let be a matrix $\textbf{A}$
$$ \textbf{A} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & \frac{x\alpha}{2H} \\ 0 & \frac{x\alpha}{2H} & 0 \\ \end{bmatrix} $$
The eigenvalues of $\textbf{A}$ are given by
$$ \det(\textbf{A} - \lambda\mathbb{I}) = \det \begin{bmatrix} -\lambda & 0 & 0 \\ 0 & -\lambda & \frac{x\alpha}{2H} \\ 0 & \frac{x\alpha}{2H} & -\lambda \\ \end{bmatrix} = -\lambda(\lambda^2-(\frac{x\alpha}{2H})^2)=0 $$ hence we find:
$$\lambda_{1}=0 \quad \lambda_{2,3} = \pm \frac{x\alpha}{2H} $$
In order to find the eigenvectors, we have to find the solution to
$$ \begin{bmatrix} -\lambda & 0 & 0 \\ 0 & -\lambda & \frac{x\alpha}{2H} \\ 0 & \frac{x\alpha}{2H} & -\lambda \\ \end{bmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $$
for each of the three values of $\lambda$.
Thus we find the second eigenvalues $$\vec{e}_2 = \begin{pmatrix}0 \\ 1 \\1\end{pmatrix} $$ and the third eigenvalue $$\vec{e}_3 = \begin{pmatrix}0 \\ 1 \\-1\end{pmatrix} $$
Here is my question:
In the solution of the exercice I have, I'm given the first eigenvector as
$$\vec{e}_1 = \begin{pmatrix}1 \\ 0 \\0\end{pmatrix} $$
However, I would tend to say that the solution $\vec{e}_1$ with eigenvalue $\lambda_1 = 0$ is given by
$$\vec{e}_1 = \begin{pmatrix}0 \\ 0 \\0\end{pmatrix} $$
What am I missing?
You are missing the fact that the null vector is never an eigenvector. An eigenvector corresponding to an eigenvalue $\lambda$ is a non-null vector $v$ such that $A.v=\lambda v$.