# An eigenvector is a non-zero vector such that…

Various sources define eigenvalues and eigenvectors in slightly different ways (context independent). For example, both of the following definitions seem not to exclude the zero-vector as an eigenvector. In Lang:

Let $V$ be a vector space over a field $K$, and let $A:V\to V$ be an operator on $V$. An element $\mathbf{v}\in V$ is called an eigenvector of $A$ if there exists $\lambda\in K$ such that $A\mathbf{v}=\lambda\mathbf{v}$. If $\mathbf{v}\ne \mathbf{0}$, then...

and, in Hoffman/Kunze

Let $V$ be a vector space over a field $F$ and let $T$ be a linear operator on $V$. A characteristic value of $T$ is a scalar $c$ in $F$ such that there is a non-zero vector $\alpha$ in $V$ with $T\alpha =c\alpha$. If $c$ is a characteristic value of $T$, then (a) any $\alpha$ such that $T\alpha = c\alpha$ is called a characteristic vector of $T$ associated with the characteristic value $c$, and (b) the collection of all $\alpha$ such that $T\alpha = c\alpha$ is called the characteristic space associated with $c$.

More typically (?), you would see explicit exclusion of the zero-vector as,

An eigenvector is a non-zero vector $x$ such that $A\mathbf{x}=\lambda\mathbf{x}$ for some scalar $\lambda$. The scalar $\lambda$ is called an eigenvalue and $x$ an associated eigenvector. The eigenspace corresponding to the eigenvalue $\lambda$ is the set of all associated eigenvectors along with the zero-vector.

I realize that the zero-vector must be explicitly excluded when defining an eigenvalue, but once that definition is made, explicitly excluding the zero-vector as being an eigenvector and then explicitly including it again to form the eigenspace seems rather artificial. Is this simply a matter of attempting to make the definitions more natural depending on their order? That is,

• eigenvalue (excluding the zero-vector), then eigenvector -> zero vector ok
• eigenvector, then eigenvalue -> zero vector not ok

or, is there anything incorrect about the definition given, for example, in Hoffman/Kunze above? If so, what sort of inconsistency would result from the definition. I'm not an algebraist, so something rather elementary (if possible) would be preferred.

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Whatever you do will create exceptions. If you exclude the zero vector as an eigenvector (which I believe is the standard definition), then eigenspaces will include a non-eigenvector. But if count the zero vector as an eigenvector, then it is no longer true that every eigenvector has a unique eigenvalue. The zero vector is just different, and there's no way 'round that fact of life. (You take the good, you take the bad, you take them both and there you have ...) –  Michael Joyce Mar 21 '13 at 23:54
For any scalar $\lambda$, the $\lambda$-eigenspace of $A$ is $\{v \in V : Av = \lambda{v}\}$. We want this to be a subspace of $V$, so it has to include the vector $0$. But we don't call the zero vector an eigenvector because that's a trivial solution and we're not interested in it (in fact $\lambda$ is an eigenvalue precisely when the $\lambda$-eigenspace is not just $0$). What's "ok" is eigenvalue --> eigenspace, not eigenvalue --> eigenvector. –  KCd Mar 21 '13 at 23:54
@MichaelJoyce Thanks, that's a good point. –  Scott H. Mar 22 '13 at 0:15
@KCd Fair enough...if you define eigenspaces from eigenvalues but not eigenvectors from eigenvalues, then it is a little less clunky. As Michael mentioned above the "eigenspace" is then a vector space but not the set of eigenvectors. I've never questioned that before, but now find it odd. There will obviously be word fiddling in either case to state things correctly, I was just wondering whether anything truly bad happens if we allow zero (while still defining an eigenvalue in the same way). –  Scott H. Mar 22 '13 at 0:31
An eigenspace is spanned by a non-zero eigenvector associated with a particular eigenvalue. The eigenspace must be at least one dimensional and therefore excludes using the zero vector as an eigenvector. –  Tpofofn Mar 22 '13 at 0:36
Put another way, suppose we are given any space $V$ over a field $F$. Then for any linear operator $T:V\to V$ and any scalar $\lambda\in F$, we define a linear transformation $S_{T,\lambda}:V\to V$ by $$S_{T,\lambda}(v)=T(v)-\lambda v.$$ We can confirm that this is always a linear operator. The kernel of $S_{T,\lambda}$ is always a subspace of $V$, and in particular clearly contains the zero vector. When such a kernel is non-trivial (has more than just the zero vector in it), we call it the eigenspace of $T$ corresponding to $\lambda$, we call $\lambda$ an eigenvalue of $T,$ and we call the kernel's elements the eigenvectors of $T$ corresponding to $\lambda$.
It's really only the non-trivial kernels of $S_{T,\lambda}$ (if any) that say anything "interesting" about $T$, so only $\lambda$ corresponding to such non-trivial kernels "deserve" the title of eigenvalue of $T$, and only members of such nontrivial kernels "deserve" the title of eigenvector. In that sense, it isn't really a problem to say that the zero vector is an eigenvector of $T$, provided that there is some $\lambda$ such that the kernel of $S_{T,\lambda}$ is non-trivial. (There may be no such $\lambda$. Consider any non-trivial rotation of the plane about the origin. This has no eigenvalues or eigenvectors.)