Can the zero vector be an eigenvector for a matrix? I was checking over my work on WolfRamAlpha, and it says one of my eigenvalues (this one with multiplicity 2), has an eigenvector of (0,0,0). How can the zero vector be an eigenvector?
 A: Note that some authors allow $0$ to be an eigenvector.  For example, in the book Linear Algebra Done Right (which is very popular), an eigenvector is defined as follows:

Suppose $T \in \mathcal L(V)$ and $\lambda \in \mathbf F$ is an
  eigenvalue of $T$.  A vector $u \in V$ is called an
  eigenvector of $T$ (corresponding to $\lambda$) if $Tu =
 \lambda u$.

The book then states,

...we see that the set of eigenvectors of $T$ corresponding to
  $\lambda$ equals $\text{null}(T - \lambda I)$.  In particular, the set
  of eigenvectors of $T$ corresponding to $\lambda$ is a subspace of
  $V$.

However, an eigenvalue is defined as follows:

a scalar $\lambda \in \mathbf F$ is called an eigenvalue of $T \in
 \mathcal L(V)$ if there exists a nonzero vector $u \in V$ such that
  $Tu = \lambda u$.  We must require $u$ to be nonzero because with $u =
 0$ every scalar $\lambda \in \mathbf F$ satisfies [the equation $Tu =
 \lambda u$].

Hoffman and Kunze, another highly esteemed Linear Algebra book, also allows $0$ to be an eigenvector.  (See the definition of "characteristic vector" in section 6.2, p. 182.)
A: As others have written, eigenvectors are usually defined (e.g. here, note the "nontrivial" part) to explicitly exclude the zero vector.
Like all other definitions in mathematics, this is of course just a convention. However, as usual there are reasons why this convention makes sense: Many of the applications of spectral theory require extracting scalar components $x_i \rightarrow \lambda_i x_i$ from a linear transformation represented by a matrix multiplication $x \rightarrow A x$ (principal component analysis in case $A$ is the covariance matrix of a random vector for some multivariate probability distribution). Here, $x_i$ is an eigenvector for the eigenvalue $\lambda_i$.
If $0$ were allowed as an eigenvector, suddenly every $\lambda \in \mathbb R$ would be an eigenvalue for it, rendering PCA meaningless because under its interpretation of the covariance eigenvectors, there would now be a "principal component" (the zero vector) with undefined variance attached.
