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Formula for unit eigenvector

  1. May 25, 2014 #1
    Last edited: May 25, 2014
  2. jcsd
  3. May 25, 2014 #2


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    ?? Take any eigenvector and divide by its length.
  4. May 25, 2014 #3
    And which is the formula for the eigenvector?
  5. May 26, 2014 #4
    ^ What do you mean by that? A (right) eigenvector of A, x, is a (nonzero) solution to Ax=λx, and λ is the corresponding eigenvalue. Any vector fulfilling the condition can be divided by its length ||x||, so that the resulting vector is still an eigenvector, since the eigenvector is not the zero vector.

    (EDIT: Ok, technically talking about ||x|| obviously means that we have to be able to define a norm, but I don't think that was the issue?)
    Last edited: May 26, 2014
  6. May 26, 2014 #5
    Again: which is the formula for the eigenvectors?
  7. May 26, 2014 #6
    What's wrong with Ax=λx? Given A and λ, it can be used to (numerically or analytically) solve x.
  8. May 27, 2014 #7


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    Your original question was about unit eigenvectors and that is what I responded to. There are a number of ways of finding eigenvectors but there is no "formula" you can just plug numbers into. Finding eigenvalues and eigenvectors is one of the harder problems in Linear Algebra.
  9. May 27, 2014 #8
    If I want to express an eigenvector like (cos(Θ), sin(Θ)), is this form good way of constraint the expression, so that Θ is function of eigenvalues?
  10. May 29, 2014 #9
    By defition:

    ##A\vec{v} =\lambda \vec{v}##
    ##(A - \lambda I)\vec{v} = \vec{0}##

    So any eigenvector ##\vec{v}## is:
    ##\vec{v} = (A - \lambda I)^{-1}\vec{0} = \vec{0}##
    ##\vec{v} = \vec{0}##

    What is wrong?
  11. May 29, 2014 #10
    A square matrix M has an inverse iff [itex]|M|\neq0[/itex]. To obtain the eigenvalues [itex]\lambda[/itex], you solve the equation [itex]|A-\lambda I|=0[/itex]. In your post, you use the expression [itex](A-\lambda I)^{-1}[/itex], which is meaningless, because the eigenvalues are exactly the values for which the inverse doesn't exist.
  12. May 29, 2014 #11


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    Also, although the math used was wrong, the 0 vector really is technically an eigenvector of all matrices...it's the trivial eigenvector, with an ill-defined eigenvalue. A*0=lambda*0 for all A and all lambda.
  13. May 29, 2014 #12


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    No, eigenvectors are defined to be non-zero vectors.




    The notion of the zero vector as a "trivial eigenvector with an ill-defined eigenvalue" doesn't have any practical (or even theoretical) value.
  14. May 29, 2014 #13


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    Alright. Well, if it's right there in the definition, then looks like I was wrong. :)
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