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So I do know how to compute the eigenvalues of a matrix. At least, that's what I thought. I got the matrix

A = \begin{bmatrix}1&-2&0\\-2&0&2\\0&2&-1\end{bmatrix}

My approach is by finding the determinant: \begin{equation} \text{det}\left(A-\lambda In\right) = 0 \end{equation}

so it becomes $$ \text{det}\left( \begin{bmatrix}1-\lambda&-2&0\\-2&0&2\\0&2&-1-\lambda\end{bmatrix} \right) = 0. $$ But this gave me 8 lambda

According to calculators, it should give me -3, 3 and 0 What did I miss?

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    $\begingroup$ In position 2,2 there should be $-\lambda$ and not $0$. $\endgroup$ – Angelo Rendina Mar 22 '15 at 20:31
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    $\begingroup$ I knew it was a stupid mistake. Thanks... $\endgroup$ – Seb Mar 22 '15 at 20:33
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Probably You have some othe mistake in the calculus of determinat.

Hint:

$$ \text{det}\left( \begin{bmatrix}1-\lambda&-2&0\\-2&-\lambda&2\\0&2&-1-\lambda\end{bmatrix} \right) = $$ $$ (1-\lambda) \text{det}\left( \begin{bmatrix}-\lambda&2\\2&-1-\lambda\end{bmatrix} \right) +2 \text{det}\left( \begin{bmatrix}-2&2\\0&-1-\lambda\end{bmatrix} \right)= $$ $$ =(1-\lambda)(\lambda + \lambda^2-4)+2(2+\lambda) $$ ......

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