The convention is set as follows: we have a semi-simple complex Lie algebra $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_\alpha\mathfrak{g}_\alpha$ as its root space decomposition. If $\alpha\notin\Delta$, where $\Delta$ is the set of roots, then $\mathfrak{g}_\alpha=\{0\}$. For any root $\alpha\in\Delta$, denote $H_\alpha\in\mathfrak{h}$ such that $\alpha(H)=\mathcal{K}(H,H_\alpha)$. We know that $H_\alpha=[E_\alpha,E_{-\alpha}]$ therefore Tr$(ad(H_\alpha))=0$ as a linear operator on $\mathfrak{g}$. For any integer $n$, the subspace $\bigoplus_n\mathfrak{g}_{n\alpha}$ is invariant under the action of $ad(H_\alpha)$ and we conclude $Tr(ad(H_\alpha))=0$ on $ \bigoplus_n\mathfrak{g}_{n\alpha}$.
From Lie's theorem we know that there exists a basis in $\mathfrak{g}_{n\alpha}$ such that $ad(H_\alpha)$ is upper-triangular and we can denote the first element of that basis as $E_{n\alpha}$ such that $ad(H_\alpha)E_{n\alpha}=n\alpha(H_\alpha)E_{n\alpha}$. We therefore could obtain the first addition term in the formula of Tr$(ad(H_\alpha))$ for each $\mathfrak{g}_{n\alpha}$.
What I fail to understand is why we could conclude that Tr($ad(H_\alpha)$)=$\sum_nn\alpha(H_\alpha)\dim\mathfrak{g}_{n\alpha}$. It would be extremely helpful if someone could offer or point me to an explanation.