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Linear Algebra Matrix Eigenvalues

  1. Oct 23, 2011 #1
    1. The problem statement, all variables and given/known data

    The matrix A has 3 distinct eigenvalues t1< t2< t3. Let vi be the unique eigenvector associated to ti with a 1 as its first nonzero component. Let

    D= [t1 0 0
    0 t2 0
    0 0 t3]

    and P= [v1|v2|v3] so that the ith column of P is the eigenvector vi associated to ti

    A=
    7 2 -8
    0 1 0
    4 2 -5

    a) Find D
    b) Find P
    c) Find P-1
    2. Relevant equations



    3. The attempt at a solution
    My thought to find D was to find the characteristic equation of A which I found to be (t-3)(t+1)=0 so the eigenvalues would be t=-3, t=1 I then plugged in these values into the matrix D so D became
    3 0 0
    0 -1 0
    0 0 0

    but it was counting it wrong. What did I mess up on?
     
  2. jcsd
  3. Oct 23, 2011 #2

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    Hi lina29 :smile:

    I'm afraid your characteristic equation is a little off.
    How did you get it?

    And your eigenvalues are not the solution of the equation you have.
    They are off by a minus sign.

    You're supposed to get 3 distinct eigenvalues.
    (They should be 3, -1, 1.)

    Oh, and according to your problem statement your eigenvalues should be sorted low to high on your diagonal.
     
  4. Oct 23, 2011 #3
    To find the determinant I did aei+bfg+cdh-ceg-bdi-afh. So I got (t-7)(t-1)(t+5)+0+0-(8)(t-1)(-4)-0-0. I simplified that to (t-7)(t-1)(t+5)-(-32)(t-1). My mistake was that I accidently crossed out the t-1 on both sides. From there how would I get to P?
     
  5. Oct 23, 2011 #4

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    Good! :smile:


    Find the eigenvectors v1, v2, v3.
    P is the matrix with these eigenvectors as columns.
     
  6. Oct 23, 2011 #5
    So the eigenvectors would be -1, 1, and 3. Then the matrix P would be
    -1
    1
    3
    is that it?
     
  7. Oct 23, 2011 #6
    or I guess that wouldn't be right since I need to find P inverse so P would have to be a square matrix
     
  8. Oct 23, 2011 #7

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    No. There's a difference between eigenvalues and eigenvectors.
    t1=-1 is your first eigenvalue.
    v1 is the vector belonging to this eigenvalue.

    To find v1 you need to solve A v1 = t1 v1.

    Note that any scalar multiple of v1 also satisfies the equation, but its direction is unique.


    Edit: I just saw that your problem statement says that v1 should have "1 as its first nonzero component".
     
    Last edited: Oct 23, 2011
  9. Oct 23, 2011 #8
    ohh I got it. Thank you so much!
     
  10. Oct 23, 2011 #9

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