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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?

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    Yes. Eigenvalues are the scalars $\lambda\in\Bbb F$ such that $\dim\ker(A-\lambda I)>0$. But if you found a root of $\det(A-tI)$ which does not satisfy that condition you've probably made some mistake along the way. – 2017-02-27
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    What do you mean? If it wasn't an eigenvalue, why were you calling it one in the first place? – 2017-02-27
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    I called $λ$ an eigenvalue because I found $λ$ as a root of $c_A(x)$ – 2017-02-27
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    @Michael If that's the case, you've made a mistake somewhere. – 2017-02-27

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It will not happen. If $\lambda $ is a root of the characteristic polynomial, then $\det (A-\lambda I)=0$. This tells you that $A-\lambda I $ is not invertible, and so it cannot be injective: an eigenvector has to exist.

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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 $\lambda$.

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    ok so it was impossible for this to happen then – 2017-02-27
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    Exactly. That is what everyone has been saying. – 2017-02-27