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Homework Help: Eigenvectors with a repeated eigenvalue

  1. Jun 6, 2010 #1
    1. The problem statement, all variables and given/known data
    For the following linear system:
    [tex]\frac{dx}{dt}[/tex] = -2x
    [tex]\frac{dy}{dt}[/tex] = -2y

    Obtain the general solution.


    2. Relevant equations



    3. The attempt at a solution
    A= -2 0
    0 -2

    Using the determinant of A-[tex]\lambda[/tex]I I got a repeated eigenvalue of -2. I am not sure how to get the eigenvectors since substituting -2 back into A-[tex]\lambda[/tex]I will give a matrix of zeroes. I checked with MATLAB and the eigenvectors should be (1,0) and (0,1) but I have no idea how to get them.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 6, 2010 #2

    Mark44

    Staff: Mentor

    I don't see that there is any need for eigenvalues and eigenvectors at all. Your system of equations is uncoupled, which means that in the matrix equation x' = Ax, A is already diagonal.

    If you have x'(t) = kx, the solution is x(t) = Aekt, right? Apply this idea to each of the differential equations in your system.
     
  4. Jun 6, 2010 #3
    Thanks :)
     
  5. Jun 6, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    As Mark44 noted, A is already diagonal, so you don't need to find eigenvectors so you can diagonalize A, but in case you're still wondering, when A-λI=0, any vector x will satisfy (A-λI)x=0, which means every vector x is an eigenvector of A. You can therefore choose any two vectors, like (1,0) and (0,1), as your eigenvectors. (Usually when you have this kind of freedom, it's a good idea to select vectors that are orthogonal to each other.)
     
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