Iterating, instead of factorising

A Krylov space for a problem that is not linear

A quadratic eigenvalue problem has no matrix to build a Krylov space out of. The recurrence that builds one anyway stores half as many numbers, returns twice as many Ritz values — and stops being a basis at twenty vectors while the answer it gives keeps improving.

Worth reading first: A parameter that counts steps · The spectrum that predicts nothing · Orthogonal is a number · A matrix that depends on its own eigenvalue.

Every method in this field builds a subspace out of A, Ab, A²b, … and asks a small projected problem for its answer. The rate the condition number predicts prices the convergence, an orthogonalisation nobody calls one is about the sweep that keeps the basis usable, and an eigenvalue that arrives twice is what happens when it stops.

A quadratic eigenvalue problem has no A. It has three coefficient matrices, and the only matrix in sight is the 2n × 2n linearisation. So the obvious method is Arnoldi on that — and it stores vectors of length 2n whose bottom half is a scaled copy of the top half of an earlier one.

Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 40 masses. Arnoldi on the 80 × 80 linearisation stores m vectors of length 80 and returns m Ritz values; the second-order recurrence stores m vectors of length 40 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 240 numbers it is 0.01379 from the dominant eigenvalue where the linearised route needs 480 to reach 0.07294. 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.33.310⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised2080storage, second-order1040Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 1 Two Krylov methods for one quadratic, priced by the numbers they hold rather than by the number of steps they take.

The recurrence that does not store the bottom half

Write the linearised matrix as [[A, B], [I, 0]] with A = −M⁻¹C and B = −M⁻¹K. A Krylov vector is [qⱼ; pⱼ], and the recurrence’s next vector has top half Aqⱼ + Bpⱼ and bottom half qⱼ. The bottom half is therefore never new information: it is what the top half was, one step ago, with the same orthogonalisation coefficients applied to it.

The second-order Arnoldi recurrence stores the top halves and updates the bottoms with the same coefficients without orthogonalising them. Same subspace, n numbers a step instead of 2n.

Then the projection can be taken onto the quadratic rather than onto the linearisation:

Mₘ = QᵀMQ,   Cₘ = QᵀCQ,   Kₘ = QᵀKQ,

an m × m quadratic eigenvalue problem, solved by the machinery of the field’s first essay, with 2m eigenvalues.

So m stored vectors of length n yield 2m approximations, where Arnoldi’s m stored vectors of length 2n yield m. A factor of four in Ritz values per number held, exactly and by construction — measured at exactly four at every subspace size, because both halves of it are counting rather than arithmetic.

And it is closer, at every size

The exactness of the storage claim makes it easy to assume the cheaper method must be worse somewhere. On this family it is not.

Measured on a chain of forty masses, distance from the dominant eigenvalue:

vectors Arnoldi on the linearisation projected quadratic
6 7.3·10⁻² 1.4·10⁻²
10 1.5·10⁻² 4.9·10⁻³
14 5.9·10⁻³ 2.3·10⁻³
20 6.4·10⁻³ 1.6·10⁻³
26 1.4·10⁻³ 1.1·10⁻³

Closer at every size, and five times closer at the smallest. The mechanism is not subtle: more approximations out of one subspace is more chances at the eigenvalue wanted, and the projected quadratic extracts 2m of them where the projected linearisation extracts m.

The honest half is the absolute rate. Twenty-six vectors buy three digits, which is slow, and it is slow for a reason that has nothing to do with either method: the chain’s spectrum is an arc with no isolated extreme, and Krylov convergence is decided by separation. This is a comparison of prices at equal work, not a demonstration of a fast method.

The two extractions are different answers

It would be tidy if the two methods produced the same Ritz values from the same subspace. They do not, and measured they differ by 9 per cent at six vectors and 32 per cent at twenty.

That is not a defect. Arnoldi’s Ritz values are the eigenvalues of the projected linearisation; the second-order route’s are the eigenvalues of the projected quadratic. Those are two different small problems asked of the same vectors, and there is no reason for their answers to coincide. What is required — and measured — is that both approach the same spectrum, and that the best Ritz value from each is a genuine eigenvalue of the quadratic to two digits or better at every size.

It is worth noticing which of the two is the more natural. Projecting the linearisation preserves the shape of the algorithm; projecting the quadratic preserves the shape of the problem. The small problem the second route solves is a quadratic eigenvalue problem, so everything the field has established about linearising, scaling and backward error applies to it — recursively, at size m, where it is cheap. A method whose subproblem is of the same kind as its problem is easier to reason about, and here it is also better.

What the subspace actually contains

A step back, because the two extractions being different raises a fair question: what is the subspace, and is it the same one in both cases?

The linearised Krylov space is span{v, Av, A²v, …} in ℝ^{2n}, with v = [q₁; 0]. Write out its elements and every one of them is a pair whose top half lies in

span{q₁, Aq₁ + Bp₁, …}

and whose bottom half lies in the span of the previous top halves. So the projection of the linearised space onto its top block is exactly the space the second-order recurrence builds, and the second recurrence is not an approximation of the first — it is a representation of it, with the redundant half not stored.

That is what makes the storage claim exact rather than heuristic, and it is also what makes the orthogonality finding a real one. The two methods are working in the same space. What differs is the coordinates they hold it in, and one set of coordinates degrades.

Why this is where the field meets the rest of the site

There is a practical reason this essay sits in the iterative field rather than in the polynomial one, and it is about the sizes involved.

Every measurement in the polynomial field so far is on a dense problem of size eight or twelve, factorised whole. The application the field comes from is not that: a finite-element model of a damped structure has n in the hundreds of thousands, M and K sparse, C often a low-rank modification of them, and the request is for twenty eigenvalues nearest zero. A 2n × 2n dense factorisation is out of the question, and everything the field has established has to survive the transition to a subspace method.

Most of it does, unchanged, and that is worth saying explicitly:

  • The scaling matters as much. The projected quadratic inherits the coefficients’ imbalance, so the two lines of the scaling that buys ten orders go in front of the recurrence rather than in front of a factorisation.
  • The backward error is the same quantity. η for the projected problem is η for a quadratic, and the residual for the large problem is the one that decides convergence.
  • The count is still 2n. A subspace of dimension m gives 2m Ritz values out of 2n eigenvalues, and no subspace method changes the arithmetic of that.

What does not survive is the closed form: a real structure has no Lambert function and no decoupling sine basis, so the reference disappears and the residual becomes the only evidence. Which is exactly why the measurements in this field are made on a family that has one.

The shift, which is the other half of every real run

Everything above starts the recurrence at the coefficients themselves, which converges to the largest eigenvalues — and the eigenvalues anybody wants from a damped structure are the smallest, since those are the modes that persist.

The standard repair is shift-and-invert: pick a σ near the region of interest and run the recurrence on Q(σ)⁻¹ rather than on the coefficients, which maps the eigenvalues near σ to the large end. That costs one factorisation of the n × n matrix Q(σ) — not of a 2n × 2n linearisation, which is the second place the second-order route saves — and every step after it is a solve with a factor already computed.

It also changes what the essay’s finding means in practice. With a shift, convergence to the wanted eigenvalues takes a handful of steps rather than twenty-six, which puts the useful subspace sizes below the size at which the basis collapses. The measurement above is therefore a statement about where the cliff is, made at a size where it is reachable; a shifted run in production stays on the safe side of it, and knows where the side is because someone measured.

The basis stops being a basis

The price arrives, and it arrives suddenly.

A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 40 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.37·10⁻¹³ throughout, and the second-order one runs from 1.29·10⁻¹² at four vectors to 1.347 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03036 to 0.001559 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors1.3·10⁻¹²at 201.3linearised, worst2.4·10⁻¹³distance to λ at 200.0016the basis fails firstand the answer does not say so
Fig. 2 ‖QᵀQ − I‖ for the two bases against the number of vectors, with the distance to the dominant eigenvalue on the same axes. One curve collapses and the other does not notice.

Measured, ‖QᵀQ − I‖ for the second-order basis:

6 vectors     5.9·10⁻¹¹
10            6.1·10⁻⁷
14            1.6·10⁻³
20            1.35

Orthogonality gone entirely by twenty vectors. The linearised Arnoldi basis on the same problem stays at 2.4·10⁻¹³ throughout.

Both are orthogonalised by the same modified Gram–Schmidt sweep. So the difference is not the orthogonaliser — it is the recurrence. The second-order sweep orthogonalises the vector r against the qs, and applies the same coefficients to the auxiliary vector s without orthogonalising it. The auxiliary sequence therefore accumulates error freely, and it feeds the next residual. What is being orthogonalised at step j is a vector built out of an already-degraded auxiliary, and no amount of care in the sweep repairs an input.

This is the same finding an orthogonalisation nobody calls one makes about classical against modified Gram–Schmidt — same algebra, different arithmetic — one level up: same orthogonaliser, different recurrence, and a loss that the orthogonaliser cannot see.

And the answer keeps improving through the collapse

Here is the part that makes it a measurement rather than a caution.

Across the range where orthogonality goes from 10⁻¹¹ to 1.35, the distance from the dominant eigenvalue falls from 1.4·10⁻² to 1.6·10⁻³. The answer gets better as the basis fails.

A run that watched only the answer would report steady progress at every step. There is no divergence, no stagnation, no residual plateau — the diagnostic that the residual the method reports recommends does not fire, because the Ritz residual is computed from the projected problem and the projected problem is consistent with itself.

What fails silently is everything downstream that assumes Q is orthonormal: the Ritz vectors are not orthogonal, the Ritz values may be repeated where the spectrum is not, and the projection is no longer a projection. That is precisely the situation an eigenvalue that arrives twice describes for Lanczos — a ghost, arriving because the basis lost the memory of a direction it had already covered — and it is the reason the fix in that essay is the fix here: re-orthogonalise, at a cost, or restart before the collapse.

What a code should do about it

Two options, and the measurement prices both.

Restart before twenty. The collapse arrives at about the same number of vectors whatever the chain’s length — measured at n = 16, 24, 32, 40 and 56 — because it is driven by the recurrence’s own growth rather than by the size of the problem. So the safe subspace size is a property of the method, and it is small. That is the same discipline restarting is a filter sets out, with an unusually clean number attached to it.

Or orthogonalise the auxiliary sequence too, which is what the two-level variant of this recurrence does: keep an orthonormal basis for the pair rather than for the top halves, and represent both halves in it. That restores the orthogonality and costs back part of the storage saving — which is the trade in one sentence, and the honest way to end a comparison whose headline was that the cheap method wins.

The other thing a projected quadratic can do

One consequence of projecting the problem rather than the matrix is worth drawing out, because it is not available to the linearised route at all.

The projected quadratic Mₘ, Cₘ, Kₘ inherits structure. If M, C and K are symmetric then so are their projections, and the small problem is a symmetric quadratic — which means the certificate of every eigenvalue real can be run on it for the cost of an m × m Cholesky, and the answer is a statement about the small problem that is often the statement wanted about the large one. If the coefficients are palindromic, the projection is palindromic, and the reciprocal pairing of a spectrum that comes in reciprocal pairs is available at size m.

The projected linearisation keeps none of that: the Hessenberg matrix an Arnoldi run produces is unsymmetric whatever the coefficients were, and the structure has been dissolved into it. So the second-order route’s real advantage may not be the factor of two in storage at all — it may be that the subproblem is still a member of the class the problem came from.

That is a claim this essay measures the storage half of and leaves the structural half of standing. It is the kind of loose end worth writing down rather than rounding off: the measurement here is about price, and the structural argument is about what a method can be told.

What is worth carrying into a large problem

Three sentences, and the third is the one that is easy to lose.

The cheaper recurrence is genuinely cheaper and genuinely better at equal work, by a factor of two in storage, four in Ritz values per number held, and a measured factor of two to five in distance to the dominant eigenvalue at every subspace size tested.

Its basis fails before its answer does, at about twenty vectors whatever the size of the problem, and the failure is invisible in every quantity the method returns.

And the failure is a property of the recurrence rather than of the orthogonaliser. That is what makes it worth a measurement: a reader who knew that modified Gram–Schmidt is the good one, which this site established years of essays ago, would conclude that a basis built with it is safe. Here it is not, because what is being orthogonalised has already been damaged before the sweep sees it.

At other settings

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. 3 Sixteen masses: the same factor of two in storage and four in Ritz values.
Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 56 masses. Arnoldi on the 112 × 112 linearisation stores m vectors of length 112 and returns m Ritz values; the second-order recurrence stores m vectors of length 56 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 336 numbers it is 0.01393 from the dominant eigenvalue where the linearised route needs 672 to reach 0.07166. 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.33.33.610⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised2912storage, second-order1456Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 4 And fifty-six, where both curves shift right because a vector is longer.
A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 24 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 4.17·10⁻¹³ throughout, and the second-order one runs from 1.81·10⁻¹² at four vectors to 1.558 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02877 to 2.12·10⁻⁴ across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors1.8·10⁻¹²at 201.6linearised, worst4.2·10⁻¹³distance to λ at 202.1·10⁻⁴the basis fails firstand the answer does not say so
Fig. 5 The collapse at twenty-four masses, arriving at the same number of vectors.
A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 56 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 4.16·10⁻¹³ throughout, and the second-order one runs from 3.62·10⁻¹³ at four vectors to 1.317 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03037 to 0.001601 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors3.6·10⁻¹³at 201.3linearised, worst4.2·10⁻¹³distance to λ at 200.0016the basis fails firstand the answer does not say so
Fig. 6 And at fifty-six.
‖QᵀQ − I‖ at every step of a Lanczos run, n = 40The departure from orthogonality of the Lanczos basis, against the step, on a logarithmic vertical axis. It sits at the level of rounding for 13 steps and then climbs by a factor of about seventeen a step for ten steps running, twelve orders of magnitude, before saturating. The dashed line marks the step at which the first Ritz value converged, which is step 11.051015202530354010⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹step‖QᵀQ − I‖first Ritz value converges√uno reorthogonalisationfullthe two events are one eventstep orthogonality crosses √u14step the first Ritz value converges11consecutive steps at ≥5× growth10a drift would grow like a square rootthis grows like a geometric series and then stops
Fig. 7 The same measurement for a Lanczos recurrence, where it was first drawn.
Copies of each eigenvalue after 80 steps on a 40×40 matrix with a simple spectrumA bar per eigenvalue that came back more than once, showing how many times. The matrix has 40 distinct eigenvalues by construction; the run returned 25 extra copies of 13 of them, the most-copied arriving 5 times. Every copy is accurate to 1.9·10⁻⁸ relative, which is why nothing but the true spectrum could detect them.λ = 105 timesλ = 9.55 timesλ = 95 timesλ = 8.55 timesλ = 2.952 timesλ = 2.92 timeseigenvalues that arrived more than once — the matrix has 40 distinct onesa spectrum with the wrong multiplicitiesextra copies, no reorthogonalisation25extra copies, full reorthogonalisation0worst relative error among the copies1.9·10⁻⁸steps taken of 80 asked for, full40no arithmetic error was madeevery one of these is right to eight digits
Fig. 8 And the ghost eigenvalues that arrive once it has happened.
QᵀQ from classical Gram–Schmidt and from Householder on the 8×8 Hilbert matrixTwo eight-by-eight tables of QᵀQ. The upper one has ones on the diagonal and entries as large as one off it; the lower one is the identity to three decimal places everywhere.A = H8 · κ = 1.5·10¹⁰ · both factorisations reconstruct A to 6·10⁻¹⁷the diagonal is 1 in both — every column is a unit vector either way1.000000000001.000000000001.000000000001.000000000001.00000.002-0.002000001.0000.125-0.13300000.0020.1251.000-1.0000000-0.002-0.133-1.0001.000classical Gram–Schmidt1.000000000001.000000000001.000000000001.000000000001.000000000001.000000000001.000000000001.000Householderclassical ‖QᵀQ − I‖1.4Householder ‖QᵀQ − I‖1.4·10⁻¹⁵largest off-diagonal 1 against 3.1·10⁻¹⁶length is not angle
Fig. 9 Orthogonality as a number, from the field that made it one.
The residual basis of conjugate gradients, at κ = 106A 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. 10 What the loss costs a Krylov solve rather than a Krylov eigensolver.
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. 11 A restart, which is the other repair.
The reported bound and the residual it bounds, over 10 cyclesThree quantities against the cycle count on a logarithmic vertical axis. The residual bound the method reports falls without limit, reaching 9.41·10⁻⁴¹. The residual it claims to bound stops at 5.68·10⁻⁵ and does not move. Recomputing the arrowhead's border entries, at one extra product with A a cycle, takes the residual to 3.81·10⁻¹⁴.1234567891010⁻⁵⁵10⁻⁴⁹10⁻⁴³10⁻³⁷10⁻³¹10⁻²⁵10⁻¹⁹10⁻¹³10⁻⁷10⁻¹cyclesizethe residualrepairedthe reported boundwhat the stopping rule readsreported at the last cycle9.4·10⁻⁴¹the residual there5.7·10⁻⁵with the border recomputed3.8·10⁻¹⁴products, cheap and repaired9a bound with nothing under itand one product a cycle to fix it
Fig. 12 And what is kept across one.
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. 13 The spectrum being approximated, and why its arc makes convergence slow.
One quadratic eigenvalue problem in 9 systems of units: what the solver reports and what the answer is worthAn overdamped chain of 8 masses, with λ replaced by γμ so that the coefficients become (γ²M, γC, K). That substitution is exact in both directions and divides the spectrum by γ exactly, so the closed form is still available and every error here is measured against it. The backward error of the eigenpair for the LINEARISED MATRIX — the residual a solver's own error analysis is about — is 6.76·10⁻¹⁶ at γ = 1 and 7.59·10⁻¹³ at γ = 108 — it moves by a factor of 1928 while the other two move by 1.68·10¹⁰. The backward error for the QUADRATIC, which is what the person who posed the problem is entitled to, grows by 8.76·10¹⁰ across the same sweep, and the forward error follows it: 7.49·10⁻¹⁴ to 0.00126. Nothing went wrong with the solver at any stop.0246810⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹log₁₀ γ, the change of unitsrelative errorforward errorη, the quadraticη, the linearisationagainst a closed formη(linearisation), worst7.6·10⁻¹³η(quadratic), worst1.2·10⁻⁴forward error, worst0.0013coefficient spread4.2·10¹⁵the solver is right at every stopabout a problem nobody asked
Fig. 14 The scaling question the projected quadratic inherits at size m.
The 16 real eigenvalues of an overdamped chain, in two groups of 8, with the certificate between themA chain of 8 masses with C = 8K, which is above the critical damping 5.75877 = 1/sin(π/2(n+1)), so the quadratic is hyperbolic and every one of its 16 eigenvalues is real. The vertical line is the μ at which Q(μ) is negative definite — a Cholesky that completes, and the whole proof that the spectrum is real. It also lies in the gap: exactly 8 eigenvalues above it and 8 below, with a gap of 0.6698 between the two groups. One factorisation answers both questions, which is the second thing a boolean would not have carried.-33-28.2505-23.5009-18.7514-14.0018-9.25229-4.502750eigenvalueQ(μ) ≺ 0one Cholesky, two answersabove the certificate8below it8the gap0.67critical β for this n5.8the spectrum is real by classnot by outcome
Fig. 15 The family the sweep is run on.
Conjugate gradients at κ = 104, 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 bound119 steps40×40, spectrum spread evenly in logbound permits 1417
Fig. 16 Krylov convergence where the spectrum does have an isolated end.
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. 17 And where the spectrum predicts nothing at all.
Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 24 masses. Arnoldi on the 48 × 48 linearisation stores m vectors of length 48 and returns m Ritz values; the second-order recurrence stores m vectors of length 24 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 144 numbers it is 0.01305 from the dominant eigenvalue where the linearised route needs 288 to reach 0.06995. 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.310⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised1248storage, second-order624Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 18 Twenty-four masses: the same exact ratios, on a shorter axis.
Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 32 masses. Arnoldi on the 64 × 64 linearisation stores m vectors of length 64 and returns m Ritz values; the second-order recurrence stores m vectors of length 32 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 192 numbers it is 0.01363 from the dominant eigenvalue where the linearised route needs 384 to reach 0.07029. 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.33.310⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised1664storage, second-order832Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 19 And thirty-two.
A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 16 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 6.83·10⁻¹³ throughout, and the second-order one runs from 3.65·10⁻¹² at four vectors to 2.828 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.02878 to 0 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors3.7·10⁻¹²at 202.8linearised, worst6.8·10⁻¹³distance to λ at 200the basis fails firstand the answer does not say so
Fig. 20 The collapse at sixteen masses, arriving at the same number of vectors.
A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 32 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 2.1·10⁻¹³ throughout, and the second-order one runs from 9.72·10⁻¹³ at four vectors to 1.475 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03006 to 0.001353 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors9.7·10⁻¹³at 201.5linearised, worst2.1·10⁻¹³distance to λ at 200.0014the basis fails firstand the answer does not say so
Fig. 21 And at thirty-two, which is the control: it is the recurrence rather than the size.
Loss of orthogonality against condition number, in binary64A log–log plot of the norm of Q-transpose-Q minus the identity against condition number. Classical Gram–Schmidt rises steeply, modified Gram–Schmidt rises gently, and Householder is flat.110²10⁴10⁶10⁸10¹⁰10¹²10⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹condition number κ(A)‖QᵀQ − I‖classicalmodifiedHouseholderκ²uκu8×8, eight seeds per κ, binary64all three reconstruct A
Fig. 22 Loss of orthogonality measured against precision rather than against steps.
What each reorthogonalisation costs and what it leaves, n = 40Two quantities for three variants on one logarithmic axis: the number of projections performed, and the departure from orthogonality left behind. Doing nothing costs nothing and leaves a basis that is not one. Full reorthogonalisation costs 820 projections and leaves 1.9·10⁻¹⁵. Selective costs 137 — 6.0 times fewer — and leaves 10·10⁻⁸, which is √u and is enough.010⁻¹⁷10⁻¹⁴10⁻¹¹10⁻⁸10⁻⁵10⁻²10¹count, and ‖QᵀQ − I‖√u — semi-orthogonalitynoneselectivefullprojections‖QᵀQ − I‖two counters, three methodsghosts, none4ghosts, selective0selective is cheaper by6the cheap repair leaves a basis that is not orthogonaland that is the claim, not a shortfall
Fig. 23 What re-orthogonalising costs, which is the first of the two repairs.
12 restarts keeping 4 of 8, on a 40×40 matrixThe residual bound of the worst wanted eigenvalue and its true error, against the number of products with A. The bound falls from 1.21 to 1.08·10⁻¹³ across 12 cycles and 140 products, and the true error reaches 7.11·10⁻¹⁵. The basis is 8 vectors at every cycle and never grows.0183654729010812610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹products with Asizeresidual boundtrue errorbounded memorybasis vectors kept8products with A140worst error in the k wanted7.1·10⁻¹⁵the bound is free and the error is notand the basis never grows
Fig. 24 And what restarting costs, which is the second.
The width of a block of 4, at every stepThree step functions against the step count. A block of 4 independent columns keeps its width at every step. A block with one column copied from another loses that column before the first product with A and keeps 3 for the rest of the run. A block whose columns all lie in the span of two eigenvectors starts at 2 and ends at 0 after one step, having found the whole subspace.01234567891001234stepcolumns in the blockindependent columnsone column repeatedinside an invariant planewhen a block narrowsrandom: width at the end4repeated: width at the end3invariant: vectors found2the mechanism worksand nothing a random start does sets it off
Fig. 25 A block Krylov method, which is the other way to see a repeated eigenvalue.
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. 26 A shift, which is how a real run reaches the eigenvalues anybody wants.
The condition number the problem has, and the one the solver's error analysis is written againstThe same eigenvalue of the same overdamped chain of 8 masses, in seven systems of units. Its condition number as an eigenvalue of the QUADRATIC — Tisseur's, with the three coefficient norms in the numerator and yᵀQ′(λ)x in the denominator — is 4.98 at γ = 1 and 4.98 at γ = 10⁶, a spread of 1 over six decades: it cannot move, because a change of units is not a change of problem. Its condition number as an eigenvalue of the LINEARISED MATRIX runs 49.66 to 7.844·10¹¹, a factor of 1.58·10¹⁰. The forward error follows the second one, and the first one is the honest description of the problem — so the substitution has manufactured an ill conditioning that belongs to the algorithm rather than to the question.012345610⁻¹10²10⁵10⁸10¹¹log₁₀ γ, the change of unitscondition numberthe linearisationthe quadraticone problem, two amplifiersκ(quadratic), first5κ(quadratic), last5κ(linearisation), last7.8·10¹¹how far the first moved1the problem is as well conditioned as everand the method is not
Fig. 27 The conditioning the projected quadratic inherits at size m.
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. 28 And the scaling that goes in front of the recurrence rather than in front of a factorisation.
Two Krylov methods for one quadratic, priced by the numbers they storeA chain of 56 masses. Arnoldi on the 112 × 112 linearisation stores m vectors of length 112 and returns m Ritz values; the second-order recurrence stores m vectors of length 56 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 336 numbers it is 0.01393 from the dominant eigenvalue where the linearised route needs 672 to reach 0.07166. 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.33.33.610⁻⁴10⁻³10⁻²10⁻¹1log₁₀ numbers storeddistance to the dominant eigenvalueArnoldi, linearisedprojected quadraticper number heldstorage, linearised2912storage, second-order1456Ritz values, linearised26Ritz values, second-order52half the storageand twice the approximations
Fig. 29 The largest chain the sweep is run on.
A basis that stops being a basis, and a Ritz value that keeps improving through it‖QᵀQ − I‖ for the two Krylov bases on a chain of 56 masses, against the number of vectors. Both are orthogonalised by the same modified Gram–Schmidt sweep, so what differs is the recurrence: the linearised basis stays at 4.16·10⁻¹³ throughout, and the second-order one runs from 3.62·10⁻¹³ at four vectors to 1.317 at twenty — orthogonality gone entirely. The third curve is the distance from the dominant eigenvalue, which falls from 0.03037 to 0.001601 across the same range — the answer improves across the collapse, so a run watching only the answer would report progress at every step.3691215182110⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹vectors held‖QᵀQ − I‖, and the distance to λsecond-order basisdistance to λlinearised basisthe symptom is absence‖QᵀQ − I‖ at 4 vectors3.6·10⁻¹³at 201.3linearised, worst4.2·10⁻¹³distance to λ at 200.0016the basis fails firstand the answer does not say so
Fig. 30 And its orthogonality, collapsing at the same number of vectors.

What links here

Computed from the collection, not written here: the essays that point at this one.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

Named objects

A flat tag is an object no other essay names yet.

ArnoldiKrylov subspaceLinearisationModified Gram–SchmidtOrthogonalityQuadratic eigenvalue problemRitz valuesSecond order arnoldi