Is the Null Space of an Operator Defined by Its Zero Eigenvalue?

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Discussion Overview

The discussion revolves around the relationship between the null space of an operator and its eigenvalues, specifically focusing on the implications of having a zero eigenvalue in a finite dimensional complex vector space. Participants explore whether the null space consists solely of eigenvectors associated with the zero eigenvalue and the implications for operators without a zero eigenvalue.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • Some participants propose that if an operator ##T## has a zero eigenvalue, then the null space consists of all eigenvectors corresponding to that eigenvalue, along with the zero vector.
  • Others argue that if ##T## does not have a zero eigenvalue, it implies that the operator is injective, suggesting that the null space would only contain the zero vector.
  • A later reply clarifies that the discussion assumes ##T## is a linear operator, which is relevant to the conclusions drawn about injectivity and the null space.

Areas of Agreement / Disagreement

Participants generally agree on the implications of having a zero eigenvalue and its relation to the null space, but there is some uncertainty regarding the broader implications for operators without a zero eigenvalue. The discussion remains unresolved regarding the completeness of the null space characterization.

Contextual Notes

Assumptions about the linearity of the operator are critical to the discussion, and the implications of the definitions of eigenvalues and null spaces are not fully explored, leaving some aspects open to interpretation.

maNoFchangE
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Suppose ##T## is an operator in a finite dimensional complex vector space and it has a zero eigenvalue. If ##v## is the corresponding eigenvector, then
$$
Tv=0v=0
$$
Does it mean then that ##\textrm{null }T## consists of all eigenvectors with the zero eigenvalue?
What if ##T## does not have zero eigenvalue? Does it mean that its null space is just the zero vector?

Thanks
 
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maNoFchangE said:
Suppose ##T## is an operator in a finite dimensional complex vector space and it has a zero eigenvalue. If ##v## is the corresponding eigenvector, then
$$
Tv=0v=0
$$
Does it mean then that ##\textrm{null }T## consists of all eigenvectors with the zero eigenvalue?
What if ##T## does not have zero eigenvalue? Does it mean that its null space is just the zero vector?

Thanks
Yes.
(Except that ##\textrm{null }T## consists of all eigenvectors with the zero eigenvalue and the 0 vector.)
 
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Samy_A said:
Yes.
So, the answer to all of my questions is affirmative?
Then, if ##T## does not have a zero eigenvalue, it's equivalent of being injective.
 
maNoFchangE said:
So, the answer to all of my questions is affirmative?
Then, if ##T## does not have a zero eigenvalue, it's equivalent of being injective.
For a linear operator, yes.
(I should have mentioned that in my first answer too, I just assumed you meant that T is a linear operator.)
 
Yes, it's linear.
 

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