The question: Let $L$ be a three dimensional simple Lie algebra over $\mathbb{C}$. Say that $x \in L$, $x \neq 0$, and that $ad_x$ has an eigenvalue $\lambda \neq 0$. Show that $L \cong \mathfrak{sl}(2, \mathbb{C})$.
Things I've tried: We can say $[h,e] = 2e$, $[h,f] = -2f$, and $[e,f] = h$. If I say $ad_x(y) = [x,y] = \lambda y$ then maybe I can have a basis $\{x,y,z\}$. I tried to just create a map $\phi(\frac{2}{\lambda}x)=h$, $\phi(y)=e$, and $\phi(z)=f$. Then I think $x$ and $y$ work but I'm not sure how to ensure $z$ works the way I want.