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 $\mathbb H$. 



Matrix representation of real *-algebras
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.


