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Linear Algebra - Raising a matrix to a power

  1. Nov 24, 2012 #1
    | x 1 0 |
    | 0 x 1 |
    | 0 0 x |

    I need to raise this matrix to the 50th power. Of course I can not solve this the extremely long way. Here is my attempt at this:

    Let A represent the matrix above, and let N represent the following matrix:

    | 0 1 0 |
    | 0 0 1 |
    | 0 0 0 |

    Then A = XI + N. Since XI * N = N * XI, we can apply the binomial theorem.

    A^50 = sum from k = 0 to 50 of (50 C k)(XI)^(50-k)N^k by the binomial theorem

    = (XI)^50 + 50N(XI)^49 + 1225N^(2)(XI)^48 + N^3

    Now if you take N^3 you will get 0, and if you take N^2, you get the following:

    | 0 0 1 |
    | 0 0 0 |
    | 0 0 0 |

    So we get:

    I(X)^50 + 50N(X)^49 + 1225N^2(X)^48 which is,

    | X^50 50X^49 1225X^48 |
    | 0 X^50 50X^49 |
    | 0 0 X^50 |

    I think this is right. If my work is right could someone help me make this more thorough? I do not want to miss any details and want to show all my work. Thanks.
     
  2. jcsd
  3. Nov 24, 2012 #2

    Ray Vickson

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    Science Advisor
    Homework Helper

    Your matrix is a so-called "Jordan Block", and raising it to large powers is a well-solved problem; see, eg., http://en.wikipedia.org/wiki/Jordan_matrix

    RGV
     
  4. Nov 24, 2012 #3

    Dick

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    Science Advisor
    Homework Helper

    Your solution looks just fine to me.
     
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