0.- Consider a three-dimensional vector space with

A,B 3-d vectors,

and orthonormal basis: {x, y, z}.

1.- Given B, neither A·B nor A×B define A in a unique way. Let us combine both defining the non-commutative "Clifford product" of two vectors, with the following two axioms:

Axiom 1: The Clifford (or matrix) product of a vector with itself is the same as the dot product:

DD = D·D = D² a scalar.

Axiom 2: The Clifford product is associative:

(AB)C = A(BC).

2.- Working directly with the basis, let us choose

D = x + y, to find immediately that:

xy = -yx,

and similarly, yz = -zy, and zx = -xz. In general,

AB + BA = 2 A·B,

for the symmetric part of the Clifford product.

On the other hand, the Clifford product xyz (= zxy = yzx) is such that:

(xyz)² = xyzxyz = -xyxzyz = yzyz = -1,

and it commutes with x, y, and z:

(xyz)x = xyzx = -xyxz = xxyz = x(xyz), and so on.

Hence, we can identify xyz = i, the imaginary unit.

3.- We can thus rewrite xy = iz, yz = ix, and zx = iy, since, e.g.

xy = xy(zz) = (xyz)z = iz.

All possible products yield the basis

{1, x, y, z, i, ix, iy, iz}

of an eight-dimensional linear space of "cliffors" allowing for linear combinations of complex scalars and complex vectors:

cliffor = a + ib + A + iB,           a,b real numbers,     A,B real 3-d vectors,

with i = √-1.

Since xy = ix×y = -yx, the antisymmetric part of the Clifford product of A and B is given by iA×B, so finally:

AB = A·B + iA×B.

In general, the Clifford product of two cliffors is:

(u + U)(v + V) = uv + uV + vU + U·V + iU×V,

where u, v are complex numbers and U, V are complex vectors.

Using this definition, a product of two cliffors is again a cliffor, i.e. the Clifford algebra is closed with respect to the product.

This product allows us to find the inverse of a vector, and in general, the inverse of a cliffor:

AA = A², so A^-1 = A(A²)^-1 = (A²)^-1A, and

(u + U)^-1 = (u - U)/(u² – U²),

whenever the (numerical) denominator does not vanish.