Projective equivalence Definition
Two projective plane curves $F$ and $G$ are projectively equivalent if there is a $\varphi_A\in PGL_2(k)$ such that
$F(x,y,z)=G(a_{00}x+a_{01}y+a_{02}z,a_{10}x+a_{11}y+a_{12}z, a_{20}x+a_{21}y+a_{22}z)$, where $A=(a_{ij})\in GL_3(k)$.
I didn't understand why $Z(G)=\varphi_A(Z(F))$. Someone told me that this is very easy to show, but I couldn't prove it.
I would like to know also where can I see more about projective/affine equivalences and affine/projective transformations, I don't know why, I didn't see these stuff in classical books such as Fulton.
Thanks in advance
 A: It seems you are already aware, but just to be explicit, in considering actions on projective space we may lift a matrix $M \in \operatorname{PGL}_2(k)$ to (by abuse of notation) $M \in \operatorname{GL}_3(k)$, as any such lift has the same action on $\mathbb{P}^2_k$. As such, all matrices below are in $\operatorname{GL}_3(k)$.
Let $M$ be a matrix representing $\varphi_A$. For $p = [x : y : z] \in \mathbb{P}^2_k$, the condition of projective equivalence can be expressed by $F(p) = G(M \cdot p)$.
To prove the equality $Z(G) = \varphi_A(Z(F))$, we simply show the two containments. To show $Z(G) \subset \varphi_A(Z(F))$, start out with $p = [x : y : z] \in \mathbb{P}^2_k$ such that $G(p) = 0$. We wish to show that there is a $q$ with $F(q) = 0$ such that $p = M \cdot q$. Since $M$ is invertible, we know what $q$ must be: $q = M^{-1} \cdot p$. It remains to verify that $F(q) = 0$. Indeed:
$$F(q) = F(M^{-1} \cdot p) = G(M \cdot M^{-1} \cdot p) = G(p) = 0.$$
It remains to show the other inclusion. This is easier, try it out yourself.
