TITLE: Do eigenvectors always form a basis?
QUESTION [11 upvotes]: Suppose we have a $n \times n $ matrix  over $\Bbb R$. 
Is it necessary that we should have $n$ linearly independent eigenvectors associated with eigenvalues so that they form a basis? 
Can you give a proof or counterexample?
How about if you have the same question over complex numbers?

REPLY [9 votes]: 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.