The two 6×6 matrices of a Sylvester equation, and the 36×36 matrix it means
At its defaults it draws the two 6×6 matrices of a sylvester equation, and the 36×36 matrix it means. A and B are 6×6 with 11 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 36×36 with 96 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 1.7·10⁻¹⁶, and that X satisfies AX + XB = C to 2·10⁻¹⁶.
sylvester-spy is one function in lib/figures/sylvester.js —
matrix equations — the n²×n² coefficient matrix nobody forms, and sep. 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 and B are 6×6 with 11 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 36×36 with 96 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 1.7·10⁻¹⁶, and that X satisfies AX + XB = C to 2·10⁻¹⁶.
n: 6, mu: 2
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A and B are 6×6 with 11 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 36×36 with 96 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 1.7·10⁻¹⁶, and that X satisfies AX + XB = C to 1.4·10⁻¹⁶.
n: 8, mu: 0
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A and B are 8×8 with 8 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 64×64 with 64 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 0, and that X satisfies AX + XB = C to 8.2·10⁻¹⁷.
n: 8, mu: 4
The arguments are the ones The elimination the matrix does not need passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A and B are 8×8 with 15 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 64×64 with 176 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 3.6·10⁻¹⁶, and that X satisfies AX + XB = C to 3.6·10⁻¹⁶.
n: 6
The arguments are the ones The form that makes it affordable passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.
A and B are 6×6 with 11 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 36×36 with 96 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 1.7·10⁻¹⁶, and that X satisfies AX + XB = C to 2·10⁻¹⁶.
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.
7 distinct claims across 5 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 size the n²×n² grid can be drawn at
a superdiagonal the pair is defined for
and it satisfies the equation
and the coefficient matrix is n² by n²
LU is for square matrices
matmul shapes agree
the Kronecker solve and Bartels–Stewart give one X agree
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.
The elimination the matrix does not need
The Kronecker form of AX + XB = C is dismissed with a hundred million entries and (2/3)n⁶ operations. Both price an elimination, and after the reduction both routes take, the matrix has exactly n³ nonzeros, none of them above the block diagonal, and nothing left to eliminate.
Eigenvalues, singular values, rankThe form that makes it affordable
One Householder reduction, done once, turns every subsequent iteration of the eigenvalue algorithm from cubic to quadratic cost. It changes no answer at all, which is why it is easy to describe as an optimisation and wrong to.