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Math Help - Eigen Diagonalization

  1. #1
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    Eigen Diagonalization

    Let M=\left( \begin{array}{ccc} 5 & 2 \\ -1 & 2 \end{array} \right).

    Find formulas for the entries of M^n, where n is a positive integer.

    M^n=\left( \begin{array}{ccc} ? & ? \\ ? & ? \end{array} \right).

    I'm not sure how to go about this. Could someone give push me in the right direction?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Thomas View Post
    Let M=\left( \begin{array}{ccc} 5 & 2 \\ -1 & 2 \end{array} \right).

    Find formulas for the entries of M^n, where n is a positive integer.

    M^n=\left( \begin{array}{ccc} ? & ? \\ ? & ? \end{array} \right).

    I'm not sure how to go about this. Could someone give push me in the right direction?
    Probably the simplest thing to do would be to calculate what multiplying M by an arbitrary 2 x 2 matrix would give:
    M = \left ( \begin{matrix} 5 & 2 \\ -1 & 2 \end{matrix} \right )

    M \times \left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} 5a + 2c & 5b + 2d \\ -a + 2c & -b + 2d \end{matrix} \right )

    Now you can set up a recursion: for M^2 a = 5, b = 2, c = -1, and d = 2. This generates the next set of a, b, c, and d.

    So a_n = 5a_{n - 1} + 2c_{n - 1}, b_n = 5b_{n - 1} + 2d_{n - 1}, etc.

    Now solve the system of recursions. (That step is beyond me, I'm afraid!)

    -Dan
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    That is not how to do it topsquark. You need to diagnolize the matrix.
    Meaning,
    AMA^{-1} = D.
    Where A is some invertible matrix and D is a diagnol matrix. Once you do that the above equation is easy to handle.
    Bring everything to power of n,
    (AMA^{-1})^n = D^n
    Thus,
    AM^nA^{-1} = D^n
    And,
    \boxed{M^n = A^{-1}D^nA}
    Where A can be found from its eigenvectors and D from its eigenvalues.
    (Note, computing a power of a diagnol is easy. Just raise each diagnol entry to that exponent).
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