# How to demonstrate a set is a real vector space (set governed by nonstandard operations)

I am really not that familiar with questions that ask you to work with a operation vector space, even less with the English terms for it. I am... quite lost. How would you prove that it is a real vector space? Furthermore, is the null vector element just a $(1_1,1_2,...,1_n)$ vector? It also asks for a opposite element to each vector... I suppose that would be $u=\{\lambda v\in \Bbb R^n_+\mid\lambda=-1\}$ since $-v\oplus v=\varnothing$ ... am I even close to understanding?

Below is the given set, all real positives:

$\Bbb R^n_+=\{x_1,x_2,x_3,...,x_n\mid x_1>0,x_2>0,...,x_n>0\}$

The question sorta says, and I remind you I am translating this, "Show that the set that is given with the indicated operations, is a real vector space."

Operations:

$(x_1,...,x_n)\oplus(y_1,...,y_n)=(x_1y_1,...,x_ny_n)$

and the scalar multiplication:

$\lambda \otimes (x_1,...,x_n)=\left(x^\lambda_1,...,x^\lambda_n\right)$

You're not quite on target with your interpretation of the "additive" inverse of a given, arbitrary $u=(u_1,...,u_n)\in\Bbb R^n_+.$ You are correct about the "additive" identity. Use the latter together with the definition of "addition" to deduce what $\ominus(u_1,...,u_n)$ must be. It may help you to recall/note/prove that $-1\otimes(u_1,...,u_n)$ is what you're looking for.
It might help you (with intuition, and possibly with proof) to notice that $\Bbb R^n_+$ is the image of $\Bbb R^n$ under the map $$(x_1,...,x_n)\mapsto\left(e^{x_1},...,e^{x_n}\right),$$ that this map is injective, and that this map is "operation-preserving" (that is, "addition" in $\Bbb R^n_+$ looks like the image of usual addition in $\Bbb R^n$ under the mapping, and "scalar multiplication" in $\Bbb R^n_+$ looks like the image of usual scalar multiplication in $\Bbb R^n$ under the mapping).
• I tried proving the inverse element axiom by expanding $v\ominus v=0$ so I got the individual operations which give us each element of the identity element such that $v_1 \ominus v_1=1 ...v_n=1$. Can you just say that by definition the operation must be $v_1 \oplus v_1^{-1}=1$? PS. Is there a way to prove this part using logarithms? – FemtoComm Dec 15 '14 at 23:17
• Well, all you have to do is show that $(u_1^{-1},...,u_n^{-1})$ is an "additive" inverse of $(u_1,...,u_n).$ There's actually no need to prove uniqueness. – Cameron Buie Dec 15 '14 at 23:20
• As for proving it using logarithms, I'd start by taking $(u_1,...,u_n)$ to $(\log u_1,...,\log u_n),$ then observing that its additive inverse is $(-\log u_1,...,-\log u_n),$ then using the observations mentioned in my post to conclude that the "additive inverse" of $(u_1,...,u_n)$ is $$\left(e^{-\log u_1},...,e^{-\log u_n}\right)=\left(u_1^{-1},...,u_n^{-1}\right).$$ Still, it's better just to note/prove the observations mentioned in my post, so long as you have a result letting that be enough to justify that $\Bbb R_+^n$is a real vector space with the given operations. – Cameron Buie Dec 15 '14 at 23:27