if det(A-xI) = x^2 - 6x + 9 must there exist two linearly independent eigenvectors? or can just the one eigenvector for the double root of 3 be enough?
The eigenvectors will be at most two because when you put your eigenvalue back in the $\displaystyle A-\lambda I$ matrix, you will reduce it down to find the numbers of vectors that form a basis for your eigenspace.
Eigenspace = $\displaystyle \left \{ \vec{a} \vert (A-\lambda I)\vec{a}=\vec{0} \right \} $
The matrix $\displaystyle \begin{bmatrix}-3 & 0 \\ 0 & -3\end{bmatrix}$ has -3 as a double eigenvalue and both [1 0] and [0 1] as eigenvectors.
$\displaystyle \begin{bmatrix}-3 & 1 \\ 0 & -3\end{bmatrix}$, on the other hand, has -3 as a double eigenvalue and only [1 0] as an eigenvector.
By the way, you shouldn't say a matrix "has one eigenvector" or "has two eigenvectors". Every matrix has an infinite number of eigenvectors corresponding to each eigenvalue. The question here is whether there are one or two independent eigenvectors.