System of differential equations Basic question

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Discussion Overview

The discussion revolves around a system of differential equations represented as W´=AW, where W consists of functions of time and A is a 3x3 matrix. Participants explore the implications of encountering a null vector as an eigenvector, particularly in the context of finding solutions to the system.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant describes their usual approach of finding eigenvalues and eigenvectors to solve the system but questions what to do when the eigenvector associated with an eigenvalue appears to be the null vector.
  • Another participant asserts that by definition, an eigenvector cannot be the zero vector and notes that the given matrix has repeated eigenvalues of ##\lambda = -1##, with a non-zero associated eigenvector.
  • A participant expresses curiosity about alternative methods to solve the problem, indicating that their current method is yielding null eigenvectors.
  • Another participant suggests the technique of generalized eigenvalues as a potential approach for obtaining eigenvectors in such cases.

Areas of Agreement / Disagreement

Participants generally agree on the definition of eigenvectors but disagree on the implications of the eigenvalue problem presented. The discussion remains unresolved regarding the best method to address the issue of null eigenvectors.

Contextual Notes

There is a lack of consensus on how to proceed when encountering null eigenvectors, and the discussion highlights the potential need for alternative methods such as generalized eigenvalues.

dumbdumNotSmart
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So I ran into an case I have not seen before. Say we have a system of 3 equations such that W´=AW, where W=(x(t),y(t),z(t)) and A is a 3x3 matrix. The way I usually approach these is by finding the eigenvalues of A to then find the eigenvectors and thus find the ¨homogenous¨ solution. What happens when the eigenvector I am looking for, given by a eigenvalue, is the null vector?

An example where this happens: A=(-1,0,1//2,-1,1//0,0,-1)
 
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dumbdumNotSmart said:
So I ran into an case I have not seen before. Say we have a system of 3 equations such that W´=AW, where W=(x(t),y(t),z(t)) and A is a 3x3 matrix. The way I usually approach these is by finding the eigenvalues of A to then find the eigenvectors and thus find the ¨homogenous¨ solution. What happens when the eigenvector I am looking for, given by a eigenvalue, is the null vector?

An example where this happens: A=(-1,0,1//2,-1,1//0,0,-1)
By definition, an eigenvector can't be the zero vector.
For the matrix you show, I get repeated eigenvalues of ##\lambda = -1##, but the associated eigenvector is not the zero vector.
 
Yes I am aware of that. That is why Iam curious to how you solve it using another method since mine is not working(giving me null eigenvectors).
 
dumbdumNotSmart said:
Yes I am aware of that. That is why Iam curious to how you solve it using another method since mine is not working(giving me null eigenvectors).
I would be curious to see your work. As already mentioned, for the matrix you showed, there is only one eigenvalue (repeated), and its eigenvector is not the zero vector.

There is a technique called generalized eigenvalues that can sometimes be used to get eigenvectors in cases like the one here. Do a web search for "generalized eigenvalues".
 

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