0
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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?

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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 $A.v=\lambda v$.

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  • $\begingroup$ Thank you @Jose Carlos Santos. So $\begin{pmatrix}0 \\ 0 \\0\end{pmatrix} $ cannot be an eigenvector. Now, if we compute $\lambda_1 = 0$ in $ \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} $, how can we obtain the first eigenvector $\vec{e}_1 = \begin{pmatrix}1 \\ 0 \\0\end{pmatrix}$? $\endgroup$ – ecjb Dec 23 '18 at 22:50
  • $\begingroup$ Solve the system $A.\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$. You'll see that one of the solutions is $x=1$ and $y=z=0$. $\endgroup$ – José Carlos Santos Dec 23 '18 at 22:58

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