Group acts with scalar multiplication Show that ($\mathbb{R}^*,\cdot$) acts on $\mathbb{R}^{n+1}$ by scalar multiplication. What are the orbits under this action?
I first went about trying to prove the first part by induction but that doesn't work. The second part seems ambiguous to me, so I don't think I'm understanding it. Any help is welcome, thank you in advance!                       
 A: I'm not sure why the question asks about $\mathbb{R}^{n+1}$, since there is no definition of $n$. I will show that $(\mathbb{R}^{\times},\cdot)\curvearrowright\mathbb{R}^n$, for all $n\in\mathbb{N}$.


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*Let $(x_1,\dots,x_n)\in\mathbb{R}^n$. Realize that $1\cdot(x_1,\dots,x_n) = (1x_1,\dots,1x_n)$ by the definition of our action, which is equal to $(x_1,\dots,x_n)$. Thus the identity of $(\mathbb{R}^{\times}, \cdot)$ acts as the identity in our action.

*Now let $a,b\in(\mathbb{R}^{\times},\cdot)$ and consider $a\cdot(b\cdot(x_1,\dots,x_n)) = a\cdot(bx_1,\dots,bx_n) = (abx_1,\dots,abx_n)$. But $(ab)\cdot(x_1,\dots,x_n) = (abx_1,\dots,abx_n)$. Thus $a\cdot(b\cdot(x_1,\dots,x_n)) = (ab)\cdot(x_1,\dots,x_n)$, and the action is compatible.


Thus our action is indeed an action. To find the orbits under it, fix $(\alpha_1,\dots,\alpha_n)\in\mathbb{R}^n$. The orbit $(\mathbb{R}^{\times},\cdot)\cdot(\alpha_1,\dots,\alpha_n) = \{x\cdot(\alpha_1,\dots,\alpha_n) $ | $ x\in(\mathbb{R}^{\times},\cdot)\}= \{(x\alpha_1,\dots,x\alpha_n) $ | $ x\in(\mathbb{R}^{\times},\cdot)\}$. Thus the orbits are sets of vectors parallel (and antiparallel) to each other.
Hope this helped!
