Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question

1 Answer 1

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.

share|improve this answer
Could you please recommend a reference for me? –  Yuan Oct 9 '11 at 1:33
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
Very good, thanks. I really need to review the linear algebra. –  Yuan Oct 9 '11 at 4:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.