Numerical Linear Algebra
where the algebra stops being the arithmetic
Generator

one-step-cost

One function in the gramian library, called 6 times across 2 essays. Below: what it draws at its defaults, what it draws at every value an essay asks for, the 9 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 removing one state costs exactly twice the hankel singular value it removed. ‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

one-step-cost is one function in lib/figures/gramian.js — the bound known before the computation — hankel singular values, and the product that halves them. 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2four systems, four ratios, one number: 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

which: "all"

The arguments are the ones The bound that is known in advance 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2four systems, four ratios, one number: 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

which: "positive residues"

The arguments are the ones The bound that is known in advance 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2positive residues: ‖ΔH‖ = 3.886·10⁻⁴, σ = 1.943·10⁻⁴, ratio 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

which: "clustered poles"

The arguments are the ones The bound that is known in advance 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2clustered poles: ‖ΔH‖ = 0.005024, σ = 0.002512, ratio 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

which: "four poles"

The arguments are the ones The bound that is known in advance 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2four poles: ‖ΔH‖ = 8.413·10⁻⁴, σ = 4.207·10⁻⁴, ratio 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

which: "wide spectrum"

The arguments are the ones The bound that is known in advance 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.

Removing one state costs exactly twice the Hankel singular value it removed‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.‖H − Hₙ₋₁‖∞ ÷ σₙpositive residues2.0000mixed residues2.0000clustered poles2.0000four poles2.0000wide spectrum2.0000‖ΔH‖ 3.89·10⁻⁴ σ 1.94·10⁻⁴‖ΔH‖ 3.92·10⁻⁴ σ 1.96·10⁻⁴‖ΔH‖ 0.00502 σ 0.00251‖ΔH‖ 8.41·10⁻⁴ σ 4.21·10⁻⁴‖ΔH‖ 4.6·10⁻⁴ σ 2.3·10⁻⁴2wide spectrum: ‖ΔH‖ = 4.602·10⁻⁴, σ = 2.301·10⁻⁴, ratio 2.0000one state, one equalityand the sum is that, repeated

‖H − Hₙ₋₁‖∞ divided by σₙ, for 5 systems that share nothing but the shape of the question: residues of one sign and of both, a clustered pair, a spectrum spanning three decades. Every one comes back at 2.0000. That is where the bound for a many-state truncation comes from — it is this equality applied once per removed state and the terms added up — and it is why the sum can only be loose in how the removals interact, never in the single step. The measurement is the whole reason to trust a bound that was computed from two Lyapunov solves and no reduced model.

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.

9 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 family this figure has been drawn for

a Lyapunov equation with a solution

a reduction that removes something and keeps something

a stable system

a truncation above the numerical rank of the Gramians

every one-step removal costs twice its own σ

Jacobi needs a symmetric matrix

LU is for square matrices

matmul shapes agree

Against the rule

It calls a factoriser without drawing a factorisation (hankelSingularValues, balancedTruncation, reductionError), so the rule is written down as not applying, with the reason: the bars ARE the ratio of a measured H∞ error to the Hankel singular value that was removed, and both components are printed beside each bar — a residual badge would restate the figure

The exemption list is the interesting half of the rule rather than an escape hatch — it is where a decision about a figure had to be argued in one line. residualcheck refuses an exemption that is not doing work, and rejected ten of the fifteen written for the expansion's figures on exactly that ground: a figure whose vertical axis is a residual satisfies the rule by construction, and touching a factoriser does not by itself require an entry.

Across the library: the rule bites on 192 of 346 generators — 174 print a residual and 18 are exempt with a published reason; 154 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.

Reduction, and what a model is for

The bound that is known in advance

Almost every error on this site is measured after the fact. Balanced truncation has one that is computable before the reduced model exists, in a norm of a function rather than of a residual — and on ordinary problems it is not an upper bound that is loose. It is attained.

 Reduction, and what a model is for

The product nobody had to form

The Hankel singular values are the square roots of the eigenvalues of PQ. Form that product and half of them stop existing, at a floor this site can predict from one number — and the fix is the one the least-squares field has had since its first essay, arriving in a place with no least-squares problem in it.

The whole library · All essays · What must fail