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[Lin Alg - Eigenvectors]

  1. Jan 24, 2012 #1
    I am little confused about the choice of eigenvectors chosen by my book. I am wondering if an eigenvalue can have multiple eigenvectors and if all are equally correct. Case in point the example below:

    1. The problem statement, all variables and given/known data
    find a fundamental matrix for the system x'(t) = Ax(t) for the given matrix A:
    A = Row1 = {3 1 -1}
    Row2 = {1 3 -1}
    Row3 = {3 3 -1}

    The attempt at a solution:
    So i go through the motions and find that th eigenvalues are 1 and -2. the eigenvector for eValue 1 = {1, 1, 3}

    When I try to solve for the eVector for eValue =2, There are 2 possibilties because:
    (A-rI), where r = 2:
    Row1 = {1 1 -1}
    Row2 = {1 1 -1}
    Row3 = {3 3 -3}

    which basically leads to the equation U1 + U2 - U3 = 0
    i.e. U3 = U1 + U2
    from here, I can say
    U1 = s, U2 =t, and U3 = s + t, leading to the eVectors:
    {s, t, s+t}
    or s * {1,0,1} + t * {0,1,1}

    OR i can say U1 = -U2 + U3, U2 =t, U3 =s and my eVectors are:
    {t - s, t, s} or t*{1, 1, 0} + s*{-1, 0, 1}

    Which is correct? the book says the 2nd one but i dont see why pick one over the other.
     
  2. jcsd
  3. Jan 24, 2012 #2

    lanedance

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    Homework Helper

    when you have degenerate eignevectors (algebraic multiplicity of eigenvalue is greater than 1 in characteristic equation), usually there is more than one eigenvector for the given eigenvalue (this is called the geometric multiplicity of the eigenvalue)

    And whenever you have more than one eigenvector for a given eigenvalue the eignevectors you solve for will not be unique, however the ones you solve for will span the same "eigenspace" which is unqiue. In this case a 2D plane

    Note that even when you have only a single eignevector it is only unique up to a given multiplicative constant and actually represents a 1D line of valid eignevectors
     
  4. Jan 24, 2012 #3
    :) I have not taken linear algebra yet (this is for diff eq class) and so most of what you said went over my head. but in lamens terms, i think this is what you are saying:

    because the eigenvalue 2 is a repeated root of the equation A-rI = 0, the eigenvectors will not be unique and so it is possible to have two different eigenvectors for the same eigenvalue. Therefore, the choice of the eigenvector does not matter and i can use either eigenvector for my fundamental matrix?
    is that correct?
     
  5. Jan 24, 2012 #4

    lanedance

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    Homework Helper

    note quite but close

    As the eigenvalue 2 is a repeated root, it can have either 1 or 2 eigenvectors. In this case it has 2.

    Now say you found eigenvectors u & v, and the book found p & q. As they "span" the same space, each set can be written as a linear combination of the other. This means you can find real numbers a,b,c,d such that
    p = au+bv
    q = cu+dv
     
  6. Jan 24, 2012 #5
    Gotcha. Thanks a lot!
     
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