Let $A$ be a square matrix with eigenvalue $\lambda$. Let $\mathbf{c}$ be such that $\mathbf{v} := (A - \lambda I) \mathbf{c}$ is an eigenvector, but $\mathbf{c}$ is not an eigenvector. That is, $\mathbf{c}$ is a generalized eigenvector of order 2.
Let $i$ be a component where $\mathbf{v}_i = 0$, which implies $(A \mathbf{c})_i = \lambda \mathbf{c}_i$. Therefore, we can say that $\mathbf{c}$ "behaves like an eigenvector" outside of the support of $\mathbf{v}$.
Let $\mathbf{c'}$ equal $\mathbf{c}$ outside the support of $\mathbf{v}$, and zero otherwise. Is $\mathbf{c'}$ an eigenvector?. This sounds like just a rephrasing of what I said above, but when trying to find a more concrete proof, I become stuck.
Any help is appreciated. Thanks!
The answer is no. For example, in the matrix $$ A = \pmatrix{0&1\\0&0}, $$ we have eigenvector $v = (1,0)$ and generalized eigenvector $c = (0,1)$. It is clear that $c' = c$ is not an eigenvector.