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Diagonalization help

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data

    [tex]
    A=\left[\begin{array}{ccc}1 & 0 & 0\\ 0 & 1 & -1\\ 0 & 0 & 2\end{array}
    [/tex]

    a) Find the eigenvalues and corresponding eigenvectors of matrix A.
    b)Find the matrix P that diagonalizes A.
    c)Find the diagonal matrix D suh that A = PDP-1, and verify the equality.
    d) Find the orthogonal matrix P that diagonalizes A.
    e) Compute A4

    2. Relevant equations

    A = PDP-1,
    AP = DP
    A-I[tex]\lambda[/tex] = 0

    3. The attempt at a solution

    First I started by finding the eigenvalues values where [tex]\lambda[/tex]=1 multipity two, 2. After this I tried finding the eigenvectors that form P and got v1=[0,-1,1] from [tex]\lambda[/tex]=2 , and {v2, v3} = {[0, 1, 0], [0, 0, 1]}. From this I constructed the P matrix and got [tex]
    P=\left[\begin{array}{ccc}0 & 0 & 0\\ -1 & 1 & 0\\ 1 & 0 & 1\end{array}
    [/tex] and [tex]
    D=\left[\begin{array}{ccc}2 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{array}
    [/tex] and this is where I get confused. The P matrix doesn't work in the form AP = PD and you can't find the inverse of P since the top row is all zeros. Once I figure this out, parts d and e should be straight-forward. Can someone point me to where I'm making a mistake here please. Thanks to everybody who helps.
     
  2. jcsd
  3. May 9, 2010 #2

    jbunniii

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    [itex]v_3 = [0, 0, 1][/itex] is not an eigenvector corresponding to [itex]\lambda = 1[/itex]. Just perform the multiplication to see that [itex]A v_3 \neq v_3[/itex].
     
  4. May 9, 2010 #3

    vela

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    Show us how you found the eigenvectors because only one actually is correct.
     
  5. May 9, 2010 #4

    jbunniii

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    Both [itex]v_1[/itex] and [itex]v_2[/itex] are correct, actually.
     
  6. May 9, 2010 #5

    vela

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    Apparently, I can't add. :)
     
  7. May 9, 2010 #6
    Alright, then there is something I'm possibly missing about eigenvectors. The first eigenvector was found by plugging [tex]\lambda[/tex] = 2 into the (A - I[tex]\lambda[/tex]) matrix producing [tex]
    \begin{bmatrix}-1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & 0\end{bmatrix}
    [/tex] which gives the equations x1 = 0 and x2 = -x3 and constructing the vector from x3 gives x3[0, -1, 1] where the first eigenvector v1 = [0, -1, 1]. I then used the other eigenvalue [tex]\lambda[/tex] = 1 and found the matrix [tex]
    \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 0 & 1\end{bmatrix}
    [/tex] and the way I'm interpreting the book's reasoning for dealing with these matrixes, the eigenvectors that form would be x2[0, 1, 0] + x3[0, 0, 1] which are also the eigenvectors. Clearly there is something different I need to do when dealing with this kind of matrix.
     
  8. May 9, 2010 #7

    vela

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    The matrix you got for the eigenvalue equal to 1 gives you the equation x3=0. It doesn't tell you anything about the other components. Do you see what the other eigenvector should be now?
     
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