# semisimple algebra question

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!

-
Yes. The hypotheses imply that $V$ breaks up into a direct sum of simple modules (take orthogonal complements). – Qiaochu Yuan Aug 25 '12 at 22:51
Well, that proves $V$ is a semisimple $A$-module; a little more is needed to show $A$ itself is a semisimple algebra ;-) – Mariano Suárez-Alvarez Aug 25 '12 at 22:54
By the way, you may not know about math.stackexchange.com which is a useful place for this. – Mariano Suárez-Alvarez Aug 25 '12 at 22:56
missed the hint earlier on orthogonal complements. Now it works. thanks! – magya_bloom Aug 26 '12 at 0:53
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 $V$. – magya_bloom Aug 28 '12 at 20:30