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Let $V_1=\{a\in \mathbb{R}\mid a>0\}$ with the multiplication as the vector addition and the scalar multiplication $\lambda \odot v=v^{\lambda}$.

I want to check if $V_1$ is a $\mathbb{R}$-vector space.

Let's symbolize with $\oplus$ the vector addition. Then we have the following:

Closure of addition: $x\oplus y=x\cdot y\in V_1$

Associativity: $(x\oplus y)\oplus z=(x\cdot y)\cdot z=x\cdot (y\cdot z)=x\cdot (y\oplus z)=x\oplus (y\oplus z)$

Existence of neutral element: $e\oplus x=x=x\oplus e \Rightarrow e\cdot x=x=x\cdot e \Rightarrow e=1$

Existence of additive inverse: $x\oplus y=e=y\oplus x \Rightarrow x\cdot y=e=y\cdot x$, so $y$ is $x^{-1}$ ? (Wondering)

Commutativity: $x\oplus y=x\cdot y=y\cdot x=y\oplus x$

So, $(V_1, \oplus)$ is an abelian group.

Is this correct?

Let $a,b\in K$ and $x\in V_1$ $(a+ b)\odot x=x^{a+ b}$ $a\odot x\oplus b\odot x=x^a\cdot x^b=x^{a+b}$

Let $a\in K$ and $x,y\in V_1$
$a\odot (x\oplus y)=a\odot (x\cdot y)=(x\cdot y)^a=x^a\cdot y^a$ $a\odot x\oplus a\odot y=x^a\cdot y^a$

Let $a,b\in K $ and $x\in V_1$ $(a\cdot b)\odot x=x^{a\cdot b}$ $a\odot (b\odot x)=a\odot x^b=(x^b)^a=x^{a\cdot b}$

Let $x\in V_1$ $1\odot x=x^1=x$

Is everything correct so far?

So, do we get that $V_1$ is a $\mathbb{R}$-vector space?

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    $\begingroup$ Looks fine. I add a spoiler as an answer. $\endgroup$ Dec 3, 2016 at 23:03
  • $\begingroup$ What I said at "Existence of additive inverse" is it correct? @AndreasCaranti $\endgroup$
    – Mary Star
    Dec 3, 2016 at 23:08
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    $\begingroup$ Just perfect, kudos. I wish everyone here posted questions like you, showing their (good) work. $\endgroup$ Dec 3, 2016 at 23:11
  • $\begingroup$ Thank you very much!!!! :-) @AndreasCaranti $\endgroup$
    – Mary Star
    Dec 4, 2016 at 8:01

1 Answer 1

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Spoiler

Consider the map $\log : V_{1} \to \mathbb{R}$. Then this is a bijective linear map, where $\mathbb{R}$ has the usual vector space structure. So your $V_{1}$ is just $\mathbb{R}$ in disguise.

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