# Decomposition of a semisimple Lie algebra.

I am struggling with the following logic which appears in Humphreys' introduction to semisimple Lie algebras.

Suppose that $\mathfrak{h}$ is a Cartan subalgebra in a semisimple $\mathfrak{g}$. Since $\mathfrak{h}$ is abelian (because it's toral), we have that $Ad_{\mathfrak{h}}$ is a commuting family of semisimple endomorphisms of $\mathfrak{g}$. This means that the family $Ad_{\mathfrak{h}}$ is simultaneously diagonalizable.

It follows, by inducting on the dimension of $\mathfrak{g}$, that $\mathfrak{g}$ is a direct sum of the subspaces $\mathfrak{g}^{\alpha} = \{x \in \mathfrak{g} | [a,x] = \alpha(a)x \; \forall a \in \mathfrak{h}\}$, where $\alpha$ ranges over $\mathfrak{h}^*$.

I understand the first paragraph. I don't understand how inducting on the dimension of $\mathfrak{g}$ gives us that $\mathfrak{g}$ is the direct sum as suggested. I think it's a result from linear algebra, but I can't find a proof of it.

Since $Ad_h$ is simultaneously diagonalizable, Let $(e_1,...,e_n)$ be a basis of $g$ where the element of $Ad_h$ are diagonalisable. We have for every $x\in h, [x,e_i]=\alpha_i(x)e_i$, $\alpha_i$ is an element of $h^*$, you deduce that $g$ is the direct sum of $g^{\alpha_i}$ where $g^{\alpha_i}$ is the linear subspace generated by $e_i$.
• Thank you for your answer. To check my understanding: I think that $(e_1,...,e_n)$ here is an eigenbasis for all elements of $Ad_h$, which is possible because they're simultaneously diagonalizable. Because these are all eigenvectors, we get that $[x,e_i] = \alpha_i(x)e_i$ for some $\alpha_i \in h^*$. This means that $e_i \in \mathfrak{g}^{\alpha_i}$, so that the linear subspace generated by $e_i$ is also in $\mathfrak{g}^{\alpha_i}$. This gives the result. Is this correct?