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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: Hey guys, 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

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    $\begingroup$ Oh come on guys. By simple I meant it is easy to explain. Of course, I couldn't solve it and that's why I posted it. $\endgroup$ – Eri May 31 '15 at 22:26
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    $\begingroup$ Eri, I know that's what you meant! ;-) But it might get you out from under the eye of the closure nazhgul if you edited in a few words about any attempts you have made. By the way, nice question,, endorsed! Good Luck with it! $\endgroup$ – Robert Lewis May 31 '15 at 22:38
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    $\begingroup$ Eri, you might look into considering the map $X \mapsto [A, X]$ as a linear operator on matrix space, trying to express it in a suitable basis, and using properties of $A$ to see what you can about its eigenstructure etc. That's what I'd do. Or research the literature; pretty sure you'll find something useful. Cheers! $\endgroup$ – Robert Lewis May 31 '15 at 22:47
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    $\begingroup$ Eri, look at this: $[A, aX + bY]= A(aX + bY) - (aX + bY)A = a[A, X] + b[A, Y]$. Cheers! $\endgroup$ – Robert Lewis May 31 '15 at 23:10
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    $\begingroup$ @RobertLewis, Oh ok, I see. Tnx $\endgroup$ – Eri May 31 '15 at 23:58
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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, $\operatorname{diag}(R) = 0$).

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  • $\begingroup$ uranix, thank you very much for your explanation. A closed form would be great. but at least I need some information to prove on X. $\endgroup$ – Eri May 31 '15 at 22:54
  • $\begingroup$ For your updated approach. I am going to investigate and try to use it. I'll let you know how it goes. Thanks again! $\endgroup$ – Eri May 31 '15 at 22:55
  • $\begingroup$ Your last edit solved my question. It is closed from enough for me. Tnx. And I "accepted" it. $\endgroup$ – Eri Jun 2 '15 at 20:37
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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 $\operatorname{tr}(C)=0$.

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    $\begingroup$ The downvotes came within seconds - I happened to be here and saw it myself. That was strange. It is too late an hour for me to investigate the matter now. Somebody is having a childish fit. I upvote this because of the bound on the dimension of the kernel (resp. image). $\endgroup$ – Jyrki Lahtonen May 31 '15 at 23:47
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    $\begingroup$ Downvotes are not mine. I just would suggest some explanation why does $spectrum(f)$ consist of pairwise difference of eigenvalues. If I am not mistaken, the $ij$ eigenvector of $f$ is a Kronecker product of $i$-th eigenvector of $A$ and $j$-th eigenvector of $A^\top$. $\endgroup$ – uranix Jun 1 '15 at 4:45
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    $\begingroup$ Orthogonality to zero eigenvector $0 = (y_i \otimes x_i)^\top \cdot \operatorname{vec}(C) = (y_i^\top \otimes x_i^\top) \cdot \operatorname{vec}(C)$ where $x_i A = \lambda_i x_i,\; y_i A^\top = \lambda_i y_i$ can be rewritten as $$x_i C y_i^\top = 0, i = 1, \dots, n$$ or $$\operatorname{diag}(\Omega_L C \Omega_R) = 0$$ where $\Omega_L, \Omega_R$ are matrices of left and right eigenvectors of $A$. $\endgroup$ – uranix Jun 1 '15 at 5:19
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    $\begingroup$ For the downvotes, this meta post is relevant. Most likely also the downvotes on uranix' answer and two downvotes to the question come from the same source. $\endgroup$ – Daniel Fischer Jun 1 '15 at 10:39

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