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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?

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

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