A basis that stops being a basis, and a Ritz value that keeps improving through it
At its defaults it draws a basis that stops being a basis, and a ritz value that keeps improving through it. ‖QᵀQ − I‖ for the two Krylov bases on a chain of 40 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.37·10⁻¹³ throughout, and the second-order one runs from 1.29·10⁻¹² at four vectors to 1.347 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03036 to 0.001559 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
basis-orthogonality is one function in lib/figures/soar.js —
a krylov space with no matrix — half the storage, twice the ritz values, and a basis that collapses. 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.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 40 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.37·10⁻¹³ throughout, and the second-order one runs from 1.29·10⁻¹² at four vectors to 1.347 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03036 to 0.001559 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
n: 40
The arguments are the ones A Krylov space for a problem that is not linear passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 40 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.37·10⁻¹³ throughout, and the second-order one runs from 1.29·10⁻¹² at four vectors to 1.347 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03036 to 0.001559 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
n: 12
The arguments are the ones The answer that arrives when the space runs out passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 12 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.73·10⁻¹³ throughout, and the second-order one runs from 8.86·10⁻¹³ at four vectors to 4 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02521 to 8.833·10⁻¹⁵ across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
n: 16
The arguments are the ones The answer that arrives when the space runs out passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 16 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 6.83·10⁻¹³ throughout, and the second-order one runs from 3.65·10⁻¹² at four vectors to 2.828 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02878 to 0 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
n: 20
The arguments are the ones The answer that arrives when the space runs out passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 20 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 1.81·10⁻¹³ throughout, and the second-order one runs from 6.59·10⁻¹³ at four vectors to 1.793 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02935 to 9.531·10⁻¹⁶ across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
n: 24
The arguments are the ones The answer that arrives when the space runs out passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
‖QᵀQ − I‖ for the two Krylov bases on a chain of 24 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 4.17·10⁻¹³ throughout, and the second-order one runs from 1.81·10⁻¹² at four vectors to 1.558 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02877 to 2.12·10⁻⁴ across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.
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.
11 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 chain long enough to have a spectrum and short enough to draw
a chain long enough to run twenty steps on
a linearisation has as many eigenvalues as it has rows
a positive mass
a reduction this file knows
and the second-order one does not
damping that removes energy rather than adding it
LU is for square matrices
matmul shapes agree
the linearised basis stays orthogonal
while the answer keeps improving
Against the rule
The rule does not apply to it. It factorises nothing, so there is no residual it could be withholding. That is worth stating rather than leaving blank: a site that reported the rule as satisfied by every generator would be counting mostly generators the rule never reached.
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 Krylov space for a problem that is not linear
A quadratic eigenvalue problem has no matrix to build a Krylov space out of. The recurrence that builds one anyway stores half as many numbers, returns twice as many Ritz values — and stops being a basis at twenty vectors while the answer it gives keeps improving.
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.