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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 $\mathbb H$.

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Do you still require that the size of the matrices be the dimension of the algebra? – L Spice Jun 3 '11 at 22:10
@L Spice: No, it's fine as long as the representation is finite dimensional. I forgot to say that the algebra is finite dimensional. – Marcos Cossarini Jun 4 '11 at 4:15
up vote 7 down vote accepted

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.

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I see. The relation $i^2=-1$ must be mantained by any morphism, so the matrix that represents $i$ should satisfy $A^2=-1$ and would have imaginary spectrum. – Marcos Cossarini Jun 4 '11 at 4:18
I think the nearest concept that behaves well with matrix representations is that of $C*$-algebra. – Marcos Cossarini Jun 4 '11 at 4:21

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.

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