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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?

dad
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    You could say the eigenspace is all of the eigenvectors plus zero, which isn't an eigenvector. – mathematician Oct 26 '14 at 23:56
  • So, eigenspace has zero element but zero element is not eigenvector? But, isn't eigenspace a set of all x which satisfies T(x)=ax ? where a is eigenvalue? – dad Oct 27 '14 at 00:03
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    $T(0)=a0=0,$ so $0$ satisfies the condition that eigenvectors must satisfy. But the definition of eigenvector excludes $0,$ so $0$ is not an eigenvector even though it satisfies the condition. Another way to say it: if $x$ satisfies $T(x)=ax$, then so does $cx,$ for any scalar $c.$ The condition still holds when the scalar is $0,$ but $cx$ is not considered an eigenvector when $c=0.$ – Will Orrick Oct 27 '14 at 03:51
  • See also this question. Some authors do allow the zero vector to be called eigenvector, but make appropriate "nonzero" provisions in related definitions, like that of eigenvalue (a.k.a. characteristic value). – Marc van Leeuwen Oct 27 '14 at 05:12

3 Answers3

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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,

Patrick Da Silva
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The eigenspace associated with an eigenvalue consists of all the eigenvectors (which by definition are not the zero vector) associated with that eigenvalue along with the zero vector.

If we allowed the zero vector to be an eigenvector, then every scalar would be an eigenvalue, which would not be desirable.

paw88789
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    Your second paragraph makes an implicit assumption about how eigenvalues are defined in terms of eigenvectors that is quite similar to the confusion in the question about the definition of eigenspaces. One could very well call $0$ an eigenvector (for any $\lambda$) while defining eigenvalues to be those values$~\lambda$ for which a nonzero eigenvector exists; a similar exception as is normally made in the definition of eigenspace (as you correctly indicate). In fact the question already gives a better motivation to exclude $0$: then a unique eigenvalue is associated to every eigenvector. – Marc van Leeuwen Oct 27 '14 at 05:03
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Here's another definition of eigenvalue (for matrices) that should shed light on your question.

Let $\mathbb{F}$ be a field. Fix $A \in \mathsf{M}_n(\mathbb{F})$ and $\lambda \in \mathbb{F}$. It is a very simple excersie to show that $$\mathsf{E}_\lambda := \{ v \in \mathbb{F}^n \mid Av = \lambda v \}$$ is a subspace of $\mathbb{F}^n$.

Now, in all but finitely many cases, $\mathsf{E}_\lambda$ will be trivial, i.e., $\mathsf{E}_\lambda = \{ \mathbf{0}\}$. If $\mathsf{E}_\lambda$ is nontrivial (i.e., $\dim\mathsf{E}_\lambda > 0$), then $\lambda$ is called an eigenvalue (of $A$). Notice that, with this definition, $\exists v \ne \mathbf{0}$ such that $Av = \lambda v$ (the vector $v$ is called an eigenvector of $A$ corresponding to $\lambda$). Thus, $\mathbf{0}$ is not an eigenvector and its exclusion does not preclude $\mathsf{E}_\lambda$ from being a subspace.

Conversely, if there is a nonzero vector $v$ such that $Av = \lambda v$, then $\mathsf{E}_\lambda$ is nontrivial. Thus, this definition is equivalent to the standard definition.

In summary, these sets are always subspaces, but are of interest when they are nontrivial.

Pietro Paparella
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