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Suppose $S$ is a subalgebra of the matrix algebra $M_n(\mathbb{C})$. If for any vector $v$ and $w$ in $\mathbb{C}$, there always exists a matrix $A$ in $S$, depending on $v$ and $w$ of course, which sends $v$ to $w$, then is $S$ necessarily $M_n(\mathbb{C})$?

I arise this technical question when I read Wallach's book Real Reductive Groups, I think the answer is yes, but I am not sure. Can anyone help me prove it or give me a counterexample if I am wrong? Thank you!

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    $\begingroup$ This is a corollary of Burnside's theorem that an irreducible subalgebra of $M_n(\mathbf{C})$ is the whole algebra. $\endgroup$
    – YCor
    Sep 9, 2014 at 9:28
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    $\begingroup$ You can also obtain this from Schur's lemma+double centralizer theorem (by Schur's lemma the centralizer of $S$ is scalar matrices, and so by the double centralizer theorem $S$ is all matrices). $\endgroup$
    – Stephen
    Sep 9, 2014 at 17:54

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The answer is Yes. By Wedderburn-Malcev [http://www.math.uni-bielefeld.de/~sek/select/RF6.pdf] $S=A\oplus J(S)$ as ${\mathbb C}$-vector spaces, where $A$ is a unital semi-simple ${\mathbb C}$-algebra and $J(S)$ is the Jacobson radical of $S$ (the maximal nilpotent $2$-sided ideal of $S$). We claim that $J(S)=0$. Suppose otherwise. Then there exists $j\in J(S)$ and $v\in{\mathbb C}^n$ such that $j(v)\ne0$. By hypothesis there exists $s\in S$ such that $sj(v)=v$. So $(sj)^n(v)=v$ for all $n\geq1$, whence $sj$ is not nilpotent. But this contradicts the fact that $sj\in J(S)$. This proves our claim.

Now $S=A$ is a semi-simple ${\mathbb C}$-algebra. So it is a direct product $\prod_{n_i} M_{n_i}({\mathbb C})$ of matrix algebras, as ${\mathbb C}$ is algebraically closed. With respect to a suitable choice of basis for ${\mathbb C}^n$, we can represent $S$ as block diagonal matrices, with blocks of size $n_1,n_2,\dots$ where $n_1+n_2+\dots=n$. The hypothesis now implies that $n_1=n$ and $S=M_n({\mathbb C})$.

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    $\begingroup$ I had not appreciated how directly Burnside's Theorem answers this question, as YCor has commented. See for example joelshapiro.org/Pubvit/Downloads/BurnsideThm/burnside.pdf $\endgroup$
    – John Murray
    Sep 9, 2014 at 9:41
  • $\begingroup$ @JohnMurray if $V$ is an invariant subspace, and $v\in V$ is a nonzero vector, then $S.v$ is an invariant subspace of $V$ which by assumption of the original question is equal to $\mathbb{C}^n$. Quite a direct corollary, I think. $\endgroup$ Sep 9, 2014 at 9:59
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This is a straightforward corollary of Burnside's theorem that an irreducible subalgebra of $M_n(\mathbf{C})$ is the whole algebra.

(of course I understand subalgebra! I don't understand the tag "lie-algebras".)

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  • $\begingroup$ Oh! The original question was about the representation of a real reductive Lie group. I reduced the problem to the above case and got stuck. After I posted my question, I forgot to remove the Lie theory tag. I am sorry. $\endgroup$ Sep 10, 2014 at 0:32

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