Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

pivot-blocks

One function in the symmetric library, called 16 times across 9 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.

tau: 1

The arguments are the ones A factorisation with nothing to pivot for 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 block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 1The matrix D from PAPᵀ = LDLᵀ, drawn as a matrix. 0 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 pivots1········1········1········1········-1.47········-0.532········-1.53········-0.105three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks1.3·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal10⁻¹⁶growth, blocks2.5growth, diagonal2.2the 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. 0 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: 0.01

The arguments are the ones A threshold between fill and growth 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 block diagonal of a Bunch–Kaufman factorisation of an 8×8 saddle-point matrix at τ = 0.01The 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 pivots0.010.749······0.749·········0.01171.4······1.4·········0.02130.549······0.549·········0.03590.614······0.614·three rules, one matrix‖PAPᵀ − LDLᵀ‖, blocks5.8·10⁻¹⁷‖PAPᵀ − LDLᵀ‖, diagonal1.6·10⁻¹⁵growth, blocks1.3growth, diagonal50the 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 How wide the block should be 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 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.

tau: 0.000001

The arguments are the ones Smoothing a whole line at once 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 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.

tau: 0

The arguments are the ones Two ends of the same arrow 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 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 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.

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 = 10⁻⁶

while diagonal pivoting completes at τ = 10⁻⁸

while none pivoting completes at τ = 0.01

while none pivoting completes at τ = 10⁻⁶

while none pivoting completes at τ = 10⁻⁸

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 90 of 174 generators — 75 print a residual and 15 are exempt with a published reason; 84 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.

Elimination, and the swap

A factorisation with nothing to pivot for

Cholesky's growth factor is not bounded by one. It is equal to one, at every size and every condition number, and the two-line reason is why the algorithm needs no pivoting at all — not "usually gets away without it". Its only failure is the square root of a non-positive number, which is exactly the test for definiteness, and in floating point that test moves with the precision.

 Sparsity, and what elimination costs

A threshold between fill and growth

One number decides how small a pivot an elimination will accept. At 0.001 the factor holds 172 entries and the matrix grows by 1,330; at 1 it holds 260 and grows by 1.2. The libraries ship 0.1, and the measurement says why.

 Eigenvalues, singular values, rank

How wide the block should be

A block narrower than the multiplicity does not converge slowly — it never returns the missing copy at all. Above the multiplicity every extra column buys iterations at about ten products with A each. And the mechanism that is supposed to make the choice unimportant never fires from a random start.

 Iterating, instead of factorising

Smoothing a whole line at once

Solve every grid line in the strong direction exactly rather than sweeping over it, and the smoothing factor goes from 0.9993 back to 0.3340 — which is the one-dimensional answer, on a problem that is not one-dimensional. The repair replaces one ε in the closed form by a one.

 Sparsity, and what elimination costs

Structure and stability stop being separable

The sparsest variable to eliminate on this matrix has a diagonal entry of 10⁻¹². Eliminating it produces the smaller factor, reproduces the matrix to 3.8·10⁻¹⁷ — better than pivoting does — and returns an answer wrong in the fifth digit.

 Where the flop count stopped predicting the time

The same arithmetic at a different price

A blocked and an unblocked elimination perform 72,568 operations each — the same operations, associated differently — choose the same pivots, and return a factorisation identical to the last bit: ‖PA − LU‖/‖A‖ = 4.487946226420872·10⁻¹⁶ in both. One of them moves 41,332 words between fast and slow memory and the other moves 19,476.

 Sparsity, and what elimination costs

Two ends of the same arrow

One matrix, one row moved from the front of the elimination order to the back, and the factor goes from completely dense to no fill at all. Both factorisations are exact to rounding, and nothing numerical chose between them.

 Sparsity, and what elimination costs

What the symbolic phase can only bound

Without pivoting, the fill can be computed from the graph and the count is exact — 233 predicted, 233 measured. With pivoting it is 233 predicted and 242 measured, and what survives is a bound that is right at every threshold and loose by 1.7 times at the largest grid drawn.

 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