Lie bracket in coordinates In $\mathbb{R}^2-\{0\}$ we consider the vector field defined by, $$V=-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}$$
I am trying to find all other vector fields $X$ that $[V,X]=0$.
My idea was to use polar coordinates $(r,\theta)$ to compute the equation $[V,X]=0$, since $V=\frac{\partial}{\partial \theta}$, and then go back to cartesian coordinates. 
If we denote $X=a\frac{\partial}{\partial r}+b\frac{\partial}{\partial \theta}$, we have that,
$$0=[V,X]=\frac{\partial a}{\partial \theta}\frac{\partial}{\partial r}+\frac{\partial b}{\partial \theta}\frac{\partial}{\partial \theta}$$
and here is where I am stuck at, all I manage to extract from this equation is that $a$ and $b$ only depend on $r$, how do I get the expression in cartesian coordinates?
 A: Since
$$
V=-y\frac{\partial}{\partial x}+x\frac{\partial }{\partial y}=\frac{\partial}{\partial\theta},
$$
if we set
$$
X=X_1\frac{\partial}{\partial x}+X_2\frac{\partial }{\partial y}=\bar{X}_1\frac{\partial}{\partial r}+\bar{X}_2\frac{\partial}{\partial \theta},
$$
we have:
$$
0=[V,X]=\frac{\partial\bar{X}_1}{\partial \theta}\frac{\partial}{\partial r}+\frac{\partial\bar{X}_2}{\partial\theta}\frac{\partial}{\partial \theta}.
$$
It follows that
$$
\frac{\partial\bar{X}_1}{\partial \theta}=0=\frac{\partial\bar{X}_2}{\partial\theta},
$$
i.e.
$$
\bar{X}_1=f(r),\quad \bar{X}_2=g(r),
$$
where $f$ and $g$ are smooth functions from $(0,\infty )$ into $\mathbb{R}$.
Now, we can find $X_1$ and $X_2$. Since
$$
X=X_1\frac{\partial}{\partial x}+X_2\frac{\partial }{\partial y}=f\frac{\partial}{\partial r}+g\frac{\partial}{\partial \theta},
$$
and
$$
x=r\cos\theta,y=r\sin\theta,
$$
we have
\begin{eqnarray}
X_1(x,y)&=&X\cdot x=\left[f(r)\frac{\partial}{\partial r}+g(r)\frac{\partial}{\partial \theta}\right]\cdot(r\cos\theta)=\bar{X}_1(r)\cos\theta-r\bar{X}_2(r)\sin\theta\\
&=&\frac{x}{\sqrt{x^2+y^2}}f\left(\sqrt{x^2+y^2}\right)-yg\left(\sqrt{x^2+y^2}\right)\\
X_2(x,y)&=&X\cdot y=\left[f(r)\frac{\partial}{\partial r}+g(r)\frac{\partial}{\partial \theta}\right]\cdot(r\sin\theta)=f(r)\sin\theta+rg(r)\cos\theta\\
&=&\frac{y}{\sqrt{x^2+y^2}}f\left(\sqrt{x^2+y^2}\right)+xg\left(\sqrt{x^2+y^2}\right).
\end{eqnarray}
Hence
$$
X=\left[\frac{x}{\sqrt{x^2+y^2}}f\left(\sqrt{x^2+y^2}\right)-yg\left(\sqrt{x^2+y^2}\right)\right]\frac{\partial}{\partial x}+\left[\frac{y}{\sqrt{x^2+y^2}}f\left(\sqrt{x^2+y^2}\right)+xg\left(\sqrt{x^2+y^2}\right)\right]\frac{\partial}{\partial y}.
$$
