1
$\begingroup$

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:

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

$\endgroup$
1
3
$\begingroup$

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.

As for what you must do, the idea is to check that the vector space axioms hold for the "addition" operation and the "scalar multiplication" operation thus defined. These should follow fairly readily from properties of arithmetic on positive real numbers. Let me know if you get stuck on any of them, or if you just want someone to check your reasoning.

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).

$\endgroup$
8
  • $\begingroup$ 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? $\endgroup$ – FemtoComm Dec 15 '14 at 23:17
  • $\begingroup$ 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. $\endgroup$ – Cameron Buie Dec 15 '14 at 23:20
  • 1
    $\begingroup$ Well now you say it like that it seems obvious what I had to do. $\endgroup$ – FemtoComm Dec 15 '14 at 23:21
  • 1
    $\begingroup$ Alas! It is all too often that way in mathematics.... $\endgroup$ – Cameron Buie Dec 15 '14 at 23:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Cameron Buie Dec 15 '14 at 23:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.