For a given invertible real matrix $G\in \mathrm{GL}_d$, we ask for a solution $B$ of the matrix exponential equation

$$ G = \exp(B) \exp\bigl(\tfrac{1}{2}(B^T-B)\bigr) . $$

Basic observation

For $G$ symmetric positive definite, the matrix logarithm of $G$ is a solution.

A different representation is to search for a pair $(S,T)$ with $S$ symmetric and $T$ skew-symmetric such that

$$ G = \exp(S-T) \exp(T) . $$

In particular, if $S$ and $T$ commute (equivalently $B$ and $B^T$), we obtain again the symmetric positive solutions.

The equation is Gauge invariant under orthogonal transposition, i.e. for any $Q\in O(d)$ it also holds

$$ Q G Q^T = \exp(QBQ^T) \exp\bigl(\tfrac{1}{2}((QBQ^T)^T-QBQ^T)\bigr) . $$

Ansatz for orthogonal case

Since a solution exists for $G$ symmetric positive, another subcase to consider is $G\in \mathrm{O}(d)$. A possible Ansatz in this case is to choose $T$ such that $e^{-T}=-G^T$, that is the skew symmetric matrix $\log(-G)$. Then, we are left with the identity

$$- I = e^{Z-\log(-G)} . $$

If a solution $Z$ exists, we can make use of invariance under change of basis and find that for any invertible $P$ it holds

$$- I =- P^{-1} P = P^{-1}e^{Z-\log(-G)} P = \exp[ P^{-1}(Z-\log(-G))P] . $$

A possible solution is to find an invertible $P$ and symmetric $S$ such that $$ P^{-1}(S-\log(-G))P = \pi \begin{pmatrix} 0 & 1 \\\ -1 & 0 \end{pmatrix} $$ Any tips and ideas for the above matrix equation, but also the general case are much appreciated.