Numerical Linear Algebra
where the algebra stops being the arithmetic
Series

Sequence stability — the series

4 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 10²10³10⁴10⁻¹⁷10⁻¹⁶10⁻¹⁵10⁻¹⁴10⁻¹³10⁻¹²steps taken‖RᵀR − AᵀA‖ ⁄ ‖AᵀA‖the bound, linear in the steps√k · uevery step safe, the chain notdrift after the run3.9·10⁻¹⁴the bound there3.3·10⁻¹³√k · u there6.1·10⁻¹⁵worst single amplification2.7worst leverage met0.69refreshes1backward stable onceand three thousand times is a different claim

    Stable once, and three thousand times

    A sliding window adds a row and removes one at every step and never looks at the data again. No single step of it amplifies by more than 2.72, no downdate fails, and after three thousand steps the triangular factor in memory is 3.9·10⁻¹⁴ from the matrix it is supposed to be a factor of — six hundred times growth from a per-step bound that says nothing about chains.

    part 1 · sequence
  2. 10²10³10⁻¹⁶10⁻¹³10⁻¹⁰10⁻⁷10⁻⁴10⁻¹steps takenlargest relative error in a coefficientcarried factorrecomputedcarried + one correction+ a second correctionHouseholder QR, same rowsdrift of the factorafter 2875 stepsκ(A) of the window2·10⁴carried factor2.8·10⁻⁸recomputed5.8·10⁻⁹carried + one correction7.1·10⁻¹³+ a second correction3.7·10⁻¹²Householder QR, same rows1.1·10⁻¹²measured against the coefficients in exact rationalsthe refresh repairs the factor, not the answer

    The repair the drift did not need

    A sliding window's carried Cholesky factor drifts 3.9·10⁻¹⁴ from its data, and multiplying by κ(AᵀA) predicts eight lost digits in the coefficients, a stream conditioned at 10¹² losing the answer, and a periodic refresh of the factor as the default repair. Measured against coefficients computed exactly in rationals, all three come out differently. On a stream made ill-conditioned by scaling, the conditioning never reaches the coefficients. On a collinear stream, a freshly recomputed factor is as wrong as the drifted one. And one correction from the window's own rows reaches Householder's accuracy for a fraction of a refresh's cost.

    part 2 · sequence
  3. noise none, mediansthe answer's step0fresh start, corrected6.2·10⁻⁹carried start, corrected1.3·10⁻¹¹0200400600800100010⁻¹²10⁻¹⁰10⁻⁸steps along the streamlargest relative coefficient errorfresh solve, one correctioncarried answer, one correctionfresh solve, two correctionsHouseholder on the rowseach correction contracts its start by the same factorthe nearer start wins

    The answer the last window left

    A sliding window that corrects its least-squares answer at every step could start each correction from the previous step's corrected answer instead of from a fresh solve: the two windows share all but one row. On a stream with any noise in it, that start is three orders worse. The window's exact answer moves by 0.79 of itself in one step at κ(A) = 3·10⁶, a fresh seminormal solve is wrong by only 1.8·10⁻⁴, and one correction contracts either start by the same factor — so the fresh start ends at 2.2·10⁻⁸ and the carried one at 3.7·10⁻⁵. Only on data that agree exactly does carrying win.

    part 3 · sequence
  4. κ(A) = 3.22·10⁶predicted ÷ fresh, noise 10.15predicted ÷ carried, noise 0110⁻¹⁴10⁻¹²10⁻¹⁰10⁻⁸10⁻⁶10⁻⁴noise in the response, relative to the rowslargest relative error010⁻⁶10⁻⁴10⁻²1carried, one correctionfresh, one correctionpredicted, one correctionfresh, two correctionszero noise drawn at the leftthe update takes the better of both starts

    The step the two rows owe

    A sliding least-squares window can start each step's correction from a fresh solve or from the answer it already has. The answer it has is three orders worse on noisy data, because the exact answer moves by most of itself in a step. The proposal was a start that moves too: the previous answer plus the change the entering and leaving rows imply, two triangular solves from the factor the window keeps. Its start lands exactly where one correction of the carried answer lands — the update is that correction, computed from two rows instead of twenty-four — and one correction after it ends 2.9 to 440 times below the fresh start at κ(A) = 3·10⁶, and level with the carried answer when the data agree exactly.

    part 4 · sequence

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