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Eigenvectors and eigenvalues - how to find the column vector

  1. Nov 19, 2012 #1
    Hi


    We have a matrix A (picture), the eigenvalues are λ1 = 4 and λ2 = 1 and the eigenvectors are

    λ1 : t(1,0,1)
    λ2 : t1(1,0,2) + t2(0,1,0)

    I have to examine if there's a column vector v that satifies :

    A*v = 2 v


    I would say no there doesn't exist such a column vector v because 2 isn't an eigenvalue:

    Av = λv

    so

    Av = 2v

    but we know that λ is 4 or 1 and not 2


    Am I wrong ?

    I would be nice if someone could give me their opinion :)
     

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    Last edited: Nov 19, 2012
  2. jcsd
  3. Nov 19, 2012 #2

    AlephZero

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    You seem to understand what eigenvalnes and vectors are, which is good.

    But you aren't quite right. There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.

    Try solving (A - 2I)v = 0, and see what you get.
     
  4. Nov 19, 2012 #3
    When I solve it I get

    v =

    (1 0)
    (0 0)
    (0 0)

    ?
     

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  5. Nov 19, 2012 #4

    Erland

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    You cannot use Maple in this way. You should use linsolve(A1,b) or LinearSolve(A1,b). Or better, do it by hand.
     
  6. Nov 19, 2012 #5

    HallsofIvy

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    I don't understand what you mean by that. Your vectors before were single columns of three numbers. How could v be two columns?

    You were close to right when you said before "doesn't exist such a column vector v because 2 isn't an eigenvalue". What is true is that [itex]\lambda[/itex] is an eigenvalue for matrix A if and only if there exist a non-trivial (i.e. non-zero) vector v such that [itex]Av= \lambda v[/itex].

    Since two is not an eigenvalue, there cannot exist a non-trivial vector but, as
    Aleph-zero said, "There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.".
     
  7. Nov 19, 2012 #6
    So the vector is (0,0,0) ?

    Or have I completely misunderstood this ?
     
  8. Nov 19, 2012 #7

    Erland

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    Right. But it is not counted as an eigenvector.
     
  9. Nov 19, 2012 #8

    AlephZero

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    Correct. The question asked you to find a column vector, not a non-zero column vector!

    You only get non-zero solutions of Av = λv when λ is an eigenvalue, but v = 0 is a solution for any value of λ.
     
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