We usually define semisimplicity of a Lie algebra $\mathfrak{g}\subset M_n ({\bf R})$ from two ways. I want to know the relation between them.
One of the definitions of semisimple Lie algebra is as follows (cf 156 page of the book : Differential geometry, Lie groups, and symmetric spaces - Helgason):
A Lie algebra $\mathfrak{g}\subset M_n ({\bf R})$ is semisimple if
(1) whenever $\mathfrak{g}$ on ${\bf R}^n$ has an invariant subspace $W$, then $W$ has a complementary invariant subspace.
(2) $\{ v\in {\bf R}^n| Xv=0$ for all $X\in \mathfrak{g}\}=\{0\}$.
Here I have a question : Semisimplicity of $\mathfrak{g}\subset M_n ({\bf R})$ is equivalent to the following :
Invariant subspace of ${\bf ad}(\mathfrak{g})$ on $\mathfrak{g}$ has complementary subspace which is invariant.
The above statement is true ? If it is true then how can we prove ?
Thank you in advance.
[Additional Question]-------------------------------------------------------
Assume that $\{ v\in {\bf R}^n|\ Xv=0$ for all $X\in \mathfrak{g} \} = \{ 0 \}$. Then a Lie algebra $\mathfrak{g} \subseteq \{ X\in M_n({\bf R}) |\ X=-X^T \}$ is semisimple on ${\bf R}^n$, that is, invariant subspace has a complementary invariant subspace. This statement is correct ? If it is right how can we prove ?