Example [view without proportional spacing]: Stage 0: commencing with | 0 1 3 6 10 15 | | -1 0 1 3 6 10 | | -3 -1 0 1 3 6 | | -6 -3 -1 0 1 3 | |-10 -6 -3 -1 0 1 | |-15 -10 -6 -3 -1 0 | clearing column 0 via pivot a_10: | 0 1 3 6 10 15 | | -1 0 1 3 6 10 | | 0 -1 -3 -8 -15 -24 | | 0 -3 -7 -18 -35 -57 | | 0 -6 -13 -31 -60 -99 | | 0 -10 -21 -48 -91 -150 | clearing row 0 via a_01: skew-symm | 0 1 0 0 0 0 | | -1 0 1 3 6 10 | | 0 -1 0 -2 -5 -9 | | 0 -3 2 0 -5 -12 | | 0 -6 5 5 0 -9 | | 0 -10 9 12 9 0 | col 1 via a_01, row 1 via a_10: skew symm | 0 1 0 0 0 0 | | -1 0 0 0 0 0 | | 0 0 0 -2 -5 -9 | | 0 0 2 0 -5 -12 | | 0 0 5 5 0 -9 | | 0 0 9 12 9 0 | Stage 1: col 2 via a_32: | 0 1 0 0 0 0 | | -1 0 0 0 0 0 | | 0 0 0 -2 -5 -9 | / (2^2) | 0 0 2 0 -5 -12 | | 0 0 0 10 25 42 | | 0 0 0 24 63 108 | row 2 via a_23: skew symm | 0 1 0 0 0 0 | | -1 0 0 0 0 0 | | 0 0 0 -2 0 0 | / (4^2) | 0 0 2 0 -10 -24 | | 0 0 0 10 0 -6 | | 0 0 0 24 6 0 | col 3 via a_23, row 3 via a_32: skew symm | 0 1 0 0 0 0 | | -1 0 0 0 0 0 | | 0 0 0 -2 0 0 | / (4^2) | 0 0 2 0 0 0 | | 0 0 0 0 0 -6 | | 0 0 0 0 6 0 | Stage 2: nugatory. Determinant = -(1 x -1 x -2 x 2 x -6 x 6) / 4^2 = 3^2; Pfaffian = (1 x -2 x -6) / 4 = 3. [Using the lower diagonal instead would also be consistent --- giving -3 here, and factor (-1)^(n/2) in general.] On 10/26/08, Fred lunnon <fred.lunnon@gmail.com> wrote:
... Given some skew-symmetric matrix [a_{ij}] of order n, reduce it thus: at stage k pivot about a pair of elements a_{2k,2k+1} = -a_{2k+1,2k}, reducing all remaining elements in both row and column 2k and 2k+1 to zero.
Should the prospective pivot be zero, permute rows and columns equally to correct this, recording the sign of the permutation. If no nonzero pivot can be found, the Pfaffian equals zero: when n is odd, the final row and column being already zero as a result of the previous stage, the Pfaffian vanishes.
Easily, each stage preserves the skew-symmetry, k = 0, 1, ... n/2-1.
Finally, the matrix is reduced to an equivalent skew-symmetric bi-diagonal form, where only a_{2k,2k+1} = -a_{2k+1,2k} are nonzero. The determinant now has only a single nonzero term (\prod_k a_{2k,2k+1})^2; the signed Pfaffian equals the accumulated permutation sign times \prod_k a_{2k,2k+1}.
Fred Lunnon