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Let $Q = \{(x,y) \in \mathbb{R}^2: x,y > 0 \}$ be a vector space over the real numbers with the operations $$+:Q\times Q \rightarrow Q$$ $$(x_1,y_1)+(x_2,y_2)=(x_1x_2,y_1y_2)$$ and $$\cdot : \mathbb{R} \times Q \rightarrow Q$$ $$c\cdot(x,y)=(x^c,y^c)$$

It's easy to show that this structure is a vector space, but what about the geometrical meaning of these operations? Is there any?

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  • $\begingroup$ what do you mean $x^c$ $\endgroup$
    – R.N
    Sep 11, 2015 at 20:44
  • $\begingroup$ $\prod_{k=1}^{c} x$ $\endgroup$ Sep 11, 2015 at 20:45

2 Answers 2

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Consider $\tilde x = \ln x$ and $\tilde y = \ln y$. Then $(\tilde x, \tilde y)$ is just the standard $\mathbb R^2$. Therefore your vector space is just the standard $\mathbb R^2$ using logarithmic coordinates instead of Cartesian coordinates.

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Let $\mathbb R_{> 0} = \{x \in \mathbb R \mid x > 0 \}$. Note that $\log\colon (\mathbb R_{>0},\cdot) \to \mathbb (\mathbb R,+)$ is an isomorphism of abelian groups. This induces an isomorphism $Q=\mathbb R_{>0}^2 \to \mathbb R^2, (x,y) \mapsto (\log x, \log y)$.

Now note that under this isomorphism, your scalar multiplication becomes the usual one on $\mathbb R^2$ since $\log(x^c) = c\log(x)$.

So $Q$ is isomorphic to $\mathbb R^2$ with the usual $\mathbb R$-vector space structure.

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