Matrix representation of real *-algebras

Question

It is a standard fact that everyt real $n$-dimensional algebra is a subalgebra of $M_n(\mathbb R)$.

The transposition map, operating in $M_n(\mathbb R)$, is an involutive ($(A^t)^t=A$) antiautomorphism (an $\mathbb R$-linear isomorphism satisfying $(AB)^t=B^tA^t$). This makes $M_n(\mathbb R)$ a real *-algebra. The same is true in every transpose-closed subalgebra of $M_n(\mathbb R)$.

Is every real *-algebra of this kind? (a transpose-closed subalgebra of $M_n(\mathbb R)$, with the * represented by transposition)

This is true at least in the famous real *-algebras $\mathbb C$ and $\mathbb H$.

Answer 1

I think that if you consider $\mathbb C$ with the ∗-structure under which every element is satisfies $ z^*=z $, then this ∗-algebra cannot be embedded into $M_n(\mathbb R)$. In other words, it is not a real $C^*$-algebra.

The reason is that the spectrum of the element $i$ is purely imaginary, and that's incompatible with the requirement that it be self-adjoint.

Answer 2

You have to add the condition $$\sum_{i} a_i^*a_i =0 \quad \Rightarrow \quad a_i =0 \quad \forall i.$$

If you consider $\ast$-algebras satisfying this condition, then $-1$ is not a sum of hermitean squares and you find a positive linear functional $\varphi$ on the algebra which satisfies $\varphi(1)=1$. You can now perform the GNS-construction (in the real setting if you want) to obtain a $\ast$-representation on a real Hilbert space. Since $A$ is finite-dimensional, you obtain a $\ast$-homomorphism to a real matrix-algebra.