Numerical Linear Algebra
where the algebra stops being the arithmetic
Concept

Orthogonal projection — where it appears

The map onto a subspace along its orthogonal complement, which is what a least-squares fit does to the right-hand side and nothing else. It is what a least-squares fit does and nothing more, so the residual it leaves is orthogonal to the column space by construction rather than by accident.

Named by 12 essays across 5 fields — each of them below, with the objects they name alongside it.

everything Ax can reachb = (1.1, 0.4, 1.5)Ax, the closest reachable pointr = b − Ax‖Aᵀr‖ / (‖A‖‖r‖)1.7·10⁻¹⁶‖b‖² − ‖Ax‖² − ‖r‖²1.3·10⁻¹⁵‖r‖1.3200 random nearby points of the plane were tried; none is closer.a 3×2 system, Householder QRperpendicularity is checked

The projection and the right angle

The least-squares solution is the one whose residual is perpendicular to everything the columns can reach. That is not a mnemonic — it is an equation, Aᵀr = 0, and the computed answer satisfies it to 10⁻¹⁶.

leastsquares · Least-squares 04812162010⁻¹10⁻⁰.⁵1target rank k‖A − Aₖ‖₂published boundrandomisedσₖ₊₁, optimalhow far apart the three areworst seed spread1.6bound / median at k = 125.9median / optimum at k = 121.960×60, 6 seeds, oversampling p = 5band is best to worst

A bound that holds with probability

Every other guarantee in this collection is deterministic. The randomised low-rank approximation offers one that holds with a probability, the seed changes the answer, and the honest figure is a band rather than a line.

randomised · Randomised 10²10².³10².⁵⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁹⁶10⁻¹10⁻⁰.⁵1rows in the sketchworst relative distortiondimension 64dimension 2565 seeds per point, band is best to worstthe dimension does not appear

The dimension does not appear

A random projection preserves the lengths of a set of vectors to within a distortion that depends on how many vectors there are and not on how many coordinates each one has. That is the fact the whole field rests on, and it is genuinely surprising.

randomised · Sketching for each previous column i, subtract the projection of column j onto qᵢclassicalr[i][j] = qᵢ · a[j] ↑ the ORIGINAL columnv = v − r[i][j] · qᵢthe three worst |qᵢ · qⱼ|:columns 7 and 8: 1columns 6 and 8: 0.13columns 6 and 7: 0.13modifiedr[i][j] = qᵢ · v ↑ what is LEFT of itv = v − r[i][j] · qᵢthe three worst |qᵢ · qⱼ|:columns 1 and 8: 4.4·10⁻⁷columns 2 and 8: 2.7·10⁻⁷columns 3 and 8: 2.4·10⁻⁸The two R factors agree to 1.2·10⁻⁶ relative. The two Q factors do not.the 8×8 Hilbert matrixone word, eight orders

Two Gram–Schmidts

One argument changes. Classical Gram–Schmidt projects the original column onto each previous direction; modified projects what is left of it. In exact arithmetic the coefficients are identical. In floating point they differ by eight orders of magnitude in the thing that matters.

orthogonality · Gram–Schmidt 00.250.50.75110⁻¹110¹share of the noise placed in the matrixleast-squares error ÷ total least-squares errorequally accuratetotal leastsquares aheadordinary leastsquares aheadthe model, not the methodadvantage, all noise in b0.28advantage, all noise in A2.5seeds at each share40the same total noise at every pointand only where it sits changes

When the matrix is wrong too

Every least-squares problem here has assumed A is exact and b is not, and moved b onto the column space of A. Where both were measured, the smallest correction that makes the system consistent moves the matrix as well — and on the problems where that answer is more accurate, it has the larger residual, by construction rather than by luck.

leastsquares · Total least-squares 012345610⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹round‖AΩ‖ ⁄ ‖A‖ at that roundredrawn each roundhalf redrawnkept: the sketch is now the zero matrixthe randomness is spentkept, round 12.6kept, round 29·10⁻¹⁵redrawn, round 23.1kept, final error0.28redrawn, final error0.11a random matrix used twiceis not random the second time

The sketch that is spent

Every other object a sequence carries has a shelf life. A random sketch has one use. Deflate what its first round found and apply it again, and it returns the zero matrix — 9.0·10⁻¹⁵ where the first round saw 2.62 — because the input has been made orthogonal to the very draw the guarantee is over.

randomised · Sketching 15731pointsfinest gridone unknown — the recursion bottoms out in a divisionthe coarse operator, two ways‖RAₕP − A₂ₕ‖/‖A₂ₕ‖10⁻¹⁸unknowns / finest grid1.7cycle cost, in fine sweeps12each coarse point reaches three fine ones½, 1, ½ — and the restriction is its transpose

The same problem on a coarser grid

Restriction, the coarse operator and interpolation are three matrices with nine distinct entries between them. Two of the three are each other's transpose, and their product with the fine operator is the coarse discretisation exactly — not approximately, entry for entry, at every level.

iterative · Multigrid the diagonal of the hat matrix, hᵢ = aᵢᵀ(AᵀA)⁻¹aᵢaverage p/m = 0.20010the leave-one-out residual: eᵢ/(1 − hᵢ), and forty refitsbars: closed form · dots: refitted without that pointone number, two fieldsΣ hᵢ, exactly p10largest leverage0.5closed form against refits8.9·10⁻¹³1 − h of the first row0.5y appears in the residualand nowhere in the leverage

Influence is decided before the data

The diagonal of the hat matrix sums to the number of columns and the response appears nowhere in it, so a fit has exactly p units of influence to hand out among m observations. The same row at h = 0.5 is a ten-fold outlier on one design and a boundary case on another, and which of those it is was settled before a single measurement was taken.

leastsquares · Leverage the invariant planesolid: beforedashed: aftersame perturbation, two questionsthe vectors turned, radians0.029the plane turned, radians7.6·10⁻⁸what left the plane5.6·10⁻⁸drawn in the unperturbed plane's own basisa radius is not determined; the circle is

The plane survives what its vectors do not

At a gap of 10⁻⁹ a perturbation of 10⁻⁶ turns the two eigenvectors through half a radian and turns the plane they span through 7.6·10⁻⁸ — a ratio of six million. Ask for the subspace instead of the vectors and a hopeless computation becomes a well-conditioned one, with no change to the arithmetic.

spectra · Invariant subspace 08162432404810⁻¹10²10⁵10⁸10¹¹10¹⁴10¹⁷10²⁰degreeκ of the design matrix1/u: past this a double holds nothingmonomialsChebyshevArnoldi on the pointsκ at degree 48monomials1.5·10¹⁷Chebyshev1.6·10⁷Arnoldi on the points1200 two intervals with a gap between themthree bases for one space of polynomials

A basis built from the points

A polynomial fit computed in monomials and in an orthogonal basis gives the same curve on exact data, and the valley essay drew the two lying on top of each other. Add 0.1% noise and they separate — by 1.7·10⁻⁵ at degree 40 and 0.004 at degree 48 — because the fitted curve moves with the basis by its condition number times the rounding times the residual. Chebyshev polynomials keep that small only on points spread like their weight; on a sample with a hole in it they reach κ = 1.55·10⁷. A basis orthogonalised against the sample points themselves stays at 1 on every set.

leastsquares · Fitting -1012345612345678xyall 32without one twinwithout boththe pair at x = 6leverage of each0.44pair's joint share0.89deleted alone, mean0.62deleted together, mean3each twin covers for the otheronly removing both shows the shift

Two observations that hide each other

Two observations at the same place, wrong by the same amount, each look harmless when deleted alone, because a fit without one still has the other. Single deletion sees the shared error cut by (1 − 2h)/(1 − h) — measured at 261 times at the far end — and only the pair's two-by-two block of the hat matrix says what the two of them hold.

leastsquares · Leverage 10³10⁶10⁹10¹²10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹κ(A), identical for both problemsrelative error in the answerthe weak directions in the constraint's null spacethe weak directions in the constraint's own rowsone κ(A), two problemsκ(A), both10¹²κ left, constrained1κ left, free10¹²error, constrained4.7·10⁻¹⁶error, free3·10⁻⁴between them6.4·10¹¹a constraint is informationand κ(A) does not know it arrived

The condition number that does not know

Two constrained fits with the same size, the same number of constraints and the same κ(A) to twelve figures. One returns 4.7·10⁻¹⁶ and the other 3.0·10⁻⁴. What separates them is the conditioning of A restricted to the constraint's null space — 1.00 against 10¹² — which every solver computes on the way and none reports.

leastsquares · Constrained least-squares

Named alongside it

The objects these essays reach for when they reach for this one.

ResidualLeast-squaresCondition numberNormal equationsOrthogonalityEckart–YoungLeverageLow-rank approximationProbabilistic boundsRandomised SVDReorthogonalisationSingular value decomposition

All concepts