Is linear space a finite set?

Question

A linear space has to agree with closure axioms.So,on the basis that,can a linear space be defined so that it is a finite set?

The zero vector forms a linear space by itself, so this is certainly possible, but that is not a very interesting example. A better depends on you telling us key details about your background. If you have never heard of finite fields (likely if this is your first encounter with vector spaces), then the trivial space is, indeed, the only example of a finite vector space. Otherwise it depends. — Jyrki Lahtonen, Dec 31 '23 at 13:50
Anyway, you are probably well advised to take a look at our abridged guide for new askers. You really want to avoid the occasions when your post is judged not to meet the expected quality standards. — Jyrki Lahtonen, Dec 31 '23 at 13:52
Does this answer your question? What kind of vector spaces have exactly one basis? — QuantumSuperfield, Dec 31 '23 at 13:52
Hello,I am new here & that's why I was not able to ask my questions in decent manner.Your guidelines for asking questions properly was helpful for me.Thanks — Sumit Roy, Dec 31 '23 at 14:55
Title: No, a linear space need not be finite as a set. But it can be finite, as explained, over finite fields. However, I suspect, that your vector spaces are defined over $\Bbb R$ (at least this is often the case when no field is named). Then this duplicate answers your question. — Dietrich Burde, Dec 31 '23 at 15:01
Does this answer your question? Is there a finite vector subspace over the reals? — Dietrich Burde, Dec 31 '23 at 15:21

sobornost · Accepted Answer · 2023-12-31T16:29:08.330

1

Yes, it's possible to have finite vector spaces ! That is the case if and only if our vector space (also called linear space) is defined over a finite field and of finite dimension. (as pointed out by Joe, that's for nonzero vector spaces, $\{ 0 \}$ is always finite no matter over which field it is defined)

For example, we can define a two dimensional space over $\mathbb{F_2}$ (its vectors will simply be

$$\begin{pmatrix}0 \\\ 0\end{pmatrix} \begin{pmatrix}1 \\\ 0\end{pmatrix} \begin{pmatrix}0 \\\ 1\end{pmatrix} \begin{pmatrix}1 \\\ 1\end{pmatrix}$$

If you are not familiar with finite fields then indeed, you cannot construct non trivial (that is non $ \\{ 0 \\} $) vector spaces over infinite fields as pointed out by Jykri Lahtonen

edited Dec 31 '23 at 16:29

answered Dec 31 '23 at 13:53

sobornost

51

2

I think in your second sentence you ought to write: a nonzero vector space is finite if and only if it is over a finite field and has finite dimension. – Joe Dec 31 '23 at 14:53
Correct, I'll update it – sobornost Dec 31 '23 at 16:28

score 1 · Answer 2 · answered Dec 31 '23 at 14:29

The answer is yes. First of all the trivial example, the vector space $\{\mathbf 0\}$ only consisting of the zero vector. More interestingly:

A vector space over a field $\mathbb F$ has a finite number of elements if and only if $\mathbb F$ is a finite field and the vector space has a finite dimension (and is non-trivial).

This means that most of the "intuitive" vector spaces you get to know first when learning algebra e.g. $\mathbb R^n$ over $\mathbb R$, the vector space contains infinitely many elements as the field $\mathbb R$ is not finite.

freakish · Answer 3 · 2023-12-31T21:21:58.627

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Linear space is not a set, it is a combination of abelian group and field related through so called scalar multiplication.

So I understand your question as: can the underlying set of vertices be finite? The answer is of course yes. The trivial vector space $\{0\}$ has exactly one element, regardless of the choice of field.

However whether other finite sets can form a vector space depends on the underlying field. If it is infinite then every non-trivial vector space is infinite.

On the other hand if a field $k$ is finite, then it is well known that it has $p^n$ elements for some prime $p$, and therefore every finite dimensional vector space over $k$ has $p^{dn}$ elements for some $d$. In particular if two different primes divide size of $X$ then it cannot form a vector space. While every set of size $p^m$ for some prime $p$ is indeed a vector space over some field (e.g. $\mathbb{Z}/p\mathbb{Z}$).

edited Dec 31 '23 at 21:21

answered Dec 31 '23 at 13:54

freakish

42,851

Sorry. But I read a definition of linear space as a set."Let V denote a non empty set of objects, called elements.The set V is called a linear space if it satisfies the following axioms such as closure axioms, axioms of addition, axioms for multiplication by scalers & existence of negatives, zero elements. -From Calculus Vol 2,Tom Apostol – Sumit Roy Dec 31 '23 at 15:00
@SumitRoy well, it is incorrect. Formally at least. However ultimately we can see that the author means the same thing, that vector space consists of three components: a set of vertices, addition function and multiplication function. It is not just the set of vertices. – freakish Dec 31 '23 at 15:17
Can u suggest me any resource from which I can verify & figure out myself that linear space is not something like set? – Sumit Roy Dec 31 '23 at 15:19
Btw, this book is quite old, the first edition is 1962 or something. For the modern definition of vector space you can see wiki: https://en.wikipedia.org/wiki/Vector_space I will search for something to recommend later. – freakish Dec 31 '23 at 15:26

Is linear space a finite set?

3 Answers3