Do eigenvectors always form a basis?
Solution 1:
No, of course not. For example, $\begin{pmatrix} 0 &1 \\ 0 &0\end{pmatrix}$ has $0$ as its only eigenvalue, with eigenspace $\begin{pmatrix} x \\ 0 \end{pmatrix}$. Thus there are not enough independent eigenvectors to form a basis.