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B System of differential equations Basic question

  1. Jun 24, 2016 #1
    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)
     
    Last edited: Jun 24, 2016
  2. jcsd
  3. Jun 25, 2016 #2

    Mark44

    Staff: Mentor

    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.
     
  4. Jun 25, 2016 #3
    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).
     
  5. Jun 25, 2016 #4

    Mark44

    Staff: Mentor

    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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