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Find the eigenvalues of the matrix below:

$\begin{pmatrix} 0 & 0 & 2\\ 0 & 0 & 0\\ 2 & 0 & 0\end{pmatrix}$

As usual, I try to solve the equation below: $\begin{vmatrix} - \lambda &0&2\\0&0&0\\2&0&-\lambda\end{vmatrix} = 0 $

But in this case the equation above give $ 0 = 0 $, and I got stuck.

What should I do?

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    $\begingroup$ The middle entry should also be $-\lambda$. $\endgroup$ – angryavian Jun 25 '18 at 21:36
  • $\begingroup$ Assuming it wasn't just a mistyping, it would be interesting to know the reason why you skipped the central value, or why you though it ought to stay zero ? So we may help correcting the deeper misunderstanding rather than just say "hey, you forget $-\lambda$ in the middle." $\endgroup$ – zwim Jun 25 '18 at 23:11
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Compute the characteristic polynomial of the matrix: $$\begin{vmatrix}-\lambda &0&2 \\0 &-\lambda & 0 \\ 2& 0 &-\lambda \end{vmatrix}=-\lambda^3+4\lambda=\lambda(4-\lambda^2).$$

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$\begin{vmatrix} - \lambda &0&2\\0&-\lambda&0\\2&0&-\lambda\end{vmatrix} = 0 $ is what you want.

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The determinant you have to compute is $$ \det\begin{bmatrix} -\lambda & 0 & 2 \\ 0 & -\lambda & 0 \\ 2 & 0 & -\lambda \end{bmatrix}= -\lambda\det\begin{bmatrix} -\lambda & 2 \\ 2 & -\lambda \end{bmatrix} =-\lambda(\lambda^2-4) $$

Alternatively, you know that $0$ is an eigenvalue because the matrix is not invertible; also $2$ is an eigenvalue, because $$ \begin{bmatrix} 0 & 0 & 2 \\ 0 & 0 & 0 \\ 2 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}= \begin{bmatrix} 2 \\ 0 \\ 2 \end{bmatrix}= 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $$ The sum of the eigenvalues is the trace, which is $0$, so the third eigenvalue is $-2$.

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