Can different linear structures in a non-empty set over a given number field have same dimensions?

Question

We know that the linear space (vector space) is a non-empty set over a number field by defining two operations to constructing the linear structure. But, we can define the different linear structures on a set. And I want to know that if the different linear spaces have the same dimensions? If so, what determines this?

Answer 1

The answer is no. If $\ V\ $ is any set of cardinality $\ 2^{\aleph_0}\ $, then there is a bijection $\ \varphi:V\rightarrow \mathbb{R}^n\ $. Define a scalar multiplication and addition on $\ V\ $ by $$ \lambda * v = \varphi^{-1}\left(\lambda\,\varphi\left(v\right)\right)\\ v_1 \oplus v_2 = \varphi^{-1}\left(\varphi\left(v_1\right)+\varphi\left(v_1\right)\right)\ ,$$ for $\ \lambda\in\mathbb{R}, v,v_1,v_2\in V\ $. We then have $$ \varphi\left( \lambda * v\right)=\lambda\,\varphi\left(v\right)\ \mbox{and}\\ \varphi\left( v_1 \oplus v_2\right)=\varphi\left(v_1\right)+\varphi\left(v_1\right)\ ,$$ so $\ \varphi\ $ is a vector space isomorphism, and $\ \left(V,*,\oplus\right)\ $ is a vector space of dimension $\ n\ $ over $\ \mathbb{R}\ $. But we could choose $\ V\ $ to be $\ \mathbb{R}^m\ $ for any $\ m\ $.

On reviewing the comments above, I see the idea used in this answer seems to have already occurred to DanielWainfleet.