Eigenvectors of $\begin{pmatrix}2&-2\\ -4&-2\end{pmatrix}$ - What am I doing wrong? Eigenvalues are $2\sqrt{3}$ and $-2\sqrt{3}$, I'll calculate the eigenvector for $2\sqrt{3}$ here
We've got:
$\begin{pmatrix}2-2\sqrt{3}&-2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I multiply the first row by negative 1, I get:
$\begin{pmatrix}-2+2\sqrt{3}&2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I add the first and second row together, I get:
$\left(-6+2\sqrt{3}\right)y_1-2\sqrt{3}y_2=0$
So the solution would be
$\left(-\sqrt{3}+1\right)y_1=y_2$
The eigenvector is $\begin{pmatrix}y\\ \left(1-\sqrt{3}\right)y\end{pmatrix}$
But that's apparently wrong according to this calculator
What's wrong here?

You did nothing wrong. What you got were that the eigenvectors corresponding to the eigenvalue $2\sqrt3$ are those of the form $\left(1,1-\sqrt3\right)^Ty$, with $y\ne0$. That calculator got the vectors of the form $\left(-\frac{\sqrt3+1}2,1\right)^Ty$ with $y\ne0$. But$$\left(-\frac{\sqrt3+1}2\right)\begin{pmatrix}1\\1-\sqrt3\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt3+1}2\\1\end{pmatrix}.$$So, it's the same answer.
