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$ \DeclareMathOperator{\ad}{ad}$ Let $L$ be a non-zero Lie algebra which is semi simple. Then $L$ contains a toral element and hence a non-trivial toral subalgebra. Let $H$ denote a maximal toral subalgebra of $L$. Then, since it can be shown that $H$ must be abelian, $\ad(H)$ is a commuting family of endomorphisms. So there exists a basis for $L$ in which all $\ad x, \ (x \in L)$ are diagonal.

My question: Why is $$ L = \bigoplus_{\alpha \in H^*} L_\alpha, $$ when $$ L_\alpha := \{ x\in L | \forall h \in H: \ad h(x) = [h,x] = \alpha(h) x \}. $$

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The root space decomposition follows, because $ad(H)$ is a commuting family of semisimple endormorphisms of $L$. By a standard result of linear algebra, this family can be simultaneously diagonalized. Hence $L$ is the direct sum of the generalised eigenspaces, which are here of the form $L_{\alpha}$, since $H$ is not only nilpotent, but abelian (since $L$ is semisimple).

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  • $\begingroup$ I am reading Humphreys, too but I do not immediately see why the sum is direct. $\endgroup$
    – user42761
    May 24, 2015 at 15:55
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    $\begingroup$ The sum is direct, because for different eigenvalues the eigenspaces are different. $\endgroup$ May 24, 2015 at 16:10

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