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In more detail, let $F$ be any field and $A$ a set of matrices, $F\cdot Id_n \subseteq A\subseteq GL_n(F)$, closed under addition and product, which is irreducible: If $\{0\}\subsetneq V \subsetneq F^n$, then $V$ is not $A$-invariant. Does it follow that the only 2-sided ideals of $A$, as an abstract $F$-algebra, are $\{0\},A$?

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    This is a consequence of Wedderburn theory; the answer is yes. –  Sep 13 '14 at 13:25
  • Thank you for your answer! But doesn't Weddeburn theorem require you to know a-priori that $A$ is semi-simple? Is there any reference for the claim? –  Sep 13 '14 at 14:05
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    Part of the Wedderburn theory is that if an algebra has a faithful irreducible representation, then it is semi-simple. –  Sep 13 '14 at 14:09

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OK, I got it. Let $V=F^n, e_1,\ldots e_n$ be the standard basis of $V$. We have a morphism of $A$-modules $F:A\rightarrow V^n$ given by $a\mapsto (a\cdot e_1,\ldots ,a\cdot e_n)$, which is injective since $V$ is faithful. Now, since $A$ (as an $A$-module) embeds in the semisimple module $V^n$, $A\simeq V^k$ for some $k$. Two-sided ideals of $A$ correspond to submodules $M\le V^k$ which are invariant under $End_A(V^k)$. It enough to use $\pi_{i,j}\in End_A(V^k)$ given by $\pi_{i,j}(v_1,\ldots v_k)=(0,0,..,v_i,..,0)$, non-zero only in the $j$'th position, to see that $M\ne\{0\}$ implies $M=V^k$.