0
$\begingroup$

This question might be very silly. I was working on two Examples of Friedberg- Insel-Spence's Linear Algebra. In example $6$, in $\mathbb{R}^2$(not 2-dimensional real vector space, consider it as the set $\mathbb{R} \times \mathbb{R})$ scalar multiplication was defined as usual, but vector addition was defined as the following:

$(a_1, b_1) +(a_2, b_2)= (a_1+b_2, a_1- b_2)$ for any $(a_1, b_1), (a_2, b_2) \in \mathbb{R}^2.$ The set $\mathbb{R}^2$ is closed addition and scalar multiplication, but it's not a vector space over $\mathbb{R}$ because $\mathbb{R}^2$ is not an abelian group under addition, for instance, this operation is neither commutative nor associative. Moreover, there is an issue with the distribution. Is there any complete list of ways(addition + scalar multiplication) such that the set $\mathbb{R}^2$ is a 2-dimensional vector space?

$\endgroup$

1 Answer 1

1
$\begingroup$

I think that you are hoping for too much.

Let $\pi:\mathbb{R}^2\to\mathbb{R}^2$ be any $1-1$ and onto map. (Any map, not necessarily linear or even additive.)

Now define on $\mathbb{R}^2$ a new addition and scalar multiplication as follows: $$ a\oplus b:=\pi^{-1}(\pi(a)+\pi(b)), \ \lambda\odot a:=\pi^{-1}(\lambda\cdot\pi(a)). $$

Then $\mathbb{R}^2$ is an $\mathbb{R}$ vector space of dimension 2 with respect to these operations. You can check the axioms, but as we have just re-labelled all the vectors it is probably "clear" that this is the case.

$\endgroup$
1
  • $\begingroup$ Thanks so much. There are uncountably many such $\pi$'s. $\endgroup$
    – Matha Mota
    Commented Jun 17, 2020 at 9:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .