Editing Exterior algebra (section)

=== Alternating product ===
The exterior product is by construction [[Alternating algebra|''alternating'']] on elements of {{tmath|V}}, which means that <math> x \wedge x = 0 </math> for all <math> x \in V, </math> by the above construction.  It follows that the product is also [[anticommutative]] on elements of {{tmath|V}}, for supposing that {{tmath|x, y \in V}},
: <math>
0 = (x + y) \wedge (x + y)
= x \wedge x + x \wedge y + y \wedge x + y \wedge y
= x \wedge y + y \wedge x
</math>
hence
: <math> x \wedge y = -(y \wedge x). </math>

More generally, if <math>\sigma</math> is a [[permutation group|permutation]] of the integers {{tmath|[1, \dots, k]}}, and {{tmath|x_1}}, {{tmath|x_2}}, ..., {{tmath|x_k}} are elements of {{tmath|V}}, it follows that
: <math>
x_{\sigma(1)} \wedge x_{\sigma(2)} \wedge \cdots \wedge x_{\sigma(k)}
= \sgn(\sigma)x_1 \wedge x_2 \wedge \cdots \wedge x_k,
</math>
where <math>\sgn(\sigma)</math> is the [[signature of a permutation|signature of the permutation]] {{tmath|\sigma}}.<ref>A proof of this can be found in more generality in {{harvtxt|Bourbaki|1989}}.</ref>

In particular, if <math>x_i = x_j</math> for some {{tmath|i \ne j}}, then the following generalization of the alternating property also holds:
: <math> x_{1} \wedge x_{2} \wedge \cdots \wedge x_{k} = 0. </math>

Together with the distributive property of the exterior product, one further generalization is that a necessary and sufficient condition for <math> \{ x_{1}, x_{2}, \dots, x_{k} \} </math> to be a linearly dependent set of vectors is that
: <math> x_{1} \wedge x_{2} \wedge \cdots \wedge x_{k} = 0. </math>