Diagonalise,
$\displaystyle A$ = $\displaystyle \left[ \begin{array}{cccc} -2 & -1 & 0 \\ -1 & 0 & 1 \\ 0 & 1 & 2 \end{array}\right]$
That is find matrices P and D, where D is diagonal such that
P-1 AP = D
Hint1: D is the matrix with eigenvalues in the diagonal.
Hint2: P is the 3 x 3 matrix formed by writing the corresponding eigenvectors as columns.
Reference: Diagonalizable matrix - Wikipedia, the free encyclopedia
Hm, for $\displaystyle \sqrt{6}$ for example :
$\displaystyle -2x-y=\sqrt{6} x$
$\displaystyle -x+z=\sqrt{6} y$
The third one will just yield the first one
From the first equation, we get :
$\displaystyle y=x(-\sqrt{6}-2)$
Substituting in the second one :
$\displaystyle -x+z=6x+2 \sqrt{6} x$
--> $\displaystyle z=x(2 \sqrt{6}+7)$
Taking $\displaystyle x=1$, $\displaystyle y=-\sqrt{6}-2$ and $\displaystyle z=2 \sqrt{6}+7$
The eigenvector associated to $\displaystyle \sqrt{6}$ is $\displaystyle \begin{pmatrix} 1 \\ -\sqrt{6}-2 \\ 2 \sqrt{6}+7 \end{pmatrix}$