Eigenvalues and corresponding eigenspace I know that $λ$ is an eigenvalue of a square matrix $A$ 
$\iff \exists X \ne0: AX=λX$
$\iff λ$ is a root of the characteristic polynomial of $A$.
Given a matrix $A\in \Bbb F^{n\times n}$, we find its characteristic polynomial and so its eigenvalues.
So by continuing to find corresponding eigenvectors of an eigenvalue $λ$, we discover that it corresponds only to the zero vector.
So now we should stop calling it an eigenvalue?

As I explained before, if it is a root of the characteristic polynomial, then there exists an eigenvector; it does not matter if you can find it or not. However, if you can prove that there is no eigenvector, then it cannot be a root of characteristic polynomial.
To elevate the confusion, let me hint at a proof.
Suppose $\lambda$ is a root of $\det(A- x I)$. This means that the matrix $(A - \lambda I)$ is not invertible (because a matrix is invertible iff its determinant is non-zero).
Now is a matrix $B$ is non-invertible, there exists a non-zero vector $X$ such that $BX = 0$. Can you prove this?
Now, since $(A - \lambda I)$ is not invertible, there exists a non-zero $X$ such that $(A-\lambda I)X = 0 \implies AX = \lambda X$ whence $X$ is an eigenvector for $\lambda$.
