antisymmetricAntiSymmetric2002-04-10 10:55:172006-09-14 08:55:11Definition\usepackage{amsmath}
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\newtheorem{proposition}{Proposition}Let $U$ and $V$ be a vector spaces over a field $K$. A bilinear mapping
$B:U\times U\rightarrow V$
is said to be \emph{antisymmetric} if
\begin{equation}
B(u,u)=0
\end{equation}
for all $u\in U$.
If $B$ is antisymmetric, then the polarization of the anti-symmetry
relation gives the condition:
\begin{equation}
B(u,v) + B(v,u) = 0
\end{equation}
for all $u,v \in U$. If the characteristic of $K$ is not 2, then
the two conditions are equivalent.
A multlinear mapping $M:U^k\rightarrow V$
is said to be \emph{totally antisymmetric}, or simply antisymmetric, if
for every $u_1,\ldots,u_k\in U$ such that
$$u_{i+1} = u_i$$
for some $i=1,\ldots,k-1$ we have
$$M(u_1,\ldots,u_k)=0.$$
\begin{proposition}
Let $M:U^k\rightarrow V$ be a totally antisymmetric, multlinear
mapping, and let $\pi$ be a permutation of $\{1,\ldots,k\}$. Then,
for every $u_1,\ldots,u_k\in U$ we have
$$M(u_{\pi_1},\ldots,u_{\pi_k}) = \mathrm{sgn}(\pi)
M(u_1,\ldots,u_k),$$
where $\mathrm{sgn}(\pi)=\pm1$ according to the parity of $\pi$.
\end{proposition}
{\em Proof.}
Let $u_1,\ldots,u_k\in U$ be given. multlinearity and anti-symmetry
imply that
\begin{align*}
0 &= M(u_1+u_2,u_1+u_2,u_3,\ldots,u_k) \\
&= M(u_1,u_2,u_3,\ldots,u_k) + M(u_2,u_1,u_3,\ldots,u_k)
\end{align*}
Hence, the proposition is valid for $\pi=(12)$ (see cycle notation).
Similarly, one can show that the proposition holds for all
transpositions
$$\pi=(i,i+1),\quad i=1,\ldots,k-1.$$
However, such transpositions
generate the group of permutations, and hence the proposition holds in
full generality.
\paragraph{Note.} The determinant is an excellent example of a totally
antisymmetric, multlinear mapping.