with no value passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The eigenvalues of a 6×6 pencil with 2 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 4 finite eigenvalues — -0.3769, 0.565, 4.383, 6.428 — sit inside the arc, and the 2 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 5.53·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.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

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 4 finite eigenvalues — -0.3769, 0.565, 4.383, 6.428 — sit inside the arc, and the 2 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 5.53·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

k: 0

The arguments are the ones An eigenvalue with no value passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The eigenvalues of a 6×6 pencil with 0 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 6 finite eigenvalues — -0.7663, -0.01384, 0.9593, 4.611, 5.61, 7.6 — sit inside the arc, and the 0 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 8.9·10⁻¹⁵, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.λ = 0λ = ∞λ = ∞-0.7663-0.013840.95934.6115.617.6no constraints: B is nonsingular and nothing is at the polescounted exactly, in rationalsfinite eigenvalues6at infinity0degree of det(A − λB)6worst residual, either kind8.9·10⁻¹⁵an eigenvalue is a ratioand a ratio has a direction, not a size

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 6 finite eigenvalues — -0.7663, -0.01384, 0.9593, 4.611, 5.61, 7.6 — sit inside the arc, and the 0 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 8.9·10⁻¹⁵, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

k: 4

The arguments are the ones An eigenvalue with no value passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The eigenvalues of a 6×6 pencil with 4 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 2 finite eigenvalues — 0.4384, 4.562 — sit inside the arc, and the 4 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.55·10⁻¹⁵, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.λ = 0λ = ∞λ = ∞0.43844.5624 eigenvalues here, and it is one placecounted exactly, in rationalsfinite eigenvalues2at infinity4degree of det(A − λB)2worst residual, either kind1.5·10⁻¹⁵an eigenvalue is a ratioand a ratio has a direction, not a size

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 2 finite eigenvalues — 0.4384, 4.562 — sit inside the arc, and the 4 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.55·10⁻¹⁵, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

k: 1

The arguments are the ones An eigenvalue with no value passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The eigenvalues of a 6×6 pencil with 1 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 5 finite eigenvalues — -0.3786, 0.08367, 3.102, 4.705, 6.487 — sit inside the arc, and the 1 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.42·10⁻¹³, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.λ = 0λ = ∞λ = ∞-0.37860.083673.1024.7056.4871 eigenvalue here, and it is one placecounted exactly, in rationalsfinite eigenvalues5at infinity1degree of det(A − λB)5worst residual, either kind1.4·10⁻¹³an eigenvalue is a ratioand a ratio has a direction, not a size

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 5 finite eigenvalues — -0.3786, 0.08367, 3.102, 4.705, 6.487 — sit inside the arc, and the 1 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.42·10⁻¹³, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

k: 3

The arguments are the ones An eigenvalue with no value passes. A value drawn at the generator's defaults instead would be a picture no essay asked for and no assertion has been run against.

The eigenvalues of a 6×6 pencil with 3 algebraic constraints, on the projective lineλ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 3 finite eigenvalues — -0.04892, 4.357, 4.692 — sit inside the arc, and the 3 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.28·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.λ = 0λ = ∞λ = ∞-0.048924.3574.6923 eigenvalues here, and it is one placecounted exactly, in rationalsfinite eigenvalues3at infinity3degree of det(A − λB)3worst residual, either kind1.3·10⁻¹⁴an eigenvalue is a ratioand a ratio has a direction, not a size

λ = α/β is a ratio, so an eigenvalue of a pencil is a direction rather than a number and it lives on a line whose two ends are the same point. Drawn as an angle φ = arctan λ, the 3 finite eigenvalues — -0.04892, 4.357, 4.692 — sit inside the arc, and the 3 infinite ones sit at its ends, which is one point and not two. Nothing about them is degenerate: each carries a residual ‖βAx − αBx‖ in the same scaling as every other, the largest being 1.28·10⁻¹⁴, and an infinite eigenvalue's residual is ‖Bx‖/‖B‖ — the statement that its eigenvector is a null vector of B. The count is not a rank decision here: it is n minus the degree of det(A − λB), computed in exact rational arithmetic.

What it checked while drawing

Every figure above asserted its own claims on the way to being drawn, and a claim that failed would have failed the build rather than drawn a wrong picture. Those assertions 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.

8 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 number of constraints the pencil can carry

a pencil small enough for the exact determinant

a regular pencil, whose characteristic polynomial is not identically zero

an integer entry, which is what makes the exact route exact

and the infinite ones are the constraints

each with a residual in one scaling

every eigenvalue accounted for, finite and infinite

matmul shapes agree

Against the rule

It draws a decomposition and prints its residual. It calls pencilEigen, 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 113 of 219 generators — 98 print a residual and 15 are exempt with a published reason; 106 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.

The whole library · All essays · What must fail