Null Space and Eigenvalues/Eigenvectors

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Discussion Overview

The discussion revolves around the relationship between the null space of a linear operator and its eigenvalues and eigenvectors. Participants explore concepts of eigenvalues, specifically focusing on the eigenvalue of zero, and the distinctions between geometric and algebraic multiplicity in the context of linear operators.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant proposes that if a linear operator has a non-trivial null space of dimension k, then it has k eigenvalues, all equal to 0, with corresponding linearly independent eigenvectors from the null space.
  • Another participant agrees with the initial claim but questions what concerns the first participant might have about it.
  • A later reply suggests a more standard terminology, stating that 0 is an eigenvalue of geometric multiplicity k, implying that its algebraic multiplicity is greater than or equal to k, and emphasizes that eigenvalues are defined by their value alone.
  • Further clarification is provided regarding the definitions of algebraic and geometric multiplicity, with a participant affirming that algebraic multiplicity is the count of an eigenvalue as a root in the characteristic equation, while geometric multiplicity is the dimension of the eigenvector subspace.
  • It is noted that algebraic multiplicity is always greater than or equal to geometric multiplicity.

Areas of Agreement / Disagreement

Participants generally agree on the definitions of eigenvalues and their multiplicities, but there is some contention regarding the phrasing of eigenvalues as "k eigenvalues" versus the concept of multiplicity.

Contextual Notes

The discussion does not resolve the terminology used for eigenvalues and multiplicities, and assumptions about the implications of multiplicity are not fully explored.

psholtz
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Suppose I have a linear operator of dimension n, and suppose that this operator has a non-trivial null space. That is:

A \cdot x = 0

Suppose the dimension of the null space is k < n, that is, I can find 0 < k linearly independent vectors, each of which yields the 0 vector when the linear operator A is applied to it.

Is it fair to say that this operator then has k eigenvalues, of value 0? and that the k eigenvectors corresponding to this eigenvalue=0 are linearly independent vectors of the null space?
 
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hi psholtz! :wink:
psholtz said:
… Is it fair to say that this operator then has k eigenvalues, of value 0? and that the k eigenvectors corresponding to this eigenvalue=0 are linearly independent vectors of the null space?

yes :smile:

(what is worrying you about that? :confused:)
 
Nothing worrying me about that..

Just wanted to make sure I had it straight.. :smile:

thanks!
 
It would be more standard to say that 0 is an eigenvalue of geometric multiplicity k (which would imply that it had algebraic multiplicity greater than or equal to k: the characteristic equation has a factor x^n for n\ge k) rather than to talk about "k eigenvalues", all of value k, as if they were different eigenvalues that happened to have the same value. Its value is the only property an eigenvalue has!
 
Yes, thanks..

So algebraic multiplicity is the number of times the eigenvalue appears as a root in the characteristic equation.

Geometric multiplicity is the dimension of the subspace formed by the eigenvectors of that particular eigenvalue..

And the algebraic multiplicity is always going to be greater than or equal to the geometric multiplicity, correct?
 

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