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I just posted this on overflow.... just can't figure it out.

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

Let V be a finite-dimensional complex Hilbert space. Let A⊂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!

On overflow it was stated that the hypotheses imply that V breaks up as a direct sum of simple modules... don't see it.

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    What is a self-adjoint subalgebra? The endomorphism ring of a finite dimensional vector space is just the matrix ring over the field, which is a semisimple algebra, which implies $V$ breaksup as a sum of simple submodules.2012-08-26
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    one that includes the adjoint of every matrix in it.2012-08-26
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    well the matrix ring is a simple algebra, but this is a question about any self-adjoint subalgebra.2012-08-26
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    I was just pointing out why $V$ is semisimple, as per your question.2012-08-26

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