Finding a Matrix to Connect Equivalence Classes in Quotient Space

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Homework Help Overview

The problem involves understanding equivalence relations in the context of linear algebra, specifically regarding the quotient space of vectors in \(\mathbb{R}^n\) under a defined equivalence relation. The original poster attempts to show that the quotient space consists of two elements based on the existence of a matrix connecting equivalence classes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Some participants suggest considering linear dependence as a simpler case, while others propose constructing a basis that includes both vectors. The original poster questions how this approach aids in finding a connecting matrix.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the construction of a basis and the potential for a matrix that connects the two vectors, but no consensus has been reached on the best approach.

Contextual Notes

There is an indication of uncertainty regarding the implications of linear dependence and the construction of a basis, as well as the challenge of demonstrating the existence of a connecting matrix.

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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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