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Background -

I am currently taking an introductory linear algebra course and we were introduced to vector spaces. Now, at first it seemed pretty much 'obvious' that the only vector space with finitely many elements is the trivial $\{ 0 \}$ as all other finite sets seemed unable to satisfy closure.

However I recently realized that using 'remainders', i.e. modular arithmatic would allow me to sidestep this problem. So, I experimented a bit and came up with the following general vector space with $(n + 1)$ elements (over the field $\mathbb{R}$)-

$\{ V , +, \cdot \}$

$V= \{ a \cdot i : a \in \mathbb{R}; n \in \mathbb{N} , 0 \leq i \leq n \}$

$\forall \ v_1, v_2 \in V \ \ v_1 + v_2 = ((\frac{v_1 + v_2}{a}) \ (\text{mod} n)) \cdot a$

$\forall \ \alpha \in \mathbb{R}, v \in V \ \ \alpha \cdot v = (\vert [ \alpha \cdot \frac{v}{\pi} ] \vert (\text{mod} n)) \cdot a$, where $[ ... ]$ denotes the greatest integer function.

It obeys closure, commutes and is associative. I had the addditive and multiplicative identities as $0$ and $1$ and I was able to define the additive inverse of $v$ as $(n-v)$.

Question -

My question relates to the fact that the the dimension of a vector space is defined as the number of elements in its basis. Before I constructed the above space, I understood that vector spaces ${\mathbb{R}}^n(\mathbb{R})$ have dimension $n$, $\mathbb{Z} (\mathbb{Z})$ has dimension 1. While $\{ 0 \}$ and {by extension) all finite element vector spaces have dimension $0$.

However, I am now unsure whether that is the case. After all is $\{ 1 \}$ not a basis for the vector space I defined? It has $1$ element.

So, my question is -

What would be the dimension of a vector space with finitely many elements like the one I defined?

It would be great if if the answer could provide some background/intuitive justification for the choice made in the case in addition to stating it.

Community
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user0
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    you are aware that your set has infinetely many elements as you are running aver all $a \in \mathbb{R}$? Also, no, there are no finite nontrivial vectorspaces over $\mathbb{R}$, you need a finite field for that, however, what you are constructing here, is most likely just the cyclic group iwht the obvious $\mathbb{Z}$ action, but surely no vector space, as you evidently have a nonzero scalar $\lambda$ such that $\lambda\dot x =0$ for some $0\neq x$ – Felix Jan 21 '19 at 15:32
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    Your construction is ill-defined so this is not a vector space nor anything else. As others have pointed out, if you take any nonzero vector space over $\mathbb{R}$, it will always have infinitely many elements. There is, however, a way to construct a nonzero vector space with finitely elements over a finite field $\mathbb{F}$. See here for more info: https://math.stackexchange.com/questions/1343523/are-there-nontrivial-vector-spaces-with-finitely-many-elements – Ben W Jan 21 '19 at 15:37

2 Answers2

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Good for you for being curious. That said, your definition does not make a vector space. I suggest you fix $n=1$ and actually write down a few "vectors" in your vector space and try adding them and multiplying by scalars. In particular, in a vector space over the real numbers, $av \ne bv$ whenever $a \ne b$ and $v \ne 0$. Is that true for your operations?

You can make finite vector spaces using modular arithmetic - if you use "remainders" for the scalars. For example, suppose the scalars are from the field $\mathbb{Z}_3$ with elements $\{0, 1, 2\}$ and arithmetic modulo $3$. Then the two dimensional vector space of pairs of scalars has $9$ elements. Sample vector arithmetic: $$ (1, 0) + (2,1) = (0,1) $$ and $$ 2(1,2) = (2,1) . $$

Technical note. You can do this for any $n$, but you get a vector space only when $n$ is prime. For composite values the vector space axioms still all work but the object is called a "module".

Ethan Bolker
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  • I see. I seem to have messed up the multiplication operation pretty badly and I can now see that changing the field to something like $\mathbb{Z}$ won't fix it either. So, does this mean that our choice of the associated fields us heavily restricted if we want to have finite vector spaces? Will be fun to see if they are required to be finite themselves, for example. Thanks for the pointers (and the example), anyway :) – user0 Jan 21 '19 at 16:22
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    For a finite vector space you must have a finite field of scalars. There is exactly one such field, with $p^k$ elements, for each prime $p$. An $n$ dimensional vector space over such a field will have $p^{kn}$ elements. – Ethan Bolker Jan 21 '19 at 16:30
  • What is $k$ in $p^k$? – user0 Jan 21 '19 at 16:41
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    An ordinary integral exponent. There is one field with $3$ elements, one with $9$, one with $27$ and so on. – Ethan Bolker Jan 21 '19 at 17:18
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A vector space $V$ over the field $ \mathbb R$ with $V \ne \{0\}$ has infinitely many elements! Reason: let $v \in V$ with $v\ne 0$. Then

$$\{tv: t \in \mathbb R\} \subseteq V$$

and $\{tv: t \in \mathbb R\}$ has infinitely many elements.

J. W. Tanner
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Fred
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  • Hi! and thanks for taking the trouble to answer. I think I understand how your argument precludes sucj vector spaces when we define multiplication in 'the usual way'. However, my motivation for this question was exactly whether we could use some kind of 'wrapping-around' to avoid this issue. I would be thankful of you could explain why that would be possible. The post that @Ben W linked to didn't explain that either. Thanks again. – user0 Jan 21 '19 at 16:10