Is the set of fixed points an algebraic variety? Let $V$ be a finite dimensional $\mathbb{C}-$vector space. The linear action of its automorphism group $GL(V)$ on $V$ induces an action on the projective space $\mathbb{P}(V)$,  i.e. 
$$
GL(V) \times \mathbb{P}(V) \to \mathbb{P}(V), \ (A,[v] ) \mapsto [Av].
$$
I am trying to understand whether for any $A \in GL(V)$ the space of fixed points $Fix(A):=\{[v]\in \mathbb{P}(V): A.[v]=[v]\}$ is an algebraic variety, i.e. $Fix(A)$ is the zero set of one or several polynomials. 
Obviously the condition on fixed points is 
$$
A.[v]=[v] \Rightarrow [Av]=[v] \Rightarrow Av = \lambda_v v \Rightarrow (A-\lambda_vI)v=0
$$
for some $\lambda_v \in \mathbb{C}^*$ that depends on $[v]$. I am not sure if I can conclude anything from this. Any idea on how to proceed further? 

Let $A\in GL(V)$. Then $A$ has a finite number of eigenvalues, $\lambda_1,\ldots,\lambda_n$. For each of the $\lambda$'s, the equation you wrote is polynomial (of degree 1). Hence, the answer is yes.
As a matter of fact, we know what such a fixed locus looks like. A fixed point in $\mathbb{P}(V)$ corresponds to a $1$-dimensional eigenspace in $V$. So for example, if all eigenvalues are distinct, then all the eigenspaces are $1$-dimensional and the fixed locus in $\mathbb{P}(V)$ is discrete. If there is a $2$-dimensional eigenspace in $V$, then the fixed locus in $\mathbb{P}(V)$ contains a line, etc. In conclusion, the fixed locus is the union of linear subspaces of $\mathbb{P}(V)$.
