Can a vector space $\{0\}$ exist?

Question

Sorry if that question sounds similar to another question asked some time ago but the book that I am using to learn linear algebra defines a vector space to be a set $V$ with addition and scalar multiplication with a field $F$.

The problem comes from the definition of a field in the same book: A field has at least two distinct elements called 0 and 1. This definition would mean that the set $\{0\}$ would not be a valid field and therefore could have no vector space over it, right?

Why would $V={0}$ have to be a field in order to be a vector space? — Randall, Nov 23 '21 at 14:10
${0}$ is a vector field, not very interesting, everybody will agree on it... — Jean Marie, Nov 23 '21 at 14:11
Since a vector space is defined over a field, atleast according to the book. Then the field would have to exist. — Folkert Kevelam, Nov 23 '21 at 14:11
The field $F$ is arbitrary. The space we have is $V=0$. Which axiom doesn't hold in your opinion? We just define $x.0=0$ for all $x\in F$. Have you checked the axioms already? — Dietrich Burde, Nov 23 '21 at 14:12
Huh? It does exist. I can make ${0}$ a vector space over any field $F$ by $x \cdot 0 = 0$. — Randall, Nov 23 '21 at 14:12
@FolkertKevelam The zero vector multiplied by any scalar is still the zero vector. You should find your example of a vector space satisfies all the axioms, regardless of your choice of field. This vector space has dimension zero. — J.G., Nov 23 '21 at 14:13
${0}\subset \mathbb{R}$ is a vector space over $\mathbb{R}$. — Matheus Nunes, Nov 23 '21 at 14:14
Why not? To show that his statement is wrong, it is enough to show a counterexample. — Matheus Nunes, Nov 23 '21 at 14:16
I am guessing that your misconception is that you believe that $F \subseteq V$ must be true. It does not. — Randall, Nov 23 '21 at 14:18
@Randall So a Vector space $V$ over a Field $F$ can be a subset of that Field? — Folkert Kevelam, Nov 23 '21 at 14:19
No. There is no relation between $F$ and $V$ at all. You only need the vector addition and scaling to meet the axioms, nothing more. — Randall, Nov 23 '21 at 14:19
Yes, it can. Take $V=F$ as $F$-vector space. But it need not be. — Dietrich Burde, Nov 23 '21 at 14:20
@DietrichBurde that makes sense, then a vector space can consists of an arbitrary amount of vectors as long as the elements of the vector exists within the field, right? — Folkert Kevelam, Nov 23 '21 at 14:27
Technically, it is a vertor spcae, but such a boring one that one usually demands that a vector space contains more than one vector to rule out this trivial space. — Peter, Nov 23 '21 at 14:29
@Peter: to rule out this trivial space --- Worth noting is that for some contexts the zero vector space is very useful. For example, when considering exact sequences of vector spaces: Why any short exact sequence of vector spaces may be seen as a direct sum? — Dave L. Renfro, Nov 23 '21 at 14:33
@md2perpe The book defines vector spaces as $F^n$ over a field $F$ with $n>0$. This would imply that that vectors inside a vector space must be at least 1 dimensional and therefore have an element — Folkert Kevelam, Nov 23 '21 at 14:37
@FolkertKevelam. That's interesting. Does the book define vector spaces as $F^n$ or as subsets of $F^n$ satisfying some conditions? — md2perpe, Nov 23 '21 at 14:41
@md2perpe it implies that a vector space must be a subset of $F^n$ with at least one vector, that would satisfy the identities such as the additive identity or multiplicative identity, which would imply that there needs to be at least a zero vector. However, $F^0$ would not be an subset of $F^n$ since $n>0$ so i fail to see how there can be a vector that is part of a valid vector space with 0 elements — Folkert Kevelam, Nov 23 '21 at 14:46
@Peter No, that isn't true. There is never a requirement that a vector space contains at least two elements. And (@ FolkertVevelam), defining vector spaces as $F^n$ makes no sense. Those are examples of vector spaces, but by no means the only ones. — Magdiragdag, Nov 23 '21 at 14:46
@FolkertKevelam. You asked about ${0}.$ Does $0$ denote the zero element in $F$ or is it perhaps the zero element of $F^n$? In the first case you can equate it with the zero element of $F^1.$ — md2perpe, Nov 23 '21 at 14:56
@md2perpe, the actual question concerning the validity of a vector space $V = {0}$ has been answered by Magdiragdag , it makes sense whether ${0}$ is an 1-dimensional zero vector or a n-th dimensionsal vector. The tidbit I'm now confused about is the comment that it is not necessary for vectors to have elements. Since a vector space is defined over a field, and a field must have two distinct elements, the vector that is a part of that vector space must have at least some length $n$, since there can be no two distinct elements? — Folkert Kevelam, Nov 23 '21 at 15:02
It's not a vector that has a dimension, it is the vector space that has a dimension. Your book really does you a disservice if it defines vector spaces as $F^n$. It makes you think that a vector has, what you call, elements and that the 'length' of a vector is the same as the dimension of the vector space. Neither is true. To give a totally different example: the set of continuous functions from ${\mathbb R}$ to ${\mathbb R}$ with pointwise addition and pointwise scalar multiplication is a vector space over ${\mathbb R}$. This is an infinite dimensional vector space. — Magdiragdag, Nov 23 '21 at 15:07

score 10 · Accepted Answer · answered Nov 23 '21 at 14:39

You are right that the set $\{0\}$ cannot be made into a field, because a field must have distinct $0$ and $1$. So, indeed, there is no such thing as a vector space over $\{0\}$; that doesn't mean anything.

However, that is different from $\{0\}$ being a vector space over some field $F$. That is possible and in fact, for every field $F$, $\{0_F\}$ is a zero-dimensional vector space over $F$. Addition and scalar multiplication are defined in the only way they can: everything always results in $0_F$.

Can a vector space $\{0\}$ exist?

1 Answers1