n = 32  ·  h = 1/33  ·  seven methods  ·  four languages  ·  65 machine checks, all PASS

One Update, Four Languages

Richardson, Jacobi, Gauss–Seidel, SOR, SSOR, incomplete (pivoted) Cholesky, two-level — on the grounded Dirichlet chain every one of them is the same line of code, $x \leftarrow x + M^{-1}(b - Ax)$, heard in four languages: a per-node formula, an operator with a spectral radius, gradient descent in transformed coordinates, and a lineman resetting voltages.

Numerics · splitting · Trefethen · Greenbaum
$x_{k+1} = M^{-1}(N x_k + b)$
Physics · statistics · relaxation
reset voltages = take conditional expectations
§1 · The stage

One rod, one matrix, one right-hand side

The system is the suite's Dirichlet chain: $n=32$ interior nodes at spacing $h=\tfrac1{33}$, stiffness $A=\mathrm{tridiag}(-1,2,-1)/h^2$. The diagonal is a constant $2/h^2 = 2178$, end rows included, because the walls are grounded — the free-vs-grounded story is the green tents' §4, and it is why "divide by the diagonal" will keep collapsing to "multiply by a scalar" below. Underneath sits the difference operator: $A = B^\top B$ with $B$ the grounded $33\times32$ incidence matrix — one row per edge, walls included as grounded rails, entries $\pm1/h = \pm33$ (report 09's $D$, Vishnoi's electrical network). Verified entrywise to 2.1×10−16.

Every experiment on this page drives one fixed right-hand side, the green-tents convention: one unit of heat in at node 8, one out at node 24 ($b_8 = +1/h$, $b_{24} = -1/h$), true solution $x^\star = A^{-1}b$. The spectrum runs from $\lambda_1 = 9.8622$ to $\lambda_{32} = 4346.14$, so $\kappa(A) = 440.69$.

Now the claim in the title. Take any of the seven methods below. It ships with a per-node loop (how you would code it), a splitting $A = M - N$ (how Trefethen and Greenbaum write it), a residual correction (how a preconditioner person writes it). These are the same iteration — for all ten $(M,\text{loop})$ pairs the three forms track each other to at worst 2.3×10−14 over five sweeps from a random start. What differs between methods is only which $M$ you can afford to invert. And each $M$ can be heard in four languages:

V1 · weight update

The per-node formula. Jacobi: $x_i \leftarrow (x_{i-1}+x_{i+1}+h^2b_i)/2$, stale neighbors. The splitting $A = M-N$, iteration matrix $T = M^{-1}N = I - M^{-1}A$, error $e_k = T^k e_0$.

V2 · operator

Residual correction $x \leftarrow x + M^{-1}(b-Ax)$: $M$ is a preconditioner, and $\rho(T) = \rho(I - M^{-1}A)$ is report 12's per-sweep unexplained fraction — exact closed forms on this chain.

V3 · coordinates

For symmetric $M = C^\top C$: one step is plain gradient descent in whitened coordinates $z = Cx$. For the sweeps (GS, SOR) no such $C$ exists — they are exact cyclic coordinate descent instead.

V4 · electrical

$A = B^\top B$: nodes carry voltages, edges carry currents. Jacobi resets every voltage to the neighbors' average at once; Gauss–Seidel sends a lineman down the rod; SOR tells him to overshoot.

Here is the whole scoreboard, exact against measured. Rates are spectral radii; "sweeps" counts iterations to relative error $10^{-8}$ on the fixed right-hand side from $x_0 = 0$; "predicted" is $\lceil \ln 10^{-8} / \ln\rho \rceil$.

methodM (the splitting)ρ, closed formρ, exactρ, measuredsweepspredicted
Richardson, α = h²/2$I/\alpha$$\cos\pi h$0.9954720.99547236094059
Jacobi$D$$\cos\pi h$0.9954720.99547236094059
Gauss–Seidel$D+L$$\cos^2\pi h$0.9909640.99096418702030
SOR, ω* = 1.826391$D/\omega+L$$\omega^*-1$0.8263910.826578 †12097 †
SSOR, same ω*$\tfrac{\omega}{2-\omega}(D/\omega+L)D^{-1}(D/\omega+U)$0.9078160.907816174191
IC, rank T = 4$R_4^\top R_4 + \mathrm{diag}\,d$$\tfrac{\kappa-1}{\kappa+1}$, κ = 415.620.9952000.99520035553829
IC, rank T = 8$R_8^\top R_8 + \mathrm{diag}\,d$κ = 306.610.9934980.99349827582825
IC, rank T = 16$R_{16}^\top R_{16} + \mathrm{diag}\,d$κ = 109.730.9819390.9819398991011
IC, rank T = 31$= A$ exactlyκ = 10011
two-level, ω = 0.769949$\big(\omega(ZA_c^{-1}Z^\top + D^{-1})\big)^{-1}$$\tfrac{\kappa-1}{\kappa+1}$, κ = 15.600.8795370.879537133144

† SOR at exactly $\omega^*$ has a defective eigenvalue pair (two Jordan-chained eigenvalues at the kink), so the error carries a polynomial factor $k\rho^k$: the measured asymptotic slope is off by 2.3×10−4 (every other method matches its $\rho$ to ≤ 6.7×10−15 relative), and it lands at 120 sweeps rather than the eigenvalue-only prediction 97. Measured slopes are renormalized power iterations on $T$; both columns come straight from results/update_views.json.

§2 · Interactive

The sweep theater

Before dissecting the languages, watch the methods run. Pick one; play, scrub, or walk a single sweep node-by-node. Gauss–Seidel and SOR update serially — the lineman walks left to right, each node reset against the already-fresh voltage of its left neighbor (SSOR walks there and back) — while Jacobi and Richardson update every node simultaneously from stale values. The trajectories here are recomputed live from the five-line component loops and verified at load against reference values baked in from results/update_views.json.

sweep theater · xk vs x* on the 32-node chain self-test …
iteration k = 0
rel. error  1.0 sweeps to 10⁻⁸ (this run)  3609 reference (JSON)  3609
Gold dashed: $x^\star$. Blue: the current iterate $x_k$, from $x_0=0$. Arrows mark the fixed source ($+1/h$ at node 8) and sink ($-1/h$ at node 24). During "walk one sweep" the node being rewritten flares red — serially for the sweeps, all at once for Jacobi/Richardson. The ω slider re-derives the whole SOR/SSOR trajectory live; at ω = 1 SOR is Gauss–Seidel. The slider's drag grid does not contain ω* exactly — snap ω* sets it to full precision and restores the JSON reference readout. Self-test at load: 15 reference numbers (two profile entries and the sweep count for each method) reproduced by the in-page loops.
§3 · Method card

Richardson

The bluntest instrument: step down the residual with one fixed scalar. Everything below is a dressed-up Richardson — the whole subject is the question "what may I multiply the residual by?"

Richardson α = h²/2 = 4.591×10⁻⁴ρ = 0.995472 = cos πhsweeps 3609M symmetric
V1 · weight update
$x_i \leftarrow x_i + \alpha\big(b_i - (2x_i - x_{i-1} - x_{i+1})/h^2\big)$

Splitting: $M = I/\alpha$, $N = I/\alpha - A$. The optimal step is $\alpha = 2/(\lambda_1+\lambda_{32}) = h^2/2$, and the error obeys $e_k = (I-\alpha A)^k e_0$.

V2 · operator

$T = I - \alpha A$ has eigenvalues $1 - \alpha\lambda_k = \cos k\pi h$ — Jacobi's spectrum exactly (see the interlude). $\rho = \cos\pi h = $ 0.995472, measured slope matches to 1.1×10−16 relative; per report 12, each sweep leaves that fraction of the worst-mode error unexplained.

V3 · transformed coordinates

Raw coordinates. The update is plain gradient descent on $J(x) = \tfrac12 x^\top A x - b^\top x$ with fixed step $\alpha$ — the whitening map is the scalar $C = \alpha^{-1/2} I$, which changes nothing directional. Every other card is this card after a change of variables.

V4 · electrical

Every node bleeds current down its own residual through one fixed conductance $\alpha$ — pure Ohm's law against the imbalance, no averaging. Equivalently (§12): plain gradient descent on the edge energy $\tfrac12\Vert Bx - y\Vert^2$ of the grounded network.

machine check — component loop == $M^{-1}(Nx+b)$ == $x + M^{-1}(b-Ax)$, 5 sweeps: max dev 5.7×10⁻¹⁶ · measured slope vs cos πh: 1.1×10⁻¹⁶ rel
§4 · Method card

Jacobi

Divide the residual by the diagonal: each node takes the conditional expectation of itself given its neighbors — all at once, from the stale board.

Jacobi M = Dρ = 0.995472 = cos πhsweeps 3609M symmetric
V1 · weight update
$x_i \leftarrow \big(x_{i-1} + x_{i+1} + h^2 b_i\big)/2$

Both neighbors read from the old sweep. Splitting: $M = D$, $N = -(L+U)$, so $T_J = -D^{-1}(L+U) = I - D^{-1}A$.

V2 · operator

Eigenvalues $\cos k\pi h$, $k = 1..32$ — symmetric about zero, so the smoothest and roughest modes tie for slowest at $|\cos\pi h|$. $\rho = \mu = $ 0.995472 (eigencheck dev 1.6×10−15); 3609 sweeps to $10^{-8}$.

V3 · transformed coordinates

$M = D = C^\top C$ with $C = D^{1/2} = (\sqrt2/h)\,I = $ 46.669$\,I$ — a global rescale. The constant diagonal means every conditional variance is already equal, so "whiten by the diagonal" turns no direction: the suite's Jacobi no-op. The theorem still holds — one plain-GD step on the whitened energy in $z = Cx$, mapped back by $C^{-1}$, equals one Jacobi sweep at every iterate: 60 sweeps, max dev 1.0×10⁻¹⁴.

V4 · electrical

Every node simultaneously resets its voltage to the average of its neighbors' stale voltages, plus the $h^2\!/2$ source lift. Checked against $D^{-1}(b - (L+U)x)$ and against the padded-average formula to 5.6×10⁻¹⁷.

machine check — splitting equivalence: 3.4×10⁻¹⁶ · ρ(T_J) = cos πh: 1.6×10⁻¹⁵ · voltage-average identity: 5.6×10⁻¹⁷ · whitened-GD equivalence: 1.0×10⁻¹⁴
Interlude — the Jacobi no-op, finally cashed out. On this chain $D = (2/h^2)I = 2178\,I$, so Richardson at $\alpha_{\rm opt} = h^2/2$ has $M = I/\alpha = D$ exactly (dev 2.1×10−16): optimally-tuned Richardson and Jacobi are the same iteration here, and both take exactly 3609 sweeps. "Divide by the diagonal" only becomes a real preconditioner when the diagonal varies; on a constant-diagonal chain it is the global rescale every other report in the suite has been calling the Jacobi no-op.
§5 · Method card

Gauss–Seidel

One word changes: fresh. Update in place, left to right, and each node's left neighbor has already moved. That word breaks the symmetry of $M$ — and switches the method's V3 language entirely.

Gauss–Seidel M = D + Lρ = 0.990964 = cos² πhsweeps 1870M nonsymmetric
V1 · weight update
$x_i \leftarrow \big(x_{i-1}^{\rm new} + x_{i+1}^{\rm old} + h^2 b_i\big)/2$

Same average, but the left neighbor is already fresh. Splitting: $M = D + L$ (lower triangular), $N = -U$; the in-place loop is the triangular solve $M x_{k+1} = b - U x_k$.

V2 · operator

$\rho(T_{GS}) = \cos^2\pi h = \mu^2 = $ 0.990964 (dev 1.1×10−16): squaring. One GS sweep does two Jacobi sweeps' worth of damage — 1870 vs 3609 sweeps, the ratio checked directly. Spectrum: sixteen values $\mu_k^2$ plus zero with multiplicity 16.

V3 · transformed coordinates

No whitening exists. $M = D+L$ is nonsymmetric — $\max|M - M^\top| = 1/h^2 = $ 1089 — so there is no $C$ with $M = C^\top C$. The honest reading: exact cyclic coordinate descent. Each micro-step is the exact line minimum of $J$ along $e_i$ ($x_i \leftarrow x_i + r_i/A_{ii}$, dev 2.2×10⁻¹⁶), leaving $\partial J/\partial x_i = 0$ at that instant (9.3×10⁻¹⁷); the 32 micro-steps compose to one GS sweep exactly. A Gibbs sweep with the noise turned off (report 09 §3).

V4 · electrical

A lineman walks the rod left to right, resetting each node's voltage against fresh upstream voltages — by the time he reaches node $i$, node $i{-}1$ already holds its new value. Serial relaxation: the original 1823 Gauss move, done by hand.

machine check — splitting equivalence: 3.3×10⁻¹⁶ · ρ = cos² πh: 1.1×10⁻¹⁶ · micro-step = coordinate minimum: 2.2×10⁻¹⁶ · gradient zero after each micro-step: 9.3×10⁻¹⁷ · micro-steps compose to the sweep: dev 0
§6 · Method card

SOR — successive over-relaxation

Gauss–Seidel's exact line minimizations are, on this problem, always too timid: the corrections of neighboring nodes are positively correlated, so when the dust of the sweep settles each node wishes it had moved further. SOR moves further on purpose.

SOR M = D/ω + L, ω* = 1.826391ρ = ω* − 1 = 0.826391sweeps 120M nonsymmetric
V1 · weight update
$x_i \leftarrow (1-\omega)\,x_i + \omega\,\big(x_{i-1} + x_{i+1} + h^2 b_i\big)/2$

Move to the GS target, then keep going: the new value is $\omega$ of the way there and beyond. Splitting: $M = D/\omega + L$, $N = D(1/\omega - 1) - U$.

V2 · operator

At $\omega^*$ every eigenvalue of $T$ sits on the circle $|z| = \omega^* - 1 = $ 0.826391 in the complex plane — no single slow mode survives. Measured slope 0.826578: the eigenpair at the kink is defective (a Jordan block), so a polynomial factor $k\rho^k$ rides on top. 120 sweeps vs GS's 1870.

V3 · transformed coordinates

No whitening — $M$ nonsymmetric again ($\max|M-M^\top| = 1089$). The reading is over-relaxed coordinate descent: each micro-step is $x_i \leftarrow x_i + \omega\, r_i/A_{ii}$ — deliberately $\omega = 1.8264$ times past the exact coordinate minimum (dev 4.4×10⁻¹⁶). An intentionally sloppy line search that beats the exact one, because it anticipates the correlated corrections still coming down the rod.

V4 · electrical

The same lineman, but he has learned the pattern: his resets are systematically too timid, every neighbor will keep drifting the same way after he leaves. So he overshoots every reset by the factor $\omega^* = 1.8264$ — and finishes the rod's paperwork fifteen times faster.

machine check — splitting equivalence: 1.1×10⁻¹⁵ · ρ(T(ω*)) = ω* − 1: 2.1×10⁻¹⁵ (defective pair; eig tol 10⁻⁶) · micro-step = ω × CD step: 4.4×10⁻¹⁶
M symmetric positive definite
Richardson · Jacobi · SSOR · IC · two-level
= plain GD on the whitened energy, $z = Cx$, $M = C^\top C$
M triangular (a sweep)
Gauss–Seidel · SOR
= exact cyclic coordinate descent — Gibbs without noise

This dichotomy is the page's second headline. "Preconditioning = change of coordinates" is only half the world. If $M$ is SPD it factors as $C^\top C$ and one preconditioned sweep is one plain gradient-descent step on the whitened problem. Eight of this page's $M$'s are symmetric; the equivalence is verified iterate-for-iterate for the seven nontrivial ones — Jacobi, SSOR, the four IC ranks, two-level (max dev 5.3×10−14 over 60 sweeps) — and Richardson's $M = I/\alpha$ is the trivial eighth, its whitener the scalar $\alpha^{-1/2}I$ of §3. But a sweep's $M$ is triangular: $M - M^\top$ has a full strict triangle of size $1/h^2 = 1089$, no $C$ exists, and no rotation of coordinates makes GS a gradient flow. Its true transformed-coordinates identity runs through coordinate, not gradient, descent: axis by axis, each update an exact conditional expectation — which is why the same loop with noise injected is precisely the Gibbs sampler (report 09 §3), and why symmetrizing the sweep (next card but one) buys the whitening reading back.

§7 · Interactive

The ω story

Why 1.8264 and not 1.5 or 1.9? The answer is Young's theory of consistently-ordered matrices (David Young's 1950 thesis — which also delivers §5's $\rho(T_{GS}) = \mu^2$ as its $\omega = 1$ case): the SOR eigenvalues $z$ solve the quadratic $(z + \omega - 1)^2 = z\,\omega^2\mu_k^2$ against each Jacobi eigenvalue $\mu_k = \cos k\pi h$. Below the optimum the largest root is real and $\mu$-limited; past it, all roots go complex with modulus exactly $\omega - 1$. The two branches cross at a kink — not a smooth minimum — at

$\omega^* \;=\; \dfrac{2}{1+\sin\pi h} \;=\; \dfrac{2}{1+\sqrt{1-\mu^2}} \;=\; 1.8263905415884214,$

the second form being the one in the user's Trefethen-notes margin — identical because $\sqrt{1-\cos^2\pi h} = \sin\pi h$ (the two evaluate within 4.4×10−16; $\sin\pi h = 0.095056$). At the kink $\rho = \omega^* - 1 = (1-\sin\pi h)/(1+\sin\pi h) = $ 0.826391, and 1870 GS sweeps collapse to 120. Note what $\omega^*$ depends on: only $\mu$. Tune Jacobi's rate and SOR's dial comes free.

ω explorer · ρ(TSOR(ω)) on (0, 2) self-test …
ω 1.4000
ρ(ω)  0.98… predicted sweeps ⌈ln 10⁻⁸ / ln ρ⌉   measured sweeps (live run)   at ω*: measured 120 (JSON)
Blue: the quadratic-root formula, evaluated live. Gold dots: the 192-point eigenvalue grid baked in from results/update_views.json (the two agree to 5.1×10−15; self-test at load re-verifies all 192 points and re-finds the kink within 5×10−5). Red: the law $\rho = \omega - 1$, exact for every $\omega \ge \omega^*$. The "measured sweeps" readout actually runs the SOR loop to $10^{-8}$ at your ω — drag toward 0 and watch the price of timidity; drag past the kink and watch $\omega - 1$ take over. The drag grid does not contain ω* exactly (nearest stops land at 125 and 113 sweeps — the defective-pair footnote of §1 explains why a hair past the kink can beat 120); snap ω* lands on it to full precision. The asymmetry is the practical moral: overshooting the optimal ω is cheap, undershooting is ruinous.
§8 · Method card

SSOR — the sweep, symmetrized

Walk the rod, then walk back. The composition of a forward and a backward SOR sweep has a symmetric $M$ — and with symmetry, the whitening language switches back on.

SSOR M = $\tfrac{\omega}{2-\omega}$(D/ω+L)D⁻¹(D/ω+U), ω = 1.826391ρ = 0.907816sweeps 174M symmetric
V1 · weight update
forward SOR sweep, then backward SOR sweep

Two triangular solves per iteration. The product $M = \tfrac{\omega}{2-\omega}(D/\omega+L)\,D^{-1}(D/\omega+U)$ is symmetric positive definite — the sweeps' triangles cancel each other's skew.

V2 · operator

Real spectrum ($M$ SPD). $\rho = $ 0.907816 at the same $\omega^*$, measured to 6.7×10−15 relative; 174 sweeps. Slower per sweep than raw SOR's 0.826391 — each iteration is two sweeps, and symmetry costs rate. What it buys: admissibility as a CG preconditioner, and the next pane.

V3 · transformed coordinates

Whitening returns, in closed form: $C = K = \sqrt{\tfrac{\omega}{2-\omega}}\;D^{-1/2}\big(D/\omega + U\big)$. Then $K^\top K = M_{\rm SSOR}$ and $K^\top = \mathrm{chol}(M_{\rm SSOR})$ — the Cholesky factor is the sweep itself (dev 3.4×10⁻¹⁶). In the sweep-whitened variables $z = Kx$, one plain-GD step equals one SSOR double-sweep at every iterate: 60 sweeps, dev 9.8×10⁻¹⁵.

V4 · electrical

The lineman walks the rod and then walks it back. Each direction alone keeps crooked books — upstream always fresher than downstream — but the round trip's ledger is symmetric, so the double pass acts like a single well-defined network operator rather than a moving wavefront.

machine check — splitting equivalence: 1.2×10⁻¹⁴ · KTK = M and KT = chol(M): 3.4×10⁻¹⁶ · whitened-GD equivalence: 9.8×10⁻¹⁵ · measured slope: 6.7×10⁻¹⁵ rel
§9 · Method card

Incomplete (pivoted) Cholesky — partial Gram–Schmidt

All methods so far invert something cheap and fixed. The factor-analysis preconditioner of report 09 §6 instead buys structure rank by rank: keep $T$ greedily-chosen pivot columns of the Cholesky factorization exactly, replace everything else by its diagonal Schur complement. At $T = n{-}1$ it silently becomes the exact solver.

IC — rank-T pivoted Cholesky M₀ = RTTRT + diag dκ_eff: 415.62 → 306.61 → 109.73 → 1sweeps 3555 / 2758 / 899 / 1M symmetric
V1 · weight update
$x \leftarrow x + \omega\,M_0^{-1}(b - Ax),\qquad \omega = 1.000$ for $T<31$

Pick $T$ pivots greedily (on $A$ they come out odd-first: 1, 3, 5, 7, …). Apply $M_0^{-1}$ by the innovations recipe of report 15: regress the pivot residuals out of the rest, divide by the conditional variances, back-substitute — the $\hat L\,\mathrm{blkdiag}(A_{PP}, \mathrm{diag}\,s)\,\hat L^\top$ form, checked to 4.2×10⁻¹⁶.

V2 · operator

$\kappa(M^{-1}A)$ falls monotonically with rank: 415.62 (T=4) → 306.61 (T=8) → 109.73 (T=16) → 1 (T=31). With optimal damping, $\rho = \tfrac{\kappa-1}{\kappa+1}$: 0.995200, 0.993498, 0.981939, 0 — sweeps 3555, 2758, 899, 1. At $T=31$ the "preconditioner" reconstructs $A$ exactly (dev 2.1×10−16) and the error is dead after one sweep.

V3 · transformed coordinates

$z = $ the innovations coordinates of report 15's truncated regression: pivot variables kept whole, every other variable replaced by its residual after regressing on the pivots. Whitened-GD equivalence at every iterate: 1.1×10⁻¹⁴ (T=4), 4.7×10−15 (T=8), 2.7×10−15 (T=16), 5.3×10−14 (T=31). On a chain the truncated regression is exact at full rank — whence the one-step finish.

V4 · electrical

Gram–Schmidt on $B$'s edge columns is Cholesky on $A = B^\top B$ (Shawe–Taylor §5.2 duality: dev 3.0×10⁻¹⁶). Stop after $T$ columns and you have partial Gram–Schmidt in edge space = incomplete Cholesky on $A$: the first 8 GS rows equal the first 8 rows of $\mathrm{chol}(A)^\top$ to 1.5×10⁻¹⁶, and the pivoted variant picks the same 16 pivots with the same rows, dev 0 (§10).

machine check — splitting equivalence, all four T: ≤ 2.3×10⁻¹⁴ · innovations/LDL form: 4.2×10⁻¹⁶ · κ falls monotonically, T=31 exact: 2.1×10⁻¹⁶ · IC(0) = chol(A) on the chain: 1.5×10⁻¹⁶
§10 · The recursion behind the card

Shawe–Taylor's Algorithm 5.12, run twice

The kernel-methods textbook (Shawe-Taylor & Cristianini, §5.2) writes pivoted partial Cholesky as a regression bookkeeping loop. Verbatim, in its notation, with $K$ the matrix being factored:

$R_{ji} = \dfrac{1}{\nu_j}\Big(K_{ji} - \textstyle\sum_{t<j} R_{tj}R_{ti}\Big), \qquad d_i \leftarrow d_i - R_{ji}^2,\qquad \nu_j = \sqrt{d_{i_j}},\qquad i_j = \arg\max_i d_i$

with $d_i = K_{ii}$ initially and $K = R^\top R$ when run to completion. The $d$ array is the whole story: after $j$ pivots, $d_i$ is exactly the diagonal of the Schur complement — the variance of variable $i$ still unexplained by regression on the pivots chosen so far (checked at every one of the 32 steps: 3.7×10−16 on $\Sigma$, 7.3×10−16 on $A$). Pivoting on $\max d$ is therefore forward-selection regression: always recruit the variable you currently understand least. Stopping after $T$ pivots is the incomplete factorization of the card above; $K = R^\top R$ holds to 2.3×10−16 at full rank, $R[:,\mathrm{pivots}]$ is upper triangular, and $PKP^\top$ reproduces pivoted Cholesky exactly.

Run it on the two natural $K$'s of this suite and it tells two different stories:

On $K = \Sigma = A^{-1}$ — the Brownian-bridge covariance $\Sigma_{ij} = h\,x_{\min}(1-x_{\max})$ of the bridge explainer (closed form checked to 1.4×10−15) — greedy variance bisects: it picks the midpoint node 17 first ($d = 0.007569$, $\nu_1 = 0.08700$; its mirror twin 16 ties to the last bit, so the recorded tape's coin-flips are float noise), then 8 and 25, then 12, 21, 4, 29 … Each pick pins the bridge where it wobbles most, and the unexplained trace collapses from 0.1665 to 0.0831 after a single pivot — half the field's variance explained by one well-chosen node. This is the eigenmodes explainer's leave-wrong ledger discovered by greed: coarse structure first, and the pivot order is a multiscale hierarchy.

On $K = A$ — the precision matrix — the same rule sweeps the odd nodes first: 1, 3, 5, …, 31, then the evens. Greedy variance on the stiffness side discovers red–black ordering, the classic parallel sweep schedule, with no one telling it about graph coloring. Precision pivots buy conditional independence; covariance pivots buy variance — same algorithm, opposite hierarchies.

And the edge-space dual, which makes "incomplete Cholesky" and "partial Gram–Schmidt" one algorithm: orthogonalizing $B$'s columns in their natural order gives $R = \mathrm{chol}(A)^\top$ (dev 3.0×10−16); orthogonalizing with pivot-on-largest-residual-column matches Algorithm 5.12 on $A$ pivot for pivot and row for row (dev 0, all 16 checked). IC(0) on a path graph has no fill-in to drop, so it equals exact Cholesky here (1.5×10−16) — report 15's truncated regression run with nothing truncated.

pivot theater · Algorithm 5.12 on the bridge covariance self-test …
rank T = 0
next pivot  node 17 ν at pick   unexplained trace Σdᵢ  0.166514 as preconditioner at this T  
Bars: the $d$ array — per-node variance still unexplained after regressing on the gold pivots; the outlined bar is the next $\arg\max d$. Watch it take the midpoint, then the quarters, then the eighths: forward-selection regression (report 09 §6) rediscovering bisection. The trace readout matches unexplained_trace_sigma in the JSON at every step; at $T \in \{4, 8, 16, 31\}$ the readout shows the resulting preconditioner's $\kappa(M^{-1}A)$ and sweep count — 415.62 / 3555, 306.61 / 2758, 109.73 / 899, 1 / 1. The bridge is mirror-symmetric, so twins (16/17, 8/25, …) tie in $d$ to the last bit and the recorded tape's tie-breaks are float noise; the theater replays the recorded sequence. Self-test at load: the in-page recursion confirms every recorded pivot attains the maximal $d$ (within 10⁻¹²) and reproduces the initial diagonal, every $\nu_j$, and the full 33-entry trace sequence.
§11 · Method card

Two-level — solve the coarse network exactly

The endgame of the pivot theater's hierarchy: don't just damp the smooth error, solve it. Eight block-average columns $Z$ make an $8\times8$ coarse stiffness $A_c = Z^\top A Z$; invert that exactly and let Jacobi handle what remains.

Two-level (additive) M₀⁻¹ = Z A_c⁻¹ ZT + D⁻¹, ω = 0.769949ρ = 0.879537, κ_eff = 15.60sweeps 133M symmetric
V1 · weight update
$r = b - Ax;\quad$ solve $A_c\,y_c = Z^\top r;\quad x \leftarrow x + \omega\,(Z y_c + D^{-1} r)$

One $8\times8$ solve plus one Jacobi reset, added, then damped by $\omega = 0.7699$. Blocks of four nodes share one coarse unknown ($Z$ columns $= \tfrac14$ on their block).

V2 · operator

Spectrum of $M_0^{-1}A$ squeezed into $[\,0.1565,\ 2.4411\,]$: $\kappa_{\rm eff} = $ 15.60 against 440.69 plain — 28×. Damped, $\rho = $ 0.879537 (measured to 2.5×10−16 rel): 133 sweeps where Jacobi needs 3609.

V3 · transformed coordinates

Coarse modes solved exactly: $\Pi = Z A_c^{-1} Z^\top A$ is the $A$-orthogonal projector onto $\mathrm{range}(Z)$ ($\Pi^2 = \Pi$, $A\Pi = \Pi^\top A$, $\Pi Z = Z$: 1.1×10−15) — the eigenmodes explainer's deflation view — plus a Jacobi-whitened fine residual. Whitened-GD equivalence 1.6×10⁻¹⁴. The honest ledger: Euclidean overlap with mode 1 is 0.989 (largest principal angle 38.29°), yet $A$-energy overlap only 0.296 — block averages are energy-rough, and the win is spectral, not modal.

V4 · electrical

Bundle the rod into 8 supernodes and solve that little grounded network exactly — its 8 voltages are set in one shot, no relaxation — while every fine node also takes one ordinary Jacobi reset against its neighbors. Long-range current bookkeeping done wholesale, retail smoothing on top.

machine check — splitting equivalence: 1.2×10⁻¹⁴ · A-orthogonal projector identities: 1.1×10⁻¹⁵ · whitened-GD equivalence: 1.6×10⁻¹⁴ · measured slope: 2.5×10⁻¹⁶ rel
§12 · The margin notes, formalized

Two lines from the Trefethen notes, made theorems on the chain

"Richardson on $X^\top X w = X^\top b$ is gradient descent on the least-squares problem for $Xw = b$." True, and nothing about it is special to this page: the gradient of $\tfrac12\Vert Xw - b\Vert^2$ is $X^\top(Xw - b)$ — exactly the normal-equations residual — so a fixed-step move down the gradient and a Richardson sweep on the normal equations are the same assignment. With $X = B$ (the grounded difference operator) and $y = Bx^\star$ the edge currents of the true solution ($B^\top y = b$ to 1.7×10−15), plain GD on $\tfrac12\Vert Bx - y\Vert^2$ tracks Richardson on $Ax = b$ for 60 iterates to 6.6×10⁻¹⁵. This is the V4 = V3 bridge: the electrical network's edge energy is the least-squares objective.

"Jacobi for $X^\top X w = X^\top b$ is diagonal Newton." Also true, stated carefully: Newton's method on the least-squares objective would divide the gradient by the full Hessian $X^\top X$; keep only the Hessian's diagonal and you get $w \leftarrow w - \mathrm{diag}(X^\top X)^{-1}X^\top(Xw-b)$, which is precisely Jacobi's $M = D$ sweep. With $X = B$: $\mathrm{diag}(B^\top B) = 2/h^2 = D$ to 2.1×10−16, and the two loops agree over 20 iterates to 4.4×10⁻¹⁶. On this chain the diagonal is constant, so "diagonal Newton" quietly degrades into optimally-scaled gradient descent — the Jacobi no-op of §4, seen through a third window.

§13 · Synthesis

The matrix of methods and languages

One row per method, one column per language. Every cell is a checked identity, not an analogy.

methodV1 · weight updateV2 · operatorV3 · coordinatesV4 · electrical
Richardson$x_i \mathrel{+}= \alpha r_i$; $M = I/\alpha$ρ = cos πh = 0.995472; 3609 sweepsplain GD, raw coordinatesleak current down the residual, fixed conductance
Jacobiaverage stale neighbors + lift; $M = D$ρ = cos πh = 0.995472; 3609GD in $z = (\sqrt2/h)x$ — global rescale, the no-opall voltages reset at once from the stale board
Gauss–Seidelsame, fresh left neighbor; $M = D{+}L$ρ = cos² πh = 0.990964; 1870exact cyclic coordinate descent — no whitening existslineman walks left→right on fresh upstream volts
SORovershoot the GS target by ω; $M = D/\omega{+}L$ρ = ω*−1 = 0.826391; 120coordinate descent, each step ω past the minimumlineman overshoots every reset by 1.8264
SSORforward sweep, then backward; $M$ SPDρ = 0.907816; 174GD in $z = Kx$, the closed-form sweep-whitenerthere-and-back walk; the round trip keeps symmetric books
IC rank Tkeep T pivots exactly + diagonal Schurκ_eff 415.62306.61109.731; sweeps 3555/2758/899/1GD in report 15's innovations coordinatespartial Gram–Schmidt on edge columns = partial Cholesky on A
two-level8×8 coarse solve + Jacobi reset, dampedρ = 0.879537, κ_eff = 15.60; 133coarse modes deflated exactly + whitened fine residualsolve the supernode network wholesale, smooth retail

Read the columns downward and the page compresses to one sentence per language. V1: everyone updates a node from its neighbors and a lift. V2: everyone is $I - M^{-1}A$ racing its spectral radius. V3: the symmetric ones are gradient descent in the right coordinates, the sweeps are coordinate descent, and the better the coordinates, the fewer the sweeps — from 3609 to 1. V4: everyone is a maintenance policy for the same grounded network, from "nudge every voltage" to "solve the substation exactly." One update, four languages — and 65 checks saying none of the translations leak.

Everything above is machine-verified (python/experiments/update_views_checks.py, 65 checks, all PASS, deterministic, 0.4 s; every quoted number and curve exported to results/update_views.json): the ten splitting equivalences component-loop == $M^{-1}(Nx{+}b)$ == $x + M^{-1}(b{-}Ax)$ (≤ 2.3×10−14); the exact rates $\rho(T_J) = \cos\pi h$ (1.6×10−15), $\rho(T_{GS}) = \cos^2\pi h$ (1.1×10−16), $\omega^* = 2/(1{+}\sin\pi h) = 2/(1{+}\sqrt{1-\mu^2})$ (4.4×10−16), $\rho(T_{SOR}(\omega^*)) = \omega^*{-}1$, the 192-point ω-grid vs the quadratic-root formula (5.1×10−15) with $\rho = \omega{-}1$ above the kink and the argmin at $\omega^*$ (5×10−5); measured asymptotic slopes vs every $\rho$ (< 1% rel; SOR's defective pair the only case above 10−14); iterations to $10^{-8}$ with Richardson == Jacobi sweep-for-sweep and GS half of Jacobi; whitened-GD == $M$-preconditioned Richardson for the seven nontrivial SPD $M$'s (Jacobi, SSOR, IC×4, two-level; Richardson's scalar $M$ is the trivial eighth; ≤ 5.3×10−14, 60 iterates) with the closed-form Jacobi scalar $\sqrt2/h$ and SSOR whitener; GS/SOR nonsymmetry ($1/h^2 = 1089$) and the exact-coordinate-descent micro-step identities (≤ 4.4×10−16); Shawe–Taylor Algorithm 5.12 on $\Sigma$ and on $A$ — $K = R^\top R$, triangular $R[:,\mathrm{pivots}]$, pivoted-Cholesky agreement, Schur-diagonal $d$ bookkeeping at every step, recorded pivot sequences (bisection on $\Sigma$, red–black on $A$); the Gram–Schmidt/Cholesky duality on the grounded $B$ — unpivoted, partial (T=8), and pivoted (dev 0) — with IC(0) = chol($A$); the electrical checks $A = B^\top B$ (±33 entries), the voltage-average identity, edge currents $B^\top y = b$, GD-on-least-squares == Richardson and Jacobi == diagonal Newton; the two-level spectrum $[0.1565, 2.4411]$, projector identities, and deflation ledger; and the $n=8$ exact small matrices. Reuses poisson.laplacian_1d and preconditioners.ic0. Companion pages: 05 · classical preconditioners · 09 · stiffness as precision · 12 · autoregressive preconditioning · 15 · preconditioning as prediction · the bridge explainer · the green tents · the eigenmodes explainer.