I'm reading some notes on Lie Algebras, and right now we are constructing a complex, 3-dimensional, simple nonabelian Lie algebra. Suppose $\mathfrak{g}$ is a 3-dimensional sub-Lie algebra of $\mathfrak{gl}(V)$ for some complex finite-dimensional vector space $V$. Simplicity implies that $$\mathfrak{g} = [\mathfrak{g},\mathfrak{g}],$$ where $[\mathfrak{g},\mathfrak{g}]$ is the subspace generated by all brackets.
Next consider the linear map ad$(H) = [H,\cdot]$ for some $H\neq 0$ in $\mathfrak{g}$. By the above, $trace(H)=0$, since $H$ is a commutator (or linear combination of commutators). It is then claimed that $trace($ad$(H)) = 0$. I can't see why this is true, and I can't understand this answer yet because it refers to material from much later. I thought I might be able to relate the minimal or characteristic polynomials of $H$ and ad$(H)$, if it were true that $p($ad$(H))=$ad$(p(H))$ for polynomials $p$, but I don't think this is true. If anyone could give an elementary explanation, that would be appreciated.
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 of $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$. So we do not need that $tr(H)=0$ here, if $\mathfrak{g}$ is simple.
In general, if $H\in \mathfrak{gl}(V)$ has the eigenvalues $a_1,\ldots ,a_n$ in $\mathbb{C}$, then $ad(H)$ has the $n^2$ eigenvalues $a_i-a_j$ for $1\le i,j\le n$. This is exercise $6$ of Humphrey's book on Lie algebras, with solution given here (really elementary). Now $0=tr(H)=\sum_i a_i$ implies $tr(ad(H))=\sum_{i,j}(a_i-a_j)=0$.