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?
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11$\begingroup$ It cannot. By definition. However, the eigenspace associated to an eigenvalue always contains the zero vector. $\endgroup$– Andrés E. CaicedoOct 25, 2014 at 6:20
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7$\begingroup$ The zero vector by convention is not an eigenvector, much in the same way that $1$ is not a prime number. If we let zero be an eigenvector, we would have to repeatedly say "assume $v$ is a nonzero eigenvector such that..." since we aren't interested in the zero vector. The reason being that $v=0$ is always a solution to the system $Av = \lambda v$. $\endgroup$– Cameron WilliamsOct 25, 2014 at 6:22
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1$\begingroup$ An eigenvalue always has at least a one-dimensional space of eigenvectors. If it has multiplicity $n$, it might still just have a one-dimensional space of eigenvectors, but it definitely has an $n$-dimensional space of generalized eigenvectors. $\endgroup$– Daniel McLauryOct 25, 2014 at 6:26
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3$\begingroup$ Regardless, Wolfram|Alpha shouldn't be reporting $(0,0,0)$ as an eigenvector. Show us your input so we can see what's going on. $\endgroup$– Daniel McLauryOct 25, 2014 at 6:29
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2$\begingroup$ @MarcvanLeeuwen If we think of a vector space $V$, with an endomorphism, as a $k[X]$-module ($k=\overline{k}$), then $v\in V$ is an eigenvector exactly when it generates a simple submodule. So $0$ isn't an eigenvector exactly because $(0)$ is not a simple module. But over $\mathbb{Z}$, $(0)$ is not a simple module because $(1)$ is not a prime ideal. The comparison seems compelling to me! $\endgroup$– Andrew DudzikOct 25, 2014 at 9:42
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2 Answers
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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.
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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.)
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$\begingroup$ If lambda consistently denotes eigenvalues, and a value is only an eigenvalue if the equality is true for a nonzero vector, then it would seem the equality says the zero vector is not an eigenvector. But the zero vector is always in the null space of a matrix, which they say is all the eigenvectors. So either they have a contradiction, or inconsistent notation (sometimes lambda is just a scalar and sometimes it is specifically an eigenvalue). $\endgroup$ Jul 20, 2017 at 23:28
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$\begingroup$ Note that the 3rd edition of Axler (Linear Algebra Done Right) does define eigenvectors to exclude 0. $\endgroup$– AtsinaOct 3, 2018 at 19:43