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Diagonalization, eigenvectors, eigenvalues

  1. Feb 26, 2008 #1
    [SOLVED] diagonalization, eigenvectors, eigenvalues

    1. The problem statement, all variables and given/known data

    Find a nonsingular matrix P such that (P^-1)*A*P is diagonal

    | 1 2 3 |
    | 0 1 0 |
    | 2 1 2 |


    2. Relevant equations
    I am at a loss on how to do this. I've tried finding the eigen values but its getting me nowhere


    3. The attempt at a solution

    i row reduced to 1 2 3
    0 1 0
    0 0 -4

    and found : (x-1)^2 and (x+4), which gives me eigen values of 1 and -4.

    Using A - Ix, when x = -4, the matrix becomes 5 2 3 0
    0 5 0 0
    2 1 6 0

    Using row reduction: 1 2/5 1/5 0
    0 5 0 0
    2 1 6 0

    1 2/5 1/5 0
    0 1 0 0
    0 1/5 28/5 0

    1 2/5 1/5 0
    0 1 0 0
    0 0 28/5 0

    1 0 0 0
    0 1 0 0
    0 0 1 0
    which means X = y = z = 0, but thats wrong


    When i use x = 1: 0 2 3 0
    0 0 0 0
    2 1 1 0
    this gives 2y = -3z
    2X + y + z = 0 ==> 2X = z/2

    If z = 4, then x = 1, y = -6

    That's one correct answer, but i cant get the second one, which is suppose to be
    X = -3, y = 0, z = 2
     
  2. jcsd
  3. Feb 26, 2008 #2

    HallsofIvy

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    Yes, the "diagonal" elements will be the eigenvalues of the matrix and the matrix P has eigenvectors corresponding to those eigenvalues as columns.


    Why in the world would you row-reduce? In general, the eigenvalues of the row-reduced matrix are not the eigenvalues of the original matrix. If it were, there wouldn't be such complicated computer routines for finding eigenvalues!

    The eigenvalue equation for your matrix is
    [tex]\left|\begin{array}{ccc} 1-\lambda & 2 & 3 \\ 0 & 1-\lambda & 0 \\ 2 & 1 & 2-\lambda \end{array}\right|= 0[/tex].

    I recommend expanding on the middle row. That gives
    [tex](1- \lambda)\left|\begin{array}{cc} 1- \lambda & 3 \\ 2 & 2-\lambda \end{array}\right|= 0[/tex]
    which is easily solved for [itex]\lambda[/itex]= 1, -1, and 4.

    Yes, it is wrong: -4 is not an eigenvalue!

    Saying that 4 (not -4) is an eigenvalue gives
    [tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 2 & 1 & -2 & 0\end{array}\right][/tex]
    subtracting 2/3 of the first row from the third row gives
    [tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & \frac{7}{3} & 0 & 0\end{array}\right][/tex]
    and clearly now adding 7/9 of the second row to the third row gives
    [tex]\left[\begin{array}{cccc}-3 & 2 & 3 & 0 \\ 0 & -3 & 0 & 0\\ 0 & 0 & 0 & 0\end{array}\right][/tex]
    (You don't really need to carry that fourth column along- it will always be 0s). That gives y= 0, z= x so any eigenvector corresponding to eigen value 4 is a multiple of < 1, 0, 1>.





    Yes, 1 is an eigenvalue, but not a double eigenvalue. -1 is the third eigenvalue.
     
    Last edited: Feb 26, 2008
  4. Feb 26, 2008 #3
    Could you explain what you did here? What do you mean my expanding the middle row?
     
  5. Dec 14, 2011 #4
    Re: [SOLVED] diagonalization, eigenvectors, eigenvalues

    You can use the zeros in a matrix to make your work easier (and less likely to introduce arithmetic mistakes - a worthy skill to develop :smile:).

    Remember that if you switch a row/column in the determinant, you change the sign of determinant? Well, imagine switching row1 with row2, then col2 with col1 (for a net sign change of +1).
    Now, if you calculate your determinant using expansion of minors, all the zeros help you out, so that you are taking (1-λ) times the determinant of the remaining, smaller matrix, as HallsofIvy showed.
    If you calculate your determinant by drawing the diagonal lines & multiplying, only the lines that go through (1-λ) will be non-zero, and you'll get the same result.
     
    Last edited: Dec 14, 2011
  6. Dec 14, 2011 #5

    micromass

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    Re: [SOLVED] diagonalization, eigenvectors, eigenvalues

    This thread is 3 years old...
     
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