Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

The block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 10⁻⁶

One function in the symmetric library, called 10 times across 3 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 21 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 block diagonal of a bunch–kaufman factorisation of an 8×8 saddle-point matrix at τ = 10⁻⁶. The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

pivot-blocks is one function in lib/figures/symmetric.js — symmetric eliminations — a growth factor of exactly one, and a diagonal that is zero. 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 block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 10⁻⁶The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

n: 8

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.

The block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 10⁻⁶The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

n: 10

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

The block diagonal of a Bunch–Kaufman factorisation of an 10×10 saddle-point matrix at τ = 10⁻⁶The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 5 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶-0.8········-0.8···········2·10⁻⁶0.9········0.9···········2·10⁻⁵1········1············-0.7········-0.73·10⁻⁶··········2·10⁻⁶0.7········0.7·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks1.2·10⁻¹⁶‖PAPᵀ − LDLᵀ‖, diagonal1.9·10⁻¹¹growth, blocks1growth, diagonal2.9·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 5 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

tau: 0.000001

The arguments are the ones When symmetry is not enough passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 10⁻⁶The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749······0.749·········1.2·10⁻⁶1.4······1.4·········2.1·10⁻⁶0.549······0.549·········3.6·10⁻⁶0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.1·10⁻¹¹growth, blocks1.3growth, diagonal5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

n: 4

The arguments are the ones When symmetry is not enough passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The block diagonal of a Bunch–Kaufman factorisation of an 4×4 saddle-point matrix at τ = 10⁻⁶The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 2 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots10⁻⁶0.749··0.749·····1.8·10⁻⁶-0.498··-0.498·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.1·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal3.6·10⁻¹⁷growth, blocks1growth, diagonal7.5·10⁵the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 2 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

tau: 0

The arguments are the ones When symmetry is not enough passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

The block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 0The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.D from PAPᵀ = LDLᵀ — the shaded pairs are 2×2 pivots·0.749······0.749··········1.4······1.4··········0.549······0.549··········0.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal∞growth, blocks1.3growth, diagonal∞the zero block is what the problem saysand one rule does not need it to be nonzero

The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 4 of its blocks are 2×2 — the shaded pairs — and the rest are single entries. A 2×2 block is taken exactly where no single diagonal entry is large enough to divide by safely, which on a matrix whose lower-right block is zero by construction is most of it. The badge carries the residual of all three pivot rules.

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.

21 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 pivot rule this routine implements

a regularisation between none and one

an even size the grid has room for

an even size, so the two blocks are square

and diagonal pivoting fails outright when the diagonal is entirely zero

and none pivoting fails outright when the diagonal is entirely zero

and the diagonal rule is no worse than it there

matmul shapes agree

orders above the block rule's

taking a 2×2 pivot for every pair of variables

taking almost no 2×2 pivots once the diagonal is as large as the coupling

the block rule factorises the saddle-point matrix

to a residual at rounding

while diagonal pivoting completes at τ = 0.01

while diagonal pivoting completes at τ = 10⁻⁴

while diagonal pivoting completes at τ = 10⁻⁶

while none pivoting completes at τ = 0.01

while none pivoting completes at τ = 10⁻⁴

while none pivoting completes at τ = 10⁻⁶

with a growth factor of order 1/τ

with a growth factor of order one

Against the rule

It draws a decomposition and prints its residual. It calls ldl, 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.

Iterating, instead of factorising

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 factorising

The division that cannot be done

Conjugate gradients divides by pᵀAp at every step, and on a matrix that is not positive definite that number can be zero or negative. The guard against it has been here from the first essay and described it as a failure. In the method that made conjugate gradients famous it is the single most valuable object the iteration can produce, and it costs six matrix–vector products.

 Elimination, and the swap

When symmetry is not enough

The matrix [[0, 1], [1, 0]] is symmetric, nonsingular and perfectly conditioned, and there is no diagonal entry to pivot on. Every factorisation restricted to symmetric interchanges and one-by-one pivots fails on it, at any depth of searching, because every entry it could search is zero. The repair is to take two variables at once.

The whole library · All essays · What must fail