# Root space $L_\alpha$ is completely contained in simple ideal?

I'm having trouble understanding a section in Humphrey's Lie algebras on page 74.

Suppose $$L$$ is a semisimple Lie algebra which decomposes as a direct sum of simple ideals $$L_1\oplus\cdots\oplus L_t$$. Let $$H$$ be a maximal toral subalgebra, so $$H=H_1\oplus\cdots\oplus H_t$$, where $$H_i=L_i\cap H$$.

If $$\alpha\in\Phi$$ is a root of $$L$$ to $$H$$, and $$L_\alpha=\{x\in L:[hx]=\alpha(h)x,\ \forall h\in H\}$$, then $$[H_iL_\alpha]\neq 0$$ for some $$i$$, and then $$L_\alpha\subset L_i$$.

I just don't understand why this forces to $$L_\alpha$$ to be completely in $$L_i$$?

We have $0\neq [H_i,L_\alpha] = \{\alpha(h)x\,\big|\, h\in H_i, x\in L_\alpha\}$. In particular there exists $h\in H_i$ such that $\alpha(h) \neq 0$. But this means $[H_i,L_\alpha]= L_\alpha$. Since $L_i\subseteq L$ is a simple ideal, we also have $[L_i,L_\alpha]\subseteq L_i$. Putting things together, we arrive at $$L_\alpha = [H_i,L_\alpha] \subseteq [L_i,L_\alpha] \subseteq L_i.$$