# Find eigenvalue of a matrix that have all diagonal elements are zeros

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?

• The middle entry should also be $-\lambda$. – angryavian Jun 25 '18 at 21:36
• 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." – zwim Jun 25 '18 at 23:11

Compute the characteristic polynomial of the matrix: $$\begin{vmatrix}-\lambda &0&2 \\0 &-\lambda & 0 \\ 2& 0 &-\lambda \end{vmatrix}=-\lambda^3+4\lambda=\lambda(4-\lambda^2).$$
$\begin{vmatrix} - \lambda &0&2\\0&-\lambda&0\\2&0&-\lambda\end{vmatrix} = 0$ is what you want.
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 $-2$.