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.

Worth reading first: A model that is a rational function · Why a Gramian can be truncated at all.

Almost every error on this site is measured after the fact. Run the method, compare against a closed form, report the difference. The three or four bounds this collection has met are each disqualified from being useful in a different way: the growth factor’s 2ⁿ⁻¹ is never attained, the conjugate-gradient rate describes a rate rather than a number, and Johnson–Lindenstrauss holds with a probability.

Balanced truncation has one that is none of those:

‖H − Hᵣ‖∞  ≤  2 (σᵣ₊₁ + … + σₙ).

It is computable before the reduced model exists. It bounds a norm of a function rather than the size of a residual. And — this is the part worth the essay — on ordinary problems it is not a loose upper bound. It is attained.

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. 1 The measured error and the bound computed before the model existed. They are the same curve.

What the σ are

They are the Hankel singular values, and getting to them costs two Lyapunov equations:

AP + PAᵀ + BBᵀ = 0,     AᵀQ + QA + CᵀC = 0,

whose solutions are the controllability and observability Gramians, followed by σₖ = √λₖ(PQ). Nothing new has to be written to solve them: an equation whose unknown is a matrix already built a Bartels–Stewart solver for AX + XB = C, and a Lyapunov equation is that with B = Aᵀ. This field asks the existing routine a question it was already able to answer, which is the whole point of having written it there.

What the two Gramians mean is worth one sentence each, because the bound is unreadable without them. P measures how much of the state space the input can reach; Q measures how much of it the output can see. A direction that is hard to reach and hard to see contributes almost nothing to the map from u to y, and is exactly what a reduced model should drop. σₖ = √λₖ(PQ) is the number that combines the two, and it is invariant under any change of state coordinates — which is what makes it a property of H rather than of how the model was written down, in the same way the McMillan degree is.

And an answer to check them against

The model here is the second-difference operator, so A = VΛVᵀ with V orthogonal and Λ a closed form. In modal coordinates the Lyapunov equation becomes n² scalar equations:

P̃ᵢⱼ = −b̃ᵢ b̃ⱼ / (λᵢ + λⱼ),      b̃ = VᵀB.

That is a formula, not a solve — n² divisions and one similarity. The Bartels–Stewart route factorises two matrices and never mentions an eigenvalue, and the two agree to 3.5·10⁻¹⁴. The equation the solve claims to have solved is separately checked by substitution, at 1.1·10⁻¹⁴.

Both checks matter and they are different checks. The second says the solver did what it said. The first says the thing it solved is the object the formula describes. A field whose central claim is about a bound needs the numbers the bound is made of to be known rather than computed, and this is the strongest form of that available here.

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. 2 The numbers the bound is a sum of, and the route that cannot see half of them.

The reduced model

The balancing transformation is the change of coordinates in which P = Q = diag(σ) — every direction equally reachable and observable — and in those coordinates truncation is just dropping the last n − r rows and columns. In practice nobody forms it that way, for a reason that is the next essay and that has a familiar shape.

What comes back is a Petrov–Galerkin projection: Aᵣ = WᵀAT, Bᵣ = WᵀB, Cᵣ = CT with WᵀT = I. The two bases are built from square-root factors of the Gramians, and the number that decides both the accuracy and the conditioning of the construction is the same one — the smallest σ kept. That is worth noticing early, because it means truncating at a σ that has not yet fallen is both inaccurate and unstable, rather than one or the other, and the two failure modes cannot be traded against each other.

The measurement

Take a twenty-state model whose transfer function has ten poles, reduce it at every order from one to seven, and measure ‖H − Hᵣ‖∞ properly — not by sampling, but by forming the difference as a system in its own right. H − Hᵣ is itself a transfer function, of the block-diagonal state matrix with the outputs subtracted, so every norm applies to it with no new code. That is what makes an error in this field measurable rather than estimated, and it is the reason the numbers below are worth reading.

r      measured error     2 Σ σ        ratio
1      1.421·10⁻²         1.421·10⁻²   1.0000
2      6.880·10⁻⁴         6.880·10⁻⁴   1.0000
3      3.953·10⁻⁵         3.953·10⁻⁵   1.0000
4      2.026·10⁻⁶         2.026·10⁻⁶   1.0000
5      9.944·10⁻⁸         9.944·10⁻⁸   1.0000
6      3.561·10⁻⁹         3.561·10⁻⁹   1.0000
7      1.181·10⁻¹⁰        1.181·10⁻¹⁰  1.0000

Seven orders of magnitude of error, and one number for the ratio. The bound is not a safety margin.

Where the equality comes from

It is not a coincidence and it is not a property of this model. It comes from a one-state result and a sum, and separating the two is what makes the whole thing legible.

Removing exactly one state costs exactly 2σₙ. Measured on four systems that share nothing but the shape of the question — residues all of one sign, residues of both signs, a clustered pair of poles, a spectrum spanning three decades — the ratio of the measured H∞ error to the σ removed is 2.0000 every time.

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. 3 Four systems, one removal each, and one number.

The bound for a truncation of many states is that result applied n − r times, each step contributing twice the σ it removed, and the terms added up. So the sum can only be loose in how the removals interact, never in the single step. On a spectrum falling as fast as this one the tail sum is dominated by its own first term and the interaction has nothing to do, which is why the ratio is one.

And the sum is loose where the σ are close together. On a model with two nearly equal Hankel singular values, removing both at once costs 1.52 times less than the two single steps suggest. That is the entire looseness of the bound, located: it lives in the interaction between removals of comparable size, and nowhere else.

What makes the single step exact

The one-state result is the load-bearing one, so it is worth saying where it comes from rather than citing it. In balanced coordinates the state being dropped is a single direction that is equally reachable and observable, with σₙ measuring both. Its contribution to the map from input to output is a rank-one term whose largest gain over all frequencies is exactly 2σₙ — the factor of two is the round trip, in and out, and the equality is a property of the coordinates rather than of the system.

That is why the number comes back at 2.0000 on four systems with nothing in common. It is not an empirical regularity that happens to hold on well-behaved problems; it is what balancing is, and the four systems are there to show that nothing about the residues, the pole spacing or the spread of the spectrum enters it.

It also explains the one place the equality does not survive. Two states with nearly equal σ are two directions the balancing transformation cannot separate cleanly, so dropping them together is not the same operation as dropping them one at a time. The measured 1.52 is the size of that effect on a deliberately clustered pair, and it is the largest looseness this bound has anywhere.

The lower bound, which is the other half

σᵣ₊₁ is a lower bound too. No reduced model of order r — not balanced truncation’s, not anybody’s — has an H∞ error below the first Hankel singular value it neglected. Measured here at r = 4, the error is 2.10 times σ₅.

Put the two together and the picture is unusual for this site:

σᵣ₊₁  ≤  ‖H − Hᵣ‖∞  ≤  2 Σ_{k>r} σₖ,

with both ends computed from one pair of Lyapunov solves and neither of them a heuristic. The question what order is worth truncating at? has an answer that can be read off a curve before any reduced model is built, and the answer is within a factor of about two of optimal.

Nothing else in this collection has that shape. Choosing without knowing is about a regularisation parameter and every published rule for choosing it is a rule of thumb; four knobs and one floor measures four such rules against each other and finds them landing within a few per cent because none of them knows anything the others do not. Here the number is not a rule of thumb. It is a bound, it is proved, and it is tight.

What it is a bound on, which is not a residual

Worth stating plainly because every other error on this site is a residual or a forward error of a vector. ‖H − Hᵣ‖∞ is the largest, over all frequencies, of the difference between two functions — equivalently the largest possible ratio of output energy to input energy that the reduced model gets wrong. It is a statement about every input the model could ever be given.

That is a much stronger object than a residual. A residual says this computed answer nearly solves this problem. This says this smaller model behaves like the larger one under every excitation, to within this number. And it is available before the smaller model exists.

The price is visible in the same sentence. Getting it requires two Lyapunov equations, each of which is a Schur factorisation of an n × n matrix and therefore O(n³) — which is exactly what a model of a hundred thousand states cannot afford. Every large-scale reduction in use is the other kind, which asks for r numbers, costs r solves, and bounds nothing at all. That trade is the field’s central one and this essay is one half of it.

The decay, which is the assumption underneath

Everything above assumes the σ fall. If they did not, 2Σσ would be of the same size as ‖H‖∞ at every order and the bound would say nothing.

They do fall, on this model and on nearly every model anybody discretises, and the reason is not obvious. A Gramian is defined by a matrix equation whose right-hand side has rank one; there is no low-rank structure anywhere in the statement. That the answer nonetheless has eigenvalues falling off a cliff is a theorem about rational approximation, and it has a rate with a closed form. Why a Gramian can be truncated at all is that essay, and it is the one that says which models this field works on before any of them is tried.

What the bound is not, and the three ways that matters

Three readings of an a-priori bound are natural and all three are wrong here, so they are worth separating.

It is not a bound on the state. Nothing above says the reduced model’s internal variables resemble the original’s, and they do not: the balancing transformation mixes every state into every other, so xᵣ is a combination of all n components and has no interpretation as a subset of them. A reader wanting to know what happened to the temperature at grid point seven will not find it, and the bound says nothing about it. What is bounded is the map from u to y, which is what the model was for.

It is not a bound on the poles. The reduced model’s poles are not a subset of the original’s, and there is no statement that each of the r kept poles is close to one of the n originals. On the model here the reduced poles at order four sit at 9.9, 39.3, 87.8 and 154.6, which happen to be close to the four slowest original poles, and that is a property of a system whose modes are cleanly separated rather than a theorem. A model with clustered poles produces reduced poles that are near neither the cluster’s members nor its centre, and the bound continues to hold at exactly the same value.

It is not a bound anybody has to trust. This is the useful one. The reduced model is r × r; the difference system is (n + r) × (n + r); measuring ‖H − Hᵣ‖∞ on it is a search over frequency and costs nothing at these sizes. So the bound can be checked on any problem small enough to build one, which is how the equality above was found rather than assumed. On a problem too large to check, the bound is what remains — and the check has been done on enough small ones to say what kind of number it is.

The third point is the one worth carrying into a working habit. A bound that is only ever quoted is a bound nobody has calibrated. This one is a factor of one from the truth on the models drawn here and about a factor of two from the best possible reduced model of the same order, and both of those statements come from measurements that a user with a small test problem can repeat in a minute.

Where the two Lyapunov solves actually go

One practical note, since the cost is the whole reason this method has a rival.

Both Gramians come from the same Schur factorisation used twice — A and Aᵀ have the same Schur vectors up to a transpose — so a careful implementation factorises once and solves two triangular systems. That halves the O(n³) and changes nothing about which side of the availability line it sits on: half of a hundred-thousand-cubed is still a hundred-thousand-cubed.

What it does change is the region where the comparison in interpolating at the model’s own poles is worth making. Up to a few thousand states the bounded method is simply available, and the argument for the cheap one is about convenience rather than necessity. The line is not sharp and it moves with the hardware, which is why this field is careful to price both in operations rather than in seconds.

At other settings

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 2.11 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.12310⁻⁵10⁻³10⁻¹order r kepterror and bounds2Σσ, and the errorσᵣ₊₁a bound that is an equalityorders3bound ÷ error, worst1spread over the sweep1error ÷ σᵣ₊₁, worst2.1computed before the modeland attained by it
Fig. 4 Three orders, where the bound and the error have not yet had room to differ.
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 2.07 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.1234510⁻⁸10⁻⁶10⁻⁴10⁻²order r kepterror and bounds2Σσ, and the errorσᵣ₊₁a bound that is an equalityorders5bound ÷ error, worst1spread over the sweep1error ÷ σᵣ₊₁, worst2.1computed before the modeland attained by it
Fig. 5 Five, and the ratio is still 1.0000.
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. 6 Nine, where the sweep stops because both quantities have reached the level the Gramians were computed at.
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
Fig. 7 One system alone, so the two numbers behind the ratio can be read.
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
Fig. 8 A clustered pair, where the single step is still exact and the sum is not.
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
Fig. 9 Four poles rather than three, and the same equality.
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
Fig. 10 A spectrum spanning three decades, where nothing about the ratio moves.
Hankel singular values, and the half of them a product cannot seeσₖ divided by σ₁, for a 14-state model of McMillan degree 10. 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 5 values and then stops, flattening at 9.19·10⁻⁹. The dashed line is σ₁√u = 2.23·10⁻⁹, 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 2.91·10³⁴ against 1.71·10¹⁷. Every σ below the line is a term in the error bound, so this is not an academic loss.13579111310⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1index kσₖ ÷ σ₁σ₁√utwo factorstheir productthe normal equations, againroutes agree to k =5product floors at9.2·10⁻⁹σ₁√u2.2·10⁻⁹κ(P)κ(Q)2.9·10³⁴√ of it1.7·10¹⁷do not form the productthe σ below the line are the bound
Fig. 11 The σ on a smaller model, falling at the same rate.
Hankel singular values, and the half of them a product cannot seeσₖ divided by σ₁, for a 34-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 8.06·10⁻¹⁰. 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 1.76·10³⁸ against 1.33·10¹⁹. 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)1.8·10³⁸√ of it1.3·10¹⁹do not form the productthe σ below the line are the bound
Fig. 12 And on a larger one, where there are more of them and the same number are visible.
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. 13 Why they fall at all, which is the assumption this whole essay rests on.
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. 14 The conditioning of the two Gramians, and the product nobody has to form.
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. 15 The function the bound is about, and the model it came from.
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. 16 How many poles it has, which is what a truncation is measured against.
The two 3×3 matrices of a Sylvester equation, and the 9×9 matrix it meansA and B are 3×3 with 5 nonzeros each. The coefficient matrix of the map X ↦ AX + XB is 9×9 with 21 nonzeros, drawn at the same scale so the ratio is visible rather than quoted. Bartels and Stewart's algorithm never forms it: two Schur reductions and a back-substitution give the same X to 5.5·10⁻¹⁷, and that X satisfies AX + XB = C to 8.8·10⁻¹⁷.A3×3B3×3I ⊗ A + Bᵀ ⊗ I9×9one equation, two objectsentries in A and B18entries in the coefficient matrix81two routes, relative gap5.5·10⁻¹⁷‖AX + XB − C‖/‖C‖8.8·10⁻¹⁷the small squares are the problemand the large one is the notation
Fig. 17 The equation both Gramians are solutions of.
sep(A, B) and the smallest |λᵢ(A) + μⱼ(B)| as the departure from normality grows, n = 6A and B are upper triangular with diagonals 1, 2, …, 6 and μ above them, so every eigenvalue sum is at least 2 at every point on the axis and the flat line is exact rather than nearly flat. sep — the smallest ‖AX + XB‖_F over ‖X‖_F = 1, and the reciprocal of the amplification a perturbation of C receives — starts equal to the gap at μ = 0, where the matrices are normal, and falls to 9.47·10⁻⁴ at μ = 8. The number a reader is invited to consult is the one that does not move.01234567810⁻⁵10⁻⁴10⁻³10⁻²10⁻¹110¹μ, the entry above the diagonalsep, and the eigenvalue gapmin |λᵢ + μⱼ|sep(A, B)the spectra never moveeigenvalue gap, throughout2sep at μ = 02sep at μ = 89.5·10⁻⁴amplification there1056solvability is the eigenvaluesand conditioning is not
Fig. 18 And the quantity that governs how well it can be solved.
Multiplications in a 2-dimensional model solve, through the eigenbasis against a dense factorisationA Kronecker sum's eigenvectors are the Kronecker products of its factors' eigenvectors, so the change of basis that diagonalises an operator with 1,024 rows is 2 changes of basis along 2 indices. The whole solve is a transform, 1,024 divisions and a transform back: 1.31·10⁵ multiplications at n = 32, against a dense factorisation's 7.16·10⁸, a factor of 5461. The fitted exponent is 3.00 against 3d = 6. The only decompositions taken are of the 2 one-dimensional factors, which on the model problem are the same matrix — so there is exactly one, of size 32. Every solve reproduces its right-hand side to 1.44·10⁻¹³.10¹10²10⁴10⁶10⁸n, points along one axismultiplicationsa dense factorisationthrough the eigenbasis2 decompositions of an n × nunknowns1024multiplications1.3·10⁵dense factorisation7.2·10⁸fitted exponent3‖Ax − b‖ ⁄ ‖b‖1.4·10⁻¹³nothing of size n^dis ever factorised
Fig. 19 The route to that equation nobody takes, and why.
Growth factor under partial pivoting to n = 12: the bound, the worst case, and realityGrowth factor against matrix size on a logarithmic vertical axis. The two-to-the-n bound rises as a straight line; Wilkinson's matrix sits exactly on it; random matrices stay near one.02468101214110¹10²10³10⁴10⁵10⁶matrix size ngrowth factor max|u| / max|a|the 2ⁿ⁻¹ boundworst of 30 randommedian randomWilkinson's matrix sits on the bound30 Gaussian matrices per sizeat n = 12: bound 2048, worst 3.6
Fig. 20 A bound that is never attained, for contrast with one that always is.
The proved bound against the error it bounds, at 24 significand bitsTwo curves against the Hilbert size on a logarithmic vertical axis: the error the computation actually made, which is known because the exact answer is the integers by construction, and the bound proved for it in interval arithmetic. The bound is above the error at every size, by 9.949 to 23.12 times. Past n = 5 the method refuses to return a bound at all.345678910111210⁻⁵10⁻³10⁻¹Hilbert size nerrorrefusedrefusedrefusedrefusedrefusedthe proved boundthe actual errorbound ÷ errorslack at n = 313slack at n = 423slack at n = 59.9the error is known, not estimatedthe exact answer is the integers 1 … n
Fig. 21 Tightness, measured, in the field where this site first asked about it.
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. 22 A bound that describes a rate rather than a number.
What extra columns buy a construction that never sees the matrix, over five seedsThe band is the range across five seeds and the line is their mean; the flat lower line is the best representation of rank 8 there is, which a decomposition of every block would find. With no oversampling the construction is 11.6 times behind it and the spread across seeds is 44 per cent of the mean. Eight extra columns bring it to 3.0× and the spread to 12 per cent. Both halves matter and only one of them is visible in a single run: what oversampling buys is a better answer and a more predictable one, and a figure drawn from one seed would have shown the first and hidden the second. No draw is ever better than the flat line, which is the check that the comparison is honest — a sample cannot beat the decomposition it is approximating.048121610⁻⁸10⁻⁷10⁻⁶10⁻⁵extra columns in the sample, p‖A − A_H‖ ⁄ ‖A‖the best rank-8 representationfive draws, mean and rangea band, not a lineexcess at p = 012excess at p = 83spread at p = 00.44spread at p = 80.12the optimum of this rank7.8·10⁻⁸one seed shows the meanand five show the risk
Fig. 23 And one that holds with a probability.
Error of the best rank-k approximation to a 12×12 matrixApproximation error against k on a logarithmic axis, with the measured error and the next singular value drawn as separate curves lying exactly on top of one another.123456789101110⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²1rank k of the approximation‖A − Aₖ‖measured, 2-normσₖ₊₁, from theorymeasured, Frobeniusthe first two agreeto 2·10⁻⁸worst |‖A−Aₖ‖₂ − σₖ₊₁| / σₖ₊₁2·10⁻⁸worst Frobenius discrepancy2·10⁻⁸κ = 10⁹; 30 random rank-3 matrices, none closerthe error is σₖ₊₁
Fig. 24 The other theorem on this site that says what the best approximation of a given rank costs.
What a rank-10 approximation can achieve, by spectrumA semi-logarithmic plot of singular value against index for three spectra — geometric decay, algebraic decay, and flat — with the rank-ten approximation error marked on each.1112131415110⁻³10⁻²10⁻¹1index jσⱼσ11geometric, 0.85ʲalgebraic, j^−0.00flatmeasured, and equal to σ₁₁rank-10 error, geometric0.2rank-10 error, algebraic1rank-10 error, flat160×60, spectrum chosen rather than the entriesthe matrix decides, not the method
Fig. 25 A spectrum with no decay, where every bound in this essay says nothing.
The share of a HOSVD core's energy on its superdiagonal, by family, at rank 10A matrix SVD hands over orthonormal factors and a diagonal middle at the same time. For three indices they come apart, and this is the half that does not survive. Every core here is all-orthogonal — the largest inner product between two slices perpendicular to a mode, relative to the core's own energy, is 3.5·10⁻¹⁶ — and none of them is diagonal. The bars are the fraction of the squared norm carried by the 10 entries on the superdiagonal: smooth 97.7%, hilbert 94.5%, wave 26.6%, noise 0.8%. Keeping only those entries costs 0.152, 0.235, 0.857, 0.996 in relative error against the full core's 1.09·10⁻¹¹, 1.44·10⁻¹², 1.33·10⁻¹⁵, 0.498.smooth97.7%hilbert94.5%wave26.6%noise0.8%share of the core's energy on its 10 superdiagonal entrieskeeping only them: 0.152 against 1.09·10⁻¹¹keeping only them: 0.235 against 1.44·10⁻¹²keeping only them: 0.857 against 1.33·10⁻¹⁵keeping only them: 0.996 against 0.498orthogonal, and not diagonalsmooth on-diagonal0.98hilbert on-diagonal0.94wave on-diagonal0.27noise on-diagonal0.0081worst slice pair3.5·10⁻¹⁶the slices are orthogonalthe core is not diagonal
Fig. 26 Energy kept against rank kept, which is the tensor field’s version of the same curve.
Singular values of a rank-4 matrix with noise of relative size 10⁻⁴Ten singular values on a logarithmic axis. The first four sit near one; the rest sit at the noise level, and the vertical distance between the two groups is the evidence for the rank.1234567891010⁻¹⁸10⁻¹⁵10⁻¹²10⁻⁹10⁻⁶10⁻³1indexsingular valuecutoff, σ₁ · 10⁻¹⁰numerical rank 10gap 820decisivetrue rank 410×10, built with 4 nonzero valuesrank is a decision
Fig. 27 Where to cut, when the curve is a decision rather than a bound.
Conjugate gradients on the three-dimensional model problem with every iterate cut to rank 1The falling curves are relative residuals: the lower one is the same method with no budget, which reaches 1.61·10⁻¹⁴ in 40 steps, and the upper one is the budgeted run, which stops at 0.119. The two step curves near the top are ranks, on their own scale: the un-truncated step asks for 5 at every step from the third onwards and the budget allows 1. The truncation is therefore not an occasional tidy-up — it is happening at every step, and the distance between the two residual curves is what it costs. The solution of this problem is itself a train of rank five, so a budget of five or more removes nothing and the two curves coincide.081624324010⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²conjugate gradient steprelative residualabove, on their own scale: rank asked 5, rank kept 1solid: the budgeted residual · light: no budgetbudget 1rank asked for5rank kept1residual, budgeted0.12residual, unbudgeted1.6·10⁻¹⁴numbers stored30the step asks for moreat every step
Fig. 28 Truncating inside an iteration instead of after it.
Numbers stored per unknown, against the size of the matrix, at 10⁻¹⁰The dense matrix stores n numbers per unknown, which is the straight line through the origin and doubles whenever n does. The hierarchical representations store 57, 87, 118, 151 and 56, 80, 108, 140, adding about 31 per doubling — a constant per doubling is a logarithm, and it is the whole claim. At n = 512 that is 30 and 27 per cent of the dense matrix, and the share falls at every size. The exponent between consecutive sizes is 1.60, 1.45, 1.35, falling towards one and never arriving, which is what a logarithm looks like from inside.56789100108216324432log₂ nnumbers stored per unknowndenseweak: every off-diagonal blockstrong: only the admissible onesa constant per doublingstrong, n = 5127.7·10⁴weak, n = 5127.2·10⁴dense, n = 5122.6·10⁵per doubling31‖A − A_H‖ ⁄ ‖A‖4.9·10⁻¹²the dense line doublesand the other two add a constant
Fig. 29 What a truncation buys, counted in numbers held.
The accuracy worth paying for in a preconditioner, against the condition number of the problemFour complete cost sweeps, each reduced to the ε that minimised the total work, against the κ of the problem it was run on. The shift that moves κ over 2.7 decades leaves every distance between every pair of clusters exactly where it was, so the ranks the partition asks for stay at 10, 9, 7, 5 throughout and nothing geometric is moving. The optimum walks 0.5 → 0.5 → 10⁻⁶ → 10⁻⁸ — six decades of accuracy, bought because the problem got harder and for no other reason. On the easiest problem the best hierarchical preconditioner in the comparison has rank one; on the hardest it is the tightest one on the sweep. How accurate an approximate inverse should be is a question with an answer, and the answer is not in the matrix's structure.10¹10²10³10⁴10⁵-9-7-5-3-11condition number of the problemlog₁₀ ε at the cheapest totalrank 1rank 1rank 6rank 5the knob answers to κκ = 210.5κ = 1.1·10⁴10⁻⁸rank at the easy end1rank at the hard end5ranks at 10⁻⁸, spread5the geometry did not moveand the right accuracy did
Fig. 30 An accuracy that has an interior optimum, which this one does not.
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. 31 The standing every number in this essay is measured against.
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 · u30×30, 20 seeds per κ; dashed is the worstthe problem worsens, not the method
Fig. 32 A residual, for contrast with a norm of a function.
How much a perturbation of the right-hand side is amplified, κ = 10⁴The cumulative distribution of the amplification factor over two hundred random perturbation directions, with the condition number marked as the upper limit.110¹10²10³10⁴10⁵00.250.50.751amplification of the input perturbationfraction of directions at or belowκ = 10000worst found 76206×6, 200 directionsmedian reaches 0.29 of κ
Fig. 33 The amplifier that turns one into the other, everywhere but here.
The error a reduced model has, and the bound computed before it existed‖H − Hᵣ‖∞ for balanced truncation of a 14-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 2.02 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. 34 The same sweep on a smaller model, where the σ fall faster.
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. 35 Twenty-six states, and the tail the bound is a sum of.

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A-priori boundBalanced truncationExact ground truthGramianHankel singular valuesLyapunov equationMatrix equationMcMillan degreeSingular valuesTransfer function