Is $V$ a vector space under these operations?

Question

Let $$V=\{x \in \mathbb{R} : x>0\}$$

For $x,y,a \in \Bbb{R}$, define $x\oplus y=xy$ and $x\odot a=x^a$. Is $V$ a vector space under these operations? Justify your answer.

I've edited your question to include MathJax, and fixed some apparent typos. Please verify that it's correct. Also, can you share your thoughts on the problem and explain what you've tried? — , Oct 10 '13 at 22:10
There exists a "MO duplicate" (which was, of course, closed): http://mathoverflow.net/questions/144608/is-v-a-vector-space-under-these-operation-justify-your-answer — Martin Sleziak, Oct 12 '13 at 08:41
It is not clear whether in your definition of $x\odot a$ the real number $x$ is considered as a scalar and $a$ as a vector, or otherwise. (If you look at the two answers posted so far, one of the posters understood this one way, the other one understood this the other way.) — Martin Sleziak, Oct 12 '13 at 08:46

score 5 · Answer 1 · answered Oct 10 '13 at 22:44

Define $\phi:V \to \mathbb{R}$ by $\phi(v) = \log v$. Note that $\phi$ is bijective.

We see that $\phi( x\oplus y) = \phi(xy)=\log x + \log y = \phi(x)+\phi(y)$, and $\phi( x \odot a) =\log(x^a) = a \log x = a \phi(x)$.

Since $(\mathbb{R}, +,\cdot)$ is a vector space, it follows that $(V,\oplus, \odot)$ is a vector space.

score 3 · Answer 2 · answered Oct 10 '13 at 22:27

To be a vector space, you must first check that it satisfies the abelian group axioms: For all $x,y,z\in V$:

$x\oplus y\in V$
$x\oplus y=y\oplus x$
$(x\oplus y)\oplus z=x\oplus (y\oplus z)$
There is an element $e\in V$ such that $x\oplus e=x$
$x\oplus x^*=e$ for some $x^*\in V$

Furthermore, it must "play nicely" with the underlying field (in this case, $\mathbb{R}$). Rigorously, this means satisfying all of these axioms, for all $x,y\in V$ and all $a,b\in\mathbb{R}$:

$x\odot a\in V$
$x\odot 1=x$
$x\odot (ab) = (x\odot a)\odot b$
$(x+y)\odot a = (x\odot a)+(y\odot b)$
$x\odot(a+b) = (x\odot a)+(x\odot b)$

where $+$ is the usual addition.

To not be a vector space, all you need is one set of $x,y,z,a,b$ that fails just one of these axioms (or a proof that no such $e$ can exist). So if one of them seems unlikely to you, you might start by looking at simple examples to see if they hold up.

score 0 · Answer 3 · answered Oct 10 '13 at 22:30

It's not hard to verify $\oplus$ and $\odot$ are closed operations. Now $\langle V,\oplus\rangle$ is associative, commutative, has an identity $1$, and inverse is possible $x^{-1}$ (we say it's an Abelian group). For $\langle V,\odot\rangle$, it satisfies two distributive laws, $(x^b)^a=x^{ab}$, and $x^1=x$. Hence, $\langle V,\mathbb{R},\oplus,\odot\rangle$ satisfies the definition of a vector space.

Is $V$ a vector space under these operations?

3 Answers3