Permutation — where it appears
Named by 18 essays across 7 fields — each of them below, with the objects they name alongside it.
Elimination is a sequence of choices
Gaussian elimination is taught as a procedure with no decisions in it. There is one decision at every step — which row to use — and every stability property the algorithm has comes from making it well.
The order decides the memory
Four elimination orderings on one matrix give factors of 1,739, 1,354, 1,413 and 1,026 entries. All four factorisations are exact, all four return the same answer, and the one with the better asymptotics is not the one that wins.
The swap that is not optional
Run elimination without a row interchange on a matrix that needs one and nothing announces a failure. There is no division by zero, no warning, and an answer of the right shape. It is simply wrong, and how wrong depends on a number you did not look at.
What the symbolic phase can only bound
Without pivoting, the fill can be computed from the graph and the count is exact — 233 predicted, 233 measured. With pivoting it is 233 predicted and 242 measured, and what survives is a bound that is right at every threshold and loose by 1.7 times at the largest grid drawn.
The same matrix, numbered twice
One symmetric permutation. The condition number is 24.3948 either way to eight digits and the Frobenius norm is 6.13996414·10³ either way to twelve. The partition that stored 27,008 numbers now finds no admissible pair anywhere and stores all 65,536, and the format that compresses regardless stores 118,208.
A test with no answer in it
A caller with no reference answer can still ask whether a routine answered the right question: reverse the columns, run it again, compare. The polar factor's two answers agree to 10⁻¹⁵ at every conditioning drawn; a QR's differ by 2.353 on matrices whose own norm is 2.449. The test has a floor, and the floor is measurable too.
The elimination the matrix does not need
The Kronecker form of AX + XB = C is dismissed with a hundred million entries and (2/3)n⁶ operations. Both price an elimination, and after the reduction both routes take, the matrix has exactly n³ nonzeros, none of them above the block diagonal, and nothing left to eliminate.
The least fill there is
Finding the elimination order with the least fill is NP-hard, and that is a statement about the hardest graph and the largest size. On a graph of twenty vertices every one of the 20! orders can be searched at once, through the million sets of vertices already eliminated, and the least fill is a number. On the 4×4, 4×5 and 3×7 grids minimum degree finds it exactly. On eighty random sparse graphs of eighteen vertices it finds it on 53 and misses by at most 7.6 per cent, and on every one of the eighty some breaking of its ties finds it.
A bit buys an octave
The outlier a block survives is exactly two raised to its significand width — 8 at three bits, 32 at five, 128 at seven, 512 at nine. Each extra bit doubles the range the block tolerates and halves the ordinary entry's error. Reordering the same numbers buys every octave at once and costs nothing.
Which of the choices is doing the work
Elimination makes n − 1 decisions and they are not worth the same. On 8×8 standard normal matrices, removing the first pivot search and leaving the other six multiplies the median growth factor by 1.624; removing the last multiplies it by 1.000. The cost falls monotonically along the run, and the worst single matrix in the sweep grows by 2,366 when one early decision goes — so the median is the wrong statistic and the tail is where pivoting earns its reputation.
Two minima that are one minimum
The order that decides the memory found the operation count behaving like the square of the fill, which leaves room for an order with slightly more fill but a shorter heaviest column to do less arithmetic. Searched exactly over every elimination order on forty graphs, that order does not exist: one order attains both minima on thirty-nine of forty, and on the fortieth the least-fill order's arithmetic is 1.0099 times the least. Minimum degree attains both on the same thirty-three graphs and neither on the same seven.
The order the greedy rule cannot choose
Wilkinson's matrix is the standard demonstration that partial pivoting's growth bound of 2^(n−1) is attained. It is attained by the row order the greedy rule picks, and not by the matrix: a single cyclic shift of the rows gives growth 2 at every size, with no multiplier above one. At n = 7 that is 64 against 2. And perturbing one entry by 10⁻¹² leaves the good order exactly where it was while putting every tie-break of the greedy rule back on 64.
Three orders and one last entry
Over the integers there is no stability to pivot for, so a fraction-free elimination swaps rows only when the pivot is zero. Choosing a pivot for length instead does change the sequence of minors — the smallest-nonzero rule makes seven exchanges where the natural order makes none and keeps the profile two bits lower through the middle. It cannot change the peak. The last entry of the elimination is the determinant, and the determinant does not know what order it was computed in.
The column that was never fixed
Every threshold-pivoting measurement so far chose the pivot row in a fixed column, and the routine's own description said that choosing the column as well would change the constants and not the argument. Measured, it changes the argument. On the 8×8 conflict grid the factor shrinks from 875 entries to 640 at the library default, and the growth factor that climbed to 2,209 as the threshold loosened stays at 2.54 at every threshold from 0.3 down to 0.001. What does most of the work is not the column but which of several equally cheap entries is taken — and on random sparse matrices, choosing the column without that makes the growth worse.
How few columns the search needs
A full row-and-column pivot search is quadratic in the active submatrix at every step, and no library performs one. Looking at a single sparsest column takes the median fill from 244 to 136 where the full search reaches 109 — four fifths of the benefit for a linear scan — and that share is 79, 83, 80, 89 and 92 per cent across five thresholds. The worst growth appears to favour the narrow search by a factor of six, and on the next draw it favours the wide one by two.
The freedom a symmetric factorisation does not have
Permuting rows and columns together leaves no column to choose, so the conflict between the sparsest pivot and the sound one should be worse rather than better. On a saddle-point matrix whose constraint rows have no diagonal entry at all, it is not there: taking the sparsest available pivot holds 70 entries against the natural order's 113 and a growth of 1.28 against 1.83 — better on both currencies at once, at every setting of the pivot test. The two-by-two blocks that make it legal cost 1.33 entries apiece.
One step ahead is one step short
Partial pivoting takes the largest entry in the column and, on Wilkinson's matrix, walks into growth of 2^(n−1) that a cyclic shift of the rows avoids entirely. A rule that chose instead the pivot whose elimination leaves the smallest trailing submatrix was expected to see the good order at the first step. It sees nothing there: every first pivot leaves a largest entry of exactly 2, and with the ties broken by 10⁻¹² it prefers the greedy row by 10⁻¹². It attains 2^(n−1) at every size. Looking two eliminations ahead, the greedy row scores 4 and every other row 2, and the growth is 2 at every size up to 24. On random matrices one step of look-ahead helps below n = 16 and is worse than greedy on more than half of them by n = 32.
The pivot is in every product
A fraction-free elimination's intermediates are minors, and Hadamard bounds a minor by the lengths of the rows it is made of — so the rule that picks the smallest pivot entry looked like a proxy for a rule that picks the shortest row. Measured, the two keep the bit-length profile equally low: 0.969 and 0.966 of the natural order's area. They part on what the arithmetic costs. Counting every multiplication and division at the product of its operands' lengths, the smallest-pivot rule costs 0.70 of the natural order and the shortest-row rule 0.84, because the pivot multiplies every entry of the step and divides every entry of the next. Three per cent of area is thirty per cent of arithmetic.
Named alongside it
The objects these essays reach for when they reach for this one.
Gaussian eliminationFill-inGrowth factorSparsityExact ground truthPartial pivotingBackward errorFill-reducing orderingLU factorisationMultipliersSymbolic factorisationElimination graph