Eigenvalues and Eigenvectors of a 3 by 3 matrix I want to find eigenvalues and eigenvectors of this matrix: 
$$
    \begin{bmatrix}
    2 & 3 & 0 \\
    3 & 6 & 1 \\
    0 & 1 & 6 \\
    \end{bmatrix}
$$
So eigenvalues:
$$
    \begin{vmatrix}
    2-\lambda_1 & 3 & 0 \\
    3 & 6-\lambda_2 & 1 \\
    0 & 1 & 6-\lambda_3 \\
    \end{vmatrix}
$$
then
$$-\lambda^3+14\lambda^2-50\lambda+16=-(8-\lambda)(\lambda^2-6\lambda+2)$$
and the eigen values are: $\lambda_1=3-\sqrt{7}, \lambda_2=3+\sqrt{7}, \lambda_3=8$
So then I want to find eigenvectors for eigenvalue $\lambda_1=3-\sqrt{7}$
I know that:
$$$$
\begin{bmatrix}
2-(3-\sqrt{7}) & 3 & 0 \\
3 & 6-(3-\sqrt{7}) & 1 \\
0 & 1 & 6-(3-\sqrt{7}) \\
\end{bmatrix}
$$$$
and that matrix I multiply with matrix:
$$$$
\begin{bmatrix}
 x \\
y \\
z\\
\end{bmatrix}
$$$$
and this gives me:
$$$$
\begin{bmatrix}
(\sqrt{7}-1)x+3y\\
3x+(\sqrt{7}+3)y+z \\
y+(\sqrt{7}+3)z \\
\end{bmatrix}
$$$$
and then I get three equation:
$$\begin{cases}
(\sqrt{7}-1)x+3y=0\\[2ex]
3x+(\sqrt{7}+3)y+z =0\\[2ex]
y+(\sqrt{7}+3)z=0
\end{cases}$$
Now, every such system will have infinitely many solutions, because if e is an eigenvector, so is any multiple of e. So our strategy will be to try to find the eigenvector with $x=?$
How I decide the $x$ value when the correct answer is:
$x=0.87, y=-0.479, z=0.085$.
Yes I know that when I set the $x=0.87$ then the equalitons gives me the correct $y$ and $z$ values. 

The strategy is that you define one of the variables  $x$, $y$, or $z$ as independent variable. Then two other variables will be dependent on it. Let's choose $z$ to be the independent. In this case your eigenvector corresponding to $\lambda_1=3-\sqrt{7}$ is $(\dfrac{(2+\sqrt{7})(4+\sqrt{7})}{3}z,-(3+\sqrt{7})z,z)$.
There is no one correct answer. Once you fix $z$ you obtain another vector. 
