Linear Map Conditions for Defining a Map on Projective Spaces

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SUMMARY

The discussion centers on the necessary and sufficient conditions for a linear map \( f : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{m+1} \) to define a well-defined map \( [f] : \mathbb{RP}^n \rightarrow \mathbb{RP}^m \). Specifically, the condition requires that the map respects the equivalence relation of projective spaces, meaning it must map lines through the origin in \( \mathbb{R}^{n+1} \) to lines through the origin in \( \mathbb{R}^{m+1} \). This ensures that the equivalence classes are preserved under the mapping.

PREREQUISITES
  • Understanding of projective spaces, specifically \( \mathbb{RP}^n \)
  • Knowledge of linear maps and their properties
  • Familiarity with equivalence relations in mathematics
  • Basic concepts of vector spaces in \( \mathbb{R}^{n+1} \)
NEXT STEPS
  • Research the properties of projective spaces and their applications
  • Study linear transformations and their effects on vector spaces
  • Explore equivalence relations in more depth, particularly in linear algebra
  • Investigate examples of linear maps that do and do not satisfy the conditions for projective mappings
USEFUL FOR

Mathematicians, students of linear algebra, and anyone studying projective geometry will benefit from this discussion, particularly those interested in the properties of linear maps in relation to projective spaces.

latentcorpse
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Let [itex]\mathbb{RP}^n= ( \mathbb{R}^{n+1} - \{ 0 \} ) / \sim[/itex] where [itex]x \sim y[/itex] if [itex]y=\lambda x, \lambda \neq 0 \in \mathbb{R}[/itex] adn the equivalence class of [itex]x[/itex] is denoted [itex][x][/itex].

what is the necessary and sufficient condition on the linear map [itex]f : \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{m+1}[/itex] for the formula [itex][f][x]=[f(x)][/itex] to define a map

[itex][f] : \mathbb{RP}^n \rightarrow \mathbb{RP}^m ; [x] \mapsto [f(x)][/itex]

not a scooby.
 
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Well, it has to respect your equivalence relation, right? Map lines through the origin into lines through the origin. How can a linear f not do that?
 

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