Finding a Matrix to Connect Equivalence Classes in Quotient Space

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Homework Statement


(part of a bigger question)
For ##x,y \in \mathbb{R}^n##, write ##x \sim y \iff## there exists ##M \in GL(n,\mathbb{R})## such that ##x=My##.
Show that the quotient space ##\mathbb{R}\small/ \sim## consists of two elements.

Homework Equations

The Attempt at a Solution


Well, it is easy to see that the first equivalence class is ##\{0\}##. This means that the second equivalence class should be ##\mathbb{R}^n - \{ 0 \} ##.
But I don't know how to show that given ##x,y\in \mathbb{R}^n##, I can find a matrix that connects between them.
Any suggestions will be great.

Thank you!
 
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If they are linearly dependent, it's easy. If not, why not make a basis with both of them?
 
fresh_42 said:
If they are linearly dependent, it's easy. If not, why not make a basis with both of them?
Thank you for the answer.
However, I still don't get how it will help me with taking them both and complete it to a basis.
 
The matrix (expressed in the coordinates of the new basis) that simply swaps the two and leaves the rest invariant shouldn't be that hard to figure out.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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