Is there a way to find $B,C$ such that $A=[B,C]$? The fact that $\mathfrak{sl}_2(\mathbb{C})$ is a simple Lie algebra implies that every $2 \times 2$-matrix $A \in \mathbb{C}^{2\times 2}$ with $\mathrm{tr}(A) = 0$ can be expressed as a commutator of two matrices.
Is there any way to find two appropriate matrices? If I give you $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, for example, is there any algorithm that can find matrices $B,C$ with $A = BC - CB$?
 A: I think you can use the Pauli matrices $\sigma_1$, $\sigma_2$, $\sigma_3$. Traceless matrix $A$ can be decomposed as $A = \vec{a}\cdot\vec{\sigma}$. Meanwhile, any $B$, $C$ can be decomposed as $B = b_0I+\vec{b}\cdot\sigma$ and $C=c_0I+\vec{c}\cdot\sigma$, with commutator $[B,C] = 2i(\vec{b}\times\vec{c})\cdot\vec\sigma$.
So I think if you find two vectors $\vec{b}$ and $\vec{c}$ with $2i(\vec{b}\times\vec{c}) = \vec{a}$, then you have found $B$ and $C$ as $B = \vec{b}\cdot\sigma$ and $C = \vec{c}\cdot\sigma$.
For your example case, $$A = -i\sigma_2 = (0,-i,0)\cdot\vec{\sigma}\,.$$ That is, $\vec{a} = (0, -i, 0)$. So take $\vec{b} = ({1\over \sqrt{2}},0,0)$ and $\vec{c} = (0,0,{1\over\sqrt{2}})$. This gives $B = {1\over\sqrt{2}}\sigma_1$ and $C={1\over\sqrt{2}}\sigma_3$, with $$[B,C] = \frac{1}{2}[\sigma_1,\sigma_3]= \frac{1}{2}(-2i\sigma_2) = A\,.$$
A: Let $K$ be any field and $A\in M_n(K)$. If $trace(A)=0$, then there are $B,C\in M_n(K)$ s.t. $A=BC-CB$. See ("On matrices of trace $0$", A. and B. Muckenhoupt). Put $A$ in  rational canonical form and reason by recurrence...
In general we cannot choose $B$ no matter how ; indeed the linear application $f:X\rightarrow BX-XB$ has eigenvalues $(\lambda_i-\lambda_j)_{i,j}$ in which $spectrum(B)=(\lambda_i)_i$. Thus, if $B$ is generic, then $rank(f)=n^2-n$ and $im(f)$ is not whole the vector space of traceless matrices (which has dimenson $n^2-1$).
The case $n=2$ is easy. Assume $characteristic(K)\not=2$. If $A\not= 0$, then there is $u$ s.t. $u,Au$ is a basis. Since $trace(A)=0$, we may assume that $A=\begin{pmatrix}0&\alpha\\1&0\end{pmatrix}$. Then we can choose $B=\begin{pmatrix}-1/2&0\\0&1/2\end{pmatrix},C=\begin{pmatrix}0&-\alpha\\1&0\end{pmatrix}$.
