Matrix–vector products before the curvature test fires, against the size of the negative eigenvalue, n = 40
At its defaults it draws matrix–vector products before the curvature test fires, against the size of the negative eigenvalue, n = 40. A 40×40 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 9. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 12 and 42 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
detection-onset is one function in lib/figures/curvature.js —
negative curvature — the division that cannot be done, as an output. 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.
A 40×40 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 9. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 12 and 42 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
n: 50
The arguments are the ones A proof that does not ask how large the matrix is passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A 50×50 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 10. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 14 and 34 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
n: 40
The arguments are the ones A proof that does not ask how large the matrix is passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A 40×40 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 9. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 12 and 42 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
n: 20
The arguments are the ones A proof that does not ask how large the matrix is passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A 20×20 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 11. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 6 and 33 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
n: 80
The arguments are the ones A proof that does not ask how large the matrix is passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A 80×80 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 10. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 24 and 34 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
n: 30
The arguments are the ones A proof that does not ask how large the matrix is passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A 30×30 matrix whose spectrum is positive over one decade apart from a single eigenvalue at −λ. At λ = 3 conjugate gradients meets a non-positive curvature after 3 products; at λ = 0.001 it takes 10. The trend is monotone and it runs the reassuring way: a Krylov space finds large eigenvalues first, so the indefiniteness that takes longest to detect is the indefiniteness that matters least. The direction that fires the test recovers between 4 and 24 per cent of λ, so it is a certificate rather than an estimate of the eigenvalue.
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.
12 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 negative eigenvalue of -3 is found at all — checked 8 times
a size the whole sweep can be run at
and a large one is found sooner than a small one
and at least one of them a substantial share
and the direction always recovers some share of the eigenvalue
Against the rule
It draws a decomposition and prints its residual. It calls
detectionSweep,
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.
A proof that does not ask how large the matrix is
Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.
Iterating, instead of factorisingThe certificate that arrives soonest is worth least
The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λₘᵢₙ = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.