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Math Help - Powers of a matrix

  1. #1
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    Powers of a matrix

    Hi!

    If I have the following matrix, call it B, (3x3 with entries separated by commas)
    8, 12, 0
    0, 8, 12
    0, 0, 8

    How can I find a matrix A such that A^3 = B?
    Since B is not diagonalizable I have no idea how to approach this problem.
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  2. #2
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    Trial and Error gives:

    2,1,-0.5
    0,2,1
    0,0,2

    whose cube is the matrix you want.
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  3. #3
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    Quote Originally Posted by Volcanicrain View Post
    Hi!

    If I have the following matrix, call it B, (3x3 with entries separated by commas)
    8, 12, 0
    0, 8, 12
    0, 0, 8

    How can I find a matrix A such that A^3 = B?
    Since B is not diagonalizable I have no idea how to approach this problem.
    "Brute strength" works. Because B is "upper triangular", you know A will also be "upper triangular" and look for a solution of the form \begin{bmatrix} a & b & c \\ 0 & d & e \\ 0 & 0 & f\end{bmatrix}.

    Now, it's not too hard to calculate that A^2= \begin{bmatrix}a^2 & ab+ be & ac+be+ ef \\ 0 & d^2 & de+ef \\ 0 & 0 & f^2\end{bmatrix} and then that
    A^3= \begin{bmatrix}a^3 & a^2b+abd+ bd^2 & a^2c+ abe+ acf+ bde+ e^f+ cf^2 \\ 0 & d^3 & de^2+ e^2f \\ 0 & 0 & f^3\end{bmatrix}
    Set that equal to B and you have 6 equations to solve for a, b, c, d, e, and f.
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  4. #4
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    Thanks!
    I had considered that before in my mind, but I wasn't sure if there wasn't an easier way.
    But it really isn't so bad!
    Thanks a lot!
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