Reduction, and what a model is for

A model that cannot be run

A stable system, reduced by matching its transfer function at three points exactly, comes back with a pole in the right half plane at four of nineteen placements — and matches at all three points to 1.3·10⁻¹⁴ while doing it. The construction did what it promised.

Worth reading first: Exact at the points that were named · The bound that is known in advance.

Exact at the points that were named prices the two ways of building a reduced model and finds them within a fraction of a per cent of each other in the norm each was designed for. It names one advantage balanced truncation has that the cost comparison leaves out — that it cannot return an unstable model, while interpolation can — and it does not show one, because the symmetric families this field is built on never produce one.

They cannot. For a symmetric state matrix with the input and output vectors related by a transpose, the two-sided projection is a Galerkin projection onto a Krylov space, WᵀAV inherits the definiteness of A, and the reduced model is stable by construction. So the field’s own models are the wrong instrument for the question, and the sentence stayed a sentence.

Here is one.

A reduced model that interpolates exactly and has a pole in the right half planeA convection–diffusion system of 30 states at Péclet number 10, whose state matrix is tridiagonal and not symmetric and whose rightmost eigenvalue is -3.49 — stable. It is reduced to order three by two-sided interpolation at {s, 2s, 4s}, and the horizontal axis is s. The curve is the rightmost pole of the reduced model. Above the zero line the model is unstable: it is at 4 of the 19 placements, reaching 24.1. Every one of those unstable models still matches the full system's transfer function at each of its own three interpolation points, to 1.27·10⁻¹⁴ — the construction did exactly what it promised. The dashed line is balanced truncation of the same system at the same order, whose rightmost pole is -0.837: stable, as it is at every order, because stability is a theorem there rather than an outcome.10⁻¹110¹10²-45-35-25-15-551525s, the smallest of the three interpolation pointsrightmost pole of the reduced modelunstable above this linebalanced truncation, order 3exact, and unusablefull system's pole-3.5placements swept19unstable models4worst pole24their interpolation1.3·10⁻¹⁴balanced truncation-0.84the conditions all holdand the model cannot be run
Fig. 1 A convection–diffusion system of thirty states, stable, reduced to order three by two-sided interpolation at {s, 2s, 4s}. The horizontal axis is s; the curve is the rightmost pole of the model that comes back.

Four of nineteen placements return a model with a pole in the right half plane, reaching +24.1 from a system whose own rightmost eigenvalue is −3.49. And the models in that band interpolate: each matches the full system’s transfer function at each of its own three points to 1.3·10⁻¹⁴.

What has and has not gone wrong

Nothing in the construction failed. The bases were built, the projection was formed, the biorthogonality of the two bases holds to the rounding level — it is asserted inside the routine — and the interpolation conditions the method exists to satisfy are satisfied.

The reduced model is a rational function of degree three that agrees with the full system’s transfer function at three prescribed points. That is exactly what was asked for and exactly what was returned. It is also a model with a pole at +24.1, which means the differential equation it stands for has a solution growing like e²⁴ᵗ, and the system it was built from has nothing of the kind.

So the failure is not in the arithmetic and not in the algorithm. It is that the specification was incomplete, and the field’s other method satisfies the missing clause without being asked.

Why the asymmetry is necessary

The state matrix here is a centred-difference convection–diffusion operator: tridiagonal, with off-diagonals ε/h² + 1/2h and ε/h² − 1/2h. The convection term is what makes the two unequal, and the relative asymmetry ‖A − Aᵀ‖/‖A‖ runs from 0.18 at Péclet 10 to 0.69 at 40.

That matters because the whole of the stability guarantee for a symmetric problem comes from the projection preserving a quadratic form. For symmetric A with W = V, xᵀ(VᵀAV)x = (Vx)ᵀA(Vx) < 0 for every x, so the reduced matrix is negative definite whenever A is. With W ≠ V nothing of the kind holds: WᵀAV is a projection of A onto one space along another, and a projection along a different space can move an eigenvalue anywhere.

This site already knows the family that makes such things happen. It is the one the direction the error leans and the convection field’s essays are built on: a non-normal operator, whose eigenvalues are real and negative and whose behaviour is not described by them. A projection of a non-normal operator is where the gap between the spectrum and the pseudospectrum turns into an outcome, which is the spectrum stops describing the matrix’s subject and is the honest explanation of this figure.

The band, and what decides it

The instability is not a rule and it is not an accident: it is a band in the shift placement, and both edges of it are visible on the figure.

At small s the three interpolation points sit far below the interesting part of the frequency response; the projection is close to a moment matching at the origin, the bases are well conditioned, and the model is stable. At large s they sit far above it, the bases converge towards the leading directions, and the model is stable again. In between — where the points are near the frequencies at which the non-normality shows — the two bases are nearly orthogonal to each other, WᵀV is ill conditioned before it is normalised, and the projected matrix is a badly conditioned object whose eigenvalues have moved.

The drag says the band moves with the Péclet number and does not close: three placements of nineteen at Pe = 4, six at 16 and at 40. So a caller cannot avoid it by a rule about where to put shifts, because where the band is depends on the system.

What balanced truncation does instead

The dashed line on the figure is balanced truncation of the same system at the same order. Its rightmost pole is −0.84, and at orders two to six the poles are −1.50, −0.84, −1.49, −2.33 and −3.27. Stable at every order, and not by luck.

Balanced truncation of a stable system is stable — it is a theorem, and its proof is the one why a Gramian can be truncated at all is built around: the balanced realisation has a Lyapunov solution which is diagonal and positive, and the leading block of a positive definite Lyapunov solution is a positive definite Lyapunov solution for the leading block of the system. The truncation inherits a certificate rather than being tested for one.

That is the shape of the difference, and it is worth stating in the site’s usual terms. Interpolation gives what was asked for, exactly, at r linear solves. Balanced truncation gives what was asked for approximately, at two Lyapunov equations and a pair of cubic factorisations, together with a guarantee nobody asked for and an error bound computable before the model exists.

Two approximants and one matrix size prices the first half of that trade and finds the cheap method within half a per cent of the expensive one. This figure is the other half of the price, and it is not a percentage: it is a failure mode that either happens or does not.

What the interpolation error looks like when the model is unstable

Worth measuring rather than assuming, because a reader’s next thought is that the model must be wrong somewhere even if it is right at the three points.

It is. The interpolation conditions are three equations and the model has nine degrees of freedom, so matching at three points constrains a small part of it — everything between and beyond the points is whatever the projection produced. On a stable placement that remainder is a reasonable approximation of the response; on an unstable one it contains a pole in the right half plane, which means the model’s transfer function has a singularity where the true one is analytic.

So the honest description is not right at three points and wrong elsewhere, which is true of every interpolant. It is: the interpolant has a feature the original does not have, and that feature is not small, does not appear in the interpolation error, and is not what an error norm would report either — the H∞ norm of the difference is computed over the imaginary axis, and a right-half-plane pole does not sit on it. A model with a pole at +24.1 can have a perfectly ordinary-looking frequency response.

That is why this failure is a stability question rather than an accuracy one, and why the check is a check on the poles rather than a check on an error.

How much of the field this changes

Not much, and saying so precisely is part of the finding.

The cost comparison stands. Two approximants and one matrix size prices interpolation at r solves against balanced truncation’s two Lyapunov equations and finds the cheap method within half a per cent in H₂. That measurement is unaffected: it was taken on symmetric systems, where the stability question does not arise, and the numbers are what they are.

The error bound stands. The 2Σσ bound the bound that is known in advance measures at a ratio of 1.0000 belongs to balanced truncation and is untouched by anything here.

What changes is the decision the comparison feeds. A reader who takes the cost comparison as a recommendation is choosing a method whose failure mode this essay measures, and the comparison did not price it. The repair is one clause: interpolation is the cheaper method and requires a check the other does not, and the check is a line of code — which is why the honest conclusion is a caveat rather than a reversal.

And one thing gets worse rather than better. On a large system the reduced model’s poles are cheap to compute and the full system’s are not, so the check confirms that the model is stable without confirming that it should have been. A reduced model with a right-half-plane pole from a system whose stability is unknown is ambiguous evidence, and the field has no cheap instrument for resolving it.

What a practitioner does about it

Four things, and the ordering is the useful part.

Check. The eigenvalues of a 3 × 3 or 10 × 10 reduced state matrix cost nothing. A method that can return an unstable model and is used without checking is a method used carelessly, and the check is one line.

Move the shifts and try again. The band is a band. Four placements of nineteen fail here, so a second attempt at a different scale usually succeeds — which is what makes this a nuisance rather than a catastrophe, and also what makes it easy to never notice.

Or use IRKA, and check anyway. The iteration that places its shifts at the mirror images of its own poles converges, when it converges, to a model satisfying a necessary condition for optimality — interpolating at the model’s own poles is that essay. A fixed point of it has poles in the left half plane by construction of the mirroring, so the converged model is stable; the intermediate ones need not be, and a run stopped early can return one of them.

Or pay for the certificate. If the reduced model is going to be simulated, or put inside a controller, or handed to somebody who will not check, the method whose stability is a theorem is worth two Lyapunov solves. The reason this figure exists is that the cost comparison in this field was quoting a price without that clause in it.

Why this is the field’s own version of a familiar shape

The machine field arrives at nearly the same sentence from the opposite direction, which is worth recording.

There, a computation satisfies every guarantee anybody stated — every run backward stable, every answer within its bound — and two of them disagree, because the specification had no clause about agreement between runs. Here, a construction satisfies every condition anybody stated — the interpolation holds, the biorthogonality holds — and the result is unusable, because the specification had no clause about stability.

In both cases the missing clause is a property nobody thought to ask for, and in both cases the more expensive alternative supplies it without being asked: order-independent summation there, balanced truncation here. And in both cases the price of the expensive option had been quoted with the extra property left out of the comparison, which is exactly the arithmetic this site keeps finding.

The general form, since this collection has now met it twice: when a cheap method and an expensive one are compared on the quantity the cheap one optimises, the cheap one wins, and the comparison is not the decision.

What the site’s own gate would have caught

A closing note about the shape of the omission, since it is one this collection makes a habit of recording.

Every generator on this site asserts something about its own arguments before it draws. The reduction field’s generators assert interpolation errors, bound ratios, Gramian agreement with a closed form — every quantity the essays argue about. None of them asserted that a reduced model was stable, because on the symmetric families used there, stability was never in question.

That is exactly the pattern the fleet’s own rules warn about: an assertion that has never rejected anything proves nothing, and an assertion nobody wrote because the case never arose is worse. The routine that builds a projection now returns its poles’ rightmost real part, and the figure asserts that the sweep contains both stable and unstable models — which would fail if either the construction stopped producing the failure or started producing nothing else.

The system, and why it is the right instrument

A note on the construction, because a demonstration that a method fails is only worth having if the case is not contrived.

The state matrix is a centred-difference discretisation of −εu″ + u′ on the unit interval: the convection–diffusion operator this site’s convection essays are built on, at thirty states. The actuator and the sensor are random vectors from a seeded stream, so the transfer function has no structure a projection could accidentally respect.

Three things make it the right instrument rather than a construction aimed at the answer.

It is stable, and that is checked rather than assumed. The rightmost eigenvalue is −3.49 at Péclet 10 and negative at every Péclet number in the sweep, checked whenever the figure is drawn. A demonstration that a projection destroys stability is worth nothing if the system did not have any.

It is a real discretisation. Nothing about it was chosen to fail; it is the operator a convection-dominated flow gives, at a Péclet number the centred difference can carry, and the same operator the site has drawn a dozen times for other reasons.

And the failure is not everywhere. Fifteen of nineteen placements return a stable model, which is the reason this is a hazard rather than a defect — and the reason the assertion published with it has to be a refusal of the claim that every placement fails.

One line

A reduced model that matches a stable system’s transfer function exactly at every point it was asked about can have a pole in the right half plane, and the method whose stability is a theorem costs two Lyapunov solves more.

At other settings

A reduced model that interpolates exactly and has a pole in the right half planeA convection–diffusion system of 30 states at Péclet number 25, whose state matrix is tridiagonal and not symmetric and whose rightmost eigenvalue is -6.888 — stable. It is reduced to order three by two-sided interpolation at {s, 2s, 4s}, and the horizontal axis is s. The curve is the rightmost pole of the reduced model. Above the zero line the model is unstable: it is at 4 of the 19 placements, reaching 614. Every one of those unstable models still matches the full system's transfer function at each of its own three interpolation points, to 3.59·10⁻¹³ — the construction did exactly what it promised. The dashed line is balanced truncation of the same system at the same order, whose rightmost pole is -3.73: stable, as it is at every order, because stability is a theorem there rather than an outcome.10⁻¹110¹10²-9213118223328433538643s, the smallest of the three interpolation pointsrightmost pole of the reduced modelunstable above this linebalanced truncation, order 3exact, and unusablefull system's pole-6.9placements swept19unstable models4worst pole614their interpolation3.6·10⁻¹³balanced truncation-3.7the conditions all holdand the model cannot be run
Fig. 2 At Péclet 25, between the two extremes of the drag.
Hankel singular values, and the half of them a product cannot seeσₖ divided by σ₁, for a 26-state model of McMillan degree 12. The lower curve is the square-root route — Cholesky-like factors of the two Gramians and one SVD of RᵀS — and it descends to the level at which the Gramians were computed. The upper curve eigendecomposes the product PQ, agrees for the first 6 values and then stops, flattening at 8.07·10⁻¹⁰. The dashed line is σ₁√u = 2.32·10⁻⁹, which is where a route that squares the conditioning must stop: the eigenvalues of PQ are σ², an absolute error of uσ₁² on them is a relative error of √u on σ. κ(PQ) is 1.58·10³⁸ against 1.26·10¹⁹. Every σ below the line is a term in the error bound, so this is not an academic loss.1357911131510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1index kσₖ ÷ σ₁σ₁√utwo factorstheir productthe normal equations, againroutes agree to k =6product floors at8.1·10⁻¹⁰σ₁√u2.3·10⁻⁹κ(P)κ(Q)1.6·10³⁸√ of it1.3·10¹⁹do not form the productthe σ below the line are the bound
Fig. 3 The singular values a truncation reads, on a larger system.
IRKA's interpolation points walking to the condition that defines themEach curve is one of 3 interpolation points across 15 iterations of σ ← −λ(Aᵣ). They start logarithmically spread over the frequency range, which is a guess, and settle at 9.91, 45.25, 156. What is checked at the end is not that they stopped moving but that where they stopped is the first-order condition: the mirrored poles of the model they produced agree with them to 4.06·10⁻¹². A fixed-point iteration that stops moving without satisfying its own condition has converged to nothing, and this one has no convergence proof to lean on — so the condition is checked rather than the movement.012345678910111213141510⁻¹110¹10²10³iterationinterpolation point σa fixed point that is a conditionorder3iterations15σ against −λ(Aᵣ)4.1·10⁻¹²H₂ error6.5·10⁻⁴it stopped movingand the condition holds there
Fig. 4 The iteration at a smaller order, whose fixed point cannot have the pole this essay is about.
Why a Gramian can be truncated at all: a rational approximation problemλₖ₊₁(P)/λ₁(P) for the controllability Gramian of a 20-state model, against Z_k², where Z_k is the smallest a degree-(k,k) rational function with poles on one side can be made on the other. The Gramian's spectrum falls from 1 to 4.91·10⁻¹⁰ in 10 steps, and the bound falls with it. Nothing about the model enters the bound except the two ends of the spectrum of −A, 9.85 and 1754, whose ratio is 178.1 — so the decay that makes model reduction possible is predictable from two numbers before the Gramian exists. That is the fact this whole field rests on and it is a fact about rational approximation.1234567891010⁻¹⁹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹kλₖ₊₁ ÷ λ₁, and the boundZ_k²the Gramianpredicted from two numbersstates20κ of the spectrum178λ11 ÷ λ₁4.9·10⁻¹⁰the bound there1.5·10⁻⁴the cliff everything rests onand the reason for it
Fig. 5 The decay, on a system of the size this essay reduces.
What the solver reported, and what it achievedFinal relative residual and final relative error against the working precision, on a logarithmic vertical axis. The residual sits at the stopping tolerance at every precision — the method converged, by every test available to it. The error runs from 1.4·10⁻¹³ at 53 bits to 0.00597 at 8, tracking the unit roundoff, which is drawn beside it.614223038465410⁻¹³10⁻¹⁰10⁻⁷10⁻⁴working significand bitsrelative sizeerrorresidualuthe gap the solver cannot seeerror ÷ residual at 24 bits1.5·10⁶error ÷ residual at 16 bits2.3·10⁷error ÷ residual at 8 bits2.4·10¹⁰every convergence test passesand the answer is wrong in proportion to u
Fig. 6 A measurement that cannot see what it is asked about, which is what an interpolation error is here.
A reduced model that interpolates exactly and has a pole in the right half planeA convection–diffusion system of 30 states at Péclet number 4, whose state matrix is tridiagonal and not symmetric and whose rightmost eigenvalue is -3.461 — stable. It is reduced to order three by two-sided interpolation at {s, 2s, 4s}, and the horizontal axis is s. The curve is the rightmost pole of the reduced model. Above the zero line the model is unstable: it is at 3 of the 19 placements, reaching 21.8. Every one of those unstable models still matches the full system's transfer function at each of its own three interpolation points, to 2.94·10⁻¹⁵ — the construction did exactly what it promised. The dashed line is balanced truncation of the same system at the same order, whose rightmost pole is -3.14: stable, as it is at every order, because stability is a theorem there rather than an outcome.10⁻¹110¹10²-6-22610141822s, the smallest of the three interpolation pointsrightmost pole of the reduced modelunstable above this linebalanced truncation, order 3exact, and unusablefull system's pole-3.5placements swept19unstable models3worst pole22their interpolation2.9·10⁻¹⁵balanced truncation-3.1the conditions all holdand the model cannot be run
Fig. 7 At Péclet 4, where the asymmetry is mild and three placements of nineteen still fail.
A reduced model that interpolates exactly and has a pole in the right half planeA convection–diffusion system of 30 states at Péclet number 40, whose state matrix is tridiagonal and not symmetric and whose rightmost eigenvalue is -11.53 — stable. It is reduced to order three by two-sided interpolation at {s, 2s, 4s}, and the horizontal axis is s. The curve is the rightmost pole of the reduced model. Above the zero line the model is unstable: it is at 6 of the 19 placements, reaching 106. Every one of those unstable models still matches the full system's transfer function at each of its own three interpolation points, to 2.57·10⁻¹³ — the construction did exactly what it promised. The dashed line is balanced truncation of the same system at the same order, whose rightmost pole is -4.53: stable, as it is at every order, because stability is a theorem there rather than an outcome.10⁻¹110¹10²-35-14728497091112s, the smallest of the three interpolation pointsrightmost pole of the reduced modelunstable above this linebalanced truncation, order 3exact, and unusablefull system's pole-12placements swept19unstable models6worst pole106their interpolation2.6·10⁻¹³balanced truncation-4.5the conditions all holdand the model cannot be run
Fig. 8 And at 40, where the band has moved and widened.
A reduced model of order 3, and the 3 places it is exact|H(s) − Hᵣ(s)| ÷ |H(s)| along the real axis, for a rational-Krylov reduction of a 24-state model at the interpolation points 1.5, 5, 16. At each of them the curve falls to 1.5·10⁻¹⁵ — the reduced function passes through the original, and its derivative does too, because the projection is two-sided. Between and beyond them it reaches 0.00915, and there is no bound on it: the method buys 6 exact conditions for 3 solves and offers nothing anywhere else. That is the trade against balanced truncation, which asks for nothing and bounds everything at a cost of two Lyapunov solves.10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints3conditions bought6worst at a point1.5·10⁻¹⁵worst away from one0.00913 points, 6 conditionsand no bound in between
Fig. 9 The conditions that hold whether or not the model can be simulated.
The error a reduced model has, and the bound computed before it existed‖H − Hᵣ‖∞ for balanced truncation of a 20-state model, against 2Σ_{k>r}σₖ — a number available from two Lyapunov solves before any reduced model is formed — and against σᵣ₊₁, which is a lower bound no reduced model of that order can beat. The measured error and the upper bound are the same curve: the ratio is 1.0000 to 1.0000 across the sweep, spread 1.0000. The bound is attained rather than approached, because on a spectrum falling this fast the tail sum is its own first term and removing one state costs exactly twice it. The measured error sits at 2.03 times the lower bound, so the whole question of what order to truncate at is answered by a curve nobody has to compute a model to see.123456710⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹order r kepterror and bounds2Σσ, and the errorσᵣ₊₁a bound that is an equalityorders7bound ÷ error, worst1spread over the sweep1error ÷ σᵣ₊₁, worst2.1computed before the modeland attained by it
Fig. 10 The bound the other method carries, computable before the model exists.
Hankel singular values, and the half of them a product cannot seeσₖ divided by σ₁, for a 20-state model of McMillan degree 12. The lower curve is the square-root route — Cholesky-like factors of the two Gramians and one SVD of RᵀS — and it descends to the level at which the Gramians were computed. The upper curve eigendecomposes the product PQ, agrees for the first 6 values and then stops, flattening at 9.67·10⁻¹⁰. The dashed line is σ₁√u = 2.32·10⁻⁹, which is where a route that squares the conditioning must stop: the eigenvalues of PQ are σ², an absolute error of uσ₁² on them is a relative error of √u on σ. κ(PQ) is 1.8·10³⁶ against 1.34·10¹⁸. Every σ below the line is a term in the error bound, so this is not an academic loss.1357911131510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1index kσₖ ÷ σ₁σ₁√utwo factorstheir productthe normal equations, againroutes agree to k =6product floors at9.7·10⁻¹⁰σ₁√u2.3·10⁻⁹κ(P)κ(Q)1.8·10³⁶√ of it1.3·10¹⁸do not form the productthe σ below the line are the bound
Fig. 11 The decay that makes the other method possible at all.
Why a Gramian can be truncated at all: a rational approximation problemλₖ₊₁(P)/λ₁(P) for the controllability Gramian of a 30-state model, against Z_k², where Z_k is the smallest a degree-(k,k) rational function with poles on one side can be made on the other. The Gramian's spectrum falls from 1 to 2.48·10⁻⁸ in 10 steps, and the bound falls with it. Nothing about the model enters the bound except the two ends of the spectrum of −A, 9.86 and 3834, whose ratio is 388.8 — so the decay that makes model reduction possible is predictable from two numbers before the Gramian exists. That is the fact this whole field rests on and it is a fact about rational approximation.1234567891010⁻¹⁹10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹kλₖ₊₁ ÷ λ₁, and the boundZ_k²the Gramianpredicted from two numbersstates30κ of the spectrum389λ11 ÷ λ₁2.5·10⁻⁸the bound there5.2·10⁻⁴the cliff everything rests onand the reason for it
Fig. 12 And the reason that decay is there.
Each method wins the norm it was designed for, and by less than anyone guessesRatios of the loser's error to the winner's, on 4 systems. In H₂ the interpolatory model wins every time, by 0.8%, 1.3%, 0.4%, 1.9% — IRKA satisfies the first-order conditions for that norm and balanced truncation does not. In H∞ balanced truncation wins every time, by 16%, 23%, 12%, 25%, which is the norm its bound is stated in. Both orderings hold on every system, so this is not a coin flip; and the H₂ margin is under two per cent, so a method costing r solves lands within two per cent of one costing two Lyapunov solves in the norm the cheap one optimises. The choice in this field is about cost.loser's error ÷ winner's error24 states, degree 8, r = 4 · H₂1.007924 states, degree 8, r = 4 · H∞1.155824 states, degree 8, r = 3 · H₂1.012624 states, degree 8, r = 3 · H∞1.232630 states, degree 10, r = 5 · H₂1.004230 states, degree 10, r = 5 · H∞1.123424 states, degree 12, r = 4 · H₂1.019424 states, degree 12, r = 4 · H∞1.2480IRKA wins the average, balanced truncation the maximum, on every systemH₂ to the interpolationH∞ to the bound
Fig. 13 The comparison this essay adds a clause to.
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
Fig. 14 What removing one state costs, in the method with the theorem.
IRKA's interpolation points walking to the condition that defines themEach curve is one of 4 interpolation points across 13 iterations of σ ← −λ(Aᵣ). They start logarithmically spread over the frequency range, which is a guess, and settle at 9.861, 40.15, 102.1, 302.8. What is checked at the end is not that they stopped moving but that where they stopped is the first-order condition: the mirrored poles of the model they produced agree with them to 2.81·10⁻¹². A fixed-point iteration that stops moving without satisfying its own condition has converged to nothing, and this one has no convergence proof to lean on — so the condition is checked rather than the movement.01234567891011121310⁻¹110¹10²10³iterationinterpolation point σa fixed point that is a conditionorder4iterations13σ against −λ(Aᵣ)2.8·10⁻¹²H₂ error4.3·10⁻⁵it stopped movingand the condition holds there
Fig. 15 The iteration that places its own shifts, whose fixed point is stable by construction.
σ_min(zI − A) over the complex plane, for a bidiagonal 6×6 matrix with every eigenvalue at 0.8 and 3 above the diagonalA square of the complex plane shaded by how small σ_min(zI − A) is. Every eigenvalue is at 0.8, marked by the crosshair; the 10⁻³ level reaches out to 1.54, well outside the unit circle drawn through the picture. A perturbation of the matrix at the level of double-precision rounding can put an eigenvalue anywhere in that region.darker is smaller: 10⁻¹⁰, 10⁻⁶, 10⁻³, 10⁻¹one spectrum, two matricesspectral radius0.8reach of the 10⁻³ level1.5eigenvalues, all at0.8the circle is |z| = 1and every eigenvalue is well inside it
Fig. 16 The spectrum that does not describe the matrix, which is the honest explanation.
‖Aᵏ‖ for a 6×6 matrix whose spectral radius is 0.8, with 3 above the diagonalTwo curves against the power. The norm of Aᵏ rises to 1.5·10⁵ at step 24 before turning over and decaying to 1.9·10⁻⁴ by step 160; ρᵏ = 0.8ᵏ, drawn beside it, falls from the start. The two horizontal lines are the Kreiss bracket: the peak is at least K = 50880 and at most e·n·K = 8.3·10⁵, both computed from the resolvent norms outside the unit circle and not from the powers at all.027548110813510⁻⁵10⁻³10⁻¹10¹10³10⁵10⁷power‖Aᵏ‖Kreiss constant 50900e · n · K‖Aᵏ‖ρᵏtwo routes to one peakspectral radius0.8peak of ‖Aᵏ‖1.5·10⁵Kreiss constant5.1·10⁴e · n · K8.3·10⁵everything here decays in the endand one of these curves says how much first
Fig. 17 And what a non-normal operator does before its eigenvalues take over.
−εu″ + u′ = 0 on 31 points, ε = 0.01, cell Péclet 1.563Three solutions of a boundary-layer problem on the same grid. The exact one rises monotonically from 0 to 1 with a layer of width ε at the right-hand end. The central-difference answer alternates at every point and leaves [0, 1] at 11 of the 31 of them, by as much as 0.2195. The upwind answer is monotone at every Pe.00.250.50.75100.250.50.751xuexactcentral differencesupwindthe oscillation is exact‖Ax − b‖/‖b‖ for the central answer1.2·10⁻¹⁶values outside [0, 1]11worst excursion0.22the dashed lines are 0 and 1, which the equation guaranteesno solver was involved
Fig. 18 The operator this essay’s system is, in the field that first drew it.
Backward and forward error against the condition numberA log–log plot over twelve decades of condition number. The backward error is a flat line at ten to the minus sixteen; the forward error rises in proportion to the condition number.110²10⁴10⁶10⁸10¹⁰10¹²10¹⁴10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹condition number κ(A)relative errorforward errorbackward errorpredicted: κ · u8×8, 20 seeds per κ; dashed is the worstthe problem worsens, not the method
Fig. 19 A condition satisfied and an answer that is wrong, which is the shape of this failure.
Backward error, forward error and the condition number, with measured valuesTwo boxes at the top — the problem posed and the nearby problem the algorithm answered exactly — and two answers below them, with the distances between all four labelled by numbers from a Hilbert solve.the problem you posedA = H10b = A·(1, 2, …, 10)the problem it answered exactlyA + δA, b + δb‖δ‖ / ‖A‖ = 2.3·10⁻¹⁷the answer you wantedx = (1, 2, …, 10), exactlythe answer you gotx̂, wrong by 2.7·10⁻⁴ relativebackward error 2.3·10⁻¹⁷forward error 2.7·10⁻⁴κ = 1.6·10¹³κ · η = 3.6·10⁻⁴, and the measured forward error is 2.7·10⁻⁴.The algorithm is not at fault. The problem is.H10, LU with partial pivotingresidual and error differ
Fig. 20 The identity, which has no term for a specification with a clause missing.
The exact solution of a 13×13 Hilbert system beside the computed oneTwo columns of numbers: the exact answer, which is the integers one to thirteen, and the answer double-precision elimination returns, with the number of correct digits beside each.H13 x = b, b formed exactly so that x = (1, 2, …, 13)12345678910111213exact1.00002.00003.00133.97865.18864.993010.46720.048921.2657-2.575319.21478.906013.5113computed6.6 correct digits4.8 correct digits3.4 correct digits2.3 correct digits1.4 correct digit0.8 correct digitno correct digitsno correct digitsno correct digitsno correct digitsno correct digits0.6 correct digit1.4 correct digitbackward error2.2·10⁻¹⁷κ = 1.7·10¹⁸The algorithm solved a neighbouring problem perfectly. That problem's answer is this one.right-hand side built in BigInt rationalsthe truth is known
Fig. 21 The standing that makes the interpolation error a measurement.
Why the product route stops: two condition numbers and their productThe two Gramians of a 22-state model of McMillan degree 10, and what each route to the Hankel singular values is charged. κ(P) = 1.26·10¹⁸ and κ(Q) = 2.16·10¹⁸. The square-root route works with RᵀS, whose condition number is their geometric mean, 1.65·10¹⁸; the route that eigendecomposes PQ works at 2.72·10³⁶, which is past 1/u = 4.5·10¹⁵ — the point at which nothing small survives at all. The Gramians themselves are right: this one agrees with its closed form to 2.15·10⁻¹³. What is lost is lost in the last step, to a product nobody had to form.condition numbers, on a logarithmic scaleκ(P)1.26·10¹⁸κ(Q)2.16·10¹⁸√(κ(P)κ(Q)) — the SVD route1.65·10¹⁸κ(P)κ(Q) — the product route2.72·10³⁶1/u4.5·10¹⁵the mean, or the productand only one of them fits
Fig. 22 The conditioning of the other method’s own construction.

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Balanced truncationConvection diffusionMoment matchingNon-normalityPetrov–GalerkinRational krylovReduced stabilityTransfer function