Can the zero vector be an eigenvector for a matrix?

Solution 1:

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.

Solution 2:

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.)