 Irreducible traceless matrix Lie algebra is semisimple

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 $R$ 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 centre.