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Math Help - Limit of matrix A^n as n approaches infinity

  1. #1
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    Limit of matrix A^n as n approaches infinity

    Is anyone familiar with how to do the following:

    If given a matrix, A, find the limit of A^n as n approaches infinity.

    I'm sorry that I don't have a specific example; can anyone explain it in a general case?

    Thanks
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  2. #2
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    Quote Originally Posted by 3leggeddog View Post
    Is anyone familiar with how to do the following:

    If given a matrix, A, find the limit of A^n as n approaches infinity.

    I'm sorry that I don't have a specific example; can anyone explain it in a general case?

    Thanks
    It depends on the matrix. For some it will be the zero matrix and for others it
    will be something else but for many it will not exist.

    RonL
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  3. #3
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    It makes no sense to talk about the limit of A^n. Because we need a metric space defined on the matrix to talk about sequences. Unless, you want to define "limit" in a natural way, i.e. the limit of each of its components.
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  4. #4
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    Quote Originally Posted by 3leggeddog View Post
    Is anyone familiar with how to do the following:

    If given a matrix, A, find the limit of A^n as n approaches infinity.
    The usual technique for doing this (in cases where the limit exists) is to diagonalise the matrix. If A = P^{-1}DP, where P is invertible and D is the diagonal matrix whose entries are the eigenvalues of A, then A^n = P^{-1}D^nP. The matrix D^n is easy to compute: it is diagonal, and its entries are those of D raised to the power n. The limit matrix will only exist if all the eigenvalues are either 1 or have absolute value less than 1. In that case, D^n will converge to a limit matrix D_0, and A^n will converge to P^{-1}D_0P.
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  5. #5
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    What if the Matrix A^n looks like

    What if im loooking for limit of A^n as n approaches infinity when A=

    [.5 0 0]
    [.5 1 .75]
    [0 2 .25]
    Last edited by Medina; October 6th 2008 at 01:55 AM. Reason: should look like a matrix
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  6. #6
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    Quote Originally Posted by Medina View Post
    What if im loooking for limit of A^n as n approaches infinity when A=

    [.5 0 0]
    [.5 1 .75]
    [0 2 .25]
    The eigenvalues of this matrix are \lambda=\tfrac12 and \lambda=\tfrac18(5\pm\sqrt{105}). Since |\tfrac18(5+\sqrt{105})|>1 it follows that the limit of A^n as n→∞ does not exist.

    [Did you perhaps intend the 2 on the bottom row of the matrix to be a 0? The usual setting in which the limit of A^n is used is when the matrix A is stochastic, in which case I would expect to see a 0 where that 2 is.]
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  7. #7
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    thankkks verryy muchh for that
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  8. #8
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    yes thats correct it should have been a 2 instead of a 0, there was a typo in the question...after i spent about 3 hours trying to diagnolise and find the limit. pays to realize the question wudnt be worth 15 marks if the limit was divergent :P
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  9. #9
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    Quote Originally Posted by Opalg View Post
    The usual technique for doing this (in cases where the limit exists) is to diagonalise the matrix. If A = P^{-1}DP, where P is invertible and D is the diagonal matrix whose entries are the eigenvalues of A, then A^n = P^{-1}D^nP. The matrix D^n is easy to compute: it is diagonal, and its entries are those of D raised to the power n. The limit matrix will only exist if all the eigenvalues are either 1 or have absolute value less than 1. In that case, D^n will converge to a limit matrix D_0, and A^n will converge to P^{-1}D_0P.
    How to obtain the limit matrix D_0 ?
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