Finding eigenvectors with square root eigenvalues I have a matrix
$$\begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$$
Its eigenvalues are $0$, $\sqrt{5}$ and $-\sqrt{5}$
(These are checked in MATLAB to be correct). I have found its eigenvectors for $0$, and also don't seem to have a problem in similar situations when the eigenvalues are not "root". But when I have root, and are not allowed to turn into descimal-numbers, I have no clue how to proceed.
Does anyone know how to find eigenvectors when the values are $\sqrt{x}$? I'd be happy to watch a youtube video, or an example of how to proceed.

If you want results to compare against, you should get for $A = \begin{bmatrix}1 &-1 &2\\2 &-2 &4\\0 &1 &1\end{bmatrix}$, the following characteristic polynomial:
$$5 \lambda - \lambda^3 = 0$$
This CP yields the following eigenvectors and eigenvalues:
$\displaystyle \lambda_1 = -\sqrt{5}, v_1 = \left(\frac{1}{2} (-1-\sqrt{5}), -1-\sqrt{5}, 1\right)$
$\displaystyle \lambda_2 = \sqrt{5}, v_2 = \left(\frac{1}{2} (-1+\sqrt{5}), -1+\sqrt{5}, 1\right)$
$\lambda_3 = 0, v_3 = (-3, -1, 1)$
Update
To find the eigenvectors, for each eigenvalue, we solve:
$$|A -\lambda_i I|v_i = 0$$
For $\lambda_3 = 0$, we have:
$$|A -\lambda_3 I|v_3 = \begin{bmatrix} 1 & -1 & 2 \\2 & -2 & 4\\0 & 1 &1\end{bmatrix}v_3 = 0$$
The RREF is:
$$\begin{bmatrix}1 & 0 & 3 \\0 & 1 & 1\\0 & 0 & 0\end{bmatrix}v_3 = 0$$
Here, we have:
$a + 3 c = 0$, and
$ b + c = 0 \rightarrow b = -c, \text{so pick}~~ c = 1 \rightarrow b = -1$
So, $a = -3c \rightarrow a = -3$
Thus for $\lambda_3 = 0$, we have $v_3 = (-3, -1, 1)$
For $\lambda_2 = \sqrt{5}$, we have:
$$|A -\lambda_2 I|v_2 = \begin{bmatrix} 1-\sqrt{5} & -1 & 2 \\2 & -2-\sqrt{5} & 4\\0 & 1 &1-\sqrt{5}\end{bmatrix}v_2 = 0$$
The RREF is:
$$\begin{bmatrix}1 & 0 & \frac{1}{2}(1-\sqrt{5} \\0 & 1 & 1-\sqrt{5}\\0 & 0 & 0\end{bmatrix}v_2 = 0$$
So, we have:
$a + \frac{1}{2}(1-\sqrt{5})c = 0$, and
$b + (1-\sqrt{5})c = 0$
So, we can choose: $c = 1 \rightarrow b = (-1+\sqrt{5})$
This leads to: $a = \frac{1}{2}(-1 + \sqrt{5})$
So, for $\lambda_2 = \sqrt{5}$, we get $v_2 = \left(\frac{1}{2} (-1+\sqrt{5}), -1+\sqrt{5}, 1\right)$.
Can you repeat this process for the last eigenvalue? If not give a yell and I will add details.
