Finding matrices $X,Y$ such that $XY - YX = \left[\begin{smallmatrix}0&1&1\\1&0&1\\1&1&0\end{smallmatrix}\right]$ 
Can anyone help me to find the matrices $X$ and $Y$ such that
$$XY - YX = \begin{bmatrix} 0 & 1 & 1 \\ 
1 & 0 & 1 \\ 
1 & 1 & 0   \end{bmatrix}$$

I was trying to randomly choose  $A , B$ and test if they work.
 A: If the field is of characteristic zero, there are infinitely many solutions. One class of solutions that can be easily obtained is given by a diagonal matrix $X$ with distinct diagonal entries. E.g.
$$
X=\pmatrix{0\\ &1\\ &&2}\Rightarrow Y=\pmatrix{\ast&-1&-\frac12\\ 1&\ast&-1\\ \frac12&1&\ast}
$$
where the entries marked by asterisks are arbitrary.
More specifically, when an $n\times n$ matrix $A$ has a zero diagonal (i.e. it is a hollow matrix) and $X=\operatorname{diag}(x_1,\ldots,x_n)$ has distinct diagonal entries, the equation $XY-YX=A$ gives rise to $(x_i-x_j)y_{ij}=a_{ij}$. Therefore $y_{ij}=a_{ij}/(x_i-x_j)$ when $i\ne j$ and $y_{ii}$ is arbitrary.
More generally, over an arbitrary field, it is known that every traceless matrix (including but not limited to hollow matrices) can be written as a commutator. See A. A. Albert and Benjamin Muckenhoupt (1957), On matrices of trace zeros, Michigan Math. J., 4(1):1-3.
A: A complementary answer to the answer by user1551.
Let $A:=\begin{bmatrix} 0 & 1 & 1 \\ 
1 & 0 & 1 \\ 
1 & 1 & 0   \end{bmatrix}$.
Let us use notation $[X,Y]:=XY-YX$.
Let $F_X : Y \mapsto [X,Y]$.
It is a linear operator from the vector space of $3 \times 3$ matrices on itself.
For any $X \ne 0$, the kernel $K$ of $F_X$ has in general a dimension 3 with basis $\{I_3, X, X^2\}.$
Indeed; or any $n$, $ \ F_X(X^n)=[X,X^n]=X^{n+1}-X^{n+1}=0$, therefore $X^n$ belongs to $K$ for any $n$ but because of Cayley-Hamilton theorem, powers above $2$ can be expressed as polynomials in $I,X,X^2$. For a completely rigorous proof, see this answer.
Therefore if $(X_0,Y_0)$ is a solution to the equation $[X,Y]=A$, $(X_0,Y_0+aI+bX_0+cX_0^2)$  is as well a solution, for any $a,b,c \in \mathbb{R}$.
Connected: https://math.stackexchange.com/q/1223984.
Remark: Operator $F_X$ could be given a form using Kronecker product.
