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.
 A: The method to find an eigenvector is the same, regardless of what the eigenvalue is.  The only difference between different eigenvalues is the appearance of the computation involved.  For example, you will need to add, multiply, divide by real numbers like $(1-\sqrt{5})$.
More details:
To compute an eigenvector, you need to solve the system $Av=\lambda v$.  For $\lambda=\sqrt{5}$, you need to solve the system $\left[\begin{smallmatrix}1 & -1 & 2\\2 & -2 & 4\\ 0 & 1 & 1\end{smallmatrix}\right]\left[\begin{smallmatrix}x \\ y \\ z\end{smallmatrix}\right]=\left[\begin{smallmatrix}\sqrt{5} x\\ \sqrt{5} y \\ \sqrt{5} z\end{smallmatrix}\right]$, which becomes the system:
$(1-\sqrt{5})x -y +2z=0\\ 2+(-2-\sqrt{5})y + 4z=0\\ 0x+y+(1-\sqrt{5})z=0$
A: 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.
A: Calling your matrix $A$, the eigenvalue $\lambda$ and the eigenvector $y$, you use the fact that $(A-\lambda I)y=0$  So $$\begin{bmatrix}1-\sqrt 5 &-1 &2\\2 &-2-\sqrt 5 &4\\0 &1 &1-\sqrt 5\end{bmatrix}\begin{bmatrix}y_1\\y_2\\y_3 \end{bmatrix}=\begin{bmatrix}0\\0\\0 \end{bmatrix}$$ and read off three equations in three unknowns, just like you did for $\lambda=0$  The equations will be dependent, so there will be a one-parameter family. I'll use the first and third for convenience.
$$(1-\sqrt 5)y_1-y_2+2y_3=0\\y_2+(1-\sqrt 5)y_3=0\\y_2=(\sqrt 5-1)y_3\\(1-\sqrt 5)y_1+(3-\sqrt 5)y_3=0\\y_1=\frac {3-\sqrt 5}{\sqrt 5-1}y_3=\frac {(3-\sqrt 5)(\sqrt 5+1)}{(\sqrt 5-1)(\sqrt 5+1)}y_3=\frac {2\sqrt 5-2}4y_3$$ So your eigenvector is any multiple of $$\begin{bmatrix}\frac {2\sqrt 5-2}4\\\sqrt 5-1\\1 \end{bmatrix}$$
