Diagonalization in Lie algebras Real symmetric matrices can be diagonalized by orthogonal matrices , i.e.
$$ OSO^{T} = D $$ where $O \in \text{O}(N)$ is an orthogonal matrix, $S$ is real and symmetric, i.e. $S^T = S$ and $D$ is diagonal. Similarly hermitian matrices can be diagonalized by unitary matrices. Now, my question is whether there a is a general relation that elements of a Lie algebra can be diagonalized via elements of their Lie group. For example does a similar statement hold for the symplectic group $\text{Sp}(N)$?
I am aware that the Lie algebra of the orthogonal matrices $O(N)$ is actually $\mathfrak{o}(N)$, the skew-symmetric matrices. Similarly, the corresponding Lie algebra of unitary matrices are the skew-hermitian matrices. Still is the fact that symmetric or hermitian matrices can be diagonalized as above only true for these two groups or is there a more general statement for maybe semi-simple Lie algebras.
Thank you very much for your answers in advance. I look forward to learning something new. Forgive me as a physicist if my answer was not posed completely mathematically rigorous. 
 A: If $G$ is a semisimple Lie group and $\mathfrak{g}$ the corresponding Lie algebra, then the general statement is that every element of $\mathfrak{g}$ is conjugate, via the adjoint action of $G$, to an element in any fixed Cartan subalgebra. If $G$ is compact and connected, then the corresponding statement about $G$ is that every element of $G$ is conjugate to an element in any fixed maximal torus. 
When $G = U(n)$ it turns out that the subalgebra of $\mathfrak{g} = \mathfrak{u}(n)$ given by diagonal matrices is a Cartan subalgebra (similarly, the diagonal subgroup of $U(n)$ is its maximal torus), so this specializes to the statement that skew-hermitian matrices can be unitarily diagonalized. Unfortunately, the diagonal subalgebra of $\mathfrak{o}(n)$ is not a Cartan subalgebra (similarly for the symplectic group / algebra), and in any case it's not true that skew-symmetric matrices can be diagonalized (over the reals). The real spectral theorem is not a statement about $\mathfrak{o}(n)$ at all. 
