Consider the natural action of $SL_3(\mathbb{C})$ on $\mathbb{P}^2$ via: $$ SL_3(\mathbb{C}) \times \mathbb{P}^2 \to \mathbb{P}^2, \ (A,[v]\mapsto [Av]). $$
It is clear that the kernel of the action is a group of scalar matrices with unit determinant, i.e. Kernel $K=C_3$ is a cyclic group generated by a third root of unity. Obviously, for any element in $K$ every point in $\mathbb{P}^2$ is a fixed point.
I want to prove that every element $A \in SL_3$ has at least 3 fixed points on $\mathbb{P}^2$ which don't lie on a projective line. To prove the statement, I just tried to find the points $[v]\in \mathbb{P}^2$ such that $$ [Av]=[v]. $$
This means that I need to find $v\in \mathbb{C}^3$ such that $Av=\lambda v$ for some $\lambda \in \mathbb{C}^*.$ The first requirement for not being on a projective line is that the vectors $v$ have to be linearly independent. But there can only be three linearly independent vectors in $\mathbb{C}^3$. Thus, I think I need to show that $$ \dim_{\mathbb{C}} Kernel(A-\lambda I) = 3 $$
where $I$ denotes the $3\times 3$ identity matrix. This means every vector in $\mathbb{C}^3$ is in the kernel. But I don't seem to reach to the conclusion. Is the claim in the bold letters correct? How can I prove it? Please help.
Theorem. Considering ${\rm GL}(V)$'s induced action on ${\Bbb P}(V)$, for any $A\in{\rm GL}(V)$ we have
$${\Bbb P}(V)^A=\bigsqcup_{\lambda\ne0}\Bbb P(V_\lambda). \tag{$\circ$}$$
Here $V$ is a finite-dimensional complex vector space and $V_\lambda$ is the $\lambda$-eigenspace for $A$ (the space of all eigenvectors of $A$ with eigenvalue $\lambda$). Note that $V=\bigoplus V_\lambda$ does not always hold true; instead it holds true with generalized eigenspaces. The equality is equivalent to $A$ being diagonalizable. Note further that different eigenspaces have different eigenvalues, so it doesn't make sense to analyze an operator $A-\lambda\,{\rm Id}_V$ for just one value of $\lambda$ (indeed if that operator were to have kernel of dimension $\dim V$, it would be the zero map and $A=\lambda\,{\rm Id}_V$).
If $[v]\in{\Bbb P}(V)$ is a fixed point under $A$ then $[v]=A[v]=[Av]$ implies $Av=\lambda v$ for some nonzero scalar $\lambda$. Thus, fixed points correspond to eigenvectors (modulo scalars) and thus the decomposition given in $(\circ)$ follows. Note this implies $\Bbb P(V)^A\subseteq{\Bbb P}(\bigoplus_{\lambda\ne0}V_\lambda)$, and thus:
Corollary. There are $\dim V$ points in $\Bbb P(V)^A$ that are not in ${\Bbb P}(W)$ for any $W$ a proper subspace, if and only if $A$ is diagonalizable, in which case such sets of points are precisely those represented by a linearly independent set of eigenvectors.
If $A$ is diagonalizable and $\dim(V_\lambda)>1$ for some $\lambda$ (equivalently, if $A$'s characteristic polynomial has a repeated root) then $|{\Bbb P}(V_\lambda)|$ will be uncountable, so it will be possible to choose countably many fixed points contained in this proper projective subspace too.