TITLE: Inner Product on Division Algebras
QUESTION [5 upvotes]: Here, Wikipedia gives a proof that the only finite dimensional associative division algebras over $\mathbb{R}$ are $\mathbb{R}, \mathbb{C}, \mathbb{H}$.
The proof proceeds by taking such a division algebra $D$, observing that multiplication by  $d \in D$ gives an endomorphism of $D$ which can be represented by a matrix corresponding to $d$. Then, we consider $V \subset D$, the subspace of $D$ corresponding to matrices with trace $0$. At some point, an inner product of $V$ given by $\langle a, b \rangle=-ab-ba$ appears. At my current understanding (undergraduate mathematics) this inner product seems mysterious. I can prove that it works but I don't understand why that is a natural choice. Can someone give me some insight?
Edit. Is there maybe a way to think about this result using Lie Theory?

REPLY [2 votes]: Let $A$ be a finite dimensional division algebra over $\mathbb R$. The set $A'$ of all $u\in A$ s.t. $u^2\leq 0$ is a subspace in $A$. The proof needs some computations. What we can write is that
$$u^2=-Q(u)\leq 0,$$
for all $u\in A'$, with $Q(u)\in\mathbb R$ s.t. $Q(u)\geq 0$. Moreover, as $A$ is a division algebra
$$u=0\Leftrightarrow Q(u)=0,$$
and $Q(au)=a^2Q(u)$. The symmetric bilinear form $B(u,v)$ in the OP is deduced from $Q(u)$ as follows:
$$B(u,v):=Q(u+v)-Q(u)-Q(v)=-(uv+vu).$$
And now? You can use $B(u,v)$ to deduce the existence of an orthonormal base of the subspace of $A$ which generates it as an algebra. This is the point of view considered in Wikipedia's proof. One works on the (finite) dimension of $A$ to classify finite dim. division algebras, arriving at the result.