A property of simple three-dimensional Lie algebras

Question

I am reading a solution of a problem to classify $3$-dimensional simple Lie algebras. First they prove that there exists $H$ such that $[H,X]=\alpha X$ for some $X\ne 0$ and $\alpha\ne 0$. Then they say "as $ad(H)$ has trace $0$ and $H$ is an eigenvector with eigenvalue $0$, there must exist $Y$ such that $[H,Y]=-\alpha Y$".

It is true indeed, but why $\mathrm{tr}\space ad(H)=0$? Thanks for any help!

Answer 1

$ad(H)$ (as $ad(anything)$) is skew-symmetric w.r.t. the Cartan-Killing form, so its trace is zero.

Answer 2

Since $ad$ is a representation, we have that $0=tr([ad(x),ad(y)])=tr(ad[x,y])$, which implies that $tr(ad(H))=0$ for all $H$ because we assume that the Lie algebra is simple, so that $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$.