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Theorem: Every irreducible $\mathbb{R}$-representation of the real algebra $\mathbb{R}(n)$ is isomorphic to $\mathbb{R}^n$, where the matrix A ∈ $\mathbb{R}(n)$ acts via left matrix multiplication.

I would like to understand this theorem. However, I do not find a good reference. Could someone advise me one?

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It looks like $\Bbb{R}(n)$ is your notation for the $n\times n$ matrix algebra, so I'll use $R$ to denote that ring. Here's an outline solution that contains useful exercises about simple Artinian rings.

Step 1: Show that any simple right $R$-module $S$ is isomorphic to a simple right ideal of $R$. This can be done by noticing there is a map $\phi:R\rightarrow S$. Since $R$ is semisimple, $R=\ker(\phi)\oplus N$, and so $S\cong R/\ker(\phi)\cong N\subset R$.

Step 2: Show that if $S$ and $T$ are simple right ideals of a ring, then $ST=0$ or $S\cong T$. Since the ring $R$ is simple, $ST=0$ can't happen, because that would imply that $0\neq ann(S)\lhd R$. Conclude that $R$ has only one type of simple right module up to isomorphism.

Step 3: Show that the module you specified in your post is simple over $R$. You can do this, for example, by demonstrating that for any $x$ and any $y$ in $M$, there is a matrix $A$ such that $xA=y$.

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    @user46128 And I wish I had the "pure representation theoretic" view on things too, sometimes :) For real representations of finite groups, anyways, I always remember that representations correspond with modules, and irreducible representations give the simple modules. I'd be happy to answer questions along these lines... we will probably both be learning.2012-11-30