Reduction, and what a model is for

Exact at the points that were named

Balanced truncation asks for nothing and bounds everything, at a cost no large model can pay. The other kind of reduction asks for r numbers, costs r solves, is exact at every one of them — and bounds nothing anywhere else. That trade is the whole of large-scale model reduction.

Worth reading first: A model that is a rational function · A basis that is the same subspace and not the same thing · The bound that is known in advance.

The bound that is known in advance computes a reduced model with an error bound available before the model exists, and the bound turns out to be an equality. It is the best-behaved error statement on this site. It is also the one nobody uses at scale, and the reason is in the first paragraph of that essay: it costs two Lyapunov equations, each a Schur factorisation of an n × n matrix, which is O(n³).

A finite-element model of a structure has n in the hundreds of thousands. O(n³) is not expensive there; it is unavailable. Every large-scale reduction in use is the other kind, and the other kind gives up the bound entirely.

A reduced model of order 4, and the 4 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, 50. At each of them the curve falls to 5.3·10⁻¹⁶ — the reduced function passes through the original, and its derivative does too, because the projection is two-sided. Between and beyond them it reaches 8.37·10⁻⁴, and there is no bound on it: the method buys 8 exact conditions for 4 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.10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints4conditions bought8worst at a point5.3·10⁻¹⁶worst away from one8.4·10⁻⁴4 points, 8 conditionsand no bound in between
Fig. 1 A reduced model of order four, and the four places it is exact. Between them there is no bound at all.

What it does instead

Pick r points σ₁ … σᵣ in the complex plane. Build

V = span{(σᵢI − A)⁻¹B},     W = span{(σᵢI − Aᵀ)⁻¹Cᵀ},

and project: Aᵣ = WᵀAV, Bᵣ = WᵀB, Cᵣ = CV, with W rescaled so WᵀV = I. Then

Hᵣ(σᵢ) = H(σᵢ)   and   H′ᵣ(σᵢ) = H′(σᵢ)

exactly, at every one of the r points. Measured here: 4.7·10⁻¹⁶ for the values and 2.0·10⁻⁸ for the derivatives, the second being the accuracy of the finite difference used to compare them rather than of the interpolation.

Two things about that are worth separating.

It is a theorem, not an approximation. The reduced function passes through the original at the points and leaves with the same slope, in exact arithmetic, for any distinct σ. There is nothing to converge and no tolerance to set.

It costs 2r shifted solves. One per point for V and one for W. On a sparse model those are exactly the operations a sparse direct solver or a preconditioned iteration already does, so the whole method is available at a scale where nothing in the previous three essays is.

The half nobody states

The derivative condition is the part usually left out of the sentence, and it is half the value.

r points buy 2r conditions, not r. A reduced model of order r has 2r degrees of freedom in its transfer function — r poles and r residues — so 2r conditions is exactly enough to determine it. Interpolating only the values would leave it underdetermined and something else would have to fix the rest.

The second solve per point is what buys it, and dropping it is not a saving. Build the same projection one-sided, with W = V, and the values still agree to 10⁻¹⁶ while the derivatives are out by 4.6·10⁻⁵ — three orders worse than the two-sided model’s, which is the whole of what the extra solve was for. Nothing in a statement about subspaces distinguishes the two constructions, and the measurement does.

That is the same shape as the plane survives what its vectors do not: a property that belongs to a subspace and a property that does not, told apart only by measuring the thing the subspace was supposed to guarantee.

What has been given up, stated plainly

Everything away from the points.

The error curve above falls to 10⁻¹⁶ at each σ and reaches 4·10⁻³ between and beyond them, and there is no bound on it anywhere. Not a loose bound — none. The method offers 2r exact conditions and makes no statement about any other frequency.

Set the two methods side by side:

  • Balanced truncation asks the user for nothing, bounds ‖H − Hᵣ‖∞ over all frequencies before the model exists, and costs O(n³).
  • Interpolation asks the user for r complex numbers with no information about where to put them, bounds nothing, and costs 2r sparse solves.

The field’s central trade is between those two sentences, and it is a trade about cost rather than about accuracy — a point the measurements in interpolating at the model’s own poles make sharper than any argument could.

Where the points have to go

The user’s r numbers are the whole method, and there is no information available for choosing them. That is not a rhetorical flourish: at the moment the choice is made, nothing about H has been computed.

Three things are nonetheless true and worth knowing.

Points inside the range of the poles work and points beyond it do not, which is the same observation the rate the condition number predicts makes about where a Krylov method spends its effort. The model here has poles running from 9.9 to 580; the points 1.5, 5, 16, 50, 160, 500 span them. Two points both far past the last pole give two solves that are both nearly B/σ and nearly parallel, and the sixth direction does not exist — the construction refuses rather than returning a projection quietly built on rounding. That refusal is deliberate and is the sort of thing this field’s routines should do.

The error between two adjacent points falls when a point is added between them, and the far end does not improve. Past the last point there is nothing left to match, so the error there is set by the model’s behaviour at infinity rather than by how many points were bought.

Points that are close together are nearly free and nearly useless. Two σ within a per cent of each other give two nearly parallel solves, which is the beginning of the same failure the sixth point above hit outright. What that means practically is that the spread of the points is the decision and their exact placement is not — which is a relief, since nothing is known about the exact placement. It is the same relief choosing without knowing finds one field over: the parameter matters by orders and the precision with which it is chosen does not.

The subspace, and the basis that must not be the obvious one

The interpolation conditions are conditions on the subspace V, not on the vectors spanning it. Any basis will do, in exact arithmetic.

In floating point the obvious basis — the moments (σI − A)⁻¹B, (σI − A)⁻²B, … at one point — is a Krylov sequence in disguise, and it loses independence at exactly the rate the power method converges. Its condition number reaches 7.7·10⁹ by the eighth vector while the same eight solves spent at eight different points stay at 1.0. A basis that is the same subspace and not the same thing is that measurement, and it is the reason every implementation orthogonalises as it builds rather than afterwards.

Why the reduced model is a projection and not a fit

It is worth being clear about what kind of object has been produced, because the word interpolation invites the wrong picture.

Nothing has been fitted. There is no least-squares problem, no residual minimised, no tolerance. A subspace was built from r solves, the original model was projected onto it, and the interpolation conditions are a consequence of what that subspace contains — (σI − A)⁻¹B is in V, so the reduced resolvent reproduces the original one there.

The consequence worth carrying is that the reduced model inherits structure from the projection. Aᵣ = WᵀAV of a stable A need not be stable — a one-sided projection of a symmetric negative-definite A is, because then W = V and the projection is congruent, but a two-sided projection of a general A is not. The interpolation conditions hold either way. So a reduced model from this family can be a correct interpolant of a stable system and be unstable itself, which is a failure mode balanced truncation does not have and which no amount of adding points fixes.

What the interpolation conditions do not say

It is worth being precise about the failure mode, because “no bound” is easy to read as “no guarantee of anything” and that is too strong.

The reduced model is a rational function of degree r that agrees with H to first order at r points. Between those points it is whatever the projection made it, and whatever the projection made it is not arbitrary — it is a rational function of degree r, so it cannot oscillate freely. What is missing is a statement about how far a degree-r rational interpolant can be from the function it interpolates, and no such statement exists without assumptions about H that nobody has.

The practical consequence shows up in the error curve. It is smooth, it has r zeros, and it rises between and beyond them in a way that looks entirely predictable once drawn. The problem is that drawing it requires H, which is the thing the reduced model exists to avoid evaluating. A user with the full model can check the reduction and does not need it; a user who needs it cannot check it.

That asymmetry is what makes the a-priori bound valuable out of proportion to its cost, and it is worth setting against the residual the method reports, which is the same problem in the iterative field: the quantity that can be computed and the quantity that matters are different quantities, and the gap between them is not bounded either.

The partial repair, and it is only partial, is to interpolate at more points than the model’s order and check the fit at the extras. That costs solves and turns the method into something closer to a fit, and it is what a careful implementation does. It still bounds nothing between the points it checked.

The cost, counted honestly

2r shifted solves for the bases, and then the projection itself: r matrix–vector products with A per basis vector, so O(r²) products in total, and an r × r eigenproblem at the end that costs nothing.

Against balanced truncation’s two Lyapunov equations at O(n³) each, that is the entire argument. At n = 24, where every figure in this field is drawn, the comparison is meaningless — both are instant. At n = 10⁵ one of them exists and the other does not.

What makes the comparison worth drawing at n = 24 anyway is that the accuracy question can be settled at a size where both methods run. The answer, measured in interpolating at the model’s own poles, is that they land within a couple of per cent of each other in the norm the cheap one optimises — so the trade really is about cost, and the cheap method is not paying for its cheapness in the way one would expect.

The one thing that is bounded

There is a guarantee in this method and it is not about accuracy. The interpolation conditions hold exactly, in floating point, at every point. Measured at 4.7·10⁻¹⁶ across five points, which is rounding — not “approximately”, not “to the tolerance of the solve”.

That matters more than it sounds. It means a reduced model built this way can be audited: hand it the same σ, evaluate both functions, and the agreement is a check on the whole construction — the solves, the orthogonalisation, the normalisation of W against V — in one number. A construction with a wrong sign in the biorthogonalisation still produces a plausible reduced model and fails that check immediately, which is how the implementation behind these figures was debugged.

So the method offers no bound on its answer and a perfect check on its own arithmetic, which is almost exactly the opposite arrangement to the rest of this site.

The structure the projection does and does not keep

A reduced model is three small matrices, and what can be said about them beyond their size is worth a section, because it is where this method’s one genuine weakness sits.

Symmetry survives a one-sided projection. If A = Aᵀ and C = Bᵀ then W = V, the projection is congruent, and Aᵣ = VᵀAV is symmetric with the same sign of definiteness. That is a theorem and it carries stability with it: a symmetric negative-definite A projects to a symmetric negative-definite Aᵣ, so the reduced model is stable whatever points were chosen.

Nothing survives a two-sided projection. Aᵣ = WᵀAV with W ≠ V is a general square matrix, and there is no statement that its eigenvalues lie in the left half-plane. So a reduced model built by the method this essay is about can be unstable while interpolating a stable system exactly at r points, and no amount of adding points removes the possibility.

That is a failure with an unusual character. The model is not inaccurate — it matches H at r points to rounding, and its error curve between them is ordinary. It is unusable, because a simulation driven by it grows without bound while the system it represents does not. A caller has to check the poles of Aᵣ, and if they are in the wrong half-plane the only recourse within this method is to choose different points and try again.

Balanced truncation does not have this failure: truncating a balanced realisation of a stable system gives a stable system, by a theorem. That is a second entry on its side of the ledger, alongside the bound, and it is one the cost comparison usually leaves out.

So the honest statement of the trade is three-sided rather than two. Interpolation is cheap and exact where it is asked; balanced truncation is expensive, bounded everywhere, and cannot return something the caller must reject. Which of those matters most depends on whether the reduced model is going to be looked at or run.

The shape of the whole field, in three sentences

A reduced model is an approximation of a function, so its error is a norm of a function and can be measured exactly rather than estimated. One method computes that error’s bound before the model exists and costs more than a large problem has; the other is exact at points the caller names, bounds nothing between them, and costs what a large problem can afford. Everything else in this field is a consequence of which of those two a problem’s size forces.

At other settings

A reduced model of order 2, and the 2 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. At each of them the curve falls to 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 0.0607, and there is no bound on it: the method buys 4 exact conditions for 2 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 askedpoints2conditions bought4worst at a point10⁻¹⁵worst away from one0.0612 points, 4 conditionsand no bound in between
Fig. 2 Two points, and an error curve with two zeros in it.
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. 3 Three, where the region between the first two has fallen by two orders.
A reduced model of order 5, and the 5 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, 50, 160. At each of them the curve falls to 8.9·10⁻¹⁶ — the reduced function passes through the original, and its derivative does too, because the projection is two-sided. Between and beyond them it reaches 2.97·10⁻⁵, and there is no bound on it: the method buys 10 exact conditions for 5 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.10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints5conditions bought10worst at a point8.9·10⁻¹⁶worst away from one3·10⁻⁵5 points, 10 conditionsand no bound in between
Fig. 4 Five, and the far end has not improved at all.
A reduced model of order 6, and the 6 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, 50, 160, 500. 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 1.51·10⁻⁷, and there is no bound on it: the method buys 12 exact conditions for 6 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.10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints6conditions bought12worst at a point1.5·10⁻¹⁵worst away from one1.5·10⁻⁷6 points, 12 conditionsand no bound in between
Fig. 5 Six, which is as many as this model’s poles will carry before two solves become parallel.
Two bases for one subspace, at an interpolation point of 20κ₂ of the moment basis (σI − A)⁻¹B, (σI − A)⁻²B, … at a single point σ = 20, against κ₂ of the same number of solves spent at points spread across the model's poles and orthogonalised as they are built. The moment basis runs 10.9 to 7.75·10⁹, multiplying by about 65.8 per vector, while the spread one stays at 1. The reason is the one behind every Krylov basis on this site: repeated application of one operator converges to a direction, so every new moment is nearly the last. The rate is set by the ratio of distances from σ to the poles, which means the user's own choice of interpolation point decides how fast their basis becomes unusable — and spending the solves at different points instead is both a better approximation and a basis that survives.2345678110²10⁴10⁶10⁸10¹⁰vectors in the basisκ₂ of the basisas derivedsolves spread outone subspace, two spanning setseight moments at one point7.7·10⁹eight points, spread1growth per vector661/u4.5·10¹⁵the same subspaceand only one of them usable
Fig. 6 The basis that spans the right subspace and cannot be projected with.
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. 7 One choice of points that is not arbitrary, and how it is reached.
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. 8 What this method and the bounded one each win.
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. 9 The bound this method does not have.
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. 10 And the numbers it is made of, which cost O(n³) to obtain.
The transfer function of a 24-state model, by a solve and by a formula|H(iω)| for the second-difference operator on 24 points with one actuator and one sensor. The curve is C(iωI − A)⁻¹B, one factorisation of a complex 24 × 24 matrix per frequency, and the crosses are Σ(cᵀvₖ)(vₖᵀb)/(iω − λₖ) with the eigenvalues and eigenvectors written down in closed form. The two share the coefficient matrices and no code. Below ω = 100 they agree to 1.05·10⁻¹⁴ relatively; above it the modal sum is computing a number far below its own largest term, because the residues sum to cᵀb and the sensor is not at the actuator, so the relative comparison stops meaning what it looks like. Measured against the size of the terms being added, the two agree to 3.85·10⁻¹⁵ at every frequency. A state dimension of 24 and a function of one variable: the gap between those is what the field is about.10⁻¹110¹10²10³10⁻¹¹10⁻¹⁰10⁻⁹10⁻⁸10⁻⁷10⁻⁶10⁻⁵10⁻⁴10⁻³10⁻²10⁻¹frequency ω|H(iω)|one function, two routesstates24relative, ω ≤ 10010⁻¹⁴against the terms3.9·10⁻¹⁵‖H‖∞0.0924 statesone function of one variable
Fig. 11 The function being interpolated.
Two rank decisions for one integer: the McMillan degree of a 40-state modelSingular values, each divided by the largest of its own set. The Loewner matrix is built from 32 samples of H on the imaginary axis and never touches A: its 6th singular value stands 9.46·10⁴ above the next, so the rank decision is not a close call. The Hankel matrix of Markov parameters CAᵏB is built from A, B and C — strictly more information — and has no cliff, because ‖A‖ is (n+1)² and its entries span 3·10³⁷. It returns 2 where the answer is 6. The definition is not the algorithm, and the route with less information is the one that works.1357911131510⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1indexsingular value ÷ the largestdegree 6from samples of Hfrom A, B and Cone integer, two routesstates40Loewner rank6its gap9.5·10⁴Markov rank2its dynamic range3·10³⁷the rank of a divided differenceand the rank that cannot be seen
Fig. 12 How many points would be needed to reproduce it exactly.
The residual basis of conjugate gradients, at κ = 104A semi-logarithmic plot against iteration count showing the loss of orthogonality among the residual vectors rising while the relative residual falls.05101520253010⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹iteration‖RᵀR − I‖ and ‖r‖/‖b‖step n‖RᵀR − I‖residual30×30, run for exactly n stepsexact arithmetic would end here
Fig. 13 The orthogonality every Krylov basis on this site eventually loses.
Reversing the columns of a 6×6 matrix: what each orthogonalisation returnsBoth grids print the entries of Q₁P − Q₂, where Q₁ orthogonalises A and Q₂ orthogonalises A with its columns reversed. If the answer permuted with the columns, both grids would be zero. The polar factor's is: ‖U₁P − U₂‖_F = 1.53·10⁻¹⁵. QR's is not: ‖Q₁P − Q₂‖_F = 2.353 on matrices whose own Frobenius norm is 2.449, so the two answers are essentially unrelated. Gram–Schmidt and Householder both build Q one column at a time, and the first column is treated differently from the last.polar: U₁P − U₂QR: Q₁P − Q₂0000000000000000000000000000000000000.56-0.15-0.21-0.330.06-0.64-0.430.310.310.120.20.73-0.130.64-0.031-0.0028-0.420.075-0.18-0.43-0.130.21-0.29-0.420.640.630.670.62-0.52-0.420.26-0.520.022-0.0870.2-0.24the same columns, reordered‖U₁P − U₂‖_F1.5·10⁻¹⁵‖Q₁P − Q₂‖_F2.4‖Q‖_F, for scale2.4κ of the matrix10the column space did not moveand one of the two answers did
Fig. 14 And the order that decides whether it does.
Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 16 masses. Arnoldi on the 32 × 32 linearisation stores m vectors of length 32 and returns m Ritz values; the second-order recurrence stores m vectors of length 16 and its projection is taken onto the QUADRATIC, which has 2m eigenvalues for an m × m problem. So at every subspace size the second route holds half as many numbers and returns twice as many approximations — a factor of four in Ritz values per number stored, exactly and by construction. Plotted against storage, its curve is the lower one at every point: at 96 numbers it is 0.01231 from the dominant eigenvalue where the linearised route needs 192 to reach 0.07122. Both converge slowly — twenty-six vectors buy three digits — because the chain's spectrum is an arc with no isolated extreme, and that is the honest half: this is a comparison of prices rather than a fast method.22.32.62.910⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised832storage, second-order416Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 15 A subspace built for a problem that is not a matrix, by the same device.
Two Krylov methods against products with A, at a kernel shift of 1Two error curves against the number of products with A, on a logarithmic vertical axis. The Arnoldi method reaches 0.1532 after 4 products and is 9.18 by the end of the run. The bidiagonal method reaches 0.1367 after 42 and degrades far more slowly.16111621263136414651566110⁻¹110¹products with Arelative errorArnoldi's best: 4Arnoldibidiagonalwhat a step buysArnoldi's best0.15products to reach it4bidiagonal's best0.14products to reach it42a tenth of the work to the same answerand no time at all spent there
Fig. 16 Where a Krylov space stops being the right subspace.
What 3 shifts do to each direction of the starting vectorAmplification of each eigen-component against its eigenvalue, on a logarithmic vertical axis. The points are measured — the component of the vector along each eigenvector, before and after — and the curve is |Π(λ − θ)| with the roots at the 3 discarded Ritz values. They agree to 3.9·10⁻¹¹. The 3 wanted directions are amplified by at least 1 and everything else by at most 0.802.1357910⁻⁵10⁻⁴10⁻³10⁻²10⁻¹1eigenvalue λamplificationkeptdiscardeda filter on the starting vectormeasured against the polynomial3.9·10⁻¹¹worst kept direction1best discarded direction0.8the roots are the discarded Ritz valuesand the vertical lines are where they sit
Fig. 17 Choosing shifts for a different reason, in the spectral field.
The condition number a sketched preconditioner leaves, against the condition number it was givenTwo curves against κ(A), both axes logarithmic. The matrix's own condition number climbs the diagonal from 100 to 10¹⁰; κ(AR⁻¹), where R comes from a QR of a 4n-row sketch, is 2.6304 at every one of them — the same number to ten digits, not a similar one. The reason is two lines of algebra: with G = SU the preconditioned singular values are those of (GᵀG)⁻¹, which has no spectrum of A in it at all.10²10⁴10⁶10⁸10¹⁰110²10⁴10⁶10⁸10¹⁰condition number of the matrixcondition number seen by the iterationκ(A), unchangedκ(AR⁻¹)a bound with no κ(A) in itκ(AR⁻¹), every κ(A)2.6κ(SU), the other route2.6κ(A), across the sweep10⁸κ(AR⁻¹), across the sweep1the sketch never sees the spectrumand the spectrum cancels out of the answer
Fig. 18 Changing a problem before solving it, which is what a projection is.
Conjugate gradients at κ = 102, against the bound κ permitsA semi-logarithmic plot of the relative A-norm error against iteration count. The measured curve falls below a smooth dashed curve showing the classical condition-number bound.0408012016020024010⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹iteration‖e‖_A / ‖e₀‖_Ameasuredκ bound58 steps40×40, spectrum spread evenly in logbound permits 142
Fig. 19 A rate that is known in advance, for contrast with an error that is not bounded at all.
GMRES on the Laplacian and on the cyclic shift, both 12×12A semi-logarithmic plot of relative residual against iteration. One curve falls steadily; the other is flat at one for every step until the last, where it drops to zero.02468101210⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1iteration‖r‖ / ‖b‖Laplaciancyclic shiftno progress at allevery eigenvalue of the shift is on the unit circleand it predicts nothing
Fig. 20 And a spectrum that predicts nothing about one.
The plane, and the two vectors inside it, at a gap of 10⁻⁸A circle representing the invariant plane, with two perpendicular radii for the eigenvectors and two more for the perturbed ones, rotated within the same circle.the invariant planesolid: beforedashed: aftersame perturbation, two questionsthe vectors turned, radians0.43the plane turned, radians7.6·10⁻⁸what left the plane6.5·10⁻⁸drawn in the unperturbed plane's own basisa radius is not determined; the circle is
Fig. 21 A property that belongs to a subspace and not to its basis.
The computed angle against the true one, two formulationsTwo curves of computed angle against true angle on logarithmic axes. One follows the diagonal all the way down; the other leaves it and flattens at a fixed level.10⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁵10⁻¹³10⁻¹¹10⁻⁹10⁻⁷10⁻⁵10⁻³the true anglethe angle computed√(2u) = 1.49·10⁻⁸arcsine of ‖(I − QQᵀ)Q₂‖arccosine of σ(Q₁ᵀQ₂)two routes, one of which has a floorfloor of the arccosine route1.5·10⁻⁸√(2u)1.5·10⁻⁸worst overstatement1.5·10⁶angles returned as exactly zero3a plane tilted by a known anglethe flat part is the instrument, not the data
Fig. 22 The angle between two subspaces, which is what a failed projection is about.
Fitting the same degree-10 polynomial in two basesOn the left, the data and two fitted curves that lie on top of each other. On the right, the condition numbers of the two design matrices, ten orders of magnitude apart.00.10.20.30.40.50.60.70.80.910.20.40.60.81xyboth fits, drawn on top of one anothercondition number of the design matrixmonomial2.1·10⁷Chebyshev2.3largest fitted coefficientmonomial176Chebyshev0.51rms residual: 4.2·10⁻⁶ and 3.1·10⁻⁷ — the data is fitted either way.30 points, degree 10, single precisionthe basis is part of the problem
Fig. 23 A basis whose conditioning decides whether a fit can be computed.
A reduced model of order 4, and the 4 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, 50. At each of them the curve falls to 5.3·10⁻¹⁶ — the reduced function passes through the original, and its derivative does too, because the projection is two-sided. Between and beyond them it reaches 8.37·10⁻⁴, and there is no bound on it: the method buys 8 exact conditions for 4 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.10⁻¹110¹10²10³10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵s, on the real axis|H − Hᵣ| ÷ |H|exact where askedpoints4conditions bought8worst at a point5.3·10⁻¹⁶worst away from one8.4·10⁻⁴4 points, 8 conditionsand no bound in between
Fig. 24 The four-point model again, for comparison with the two-norm measurements.
Two bases for one subspace, at an interpolation point of 3κ₂ of the moment basis (σI − A)⁻¹B, (σI − A)⁻²B, … at a single point σ = 3, against κ₂ of the same number of solves spent at points spread across the model's poles and orthogonalised as they are built. The moment basis runs 13.5 to 7.08·10¹¹, multiplying by about 148 per vector, while the spread one stays at 1. The reason is the one behind every Krylov basis on this site: repeated application of one operator converges to a direction, so every new moment is nearly the last. The rate is set by the ratio of distances from σ to the poles, which means the user's own choice of interpolation point decides how fast their basis becomes unusable — and spending the solves at different points instead is both a better approximation and a basis that survives.2345678110²10⁴10⁶10⁸10¹⁰10¹²vectors in the basisκ₂ of the basisas derivedsolves spread outone subspace, two spanning setseight moments at one point7.1·10¹¹eight points, spread1growth per vector1481/u4.5·10¹⁵the same subspaceand only one of them usable
Fig. 25 The same collapse at a point near the bottom of the spectrum.
Two bases for one subspace, at an interpolation point of 800κ₂ of the moment basis (σI − A)⁻¹B, (σI − A)⁻²B, … at a single point σ = 800, against κ₂ of the same number of solves spent at points spread across the model's poles and orthogonalised as they are built. The moment basis runs 38.9 to 3.51·10⁸, multiplying by about 16.4 per vector, while the spread one stays at 1. The reason is the one behind every Krylov basis on this site: repeated application of one operator converges to a direction, so every new moment is nearly the last. The rate is set by the ratio of distances from σ to the poles, which means the user's own choice of interpolation point decides how fast their basis becomes unusable — and spending the solves at different points instead is both a better approximation and a basis that survives.2345678110²10⁴10⁶10⁸vectors in the basisκ₂ of the basisas derivedsolves spread outone subspace, two spanning setseight moments at one point3.5·10⁸eight points, spread1growth per vector161/u4.5·10¹⁵the same subspaceand only one of them usable
Fig. 26 And at one past the top of it, where it is fastest.
How far the coefficients can move without changing the fit, degree 10Relative increase in the residual against relative change in the coefficients, along the least determined direction. The residual does not move measurably until the coefficients have changed by more than a factor of one.10⁻⁴10⁻³10⁻²10⁻¹110¹10²10³10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹relative change in the coefficients, along the worst directionrelative increase in the residualcoefficients doubled245% change, fit unmoved in the sixth digit308×: the third digit movesκ(A) = 2.2·10⁷. Exact arithmetic would pick one point on this floor. It would not raise it.24 points, degree 10, monomial basisthe data leaves them free
Fig. 27 A fit with many equally good answers, for contrast with an interpolation with one.
Singular values and |uₖᵀb|, with and without 0.10% noiseThree curves against the index on a logarithmic vertical axis. The singular values fall exponentially to the level of rounding. With an exact right-hand side the coefficients fall faster and every term of the solution stays bounded. With noise they flatten at index 32, and from there on each term is noise divided by a σ of 6.8·10⁻⁴.081624324048566410⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹index kmagnitudethe floor: k = 32best truncation: k = 28σₖ|uₖᵀb| exact|uₖᵀb| with noisetwo different indicesthe crossing, from the data alone32the truncation that is actually best28relative error there0.11the exact coefficients never flattenthe noisy ones stop at ‖e‖/√n
Fig. 28 Where the data stops determining the answer, in the field that names the problem.
What 12 members of a sequence cost, by what changes between themEvery bar is 12 solves of an 120×120 system, costed in multiplications with a dense Cholesky at n³/3 and a solve at 4n². When nothing changes the answer is reused and the cost is one factorisation. When only the right-hand side changes, one factorisation serves every member — 16.5 per cent of the independent cost, with every solve still at a relative residual of 1.54·10⁻¹³. When the matrix drifts, the factorisation is reusable for a while and the while has to be measured: 4 factorisations for 12 members. When everything changes, nothing carries.nothing8.3%the right-hand side16.5%the matrix, slowly85.9%everything100.0%what can be reused: the answerwhat can be reused: the factorisationwhat can be reused: the factorisation, for a whilewhat can be reused: nothingshare of the cost of a sequence that shares nothingwhat 12 members cost in factorisationssame: factorisations1rhs: factorisations1drift: factorisations4independent: factorisations12what changes between the membersdecides what may be carried
Fig. 29 The economics of paying once and using many times.
Multiplications in a hierarchical solve and in a dense factorisation, and where they crossBoth counted rather than estimated: the recursion carries a counter and reports what it actually did. The dense factorisation is n³/3 and rises at exactly three per doubling. The hierarchical solve rises at 2.18, 1.93, 1.78 — falling, because the cost is n log²n and its exponent is on its way to one. The two cross between n = 128 and n = 256: below it the format is the more expensive way to solve the system, at 1.97 times the dense count, and at n = 512 it is 4.4 times cheaper. Every point returns an answer at a backward error of about 1.1·10⁻¹², so the comparison is between two ways of getting the same thing.567891010⁴10⁵10⁶10⁷10⁸10⁹log₂ nmultiplicationsdense factorisation, n³⁄3the recursion, countedwhere the format starts payingratio at n = 642ratio at n = 5120.23exponent, first doubling2.2exponent, last doubling1.8backward error1.1·10⁻¹²cheaper is a sizenot a property
Fig. 30 And the cost of the one operation this whole method is built from.
Steps to a relative error of 10⁻⁶, on one 60×12 problem at five condition numbersTwo curves against the condition number on a logarithmic horizontal axis, with four seeds drawn at each. The two recurrences compute the same iterates in exact arithmetic. At κ = 10² they cost 16 and 16 steps; at κ = 10¹⁰ they cost 110 and 209.10²10⁴10⁶10⁸10¹⁰0285684112140168196224condition numberstepsnormal equationsbidiagonalisationone sequence, two costssteps at κ = 10², both16at κ = 10⁶, ratio1.1at κ = 10¹⁰, ratio1.9the same iterates in the algebraand twice the work at κ = 10¹⁰
Fig. 31 Two recurrences for one sequence, which is the shape of the one-sided comparison.
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. 32 The decay the bounded method depends on and this one never asks about.

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A-priori boundBalanced truncationCondition numberFlop countHermite interpolationKrylov subspaceMoment matchingPetrov–GalerkinRational krylovTransfer function