TITLE: Matrix representation of real *-algebras
QUESTION [6 upvotes]: 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 $\mathbb H$.

REPLY [11 votes]: 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.