The eigenvalue problem that is not linear

Two approximants and one matrix size

A polynomial approximant linearises to nd rows and a rational one to n(m+1), so the fair contest fixes the matrix and varies the basis. On an easy target set the two are indistinguishable and the ordering flips with the noise; on one that reaches a branch point the rational pulls away by two orders.

Worth reading first: The problem the solver was actually given · Why a Gramian can be truncated at all · The roots are not the coefficients · A matrix that depends on its own eigenvalue.

An error committed before the arithmetic settles the region and the pole placement. What is left is the choice everybody thinks of first: rational or polynomial.

The answer is that it depends on the region, and the measurement that shows so has to be set up carefully, because the two approximants are not naturally comparable.

Two approximants at equal linearisation size, at a reach of 0.06‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 1.2, 2.9, 8.0, 23.7, 73.2 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻⁵10⁻³10⁻¹rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.06largest size drawn78rational there2·10⁻⁵polynomial there0.0015the ratio73no winner on an easy targetand two orders on a hard one
Fig. 1 Error against the size of the matrix the eigensolver is handed, for two bases. The slider moves the target set.

Making the contest fair

A rational approximant with m poles linearises to a matrix of n(m+1) rows. A polynomial of degree d linearises to nd. So degree is not the quantity to fix — a polynomial of degree twenty against a rational with ten poles is a comparison of a 120 × 120 matrix with a 66 × 66 one, and the larger matrix wins for reasons that have nothing to do with the basis.

The eigensolve is what the whole chain is charged for, and it is O(size³). So the fair contest fixes the size, which means d = m + 1, and everything else — the sample points, the fitting procedure, the error measure — is held identical. Same geometric grid, same least-squares QR, same maximum over the same interval.

The result

Ratio of the polynomial’s error to the rational’s, at equal size:

rows 30 42 54 66 78
target set clear of the cut 0.45 1.4 4.4 1.5 0.04
target set reaching towards it 1.2 2.9 8.0 24 73

Two different pictures.

On a target set that stands clear of the branch point, the two are the same measurement. The ratios are near one, they do not increase, and they flip either side of one as the size grows — which is what a comparison between two equally good methods looks like. Reporting a winner from that row would be reporting rounding.

On a target set that reaches towards it, the rational pulls away monotonically, to two orders by seventy-eight rows and still widening.

The mechanism is the one the previous essay set out. Away from the singularity √ is analytic on a neighbourhood of the target set and a polynomial converges to it geometrically — as fast as a rational, because there is nothing for the poles to do. Approaching the singularity the polynomial’s rate degrades towards algebraic while the rational’s stays geometric, because the poles can sit on the cut and the polynomial has nothing to sit anywhere.

Neither method is better, which is the same verdict six routes to one spectrum reaches about six linearisations of a quadratic and for a related reason: what varies between them is small beside what varies with the decision made before either was chosen. The question is different, and which question it is was decided by whoever said where the eigenvalues might be.

The polynomial’s linearisation, which is not a companion matrix

The polynomial approximant is fitted in the Chebyshev basis, and it linearises to a comrade matrix rather than a companion one. The two constructions are identical except for the three-term recurrence in the corner: a companion matrix has the monomial recurrence xᵏ⁺¹ = x·xᵏ and a comrade matrix has 2xTₖ = Tₖ₊₁ + Tₖ₋₁.

Which one a code uses is decided by which basis the coefficients arrived in, and the difference between the two bases is the roots are not the coefficients: a polynomial of degree twenty in the monomial basis has coefficients spanning twenty orders and one in the Chebyshev basis does not.

Using the monomial basis here would confound the comparison badly. The polynomial approximant would lose, and it would lose for a reason about coefficient representation rather than about approximation power — which is exactly the sort of thing this site insists on separating. So the polynomial is given its best available form and still loses where the rational wins.

The interior optimum

Adding poles improves the approximation and enlarges the matrix. Both of those cost accuracy in opposite directions, so the total error has a minimum somewhere in between.

It is visible in the measurements. On a target set clear of the cut, the rational approximant’s error runs 1.9·10⁻⁴, 2.9·10⁻⁶, 4.3·10⁻⁸, 6.8·10⁻⁹ — and then 1.2·10⁻⁸ at twelve poles, which is worse than at ten. The fit’s basis has become ill conditioned: twelve basis functions 1/(λ − ξⱼ) on a region that supports about ten independent directions is a least-squares problem whose coefficients are no longer determined.

The same effect appears at the other end of the chain. The residual against the approximant — the free one from the problem the solver was actually given — stays at 10⁻¹⁵ at two poles and rises to 1.5·10⁻¹² at seven, because the linearisation is larger and worse scaled. So the arithmetic term grows while the approximation term falls, and the crossing is where more poles stop buying anything.

That is a familiar shape from the last digit is the cheapest inverted: there each additional digit costs less than the last, and here each additional pole buys less than the last and eventually buys negative. What is unusual is that both curves are computable in advance — ‖g − r‖ from the fit, and the linearisation’s conditioning from its own norms — so the optimum can be located without running the eigensolve at every degree.

What the choice costs when it is made wrong

Both directions are worth pricing.

Rational on an easy target set costs nothing in accuracy and costs the machinery: poles have to be placed, a rule for placing them has to be chosen, and the fit’s conditioning has to be watched. A polynomial of the same size does as well with none of that. So on a problem whose eigenvalues are known to lie well away from the singularity, the polynomial is the right answer for reasons of simplicity rather than of accuracy.

Polynomial on a hard target set costs two orders at seventy-eight rows, and the loss grows with the size. Doubling the matrix to recover them is an eightfold cost in the eigensolve, which is the expensive step. So this direction is the one worth being careful about, and it is the one a code defaults into: a polynomial approximant needs no decisions and is therefore what gets written first.

The asymmetry says which way to lean. When the region is known to be clear of the singularity, the simpler approximant. When it is not known, the rational one — because the cost of being wrong is much larger in one direction than the other, and not knowing where the eigenvalues are is the ordinary situation — which is the same asymmetry the swap that is not optional prices for pivoting, where the cheap precaution is taken always because the case it guards against cannot be detected in advance.

The comrade matrix, written out

It is worth writing the construction down once, because it is short and because the way it is built from a recurrence rather than from a shift is the whole difference between the two bases.

For a scalar polynomial p(t) = Σ cₖ Tₖ(t), the comrade matrix is d × d with a symmetric tridiagonal skeleton — a 1 in the first row’s second entry, and ½ on both off-diagonals below — and the last row corrected by −cⱼ/(2c_d) for j = 0 … d−1. Its eigenvalues are the roots of p. The companion matrix has the same shape with the tridiagonal replaced by a subdiagonal of ones, which is the monomial recurrence.

For a matrix polynomial the same construction goes through blockwise, with each scalar coefficient replaced by its coefficient matrix. Here every coefficient is a multiple of the identity except the constant term, which carries A, and the linear term, which carries the −λ from the eigenvalue problem itself. The result is nd × nd and its eigenvalues, mapped back through the affine change of variable that put the target set on [−1, 1], are the eigenvalues of the approximated problem.

Two details are load-bearing and easy to get wrong. The affine map has to be folded into the coefficients before the comrade matrix is formed — q(t) = p(t) − αt − β, where λ = αt + β — rather than applied to the eigenvalues afterwards, because the eigenvalue problem’s own λ is not in the Chebyshev variable. And the leading coefficient divides the correction row, so a fit whose top Chebyshev coefficient has been driven to rounding produces a comrade matrix with an enormous last row: a degree chosen higher than the fit can support fails here rather than in the approximation.

That second point is the interior optimum arriving from a third direction. The approximation error falls with degree, the linearisation’s size grows with degree, and the construction of the linearisation degrades with degree once the fit stops determining its own top coefficient.

What neither of them fixes

Both approximants have the same relationship to the spurious eigenvalues that come back with the answer. A rational approximant with m poles produces n(m+1) eigenvalues of which n are wanted; a polynomial of degree d produces nd of which n are wanted. Neither construction returns fewer, neither labels them, and the ratio of noise to answer grows with the size in both cases. The eigenvalues that are answers to nothing is that measurement, and it applies to both bases equally.

And neither of them changes what the residual against the approximant means. It is a backward error for whichever object was linearised, it is at rounding in both cases, and it is silent about the approximation in both cases. The choice of basis moves one term in the error decomposition and leaves the reporting problem exactly as it was.

Why this comparison is worth making at all

A reader might reasonably ask why anyone would run both, given that the answer depends on a region whose position is exactly what is not known.

The reason is that the comparison is cheap and the decision is not otherwise available. Fitting both approximants at one size costs two least-squares problems on a few hundred sample points — call it a millisecond — against an eigensolve of an n(m+1) matrix, which is the whole cost of the method. So a code can fit both, compare the two maxima, and take the better one, at a cost that does not appear in any accounting of the method.

That is a better answer than choosing by rule, because the rule requires knowing where the region sits relative to the singularity, and the singularity’s position is sometimes as unknown as the eigenvalues’. Fitting both measures the thing the rule would have had to be told.

It also gives a diagnostic. If the two maxima are within a factor of two of each other, the target set is clear of whatever difficulty the function has and the problem is easy in the sense that matters. If they differ by an order, the region is close to a singularity, and that is worth knowing independently of which approximant gets used — because it is also the signal that the target set should be narrowed, which an error committed before the arithmetic shows is worth more than either choice.

This is the pattern this site keeps arriving at and it is worth naming once more: two routes to the same quantity, run together, cost almost nothing and answer a question neither of them was asked. Here the question is how hard the problem is, and the answer is a ratio of two numbers that were going to be computed anyway.

The size that is the real cost

One accounting note, because “equal linearisation size” has been doing a lot of work and its consequences are worth spelling out.

An eigensolve of an N × N matrix is O(N³), and N = n(m+1) here. So doubling the number of poles is roughly an eightfold cost, and the whole method is dominated by that one step: the fit is milliseconds, the linearisation is O(N²) to assemble, and the factorisation is everything else.

That changes what “a better approximation” means in practice. Going from ten poles to twelve on the easy target set buys nothing in accuracy — the interior optimum is already passed — and costs a factor of about 1.7 in the eigensolve. Going from four poles to ten on the hard one buys four orders and costs a factor of about eight. The second is obviously worth it and the first obviously is not, and the point is that both are decidable from the fit alone, before the expensive step.

For a large sparse problem the accounting is different again, because nobody forms an n(m+1) matrix and factorises it — the linearisation is solved by a Krylov method, whose cost is linear in N per iteration and whose iteration count depends on the linearisation’s conditioning. Then the growth term is not the cube but the degradation of the conditioning with m, which is the same quantity the rising residual measures and which this essay has priced only indirectly. That is the honest limit of what is measured here: the dense accounting is exact and the sparse one is inferred.

The one line

Fix the matrix size, not the degree. Fit both bases, compare the two maxima, and take the better one — it costs a millisecond against an eigensolve. If they agree, the region is easy and the simpler approximant is right; if they differ by an order, the region is close to a singularity and that is worth knowing for its own sake.

At other settings

Two approximants at equal linearisation size, at a reach of 1‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 0.4, 1.4, 4.4, 1.5, 0.0 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻¹⁰10⁻⁸10⁻⁶10⁻⁴rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach1largest size drawn78rational there1.2·10⁻⁸polynomial there5.2·10⁻¹⁰the ratio0.043no winner on an easy targetand two orders on a hard one
Fig. 2 A target set clear of the branch point, where the two curves cross more than once.
Two approximants at equal linearisation size, at a reach of 0.3‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 0.9, 2.5, 7.7, 25.6, 89.4 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻⁸10⁻⁶10⁻⁴10⁻²rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.3largest size drawn78rational there8.7·10⁻⁸polynomial there7.8·10⁻⁶the ratio89no winner on an easy targetand two orders on a hard one
Fig. 3 Thirty per cent of the way, where the rational has begun to pull away.
Two approximants at equal linearisation size, at a reach of 0.02‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 1.3, 2.6, 6.5, 17.6, 49.2 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻⁴10⁻²rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.02largest size drawn78rational there1.6·10⁻⁴polynomial there0.0079the ratio49no winner on an easy targetand two orders on a hard one
Fig. 4 Two per cent, where both are struggling and one is struggling less.
Two approximants at equal linearisation size, at a reach of 0.008‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 1.4, 2.3, 5.2, 12.7, 32.7 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻⁴10⁻²1rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.008largest size drawn78rational there5.4·10⁻⁴polynomial there0.018the ratio33no winner on an easy targetand two orders on a hard one
Fig. 5 And under one per cent.
The error that happens before any arithmetic: |g − r| on the target setγ√(λ + c) minus its approximant, plotted against the distance from the branch point at −0.4. The target set runs from -0.2 to 8.99, and its left end is 0.2 from the branch point — a choice made before anything is computed. The rational approximant with 8 poles clustered on the cut reaches 8.46·10⁻⁴; the polynomial of degree 9, which linearises to a matrix of the same size, reaches 0.00676. Both errors are committed at this stage. No residual computed later contains either of them, because a solver's residual is taken against whichever of these two functions it was handed.110¹10⁻⁷10⁻⁵10⁻³distance from the branch point, λ + c|g − r|the eigenvaluesdegree 98 poleschosen, not computedreach0.06left end, from the cut0.2rational, worst8.5·10⁻⁴polynomial, worst0.0068linearisation size, both54committed before the solveand invisible to it
Fig. 6 The two error curves on the interval, rather than as maxima against size.
The error that happens before any arithmetic: |g − r| on the target setγ√(λ + c) minus its approximant, plotted against the distance from the branch point at −0.4. The target set runs from 2.94 to 8.99, and its left end is 3.34 from the branch point — a choice made before anything is computed. The rational approximant with 8 poles clustered on the cut reaches 4.3·10⁻⁸; the polynomial of degree 9, which linearises to a matrix of the same size, reaches 1.91·10⁻⁷. Both errors are committed at this stage. No residual computed later contains either of them, because a solver's residual is taken against whichever of these two functions it was handed.10¹10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶distance from the branch point, λ + c|g − r|the eigenvaluesdegree 98 poleschosen, not computedreach1left end, from the cut3.3rational, worst4.3·10⁻⁸polynomial, worst1.9·10⁻⁷linearisation size, both54committed before the solveand invisible to it
Fig. 7 And on the easy target set, where they lie on top of each other.
Three residuals, and the one a solver returns is the one about nothingFor a rational approximant with m poles, on a target set reaching to 0.3: ‖T̃(λ)x‖ — the residual against the problem the eigensolver was handed — stays at 1.2·10⁻¹⁵ to 1.5·10⁻¹² — it does not fall with m, and past a point it slowly rises, because each added pole makes the fit's own basis worse conditioned; ‖T(λ)x‖, which costs one further evaluation of γ√(λ + c), falls with the approximation; and the forward error against the closed form falls with it, staying 20.5 to 24.8 times larger. The free number rises by 1306 across the sweep while the answer improves by 885. So the free number says nothing and the nearly-free number says almost everything. Evaluate the residual against the function you asked about, not against the one you handed over is the whole practical content of this field.23456710⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²poles in the approximantresiduals and errorforward error‖T(λ)x‖‖T̃(λ)x‖one extra evaluationagainst the approximant1.5·10⁻¹²against the problem asked1.9·10⁻⁶forward error4.8·10⁻⁵‖g − r‖ there8.5·10⁻⁵the free residual is flatand the answer is not
Fig. 8 What the approximation term costs downstream.
36 eigenvalues come back, 6 of them mean somethingThe real eigenvalues of the n(m+1) = 36 that a rational approximant with 5 poles produces, on the real line, with the branch point at −0.4 marked. The 6 discs to the right are the answers, agreeing with the closed form to 7.74·10⁻⁴. The 6 crosses to the left are exact eigenvalues of the approximant lying where γ√(λ + c) is not a real number at all — so the residual that says nothing about accuracy is decisive as a filter: against T̃ all 12 look perfect, and against T the spurious ones do not have a residual. The remaining 24 are complex and are discarded the same way.-10123456789-101λbranch point at −0.4the closed formwhich of these is an answereigenvalues returned36wanted6past the branch point6complex24worst against the closed form7.7·10⁻⁴all of them exactfor a problem nobody asked
Fig. 9 And what comes back besides the answer, in both bases equally.
The roots of ∏(x − i) up to degree 10, computed from the coefficients that define themA polynomial whose roots are the integers 1 to n, expanded exactly in BigInt and then handed to the route every library takes: the eigenvalues of the companion matrix. The measured relative error runs 1.711·10⁻¹³ at degree 6 to 1.127·10⁻¹⁰ at degree 10 — the third digit of an answer whose truth is an integer. The other curve is the root condition number times the unit roundoff, which is a prediction made from the coefficients alone, and it tracks the measurement to within two orders at every degree. Every coefficient here is still exactly representable, which is what makes the failure a conditioning one rather than a storage one.5791110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²degreerelative error of the worst rootmeasuredκ × ua polynomial is its coefficientsworst root, measured1.1·10⁻¹⁰predicted, κ × u1.4·10⁻¹⁰root condition number6.1·10⁵largest coefficient1.3·10⁷the roots are integersand the coefficients are not the roots
Fig. 10 Why the polynomial is fitted in the Chebyshev basis rather than the monomial one.
The roots of ∏(x − i) up to degree 20, computed from the coefficients that define themA polynomial whose roots are the integers 1 to n, expanded exactly in BigInt and then handed to the route every library takes: the eigenvalues of the companion matrix. The measured relative error runs 1.711·10⁻¹³ at degree 6 to 0.007614 at degree 20 — the third digit of an answer whose truth is an integer. The other curve is the root condition number times the unit roundoff, which is a prediction made from the coefficients alone, and it tracks the measurement to within two orders at every degree. Above degree 19 the coefficients are integers larger than 2⁵³, so the polynomial handed to the routine is not the polynomial that was written down — the problem is damaged before any algorithm runs.57911131517192110⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²degreerelative error of the worst rootmeasuredκ × ua polynomial is its coefficientsworst root, measured0.0076predicted, κ × u0.0023root condition number10¹³largest coefficient1.4·10¹⁹the roots are integersand the coefficients are not the roots
Fig. 11 And what happens at degree twenty in the basis that was not chosen.
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. 12 A basis whose conditioning decides whether a fit is determined.
After reflector 1 of 5: the bulge in column 0A square matrix drawn as a grid with one nonzero subdiagonal. A small triangle of entries below that subdiagonal is highlighted, and moves down the matrix from panel to panel.0.461.90.14-0.0070.2-0.18-1.20.78-1.40.54-0.18-0.0791.40.440.120.870.19-0.077-0.49-0.320.932.30.90.0380000.982.11.500001.52.3after reflector 1the highlighted entries are the bulge — the only thing that is not Hessenbergentries below the subdiagonal3reflectors used1reflectors in a whole step5the shifts are never formed — only their sum and productand both of those are real
Fig. 13 The step inside the factorisation both routes end in.
The real Schur form with 2 conjugate pairs: 2 blocks that cannot be splitA square matrix drawn as a grid. Everything below the diagonal is zero except for a small number of two-by-two boxes on the diagonal, which are highlighted.3000000120000-21000000-0.5-1.500001.5-0.5000000-2T = ZᵀAZthe highlighted boxes each hold one conjugate pair, and no real rotation removes themthe form, and that it is one‖A − ZTZᵀ‖/‖A‖1.8·10⁻¹⁵‖ZᵀZ − I‖2.5·10⁻¹⁵worst eigenvalue error2.7·10⁻¹⁵surviving subdiagonal26×6, spectrum chosen before the matrix was builtquasi-triangular is as far as the reals go
Fig. 14 And the form it produces.
Householder reduction to Hessenberg form, on a symmetric 6×6Two matrices side by side. The first is full; the second has zeros everywhere below the first subdiagonal, and being symmetric is tridiagonal.-0.16-0.63-0.493.43-0.025-0.632.3-0.69-0.95-1.70.076-0.49-0.694.3-1.6-1.41.93.4-0.95-1.65.40.0141.63-1.7-1.40.0144.60.055-0.0250.0761.91.60.0553.1A, symmetric-0.164.600004.65.92.500002.51-1.20000-1.24.90.3800000.385.80.2400000.242H = QᵀAQ, tridiagonal‖A − QHQᵀ‖/‖A‖1.1·10⁻¹⁵below the subdiagonal0worst eigenvalue movement7.1·10⁻¹⁵a similarity, so the spectrum is untouched — and every later step is O(n²) rather than O(n³)one reduction, then every iteration is cheapthe eigenvalues did not move
Fig. 15 The reduction that precedes it, whose cost is cubic in the size this essay fixes.
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. 16 An interior optimum in another field, from the same two curves crossing.
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. 17 Cost against accuracy, where the trade is monotone.
The rational function 6 ADI shifts amount to, chosen two ways|∏(λ − pⱼ)/(λ + pⱼ)| over the spectrum of −A for a 30-point discretisation, whose ends are 9.86 and 3834 and whose condition number is 388.8. The lower curve uses 6 geometrically spaced shifts and touches zero at each of them, equioscillating between with a maximum of 0.1187. The upper curve uses 6 equally spaced ones: every shift sits in the top decade of the spectrum, the small end is never approached, and the maximum is 0.8922 — a factor of 7.52 worse, which the iteration pays squared. The two runs are the same algorithm, the same number of solves, and one different line where the shifts are chosen.10¹10²10³10⁻⁹10⁻⁷10⁻⁵10⁻³10⁻¹λ, over the spectrum of −A|r(λ)|geometric maxequally spacedgeometricone line of codeshifts6κ of the spectrum389geometric max0.12equally spaced max0.89the factor between7.5the same k solvesand one choice of where
Fig. 18 The pole placement this essay takes as given.
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. 19 Where the same rational approximation question comes from in the reduction field.
Two routes to e^A as a 8×8 bidiagonal matrix walks up to a defective oneA_δ is upper bidiagonal with diagonal λ, λ+δ, …, so its eigenvalues are exact and its eigenvectors have a closed form. As δ falls the eigendecomposition route's relative error rises to 2.92·10⁶⁵ while scaling and squaring's falls to 4.09·10⁻¹², which is the distance ‖A_δ − A₀‖ = 7·10⁻¹² to the matrix whose exponential is known in closed form. The two agree at the right of the axis, where the matrix really is far from the limit; the divergence is κ(V), which reaches 3.3·10⁸².10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴10⁻²10⁻¹⁴10⁻¹10¹²10²⁵10³⁸10⁵¹10⁶⁴δ, the gap between consecutive eigenvaluesrelative error in e^Aan answer with no correct digitsV f(Λ) V⁻¹scaling and squaring‖A_δ − A₀‖exact eigenvalues throughoutκ(V) at the smallest δ3.3·10⁸²eigen route2.9·10⁶⁵scaling and squaring4.1·10⁻¹²distance to the limit7·10⁻¹²the eigenvalues are the diagonaland they are exact at every stop
Fig. 20 Rational against series for a matrix function, one field over.
‖Aᵏ/k!‖ against ‖e^A‖ for Moler and Van Loan's 2×2, scaled by 0.25The terms of the Taylor series rise to 81.6 at k = 4 and fall back; the sum they produce has norm 5.37, drawn as the flat line. The ratio is 15.2, so that many digits are cancelled away before truncation is a question, and the computed exponential has a relative error of 1.87·10⁻¹⁵ after 36 terms. Scaling and squaring on the same matrix returns 1.65·10⁻¹⁴.0510152025303510⁻⁶10⁻⁴10⁻²110²k‖Aᵏ/k!‖‖e^A‖ = 5.4largest term 82what the series throws awaylargest term82‖e^A‖5.4digits cancelled away15error after the sum1.9·10⁻¹⁵every term is computed correctlyand the sum has lost seven digits
Fig. 21 And what the series costs there.
Where the missing eigenvalues went: an 6 × 6 quadratic with 1 of its masses removedA chain of 6 masses with the first 1 of them set to zero, so M is singular of rank 5. det Q(λ) is a polynomial of degree at most 12; interpolated exactly in BigInt rationals at 13 nodes it has degree 11, so 1 of the 12 eigenvalues are at infinity — the same object a descriptor pencil has, arriving here because a degree of freedom with no inertia is an algebraic constraint. The float route counts the singular values of M judged to be zero and reaches 1, backed by a gap of ∞ between consecutive singular values. One integer, two routes, and only the second of them is a decision.finite eigenvalues (degree of det Q)11at infinity (2n − degree)1at infinity, by the rank of M12n, if M were nonsingular12a degree, not a decisiondegree of det Q11at infinity1by the rank of M1singular-value gapthe count is a degreeand the other route is a judgement
Fig. 22 A count that is a degree, in the field’s first essay.
The 16 eigenvalues of an 8 × 8 quadratic eigenvalue problem, computed and in closed formλ²M + λC + K for a chain of 8 masses with C = 0.3M + 0.1K. The crosses are the closed form — one scalar quadratic per eigenvalue of K, whose roots are known exactly — and the discs are the eigenvalues a real Schur factorisation returns from the 16 × 16 first companion linearisation. There are 16 of them for a matrix with 8 rows, of which 16 are complex and arrive in conjugate pairs, so the eigenvectors cannot be independent: 9 vectors in 8 dimensions never are. The worst disagreement between the two routes is 2.12·10⁻¹⁵, and the routes share nothing but the three coefficient matrices.00.0939569-2-1.24498-0.4899610.2650581.020081.7751real partimaginary parttwo routes, one spectrumeigenvalues16rows8complex16against the closed form2.1·10⁻¹⁵n rows and 2n eigenvaluesso the eigenvectors are not a basis
Fig. 23 And the spectrum it counts.
Six ways of computing one spectrum, at a change of units of 106Three linearisations — the first companion form, the second, and the symmetric member of the DL(Q) family — each reduced to a standard eigenvalue problem in two ways: by inverting the leading coefficient, whose eigenvalues are λ, and by inverting the trailing one, whose eigenvalues are 1/λ. All six have exactly the eigenvalues of the quadratic in exact arithmetic. Measured against the closed form at γ = 106, the best is first/leading at 5.473·10⁻⁶ and the worst second/trailing at 2.234·10⁻⁴, a spread of 40.81. Under the leading reduction the symmetric form and the first companion form are THE SAME MATRIX — −A₁⁻¹A₀ is [[−M⁻¹C, −M⁻¹K], [I, 0]] for both — so the distinction between them is one the inversion discards, and it survives only under the trailing one.first · leading5.47·10⁻⁶first · trailing4.62·10⁻⁵second · leading4.27·10⁻⁵second · trailing2.23·10⁻⁴symmetric · leading5.47·10⁻⁶symmetric · trailing4.05·10⁻⁵all six are the same algebrabest route5.5·10⁻⁶worst route2.2·10⁻⁴spread across the six41condition of the linearisation8.3·10¹²the spectra agreeand the arithmetic does not
Fig. 24 Six linearisations of one quadratic, which is the same question one degree down.
What two lines of scaling are worth: the same quadratic in nine systems of unitsThe forward error against the closed form for an overdamped chain of 8 masses, before and after Fan–Lin–Van Dooren scaling. The unscaled curve runs 7.49·10⁻¹⁴ to 0.001258 — every digit gone by the far end — and the scaled one runs 1.26·10⁻¹³ to 8.23·10⁻¹⁴, flat to within a factor of 2.01. The scaling is γ = √(‖K‖/‖M‖) and δ = 2/(‖K‖ + γ‖C‖), computed from three norms and nothing else, and the map back is λ = γμ with no rounding in the statement. Flatness is the half that matters: after scaling every stop of the sweep IS the same problem, so there is nothing left for the change of units to do.0246810⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹log₁₀ γ, the change of unitsforward erroras writtenafter scalingone change of variableunscaled, worst0.0013scaled, worst1.7·10⁻¹³orders recovered10scaled coefficient spread4.5the answer was never the problemthe units were
Fig. 25 And the decision that dominates all of them.
A spectrum with no last eigenvalue: 12 of the infinitely many solutions of a delay problemT(λ) = A − λI + 0.5e^−λI for the model matrix at n = 4. The problem decouples in A's eigenvector basis, so each of its 4 eigenvalues μ contributes the scalar equation μ − λ + γe^−λ = 0, whose solutions are λ = μ + W_k(γe^−μ) — one for every branch of the Lambert W function, and therefore countably infinitely many, running off to the left along two curves. The marks are the closed form, each checked by substitution to 1.2·10⁻¹⁵. There is no computation that returns this set: the only well-posed question is which eigenvalues lie inside a region, and the circle of radius 4 contains 4 of them.-3-2-101234-5-3-1135real partimaginary part4 insidea countable spectruminside the contour4drawn12existingworst branch residual1.2·10⁻¹⁵there is no last eigenvalueso the question has to change
Fig. 26 A nonlinearity with no branch point, where the contest disappears.
Three filters at their own optima, on one problemFilter factors against the singular-value index. Truncation at K = 21 is a step from one to zero; Tikhonov at λ = 0.0215 is a smooth descent; the 20-step conjugate gradient filter rises above one and changes direction 5 times. The three reach errors within a few per cent of each other while differing by up to 0.99 in individual factors.081624324048566400.250.50.7511.25index kfilter factortruncation K = 21TikhonovCGLS, 20 stepsone floor, three weightingstruncation vs Tikhonov0.77truncation vs CGLS0.99reversals in the CGLS filter5the same error, to a few per centby three different weightings
Fig. 27 Two filters compared at equal cost, in the regularisation field.
A 6.25-bit block format against 8-bit E4M3, read two waysFour curves of relative error against the spread of the data. Two, for the block format, rise steeply when read as a median and gently when read as a norm; the other two are nearly flat.048121610⁻²10⁻¹1octaves of spread within a blockrelative errorblock, medianblock, 2-normE4M3, medianE4M3, 2-normthe same data, two readingsblock format, bits a value6.3norm across 16 octaves1.8median across 16 octaves62entries deleted at the wide end383640 values, blocks of 32a norm is dominated by what a block format keeps
Fig. 28 Two methods crossing as a parameter moves.
One subspace, solved and regularised, at 0.50% noiseTwo error curves against the number of bidiagonalisation steps, on a logarithmic vertical axis. The unregularised iterate reaches 0.1221 at step 22 and then climbs to 80.4 — a factor of 658. The iterate with a penalty on the projected problem reaches 0.1331 and stays within 0.00% of it for the rest of the run.1591317212529333710⁻¹110¹10²bidiagonalisation stepsrelative errorleast without: 22no penaltypenalty insidewhat stopping is worthbest without a penalty0.12and at step 4080best with one0.13and at step 400.13the same floor, reached twiceand only one run stays on it
Fig. 29 And two more, where the crossing is the whole finding.
Two approximants at equal linearisation size, at a reach of 0.06‖g − r‖ against the size of the matrix the eigensolver is then handed: n(m+1) for a rational approximant with m poles and nd for a polynomial of degree d, so d = m + 1 is the fair comparison. At this target set the ratio of the polynomial's error to the rational's runs 1.2, 2.9, 8.0, 23.7, 73.2 across the five sizes. On a target set that stands clear of the branch point the two are the same measurement and the ordering flips with the noise; on one that reaches it the rational pulls away, because a polynomial converges algebraically against an algebraic singularity. Neither method is better — the question is different, and which question it is was decided by whoever said where to look.263850627410⁻⁵10⁻³10⁻¹rows in the linearisation‖g − r‖ on the target setpolynomialrationalequal cost, two basesreach0.06largest size drawn78rational there2·10⁻⁵polynomial there0.0015the ratio73no winner on an easy targetand two orders on a hard one
Fig. 30 The hero again, beside the error curves it summarises.

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Approximation before linearisationBranch pointChebyshev basisCompanion formComrade matrixCondition numberFlop countLinearisationRational approximationSpurious spectrumTarget set