Proving a specific vector space with two operations

Question

Prove that $$V = \{ (a,b) \in \Bbb R^2 : a,b > 0 \}$$ is a vector space with the operations $$(a,b) \oplus(c,d) = (ac,bd) \,\,\,\forall (a,b),(c,d) \in V$$ and $$\alpha(a,b) = (a^\alpha,b^\alpha)\,\,\,\forall\alpha \in \Bbb R\,\,\,\forall(a,b) \in V.$$

I know that I must show the 8 properties but this $(ac,bd)$ is confusing me in two properties and the $\alpha$ exponent also is driving me crazy to show the multiplication properties.

Could anyone help?

Answer 1

We need closure first of all. Do the two operations in question produce vectors that are still in the space? The answer is yes; the multiplication of two positive numbers is positive, and a positive number raised to any real number is also positive. Check. Then we move on to:

1. Associativity of addition: \begin{align*} (a,b)\oplus((c,d)\oplus(e,f))&=(a,b)\oplus(ce,df)\\ &=(a(ce),b(df))\\ &=((ac)e,(bd)f)\\ &=(ac,bd)\oplus(e,f)\\ &=((a,b)\oplus(c,d))\oplus(e,f), \end{align*} as required.
2. Commutivity of addition. Perform a similar calculation as above; the result follows from the commutivity of real multiplication.
3. Identity element of addition. I think you'll find this is $(1,1):$ \begin{align*} (a,b)\oplus(1,1)&=(a\cdot 1, b\cdot 1)\\ &=(a,b)\\ &=(1\cdot a,1\cdot b)\\ &=(1,1)\oplus(a,b). \end{align*}
4. Inverse elements of addition. I think you'll find that, for vector $(a,b),$ its inverse is $(1/a,1/b).$ Note that $a,b>0,$ so this is defined. We'd have \begin{align*} (a,b)\oplus(1/a,1/b)&=(a/a,b/b)\\ &=(1,1), \end{align*} the additive inverse. And you can check that the other direction of multiplication works as well.
5. Compatibility of scalar multiplication. This says that we need $a(b(c,d))=(ab)(c,d),$ where $a$ and $b$ are scalars. We have \begin{align*} a(b(c,d))&=a(c^b, d^b)\\ &=((c^b)^a,(d^b)^a)\\ &=(c^{ba},d^{ba})\\ &=(ba)(c,d)\\ &=(ab)(c,d), \end{align*} as required.
6. Identity element for scalar multiplication. This'd be $1:$ $$1(a,b)=(a^1,b^1)=(a,b). $$
7. Distribution over vector addition. We need $$a((b,c)\oplus(d,e))=(a(b,c))\oplus(a(d,e)). $$ So, we have \begin{align*} a((b,c)\oplus(d,e))&=a(bd,ce)\\ &=((bd)^a,(ce)^a)\\ &=(b^ad^a,c^ae^a)\\ &=(b^a,c^a)\oplus(d^a,e^a)\\ &=(a(b,c))\oplus(a(d,e)), \end{align*} as required.
8. I'll let you show that $(a+b)(c,d)=a(c,d)+b(c,d).$

Answer 2

Consider the map $V \to \mathbb{R} \oplus \mathbb{R}$ that sends $(a, b)$ to $(\ln a, \ln b$). This map converts coordinatewise products into coordinatewise sums and coordinatewise powers into scalar multiples, and is a bijection. It follows that since the codomain is a vector space, so is the domain, and the map is in fact an isomorphism of vector spaces.