The question is as in the title.


I know that a traceless matrix can be written as a commutator of two matrices.


Then, let $v : \mathbb{R}^3 \to \mathbb{R}^3$ be a divergence-free smooth vector field. That is, $\nabla \cdot v=0$.


Then, the matrix $A(x)=\bigl[\partial\_i v\_j(x) \bigr]$ is smooth as a mapping from $\mathbb{R}^3$ into $M\_{3 \times 3}(\mathbb{R})$ and tracelss for each $x$.


Then, what does any two matrices $B(x)$ and $C(x)$ that satisfy $A(x)=B(x)C(x)-C(x)B(x)$ look like?


This is quite new and original for me..Could anyone please help me?



For a divergence free smooth vector field $v : \mathbb{R}^3 \to \mathbb{R}^3$, how to find the commutator form of the matrix $A=(\partial_i v_j)$?
You might want to first carry out a unitary transformation $A(x)\mapsto U(x)A(x)U^\top(x)$, such that all diagonal elements are zero.$^\ast$  

Then $A(x)=BC(x)-C(x)B$ with
$$B=\begin{pmatrix}
1&0&0\\
0&2&0\\
0&0&3
\end{pmatrix},\;\;
C\_{ij}(x)=\begin{cases}
0&\text{if}\;i=j\\
\frac{A\_{ij}(x)}{B\_{ii}-B\_{jj}}&\text{if}\;\;i\neq j.
\end{cases}
$$




---


$^\ast$ This is always possible for a traceless $A$ ([proof](https://math.stackexchange.com/q/267192/87355)). 
For a $3\times 3$ matrix the unitary has the two-parameter form
$$U(x)=\begin{pmatrix}
\cos\alpha(x)&\sin\alpha(x)&0\\
-\sin\alpha(x)&\cos\alpha(x)&0\\
0&0&1
\end{pmatrix}
\begin{pmatrix}
1&0&0\\
0&\cos\beta(x)&\sin\beta(x)\\
0&-\sin\beta(x)&\cos\beta(x)
\end{pmatrix}
.$$
You can solve first for $\alpha(x)$,
$$(\partial\_1 v\_1) \cos ^2\alpha+(\partial\_2v\_2) \sin ^2\alpha+ (\partial\_1v\_2+\partial\_2 v\_1)\sin\alpha\cos\alpha=0,$$
and then for $\beta(x)$.
