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The Lie algebra of all the $n \times n$ matrices is not semi-simple. However, if we restrict ourselves to traceless $n\times n$ matrices, we do obtain a semi-simple (in fact, simple) Lie algebra which is called $\mathrm{su}(n)$. How do I prove it?

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What have you tried? This is very commonly an exercise, and it can be verified by direct computations (oh, and it of course has to be over a field of characteristic 0 to be true in general). –  Tobias Kildetoft Mar 22 '12 at 14:37

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