Field
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 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.
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.
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.