Is the zero vector not being a possible eigenvector just an arbitrary decision? I understand that if 0 is allowed to be an eigenvector, then it would be an eigenvector of every matrix (corresponding to some λ).  
But why is that a bad thing?  Does allowing this cause any contradiction? Or this is really just some convention for convenience? In that case, what is this convenience?  
Thanks.
 A: [Made my comment an answer]
It's just not useful. Eigenvectors tell you something about the operator (many times, they tell you everything). The zero vector tells you absolutely nothing about it - not even an eigenvalue, since $L0=\lambda 0$ for any $\lambda$ and any $L$.
A: Arguably the fundamental concept is that of eigenspaces, not eigenvectors. The $\lambda$-eigenspace is the subspace of all vectors $v$ such that $Av = \lambda v$. You can give it a basis consisting of $\lambda$-eigenvectors (and note that a vector in a basis must be nonzero) but this is an extra choice that you may or may not want to make in some situations. 
The zero vector is a member of every eigenspace, including the ones that have no other vectors in them, so as Nick says it doesn't tell you anything useful. In nice cases, the vector space $V$ on which $A$ acts decomposes as a sum of eigenspaces; this really tells you something useful. 
A: $\{\mathbf{0}\}$ is not linearly independent (by itself or with any other set of vectors), so we shouldn't count it in any bases. It spans only a zero-dimensional subspace ($\{\mathbf{0}\}$ is essentially the identity of the sum operation on subspaces). This is, if you will, rather like not counting $1$ as a prime number so that unique prime factorisation has a simple statement.
There's another problem, of course: what eigenvalue does it have? $0\mathbf{0} = 1 \mathbf{0}$, so then $1$'s an eigenvalue, $0$'s an eigenvalue... &c., which is not very helpful.
A: Because it would be an eigenvector corresponding to EVERY $\lambda$, so the notion of eigenvalue would be quite useless if every matrix could have any possible $\lambda$ as eigenvalue.
