For a given invertible real matrix $G\in \mathrm{GL}_d$ with $\det G>0$, 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
$$ Q G Q^T = \exp(QBQ^T) \exp\bigl(\tfrac{1}{2}((QBQ^T)^T-QBQ^T)\bigr) . $$
Solution for special orthogonal case in $d=2$
Since a solution exists for $G$ symmetric positive, another subcase to consider is $G\in \mathrm{SO}(d)$.
Denote with $\hat T =\left(\begin{smallmatrix} 0 & 1 \\ -1 & 0 . \end{smallmatrix}\right)$. We rewrite the equation as $$ G e^{-T} = e^{S-T} $$ Since $G$ is orthogonal with determinant $1$, we have $G= e^{\theta \hat S}$ for $\theta \in [0,2\pi)$, we can choose $T$ skew such that $e^{-T}=-G^T$, so $T= t \hat T$ for $t= \pi + \theta \in [\pi,3\pi)$ is a possible solution (here any multiple of $2\pi$ can be added). Hence, we get the equation $$ - \mathrm{Id} = e^{S- t \hat T} , $$ which comes with freedom of a change of basis for any invertible $P$ $$ - \mathrm{Id} = P^{-1} e^{S- T} P = \exp(P^{-1}(Z-t \hat T) P) . $$ A sufficient condition to have a solution is, if we find an invertible $P$ and symmetric $S$ such that $$ P^{-1} (S-t \hat T) P = \pi \hat T, $$ This can be rewritten $$ S = \pi P \hat T P^{-1} + t \hat T . $$ Let us denote $\hat P = P \hat T P^{-1}$, then it is enough to satisfy the identity $$ \pi \hat P_{12} + t = \pi \hat P_{21} - t $$ By explicit calculation, this becomes $$ \pi( P_{11}^2 + P_{22}^2 + P_{12}^2 + P_{21}^2) = 2 t ( P_{11} P_{22} - P_{12}P_{21}) , $$ which can be expressed in terms of the singular values $\sigma_1,\sigma_2$ of $P$ and choosing $\det P <0$ as $$ \frac{\sigma_1}{\sigma_2} + \frac{\sigma_2}{\sigma_1} = \frac{2t}{\pi}, $$ which always has a solution by our choice of $t\in [\pi, 3\pi)$. Hence, any $P= U D_\sigma V^T$ with $D_\sigma = \mathrm{diag}(\sigma_1,\sigma_2)$ with singular values satisfying the above relation and $U,V$ orthogonal with $\det U = - \det V$ is a solution. The effective parameter regime for the choice of $P$ is $2$, one for each of the orthogonal matrix $U,V$.
Question
Can those two constructions be combined to give a solution for $G\in \mathrm{GL}_2$ with $\det G>0$?
