Let $G$ be an algebraic (zariski closed) subgroup of $SL(n,C)$ for some algebraically closed field $C$. Now $G$ acts on an $n$-dimensional vector space $V$ over $C$ where $V$ is a solution space of a linear homogeneous differential equation of order $n$ and is contained in a field, say, K (Picard-Vessiot extension). Induced by this action we see that $G$ is acting on the projective space $\mathbb{P}(V)$. Let $(v_1,\ldots,v_n)$ be the basis of $V$ such that the basis elements satisfy a non-zero homogeneous polynomial $F(x_1,\ldots,x_n)$ with coefficients in $C$, i.e, $F(v_1,\ldots,v_n)=0$. Now does that imply that $G$ leaves a proper algebraic subset of $\mathbb{P}(V)$ invariant ? If yes, please help me in realising this.
I am stuck at this and could not realise this after some attempts. Please provide a small outline of the proof if this is true. Any help will be appreciated.
It follows from linear algebra that $G$ always acts transitively on $P(V)$. (For simplicity take $V=\mathbb K^n$. Given a one-dimensional subspace in $V$, take a non-zero vector $w_1$ in that space and extend it to a basis of $\mathbb K^n$. Then the matrix having the basis elements as columns sende the line spanned by the first vector in the standard basis to the line spanned by $w_1$.) So you see that there is just one $G$-orbit in $P(V)$ and there are no non-trivial invariant subsets. There is simple relation between $F(v_1,\dots, v_n)$ and $F(Av_1,\dots,Av_n)$, which would allow you to make deductions.
This corresponds to the fact that in $\mathbb R^n$ you can of course choose a basis of vectors of unit length, but this does not say anything about the orbits of $GL(n,\mathbb R)$.