Finding a Matrix to Connect Equivalence Classes in Quotient Space

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SUMMARY

The discussion centers on the mathematical concept of equivalence classes in quotient spaces, specifically for vectors in ##\mathbb{R}^n##. It establishes that the quotient space ##\mathbb{R}/\sim## consists of two elements: the equivalence class containing the zero vector, ##\{0\}##, and the class containing all non-zero vectors, ##\mathbb{R}^n - \{0\}##. Participants suggest using a basis to connect linearly independent vectors through a matrix in the general linear group, ##GL(n,\mathbb{R})##, which can facilitate the transformation between these vectors.

PREREQUISITES
  • Understanding of equivalence relations in linear algebra
  • Familiarity with quotient spaces in topology
  • Knowledge of the general linear group, ##GL(n,\mathbb{R})##
  • Ability to work with vector spaces and bases
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  • Study the properties of equivalence relations in vector spaces
  • Learn about quotient spaces and their applications in linear algebra
  • Explore the structure and properties of the general linear group, ##GL(n,\mathbb{R})##
  • Investigate methods for constructing bases from sets of vectors
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Mathematicians, students of linear algebra, and anyone interested in advanced topics in vector spaces and quotient spaces.

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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.
 

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