problem arriving at 0 eigenvector

Question

I am trying to find the eigenvectors and eigenvalues for my covariance matrix: \begin{bmatrix}2&\sqrt2/2\\\sqrt2/2&1\\\end{bmatrix}

I found the eigenvalues:

$\frac{3+\sqrt3}2$

$\frac{3-\sqrt3}2$

Now, my problem is trying to find the eigenvectors for $\frac{3+\sqrt3}2$ , I arrive to:

$\frac{1-\sqrt3}2X_1=\frac{-\sqrt2}2X_2$

$\frac{\sqrt2}2X_1=\frac{1+\sqrt3}2X_2$

And the above takes me to an eigenvector (0,0); which does not make much sense...

I do not know where I made my mistake? I checked my math several times... Could it be something wrong in my covariance matrix or the way I calculated the eigenvalues?

Answer 1

As you indicated, the eigenvectors for eigenvalue $\dfrac{3+\sqrt3}2$ are non-zero vectors $(X_1,X_2),$

where $\dfrac{\sqrt2}2X_1=\dfrac{1+\sqrt3}2X_2$; i.e., $X_1=\dfrac{1+\sqrt3}{\sqrt2}X_2.$

Thus, the eigenvectors are non-zero multiples of $\left(\dfrac{\sqrt2+\sqrt6}2,1\right)$;

that is, vectors $\left(\dfrac{(\sqrt2+\sqrt6)c}2,c\right)$ with $c\ne0$.