Proving KerD^2=KerD and ImD=ImD^2 with Linear Transformations

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Discussion Overview

The discussion centers around proving the relationships between the kernel and image of a linear transformation \( D: A \to A \), specifically that \( \text{Ker} D^2 = \text{Ker} D \) and \( \text{Im} D = \text{Im} D^2 \). Participants explore the implications of dimensionality and isomorphism in the context of linear transformations.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant suggests that proving \( \text{Ker} D^2 = \text{Ker} D \) and \( \text{Im} D = \text{Im} D^2 \) could be simplified by stating that \( \dim A = \dim A \), implying isomorphism and leading to \( \text{Ker} D = \{0\} \) and \( \text{Im} D = A \).
  • Another participant questions the reasoning behind the claim that \( \dim A = \dim A \) implies isomorphism, seeking clarification on how this conclusion is reached.
  • A different participant challenges the assumption that \( A \) being isomorphic to itself means every linear map \( D: A \to A \) is an isomorphism, indicating a misunderstanding of the concept.
  • One participant references a theorem stating that two vector spaces of the same dimension are isomorphic, suggesting that if \( B \) is a basis for \( A \), then \( \text{Im} D = A \) and \( D \) is injective, leading to \( \text{Ker} D = \{0\} \).
  • Another participant points out that the theorem requires \( D \) to be an isomorphism, which was overlooked in earlier claims.
  • A participant acknowledges their earlier misunderstanding regarding isomorphism and thanks others for their input, indicating a shift in their perspective.

Areas of Agreement / Disagreement

Participants express differing views on the implications of dimensionality and isomorphism in relation to linear transformations. There is no consensus on the validity of the initial claims regarding the kernel and image of \( D \).

Contextual Notes

Some participants highlight the importance of specific hypotheses for theorems regarding isomorphisms, indicating that assumptions may not have been fully addressed in the discussion.

ergonomics
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If i am given a linear transformation D:A->A,that is followed by
A=ImD(+)kerD
and i am asked to prove that kerD^2=kerD and imD=imD^2.

instead of trying to work it out the hard way by showing that every element of KerD is an element of kerD^2 , both directions.

would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic which means that KerD={0} and ImD=A.

same goes for D^2:A->A
KerD^2={0}
ImD^2=A

=> therefore KerD^2=KerD and ImD^2=ImD ?
 
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ergonomics said:
would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic which means that KerD={0} and ImD=A.
Why does it mean that?
 
would it not be easier to just say that dimA=dimA and hence the two structures are isomorphic
What two structures?
which means that KerD={0} and ImD=A.
Huh? I'm not sure what in the world you're doing, but just because A is isomorphic to A doesn't mean that every linear map D:A->A is an isomorphism. Is that what you were thinking?
 
According to my book two vector spaces of the same dimension are isomorphic to each other.
and the proof also apparently seems to be pretty simple.
If B is a basis for A, then we can easily show that ImD=A
and that T is injective and if T is injective then KerD={0}
 
If B is a basis for A, then we can easily show that ImD=A
and that T is injective and if T is injective then KerD={0}
You're forgetting one of the hypotheses for the theorem -- you need D to be an isomorphism.
 
yes akg unfortunately that is what i was thinking, that if the two were isomorphic to each other, then the map would necessairly be an isormophism.

just a few minutes before hurkyl put his post up, i was about to say that i went over the theorems in my book again, and that my line of thought was incorrect.

anyway, thank you all.
 
Last edited:

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