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Math Help - Diagonalization, soluton check

  1. #1
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    Diagonalization, soluton check

    Was just wondering if I did this problem correctly

    Diagonalize the matrix B
    3 0 0
    1 2 0
    1 −1 4
    That is, find matrices P and D so that D is diagonal
    and P(inverse)BP = D.

    So to find P, I found the eigenvalues, 2,3,4,
    then fond the eigenvectors by plugging in 2,3,4 respectively, so
    when eigenvalue is 2,
    1 0 0
    1 0 0
    1 -1 2
    which gave me the eigen vector [0 .5 1]
    when eigenvalue is 3,
    0 0 0
    1 -1 0
    1 -1 1
    which gave me the eigen vector [1 1 0]
    when eigenvalue is 4,
    -1 0 0
    1 -2 0
    1 -1 2
    which gave me the eigen vector [0 0 1]
    so P =
    0 1 0
    .5 1 0
    1 0 1
    Then P inverse has to be
    -2 2 0
    1 0 0
    2 -2 1
    and so D =
    2 0 0
    0 3 0
    0 0 4

    Did I do this problem correctly?
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  2. #2
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    Quote Originally Posted by daklutz View Post
    Was just wondering if I did this problem correctly

    Diagonalize the matrix B
    3 0 0
    1 2 0
    1 −1 4
    That is, find matrices P and D so that D is diagonal
    and P(inverse)BP = D.

    So to find P, I found the eigenvalues, 2,3,4,
    then fond the eigenvectors by plugging in 2,3,4 respectively, so
    when eigenvalue is 2,
    1 0 0
    1 0 0
    1 -1 2
    which gave me the eigen vector [0 .5 1]
    when eigenvalue is 3,
    0 0 0
    1 -1 0
    1 -1 1
    which gave me the eigen vector [1 1 0]
    when eigenvalue is 4,
    -1 0 0
    1 -2 0
    1 -1 2
    which gave me the eigen vector [0 0 1]
    so P =
    0 1 0
    .5 1 0
    1 0 1
    Then P inverse has to be
    -2 2 0
    1 0 0
    2 -2 1
    and so D =
    2 0 0
    0 3 0
    0 0 4

    Did I do this problem correctly?
    The eigenvector I got for  \lambda = 2 is {\bf e_1} = \left(\begin{array}{c}0\\1\\0.5\end{array}\right). This then changes P and it's inverse.
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