How to solve $M - EV - (EV)^T = 0$ against $E$? I have a matrix equation:
$$M - EV - (EV)^T = 0$$
which i want to solve against $E$. Is this possible? How to do that? 
Remarks: matrices are square, $E$, $M$ are symmetric, $V$ is invertibile.
Regards,
Marek
 A: In one dimensional case
$$
M - EV - (EV)^T = M - 2EV = 0 \implies E = \frac{M}{2V}.
$$
So, maybe you would like to check if
$$
E = \frac{1}{2} MV^{-1}
$$
in general case?
A: We can write this equivalently as
$$
EV + (EV)^T = M.  
$$
Because the (linear) operator $E \mapsto EV + (EV)^T$ is not invertible, this equation will have infinitely many solutions.  In particular, we can always take $E = \frac 12 MV^{-1}$.  Indeed, plugging this $E$ in yields
$$
EV + (EV)^T = \frac 12 MV^{-1}V + (\frac 12 MV^{-1}V)^T = \frac 12 M + \frac 12 M^T = M
$$
As it turns out, every solution to this equation can be written in the form
$$
E = \frac 12 (M + S)V^{-1}
$$
where $S$ is skew symmetric, which is to say that $S$ satisfies $S^T = -S$.

Now, requiring that $E$ is symmetric amounts to requiring that
$$
[(M + S)V^{-1}]^T = (M + S)V^{-1} \implies\\
V^{-T}(M - S) = (M + S)V^{-1} \implies\\
V^{-T}S + SV^{-1} = V^{-T}M - MV^{-1}.
$$
In other words: there exists a symmetric solution $E$ if and only if the Sylvester equation (more specifically Lyapunov equation) $V^{-T}S + SV^{-1} = V^{-T}M - MV^{-1}$ has a solution $S$ in which $S$ is also skew-symmetric. I don't see a way of simplifying this condition.
It is notable that if the eigenvalues of $V$ have all positive real part or negative real part, then there exists a unique solution $S$.  If the eigenvalues of $V$ have only negative real part, then this unique $S$ can be written explicitly as the integral
$$
S = \int_0^\infty e^{V^{-T}\tau}[V^{-T}M - MV^{-1}]e^{V^{-1}\tau} d\tau.
$$
This $S$ is necessarily skew-symmetric, which is to say that our original equation has a (unique) symmetric solution $E$ in this case.

Another approach: if we're looking specifically for a symmetric solution $E$, then we can rewrite the original equation as
$$
EV + V^TE^T = M \implies V^TE + EV = M.
$$
Now, if the eigenvalues of $V$ all have negative real parts, then this equation has a unique solution that can be written as
$$
E = \int_0^\infty e^{V^{T}\tau}\,M\,e^{V\tau} d\tau,
$$
which is necessarily a symmetric matrix.
