Diagonalizable endomorphism has trivial null space

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Discussion Overview

The discussion revolves around the properties of diagonalizable endomorphisms, specifically focusing on the relationship between diagonalizability and the kernel of the operator. Participants explore whether a diagonalizable endomorphism necessarily has a trivial null space and consider examples such as the zero operator.

Discussion Character

  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions if a diagonalizable endomorphism always has a kernel equal to {0}.
  • Another participant asks whether the zero operator qualifies as a diagonalizable endomorphism.
  • A participant expresses confusion regarding the topic.
  • It is noted that the zero map is diagonalizable but has a kernel that is the entire space.
  • A participant suggests that if all eigenvalues of an operator are non-zero, then the operator has a trivial kernel.
  • A later reply acknowledges this point and expresses gratitude for the clarification.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether all diagonalizable endomorphisms have a trivial null space, as the example of the zero operator introduces a counterpoint. Multiple competing views remain regarding the conditions under which the kernel is trivial.

Contextual Notes

The discussion highlights the dependence on the nature of eigenvalues and the definitions of diagonalizability and kernel, with unresolved aspects regarding specific cases and examples.

Daaavde
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Is it correct to state that a diagonalizable endomorphism has always kernel = {0}?
 
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Is the zero operator a diagonalizable endomorphism?
 
I still don't understand.
 
R13's point was that the zero map is diagonalizable and has kernel the whole space.
 
Right, that's the case of the zero operator. But what if my eigenvalues are all non-zero (hence my eigenvectors are all linear indipendent)?
 
In that case your operator has trivial kernel.
 
Perfect. Thank you.
 

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