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Eigenvector sign

  1. Apr 15, 2014 #1

    939

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    1. The problem statement, all variables and given/known data

    Find eigenvector for the root -7 of:

    |2 3|
    |3 -6|

    2. Relevant equations

    |2 3|
    |3 -6|

    3. The attempt at a solution

    I got
    1
    -3

    But my books says
    -1
    3

    I am only wondering if this is possibly the same answer, because when I check my answer by multiplying the eigenvector by the original matrix and the root by the eigenvector the answer appears correct.

    I.e.
    |2 3|
    |3 -6| times (1/-3) = (-7)(1/-3) = -7/21, while if you do the same for (-1/3), allbeit a different answer, but the condition still holds.

    Is this correct? Is this condition truly the indicator of if you got the correct answer? (original vector * eigenvector) = (root * eigenvector)
     
  2. jcsd
  3. Apr 15, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, any "eigenvector", v, corresponding to a given eigenvalue (I would not say "root") [itex]\lambda[/itex], of matrix A, has the property that [itex]Av= \lambda v[/itex].

    What you are missing is that if [itex]v[/itex] is such an eigenvector then [itex]av[/itex], for any number a, is also an eigenvector, corresponding to eigenvalue [itex]\lambda[/itex]: [itex]A(av)= a (Av)= a(\lambda v)= \lambda (av)[/itex].

    In particular, with a= -1, if v is an eigenvalue then so is -v. The eigenvector is NOT unique.
     
  4. Apr 15, 2014 #3

    939

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    Thanks a lot.

    Just to confirm, the signs in this case do not matter and both answers are correct? I am such a noobie...
     
    Last edited: Apr 15, 2014
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