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I know this is not a research level question, but may be I can get an idea...

Is there a direct proof of the following without going through composition series or Artin-Wedderburn theorem?

Let $V$ be a finite-dimensional complex vector space. Let $A \subset End(V)$ be a self-adjoint subalgebra. Then $A$ is semisimple. I am using the following definition of semisimple algebra: semisimple algebra is a direct sum of simple algebras, and a simple algebra is one with no two-sided ideals other than $\{0\}$ and itself. thanks!

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    $\begingroup$ Yes. The hypotheses imply that $V$ breaks up into a direct sum of simple modules (take orthogonal complements). $\endgroup$
    – Qiaochu Yuan
    Aug 25, 2012 at 22:51
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    $\begingroup$ Well, that proves $V$ is a semisimple $A$-module; a little more is needed to show $A$ itself is a semisimple algebra ;-) $\endgroup$
    – Mariano Suárez-Álvarez
    Aug 25, 2012 at 22:54
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    $\begingroup$ By the way, you may not know about math.stackexchange.com which is a useful place for this. $\endgroup$
    – Mariano Suárez-Álvarez
    Aug 25, 2012 at 22:56
  • $\begingroup$ missed the hint earlier on orthogonal complements. Now it works. thanks! $\endgroup$
    – magya_bloom
    Aug 26, 2012 at 0:53
  • $\begingroup$ Then one can decompose $V$ into a direct sum of irreducible $A$-modules. Then $A$ is by definition semi-simple as it is defined by its action on $V$. $\endgroup$
    – magya_bloom
    Aug 28, 2012 at 20:30

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