Linear Lie algebra generated by $\mathbb{R}$-diagonalizable matrices If $\mathfrak{gl}_n(\mathbb{R})$ denotes the Lie algebra of real $(n \times n)$-matrices, then the $\mathbb{R}$-diagonalizable matrices generate $\mathfrak{gl}_n(\mathbb{R})$ as a Lie algebra.
If $\mathfrak{g}<\mathfrak{gl}_n(\mathbb{R})$ is an arbitrary Lie subalgebra, then it need not be generated by the $\mathbb{R}$-diagonalizable matrices which it contains. In fact, $\mathfrak{g}$ might not contain any diagonalizable matrices (e.g. if it is generated by a single non-diagonalizable matrix).
What are criteria which guarantee that $\mathfrak{g}$ is generated by $\mathbb{R}$-diagonalizable matrices? In particular, is it sufficient for $\mathfrak{g}$ to be simple and non-compact?
 A: Yes: if $K$ is a field of characteristic zero and $\mathfrak{g}$ is semisimple with no $K$-anisotropic factor (when $K=\mathbf{R}$, $K$-anisotropic means compact) then it is generated (and even linearly spanned) by its $K$-diagonalizable elements. And more generally if $\mathfrak{g}$ is perfect (trivial abelianization) with no compact factor.
Since $\mathfrak{g}$ is perfect, it is the Lie algebra of some algebraic subgroup of $\mathrm{GL}_n$.
The set of $K$-diagonalizable elements in $\mathfrak{g}$ is stable under scalar multiplication, and under the action of $G$. Hence its span is a $G$-invariant subspace of $\mathfrak{g}$, hence is an ideal $I$.
If by contradiction $I\neq \mathfrak{g}$, let $J$ be a maximal proper ideal of $\mathfrak{g}$ containing $I$. Hence $\mathfrak{g}/J$ is either 1-dimensional or simple. The first case is excluded since $\mathfrak{g}$ is perfect. In the second case, the quotient is $K$-isotropic by assumption. It is a split quotient (by standard structure theory), but then we have a simple $K$-isotropic subalgebra $\mathfrak{s}\subset\mathfrak{g}$ with no nontrivial $K$-diagonalizable element. This is a contradiction: indeed a maximal $K$-split abelian subalgebra of $\mathfrak{s}$ is positive-dimensional and consists of $K$-diagonalizable elements. 

Addendum:
In the above setting, every element is a sum of $K$-diagonalizable elements. Hence   every element of $\mathfrak{g}$ is a sum of $m$ $K$-diagonalizable elements, for some $m$ (e.g., $m=\dim(\mathfrak{g})$ works). One can wonder whether one can take $m$ small, e.g., $m=2$? It seems we can't expect some universal $m$: In $\mathfrak{so}(n,1)$, if I'm correct the dimension of those $\mathbf{R}$-diagonalizable elements is $\le 2n$ and hence $m\ge (n+1)/4$.
