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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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  • $\begingroup$ 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. $\endgroup$ – rschwieb Aug 26 '12 at 0:45
  • $\begingroup$ one that includes the adjoint of every matrix in it. $\endgroup$ – magya Aug 26 '12 at 0:48
  • $\begingroup$ well the matrix ring is a simple algebra, but this is a question about any self-adjoint subalgebra. $\endgroup$ – magya Aug 26 '12 at 13:47
  • $\begingroup$ I was just pointing out why $V$ is semisimple, as per your question. $\endgroup$ – rschwieb Aug 26 '12 at 19:31
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Let $V$ be a finite-dimensional Hilbert space over $\mathbb{C}$, and let $A \subset \mathrm{End}(V)$ be a self-adjoint subalgebra. To show that $V$ is semisimple, it suffices to show that any submodule $W \subset V$ (that is, sub $A$-module) has a complement: that is, that there exists a submodule $W^{\perp} \subset V$ such that $W \oplus W^{\perp} = V$. (If we knew this, we could take $W$ to be the sum of the simple submodules of $V$; if $W$ were not $V$ we could take a simple submodule of $W^{\perp}$ to get a contradiction.) Using the self-adjointness hypotheses, it suffices in this case to take $W^{\perp}$ to be the orthogonal complement of $W$.

This point of view can be used to prove that a continuous representation of a compact (e.g. finite) group over $\mathbb{C}$ is automatically semisimple, because one can always choose an invariant inner product: choose any inner product first and then average over the group.

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