Dimension & Linear Maps: Does U=V?

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Discussion Overview

The discussion revolves around the properties of dimensions under linear maps, specifically whether the dimension of the image of a linear map equals the dimension of the original space. Participants explore the implications of the rank-nullity theorem in this context.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions if the dimension remains unchanged under a linear map, asking if dim(U) equals dim(img(f)).
  • Another participant asserts that this is not necessarily true, suggesting that the zero map serves as a trivial counterexample.
  • A participant mentions the rank-nullity theorem as relevant to the discussion, indicating it might provide insight into the relationship between dimensions.
  • One participant acknowledges their initial assumption about the dimension equality and expresses intent to use it in proving the rank-nullity theorem, indicating a reconsideration of their understanding.
  • Another participant explains that if L is a linear map, the dimension of L(U) cannot exceed that of U and can be lower if the kernel of L is non-empty, referencing the rank-nullity theorem.

Areas of Agreement / Disagreement

Participants do not reach a consensus; there are multiple competing views regarding the relationship between dimensions under linear maps, and the discussion remains unresolved.

Contextual Notes

Participants reference the rank-nullity theorem, but there are unresolved assumptions regarding the conditions under which dimensions may change, particularly concerning the kernel of linear maps.

FunkyDwarf
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Hey guys,

Does dimension remain unchanged under a linear map? Ie if i have a map f:U->V does dim(U) = dim(img(f))?

Cheers
-Z
 
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Not necessarily.
 
For a trivial example, take the zero map on a space of positive dimension.

Something that might be of interest to you is the rank-nullity theorem.
 
Yeh i kinda figured it didnt hold. Actually i was going to use it to prove the RL theorem if it was true

cheers
Z
 
If L is linear, L(U) cannot have dimension higher than U. It can have dimension lower than U. Of course, to do that, L must map many vectors to the 0 vector: its kernel is not empty. The "nullity-rank" theorem morphism mentioned says that the dimension of L(U) plus the dimension of the kernel of U is equal to the dimension of U.
 

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