0.- Consider a three-dimensional vector space with

A,B 3-d vectors,

and orthonormal basis: {e₁, e₂, e₃}.

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 = e₁ + e₂, to find immediately that:

e₁e₂ = -e₂e₁,

and similarly, e₂e₃ = -e₃e₂, and e₃e₁ = -e₁e₃. In general,

AB + BA = 2 A·B,

for the symmetric part of the Clifford product.

On the other hand, the Clifford product e₁e₂e₃ (= e₃e₁e₂ = e₂e₃e₁) is such that:

(e₁e₂e₃)² = e₁e₂e₃e₁e₂e₃ = -e₁e₂e₁e₃e₂e₃ = e₂e₃e₂e₃ = -1,

and it commutes with e₁, e₂, and e₃:

(e₁e₂e₃)e₁ = e₁e₂e₃e₁ = -e₁e₂e₁e₃ = e₁e₁e₂e₃ = e₁(e₁e₂e₃), and so on.

Hence, we can identify e₁e₂e₃ = i, the imaginary unit.

3.- We can thus rewrite e₁e₂ = ie₃, e₂e₃ = ie₁, and e₃e₁ = ie₂, since, e.g.

e₁e₂ = e₁e₂(e₃e₃) = (e₁e₂e₃)e₃ = ie₃.

All possible products yield the basis

{1, e₁, e₂, e₃, i, ie₁, ie₂, ie₃}

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 e₁e₂ = ie₁×e₂ = -e₂e₁, 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.