Given a matrix $A$ and its eigenvalue $\lambda$ and its eigenvector $x$.
Problem:
From the definition that $Ax=\lambda x$, can I say that eigenvectors $x$ must be all in $A$'s column space and thus say that the rank of matrix $A$ must be not less than the number of $A$'s independent eigenvectors?
If $\lambda \ne 0$ then $x=\lambda^{-1}Ax$ is certainly in the column space. Otherwise it is in the null space. So, no the eigenvector needn't be in the column space: e.g.
\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} has eigenvector \begin{pmatrix} 0 \\ 1 \end{pmatrix} with eigenvalue zero but it is not in the column space.
Since we have an independent eigenvector \begin{pmatrix} 1 \\ 0 \end{pmatrix} but the rank of the matrix is 1, we also contradict the other part.