## classification/description of the subalgebras of $M_n(\mathbb{R})$

Let $n\in\mathbb{N}$ and let $M_n(\mathbb{R})$ denote the algebra of the real $n\times n$ matrices. Is there a classification or characterization of the subalgebras of $M_n(\mathbb{R})$ for any fixed $n\in\mathbb{N}$?

Let $m\in\mathbb{N}$, $m< n$, and let $\mathcal{A}\subseteq M_m(\mathbb{R})$ be a subalgebra. Then $\mathcal{A}$ can be trivially embedded in $M_n(\mathbb{R})$ for instance via $\mathrm{diag}(\cdot\ ,\underbrace{1,\dots,1}_{n-m})$. Also, if $Q\in GL_m(\mathbb{R})-U(\mathcal{A})$, then conjugation with $Q$ delivers a different subalgebra, which, however, is an isomorphic copy of $\mathcal{A}$. Nothing deep yet, but it seems to me that a reasonable classification would be a classification at most (or at least?) up to algebra-isomorphisms within $M_m(\mathbb{R})$ or embeddings like the aforementioned.

I am starting reading through "Elements of representation theory of associative algebras" by Assem, Simon and Skowronski. However, they assume the underlying field to be algebraically closed, so I guess one should first look at the subalgebras of $M_n(\mathbb{C})$ and then try to filter out the real ones (pun not intended), but so far I don´t have really an idea how this can be done. On the other hand, subalgebras $\mathcal{B}\subseteq M_n(\mathbb{C})$ can be also trivially viewed as real subalgebras of $M_{2n}(\mathbb{R})$ via block matrices.

A few words to the motivation for this question: Complex numbers, quaternions (octanions etc.) are fundamental examples of structures that can be represented by real matrices. Additionally, since $M_n(\mathbb{C})\hookrightarrow M_{2n}(\mathbb{R})$ as mentioned above, it appears to me that $M_n(\mathbb{R})$ is not simply the more general setting, but also the more natural setting.

I hope my question is not too elementary. My background in algebra is not really very strong. I would be also thankful for references dealing specifically with the matter of such classification.

This is a duplicate of mathoverflow.net/questions/27376/…. Since you are going to read Assem et al, you might as well observe that they wrote 3 volumes of very intricate theory in order to understand at least $\textit{some}$ finite-dimensional associative algebras, so no elementary approach like "classifying subalgebras of $M_n(K)$" (whatever $K$ is: algebraically closed case is easier) is feasible. – Victor Protsak Jun 16 2010 at 13:02
I was thinking that taking $K=\mathbb{R},\mathbb{C}$ instead of an arbitrary algebraically closed field may actually simplify things somewhat since $M_n(\mathbb{R})$ has the additional structure of being an Operator-Algebra over a finite-dimensional Hilbert space. Hence the tag "Operator-Algebras". Maybe my post left wrong impression, but I am nowhere asking for a strictly algebraic type of classification. – ex falso quodlibet Jun 16 2010 at 13:23
In the finite-dimensional setting, operator algebras are algebraic, so the classification is the same (i.e. hopeless). If your operator algebra is $*$-closed (hence a von Neumann algebra) then it is semisimple, so Wedderburn-Artin applies. In general, the situation over the algebraic closure $L$ of $K$ is easier because if $A\simeq B$ over $K$ then certainly $A\otimes_K L\simeq B\otimes_K L$, but the converse is not true. – Victor Protsak Jun 16 2010 at 14:11