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Lower triangular matrix eigenvectors problem

  1. May 3, 2012 #1
    Ok, this is starting to come back to me, but I'm stuck again

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


    M=\begin{bmatrix}
    (1-\frac{4}{3}) & 0 \\
    -\frac{c}{3} & -c \\
    \end{bmatrix}


    Find eigenvectors and eigenvalues.
    2. Relevant equations



    3. The attempt at a solution

    Eigenvalues are [itex] λ_1= (1-\frac{4}{3})>0[/itex] and [itex] λ_2=-c>0 [/itex]

    Eigenvector for [itex] λ_2 [/itex] is <0 1>
    For [itex]λ_1[/itex], I should get [itex]<(\frac{c}{3}-1) (\frac{4}{3})> [/itex]

    However, I end up with (without writing out the matrix again, just giving the equation mind you)

    [itex] -c*e_2 = e_1 - \frac{4}{3}*e_1 +\frac{c}{3}*e_1 [/itex]

    This is [itex] -c*e_2 = (1-c)*e_1 [/itex]which gets an eigenvector of something like <(1-c) c>

    Anyone see my error? Thanks
     
  2. jcsd
  3. May 3, 2012 #2

    Office_Shredder

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    Staff Emeritus
    Science Advisor
    Gold Member

    The clearly negative eigenvalue being allegedly positive is the first thing that struck me as odd, but not really a big deal in the grand scheme of things

    I don't understand what you did to get that equation (the left hand side looks like Me2, and the right hand side is completely unclear to me where it came from, but the fact that you have a scaling of e1= a scaling of e2 seems pretty suspicious), or what you are claiming the eigenvector for the first eigenvalue is since it only has one entry as written - is it supposed to be ( (c/3-1), 4/3)?
     
  4. May 4, 2012 #3
    "The clearly negative eigenvalue being allegedly positive is the first thing that struck me as odd, but not really a big deal in the grand scheme of things"

    That's a typo on my part. λ_2 should be negative i.e. -c<0

    "is it supposed to be ( (c/3-1), 4/3)?"

    Yes

    To show explicitly what I did, using the standard eigenvector equation with v_i instead of e_i


    \begin{bmatrix}
    (1-(4/3)c) & 0 \\
    -c/3 & -c \\
    \end{bmatrix} *\begin{bmatrix}
    v_1 \\
    v_2 \\
    \end{bmatrix}

    = \begin{bmatrix}
    (1-4c/3)*v_1 \\
    (1-4c/3)*v_1\\
    \end{bmatrix}

    When you multiply this matrices (as sort of shown....latex skills withstanding) then you get the equation I've written previously.
     
  5. May 4, 2012 #4
    "but the fact that you have a scaling of e1= a scaling of e2 seems pretty suspicious"

    When finding eigenvectors I always get equations like this.
     
  6. May 4, 2012 #5
    I'm sorry everyone, that should be multiply by

    \begin{pmatrix}
    (1-4c/3)v_1\\
    (1-4c/3)v_2\\
    \end{pmatrix}
    which does indeed give the right answer,

    (c/3)*v_1 = (c/3 -1)*v_2

    Sorry for wasting everyone's time, I've finally got it now. Thank you!
     
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