Is there vector space $\mathbb R $ over $\mathbb R $ with dimension $m\neq1$?

Question

I am searching a new scalar product that with this scalar product the vector space $\mathbb R $over $\mathbb R $ with ordinary vector sum has dimension $m$ , which $m$ is any arbitrary natural number.

hint: scalar product over $R-module M$ ( $R-vector space M$ ) is a two variable function like $$f:\ R\times M \rightarrow M$$ such that

1. $f(r_1+r_2,m)=f(r_1,m)+f(r_2,m)$
2. $f(r,m_1+m_2)=f(r,m_1)+f(r,m_2)$
3. $f(r_1,f(r_2,m))=f(r_1r_2,m)$

for simplicity we write $f(r,m)=r*m$

I have tried it , but with no result. I even don't know how to start to solve problem.

thanks.

Answer 1

$\mathbb R $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. and also $\mathbb R^n $ over $\mathbb Q$ is a vector space with dimension $2^{\mathbb N}$. So the additive group of $\mathbb R$ is isomorphic with additive group of $\mathbb R^n$. for example consider $f:\mathbb R\rightarrow \mathbb R^n$ be such isomorphism , now define new scalar product as follow: $$q*r=f^{-1}(q . f(r))$$ which $q,r\in\mathbb R$ and $q . f(r)$ means ordinary scalar product of $\mathbb R^n$ .

now because $$f(q*r)=q . f(r)$$ so linear space $\mathbb R $ over $\mathbb R$ with new scalar product is linear isomorphic with space $\mathbb R^n $ over $\mathbb R$ so $\dim_\mathbb R\mathbb R$ with new scalar product is the same as $\dim_\mathbb R\mathbb R^n=n$.