<s>[INST] 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?
 [/INST] It is not clear that your answer disagrees with the source you cite.
\begin{align}
& \left[ \begin{array}{c} 1 \\ 1- \sqrt 3 \end{array} \right] = \frac 1 {1+\sqrt3} \left[ \begin{array}{c} (1+\sqrt3\,)\cdot1 \\[4pt] (1+\sqrt3\,) (1- \sqrt 3\,) \end{array} \right] \\[15pt]
= {} & \frac 1 {1+\sqrt3} \left[\begin{array}{c} 1+ \sqrt 3 \\[4pt] -2 \end{array} \right] = \frac {-2} {1+\sqrt3} \left[\begin{array}{c} (-1- \sqrt 3\,)/2 \\[4pt] 1 \end{array} \right]
\end{align}
</s>