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Find the eigenvectors of the matrix

$$\displaystyle\begin{bmatrix} 0 &2 &3 \\ -2 &0 &5 \\ -3 &-5 &0 \end{bmatrix}.$$

So I start with $|A-\lambda I|=0$

$$\begin{bmatrix} -\lambda &2 &3 \\ -2 &-\lambda &5 \\ -3 &-5 &-\lambda \end{bmatrix}=0$$

$$-\lambda(\lambda^2+25)-2(2\lambda+15)+3(10-3\lambda)=0 $$

$$\lambda^3+38\lambda=0$$

The given eigenvalue is -2, which is later used for finding the eigenvector. My equation won't give -2. This is also a part of a multiquestion series, so I cannot think that the question has something wrong in it.

What did I do wrong here ?

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    $\begingroup$ Well, the given eigenvalue is wrong for that matrix, I'm getting $0, i\sqrt{38}, -i\sqrt{38}$ as the eigenvalues. $\endgroup$ – Silynn Aug 25 '14 at 9:47
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Wolfram alpha confirms that $\lambda = -2$ is not an eigenvalue and your characteristic polynomial seems correct. Are you sure it is written correctly?

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Your characteristic polynomial is right, it is $\lambda^3 + 38 \lambda = 0$. But $-2$ is not a root of it, since $(-2)^3 + 38 \cdot (-2) = -84$. If you transform the equation like this $\lambda^3 + 38 \lambda = \lambda ( \lambda^2+38)=0$, then you find that $\lambda_1 = 0$ is a real eigenvalue, and there are two complex $\lambda_{2,3} = \pm i \sqrt{38}$. I let you the rest to solve.

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You have not done anything wrong. I have checked your answer with gp (available from http://pari.math.u-bordeaux.fr/). I am awful in numerical calculations! Check whether you have copied the matrix correctly.

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