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Find the eigenvectors from general equation

  1. Sep 29, 2011 #1

    sharks

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    Gold Member

    1. The problem statement, all variables and given/known data
    I know the eigenvalues, and i have derived the general equation as follows:
    -x1 + x2 = 0
    x1 - x2 = 0
    0.x3 = 0

    What is the eigenvector?

    The attempt at a solution

    I am trying to find the exact procedure to solve this, but there seems to be none.
    As a general rule of thumb (i think) i try to express both x1 and x2 in terms of x3.

    so,
    x1 = x2.
    x2 = x1
    x3 = 0

    The eigenvector is: [1 1 0] ??

    In the book, it says there are 2 eigenvectors for the same eigenvalue used for the general equation above.
     
  2. jcsd
  3. Sep 29, 2011 #2

    Mark44

    Staff: Mentor

    There are actually two free variables here, x2 and x3.

    Your system above could be simplified, by eliminating the second equation (which is equivalent to the first). The third equation places no restrictions on x3, which can be any value.
    Write them this way.
    x1 = x2
    x2 = x2 (obviously true)
    x3 = ...... x3 (also obviously true)

    This means that a vector x with components x1, x2, and x3, can be written as the sum of scalar multiples of two vectors: <1, 1, 0>T and <0, 0, 1>T. These will do for your eigenvectors for whatever eigenvalue you're working with in this problem.
     
  4. Sep 29, 2011 #3

    sharks

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    Thanks, Mark44. Your advice is enlightening. :)

    So, as a general rule, from now onwards (i don't want to rely on visual interpretation to find the free variables) i will do row echelon on the system of equations, which will give me the free variables, and from there, i can find the eigenvector/s by expressing all the pivot variables in terms of the free variables. I recall that i did the same thing to find the special solution/s or null space.
    So, am i correct in adopting this way of thinking about finding eigenvectors?
     
  5. Sep 29, 2011 #4

    Mark44

    Staff: Mentor

    Yes, that's pretty much it. If you get your matrix in reduced row echelon form, all of the entries above or below a leading entry will be 0. From that form you can write equations that give your pivot variables in terms of the free variables.

    I should add that in reduced row echelon form, the system you showed in post #1 looks like this:
    [1 1 0 |0]

    From this I can get the equations I showed earlier.
     
  6. Sep 29, 2011 #5

    vela

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    Just thought I'd point out when you're looking for the eigenvectors, that's exactly what are you doing. You are solving (A-λI)x=0, that is, finding the null space of (A-λI).
     
  7. Sep 29, 2011 #6

    sharks

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    Thanks for pointing out this aspect of eigenvectors, vela. :) No more cowering whenever i see an eigenvector problem.
     
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