more than n eigenvectors I am learning diagonalization of matrices. We are given the following theorem:

If $A$ is an $n\times n $ matrix with $n$ distinct eigenvalues, then $a$ is diagonalizable 

Now the proof is:

let $v_1,v_2,\dots,v_n$ be eigenvectors corresponding to the $n$ distinct eigenvalues of $a$. (Why could there not me more than $n$ such eigenvectors?) $v_1, v_2,\ldots,v_n$ are linearly independent, so by (a specific theorem) A is diagonalizable.

Now the question posed in the proof has gotten my thinking, why cant there be more than n eigenvectors correspdonign to n distinct eigenvalues? because an eigenvalue can correspond to only  one eigenvalues? because they are distinct? 
 A: Who said there can't be more than $n$ eigenvectors? Nothing in the passage you quoted depends on there being exactly $n$ eigenvectors; all that's required is that there are (at least) $n$ eigenvectors. 
Having said that, eigenvectors belonging to distinct eigenvalues are linearly independent. If you have more than $n$ eigenvectors, then any two belonging to the same eigenvalue will be linearly dependent, and so of no use for diagonalization. 
A: Why could there not be more than n INDEPENDENT eigenvectors for a n×n matrix ? 
A: this is late, and the answers given here are great, but based on the poster's last comment, and the comment left by epoweritheta, I think there is still some misunderstanding here.
It is true that you can get more than one eigenvector for a single eigenvalue: consider the matrix $$A = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 2 & 4 \end{pmatrix}$$
However, you can have at most $n$ linearly independent eigenvectors for a given matrix. There are many ways to see why this has to be so, but here's one: An $n \times n $ represents a linear operator, that is, it is a linear transformation from an $n$-dimensional vector space to itself. Now, in an $n$-dimensional vector space, any set of vectors having more than $n$ vectors is DEPENDENT. To see this, you could set up the linear independence equation, $$c_1\vec{v_1} + c_2 \vec{v_2} + \cdots + c_n \vec{v_n} + \cdots + c_{n+m} \vec{v_{n+m}} = \vec{0}$$ You will find that when you augment the matrix for that system, you will get an $n \times (n+m)$ matrix and hence you will have less pivots than there are columns upon performing Gaussian elimination. This gives rise to a solution that depends on parameters, which means you won't have a unique solution and hence you get dependence. 
So having $n$ eigenvalues, each will have at least one eigenvector (by construction--if you want to know why each eigenvalue will have at least one, you can ask me--the short answer is we find eigenvalues by setting up a homogeneous system whose coefficient matrix has determinant zero, hence by the equivalence theorem, this system is guaranteed to have a non-zero vector-solution, in fact, it will have infinitely many such solutions), but there will be only $n$ such vectors that are linearly independent. And then you can apply the theorem that says having $n$ linearly independent vectors implies being diagonalizable (and vice versa).
