antisymmetric


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antisymmetric

(mathematics)
A relation R is antisymmetric if,

for all x and y, x R y and y R x => x == y.

I.e. no two different elements are mutually related.

Partial orders and total orders are antisymmetric. If R is also symmetric, i.e.

x R y => y R x

then

x R y => x == y

I.e. different elements are not related.
References in periodicals archive ?
The skewon part is characterized by the modified bidyadic [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] which is the antisymmetric part of [[??].sub.m], while the principal modified bidyadic [[??].sub.m1] = [e.sub.N][[??].sub.1] is symmetric.
Lemma 2.7 Let [less than or equal to] [sub.1] and [less than or equal to] [sub.2] be two antisymmetric relations on A such that a [less than or equal to] [sub.2] b implies [[not less than or equal to].sub.1] or b = a.
Let x [member of] M; since [([A.sub.[nabla]f]).sub.x] is diagonalisable and [J.sub.x] is antisymmetric anti-commuting with [([A.sub.[nabla]f]).sub.x], hence [([A.sub.[nabla]f]).sub.x] is of even rang.
The matrix S has a 2 x 2 block structure, where the diagonal blocks are symmetric and the off-diagonal blocks are antisymmetric. This makes the analysis of the stability of the reconstruction procedure more difficult than in the one-channel case.
This spectrum reveals the presence of calcite showing bands at 1 467/cm (with a weak shoulder at 1 430/cm) (antisymmetric stretching, [v.sub.3]), 871/cm (out of plane deformation, [v.sub.2]), and 708/cm (in plane deformation, [v.sub.4]) (Ross 1972) and the presence of opal bands at 1 100-1 060, 780 and 455/cm (Kamatani 1971, Perry 1989).
Clearly, it is in addition antisymmetric; hence a (partial) order.
The radial TEM waves associated with the structure in Figure 5 will be antisymmetric, with respect to the split, and will be of the circumferentially varying type.
In contrast to the present example, where zero-stiffness postbuckling is a special case of symmetric bifurcation, it may also be a special case of antisymmetric bifurcation (Steinboeck et al.
Since [d.sub.p] is non-negative value, we have [d.sub.p] (x, y) = 0 = [d.sub.p] (y, x), this shows that x = y,that is, "[less than or equal to]" is antisymmetric. Now for x,y,z [member of] X,let x [less than or equal to] y and y [less than or equal to] z, then,
The difference is revealed in the asymmetry of the turbulent (Reynolds) stress tensor with the symmetric constituent describing the interaction between the average flow and the turbulence constituent with energy [K.sup.0] and with the antisymmetric constituent describing the interaction between the average flow and the turbulence constituent with M, [OMEGA], and energy [K.sup.[OMEGA]].
Let S be the subspace of [R.sup.nxn] formed by the symmetric matrices and let A be the subspace of [R.sup.nxn] formed by the antisymmetric matrices.
The absorption band at 2925 [cm.sup.-1] and 2850 [cm.sup.-1] corresponding to antisymmetric and symmetric stretching [CH.sub.2] were found in all IR spectra of intercalated montmorillonites (Figs.

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