Re: [math-fun] Pfaffians
Gene asked: << The determinant of a matrix is a polynomial in its elements. For an antisymmetric matrix of odd order, the determinant is zero, while for an antisymmetric matrix of even order, this polynomial factors into the square of another polynomial, called the Pfaffian. Does anyone know of a simple, easy to follow, proof that the determinant factors as asserted?
Here's a cite for what appears to be an elementary 1-page proof: Skew-Symmetric Determinants Author(s): S. Parameswaran Source: The American Mathematical Monthly, Vol. 61, No. 2 (Feb., 1954), p. 116 I believe there is a more informative but more advanced proof in "Characteristic Classes" by J. Milnor (Princeton orange series). --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
The Parameswaran argument is summarised at http://en.wikipedia.org/wiki/Pfaffian WFL On 10/23/08, Dan Asimov <dasimov@earthlink.net> wrote:
Gene asked:
<< The determinant of a matrix is a polynomial in its elements. For an antisymmetric matrix of odd order, the determinant is zero, while for an antisymmetric matrix of even order, this polynomial factors into the square of another polynomial, called the Pfaffian. Does anyone know of a simple, easy to follow, proof that the determinant factors as asserted?
Here's a cite for what appears to be an elementary 1-page proof:
Skew-Symmetric Determinants Author(s): S. Parameswaran Source: The American Mathematical Monthly, Vol. 61, No. 2 (Feb., 1954), p. 116
I believe there is a more informative but more advanced proof in "Characteristic Classes" by J. Milnor (Princeton orange series).
--Dan
_____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
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On 10/23/08, Fred lunnon <fred.lunnon@gmail.com> wrote:
The Parameswaran argument is summarised at http://en.wikipedia.org/wiki/Pfaffian WFL
I didn't find the argument presented in that Wikipedia page particularly perspicacious; the following seems more helpful, and incidentally clarifies the matter of the ambiguous sign. 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 sgined Pfaffian equals the accumulated permutation sign times \prod_k a_{2k,2k+1}. Fred Lunnon
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
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Fred lunnon