TITLE: Why an eigenspace is a linear subspace, if the zero vector is not an eigenvector?
QUESTION [11 upvotes]: 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?

REPLY [8 votes]: 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,