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Sparsity, and what elimination costs

Eliminating a variable couples everything it touched to everything else it touched, and every one of those couplings is an entry that was zero in the matrix and is not zero in its factor. On the same matrix one elimination order gives a factor of a thousand entries and another gives ten thousand — the two factorisations equally accurate, one of them fitting in memory. Nothing numerical chooses between them.
the matrix, lower triangle408 entriesits Cholesky factor1739 entries · 1331 created‖A − LLᵀ‖/‖A‖1.4·10⁻¹⁶fill, symbolic1331fill, numeric1331n = 144 · density 3.2% · bandwidth 12same matrix, renumberedthe answer is identical to rounding

The factor is not sparse

A sparse matrix has a factor that is not sparse, and the gap between them is the entire reason iterative methods exist. The entries elimination creates can be counted before any arithmetic runs, from the graph alone.

natural1739reverse Cuthill–McKee1354minimum degree1026nested dissection1413matrix: 408 entries · dense factor: 10440bandwidth 12 · 4.26× the matrixbandwidth 12 · 3.32× the matrixbandwidth 123 · 2.51× the matrixbandwidth 108 · 3.46× the matrixn = 144, five-point stencilevery ordering fills in; none avoids it

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 matrix43 entriestip eliminated first253 entriestip eliminated last43 entries‖A − LLᵀ‖/‖A‖, tip first1.4·10⁻¹⁶‖A − LLᵀ‖/‖A‖, tip last0dense factor is n(n+1)/2 = 253 · sparse factor is 2n − 1 = 43one row swapped to the endnothing numerical chose between them

Two ends of the same arrow

One matrix, one row moved from the front of the elimination order to the back, and the factor goes from completely dense to no fill at all. Both factorisations are exact to rounding, and nothing numerical chose between them.

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