Is there any inverse-commutator for matrices My question is very simple. Given a symmetric real matrix $A$, and a square real matrix $C$, how can one solve the equation $[A,X]=C$, where $[A,B]$ is commutator of $A$ and $B$, i.e., $[A,B]=AB-BA$. We are more interested in symmetric matrices for $X$.
Thanks.
Update: Uranix' approach in his last edit is a great solution. Since it is closed-form enough, it can be used to prove any properties of interest for your application as it was the case for me. Tnx
 A: Not sure if there is a closed-form answer, but 
$$
AX-XA = C
$$
is a linear system of $n\times n$ equations of $n \times n$ unknowns.
One approach is using the vectorization operator $$\operatorname{vec} X = 
\begin{pmatrix}
x_{11}\\
x_{21}\\
\vdots\\
x_{n1}\\
x_{12}\\
\vdots\\
x_{nn}
\end{pmatrix}$$
Using Kronecker product
$$
\operatorname{vec} (AXB) = (B^\top \otimes A) \operatorname{vec} X
$$
the system becomes
$$
(I \otimes A - A^\top \otimes I)\operatorname{vec} X= \operatorname{vec} C\\
\operatorname{vec} X = (I \otimes A - A^\top \otimes I)^{-1}\operatorname{vec} C
\mkern{-227mu}\frac{\phantom{\operatorname{vec} X = (I \otimes A - A^\top \otimes I)^{-1}\operatorname{vec} C}}{\phantom{b}}
$$
Edit. As kindly pointed by @loup blanc the matrix $I \otimes A - A^\top \otimes I$ is always degenerate so there is either no solution to the equation or infinite number of the solutions.
The another approach may be the following: let $\Omega \Lambda \Omega^{-1}$ be the eigendecomposition for $A$.
$$
[A,X] = \Omega \Lambda \Omega^{-1} X - X \Omega \Lambda \Omega^{-1} =
\Omega [\Lambda, \Omega^{-1}X\Omega] \Omega^{-1} = C\\
[\Lambda, \Omega^{-1}X\Omega] = \Omega^{-1}C\Omega
$$
Denoting $R = \Omega^{-1}C\Omega, Y = \Omega^{-1}X\Omega$ the equation becomes
$$
[\Lambda, Y] = R
$$
with indices that is
$$
\lambda_i Y_{ij} - \lambda_j Y_{ij} = R_{ij}\\
Y_{ij} = \begin{cases}
\dfrac{1}{\lambda_i - \lambda_j}R_{ij}, &\lambda_i \neq \lambda_j\\
\text{any}, &\lambda_{i} = \lambda_j
\end{cases}
$$
provided that $R_{ij} = 0$ for every $\lambda_i = \lambda_j$ (at least,  $\operatorname{diag}(R) = 0$).
A: Let $\operatorname{spectrum}(A)=(\lambda_i)_i$ and $f:A\rightarrow I\otimes A-A^T\otimes I$. Then $\operatorname{spectrum}(f)=\{\lambda_i-\lambda_j\mid i,j\}$. Assume that we are in the generic case, which implies that the $(\lambda_i)_i$ are pairwise distinct. Thus $\dim(\ker(f))=n$ and $\dim(\operatorname{im}(f))=n^2-n$. Thus there are solutions in $X$ if $C$ is in the orthogonal of $\ker(f)$ (in a sense to be defined); in particular, a necessary condition is $\operatorname{tr}(C)=0$.
