Krylov subspace — where it appears
Named by 39 essays across 11 fields — each of them below, with the objects they name alongside it.
A parameter that counts steps
The regularisation field's knob is a positive real number chosen by one of three rules. The iterative field's is an integer nobody called a knob — where to stop. On the same problem the best step is 20 and the best λ is 0.025, and they reach 0.1426 and 0.1406.
The rate the condition number predicts
Conjugate gradients converge at a rate governed by the square root of the condition number. That is a bound rather than an estimate, it is provable, and it is loose enough that provisioning iterations from it wastes nine out of ten.
The accuracy that is thrown away
A Newton step is the exact answer to a linearised problem, and the linearisation is wrong at second order. So there is a floor under how close the step can land, the floor is the square of where it started, and eleven decades of inner tolerance below it buy the same four digits at four times the price.
An orthogonalisation nobody calls one
Conjugate gradients are derived as a minimisation and behave as an orthogonalisation, which is why the finite-termination property in every textbook is not a property the method has in floating point.
A tolerance that reads its own residual
The cheapest constant forcing term costs 980 inner iterations and arrives with a hundred times the forward error of the dearest, which costs 9,358. A rule that sets each step's tolerance from the ratio of the last two residuals costs 1,009 and arrives with neither problem — and it is not a constant, so it does not appear on the curve the constants are compared on.
Three eigenvalues, and two are the golden ratio
Precondition a saddle-point system by the block diagonal of its own two definite pieces and the preconditioned matrix has exactly three distinct eigenvalues — 1, and the two roots of λ² − λ − 1. A minimal polynomial of degree three means three steps, at every conditioning, and the preconditioner nobody can afford turns out to be the statement the affordable ones are measured against.
The basis decides what a filter is
The vocabulary of regularisation is spectral — a method keeps a component or discards it, and the weights are a function of the singular value. Row-normalising a symmetric blur so that it preserves a constant makes it 8.6% asymmetric, and that is enough to move GMRES's weights from 7·10⁻¹⁴ off a function of σ to 4.4·10⁻².
Exact at the points that were named
Balanced truncation asks for nothing and bounds everything, at a cost no large model can pay. The other kind of reduction asks for r numbers, costs r solves, is exact at every one of them — and bounds nothing anywhere else. That trade is the whole of large-scale model reduction.
The spectrum that predicts nothing
For a symmetric matrix the eigenvalues govern how fast an iteration converges. Drop symmetry and they stop governing anything — there is a matrix whose eigenvalues are as evenly spread as eigenvalues can be, on which GMRES makes no progress at all until the last possible step.
The sketch that is not the answer
Sketch-and-solve throws away the original problem and keeps the small one's answer, which is why its answer moves with the seed. Use the same sketch as a preconditioner instead and the condition number the iteration sees is the same number at every κ from a hundred to ten billion — identically the same, to nine digits, because the spectrum cancels out of it.
A basis that is the same subspace and not the same thing
The interpolation conditions are conditions on a subspace, so any basis of it will do. The one a derivation writes down reaches a condition number of 7.7·10⁹ by its eighth vector, and the rate at which it gets there is set by a number the user chose with no information.
A rate that is known in advance
On the model problem, Jacobi contracts by cos(π/(n+1)) per step, Gauss–Seidel by its square, and optimally relaxed SOR by a number given in closed form. Three rates, all known before anything runs, and all measurable against what runs.
An expiry date the noise does not move
The polynomial description of conjugate gradients leaves the level of rounding at step 17 or 18 on this operator, at every noise level from 10% to 0.1%. The step worth stopping at moves from 3 to 44 across the same range. They coincide at about 1% noise, which is where the coincidence was first read, and it is a fact about the noise rather than about the method.
One eigenvalue and two steps
Put the off-diagonal block back into a block-diagonal saddle-point preconditioner and every eigenvalue of the preconditioned matrix becomes exactly one. GMRES still needs two steps, because the matrix is the identity plus a nilpotent part of norm 54, and a computed eigenvalue at one comes back as a ring of radius 8·10⁻⁸ — the square root of the rounding, not the rounding. With an approximate Schur complement the triangular form leaves one copy of each value where the diagonal form leaves two, and the step count halves.
The last digit is the cheapest
Every cost curve measured here has the same shape: the first digits are cheap and the last ones are not. One method inverts it. Doubling the work buys twice as many digits as the previous doubling did, so the price of a digit halves every time it is paid.
A stable block is not a stable basis
Block Gram–Schmidt orthogonalises twice over — between blocks, and inside each one. Householder inside the blocks does not stop the classical between-block step losing orthogonality like κ², 4.2·10⁻³ at κ = 4.3·10⁷, and a second pass does not stop Cholesky QR inside the blocks breaking down at κ = 10⁸. Each level fails only on ill-conditioning placed at its own level, and one variant holds 3·10⁻¹⁵ on every placement.
A step that is not a unit of work
Landweber's iteration reaches conjugate gradients' best answer on the same deconvolution — 0.1414 against 0.1426 — at step 1,778 instead of step 20, and at 0.1% noise at step 56,234 instead of 44. Each step costs the same two products. And within 10% of its best it runs from step 7 to step 6,310, where conjugate gradients runs from 4 to 26: the slow method is the one that forgives a late stop.
Deciding that a zero has arrived
The previous tolerances were offers — accept this much error, save this much work. A detection threshold is not an offer, because both directions are failures. One matrix here has three genuinely near-invariant subspaces, and the constant somebody typed decides which of them the recurrence stops at; at eight significand bits the same kind of constant produces a proof of something false.
Restarting is a filter
A restart throws away the Ritz values it does not want and begins again from a new starting vector. Written in the eigenbasis, that vector's components have been multiplied by a polynomial with its roots at the discarded values — measured component by component, and agreeing with the polynomial to rounding.
An eigenvalue one vector cannot see
A matrix with an exactly doubled eigenvalue at 10. Twelve Lanczos steps find it once; twenty-four find it once, on a Krylov space of dimension 23 in a 24-dimensional problem. A block of two vectors finds it twice. This is not slow convergence — the second copy is not in the space.
A guess worth two per cent
The previous Newton step looks like a free guess at the next one, and it is worth nothing. Started from it unscaled, the inner solve costs 4 to 61 per cent more than starting from zero, because the guess is 15 to 209 times too large. Scaled by the ratio of the two residual norms it is the right size and halves the starting residual — which buys a constant handful of inner iterations, not a share, because conjugate gradients costs the logarithm of its tolerance.
How wide the block should be
A block narrower than the multiplicity does not converge slowly — it never returns the missing copy at all. Above the multiplicity every extra column buys iterations at about ten products with A each. And the mechanism that is supposed to make the choice unimportant never fires from a random start.
One sequence and two recurrences
CGLS and LSQR compute the same iterates — the minimiser over a space is unique, so there is nothing to choose between them in the algebra. At κ = 10⁶ they cost 42 steps and 47. At κ = 10¹⁰ they cost 110 and 209, across four seeds, and the quantity that separates them is the orthogonality of a basis neither of them keeps.
The vector was what was wanted
Nobody who computes a matrix exponential wants the matrix. They want eᴬᵗb — one vector, the state of a system at a later time. Twenty matrix–vector products get it to sixteen digits on a hundred-by-hundred problem, without ever forming a hundred-by-hundred exponential, and the exponential that does get computed is twenty by twenty.
An operator with no entries
At the sizes where linear algebra is expensive the matrix does not exist. What exists is a subroutine that returns Av. Every Krylov method survives that unchanged; every algorithm that reads an entry disappears. And the derivative such a code computes is accurate to ten digits instead of sixteen, which turns out to cost nothing at all.
The zero that means it is finished
Every Krylov method ends by dividing by a number the previous step produced, and when that number is zero the recurrence stops. In Arnoldi the stop is the answer — the subspace has closed, the solution is inside it, and the residual is at the unit roundoff. The literature calls it a lucky breakdown, and the adjective is doing real work.
The same zero, and nothing was found
Change the recurrence by two lines and the divisor stops being a norm. It becomes an inner product of two vectors from two different sequences, and an inner product of two different vectors is zero on a whole hyperplane — with neither vector anywhere near zero, nothing invariant, and nothing converged. The arithmetic event is identical and the meaning is opposite.
The division that cannot be done
Conjugate gradients divides by pᵀAp at every step, and on a matrix that is not positive definite that number can be zero or negative. The guard against it has been here from the first essay and described it as a failure. In the method that made conjugate gradients famous it is the single most valuable object the iteration can produce, and it costs six matrix–vector products.
An eigenvalue that arrives twice
A matrix with forty distinct eigenvalues, handed to Lanczos for eighty steps, returns twenty-five extra copies of thirteen of them — the largest arriving five times. Every copy is accurate to 1.9·10⁻⁸ relative. No arithmetic error was made, nothing overflowed, and a caller counting eigenvalues gets the wrong multiplicity from a computation in which no individual number is wrong.
The residual the method reports
Conjugate gradients prints a relative residual of 6.9·10⁻²¹. The unit roundoff is 1.1·10⁻¹⁶, so that is not a small residual and not a large one — it is not a residual. The vector the method is holding at that step has ‖b − Ax‖/‖b‖ = 5.1·10⁻¹⁰, and nothing in the run says so.
The number that is re-derived
GMRES prints a residual it never computes from its answer either. On the matrix that sends a conjugate gradient recurrence 7.3·10¹⁰ wrong, and on two others chosen to be worse, its number is never more than a factor of 2.86 out — while the basis it is computed from has lost orthogonality entirely. The disease is not iterative methods, and it is not floating point.
An iterate that must be made smaller
Applying a Kronecker-sum operator to a low-rank iterate multiplies its ranks by d and adding two of them adds their ranks, so a solver in a compressed format cannot keep what it produces. Every step is followed by a truncation — and whether that truncation is a floor on the residual depends on the right-hand side rather than on the truncation.
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.
A run that is over at step five
A conjugate gradient whose every iterate is cut to a rank budget reaches the floor that budget allows at step 5, 36, 42 or 59, and then does nothing for the rest of the run. Four times the iterations move the floor by a factor of 1.8, and past the answer's own rank they move it the wrong way.
The same budget, spent five ways
A restarted method has one budget — products with A — and two ways to spend it, in many short cycles or a few long ones. At about a hundred and forty products the answer is the same to a factor of seven whichever split is chosen, and the residual bound the method reports spans ten orders of magnitude across the same five runs.
A proof that does not ask how large the matrix is
Proving a Hessian indefinite costs three matrix–vector products when the negative eigenvalue is 3 and nine to eleven when it is a thousandth, and that pair of numbers barely moves across a fourfold range in n. The factorisation that settles the same question costs a third of n³, which grows by a factor of sixty-four over the same range.
A walk needs a length
The gap between the two residuals grows as the square root of something, and a square root needs a length. Two quantities are candidates — how far the iterates travelled and how many steps were taken — and only a second sweep separates them. Across a fourfold change in size the iteration count goes from 39 to 96 and the gap goes from 5.04·10⁻¹⁵ to 5.33·10⁻¹⁵.
The answer that arrives when the space runs out
A second-order Krylov recurrence holds vectors of length n for a problem with 2n eigenvalues, so it is exact at n steps where the linearised route needs 2n. The machine-precision reading at forty-four vectors on a chain of forty is that exhaustion rather than convergence, and it arrives through a basis whose ‖QᵀQ − I‖ is above one.
The certificate that arrives soonest is worth least
The more negative a Hessian's smallest eigenvalue, the sooner conjugate gradients meets a direction of negative curvature — and the less of the exact trust-region decrease that direction turns out to be worth. At λₘᵢₙ = −10 the step arrives after two products and gets 39.6 per cent; at −10⁻³ the same two products get 89.8, and the whole sweep costs eight.
Named alongside it
The objects these essays reach for when they reach for this one.
Conjugate gradientsCondition numberOrthogonalityResidualRitz valuesStopping criterionFlop countReorthogonalisationInvariant subspaceLoss of orthogonalityArnoldiFilter factors