Eigenvector basis I have this matrix:
$$
\begin{pmatrix}
3 &  \sqrt{2} \\
\sqrt{2}  &  2 \\
\end{pmatrix}
$$
I found the eigenvalues ( that are $1$ and $4$) but I find it really difficult to find the bases... ($(cI-A)x=0$ )
I think maybe one is $(1,1)$ but I'm not sure..

Indeed, if you solve $(A-I)x = 0$, i.e.
$$ \begin{pmatrix} 2 & \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} x = 0$$
you find that $x = (1, -\sqrt{2})^{T}$ is an eigenvector.
For $(A - 4I)x = 0$, i.e.
$$ \begin{pmatrix} -1 & \sqrt{2} \\ \sqrt{2} & -2 \end{pmatrix} x = 0$$
you find that  $x = (\sqrt{2}, 1)^{T}$ is another eigenvector.
