Solution to matrix equation with exponentials Assuming
$$
A, M \in \mathcal{R}^{n \times n }
$$
with A, M being invertible. Could someone tell me if it is possible to find a closed form solution for the matrix $M$ in the following equation and if so, provide me with some hints on how to obtain it
$$
A = e^M + e^{M^T}.
$$
I have tried a few things but can´t seem to make it work.
Thanks.
 A: If $A=e^M + e^{M^t}$ then $A$ is a real symmetric matrice.
You can use spectral theorem to diagonalize it that is there is $D$ a diagonal matrix and $P$ a invertible matrice, such that $A=PDP^{-1}$.
Then you can rewrite you equation as $D=e^{P^{-1}MP}+e^{P^{-1}M^{t}P}$.
If $D$ have positive coefficients on the diagonal you can search N=P^{-1}MP also as a diagonal matrice whose coefficient will be $n_{i,j}=1/2 log(d_{i,j})$. Then $M=PNP^{-1}$.
I am not there is a solution if one coefficient of $D$ is negative.
A: Given that $A$ is a real symmetric matrix and $e^{M^T} = (e^M)^T$, one find a solution for $M$ given one additional assumption: the diagonal of $A$ only contains non-zero elements.
Set $e^M$ equal to the lower triangular matrix of $A$ minus half the diagonal of $A$. So if $$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, ~~ \text{then} ~~ e^M = \begin{bmatrix} 1 & 0 \\ 2 & 3 \end{bmatrix} - \begin{bmatrix} 0.5 & 0 \\ 0 & 1.5 \end{bmatrix}$$
Given this structure of $e^M$, $A = e^M + (e^M)^T$, so the equation still holds. In order to solve $M$ for $e^M$, one has to take the matrix logarithm. Depending on the resulting properties of $e^M$, the methods could vary.
A: The equation implies that $A/2$ is the symmetric part of $e^M$. In other words, $\frac12A+K$ has a real logarithm $M$ for some appropriate skew-symmetric matrix $K$.
Hence it is not always solvable. For example, when $n$ is odd, every skew-symmetric $n\times n$ matrix $K$ is singular. Therefore the equations $e^M=K$ and  $e^M+e^{M^T}=0$ are not solvable.
However, given that $A$ is real symmetric, the equation $e^M+e^{M^T}=A$ is solvable when $n$ is even or when $A$ has at least one positive eigenvalue. By an orthogonal similarity transform, we may assume that $A$ is a real diagonal matrix. Now:

*

*When $n$ is even, let $K$ be the direct sum of $n/2$ copies of $\pmatrix{0&-1\\ 1&0}$. Then $\frac12A+cK$ has not any zero or negative eigenvalues when $c\in\mathbb R$ is sufficiently large. Hence $\frac12A+cK$ has a real logarithm $M$ and $e^M+e^{M^T}=A$.

*When $n$ is odd but $A$ has a positive eigenvalue $a$, let $A=a\oplus A'$. Since the size of $A'$ is even, our previous argument shows that there exists a skew-symmetric matrix $K'$ such that $\frac12A'+K'$ has a real logarithm. Hence $\frac a2\oplus(\frac12A'+K')$ has a real logarithm $M$ and $e^M+e^{M^T}=A$.

