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[parent] vector product in general vector spaces (Definition)

The vector product can be defined in any finite dimensional vector space $ V$ with $ \dim V=n$. Let $ v_1,\dots,v_n$ be a basis of $ V$, we then define the vector product of the vectors $ w_1,\dots,w_{n-1}$ in the following way:

$\displaystyle w_1\times\dots\times w_{n-1}=\sum_{j=1}^nv_j\det(w_1,\dots,w_{n-1},v_j).$
One can easily see that some of the properties of the vector product are the same as in $ \mathbb{R}^3$:



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Cross-references: perpendicular, Euclidean vector space, linearly dependent, properties, vectors, basis, vector space, finite dimensional
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This is version 2 of vector product in general vector spaces, born on 2004-08-09, modified 2004-08-09.
Object id is 6089, canonical name is VectorProductInGeneralVectorSpaces.
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AMS MSC: 15A72 (Linear and multilinear algebra; matrix theory :: Vector and tensor algebra, theory of invariants)
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A (better?) definition by stevecheng on 2005-07-13 18:29:38
Hello,
I think the following definition is slightly better than the current one (though equivalent for orthnormal {v_i})

For a n-dimensional inner product space,
the cross product $z = v_1 \times \dotm \times v_{n-1}$
is the vector such that

$\langle z, w \rangle = \det(v_1, \dotsb, v_{n-1}, w$

for all vectors $w$.

The angle $z$ exists and is unique by the (finite-dimensional) Riesz Representation Theorem. Or it could be constructed simply by setting $w = e_1, e_2, \dotsc$
(i.e. the basis vectors) successively.

By the way, this comes from Spivak's Calculus on Manifolds book.

I think this definition is better because it avoids the use of a basis representation,
and most of the properties for the cross product that have to be painstakingly
proven for R^3 become almost trivial with this definition.
(e.g. the triple product formula, invariance under rotations and under change
of bases of the same orientation, etc.)
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