Existence of a basis of common eigenvectors

Suppose that L is a complex semisimple Lie algebra containing an abelian subalgebra H consisting of semisimple elements. I am wondering how to see that L has a basis of common eigenvectors for the elements of ad(H).

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Commutative diagonalizable operators over algebraically closed field have simultaneous eigenspace decomposition. This is a standard fact from linear algebra. In this case, $ad(H)$ exactly consists of commutative semisimple operators.
A reference for the linear algebra fact? In fact you can do it yourself quite easily, for simplicity let's consider two commuting diagonalizable maps $S,T:V \to V$. Do the eigenspace decomposition of $S$. For each eigenspace of $S$, say $E_i$, note that each $E_i$ is $T$-invariant, and that the restriction is still diagonalizable. Then do the eigenspace decomposition, with respect to $T$, for each $E_i$. The general situation (arbitrary number of commuting diagonalizable operators) is dealt with in the same way. – Soarer Oct 9 '11 at 1:42