# Is the linear combination of vectors in a vector space subject to the rules of addition/multiplication of that vector space?

I'll explain with this example that I'm working on:

Let vector space $$V = \{ x \in R| x > 0\}$$ with addition defined by $$x + y = xy$$ and scalar multiplication as $$a * x = x^a$$. Find a basis.

To do this, I should find that some arbitrary vector in V can be written as a linear combination of vectors in V that are linearly independent, and span V.

So let $$u \in V$$, so that $$u > 0$$, and $$u = \alpha_1x_1+\alpha_2x_2+...+\alpha_nx_n$$

Then

$$u = {x_1}^{\alpha_1}+{x_2}^{\alpha_2}+...+{x_n}^{\alpha_n}$$ $$= {x_1}^{\alpha_1}{x_2}^{\alpha_2}...{x_n}^{\alpha_n}$$

Then I can take $$x_1 = 2$$, and see that $$2^\alpha$$ for $$\alpha \in R$$ will be in the span of V. So {2} is a basis of V, dimension 1. This feels wrong to me, which is why I ask: is the linear combination of vectors in a vector space subject to the rules of addition/multiplication of that vector space, as I have applied them here?

• Yes, this is correct. – rubikscube09 Oct 2 '19 at 5:16
• I guess with $R$ you mean $\mathbb R$. – principal-ideal-domain Oct 2 '19 at 5:17
• Wow, I just realized that linear independence is defined in respect to the identity in the vector space. – Javi maxwell Oct 2 '19 at 6:25

The map $$\log : \mathbb R^+ \to \mathbb R$$ is a vector space isomorphism since you have $$\log(x^{\lambda}y^{\mu}) = \lambda \log(x)+\mu\log(y)$$ and the map is bijective. So your vector space is one dimensional and every $$x\in\mathbb R^+\setminus\{1\}$$ is a possible basis. In particular, $$x=2$$ is a basis.