Determinants are more fundamental than exterior/wedge products. Specifically,
wedge products and k-vectors emerge very naturally as a way of
determining whether the columns of a non-square matrix are linearly
independent, generalising the determinant, which only shows whether the columns
of a square matrix are linearly independent. All of this actually comes out of
an attempt to derive equations for planes and other linear spans of vectors in
a Cartesian space; these equations end up being the perfect intuition for
understanding how a k-vector can represent a plane or linear
subspace.
Historically, the wedge product has been described using axioms, using tensor products, or by simply declaring what the rules are for multiplying vectors, and in contemporary discussion the cross product and the determinant are claimed to be possible to derive as special cases of this wedge product. These approaches are consistent, and give the correct mathematical results, but by focussing first on the problem of finding equations for representing a linear subspace, we are able to show very clearly why the formulae make sense, and how they relate to the geometry of planes and other linear subspaces.
Given a finite set of vectors u_i in some Cartesian space, the
line, plane, or other linear subspace that is spanned by those vectors is a
natural geometric object that we might want to understand. The parameterised
definition of a span is the set of arbitrary linear combinations of the vectors
involved. These will take the form sum_i(a_i*u_i), where
a_i are arbitrary scalar values. Arranging a_i into a
column vector v, this is just the image of the matrix
A whose columns are the vectors u_i. This is the
parameterised definition of the span, but when we represent lines and planes,
we usually do so as one or more algebraic equations that can be tested
directly, rather than a single vector equation with some number of free
parameters. It would be nice to look at how these free parameters can be
eliminated, in more systematic ways, given how fundamental the span and linear
subspaces are to areas of algebra and geometry.
Related to the span is the concept of linear independence. A set of vectors is
linearly independent if none of the vectors are in the span of any other
vectors in the set, or if no linear combination of them is the zero vector,
except for the case where a_i are all zero anyway. In the matrix
interpretation, this means that the only vector in the null space of
A is the zero vector, or the nullity is zero. If our vectors are
linearly independent, then a vector v is in their span if and only
if A is still linearly independent with v included.
We already have a tool for understanding the preimages and null space of
matrices, which is Gaussian elimination. Performing Gaussian elimination on our
matrix A, we can derive a PLU decomposition, and if
A has n rows, and k columns, then
P will be an n by n permutation matrix,
L an invertible n by n lower triangular
matrix, and U will be A in row echelon form. Say that
U has r non-zero rows, followed by n - r
rows of zero, then A has rank r, and nullity
k - r. Since U is a matrix whose span must be exactly
the vectors e_1 through to e_r, the matrix
PL must map theses basis vectors to some new set of vectors
w_i whose span is the same as A. In other words, the
first r columns of PL have the same span as
A. Now a vector v will be in the span of these
w_i if the equation v = sum_i(a_i*PLe_i) holds, which
will be the case if and only if the inverse of PL maps
v to a vector whose last n - r coordinates are all
zero. This gives us n - r equations that must be satisfied by a
vector simultaneously in order to be in the span of A.
Just as when analysing the invertibility of matrices, we find that less
direct methods of calculating the determinant and inverse lead to a closed form
polynomial formula, giving something much more amenable to reasoning and
algebraic analysis, we also find that by using the determinant to derive
equations for spans, we can once again achieve polynomial equations which will
benefit from some of the algebraic properties of the determinant. If our
vectors u_i are not linearly independent, then we can use Gaussian
elimination, or simple inspection, to find a subset of these vectors with the
same span, such as the w_i columns found above. Then once
u_i are linearly independent, we will know that a vector
v is in their span if and only if u_i and
v are together no longer linearly independent. In the case where
there are n - 1 vectors u_i, we can represent this
condition as a single equation, by collecting these n vectors into
a matrix, and checking whether their determinant is zero. If their determinant
is zero then v must be in the span of u_i.
In general we might have fewer than n - 1 vectors,
k - 1, say, giving us a matrix with k columns. We
don't have a determinant for such a matrix, but we could take the determinant
of the first k rows, and get a number. If this number is non-zero,
then we know that our kth vector v is not in the span
of the first k - 1 vectors, but if it is zero, then we can't say
much about this vector. In fact we can select any k rows, giving a
total of n choose k determinants that we can evaluate, and if any
of them are non-zero, then v is not in the span of
u_i. What if all of these determinants are zero? Is it possible
for v to be outside of the span of u_i, and yet for
none of these determinants to be zero? Looking at the PLU
decomposition from earlier, but applied to the matrix with v
included, we see that this is impossible. If v is not in the span
of u_i, then we will have a matrix of rank k, whose
span is the same as the span of the first k columns of
PL. The first k rows and columns of L
will also form a square, invertible lower triangular matrix, and so if
P is simply the identity matrix, then the determinant of the first
k rows of PLU should be non-zero. If P is
not the identity, then we can look at its first k columns to see
where these linearly independent coordinates will be mapped, again giving a
non-zero determinant. This is how we know that such a non-zero determinant will
exist, but we don't need to perform Gaussian elimination to find this
determinant, we can simply calculate all n choose k determinants,
and know that v will be in the span of u_i if and
only if all of these determinants are zero.
Perhaps unwittingly, we have derived a formula for the exterior/wedge product
of k vectors. (Forming, confusingly, a k-vector, but
more specifically a k-blade.)
Just as the determinant tells us whether vectors are linearly independent, and
tells us about their geometric/volumetric properties when they are linearly
independent, these n choose k determinants of course tell use
about some geometric properties of the k vectors we started with,
when projected onto different axis-aligned subspaces. What a non-zero
k-blade offers above a determinant, however, is information
sufficient to recover the orientation of the span of our starting vectors. To
see this, we need to derive a formula for the wedge product of a
k-vector with a single vector. The result of this calculation
should be a k+1-vector, a list of n choose k + 1
determinant, of each of the different possible choices of k + 1
rows from a matrix formed by k + 1 vectors. By performing a
cofactor expansion in the last column of these determinants, we can derive
formulas for the final k+1-vector, in terms of the coordinates of
the k+1th vector, and the determinants of different choices of
k rows of the first k vectors. This will therefore be
exactly the formula we are looking for, for taking the wedge of a
k-vector with a single vector. For example, the
{e_1, e_2, e_3} component of a 2-vector wedged with a vector will
be v_1 * d_12 - v_2 * d_13 + v_3 * d_23, where v_i
are the coordinates of the vector, and d_ij are the coordinates of
the 2-vector.
Since we can find the wedge of k + 1 vectors, given only the wedge
of the first k vectors u_i, and the final vector
v, we can apply this to our span-testing problem, and calculate
the determinants of all the different combinations of k rows of
our vectors u_i, giving a wedge
u_1∧u_2∧...∧u_k, and then simply test if a
vector v is in their span, by testing
(u_1∧u_2∧...∧u_k)∧v = 0. This is a system
of n choose k + 1 linear equations, or a single
k+1-vector equation. If we can determine whether a vector is in a
subspace, using only the wedge of its basis vectors, then that means that the
wedge somehow contains all of the information relevant to the orientation of
this subspace, in addition to the volumetric information associated with the
determinants with which we derived it. This will turn out to be the basis of
exterior algebra, and later, geometric algebra, but on its own is also an
interesting way of testing linear independence and spans, as well as an
interesting way of memoizing the cofactors involved in calculating determinants
algebraically.