Introducing some notation

• Let D be the division algebra in question.
• Let n be the dimension of D.
• We identify the real multiples of 1 with R.
• When we write a ≤ 0 for an element a of D, we imply that a is contained in R.
• We can consider D as a finite-dimensional R-vector space. Any element d of D defines an endomorphism of D by left-multiplication, we identify d with that endomorphism. Therefore, we can speak about the trace of d, and its characteristic- and minimal polynomials.
• For any z in C define the following real quadratic polynomial:

${\displaystyle Q(z;x)=x^{2}-2\operatorname {Re} (z)x+|z|^{2}=(x-z)(x-{\overline {z}})\in \mathbf {R} [x].}$

Note that if z ∈ C ∖ R then Q(z; x) is irreducible over R.

The claim

The key to the argument is the following

Claim. The set V of all elements a of D such that a² ≤ 0 is a vector subspace of D of dimension n − 1. Moreover D = R ⊕ V as R-vector spaces, which implies that V generates D as an algebra.

Proof of Claim: Pick a in D with characteristic polynomial p(x). By the fundamental theorem of algebra, we can write

${\displaystyle p(x)=(x-t_{1})\cdots (x-t_{r})(x-z_{1})(x-{\overline {z_{1}}})\cdots (x-z_{s})(x-{\overline {z_{s}}}),\qquad t_{i}\in \mathbf {R} ,\quad z_{j}\in \mathbf {C} \setminus \mathbf {R} .}$

We can rewrite p(x) in terms of the polynomials Q(z; x):

${\displaystyle p(x)=(x-t_{1})\cdots (x-t_{r})Q(z_{1};x)\cdots Q(z_{s};x).}$

Since z_j ∈ C ∖ R, the polynomials Q(z_j; x) are all irreducible over R. By the Cayley–Hamilton theorem, p(a) = 0 and because D is a division algebra, it follows that either a − t_i = 0 for some i or that Q(z_j; a) = 0 for some j. The first case implies that a is real. In the second case, it follows that Q(z_j; x) is the minimal polynomial of a. Because p(x) has the same complex roots as the minimal polynomial and because it is real it follows that

${\displaystyle p(x)=Q(z_{j};x)^{k}=\left(x^{2}-2\operatorname {Re} (z_{j})x+|z_{j}|^{2}\right)^{k}}$

Since p(x) is the characteristic polynomial of a the coefficient of x^{2k − 1} in p(x) is tr(a) up to a sign. Therefore, we read from the above equation we have: tr(a) = 0 if and only if Re(z_j) = 0, in other words tr(a) = 0 if and only if a² = −|z_j|² < 0.

So V is the subset of all a with tr(a) = 0. In particular, it is a vector subspace. The rank–nullity theorem then implies that V has dimension n − 1 since it is the kernel of ${\displaystyle \operatorname {tr} :D\to \mathbf {R} }$. Since R and V are disjoint (i.e. they satisfy ${\displaystyle \mathbf {R} \cap V=\{0\}}$), and their dimensions sum to n, we have that D = R ⊕ V.

The finish

For a, b in V define B(a, b) = (−ab − ba)/2. Because of the identity (a + b)² − a² − b² = ab + ba, it follows that B(a, b) is real. Furthermore, since a² ≤ 0, we have: B(a, a) > 0 for a ≠ 0. Thus B is a positive-definite symmetric bilinear form, in other words, an inner product on V.

Let W be a subspace of V that generates D as an algebra and which is minimal with respect to this property. Let e₁, ..., e_n be an orthonormal basis of W with respect to B. Then orthonormality implies that:

${\displaystyle e_{i}^{2}=-1,\quad e_{i}e_{j}=-e_{j}e_{i}.}$

If n = 0, then D is isomorphic to R.

If n = 1, then D is generated by 1 and e₁ subject to the relation e²
₁ = −1. Hence it is isomorphic to C.

If n = 2, it has been shown above that D is generated by 1, e₁, e₂ subject to the relations

${\displaystyle e_{1}^{2}=e_{2}^{2}=-1,\quad e_{1}e_{2}=-e_{2}e_{1},\quad (e_{1}e_{2})(e_{1}e_{2})=-1.}$

These are precisely the relations for H.

If n > 2, then D cannot be a division algebra. Assume that n > 2. Let u = e₁e₂e_n. It is easy to see that u² = 1 (this only works if n > 2). If D were a division algebra, 0 = u² − 1 = (u − 1)(u + 1) implies u = ±1, which in turn means: e_n = ∓e₁e₂ and so e₁, ..., e_n−1 generate D. This contradicts the minimality of W.