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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?

user83039
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    It cannot. By definition. However, the eigenspace associated to an eigenvalue always contains the zero vector. – Andrés E. Caicedo Oct 25 '14 at 06:20
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    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 $Av = \lambda v$. – Cameron Williams Oct 25 '14 at 06:22
  • So eigenvalues with multiplicity 2 don't always have 2 eigenvectors? – user83039 Oct 25 '14 at 06:24
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    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. – Daniel McLaury Oct 25 '14 at 06:26
  • The simple example is the matrix $\begin{pmatrix} 0 & 1 \ 0 & 0 \end{pmatrix}$. The only eigenvalue is zero, with multiplicity two. It has a one-dimensional space of zero-eigenvectors, spanned by $(1,0)$. $(0,1)$ is not a zero-eigenvector, but it is a generalized zero-eigenvector: if we hit it with the matrix it goes to $(1,0)$, which then goes to $(0 0)$ if we hit it again. – Daniel McLaury Oct 25 '14 at 06:28
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    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. – Daniel McLaury Oct 25 '14 at 06:29
  • The eigenpolice are coming... – copper.hat Oct 25 '14 at 06:31
  • @copper.hat: Oooooooh Nooooooo! Mr. Bill! *Not* the eigenpolice! Actually, I just saw them here on Telegraph arresting a kid for having the wrong eigenvalue! (I think eigen is German for "proper", so this all makes sense!) Live from the Caffe Med . . . Regards! – Robert Lewis Oct 25 '14 at 06:35
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    @CameronWilliams: I don't quite agree with the comparison with prime numbers. The main reason to exclude zero as eigenvector is to have eigenvalues be those values for which any eigenvectors exist, and for them to be uniquely associated to any eigenvector. But one could define $0$ to be an eigenvector (for any$~\lambda$) but define eigenvalues as those $\lambda$ for which $\dim(\ker(A-\lambda I))>0$, and I think it would in many cases be easier in practice (I find myself saying "element of the eigenspace for $\lambda$" fairly often, to avoid "eigenvector for $\lambda$ or zero"). – Marc van Leeuwen Oct 25 '14 at 09:01
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    @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! – Andrew Dudzik Oct 25 '14 at 09:42
  • @Slade: You are right (but only because you added the requirement that $k$ be algebraically closed, artificial especially when considering $k[X]$ modules), but the comparison with prime numbers is still is very incomplete. There is no decomposition in simple modules in general. And these considerations are way above the level where one introduces eigenvectors. For what it's worth, I define eigenvectors of $\phi$ to be vectors that span a $1$-dimensional $\phi$-stable subspace, which makes it fairly natural why $0$ is excluded. – Marc van Leeuwen Oct 25 '14 at 10:11
  • @MarcvanLeeuwen It's only natural if excluding dimension $0$ is natural. It hardly seems high-level to observe that this is natural precisely because zero-dimensional spaces don't play well with decomposition. Also, the condition $k=\overline{k}$ may be unnatural, but without it, eigenvectors are not terribly natural either—a course focusing on linear algebra over $\mathbb{R}$ typically has a hard time conveying to students why, for example, the real and imaginary parts of complex eigenvectors should appear in pairs to "diagonalize" matrices into standard $2\times 2$ blocks. – Andrew Dudzik Oct 25 '14 at 10:26
  • Instead of this discussion, maybe better give answers at this related question. – Marc van Leeuwen Oct 25 '14 at 10:44
  • Here's an example, where WolfRamAlpha says Eigenvector is zero vector: http://www.wolframalpha.com/input/?i=eigenvalues+of+%5B%5B0%2C1%2C0%5D%2C%5B0%2C0%2C1%5D%2C%5B2%2C-5%2C4%5D%5D – User Oct 25 '14 at 18:47

2 Answers2

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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.)

littleO
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  • 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). – Joseph Garvin Jul 20 '17 at 23:28
  • Note that the 3rd edition of Axler (Linear Algebra Done Right) does define eigenvectors to exclude 0. – Atsina Oct 03 '18 at 19:43