Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is not hard to see that $(\mathbb R^2,+)$ with this product

$ {r\cdot(x,y)=(rx,ry) } $

is vector space over field $\mathbb R$.

I'm looking for another product that $(\mathbb R^2,+)$ is vector space over $\mathbb R$. I know

$n*(x,y)=\underbrace{(x,y)\oplus (x,y) \oplus \cdots \oplus (x,y)}_n=(nx,ny)$ but I have no idea for arbitrary element of $\mathbb R$. Any Suggestion. Thanks

share|cite|improve this question
You are trying to find other definitions for scalar multiplications? – Asaf Karagila Jun 23 '12 at 18:23
Aren't the two products equivalent, at least for natural $n$? – talmid Jun 23 '12 at 18:30
@AsafKaragila yes I'm trying find another definitions for scalar multiplications – Babak Miraftab Jun 23 '12 at 18:36
You point out that $n\cdot (x,y) = (n\cdot x, n\cdot y)$ for $n\in \Bbb N$. It's worth pointing out that you can extend this argument to show that $q\cdot (x,y) = (q\cdot x, q\cdot y)$ for $q\in \Bbb Q$. And, if you have just a bit more structure (like say, considering inner product space structures of $(\Bbb R^2,+)$ over $\Bbb R$), you can extend the argument further to any $r\in \Bbb R$. – Managu Jun 23 '12 at 18:53

If you are allowed to use the axiom of choice, then the abelian groups $(\mathbb R, +), (\mathbb R^2, +), (\mathbb R^3,+), \ldots$ are isomorphic. In fact they are all isomorphic to a direct sum of continuum-many copies of $(\mathbb Q,+)$. So giving a vector space structure to any of them is the same thing. In particular, we know how to give $(\mathbb R^3,+)$ a structure of $\mathbb R$-vector space which is of dimension $3$, thus by transporting it to $\mathbb R^2$, we can give a vector space structure to $\mathbb R^2$ making it into a 3-dimensional $\mathbb R$-vector space (and similarly for any dimension you wish, as long as it's not more than the continuum cardinality).

In fact, even without the axiom of choice, giving a finite dimensional $\mathbb R$-vector space structure to any abelian group $G$ is the same as finding a group isomorphism from $G$ to $\mathbb R^n$ and transporting the natural vector space structure of $\mathbb R^n$ back to $G$. So the real question is how to find those group isomorphisms.

Sadly, I don't think it is possible to find any nontrivial one without the axiom of choice, so all the nontrivial vector space structure you can put on $\mathbb R^2$ need you to use it, and are not constructive.

share|cite|improve this answer
"not too big"=cardinality at most that of the continuum? – Managu Jun 23 '12 at 19:03
@ Managu : yes, I wasn't sure so I needed to check that a direct sum of $\kappa$-many direct sums of $\kappa'$-many copies of $\mathbb Q$ is a direct sum of max $(\kappa,\kappa')$-many copies, and of cardinal max $(\kappa,\kappa',\aleph_0)$ as a result. – mercio Jun 23 '12 at 19:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.