Why an eigenspace is a linear subspace, if the zero vector is not an eigenvector? I've started studying Eigenvector and Eigenvalue. 
It says in my book that 0 is excluded from being an eigenvector because it breaks the uniqueness of eigenvalue associated with each eigenvector. 
But, there is a proof in my book showing that Eigenspace is a subspace. In order for it to be subspace, does that mean that there must be a zero element? But, Eigenvector can not be zero... Am I misunderstanding something? 
 A: The eigenspace associated with an eigenvalue consists of all the eigenvectors (which by definition are not the zero vector) associated with that eigenvalue along with the zero vector.  
If we allowed the zero vector to be an eigenvector, then every scalar would be an eigenvalue, which would not be desirable.
A: You have two solutions to this. 
Either you call a non-zero $v \in V$ (your vector space) an eigenvector of $A : V \to V$ if and only if there exists $\lambda$ such that $Av = \lambda v$, in which case you say $\lambda$ is an eigenvalue of $A$ associated to $v$.
Or 
You call $\lambda$ an eigenvalue of $A$ if $\dim \ker(A - \lambda \mathrm{id}) > 0$, and you define eigenvectors associated to the eigenvalue $\lambda$ as non-zero elements of $\ker(A-\lambda \mathrm{id}_V)$.
In both cases you have to exclude $0$ as an eigenvector in some way. Also in both cases, the eigenspace is defined as $\ker(A- \lambda \mathrm{id}_V)$, so of course $0$ is naturally included in it. 
Hope that helps,
A: Here's another definition of eigenvalue (for matrices) that should shed light on your question.
Let $\mathbb{F}$ be a field. Fix $A \in \mathsf{M}_n(\mathbb{F})$ and $\lambda \in \mathbb{F}$. It is a very simple excersie to show that $$\mathsf{E}_\lambda := \{ v \in \mathbb{F}^n \mid Av = \lambda v \}$$ is a subspace of $\mathbb{F}^n$.
Now, in all but finitely many cases, $\mathsf{E}_\lambda$ will be trivial, i.e., $\mathsf{E}_\lambda = \{ \mathbf{0}\}$. If $\mathsf{E}_\lambda$ is nontrivial (i.e., $\dim\mathsf{E}_\lambda > 0$), then $\lambda$ is called an eigenvalue (of $A$). Notice that, with this definition, $\exists v \ne \mathbf{0}$ such that $Av = \lambda v$ (the vector $v$ is called an eigenvector of $A$ corresponding to $\lambda$). Thus, $\mathbf{0}$ is not an eigenvector and its exclusion does not preclude $\mathsf{E}_\lambda$ from being a subspace.
Conversely, if there is a nonzero vector $v$ such that $Av = \lambda v$, then $\mathsf{E}_\lambda$ is nontrivial. Thus, this definition is equivalent to the standard definition.
In summary, these sets are always subspaces, but are of interest when they are nontrivial.
