Can any eigenvalue be matched with several eigenvectors? Is there any case where there is an eigenvalue of matrix $A$ that have more than one eigenvector? (So one eigenvalue can be matched with some different eigenvectors.)
Edit: what about matrices other than zero and identity matrix?
 A: Yes, if you take $A$ to be (for example) a $3$ dimensional identity matrix, all vectors are eigenvectors with eigenvalue $1$
For the edit, yes, it is still possible.  If you (can) diagonalize the matrix and find two elements the same, you will have a repeated eigenvalue and there will be a two dimensional space of vectors with that eigenvalue.  For example, $$\begin {pmatrix} 2&0&0\\0&2&0\\0&0&1 \end {pmatrix}$$
has all vectors in the $xy$ plane as eigenvectors with eigenvalue $2$ (plus the unit vector along $z$ with eigenvalue $1$).  Now you can apply your favorite similarity transformation to make it no longer diagonal.
A: You should note that when talking about eigenvectors, it is more proper to rather talk about eigenspaces.
Because if $Av=\lambda v$, then $A(\alpha v)=\lambda (\alpha v)$ for any nonzero $\alpha\in\mathbb R$: every time you have an eigenvector for some eigenvalue $\lambda$, there is uncountably many of them. 
But it is more interesting to think about eigenspaces. An eigenspace for some eigenvalue can have dimension greater than $1$ (as in Ross' example). 
If you think of your matrix in its Jordan form, then different Jordan blocks with the same eigenvalue (as in Ross' example) are usually considered to  correspond to different eigenspaces. 
