= Irreducible traceless matrix Lie algebra is semisimple
{title2=$L\le\mathfrak{sl}(V),\quad V\text{ irreducible}\ \Longrightarrow\ \operatorname{rad}L=0$}
Let $V$ be a finite-dimensional complex <vector space> and let $L$ be a <Lie subalgebra> of its traceless <endomorphisms> acting irreducibly. Its <solvable radical> $R$ is simultaneously triangularizable by <Lie's theorem>. The resulting finite set of diagonal <characters> is invariant under conjugation by $e^{tx}$ for $x\in L$, because this conjugation preserves the <ideal of a Lie algebra>[ideal] $R$ and gives an equivalent <representation>. Each <character> varies continuously with $t$ in a finite set, hence is constant; differentiation gives $\lambda([x,r])=0$.
A nonzero common $R$-<eigenspace> exists by <Lie's theorem>. The displayed vanishing makes it invariant under all of $L$: $r(xv)=\lambda(r)xv+\lambda([r,x])v=\lambda(r)xv$. Irreducibility therefore makes it all of $V$. Thus $R$ acts by scalars, and the zero-<trace> condition forces those scalars to vanish. The <matrix> <representation> is faithful, so $R=0$. The traceless condition matters: adjoining scalar <matrices> to an irreducible semisimple algebra gives a reductive algebra with nonzero <center of a Lie algebra>[centre].
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