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Solving 3x3 Matrix of DEs

  1. Apr 13, 2008 #1
    [tex]X'=\left[\begin{array}{ccc}4 & 1 & 4\\ 1 & 7 & 1\\ 4 & 1 & 4\end{array}\right]X[/tex]

    After evaluating the determinant I get from the characteristic equation [itex]\lambda=0[/itex]
    [tex]\lambda=\frac{15\pm \sqrt{119}*i}{2}[/tex]

    Now here is where I am not sure where to go. I need to create an Eigenvector. I stared it off by plugging in for [itex]\lambda=0[/itex] giving the relationships:


    Now usually, in the case of a 2x2 matrix, I would solve for let's say k1 in terms of k2 and then just PICK some value for k2 and that would establish my Eigenvector.

    I am a little confused as to how to do this with a 3x3. It would appear I have some sort of special case here since row 1 is the same as row 3, but I am unsure how to use that to my advantage?

    Could someone just point me toward the next step?

  2. jcsd
  3. Apr 13, 2008 #2
  4. Apr 13, 2008 #3
    Multiply the middle equation by 4 and then subtract from the top equation, and you see that

    [tex]k_2 = 0[/tex] which implies that
    [tex] k_1 = -k_3[/tex]

    And there you have your vector defined up to one parameter, which is common. You are constructing U to go [tex]A=U^{\dagger}DU[/tex] or something like that right? So that you can go [tex]e^{tA} = U^{\dagger}e^{tD}U[/tex] arg I don't know if I have that right, but

    you need to normalize your eigenvectors to do that and so you should fix your parameter by requiring that eigenvector to have unit norm.
  5. Apr 13, 2008 #4
    See, I wish I knew what David was saying, but I don't. In this class, we tend to skip over terminology like 'normalization' and the proofs of our approaches, but we don't. We are dealt these 'cookie cutter' methods (I am sure these are common of Engineering programs) and are expected to just work with them.

    It works fine and all until we run into something that does not fit the mold of what we have been dealing with.

    Anyone else able to elaborate on what Davis was saying?
  6. Apr 13, 2008 #5
  7. Apr 13, 2008 #6
    To construct U, it's not just laying out any old eigenvectors side by side, but using the eigenvectors with a norm or length/magnitude of unity or 1. That is to say each eigenvector [tex]x[/tex] for that matrix needs to satisfy

    [tex]x^T x = 1[/tex]

    For example if you found [tex]x = (-1, 1)[/tex] you will actually want to use [tex]x = 1/\sqrt{2}(-1, 1)[/tex] instead. Sorry for throwing the jargon around.
    Last edited: Apr 13, 2008
  8. Apr 14, 2008 #7
    Yeah, see. Not even sure what this U is supposed to represent....I can see this is not going to go anywhere. This is my professor's methodology though. Not anyone else's.
  9. Apr 14, 2008 #8
    Oh well why don't you just tell me what method you use. I thought you were diagonalizing a matrix, but I guess not??
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