Let $\mathfrak{g} \subseteq \mathfrak{su}(n)$ be a linear Lie algebra represented by skew-symmetric $n\times n$ matrices. Let $C \in SU(n)$ be a special unitary matrix where it is known that $C = exp\{G\}$ for some $G \in \mathfrak{g}$.
Given $C$ and a basis $\{G_k\}$ for $\mathfrak{g}$, suppose we wish to find a matrix $G$ expressed in terms of its real basis coefficients $\vec{a}$:
$$G = \sum a_k G_k$$
such that $C = \exp\{ G\}$.
If more than one such matrix exists, we wish to find the one where $\sum |a_k|^2$ is minimized.
I know that if we diagonalize $C = P^{-1}\Lambda P$ where $$\Lambda = \left( {\begin{array}{*{20}{c}} {{\lambda _1}}&{}&{}\\ {}& \ddots &{}\\ {}&{}&{{\lambda _n}} \end{array}} \right)$$
then is suffices to find the vector $\vec{a}$ and set of integers $\{z_k\}$ such that
$$\sum a_k PG_kP^{-1} = \left( {\begin{array}{*{20}{c}} {\log {\lambda _1}}&{}&{}\\ {}& \ddots &{}\\ {}&{}&{\log {\lambda _n}} \end{array}} \right) + 2\pi i\left( {\begin{array}{*{20}{c}} {{z_1}}&{}&{}\\ {}& \ddots &{}\\ {}&{}&{{z_n}} \end{array}} \right)$$
where $\log \lambda$ is the (imaginary) principle log of $\lambda$.
But while it's possible to iterate through guesses for the $z$'s, checking to see whether the equation is solvable, this is a terribly inefficient way of solving the problem.
Is there a faster, more scalable way to find $\vec{a}$ for a given $C$? Does this problem have a name?
We assume that the $(\log(\lambda_i))$ are distinct, cf. my comment above.
For every choice of the $(a_k)_{k\leq p}$, $K=\sum_k a_kPG_kP^{-1}$ is skew-hermitian. Then the main task to do is to seek the $(a_k)$ s.t. $K$ is diagonal. We must solve a homogeneous LINEAR system of $n^2-n$ real equations in the $p$ unknowns $(a_k)$.
Of course, these equations are not linearly independent; yet, I think there should not remain a lot of parameters after solving this system.