Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

Three iterations to the orthogonal polar factor, κ = 10^4

One function in the polar library, called 5 times across 1 essay. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 13 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 three iterations to the orthogonal polar factor, κ = 10^4. Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 57 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 28 steps to get to the same place.

polar-iterations is one function in lib/figures/polar.js — the polar factor — the nearest orthogonal matrix, and two ways to it without an svd. 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.

Three iterations to the orthogonal polar factor, κ = 10^4Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 57 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 28 steps to get to the same place.159131721252910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps7Newton–Schulz, steps28Newton at step 657scaled Newton at step 64.4·10⁻¹³a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 57 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 28 steps to get to the same place.

logKappa: 4

The arguments are the ones An iteration that only multiplies passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Three iterations to the orthogonal polar factor, κ = 10^4Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 57 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 28 steps to get to the same place.159131721252910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps7Newton–Schulz, steps28Newton at step 657scaled Newton at step 64.4·10⁻¹³a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 57 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 28 steps to get to the same place.

logKappa: 1

The arguments are the ones An iteration that only multiplies passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Three iterations to the orthogonal polar factor, κ = 10^1Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 1.9·10⁻⁶ after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 5. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 11 steps to get to the same place.1471013161910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps5Newton–Schulz, steps11Newton at step 61.9·10⁻⁶scaled Newton at step 67.5·10⁻¹⁶a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 1.9·10⁻⁶ after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 5. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 11 steps to get to the same place.

logKappa: 2

The arguments are the ones An iteration that only multiplies passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Three iterations to the orthogonal polar factor, κ = 10^2Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 0.28 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 6. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 16 steps to get to the same place.1471013161910⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps6Newton–Schulz, steps16Newton at step 60.28scaled Newton at step 61.1·10⁻¹⁵a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 0.28 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 6. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 16 steps to get to the same place.

logKappa: 5

The arguments are the ones An iteration that only multiplies passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Three iterations to the orthogonal polar factor, κ = 10^5Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 563 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 34 steps to get to the same place.1611162126313610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps7Newton–Schulz, steps34Newton at step 6563scaled Newton at step 61.1·10⁻¹²a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 563 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 7. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 34 steps to get to the same place.

logKappa: 3

The arguments are the ones An iteration that only multiplies passes. A value nobody placed would be a picture no essay asked for and no claim was ever checked against.

Three iterations to the orthogonal polar factor, κ = 10^3Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 5.5 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 6. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 22 steps to get to the same place.14710131619222510⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steprelative error in the orthogonal factorNewtonNewton, scaledNewton–Schulzone fixed point, three costsscaled Newton, steps6Newton–Schulz, steps22Newton at step 65.5scaled Newton at step 61.9·10⁻¹⁴a Newton step needs an inverseand a Schulz step needs two products

Newton's iteration, X ← (X + X⁻ᵀ)/2, halves its error per step while it is far away and only becomes quadratic near the end: it is at 5.5 after six steps. Higham's scaling costs two norms and no extra factorisation and reaches thirteen digits in 6. Newton–Schulz, X ← X(3I − XᵀX)/2, uses no inverse at all — two matrix products a step and nothing that reads an entry — and needs 22 steps to get to the same place.

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.

13 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.

the constructed matrix has κ = 10000 — checked 5 times

a conditioning all three iterations still reach

a power of ten rather than an exponent literal

a size the repeated inversions can afford

and Newton–Schulz gets there too

in more of them, which is what it charges for using no inverse

LU is for square matrices

matmul shapes agree

scaled Newton reaches thirteen digits within eight steps

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.

Orthogonality, measured

An iteration that only multiplies

Newton's iteration for the polar factor needs an inverse every step. Newton–Schulz needs only matrix products — nothing that reads an entry, nothing that pivots — and it converges if and only if every singular value is below √3. At 1.73205 it converges and at 1.73206 it returns an orthogonal matrix that is not the answer, with a residual of 5·10⁻¹⁶ and nothing to say so.

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