How to change coordinates of a vectorfield $X$ on $\mathbb{R}P^2$ Let $\phi_1:U_1=\lbrace[x:y:z]\in  \mathbb{R}P^2:x\ne0\rbrace\longrightarrow\mathbb{R}^2, \,\, \phi_1([x:y:z])=(y/x,z/x)=(u_1,u_2)$ be a coordinate map on $\mathbb{R}P^2$.
And $\phi_2:U_2=\lbrace[x:y:z]\in  \mathbb{R}P^2:y\ne0\rbrace\longrightarrow\mathbb{R}^2, \,\, \phi_2([x:y:z])=(x/y,z/y)=(v_1,v_2)$ another coordinate map.
Also consider the vectorfield: $X|_{U_1}=u_1\frac{\partial}{\partial u_1}+2u_2\frac{\partial}{\partial u_2} $. How does this vecotrfield look in the coordinate map $(U_2,\phi_2)$?
My problem ist, I know how to change the basisvectors, the vectorfield should look like:
$X|_{U_2}=a(\frac{\partial v_1}{\partial u_1}\frac{\partial}{\partial v_1}+\frac{\partial v_2}{\partial u_1}\frac{\partial}{\partial v_2})+b(\frac{\partial v_1}{\partial u_2}\frac{\partial}{\partial v_1}+\frac{\partial v_2}{\partial u_2}\frac{\partial}{\partial v_2})$ with smooth coefficients $a,b \in C^{\infty}(\mathbb{R}P^2)$.
But how do I compute the coefficients?
 A: The key point is that $$v_1=\frac{1}{u_1},\; v_2=\frac{u_2}{u_1}$$ From there you get $$\frac{\partial }{\partial u_1}=-\frac{1 }{u_1^2}\frac{\partial }{\partial v_1}+(-\frac{u_2 }{u_1^2})\frac{\partial }{\partial v_2},\;\operatorname {so that}\; u_1 \frac{\partial }{\partial u_1}=-\frac{1 }{u_1}   \frac{\partial }{\partial v_1} -\frac{u_2}{u_1}\frac{\partial }{\partial v_2}$$
$$\frac{\partial }{\partial u_2}=0.  \frac{\partial }{\partial v_1} +\frac{1 }{u_1}   \frac{\partial }{\partial v_2},  \;   \operatorname {so that}\; 2u_2 \frac{\partial }{\partial u_2}=2\frac{u_2 }{u_1}   \frac{\partial }{\partial v_2} $$
Hence $$X=   -\frac{1 }{u_1}   \frac{\partial }{\partial v_1} -\frac{u_2}{u_1}\frac{\partial }{\partial v_2}+2\frac{u_2 }{u_1}   \frac{\partial }{\partial v_2}=  -\frac{1 }{u_1}   \frac{\partial }{\partial v_1} +\frac{u_2}{u_1}\frac{\partial }{\partial v_2}$$
The desired result is thus finally $$X=-v_1\frac {\partial}{\partial v_1} +    v_2\frac {\partial}{\partial v_2}  .                $$
