Problem: Given $L$ a Lie algebra, over an algebraically closed field F of characteristic zero, prove that $L$ is solvable if and only if $[L,L]$ lies in the radical $R$ of the Killing form $k$, where the radical $R= \{x \in L : k(x,y) = 0, \forall y \in L\}$.
Solution First implication: if $[L,L] \subseteq R$, then $L$ is solvable, because that means that $\forall x, y, z \in L$, $k([x,y],z) = tr(ad([x,y])ad(z))= 0$. So, by Cartan's criterion, it is true that $L$ is solvable.
Second implication: Let $[x,y] \in [L,L]$ and $z \in L$, then I have to prove that $k([x,y]z) = tr(ad([x,y])ad(z)) = 0$. I know that $L$ is solvable, so $ad(L)$ is solvable too. Now, I know that $ad(L) \subseteq gl(V)$, so there exists a vector $v \in V$ such that $ad(x)v= \lambda v$, with $\lambda \in F$ , for every $x \in L$. That means that $tr(ad[x,y]) = 0$, because all the eigenvalues on the diagonal are the same. So $k([x,y]z) = 0$.
Is this correct? Thank you.